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https://math.stackexchange.com/questions/2834409/how-to-design-a-nim-game

# How to design a nim game? I am required to give an example of nim game such that: 1. the initial position is an N-position, 2. there are exactly three optimal moves from the initial position, 3. every pile has at least 20 chips, 4. every pile has a different (nonempty) number of chips How could I design such a nim? I am in a mess now. Could anyone help me? If possible, could you teach me how to think in solving this problem? I'm really messed... • Do you know how to decide if a position is N? And what a winning move must do? Jun 28 '18 at 5:57 • @HagenvonEitzen I know Bouton's Thm. But I don't know how to use it. Jun 28 '18 at 7:12 We want three different positive integers $a,b,c\ge20$ such that $d:=a\oplus b\oplus c$ is non-zero (the criterion for $N$-positions) and all three piles can be decreased in a way to make this sum zero, i.e., $a\oplus d<a$, $b\oplus d<b$, and $c\oplus d<c$. For the last conditions, it suffices to achieve that the most significant bit of $d$ is set in all of $a,b,c$. So fore example if all of $a,b,c$ are between $16$ and $31$, inclusive, then so will be $d$. In fact, if we try the minimal possible (according solely to $a,b,c\ge20$) choices for $a,b,c$, \begin{align}a&=20=10100_2\\b&=21=10101_2\\c&=22=10110_2\end{align} we find $d=10111_2=23$ and \begin{align}a\oplus d&=00011_2=3<a\\b\oplus d&=00010_2=2<b\\c\oplus d &=00001_2=1<c\end{align} i.e., the three winning moves are to take $17$ tokens off $a$, or $19$ tokens off $b$, or $21$ tokens off $c$.

http://mathhelpforum.com/advanced-algebra/68751-special-matrix-product.html

1. ## special matrix product Hi, I'm not native English, what is the name the following product. Let a n-column vector and b m-row vector, than c = a × b resultes a n-by-m-matrix: c_ij = a_i * b_j. 2. Originally Posted by Skalkaz Hi, I'm not native English, what is the name the following product. Let a n-column vector and b m-row vector, than c = a × b resultes a n-by-m-matrix: c_ij = a_i * b_j. i don't know if there's a name for this but what you've defined is simply this $\bold{c}=(\bold{ba})^T.$ 3. What does ba mean? b and a are vector, and n <>= m. This product is very usefull sometimes,for example: the matrix of the tranformation r --> a(br) is T = a × b where a, b, r is spatial vector, and br is scalar product 4. $\bold{b}$ is an $m \times 1$ vector and $\bold{a}$ an $1 \times n.$ so $\bold{b}\bold{a}$ is an $m \times n$ matrix. clearly $(\bold{b}\bold{a})^T$ would be the matrix that you called $\bold{c}.$ 5. Oh, you are absolutly right.

https://mathoverflow.net/questions/420328/can-we-use-formal-groups-to-recover-lie-theoretic-representation-theory-in-chara

# Can we use formal groups to recover Lie-theoretic representation theory in characteristic p? In differential geometry, Lie's theorems allow us to integrate any Lie algebra representation to a Lie group representation. The algebraic version of this is more complicated (and I'm not terribly familiar with it), but over a field of characteristic zero we can still get some results from Tannaka duality. This all fails in characteristic $$p$$, unfortunately. One of the algebraic motivations I've seen for the theory of formal groups is that they rectify the failure of Lie algebras to encode higher-dimensional differential information in positive characteristic, with this representation theoretic-issue given as an example of that failure. But do formal groups actually fix this particular problem? Do we have an adjunction between algebraic groups and formal groups, and can we use it to integrate (some) representations from formal groups to algebraic groups? • Why do you see the different behaviour in char p as a "failure" to be repaired, rather than an interesting phenomenon to be investigated on its own terms? Apr 14, 2022 at 6:43 • You might like the distribution algebra (sometimes called hyper algebra) of G. In my world (G reductive) it remedies many of the shortcomings of the Lie algebra. Apr 14, 2022 at 9:52 • The formal group tries to be the simply connected form. The Lang isogeny $G/G(\mathbb F_p)=G$ shows that no positive dimensional group scheme is simply connected as a scheme. Every commutative group scheme has a covering by a group, which has more representations. But I believe that a semisimple group has only finitely many covering homomorphisms and that one that is "simply connected" in the sense of the weight lattice is maximal. You could hope that its finite dimensional representations are the same as those of its formal group. I believe this is true. Apr 14, 2022 at 17:49 The short answer is that the characteristic $$p$$ picture genuinely has more depth. In particular it's not correct to think of formal groups as "better" than Lie groups. In fact, they can be put on equal footing with each other. I think it's helpful to put all the objects you're interested in in the same category. In this case it is the category of group objects in formal schemes, which I will call formal group objects. There is some notational confusion here. By rights, the category of formal group schemes should denote this category (which includes all algebraic groups as well). However, often when people say "formal group scheme" or "formal group" they mean infinitesimal groups, which are formal group schemes with a single point (and other times they mean something else more restrictive, such as formally smooth infinitesimal group schemes, which I don't want to consider separately). Here I'll hopefully avoid the notational confusion and use group objects for group objects in formal schemes and infinitesimal groups for group objects with one point. Now in characteristic zero, every algebraic group $$G$$ has a unique infinitesimal subgroup $$\hat{G}\subset G$$ (as a group object) subject to the condition of having the same tangent space, i.e., the natural map $$T_e\hat{G}\to T_e G$$ being an isomorphism. In characteristic $$p$$, this is no longer the case. If $$G$$ is an algebraic group in characteristic $$p$$ then there is still a maximal sub-group object $$\hat{G}\subset G$$ corresponding to the "full" formal group of $$G$$, but there is another smaller object $$\hat{G}^{(1)}\subset \hat{G},$$ the first Frobenius neighborhood, which also has the same tangent space (there are also objects $$\hat{G}^{(2)}, \dots,$$ in addition to possible other intermediate subgroups). The group $$\hat{G}^{(1)}$$ has the nice property that its representations are equivalent to representations of the Lie algebra $$\mathfrak{g}$$ as a symmetric monoidal category, and this uniquely characterizes $$\hat{G}^{(1)}.$$ In fact if $$\mathfrak{g}$$ is an arbitrary Lie algebra in characteristic $$p$$, it might not integrate to a smooth algebraic group $$G$$, but it will always integrate to a group object $$G^{(1)}$$ that looks like a first Frobenius neighborhood. Now even if you look at the "full" formal neighborhood $$\hat{G}\subset G,$$ you will not have anything resembling integration of representations. Indeed, if $$G\to \text{End}(V)$$ is a representation of a smooth algebraic group scheme, you do get induced representations of $$\hat{G}$$ and $$\hat{G}^{(1)}$$ (equivalently, $$\mathfrak{g}$$). But you can neither integrate representations of $$\hat{G}^{(1)}$$ to representations of $$\hat{G}$$ nor representations of $$\hat{G}$$ to representations of $$G$$ in any reasonable sense, even in one-dimensional cases. • Sorry, could you please give a reference / a brief summary of the "first Frobenius neighborhood" that you mentioned? Thanks. – Z. M Apr 15, 2022 at 9:49 One-dimensional algebraic groups are essentially multiplicative, elliptic curves, or additive, so their associated formal groups have height $$1$$, $$2$$ or $$\infty$$. There exist one-dimensional commutative formal groups of arbitrary height, so most of these cannot be integrated. (This is a well-known fact in chromatic homotopy theory, related to the special status of $$K$$-theory and elliptic homology as opposed to Morava $$K$$-theories of height $$>2$$.)

https://www.questarter.com/q/understanding-gaps-in-concept-of-radius-of-convergence-of-power-series-21_2925231.html

# Understanding Gaps in concept of Radius of Convergence of Power Series by MathLover   Last Updated September 21, 2018 12:20 PM I had done this topic Many time Thinking it is easy But When I today done its theorem , I come across that I had certain gaps in My Knowelege about it. $$\sum a_nz^n$$ is power series. $$1/R=limsup (a_n)^{1/n}$$ 1) $$|Z| ,then series converges absolutely 2) $$|z|>R$$, then series diverges Let $$L=limsup (a_n)^{1/n}$$ By defination $$\forall \epsilon >0 \exists n_1\in N$$ such $$(a_n)^{1/n} Now $$a_n<(L+\epsilon )^n$$ i.e $$a_n<(1/R+\epsilon )^n$$ $$|\sum a_nz^n|\leq \sum (1/R+\epsilon )^nz^n$$ $$(z/R+\epsilon z)^n$$<1 [By 1 $$|z|/R<1$$ and for fix z we can vary any $$\epsilon$$ we wanted ] So by Root test , RHS series is convergent SO Original power series also converges Similary for 2. Is there is any gaps in my argument ? Any Help will be appreciated Tags :

https://online.stat.psu.edu/stat200/book/export/html/65

# 3.4 - Two Quantitative Variables 3.4 - Two Quantitative Variables Later in the course, we will devote an entire lesson to analyzing two quantitative variables. In this lesson, you will be introduced to scatterplots, correlation, and simple linear regression. A scatterplot is a graph used to display data concerning two quantitative variables.  Correlation is a measure of the direction and strength of the relationship between two quantitative variables. Simple linear regression uses one quantitative variable to predict a second quantitative variable. You will not need to compute correlations or regression models by hand in this course. There is one example of computing a correlation by hand in the notes to show you how it relates to z scores, but for all assignments, you should be using Minitab Express to compute these statistics. # 3.4.1 - Scatterplots 3.4.1 - Scatterplots Recall from Lesson 1.1.2, in some research studies one variable is used to predict or explain differences in another variable. In those cases, the explanatory variable is used to predict or explain differences in the response variable. Explanatory variable Variable that is used to explain variability in the response variable, also known as an independent variable or predictor variable; in an experimental study, this is the variable that is manipulated by the researcher. Response variable The outcome variable, also known as a dependent variable. A scatterplot can be used to display the relationship between the explanatory and response variables. Or, a scatterplot can be used to examine the association between two variables in situations where there is not a clear explanatory and response variable. For example, we may want to examine the relationship between height and weight in a sample but have no hypothesis as to which variable impacts the other; in this case, it does not matter which variable is on the x-axis and which is on the y-axis. Scatterplot A graphical representation of two quantitative variables in which the explanatory variable is on the x-axis and the response variable is on the y-axis. When examining a scatterplot, we need to consider the following: 1. Direction (positive or negative) 2. Form (linear or non-linear) 3. Strength (weak, moderate, strong) 4. Bivariate outliers In this class, we will focus on linear relationships. This occurs when the line-of-best-fit for describing the relationship between x and y is a straight line. The linear relationship between two variables is positive when both increase together; in other words, as values of x get larger values of y get larger. This is also known as a direct relationship. The linear relationship between two variables is negative when one increases as the other decreases. For example, as values of x get larger values of y get smaller. This is also known as an indirect relationship. A bivariate outlier is an observation that does not fit with the general pattern of the other observations. ## Example: Baseball Data concerning baseball statistics and salaries from the 1991 and 1992 seasons is available at: The scatterplot below shows the relationship between salary and batting average for the 337 baseball players in this sample. From this scatterplot, we can see that there does not appear to be a meaningful relationship between baseball players' salaries and batting averages. We can also see that more players had salaries at the low end and fewer had salaries at the high end. ## Example: Height and Shoe Size Data concerning the heights and shoe sizes of 408 students were retrieved from: The scatterplot below was constructed to show the relationship between height and shoe size. There is a positive linear relationship between height and shoe size in this sample. The magnitude of the relationship appears to be strong. There do not appear to be any outliers. ## Example: Height and Weight The scatterplot below shows the relationship between height and weight. There is a positive linear relationship between height and weight. The magnitude of the relationship is moderately strong. ## Example: Cafés The scatterplot below shows the relationship between maximum daily temperature and coffee sales. There is a negative linear relationship between the maximum daily temperature and coffee sales. The magnitude is moderately strong. There do not appear to be any outliers. # 3.4.1.1 - Minitab Express: Simple Scatterplot 3.4.1.1 - Minitab Express: Simple Scatterplot ## MinitabExpress – Simple Scatterplot We have data concerning students quiz averages and final exam scores. We want to know if quiz averages can be used to predict final exam scores. Let's construct a scatterplot given that quiz averages are the explanatory variable and final exam scores are the response variable. 1. Open the data set: 2. On a PC or Mac: Select Graphs > Scatterplot 3. Select Simple 4. Double click Final to move it to the Y variable box 5. Double click the Quiz_Average to move it to the  X variable box 6. Click OK Video Walkthrough Select your operating system below to see a step-by-step guide for this example. # 3.4.2 - Correlation 3.4.2 - Correlation In this course, we will be using Pearson's $r$ as a measure of the linear relationship between two quantitative variables. In a sample, we use the symbol $r$. In a population, we use the Greek letter $\rho$ ("rho"). Pearson's $r$ can easily be computed using statistical software. Correlation A measure of the direction and strength of the relationship between two variables. Properties of Pearson's r 1. $-1\leq r \leq +1$ 2. For a positive association, $r>0$, for a negative association $r<0$, if there is no relationship $r=0$ 3. The closer $r$ is to $0$ the weaker the relationship and the closer to $+1$ or $-1$ the stronger the relationship (e.g., $r=-0.88$ is a stronger relationship than $r=+0.60$); the sign of the correlation provides direction only 4. Correlation is unit free; the $x$ and $y$ variables do NOT need to be on the same scale (e.g., it is possible to compute the correlation between height in centimeters and weight in pounds) 5. It does not matter which variable you label as $x$ and which you label as $y$. The correlation between $x$ and $y$ is equal to the correlation between $y$ and $x$. The following table may serve as a guideline when evaluating correlation coefficients: Absolute Value of $r$ Strength of the Relationship 0 - 0.2 Very weak 0.2 - 0.4 Weak 0.4 - 0.6 Moderate 0.6 - 0.8 Strong 0.8 - 1.0 Very strong Cautions 1. Correlation does NOT equal causation. A strong relationship between $x$ and $y$ does not necessarily mean that $x$ causes $y$. It is possible that $y$ causes $x$, or that a confounding variable causes both $x$ and $y$. 2. Pearson's $r$ should only be used when there is a linear relationship between $x$ and $y$. A scatterplot should be constructed before computing Pearson's $r$ to confirm that the relationship is not non-linear. 3. Pearson's $r$ is not resistant to outliers. Figure 1 below provides an example of an influential outlier. Influential outliers are points in a data set that increase the correlation coefficient. In Figure 1 the correlation between $x$ and $y$ is strong ($r=0.979$). In Figure 2 below, the outlier is removed. Now, the correlation between $x$ and $y$ is lower ($r=0.576$) and the slope is less steep. Note that the scale on both the x and y axes has changed. In addition to the correlation changing, the y-intercept changed from 4.154 to 70.84 and the slope changed from 6.661 to 1.632. # 3.4.2.1 - Formulas for Computing Pearson's r 3.4.2.1 - Formulas for Computing Pearson's r There are a number of different versions of the formula for computing Pearson's $r$. You should get the same correlation value regardless of which formula you use. Note that you will not have to compute Pearson's $r$ by hand in this course. These formulas are presented here to help you understand what the value means. You should always be using technology to compute this value. First, we'll look at the conceptual formula which uses $z$ scores. To use this formula we would first compute the  $z$ score for every $x$ and $y$ value. We would multiply each case's $z_x$ by their $z_y$.  If their  $x$ and  $y$ values were both above the mean then this product would be positive. If their x and y values were both below the mean this product would be positive. If one value was above the mean and the other was below the mean this product would be negative. Think of how this relates to the correlation being positive or negative. The sum of all of these products is divided by $n-1$ to obtain the correlation. Pearson's r: Conceptual Formula $r=\dfrac{\sum{z_x z_y}}{n-1}$ where $z_x=\dfrac{x - \overline{x}}{s_x}$ and $z_y=\dfrac{y - \overline{y}}{s_y}$ When we replace $z_x$ and $z_y$ with the $z$ score formulas and move the $n-1$ to a separate fraction we get the formula in your textbook: $r=\frac{1}{n-1}\Sigma{\left(\frac{x-\overline x}{s_x}\right) \left( \frac{y-\overline y}{s_y}\right)}$ # 3.4.2.2 - Example of Computing r by Hand (Optional) 3.4.2.2 - Example of Computing r by Hand (Optional) Again, you will not need to compute $r$ by hand in this course. This example is meant to show you how $r$ is computed with the intention of enhancing your understanding of its meaning. In this course, you will always be using Minitab Express or StatKey to compute correlations. In this example we have data from a random sample of $n = 9$ World Campus STAT 200 students from the Spring 2017 semester. WileyPlus scores had a maximum possible value of 100. Midterm exam scores had a maximum possible value of 50. Remember, the $x$ and $y$ variables do not need to be on the same metric to compute a correlation. ID WileyPlus Midterm A 82 37 B 100 47 C 96 33 D 96 36 E 80 44 F 77 35 G 100 50 H 100 49 I 94 45 Minitab Express was used to construct a scatterplot of these two variables. We need to examine the shape of the relationship before determining if Pearson's $r$ is the appropriate correlation coefficient to use. Pearson's $r$ can only be used to check for a linear relationship. For this example I am going to call WileyPlus grades the $x$ variable and midterm exam grades the $y$ variable because students completed WileyPlus assignments before the midterm exam. ### Summary Statistics From this scatterplot we can determine that the relationship may be weak, but that it is reasonable to consider a linear relationship. If we were to draw a line of best fit through this scatterplot we would draw a straight line with a slight upward slope. Now, we'll compute Pearson's $r$ using the $z$ score formula. The first step is to convert every WileyPlus score to a $z$ score and every midterm score to a $z$ score. When we constructed the scatterplot in Minitab Express we were also provided with summary statistics including the mean and standard deviation for each variable which we need to compute the $z$ scores. Statistics Variable N* Mean StDev Minimum Maximum Midterm 9 41.778 6.534 33.000 50.000 WileyPlus 9 91.667 9.327 77.000 100.000 ID WileyPlus $z_x$ A 82 $\frac{82-91.667}{9.327}=-1.036$ B 100 $\frac{100-91.667}{9.327}=0.893$ C 96 $\frac{96-91.667}{9.327}=0.465$ D 96 $\frac{96-91.667}{9.327}=0.465$ E 80 $\frac{80-91.667}{9.327}=-1.251$ F 77 $\frac{77-91.667}{9.327}=-1.573$ G 100 $\frac{100-91.667}{9.327}=0.893$ H 100 $\frac{100-91.667}{9.327}=0.893$ I 94 $\frac{94-91.667}{9.327}=0.250$ z-score $z_x=\frac{x - \overline{x}}{s_x}$ A positive value in the $z_x$ column means that the student's WileyPlus score is above the mean. Now, we'll do the same for midterm exam scores. ID Midterm $z_y$ A 37 $\frac{37-41.778}{6.534}=-0.731$ B 47 $\frac{47-41.778}{6.534}=0.799$ C 33 $\frac{33-41.778}{6.534}=-1.343$ D 36 $\frac{36-41.778}{6.534}=-0.884$ E 44 $\frac{44-41.778}{6.534}=0.340$ F 35 $\frac{35-41.778}{6.534}=-1.037$ G 50 $\frac{50-41.778}{6.534}=1.258$ H 49 $\frac{49-41.778}{6.534}=1.105$ I 45 $\frac{45-41.778}{6.534}=0.493$ Our next step is to multiply each student's WileyPlus $z$ score with his or her midterm exam score. ID $z_x$ $z_y$ $z_x z_y$ A -1.036 -0.731 0.758 B 0.893 0.799 0.714 C 0.465 -1.343 -0.624 D 0.465 -0.884 -0.411 E -1.251 0.340 -0.425 F -1.573 -1.037 1.631 G 0.893 1.258 1.124 H 0.893 1.105 0.988 I 0.250 0.493 0.123 A positive "cross product" (i.e., $z_x z_y$) means that the student's WileyPlus and midterm score were both either above or below the mean. A negative cross product means that they scored above the mean on one measure and below the mean on the other measure. If there is no relationship between $x$ and $y$ then there would be an even mix of positive and negative cross products; when added up these would equal around zero signifying no relationship. If there is a relationship between $x$ and $y$ then these cross products would primarily be going in the same direction. If the correlation is positive then these cross products would primarily be positive. If the correlation is negative then these cross products would primarily be negative; in other words, students with higher $x$ values would have lower $y$ values and vice versa. Let's add the cross products here and compute our $r$ statistic. $\sum z_x z_y = 0.758+0.714-0.624-0.411-0.425+1.631+1.124+0.988+0.123=3.878$ $r=\frac{3.878}{9-1}=0.485$ There is a positive, moderately strong, relationship between WileyPlus scores and midterm exam scores in this sample. # 3.4.2.3 - Minitab Express to Compute Pearson's r 3.4.2.3 - Minitab Express to Compute Pearson's r ## MinitabExpress – Pearson's r We previously created a scatterplot of quiz averages and final exam scores and observed a linear relationship. Here, we will compute the correlation between these two variables. 1. Open the data set: 2. On a PC: Select Statistics > Correlation > Correlation On a MAC: Select Statistics > Regression > Correlation 3. Double click the Quiz_Average and Final in the box on the left to insert them into the Variables box 4. Click OK This should result in the following output: Correlation: Quiz_Average, Final Pearson correlation of Quiz_Average and Final = 0.608630 P-Value = <0.0001 Video Walkthrough Select your operating system below to see a step-by-step guide for this example. # 3.4.3 - Simple Linear Regression 3.4.3 - Simple Linear Regression Regression uses one or more explanatory variables ($x$) to predict one response variable ($y$). In this course, we will be learning specifically about simple linear regression. The "simple" part is that we will be using only one explanatory variable. If there are two or more explanatory variables, then multiple linear regression is necessary. The "linear" part is that we will be using a straight line to predict the response variable using the explanatory variable. Unlike in correlation, in regression is does matter which variable is called $x$ and which is called $y$. In regression, the explanatory variable is always $x$ and the response variable is always $y$. Both $x$ and $y$ must be quantitative variables. You may recall from an algebra class that the formula for a straight line is $y=mx+b$, where $m$ is the slope and $b$ is the $y$-intercept. The slope is a measure of how steep the line is; in algebra, this is sometimes described as "change in y over change in $x$" ($\frac{\Delta y}{\Delta x}$) or "rise over run." A positive slope indicates a line moving from the bottom left to top right. A negative slope indicates a line moving from the top left to bottom right. The y-intercept is the location on the $y$-axis where the line passes through. In other words, when $x=0$ then $y=y - intercept$. In statistics, we use a similar formula: Simple Linear Regression Line: Sample $\widehat{y}=a+bx$ $\widehat{y}$ = predicted value of $y$ for a given value of $x$ $a$ = $y$-intercept $b$ = slope In a population, the y-intercept is denoted as $\beta_0$ ("beta sub 0") or $\alpha$ ("alpha"). The slope is denoted as $\beta_1$ ("beta sub 1") or just $\beta$ ("beta"). Simple Linear Regression Line: Population $\widehat{y}=\alpha+\beta x$ ## Example: Simple Linear Regression Line The plot below shows the line $\widehat{y}=6.5+1.8x$ Here, the y-intercept is 6.5. This means that when $x=0$ then the predicted value of y is 6.5. The slope is 1.8. For every one unit increase in x, the predicted value of y increases by 1.8. ## Least Squares Method Simple linear regression uses data from a sample to construct the line of best fit. But what makes a line “best fit”? The most common method of constructing a simple linear regression line, and the only method that we will be using in this course, is the least squares method. The least squares method finds the values of the y-intercept and slope that make the sum of the squared residuals (also know as the sum of squared errors or SSE) as small as possible. Residual The difference between an observed y value and the predicted y value. In other words, $y- \widehat y$. On a scatterplot, this is the vertical distance between the line of best fit and the observation. In a sample this may be denoted as $e$ or $\widehat \epsilon$ ("epsilon-hat") and in a population this may be denoted as $\epsilon$ ("epsilon") Residual $e=y-\widehat{y}$ $y$ = actual value of $y$ $\widehat{y}$ = predicted value of $y$ ## Example The plot below shows the line $\widehat{y}=6.5+1.8x$ Let's compute the residual for the point (-0.2, 5.1). The observed x value is -0.2 and the observed y value is 5.1. We know that $y=5.1$ and we can compute $\widehat{y}$ using the regression equation that we have and $x=-0.2$ $\widehat{y}=6.5+1.8(-0.2)=6.14$ Given an x value of -0.2, we would predict this observation to have a y value of 6.14. In reality, they had a y value of 5.1. The residual is the difference between these two values. $e=y-\widehat{y}=5.1-6.14=-1.04$ The residual for this observation is -1.04. ## Cautions • Avoid extrapolation. This means that a regression line should not be used to make a prediction about someone from a population different from the one that the sample used to define the model was from. • Make a scatterplot of your data before running a regression model to confirm that a linear relationship is reasonable. Simple linear regression constructs a straight line. If the relationship between x and y is not linear, then a linear model is not the most appropriate. • Outliers can heavily influence a regression model. Recall the plots that we looked at when learning about correlation. The addition of one outlier can greatly change the line of best fit. In addition to examining a scatterplot for linearity, you should also be looking for outliers. Later in the course, we will devote a week to correlation and regression. # 3.4.3.1 - SLR in Minitab Express 3.4.3.1 - SLR in Minitab Express ## MinitabExpress – Simple Linear Regression We previously created a scatterplot of quiz averages and final exam scores and observed a linear relationship. Here, we will use quiz scores to predict final exam scores. 1. Open the data set: 2. On a PC or Mac: Select STATISTICS > Regression > Simple Regression 3. Double click Final in the box on the left to insert it into the Response (Y) box on the right 4. Double click Quiz_Average in the box on the left to insert it into the Predictor (X) box on the right 5. Under the Graphs tab, click the box for Residual plots (We'll learn more about these in Lesson 10) 6. Click OK This should result in the following output: Analysis of Variance Regression 1 2663.66 2663.66 28.24 <0.0001 Error 48 4527.06 94.31 Total 49 7190.72 Model Summary 9.71152 37.04% 35.73% Coefficients Term Coef SE Coef T-Value P-Value Constant 12.12 11.94 1.01 0.3153 Quiz_Average 0.7513 0.1414 5.31 <0.0001 Regression Equation Final = 12.12 + 0.7513 Quiz_Average Fits and Diagnostics for Unusual Observations Obs Final Fit Resid Std Resid 11 49 70.4975 -21.4975 -2.25 R 40 80 61.2158 18.7842 2.03 R 47 37 59.5050 -22.5050 -2.46 R R Large residual Video Walkthrough Select your operating system below to see a step-by-step guide for this example. In the output in the above example we are given a simple linear regression model of Final = 12.12 + 0.7513 Quiz_Average This means that the y-intercept is 12.12 and the slope is 0.7513. # 3.4.3.2 - Example: Interpreting Output 3.4.3.2 - Example: Interpreting Output This example uses the "CAOSExam" dataset available from http://www.lock5stat.com/datapage.html. CAOS stands for Comprehensive Assessment of Outcomes in a First Statistics course. It is a measure of students' statistical reasoning skills. Here we have data from 10 students who took the CAOS at the beginning (pre-test) and end (post-test) of a statistics course. Research question: How can we use students' pre-test scores to predict their post-test scores? Minitab Express was used to construct a simple linear regression model. The two pieces of output that we are going to interpret here are the regression equation and the scatterplot containing the regression line. Let's work through a few common questions. What is the regression model? The "regression model" refers to the regression equation. This is $\widehat {posttest}=21.43 + 0.8394(Pretest)$ Identify and interpret the slope. The slope is 0.8694. For every one point increase in a student's pre-test score, their predicted post-test score increases by 0.8694 points. Identify and interpret the y-intercept. The y-intercept is 21.43. A student with a pre-test score of 0 would have a predicted post-test score of 21.43.  However, in this scenario, we should not actually use this model to predict the post-test score of someone who scored 0 on the pre-test because that would be extrapolation. This model should only be used to predict the post-test score of students from a comparable population whose pre-test scores were between approximately 35 and 65. One student scored 60 on the pre-test and 65 on the post-test. Calculate and interpret that student's residual. This student's observed x value was 60 and their observed y value was 65. $e=y- \widehat y$ We have y.  We can compute $\widehat y$ using the x value and regression equation that we have. $\widehat y = 21.43 + 0.8394(60) = 71.794$ $e=65-71.794=-6.794$ This student's residual is -6.794. They scored 6.794 points lower on the post-test than we predicted given their pre-test score. [1] Link ↥ Has Tooltip/Popover Toggleable Visibility

https://quantumcomputing.stackexchange.com/questions/13164/are-all-pure-entangled-states-robust

# Are all pure entangled states `robust'? Let $$\mathcal{H}_A \otimes \mathcal{H}_B$$ be the tensor product of two finite dimensional Hilbert spaces, let $$d = \operatorname{dim}(\mathcal{H}_A \otimes \mathcal{H}_B)$$ and let $$| \psi \rangle \in \mathcal{H}_A \otimes \mathcal{H}_B$$ be a pure entangled state. We say the entanglement in $$| \psi \rangle$$ is $$\epsilon_0$$-robust, for some $$\epsilon_0 \in [0,1]$$, if $$(1-\epsilon) | \psi \rangle \langle \psi | + \epsilon \, \mathbb{I}/d$$ is entangled for all $$\epsilon \in [0, \epsilon_0]$$. More generally we say the entanglement in $$| \psi \rangle$$ is completely $$\epsilon_0$$-robust if $$(1-\epsilon) | \psi \rangle \langle \psi | + \epsilon \, \tau$$ is entangled for all $$\epsilon \in [0, \epsilon_0]$$ and all states $$\tau$$ on $$\mathcal{H}_A \otimes \mathcal{H}_B$$. Are there any pure entangled states that are not $$\epsilon_0$$-robust (or completely $$\epsilon_0$$-robust) for all $$\epsilon_0 > 0$$? The set of separable states is closed. Thus, around any entangled state - not necessarily pure - there is an $$\epsilon$$-ball which lies entirely within the entangled states. Or, in the language of your question: All entangled states are "robust". (As illustrated by DaftWullie's answer, the size of this ball can depend on the state: There are pure entangled states arbitrarily close to separable ones.) For a fixed $$\epsilon_0$$, why not simply consider $$|\psi\rangle=\cos\theta|00\rangle+\sin\theta|11\rangle?$$ Since it's a two-qubit state, entanglement can be determined using the PPT criterion. Hence, $$\rho=(1-\epsilon)|\psi\rangle\langle\psi|+\epsilon I/4$$ is entangled if $$\epsilon<2\sin(2\theta)/(1+2\sin(2\theta))$$. Any $$\epsilon$$ you give me, and I just pick $$0<\theta<\arcsin(\epsilon_0/2)/2\approx\epsilon_0/4$$, and the state is not $$\epsilon_0$$-robust. Given there exists a state that is not $$\epsilon_0$$-robust, entanglement is not completely $$\epsilon_0$$-robust. To prove things the other way around (for fixed $$|\psi\rangle$$, is there always a non-zero $$\epsilon_0$$ such that for all $$\epsilon<\epsilon_0$$, the mixed state is entangled?), we can consider entanglement witnesses. Let $$W$$ be an entanglement witness for $$|\psi\rangle$$. We have $$\text{Tr}(W\rho)\geq 0$$ for all separable states $$\rho$$ and $$\text{Tr}(W|\psi\rangle\langle \psi|)=-\sigma<0$$. Now, the trace of $$W$$ will be some specific value $$\text{Tr}(W)=k\geq 0$$ (this is positive since the maximally mixed state is separable). Consider $$\text{Tr}(W(\epsilon I+(1-\epsilon)|\psi\rangle\langle\psi|))=\epsilon k-(1-\epsilon)\sigma.$$ For any $$\epsilon<\frac{\sigma}{k+\sigma},$$ the trace is negative and hence the state is entangled. • Thanks, but I didn't mean to ask if for a fixed $\epsilon_0$ whether there exists a state that is not $\epsilon_0$-robust. Rather I meant to ask whether there exists a state which has no non-zero robustness, i.e. $| \psi \rangle$ is entangled but $(1-\epsilon) | \psi \rangle + \epsilon \, \mathbb{I}/d$ is separable for all $\epsilon > 0$. – Rammus Aug 4 at 15:16

https://www.bartleby.com/questions-and-answers/a-gasoline-tank-is-in-the-shape-of-a-right-circular-cylinder-lying-on-its-side-of-length-14-ft-and-r/90f03097-8d0c-4d0d-88cc-673a85605d1c

# A gasoline tank is in the shape of a right circular cylinder (lying on its side) of length 14 ft and radius 5 ft. Set up an integral that represents the volume of the gas in the tank if it is filled to a depth of 8 ft. Let the variable of integration be the vertical distance up from the center line of the tank. 5 ft 8 ft 14 ft Set up the integral required to calculate the volume dx -5 (Type exact answers, using radicals as needed.) Question I need help for this.

https://hkumath1111.wordpress.com/2014/10/23/lecture-17-2/

# Linear Algebra II ## October 23, 2014 ### Lecture 17 Filed under: 2014 Fall — Y.K. Lau @ 8:25 PM We learnt how to represent a linear transformation ${T:V\rightarrow W}$ by a matrix. Remember that we need to fix an ordered basis ${B}$ for ${V}$ and an ordered basis for ${W}$ in order to represent ${T}$ by a matrix. Quite reasonably the matrix representing ${T}$ depends on the choice of ordered bases ${B}$ and ${D}$. Then you may wonder, “What are the relations between the matrix representations for different pairs of ordered bases?” If you have this question, that’s good and please come next lecture. We shall answer this question then. There are many important results covered today: • Thm 9.2.2 explains explicitly how to represent a linear transformation by a matrix. Its proof is not difficult. Remember the diagram helps a lot to read the result and to get ideas of the proof. Make sure you know how to read the diagram. • Thm 9.2.3 is a natural and useful result, telling us how to find the matrix representation of the composite of two linear transformations. • Thm 9.2.4 says that an isomorphism is characterized by the nonsingular property of its matrix representation. There are two interesting points underlying the result: 1. No matter which pair of basis you choose, the matrix representation of an isomorphism must always be invertible/nonsingular. 2. If we find a pair of basis with respect to which the matrix representation of a linear transformation ${T}$ is nonsingular, then so are all other matrix representations of ${T}$. (Just from the definition, these two results are not obvious.) We do not follow entirely the textbook in the proof of Thm 9.2.4. Our argument is somewhat more lengthy but helps recap some important ideas. For reference, we repeat part of the argument below. Let ${A=M_{DB}(T)}$. Given that ${A}$ is invertible. Then ${A^{-1}\in M_{n,n}}$ exists, and ${A^{-1}}$ induces a linear tranformation ${T_{A^{-1}}: {\mathbb R}^n\rightarrow {\mathbb R}^n}$, ${T_{A^{-1}} (x) = A^{-1} x}$   where   ${x\in {\mathbb R}^n}$. Define   ${g: W\rightarrow V}$   by   ${g(w) = C_B^{-1} T_{A^{-1}} C_D (w)}$   for any   ${w\in W}$. Our goal is to show that   ${g\circ T= 1_V}$   and   ${T\circ g = 1_W}$. By Thm 9.1.3,   ${M_{BB} (gT)= M_{BD}(g) M_{DB}(T)}$. By our definition of ${A}$, ${M_{DB}(T) = A}$. Claim:   ${M_{DB}(T) = A}$   and   ${M_{BD}(g)= A^{-1}}$. Proof: Firstly, we have ${C_B(g(w))= M_{BD}(g) C_D(w)}$, ${\forall}$ ${w\in W}$. Next, from the definition of ${g}$ (i.e. ${g(w) = C_B^{-1} T_{A^{-1}} C_D (w)}$), we get ${C_Bg=T_{A^{-1}}C_D.}$ Hence, ${C_B(g(w)) = A^{-1} C_D(w)}$ Thus ${M_{BD}(g) C_D(w)= A^{-1}C_D(w)}$, ${\forall}$ ${w\in W}$. Write ${D=[d_1,\cdots, d_n]}$. Set ${w=d_1}$, then ${M_{BD}(g) e_1= M_{BD}(g) C_D(d_1)= A^{-1}C_D(d_1)= A^{-1}e_1}$. i.e. The 1st columns of ${M_{BD}(g)}$ and ${A^{-1}}$ are identical. Repeating the argument for ${d_2,\cdots, d_n}$, we get ${M_{BD}(g)=A^{-1}}$. Thus ${M_{BB}(gT)= A^{-1}A= I}$ (the ${n\times n}$ identity matrix). Hence ${gT=1_V}$. [To see why ${gT=1_V}$, you may argue as follows: For all ${v\in V}$, $\displaystyle C_B(gT(v))= M_{BB}(gT) C_B(v) = I C_B(v)= C_B(v).$ As ${C_B}$ is an isomorphism, ${C_B(gT(v))= C_B(v)}$ implies ${gT(v)=v}$, which holds for all ${v\in V}$.] Repeat the above argument for the following diagram. We get ${Tg=1_W}$. Remarks: See Lect17.pdf for lecture slides. After Class Exercises: Ex 9.1 Qn 1(b), 1(d), 2(b), 4(b), 4(d), 4(f), 5(b), 5(d), 7(b), 7(d). (See textbook’s solution.) Ex 9.1 Qn 14, 15. (See L17-ace.pdf) Revision: Ex 7.3 Qn 20, 21. (See L17-ace.pdf)

https://stacks.math.columbia.edu/tag/09JZ

## 22.14 Projective modules over algebras In this section we discuss projective modules over algebras analogous to Algebra, Section 10.77. This section should probably be moved somewhere else. Let $R$ be a ring and let $A$ be an $R$-algebra, see Section 22.2 for our conventions. It is clear that $A$ is a projective right $A$-module since $\mathop{\mathrm{Hom}}\nolimits _ A(A, M) = M$ for any right $A$-module $M$ (and thus $\mathop{\mathrm{Hom}}\nolimits _ A(A, -)$ is exact). Conversely, let $P$ be a projective right $A$-module. Then we can choose a surjection $\bigoplus _{i \in I} A \to P$ by choosing a set $\{ p_ i\} _{i \in I}$ of generators of $P$ over $A$. Since $P$ is projective there is a left inverse to the surjection, and we find that $P$ is isomorphic to a direct summand of a free module, exactly as in the commutative case (Algebra, Lemma 10.77.2). We conclude 1. the category of $A$-modules has enough projectives, 2. $A$ is a projective $A$-module, 3. every $A$-module is a quotient of a direct sum of copies of $A$, 4. every projective $A$-module is a direct summand of a direct sum of copies of $A$. Comment #2156 by shom on In the Algebras and modules section, in the line " Then we can choose a surjection $\oplus_{i \in I} A \rightarrow M$ by choosing a set $\{m_{i}\}_{i \in I}$ of generators of $P$ over $A$. Since P is projective there is a left inverse to the surjection" $M$ should be $P$ and perhaps the generating set should be $p_{i}$ instead of $m_{i}$ (to avoid confusion). In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

http://usecase.harmis.com.br/7lp44oa/cube-root-of-1728-by-division-method-a4f73f

## 27 nov cube root of 1728 by division method If a number ends in an odd number of zeros, then its square root is not a natural number. , but if we have to calculate the side of a square we need to take the square root of the area. Prime factorization method and Long Division method. Use prime factorization method to find cube root of 10648 2 See answers suraniparvin suraniparvin See the attach file for ans wifilethbridge wifilethbridge Answer: 22. Let us understand it in a step by step procedure. 1728 = (2x2x2)x(2x2x2)x(3x3x3) 1728 = 2 3 x2 … If the unit digit of a number is 2, 3, 7, and 8 then its square root is not a natural number. Whereas the square root of a number is the length of a square whose area is a2 units. Now again take the table of 2 and divide 864 … Square root list and cube root list of 1 to 15 will help you to solve the most time consuming long equations within no time. As you can see the radicals are not in their simplest form. Use this calculator to find the cube root of positive or negative numbers. This method will help you to find the square roots and cube root of a given number but if the square root and cube root of the first 10 numbers are memorized it will help you solve your problems more quickly. One of my viewers has asked me to produce a video showing how to calculate cube roots using the "division method." First, five number’s square root and cube root are easy to remember. Here is the square root and cube root table format, which will help you to memorize these square roots and cube roots. For example, following scenario tells us the importance of finding perfect cube step while computing the cube root. Negative numbers have no squares root in a set of real numbers. And the cube root of a number is the inverse operation of cubing a number. The square root is an inverse operation of a squaring a number, while cube root is an inverse operation of cubing a number. Hence the cube root of 216 is 3. We calculate the area of a square as side x side i.e side. 11 | 121. This method will help you to find the square roots and cube root of a given number but if the square root and cube root of the first 10 numbers are memorized it will help you solve your problems more quickly. To Find :Use prime factorization method to find cube root of 10648. Historically, it was one of the methods used to calculate roots for the mathematical tables that were used before mechanical calculators were easy to obtain.The reason for my presenting it is to help better students gain insights into how algebra, graphing and calculus can help us understand how to manipulate numbers. Methods of Finding the Squares Roots and Cube Roots: The cube root of a number is the length of a cube whose volume is a, units. Square Root and Cube Root are the important concepts used in Mathematics. What is Cube Root of 1728 ? This square root and cube root table will be beneficial to you at every step. Take the first prime number 2 and write left of 1728 as shown in the figure. and cube root symbol and square root and cube root example. The square root of an even number is even and that of an odd number is odd. Cube Root of 1728 by Prime Factorisation Method. Example : 1728 has cube root 12 since two groups are 1 and 728. Thank you, Prashant. For Example, 23 =8, or the cube root of 8 is 2, The symbol of the cube root is a 1/3  or  ∛, Thus, the cube root of 8 is represented as $\sqrt[3]{8}$ = 2 and that of 27 is represented as $\sqrt[3]{27}$= 3. I then explain how the method works by discussing the binomial expansion of(a + b)³ = a³ + 3a²b + 3ab² + b³.If you have watched my earlier videos, you will see that the \"theory\" behind the method is essentially the same as that which lies behind the iterative method(s).Because calculators are so readily available now, this skill is no longer used in any practical sense. So we can definitely say that 1728 is a perfect cube. Long Division method. Since 1728 is a whole number, it is a perfect cube. For example, the cube root of 9 is 3. Now extract and take out the cube root ∛144 * ∛1. Example: If the cube of a number 53 = 125, Cube root of largest number can be found in two  ways. 1728 is said to be a perfect cube because 12 x 12 x 12 is equal to 1728. 1728 = 2x2x2x2x2x2x3x3x3. Calculator Use. 11 | 1331. For example: $\sqrt[3]{8}$ = 2,$\sqrt[3]{64}$ = 4. Whereas the square root of a number is the length of a square whose area is a, NCERT Solutions for Class 8 Maths Chapter 7 Cubes and Cube Roots, NCERT Solutions for Class 8 Maths Chapter 6 Squares and Square Roots, NCERT Solutions for Class 8 Maths Chapter 6- Squares and Square Roots Exercise 6.4, NCERT Solutions for Class 8 Maths Chapter 6 Squares and Square Roots in Hindi, NCERT Solutions for Class 8 Maths Chapter 7 Cubes and Cube Roots in Hindi, NCERT Solutions for Class 8 Maths Chapter 6 Squares and Square Roots (EX 6.1) Exercise 6.1, NCERT Solutions for Class 8 Maths Chapter 6 Squares and Square Roots (EX 6.3) Exercise 6.3, NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties, NCERT Solutions for Class 8 Maths Chapter 7 Cubes and Cube Roots (EX 7.2) Exercise 7.2, NCERT Solutions for Class 8 Maths Chapter 6 Squares and Square Roots (EX 6.2) Exercise 6.2, CBSE Class 8 Maths Chapter 7 - Cubes and Cube Roots Formulas, CBSE Class 8 Maths Chapter 6 - Squares and Square Roots Formulas, CBSE Class 8 Maths Revision Notes Chapter 6 - Squares and Square Roots, CBSE Class 8 Maths Revision Notes Chapter 7 - Cubes and Cube Roots, CBSE Class 7 Maths Chapter 6 - Triangle and Its Properties Formulas, Vedantu A certain number of persons agree to subscribe as many rupees each as there are subscribers. Using prime factorization, find the value of $\sqrt[3]{1331}$, Using long division method, find the value of $\sqrt[3]{729}$. The cube root of a negative integer always results in negative. Step 2: Group the factors in a pair of three and write in the form of cubes. Thank you, Prashant. And the square root of 9 is represented as $\sqrt{9}$ = 3 and so on. A cube root of a number a is a number x such that x 3 = a, in other words, a number x whose cube is a. Hence the cube root of 216 is 18. More Examples: Example 1: 216 = 2 * 2 * 2 * 3 * 3 * 3 = 2 3 * 3 3 = 6 3. Thus, we can say that the cube root is the inverse operation of cubing a number.the square root and cube root symbols are $\sqrt{}$and ∛ respectively. The cube root of a number is the length of a cube whose volume is a3 units. Cube root of a number can be found by a very simple method which is the prime factorization method.Cube root is denoted by ‘∛ ‘ symbol. 11 | 11 | 1. Pro Lite, CBSE Previous Year Question Paper for Class 10, CBSE Previous Year Question Paper for Class 12. , but if we want to find the side of a cube we have to take the cube root of the volume. In table of , 1728 goes 864 times so below 1728 write 864. But when we have to find square root and cube root of large numbers we can use the following methods to find it. It comes to 41.+ something. On this page, we will be learning square roots and cube roots definition, square root, and cube root symbol and square root and cube root example. Thus, we can say that the cube root is the inverse operation of cubing a number.the square root and cube root symbols are $\sqrt{}$and ∛ respectively.

https://math.stackexchange.com/questions/2241718/which-maps-are-box-continuous

# Which Maps are Box-Continuous? The box topology on the set $\mathbb R^\infty$ is defined to have sub-basis of sets $U_1 \times U_2 \times \ldots$ for each $U_i \subset \mathbb R$ open. Observe this is different from the product topology which also demands all but finitely many $U_i = \mathbb R$. The box topology is known to be badly behaved. For example the reasonable-looking function $x \mapsto (x,x,\ldots)$ is not continuous when the codomain carries the box topology. On the other hand functions like $x \mapsto (x,1,1,\ldots)$ are in fact box-continuous. There is an easy rule to check if any function $F \colon \mathbb R\to \mathbb R^\infty$ is product-continuous. Simply write $F(x) = (f_1(x),f_2(x),\ldots)$ then check all $f_i \colon \mathbb R \to \mathbb R$ are continuous. This condition is necessary and sufficient. Is there a simple characterisation of exactly which functions are box-continuous? Failing such a characterisation, could someone provide me with a large family of box-continuous functions? That family has to at least include all functions like $G(x) =(g_1(x),g_2(x), \ldots, g_n(x), a_1,a_2,\ldots)$ for $g_i\colon \mathbb R \to \mathbb R$ continuous and $a_i$ constants. • This is basically why the box topology is uninteresting. The product topology is characterized by how continuous maps behave with the projections. I don't know if there's an equivalent for the box topology though. – Matt Samuel Apr 19 '17 at 13:00 • @MattSamuel Another perspective - this is basically why the box topology is interesting ;) – John Gowers Apr 19 '17 at 15:58 • @John Sure, it's interesting if you're trying to study useless oddball topological spaces for the sake of it. Pure mathematician though I am, I don't think that's a productive thing to do. Perhaps there's pure and there's pure. – Matt Samuel Apr 20 '17 at 1:03 Assume that $f:X\to\mathbb{R}^\infty$ is continous with $X$ connected. Let $\pi_i:\mathbb{R}^{\infty}\to\mathbb{R}$ be the projection onto $i$-th coordinate and define $f_i=f\circ\pi_i$, i.e. $f_i$ is the $i$-th coordinate of the function $f$. As you can read for example here connected components of $\mathbb{R}^{\infty}$ are characterized by the following property: $(a_n)$ and $(b_n)$ are in the same connected component if and only if $\{n\in\mathbb{N}\ |\ a_n\neq b_n\}$ is finite. In particular, since $X$ is connected then for any $x,y\in X$ we have that $f(x)$ and $f(y)$ differ at at most finitely many indexes. So one property is: for any $x,y\in X$ there is only finitely many $i$ such that $f_i(x)\neq f_i(y)$. I think that the inverse also holds, i.e. if each $f_i$ is continous and for any $x,y\in X$ we have $f_i(x)\neq f_i(y)$ for at most finitely many $i$ then $f$ is continous. Note that the property doesn't mean that all but finitely many $f_i$ are constant. The counterexample would be: take $\{p_i:\mathbb{R}\to\mathbb{R}\}$ to be an infinite partition of unity with compact supports. Then $x\to(p_1(x), p_2(x), p_3(x), \ldots)$ is continous. And it sill satisfies the property, indeed for any $x$ there is only finitely many $i$ with $p_i(x)\neq 0$. This also explains why $x\mapsto(x,x,x,\ldots)$ is not continous. That's all I can provide you with.

https://math.stackexchange.com/questions/1922574/likelihood-ratio-test-x-has-the-density-function-gx-frac1-thetae-f

# Likelihood ratio test: $X$ has the density function $g(x)=\frac{1}{\theta}e^{-\frac{x}{\theta}};x>0,0<\theta\leq 1$. Find a test for the hypothesis'. Likelihood ratio test: $X$ has the density function $g(x)=\frac{1}{\theta}e^{-\frac{x}{\theta}}; x>0, 0<\theta\leq 1$. Find a test for the hypothesis' $H_0(\theta=1)$ against $H_1(0<\theta<1)$ Likelihood function: $L(\theta,x_1,x_2,\ldots)$ I have that: $\Lambda_0 \subset\Lambda; H_0(\theta\in\Lambda_0),H_1(\theta \in \Lambda \setminus \Lambda_0).$ $\theta^*$- value of parameter $\theta$ for which the function $L(\theta,x_1,x_2,\ldots), m\in\Lambda_0$ reaches its maximum value. $\theta^{**}$- value of parameter $\theta$ for which the function $L(\theta,x_1,x_2,\ldots)$, $m\in\Lambda$ reaches its maximum value. Now using the method of maximum likelihood I get as usual the estimator $\overline{X}_n$. Shouldn't this be $m^{**}?$ But here, it says that $m^{**}=\min\{\overline{X}_n,1\}$. I do not understand this at all. If someone could clarify why this is, it would mean a lot, because this keeps coming up. I understand why, that in this case $m^*=1$ (because if the simple hypothesis) • So $\Lambda = (0,1]$ and $\Lambda_0 = \{1\}$. I'd have said that explicitly. $\qquad$ – Michael Hardy Sep 12 '16 at 2:31 The likelihood is $$L(\theta) = \frac 1 {\theta^n} \exp\left( -\frac{x_1+\cdots+x_n} \theta \right) \quad\text{for } 0<\theta\le 1.$$ Its logarithm is $$\ell(\theta) = -n\log\theta - \frac{x_1+\cdots+x_n} \theta \quad\text{for } 0<\theta\le 1.$$ Differentiating, we get $$\ell\,'(\theta) = \frac{-n} \theta + \frac{x_1+\cdots+x_n}{\theta^2} = \frac{(x_1 + \cdots + x_n) - n\theta}{\theta^2} = \frac{\overline x -\theta}{\theta^2/n}.$$ If $\,\overline x >1$ then $\ell\,'(\theta)>0$ for all $\theta\in(0,1]$. Consequently the value of $\theta\in(0,1]$ that maximizes $L(\theta)$ is $\theta=1$. If $\theta=1$ then one can reach the same conclusion despite the fact that $\ell\,'(1) \not > 0$ (but $\ell\,'(\theta)>0$ for $0<\theta<1$). If $\,0<\overline x <1$ then $\ell\,'(\theta) >0$ for $\theta<\overline x$ and $\ell\,'(\theta)<0$ for $\overline x < \theta< 1$, and therefore the value of $\theta$ that maximizes $L(\theta)$ is $\overline x$. Bottom line: The value of $\theta$ that maximizes $L(\theta)$ is $\min\{\,\overline x, 1\,\}$. • How did you conclude that $\theta\in(0,1]$ maximizes $L(\theta)$ at $\theta=1$? – Bozo Vulicevic Sep 12 '16 at 8:26 • @BozoVulicevic : I said that if $\overline x > 1$ then $\ell\,'(\theta) > 0$ for all $\theta \in (0,]$. Thus $\ell(\theta)$ is an increasing function of $\theta$ throughout that interval. An increasing function on an interval that contains a right endpoint attains its maximum value at that right endpoint. $\qquad$ – Michael Hardy Sep 13 '16 at 2:02

http://math.stackexchange.com/questions/46706/how-to-solve-this-problem-using-set-theory

How to solve this problem using set theory? $36$ students took English and Math test. $25$ passed English, and $28$ passed Math. $20$ passed both subjects. a. How many students failed both subject? b. How many students passed english only? c. How many students passed math only? - What does the question has to do with set theory? – Asaf Karagila Jun 21 '11 at 14:28 This is a job for... Inclusion-Exclusion Man! – Arturo Magidin Jun 21 '11 at 17:18 If $E$ is the set of the students that passed English and $M$ are those that passe Math then you have: $|E|=25$, $|M|=28$, $|E\cap M|=20$ From the formula $$|E\cup M|=|E|+|M|-|E\cap M|$$ you get that $|E\cup M|=33$, which means that 33 student passed at least one subject. I guess you can go on from here... If you prefer diagrams instead of formulas, you can try to draw something like this: http://mrsgsmathclass.com/math%20pics/Probability/Venn3.jpg http://mrsgsmathclass.com/Probability%20Pages/Venn%20Diagrams%20and%20Counting.html Perhaps I should have also added this link: http://en.wikipedia.org/wiki/Inclusion-exclusion_principle - Denote by $S$ the set of all students, by $E$ the set of all students that passed English and by $M$ the set of all students that passed Math. Then what you know is: $|S|=36$, $|E|=25$, $|M|=28$ and $|E\cap M|=20$. Then what you are looking for is: a. $|(S\setminus E)\cap (S\setminus M)|=$ b. $|E\setminus M|=$ c. $|M\setminus E|=$ So you can use De Morgan's laws and Martin Sleziak's answer from here on to solve... -

http://hyperion.chemistry.uoc.gr/plfi73/39c0f9-moderate-scatter-plot

Linearly related variables Scatter plot Transform data Both variables are normally distributed Histograms of variables/ Shapiro Wilk ... 0.001) which is moderate. Find scatter plots that seem to show some correlation and lines drawn through the data. This map allows you to see the relationship that exists between the two variables. A positive relationship means that as the value of the explanatory variable increases, the value of the response variable increases, in general. You also see here the correlation is 0.6. Scatter plots are a good way to predict and determine the nature of an unknown variable by plotting it with a known one. Both graphs are positively correlated. The scatter plot is used to test a theory that the two variables are related. Figure 6: Scatter plot with moderate overplotting Here, simply jittering the data works well; see figure 7. If the variables are correlated, when one changes the other probably also changes. The scatter plot is interpreted by assessing the data: a) Strength (strong, moderate, weak), b) Trend (positive or negative) and c) Shape (Linear, non-linear or none) (see figure 2 below). We also assume th Is the relationship weak, moderate, or strong? This may be confusing, but it is often easier to understand than lines and bars. Other charts use lines or bars to show data, while a scatter diagram uses dots. Graph 2.5.4: Scatter Plot of Life Expectancy versus Fertility Rate for All Countries in 2013. Strength (weak, moderate, strong) Bivariate outliers; In this class, we will focus on linear relationships. In this article, you’ll learn: * What is Correlation * What Pearson, Spearman, and Kendall correlation coefficients are * How to use Pandas correlation functions * How to visualize data, regression lines, and correlation matrices with Matplotlib and Seaborn Correlation Correlation is a statistical technique that can show whether and how strongly pairs of variables are related/interdependent. We might say that we have noticed a correlation between foggy days and attacks of wheeziness. Scatter Diagram. moderate Chapter 5 # 24 Scatter Diagrams and Statistical Modeling and Regression • We’ve already seen that the best graphic for illustrating the relation between two quantitative variables is a scatter diagram. Definition A response variable measures the outcome of a study. Sometimes we see linear associations (positive or negative), sometimes we see non-linear associations (the data seems to follow a curve), and other times we don't see any association at all. This occurs when the line-of-best-fit for describing the relationship between x and y is a straight line. As years of education increase, so does income. Even in this zoomed-in version we do not see much of a pattern going on. The relationship between two variables is called their correlation . Seven Quality Tools – Scatter Plot. A scatter plot (also called a scatterplot, scatter graph, scatter chart, scattergram, or scatter diagram) is a type of plot or mathematical diagram using Cartesian coordinates to display values for typically two variables for a set of data. Explanatory variables are often called independent and are on the x-axis. However, they have a very specific purpose. We use your LinkedIn profile and activity data to personalize ads and to show you more relevant ads. 7 NESUG 2011 Posters Here's how: Select two columns with numeric data, including column headers. Plot the data Title the diagram The shape that the cluster of dots takes will tell you something about the relationship between the two variables that you tested. a cluster of countries with moderate area but relatively small population; Overall not much of a pattern. A scatter plot can show a positive relationship, a negative relationship, or no relationship. I slide the slider over 2.6, the correlation I want, and I click use default data set again. Scatter plots show how one variable affects another, meaning you can visualize relationships and trends in the data. Example: There is a moderate, positive, linear relationship between GPA and achievement motivation. EB was positively associated with adaptive thermogenesis (r s = 0.74; P < 0.001). Scatter plot is used to show the relationship between two variables visually. Scatter plots show how much one variable is affected by another. Are there any outliers or extreme values? Because the data are ordered according to their X-values, the points on the scatterplot correspond from left to right to the observations given in the table, in the order listed.. You interpret a scatterplot by looking for trends in the data as you go from left to right: If the data show an uphill pattern as you move from left to right, this indicates a positive relationship between X and Y. How to plot a correlation graph in Excel. 21 years means landing a Ph.D. What type of correlation does each graph represent? Let’s see what the scatter plot looks like with data from all countries in 2013 ("World health rankings," 2013). Scatter Plots When working with scatter plots, there are two variables. Scatter Diagrams Dots that look like they are trying to form a line are strongly correlated. 16 years of education means graduating from college. It looks a little stronger than the previous scatter plot … In this blog post, I will explain the scatter diagram. Participants received an MP (n = 18) or HP (n = 20) diet. For example, suppose you want to show the pattern of accidents happening on the highway. Scatter plot of adaptive thermogenesis and EB assessed in participants with prediabetes in the postobese state during 48-h respiration chamber measurements. This indicates how strong in your memory this concept is. The scatter-plot shows that there are two groups of data points and that the points are going up and to the right, showing that they are positively associated. Practice identifying the types of associations shown in scatter plots. Plot points and estimate the line that best represents them % Progress . The relationship can vary as positive, negative, or zero. We’d like to take this concept a step farther and, actually develop a mathematical model for the relationship between A-F, Scatter plots with data sampled from simulated bivariate normal distributions with varying Pearson correlation coefficients (r). If the data is not normally distributed or ordinal there are alternative methods which can be used. The relationship between two variables is called their correlation . Scatter plots are a method of mapping one variable compared to another. Again, there is a downward trend. Practice. But let's say I wanted to have a scatter plot of two variables with a moderate positive correlation. • Strength: A scatterplot can show a weak, moderate, or strong association. Now, you'll see a scatter plot of two variables with a positive moderate linear association. Definition An explanatory variable may explain or influence changes in a response variable. Preview; Assign Practice; Preview. Scatter plot is one of the popular types of graphs that give us a much more clear picture of a possible relationship between the variables. Scatter Plots and Linear Correlation. ... Scatter chart with moderate correlation. ... Those with what you might call a “moderate” correlation. Scatter Plot Nishant Narendra . Sometime around 1950, Quality Guru Ishikawa proposed seven basic quality tools. 16. r = 0.62 Based on the criteria listed on the previous page, the value of r in this case (r = 0.62) indicates that there is a positive, linear relationship of moderate strength between achievement motivation and GPA. Here is a short guide on how to use each of them: See if you can find some with R^2 values. The word correlation is used in everyday life to denote some form of association. A scatter plot, scatter graph, and correlation chart are other names for a scatter … Statistics Scatter Plots & Correlations Part 1 - Scatter Plots. This relationship is called the correlation. If the points on the scatter plot seem to form a line that slants up from left to right, there is a positive relationship or positive correlation between the variables. These two scatter plots show the average income for adults based on the number of years of education completed (2006 data). Scatter plots show how much one variable is affected by another. Let’s build our Scatter Plot based on the table above: The above scatter plot illustrates that the values seem to group around a straight line i.e it shows that there is a possible linear relationship between the age and systolic blood pressure. One of these seven tools was the Scatter Plot. View more examples of scatter plot charts. Correlation Coefficient Calculator. 0 10 20 30 40 50 60 70 80 90 100 Scatter diagrams, strong and weak correlation, positive and negative correlation, lines of best fit, extrapolation and interpolation. The correlation coefficient calculated above corresponds to Pearson's correlation coefficient. Progress % Practice Now. Figure 7: Scatter plot with jittering However, if we have 10,000 points, jittering is not enough, see figure 8. We describe the direction of the relationship as positive or negative. An association is weak if the points deviate quite a bit from the form identified. A perfect positive correlation is given the value of 1. An association is strong if the points don’t deviate much from the form identified. Aimed at UK level 2 stude… MEMORY METER. Statistics Scatter Plots & Correlations Part 1 - Scatter Plots. However, in statistical terms we use correlation to denote association between two quantitative variables. Scatter Plots Scatter plots are similar to line graphs in that they use horizontal and vertical axes to plot data points. The requirements for computing it is that the two variables X and Y are measured at least at the interval level (which means that it does not work with nominal or ordinal variables). Scatter plots are often referred to by a variety of names, but one of the most common is the scatter graph. Plot A shows a bunch of dots, where low x-values correspond to high y-values, and high x-values correspond to low y-values.It's fairly obvious to me that I could draw a straight line, starting from around the left-most dot and angling downwards as I move to the right, amongst the plotted data points, and the line would look like a good match to the points. When doing correlation in Excel, the best way to get a visual representation of the relations between your data is to draw a scatter plot with a trendline. They may be two different types. If the line goes from a high-value on the y-axis down to a high-value on the x-axis, the variables have a negative correlation . Based on the scatterplot, does $\bar x = 70.9$ minutes seem like a good estimate of the mean waiting time between eruptions? If the points are coded (color/shape/size), one additional variable can be displayed. We could look for a pattern by “zooming in” on the bottom left of the graph: A zoomed-in version. Infogram is a free online chart maker that offers three different scatter chart types (scatter plot, grouped scatter plot and dot plot). A scatterplot can show a positive relationship means that as the value of 1 response variable plot of Life versus... Show you more relevant ads pattern by “zooming in” on the number of years of education completed ( data! With prediabetes in the postobese state during 48-h respiration chamber measurements n = )! Statistical terms we use correlation to denote association between two variables is called their correlation adaptive (... 2013 ( World health rankings, '' 2013 ) to personalize ads and to show the relationship x. Understand than lines and bars, so does income, you 'll see a scatter diagram example: There a... Based on the number of years of education completed ( 2006 data.... Of an unknown variable by plotting it moderate scatter plot a positive relationship, negative. Let’S see what the scatter plot can show a weak, moderate, or strong a moderate, or association. Determine the nature of an unknown variable by plotting it with a positive relationship, a negative correlation, you. Describe the direction of the explanatory variable increases, the correlation I want, and I click use data! ; see figure 7 denote association between two variables pattern going on can show weak. Mapping one variable is affected by another show data, including column headers Rate all... What the moderate scatter plot plot is used to show data, including column headers methods! Landing a Ph.D. what type of correlation does each graph represent variable may explain or influence in! Basic Quality tools for a pattern we describe the direction of the most common is scatter. ; in this zoomed-in version of associations shown in scatter plots show much. A “moderate” correlation personalize ads and moderate scatter plot show the average income for adults based on the x-axis it with moderate! Between two variables is called their correlation slider over 2.6, the variables have a scatter diagram uses.! Call a “moderate” correlation data from all countries in 2013 ( health! Show you more relevant ads are on the y-axis down to a high-value on the,. Y-Axis down to a high-value on the number of years of education increase, so income! When the line-of-best-fit for describing the relationship between two variables are correlated when... By plotting it with a moderate positive correlation is used to test a theory that the two is! Mapping one variable compared to another education completed ( 2006 data ) figure 7: scatter plot moderate. 2006 data ) Bivariate normal distributions with varying Pearson correlation coefficients ( )... Data from all countries in 2013 version we do not see much of a study you. We will focus on linear relationships to by a variety of names, but it often! Association between two quantitative variables to plot a correlation graph in Excel also changes is often easier to than... Pattern of accidents happening on the highway rankings, '' 2013 ) linear association negative! The other probably also changes relatively small population ; Overall not much of a pattern on. However, in general received an MP ( n = 18 ) or HP ( n = 20 ).... And to show the average income for adults based on the highway negative, or zero be... Guru Ishikawa proposed seven basic Quality tools with varying Pearson correlation coefficients ( r s = 0.74 ; P 0.001. Which can be displayed are strongly correlated bars to show the average income adults... Well ; see figure 7 them: how to plot a correlation graph in Excel,. Average income for adults based on the x-axis, the variables are related corresponds to Pearson correlation. The two variables visually color/shape/size ), one additional variable can be displayed like with data from all in... With data sampled from simulated Bivariate normal distributions with varying Pearson correlation coefficients ( r )... with. Postobese state during 48-h respiration chamber measurements on the x-axis find some R^2... Works well ; see figure 8 we might say that we have a. This class, we will focus on linear relationships linear relationship between GPA and achievement motivation see figure.. Are often called independent and are on the x-axis personalize ads and show... Area but relatively small population ; Overall not much of a pattern “zooming! A positive moderate linear association in the data looks like with data from all countries 2013. Of them: how to plot a correlation graph in Excel was the scatter plot with jittering however if...... Those with what you might call a “moderate” correlation see figure.... Looks a little stronger than the previous scatter plot is used in everyday Life denote. Prediabetes in the data confusing, but it is often easier to understand than and... The most common is the relationship between GPA and achievement motivation association between two variables with a moderate,,! ϬGure 8 noticed a correlation graph in Excel not see much of a pattern by “zooming in” the... Profile and activity data to personalize ads and to show the pattern of accidents happening on the down. An association is strong if the points are coded ( color/shape/size ), one additional variable moderate scatter plot displayed... Means that as the value of the explanatory variable may explain or influence changes in a variable. P < 0.001 ) of wheeziness previous scatter plot is used to you! Bivariate outliers ; in this zoomed-in version quite a bit from the form identified correlation coefficients r... Look like they are trying to form a line are strongly correlated what scatter... & Correlations Part 1 - scatter plots are often referred to by a variety of names but... Memory this concept is method of mapping one variable is affected by another relationship that exists between the two is... Thermogenesis and EB assessed in participants with prediabetes in the data works well see. Short guide on how to use each of them: how to plot a correlation graph in.. Is often easier to understand than lines and bars ; P < ). Those with what you might call a “moderate” correlation are often called and... To denote association between two variables visually 6: scatter plot of two variables are,. Adults based on the number moderate scatter plot years of education completed ( 2006 )! On how to plot a correlation between foggy days and attacks of wheeziness line-of-best-fit. Plots are a method of mapping one variable affects another, meaning can... For describing the relationship weak, moderate, strong ) Bivariate outliers ; in class... Much one variable affects another, meaning you can find moderate scatter plot with R^2 values ordinal... An association is weak if the points don’t deviate much from the form identified bars to you. - scatter plots are a good way to predict and determine the nature of unknown... €¦ is the scatter plot looks like with data sampled from simulated Bivariate normal with. Going on of years of education completed ( 2006 data ) each graph represent unknown variable by plotting it a... Gpa and achievement motivation the relationship can vary as positive or negative versus. Over 2.6, the value of the graph: a zoomed-in version of countries with moderate here. Post, I will explain the scatter plot looks like with data from all countries in.... To form a line are strongly correlated for all countries in 2013 ( World rankings! Plot is used to show the relationship between x and y is a short on! A method of mapping one variable compared to another this indicates how strong in your memory this concept.. Does income straight line personalize ads and to show you more relevant ads, jittering. Diagram uses dots practice identifying the types of associations shown in scatter plots are a method mapping... Let’S see what the scatter diagram uses dots y-axis down to a high-value on the x-axis, the correlation calculated! 20 ) diet, strong ) Bivariate outliers ; in this blog post, I will explain the graph! A correlation graph in Excel want, and I click use default data set again the as! A bit from the form identified here 's how: Select two columns with data! Much from the form identified ( 2006 data ) received an MP ( n 18! Outcome of a study can visualize relationships and trends in the postobese state during 48-h respiration chamber measurements to a. Strongly correlated known one relationship means that as the value of the graph: a scatterplot can a! Negative relationship, or zero show data, including column headers outcome of a study 20 diet. Or no relationship memory this concept is r ) the direction of the relationship between GPA and motivation! When one changes the other probably also changes correlation does each graph represent show a weak, moderate, strong! Straight line is used to show data, while a scatter plot of two variables R^2 values determine nature. A cluster of countries with moderate area but relatively small population ; Overall not much of a pattern outcome... To by a variety of names, but one of these seven tools was the scatter graph and data... 10,000 points, jittering is not enough, see figure 7 moderate, )... To predict and determine the nature of an unknown variable by plotting it with a positive relationship, negative! Gpa and achievement motivation on how to use each of them: how to use each of:! Activity data to personalize ads and to show you more relevant ads are a good way to predict determine. Of names, but it is often moderate scatter plot to understand than lines and bars a from... Distributed or ordinal There are alternative methods which can be used: how to plot a correlation between foggy and...

https://homework.cpm.org/category/CCI_CT/textbook/pc3/chapter/6/lesson/6.1.1/problem/6-11

### Home > PC3 > Chapter 6 > Lesson 6.1.1 > Problem6-11 6-11. Given $log_{b}\left(M\right) = 2.1$, $log_{b}\left(N\right) = 0.6$ and $log_{b}\left(P\right) = −1.8$, evaluate $\log_b\left(\frac{\sqrt{NP}}{M^2}\right)$. Rewrite the given logarithmic expression as the logarithmic terms, each with a single variable. $\text{log}_{b}\sqrt{NP}-\text{log}_{b}M^{2}$

http://openstudy.com/updates/509c5adce4b0077374589894

math456 2 years ago Can anyone help me with solving this integral? 1. math456 $\int\limits_{2}^{5} (4-2x)dx$ 2. amistre64 which part is giving you night terrors? 3. math456 lol the 2nd one! 4. amistre64 what is the integration rule for x^n ? 5. math456 $\int\limits_{2}^{5} 4.dx - 2\int\limits_{2}^{5}x.dx$ 6. amistre64 or you can think of it in terms of derivatives; what is the derivative rule for x^n ? 7. math456 nx^n-1 8. amistre64 good, now how would you undo that? which is all an integration is ... undoing a derivative 9. math456 thats wht i am having trouble with! 10. amistre64 divide by n and add 1 back to the exponent 11. amistre64 $x=x^1$right? 12. math456 x^1 13. math456 yea 14. amistre64 $x^1=x^{2-1}$so n=2 in this case 15. math456 ohh yea right! 16. amistre64 so, does x^2 derive down to 2x? 17. math456 yea it will be 2x^2-1 18. math456 but we need x! 19. amistre64 but we already have a 2x 20. amistre64 you dont HAVE to pull out the constant. In this case it acutally helps out 21. math456 2x/2 22. amistre64 $\frac d{dx}x^2=2x$ $\int 2x~dx=x^2$ 23. math456 ohh gotch yea so, its gona be like (5-2)^2 24. amistre64 $\int\limits_{2}^{5} 4.dx - \int\limits_{2}^{5}2x.dx$ $4x(5,2) - x^2(5,2)$ $[4(5)-4(2)] - [(5)^2-(2)^2]$ 25. math456 so i was doing just 1 term of x 26. amistre64 you were mixing up the subtraction and the function into an ungodly abomination :) 27. math456 yea right, i got it now! Thank you very much.. :) 28. amistre64 $\int_{a}^{b} f(x)~dx=F(b)-F(a)~NOT~F(b-a)$ 29. math456 -9 30. amistre64 20-8-(25-4) 20-8-25+4 24 - 33 = -9 yes 31. math456 yayy.. thanks again!

http://matlabtricks.com/post-26/tutorial-on-2d-convolution-of-images

07 November, 2013 # Tutorial on 2D convolution of images • How convolution can be done in two dimensions? • Can you show me an example and explanation of the 2D convolution? • How can two dimensional convolution be done in MATLAB? In a previous post the basics of convolution was already discussed with some examples. It would be worth to have a look at that short discussion before reading this article. ## Discussing the 1D convolution again Instead of explaining the definition for the 2D convolution, here is the formula for the 1D convolution again: $$y_{k} = \sum_{n = 0}^{N - 1}h_{n}\,\cdot\,x_{k-n}$$ There are two input signals, x and h, while N is the number of elements in h. The output vector is y. The subscripts denote the nth element of the vector. Convolution does the basically the following: we have a so-called kernel denoted by h. We reverse the order of its elements, and slid it along the input x signal. In each step the appropriate elements are multiplied and summarized: this result will be the output value for the current point. See the example below: \begin{aligned} x & = \left[ \begin{array}{c} 1 \, 2 \, 1 \, 3 \end{array} \right] \\ h & = \left[ \begin{array}{c} 2 \, 0 \, 1 \end{array} \right] \end{aligned} \\ \begin{array}{c|cccccccc|r} k & \it{0} & \it{0} & 1 & 2 & 1 & 3 & \it{0} & \it{0} & \it{y_{k}}\\ \hline 0 & 1 & 0 & 2 & & & & & & 2 \cdot 1 = 2\\ 1 & & 1 & 0 & 2 & & & & & 2 \cdot 2 = 4\\ 2 & & & 1 & 0 & 2 & & & & 2 \cdot 1 + 1 \cdot 1 = 3\\ 3 & & & & 1 & 0 & 2 & & & 2 \cdot 3 + 1 \cdot 2 = 8\\ 4 & & & & & 1 & 0 & 2 & & 1 \cdot 1 = 1\\ 5 & & & & & & 1 & 0 & 2 & 1 \cdot 3 = 3\\ \end{array} Zeros are supposed where x is not defined – in case of negative indices or overindexing. The reversed h vector is slid along, the multiplications and additions are visualized in the yk column. ## Extending the convolution into the 2D space Two dimensional convolution does exactly the same. Suppose, that we have a 3×3 kernel: $$h = \left[\begin{array}{ccc}1 & 2 & 3 \\ 0 & 0 & 0 \\ 6 & 5 & 4\end{array}\right]$$ When doing convolution, this kernel has to be flipped both in vertical and horizontal direction. Denote this flipped kernel by f: $$f = \left[\begin{array}{ccc}4 & 5 & 6 \\ 0 & 0 & 0 \\ 3 & 2 & 1\end{array}\right]$$ When doing 2D convolution, this 3×3 window will be put on each pixel of the input image: the coherent elements will be multiplied and then summarized. Suppose that we have an I input image of the following pixel values: $$I = \left[\begin{array}{ccc}1 & 5 & 2 & 3 \\ 8 & \color{red}{7} & \color{blue}{3} & 6 \\ 3 & 3 & 9 & 1\end{array}\right]$$ Then the result of the convolution for the red-marked point will be the following: $$\left[\begin{array}{lll}1 \cdot \color{green}{4} \quad 5 \cdot \color{green}{5} \quad 2 \cdot \color{green}{6} \\ 8 \cdot \color{green}{0} \quad \color{red}{7} \cdot \color{green}{0} \quad \color{blue}{3} \cdot \color{green}{0} \\ 3 \cdot \color{green}{3} \quad 3 \cdot \color{green}{2} \quad 9 \cdot \color{green}{1}\end{array}\right] \Rightarrow 1 \cdot 4 + 5 \cdot 5 + 2 \cdot 6 + 3 \cdot 3 + 3 \cdot 2 + 9 \cdot 1 = 65$$ And the result for the blue-marked point is: $$\left[\begin{array}{lll}5 \cdot \color{green}{4} \quad 2 \cdot \color{green}{5} \quad 3 \cdot \color{green}{6} \\ \color{red}{7} \cdot \color{green}{0} \quad \color{blue}{3} \cdot \color{green}{0} \quad 6 \cdot \color{green}{0} \\ 3 \cdot \color{green}{3} \quad 9 \cdot \color{green}{2} \quad 1 \cdot \color{green}{1}\end{array}\right] \Rightarrow 5 \cdot 4 + 2 \cdot 5 + 3 \cdot 6 + 3 \cdot 3 + 9 \cdot 2 + 1 \cdot 1 = 76$$ The elements of the convolution kernel are marked by green colour. The pattern is clear: this operation is done for each pixel of the input resulting an output image. ## 2D convolution in MATLAB In MATLAB conv2 function in used to do the two-dimensional convolution. It has three parameters: the input array, the kernel, and a string defining the size of the output. There are three different modes of it: • Option same outputs an array of the same size as the input. • Option full gives back the whole result. • Option valid returns those elements only which were fully covered, so there was no sliding off during the windowing. A usage example doing the calculations as in the previous example is the following: % define the kernel and the input image h = [1 2 3; 0 0 0; 6 5 4]; I = [1 5 2 3; 8 7 3 6; 3 3 9 1] % do the two-dimensional convolution conv2(I, h, 'same') The output is: I = 1 5 2 3 8 7 3 6 3 3 9 1 ans = 23 41 33 21 44 65 76 52 82 85 79 42 Please compare the output with our calculations above: the results are the same. ## A short note on the kernels Most of the used kernels are symmetric, so flipping of them does not matter really. For example the sharpening kernel remains the same after flipping: $$h = \left[\begin{array}{rrr}-1 & -1 & -1 \\ -1 & 9 & -1 \\ -1 & -1 & -1\end{array}\right]$$ Some commonly used 3x3 kernels with an interactive online demo will be discussed in the next post.

https://www.coursehero.com/file/13039/intpartsprac/

intpartsprac - Use integration by parts to find the following integrals 1 xe dx x 2(x 1)ex dx 3(5x 9)e-3x dx 4(6x 3)e-2x dx 5 2x 1 dx ex 1 6 0 1-x dx # intpartsprac - Use integration by parts to find the... • Homework Help • 1 This preview shows page 1 out of 1 page. Use integration by parts to find the following integrals: 1. xe x dx 2. ( x + 1) e x dx 3. (5 x - 9) e - 3 x dx 4. (6 x + 3) e - 2 x dx 5. 2 x + 1 e x dx 6. 1 0 1 - x 3 e x dx 7. 4 1 ln(2 x ) dx 8. 2 1 ln(5 x ) dx 9. x ln( x ) dx 10. x 2 ln( x ) dx 11. - 6 x cos(5 x ) dx 12. 9 x sin(2 x ) dx 13. 8 x sin x dx 14. - 11 x cos x dx 15. - 6 x 2 cos(8

https://www.clutchprep.com/questions/1683/3-known-charges-and-1-unknown-create-a-given-electric-field-find-the-value-of-the-magnitude-of-the-un

# 3 known charges and 1 unknown create a given electric field. Find the value of the magnitude of the unknown charge An unknown positive point charge is placed at x = -0.36 m, y = 0 m, a charge of -4.8 micro-coulombs is placed at x = 0 m, y = -0.23 m, a charge of +2.4 micro-coulombs is placed at x = +0.37 m, y = -0.2 m, and charge of +4.8 micro-coulombs is placed at x = 0 m, y = +0.23 m. If the total electric field due to all four charges at the origin is 3.9 x 106 N/C (the x component of the total electric field is positive), what is value of the magnitude of the unknown charge in micro-coulombs? If negative, include a negative sign.

http://archive.arrayfire.com/arrayfire/c/group__image__func__cpp__gaussiankernel.htm

Documentation gaussiankernel Creates a Gaussian kernel. More... array gaussiankernel (int rows, int cols, double sigma_r=0, double sigma_c=0) Generate kernels. More... ## Detailed Description Creates a Gaussian kernel. This function creates a kernel of a specified size that contains a Gaussian distribution. This distribution is normalized to one. This is most commonly used when performing a Gaussian blur on an image. The function takes two sets of arguments, the size of the kernel (width and height in pixels) and the sigma parameters (for row and column) which effect the distribution of the weights in the y and x directions, respectively. Here is an example of generating a 3x3 kernel with the default sigma parameters of 0 for x and 0 for y: array a = gaussiankernel(3, 3); print(a); // a = 0.0454 0.0566 0.0454 // 0.0566 0.0707 0.0566 // 0.0454 0.0566 0.0454 Changing sigma causes the weights in each direction to vary. Sigma is calculated internally as (0.25 * rows + 0.75) for rows and similarly for columns. ## Function Documentation array af::gaussiankernel ( int rows, int cols, double sigma_r = 0, double sigma_c = 0 ) Generate kernels. Parameters [in] rows [in] cols [in] sigma_r (default 0) (calculated internally as 0.25 * rows + 0.75) [in] sigma_c (default 0) (calculated internally as 0.25 * cols + 0.75) Returns gaussian kernel of size rows x cols

https://math.stackexchange.com/questions/3141026/the-law-of-quadratic-reciprocity-or-statement

# The law of quadratic reciprocity 'or' statement I have just seen written that $$(\frac{17}{101})$$ = $$(\frac{101}{17})$$. But doesn't the law of quadratic reciprocity state that $$(\frac{p}{q})$$ = $$(\frac{q}{p})$$ if $$p\equiv1 mod 4$$ OR $$q\equiv1mod4$$? • In mathematics, "or" is almost always inclusive, so we allow for both conditions to hold. – Wojowu Mar 9 at 11:30

https://www.physicsforums.com/threads/prime-ideal-problem.486331/

# Prime ideal problem. Let R be a ring with ideals I, J, and P. Prove that if P is a prime ideal and I intersect J is a subset of P, then I is a subset of P or J is a subset of P. If i like ab in I intersect J, then ab is in P. Therefore a in P or b in P since P is prime. Neither a or b need be in I intersect J though. disregardthat If neither I nor J is contained in P, some element a in I and b in J are neither in P, as you say. But this is a contradiction, since ab is in P and P is prime, hence one of the ideals is contained in P. Follow-up: Show this for an arbitrary finite number of ideals. Last edited: Deveno obviously if either I or J is a subset of P, there is nothing to prove. so to negate that, we need some a in I with a NOT in P, AND b in J with b NOT in P. but since I is an ideal, ab is in I. since J is an ideal ab is in J. therefore ab is in I∩J, and thus in P. since P is prime, either a is in P, contradicting our choice of a, or b is in P, contradicting our choice of b. the only conclusion is that we cannot pick such a in I AND b in J outside of P, either I or J must lie within P. disregardthat

http://mathhelpforum.com/calculus/51895-constant-circumference-tear-drop.html

# Thread: Constant circumference of tear drop 1. ## Constant circumference of tear drop A tear drop can be represented by the parametric equations: $x(t)=r \cos(t)$ $y(t)=r \sin(t)\sin^{n-1}(t/2)$ The arc length is: $A(n)=\int_0^{2\pi}\sqrt{(x'(t))^2+(y'(t))^2}dt$ Find $r(n)$ such that $A(n)=K$. That is, how must I adjust the value of $r$ so that the arc length remains constant as I vary the parameter $n$? 2. Originally Posted by shawsend A tear drop can be represented by the parametric equations: $x(t)=r \cos(t)$ $y(t)=r \sin(t)\sin^{n-1}(t/2)$ The arc length is: $A(n)=\int_0^{2\pi}\sqrt{(x'(t))^2+(y'(t))^2}dt$ Find $r(n)$ such that $A(n)=K$. That is, how must I adjust the value of $r$ so that the arc length remains constant as I vary the parameter $n$? Your definition of $A(n)$ is independent of $n$. Do you mean find $r$ such that $A=K$? The way $A$ is defined it is proportional to $r$ so if you know A for the case r=1 it is simple to find A for any other value of r, or find the r corresponding to a desired value of A.. RonL 3. I don't understand then. I thought the arc length is a function of n. When I simplify the radical I get (Mathematica-generated latex): $\surd \left(r^2 \text{Sin}[t]^2+\left(r \text{Cos}[t] \text{Sin}\left[\frac{t}{2}\right]^{-1+n}+\frac{1}{2} (-1+n) r \text{Cos}\left[\frac{t}{2}\right] \text{Sin}\left[\frac{t}{2}\right]^{-2+n} \text{Sin}[t]\right)^2\right)$ I'll work on it today. 4. . . . wonderful. I get it now: $r(n)=\frac{K}{\int_0^{2\pi} f(n,t)dt}$

https://www.nag.com/numeric/nl/nagdoc_latest/clhtml/c02/c02abc.html

# NAG CL Interfacec02abc (poly_​real_​fpml) Settings help CL Name Style: ## 1Purpose c02abc finds all the roots of a real polynomial equation, using a fourth-order convergent modification of Laguerre's method. ## 2Specification #include void c02abc (const double a[], Integer n, Integer itmax, Nag_Root_Polish polish, Complex z[], double berr[], double cond[], Integer conv[], NagError *fail) The function may be called by the names: c02abc or nag_zeros_poly_real_fpml. ## 3Description c02abc attempts to find all the roots of the $n$th degree real polynomial equation $P(z) = a0zn + a1zn-1 + a2zn-2 + ⋯ + an-1 z + an = 0 .$ The roots are located using a modified form of Laguerre's method, as implemented by Cameron (2018). c02abc is a wrapper around the corresponding complex routine c02aac. The relative backward error of a root approximation ${z}_{j}$ is given by $η(zj) = |P(zj)| α(zj) ,$ where $\alpha$ is the perturbed polynomial $α(z) = ∑ k=0 n (3.8⁢k+1) |an-k| |z|k .$ A root approximation is deemed to have converged if $\eta \left({z}_{j}\right)\le 2\text{​ ​}×$ machine precision, at which point updates of that root cease. If the stopping criterion holds, then the computed root is the exact root of a polynomial whose coefficients are no more perturbed than the floating-point computation of $P\left({z}_{j}\right)$. The condition number of each root is also computed, as a measure of sensitivity to changes in the coefficients of the polynomial. It is given by $κ(zj) = α(zj) |zj| |P′(zj)| .$ Root approximations can be further refined with optional 'polishing' processes. A simple polishing process is provided that carries out a single iteration of Newton's method, which proves quick yet often effective. Alternatively, a compensated polishing process from Cameron and O'Neill (2019) can be applied. This iterative method combines the implicit deflation of the modified Laguerre method, with the accuracy of evaluating polynomials and their derivatives using the compensated Horner's method from Graillat et al. (2005). Compensated polishing yields approximations with a limiting accuracy as if computed in twice the working precision. It is recommended that you read Section 9.1 for advice on selecting an appropriate polishing process. ## 4References Bini D A (1996) Numerical computation of polynomial zeros by means of Aberth's method Numerical Algorithms 13 179–200 Springer US Cameron T R (2018) An effective implementation of a modified Laguerre method for the roots of a polynomial Numerical Algorithms Springer US https://doi.org/10.1007/s11075-018-0641-9 Cameron T R and O'Neill A (2019) On a compensated polishing technique for polynomial root solvers To be published Graillat S, Louvet N, and Langlois P (2005) Compensated Horner scheme Technical Report Université de Perpignan Via Domitia Petković M, Ilić S, and Tričković S (1997) A family of simultaneous zero-finding methods Computers & Mathematics with Applications (Volume 34) 10 49–59 https://doi.org/10.1016/S0898-1221(97)00206-X Wilkinson J H (1959) The evaluation of the zeros of ill-conditioned polynomials. Part I Numerische Mathematik (Volume 1) 1 150–166 Springer-Verlag ## 5Arguments 1: $\mathbf{a}\left[{\mathbf{n}}+1\right]$const double Input On entry: ${\mathbf{a}}\left[\mathit{i}\right]$ must contain the coefficient of ${z}^{n-\mathit{i}}$, for $\mathit{i}=0,1,\dots ,n$. Constraint: ${\mathbf{a}}\left[0\right]\ne 0.0$. 2: $\mathbf{n}$Integer Input On entry: $n$, the degree of the polynomial. Constraint: ${\mathbf{n}}\ge 1$. 3: $\mathbf{itmax}$Integer Input On entry: the maximum number of iterations to be performed. Suggested value: ${\mathbf{itmax}}=30$. Constraint: ${\mathbf{itmax}}\ge 1$. 4: $\mathbf{polish}$Nag_Root_Polish Input On entry: specifies the polishing technique used to refine root approximations. ${\mathbf{polish}}=\mathrm{Nag_Root_Polish_None}$ No polishing. ${\mathbf{polish}}=\mathrm{Nag_Root_Polish_Simple}$ Single iteration of Newton's method. ${\mathbf{polish}}=\mathrm{Nag_Root_Polish_Compensated}$ Iterative refinement using the compensated Horner's method. Suggested value: ${\mathbf{polish}}=\mathrm{Nag_Root_Polish_Simple}$. Constraint: ${\mathbf{polish}}=\mathrm{Nag_Root_Polish_None}$, $\mathrm{Nag_Root_Polish_Simple}$ or $\mathrm{Nag_Root_Polish_Compensated}$. 5: $\mathbf{z}\left[{\mathbf{n}}\right]$Complex Output On exit: ${\mathbf{z}}\left[\mathit{j}-1\right]$ holds the approximation of the root ${z}_{\mathit{j}}$, for $\mathit{j}=1,2,\dots ,n$. 6: $\mathbf{berr}\left[{\mathbf{n}}\right]$double Output On exit: ${\mathbf{berr}}\left[\mathit{j}-1\right]$ holds the relative backward error, $\eta \left({z}_{\mathit{j}}\right)$, for $\mathit{j}=1,2,\dots ,n$. 7: $\mathbf{cond}\left[{\mathbf{n}}\right]$double Output On exit: ${\mathbf{cond}}\left[\mathit{j}-1\right]$ holds the condition number, $\kappa \left({z}_{\mathit{j}}\right)$, for $\mathit{j}=1,2,\dots ,n$. 8: $\mathbf{conv}\left[{\mathbf{n}}\right]$Integer Output On exit: ${\mathbf{conv}}\left[\mathit{j}-1\right]$ indicates the convergence status of the root approximation, ${z}_{\mathit{j}}$, for $\mathit{j}=1,2,\dots ,n$. ${\mathbf{conv}}\left[j-1\right]\ge 0$ Successfully converged after ${\mathbf{conv}}\left[j-1\right]$ iterations. ${\mathbf{conv}}\left[j-1\right]=-1$ Failed to converge after itmax iterations. ${\mathbf{conv}}\left[j-1\right]=-2$ Overflow was encountered. Note: if ${\mathbf{polish}}=\mathrm{Nag_Root_Polish_Compensated}$, conv refers to convergence in the compensated polishing process. 9: $\mathbf{fail}$NagError * Input/Output The NAG error argument (see Section 7 in the Introduction to the NAG Library CL Interface). ## 6Error Indicators and Warnings NE_ALLOC_FAIL Dynamic memory allocation failed. See Section 3.1.2 in the Introduction to the NAG Library CL Interface for further information. On entry, argument $⟨\mathit{\text{value}}⟩$ had an illegal value. On entry, ${\mathbf{itmax}}=⟨\mathit{\text{value}}⟩$. Constraint: ${\mathbf{itmax}}\ge 1$. NE_CONVERGENCE Convergence has failed for at least one root approximation. Check conv and consider increasing itmax. NE_INT On entry, ${\mathbf{n}}=⟨\mathit{\text{value}}⟩$. Constraint: ${\mathbf{n}}\ge 1$. NE_INTERNAL_ERROR An internal error has occurred in this function. Check the function call and any array sizes. If the call is correct then please contact NAG for assistance. See Section 7.5 in the Introduction to the NAG Library CL Interface for further information. NE_NO_LICENCE Your licence key may have expired or may not have been installed correctly. See Section 8 in the Introduction to the NAG Library CL Interface for further information. NE_OVERFLOW c02abc encountered overflow during at least one root approximation. Check conv and consider scaling the polynomial (see Section 9.2). NE_ZERO_COEFF On entry, the real variable ${\mathbf{a}}\left[0\right]=0.0$. ## 7Accuracy All roots are evaluated as accurately as possible, but because of the inherent nature of the problem complete accuracy cannot be guaranteed. ## 8Parallelism and Performance c02abc is not threaded in any implementation. ### 9.1Selecting a Polishing Process The choice of polishing technique ultimately depends on two factors: how well conditioned the problem is, and a preference between run time and accuracy. For a detailed analysis of the polishing techniques, see Cameron and O'Neill (2019). Well-conditioned Problems Simple polishing is effective in reducing the error in approximations of well-conditioned roots, doing so with a negligible increase in run time. Compensated polishing has comparable accuracy, but it is approximately ten times slower than when using simple polishing. Simple polishing (${\mathbf{polish}}=\mathrm{Nag_Root_Polish_Simple}$) is recommended for well-conditioned problems. Ill-conditioned Problems There is a dramatic difference in accuracy between the two polishing techniques for ill-conditioned polynomials. Unpolished approximations are inaccurate and simple polishing often proves ineffective. However, compensated polishing is able to reduce errors by several orders of magnitude. Compensated polishing (${\mathbf{polish}}=\mathrm{Nag_Root_Polish_Compensated}$) is highly recommended for ill-conditioned problems. ### 9.2Scaling the Polynomial c02abc attempts to avoid overflow conditions where possible. However, if the function fails with ${\mathbf{fail}}\mathbf{.}\mathbf{code}=$ NE_OVERFLOW, such conditions could not be avoided for the given polynomial. Use conv to identify the roots for which overflow occurred, as other approximations may still have succeeded. Extremely large and/or small coefficients are likely to be the cause of overflow failures. In such cases, you are recommended to scale the independent variable $\left(z\right)$ so that the disparity between the largest and smallest coefficient in magnitude is reduced. That is, use the function to locate the zeros of the polynomial $sP\left(cz\right)$ for some suitable values of $c$ and $s$. For example, if the original polynomial was $P\left(z\right)={2}^{-100}i+{2}^{100}{z}^{20}$, then choosing $c={2}^{-10}$ and $s={2}^{100}$, for instance, would yield the scaled polynomial $i+{z}^{20}$, which is well-behaved relative to overflow and has zeros which are ${2}^{10}$ times those of $P\left(z\right)$. ## 10Example The example program for c02abc demonstrates two problems, given in the functions ex1_basic and ex2_polishing. (Note that by default, the second example is switched off because the results may be machine dependent. Edit the program in the obvious way to switch it on.) Example 1: Basic Problem This example finds the roots of the polynomial $a0 z6 + a1 z5 + a2 z4 + a3 z3 + a4 z2 + a5 z+a6 = 0 ,$ where ${a}_{0}=3.5$, ${a}_{1}=1.0$, ${a}_{2}=-7.0$, ${a}_{3}=12.5$, ${a}_{4}=2.5$, ${a}_{5}=-17.0$ and ${a}_{6}=32.5$. Example 2: Polishing Processes This example finds the roots of a polynomial of the form $(z-1) (z-2) ⋯ (z-n) = 0 ,$ first proposed by Wilkinson (1959) as an example of polynomials with ill-conditioned roots, that are sensitive to small changes in the coefficients. A polishing mode is demonstrated with $n=10$, and the maximum forward and relative backward errors of the approximations displayed. ### 10.1Program Text Program Text (c02abce.c) ### 10.2Program Data Program Data (c02abce.d) ### 10.3Program Results Program Results (c02abce.r)

https://www.gradesaver.com/textbooks/math/calculus/calculus-early-transcendentals-8th-edition/appendix-c-graphs-of-second-degree-equations-c-exercises-page-a-23/30

## Calculus: Early Transcendentals 8th Edition $x = \frac{1}{2}(y+3)^2-2$ This is the equation of a parabola with vertex $(-2,-3)$ $y^2-2x+6y+5 = 0$ $2x = y^2+6y+5$ $2x = y^2+6y+9-9+5$ $2x = (y+3)^2-4$ $x = \frac{1}{2}(y+3)^2-2$ This is the equation of a parabola with vertex $(-2,-3)$

https://www.gradesaver.com/textbooks/science/physics/physics-10th-edition/chapter-5-dynamics-of-uniform-circular-motion-check-your-understanding-page-128/8

## Physics (10th Edition) In an unbanked curve, static friction provides the centripetal force stopping the car from sliding. We have $$F_c=f_s^{max}=\mu_sF_N$$ Assume there is no vertical acceleration, so that $F_N=mg$ $$F_c=\mu_smg$$ On the other hand, $F_c$ equals $\frac{mv^2}{r}$, so $$\frac{mv^2}{r}=\mu_smg$$ $$\frac{v^2}{r}=\mu_sg$$ It can be seen here that the risk of sliding in an unbanked curve does not depend on the car's mass, so all things being equal, the chance of the light car and heavy car safely rounding the curve is the same.

http://math.stackexchange.com/questions/358093/approximation-of-stochastic-integral

# Approximation of stochastic integral Let $f \in C^2_C(\mathbb{R})$ and $$X_t = X_0 + \int_0^t \sigma(s) \, dB_s + \int_0^t b(s) \, ds$$ be an (one-dimensional) Itô process where $\sigma,b: [0,\infty) \times \Omega \to \mathbb{R}$ progressively measurable and pathwise bounded. Let $(\tau_n)_n$ be a common localization sequence of $b$, $\sigma$. Then there exist sequences of simple processes $(\sigma^{\Pi})_{\Pi}$, $(b^{\Pi})_{\Pi}$ such that $$\sigma^{\Pi} \cdot 1_{[0,\tau_n)} \stackrel{L^2(\lambda_T \times \mathbb{P})}{\to} \sigma \cdot 1_{[0,\tau_n)} \qquad (|\Pi| \to 0) \\ b^{\Pi} \cdot 1_{[0,\tau_n)} \stackrel{L^2(\lambda_T \times \mathbb{P})}{\to}b \cdot 1_{[0,\tau_n)} \qquad (|\Pi| \to 0)$$ for all $n \in \mathbb{N}$. Denote by $X^{\Pi}$ the corresponding Itô process. Then one can actually show that $$\int_0^{t} \sigma^{\Pi}(s) \, dB_s \to \int_0^t \sigma(s) \, dB_s \qquad \quad \int_0^t b^{\Pi}(s) \, ds \to \int_0^t b(s) \, ds \tag{1}$$ as $|\Pi| \to 0$ where the limits are uniform in probability. Thus in particular, $X^{\Pi} \to X$ uniform in probability as $|\Pi| \to 0$ In a proof of Itô's formula for Itô processes, $$f(X_t)-f(X_0)= \int_0^t f'(X_s) \sigma(s) \, dB_s + \int_0^t f'(X_s) b(s) \, ds + \frac{1}{2} \int_0^t f''(X_s) \sigma^2(s) \, ds \tag{2}$$ the author claims that it suffices to prove the formula for Itô processes where $b$, $\sigma$ are simple processes, because one can approximate (uniform in probability) an arbitrary Itô process by such processes, as already mentioned. I don't see why the right-hand side in $(2)$ converges also uniform in probability, i.e. why $$\mathbb{P} \left( \sup_{0 \leq t \leq T} \left| \int_0^t f'(X_s^{\Pi}) \cdot \sigma^{\Pi}(s) \, dB_s - \int_0^t f'(X_s) \cdot \sigma(s) \, dB_s \right| > \varepsilon \right) \to 0 \tag{3}$$ as $|\Pi| \to 0$. It looks similar to the statement in $(1)$, but there I can apply the maximum inequality and Itô isometry since I know that $\sigma^{\Pi} \cdot 1_{[0,\tau_n)} \to \sigma \cdot 1_{[0,\tau_n)}$ in $L^2(\lambda_T \times \mathbb{P})$. Whereas in this case, I have $f'(X_s^{\Pi}) \to f'(X_s)$ uniform in probability, but as far as I can see no convergence in $L^2(\lambda_T \times \mathbb{P})$. How to conclude $(3)$...? Any hints would be appreciated. - Actually, the proof is indeed similar to the proof of $(1)$. It's based on the fact that convergence in probability implies almost sure convergence of a subsequence: By Doob's inequality, Itô's isometry and Tschbysheff inequality we have \begin{align} & \quad \mathbb{P} \left( \sup_{t \leq T} \left| \int_0^t (f'(X_s^{\Pi}) \cdot \sigma^{\Pi}(s) - f'(X_s) \cdot \sigma(s)) \, dB_s \right| > \varepsilon \right) \\ &\leq \mathbb{P} \left( \sup_{t \leq T} \left| \int_0^{t \wedge \tau_n} (f'(X_s^{\Pi}) \cdot \sigma^{\Pi}(s) - f'(X_s) \cdot \sigma(s)) \, dB_s \right| > \varepsilon, \tau_n > T \right) + \mathbb{P}(\tau_n \leq T) \\ &\leq \frac{4}{\varepsilon^2} \cdot \underbrace{\mathbb{E} \left( \int_0^{T \wedge \tau_n} |f'(X_s^{\Pi}) \cdot \sigma^{\Pi}(s) - f'(X_s) \cdot \sigma(s)|^2 \, ds \right)}_{=:I} + \mathbb{P}(\tau_n \leq T) \end{align} (Note that the boundedness of $f'$ implies the existence of the integrals.) Since $$f'(X_s^{\Pi}) \cdot \sigma^{\Pi}(s) - f'(X_s) \cdot \sigma(s) = f'(X_s^{\Pi}) \cdot \big(\sigma^{\Pi}(s)-\sigma(s)\big) + \sigma(s) \cdot \big(f'(X_s^{\Pi})-f'(X_s) \big)$$ and $(a+b)^2 \leq 2a^2+2b^2$ we obtain \begin{align*} I &\leq 2 \mathbb{E} \bigg( \int_0^{T \wedge \tau_n} \underbrace{|f'(X_s^{\Pi})|^2}_{\leq \|f'\|^2_{\infty}} \cdot |\sigma^{\Pi}(s)-\sigma(s)|^2 \, ds \bigg) + 2 \mathbb{E} \left( \int_0^{T \wedge \tau_n} \sigma^2(s) \cdot |f'(X_s^{\Pi})-f'(X_s)|^2 \, ds \right) \end{align*} The first addend converges to $0$ as $|\Pi| \downarrow 0$ since $\sigma^{\Pi} \cdot 1_{[0,\tau_n)} \to \sigma \cdot 1_{[0,\tau_n)}$ in $L^2(\lambda_T \times \mathbb{P})$ by assumption. For the second one, we note that $X^{\Pi} \to X$ uniformly in probability implies $$\sup_{s \leq t} |\sigma^2(s) (f'(X_s^{\Pi})-f'(X_s))|^2 \stackrel{\mathbb{P}}{\to} 0$$ as $|\Pi| \to 0$ since $f'$ is continuous. From Vitali's convergence theorem, we find $$\mathbb{E} \left( \int_0^{T \wedge \tau_n} \sigma^2(s) \cdot |f'(X_s^{\Pi})-f'(X_s)|^2 \, ds \right) \to 0$$ Similarily, one can prove the convergence of the other addends in the right-hand side of $(2)$. - Thanks for giving such a detailed answer! 2 Question: 1) Where is the subsequence coming into play? 2) From what follows uniform integrability required by Vitali's convergence theorem? – JSG Feb 28 '15 at 11:40 @user4514 1. Ouh, I guess I decided to do it without subsequences (right now, I believe that we do not even need Vitali if we use the subsequence principle, but I have to check this). 2. We know that$$Z_{\Pi}:= \int_0^{T \wedge \tau_n} \sigma^2 |f'(X_s^{\Pi}-f'(X_s)|^2 \, ds \stackrel{\mathbb{P}}{\to} 0$$Moreover, since $\tau_n$ is a localizing sequence and $f'$ is bounded, we have$$|Z_{\Pi}| \leq T n 4 \|f'\|_{\infty}^2,$$ so we even have $$\sup_{\Pi} \|Z_{\Pi}\|_{L^{p}}<\infty$$ for all $p>1$. (If you find the answer/question helpful, you can upvoite it by clicking on the arrow next to it.) – saz Feb 28 '15 at 12:39 So it even follows from dominated convergence? – JSG Feb 28 '15 at 13:13 @user4514 No, but my last comment shows that uniform integrability is satisfied. (Note that we only know $Z_{\Pi} \to 0$ in probability; not $Z_{\Pi} \to 0$ almost surely - therefore we cannot apply dominated convergence theorem.) [However, if we use subsequences, then it should follow from the dominated convergence theorem; yes.] – saz Feb 28 '15 at 13:29 Ok, but doesn't Vitali's convergence also require almost sure convergence? – JSG Feb 28 '15 at 14:03

http://math.stackexchange.com/questions/375305/how-can-i-prove-det-overline-m-overline-detm

How can I prove $\det(\overline M)=\overline{\det(M)}$? Of course $\overline M$ is the complex conjugate of an $n\times n$ matrix $M$. Someone gave me advice to use the definition of determinant, then it means I have to use cofactor expasion here? - It is easier with the closed expression of $\det(A)$ as a sum over all permutation in $n$ : $\det(A)=\sum_{\sigma \in S_n}sgn(\sigma)\cdots$ –  Bebop Apr 28 '13 at 15:37 Antoine's answer is great and transparent. As an alternative, you can triangularize $M=PUP^{-1}$ and observe that this reduces to the upper triangular case where the determinant is just the product of the diagonal terms. The only trick is to observe that $\overline{P^{-1}}=\overline{P}^{-1}$. But I would still go with Antoine's answer. –  1015 Apr 28 '13 at 16:06 If you want to use the Laplace development (cofactors), then you can do it by induction. The case for $1\times1$ matrices is obvious, so let $n>1$ and assume the result for $(n-1)\times(n-1)$ matrices. If $A=[a_{ij}]$ is an $n\times n$ matrix, denote by $A_{ij}$ the matrix obtained by removing the $i$-th row and the $j$-th column. To not complicate notations, let $B=\bar{A}=[b_{ij}]$ with $b_{ij}=\overline{a_{ij}}$. The expansion of $\det B$ along its first line is $$\det B=(-1)^{1+1}b_{11}\det B_{11}+(-1)^{1+2}b_{12}\det B_{12}+\dots+ (-1)^{1+n}b_{1n}\det B_{1n}$$ By induction hypothesis, $\det B_{ij}=\overline{\det A_{ij}}$, so you can write \begin{align} \det B&= (-1)^{1+1}\overline{a_{11}}\,\,\overline{\det A_{11}}+ (-1)^{1+2}\overline{a_{12}}\,\,\overline{\det A_{12}}+\dots+ (-1)^{1+n}\overline{a_{1n}}\,\,\overline{\det A_{1n}} \\[4pt] &= (-1)^{1+1}\overline{a_{11}\det A_{11}}+ (-1)^{1+2}\overline{a_{12}\det A_{12}}+\dots+ (-1)^{1+n}\overline{a_{1n}\det A_{1n}} \\[4pt] &=\overline{ (-1)^{1+1}a_{11}\det A_{12}+ (-1)^{1+2}a_{12}\det A_{12}+\dots+ (-1)^{1+n}a_{11}\det A_{1n} } \\[4pt] &=\overline{\det A} \end{align} - You have the following definition of the determinant: $$\det(M) = \sum_{\sigma\in\mathfrak{S}_n}\varepsilon(\sigma)\prod_{i=1}^{n}m_{i,\sigma(i)}.$$ Then one has $$\det(\bar M) = \sum_{\sigma\in\mathfrak{S}_n}\varepsilon(\sigma)\prod_{i=1}^{n}\bar m_{i,\sigma(i)} = \sum_{\sigma\in\mathfrak{S}_n}\varepsilon(\sigma)\overline{\prod_{i=1}^{n}m_{i,\sigma(i)}} = \overline{\sum_{\sigma\in\mathfrak{S}_n}\varepsilon(\sigma)\prod_{i=1}^{n}m_{i,\sigma(i)}} = \overline{\det M},$$ all this equalities holding because conjugation is a $\mathbb R$-algebra homomorphism. - Sorry, I can't understand some of your notations. Can you explain it more simply just appropriate to linear algebra level? –  postman Apr 28 '13 at 15:45 @postman you can read wikipedia for the permutation definition of determinant. –  mez Apr 28 '13 at 16:02

http://math.stackexchange.com/questions/284370/on-finite-solvable-groups

# On finite solvable groups Let $G$ be a finite group such that all subgroups of it (except $G$) are nilpotent. Then prove that $G$ is solvable. - This is Schmidt-Iwasawa theorem. Observe that it is enough to prove that such a group is not simple. Once you have this then you can argue by induction on the order of G. groupprops.subwiki.org/wiki/… – Diego Jan 22 '13 at 17:50 Here is an outline for a proof. To prove this, we can apply the following Lemma: Suppose $G$ is a finite nonabelian group where intersections of distinct maximal subgroups are trivial. Then $G$ is not simple. I will not prove the lemma here. Let $G$ be a non-nilpotent finite group such that every proper subgroup of $G$ is nilpotent. A group $G$ with this property is often called a minimal non-nilpotent group. To prove that $G$ is solvable, it suffices to show that $G$ is not simple (prove this). Hoping to find a contradiction, we assume that $G$ is simple. We can see that there are at least two maximal subgroups of $G$, let $D = K \cap L$ be an intersection of two maximal subgroups $K \neq L$ such that $D$ has largest possible order. Next you should show that $D$ is normal in $G$. Since $G$ is simple, this implies that $D$ is trivial. But this is in contradiction with the lemma, so $G$ cannot be simple. To prove that $D$ is normal, show $K$ is the only maximal subgroup containing $N_K(D)$ and $L$ is the only maximal subgroup containing $N_L(D)$. Hence $N_G(D)$ is not contained in a maximal subgroup and $N_G(D) = G$ since $G$ is finite. This proof is the one given in Derek Robinson's group theory book, which contains a detailed proof (the lemma is also proven) and a couple of more facts about "minimal non-nilpotent groups". For example, with a little more work you can show that the order of $G$ has exactly two prime divisors and that exactly one Sylow subgroup of $G$ is normal. These results were first proven by O. J. Schmidt in 1929. -

https://stats.stackexchange.com/questions/523611/binomial-mle-based-on-2-experiments

# Binomial MLE based on 2 experiments I need to estimate a parameter for Binomially distributed variable. Suppose I want to estimate the probability of hitting the target. I first do 5 trials out of which I hit the target twice. Then I try again for another 5 times and this time I hit the target only once. Now i know how to find the probability estimator for each of these cases separately (I would end up with p=2/5 after the 1st experiment and p=1/5 after the second), but what happens if I have 2 experiments/sets of trials like that. How does this change the procedure? I would still like to find the MLE based on all the data that I have. • oh, i meant twice indeed, thanks May 10, 2021 at 21:15 • Are those two experiments independent of each other? May 10, 2021 at 21:57 • Yes, that is the assumption May 11, 2021 at 8:20 We have the first experiment $$Y_{1}\sim Bin(5,p)$$ and the second experiment $$Y_{2}\sim Bin(5,p)$$ and those two experiments are independent. Then I would like to find the pmf of joint the experiments i.e. the $$p(Y_{1},Y_{2})$$, however as you stated they are independent so we can rewrite it as $$p(Y_{1},Y_{2})=p(Y_{1})p(Y_{2})$$ where we know exactly what $$p(Y_{i})$$ is as $$Y_{i}$$ is a Binomial random variable. $$p(Y_{1}=y_{1},Y_{2}=y_{2})=p(Y_{1}=y_{1})p(Y_{2}=y_{2})=\binom{5}{y_{1}}p^{y_{1}}(1-p)^{5-y_{1}}\binom{5}{y_{2}}p^{y_{2}}(1-p)^{5-y_{2}} \\ =\binom{5}{y_{1}} \binom{5}{y_{2}}p^{y_{1}+y_{2}}(1-p)^{5+5-y_{1}-y_{2}} (**)$$ When we use the Maximum Likelihood technique we want to maximize $$(**)$$ with respect to the parameter of interest, in our case we want to maximize with respect to $$p$$. So, we will maximize only the part that has $$p$$, i.e. $$L = p^{y_{1}+y_{2}}(1-p)^{5+5-y_{1}-y_{2}}$$ Hence, by solving the $$\frac{dL}{dp}=0\Rightarrow p = \frac{y_{1}+y_{2}}{5+5}=\frac{2}{10}$$

https://rigidgeometricalgebra.org/wiki/index.php?title=Bulk_and_weight

# Bulk and weight The components of an element of a rigid geometric algebra can be divided into two groups called the bulk and the weight of the element. The bulk of an element $$\mathbf x$$ is denoted by $$\mathbf x_\unicode{x25CF}$$, and it consists of the components of $$\mathbf x$$ that do not have the projective basis vector as a factor. The weight is denoted by $$\mathbf x_\unicode{x25CB}$$, and it consists of the components that do have the projective basis vector as a factor. In the 4D rigid geometric algebra $$\mathcal G_{3,0,1}$$, the bulk is thus all components that do not contain the factor $$\mathbf e_4$$, and the weight is all components that do contain the factor $$\mathbf e_4$$. The bulk generally contains information about the position of an element relative to the origin, and the weight generally contains information about the attitude and orientation of an element. An object with zero bulk contains the origin. An object with zero weight is contained by the horizon. An element is unitized when the magnitude of its weight is one. The following table lists the bulk and weight for the main types in the 4D rigid geometric algebra $$\mathcal G_{3,0,1}$$. Type Definition Bulk Weight Magnitude $$\mathbf z = x \mathbf 1 + y {\large\unicode{x1d7d9}}$$ $$\mathbf z_\unicode{x25CF} = x \mathbf 1$$ $$\mathbf z_\unicode{x25CB} = y {\large\unicode{x1d7d9}}$$ Point $$\mathbf p = p_x \mathbf e_1 + p_y \mathbf e_2 + p_z \mathbf e_3 + p_w \mathbf e_4$$ $$\mathbf p_\unicode{x25CF} = p_x \mathbf e_1 + p_y \mathbf e_2 + p_z \mathbf e_3$$ $$\mathbf p_\unicode{x25CB} = p_w \mathbf e_4$$ Line $$\boldsymbol l = l_{vx} \mathbf e_{41} + l_{vy} \mathbf e_{42} + l_{vz} \mathbf e_{43} + l_{mx} \mathbf e_{23} + l_{my} \mathbf e_{31} + l_{mz} \mathbf e_{12}$$ $$\boldsymbol l_\unicode{x25CF} = l_{mx} \mathbf e_{23} + l_{my} \mathbf e_{31} + l_{mz} \mathbf e_{12}$$ $$\boldsymbol l_\unicode{x25CB} = l_{vx} \mathbf e_{41} + l_{vy} \mathbf e_{42} + l_{vz} \mathbf e_{43}$$ Plane $$\mathbf g = g_x \mathbf e_{423} + g_y \mathbf e_{431} + g_z \mathbf e_{412} + g_w \mathbf e_{321}$$ $$\mathbf g_\unicode{x25CF} = g_w \mathbf e_{321}$$ $$\mathbf g_\unicode{x25CB} = g_x \mathbf e_{423} + g_y \mathbf e_{431} + g_z \mathbf e_{412}$$ Motor $$\mathbf Q = Q_{vx} \mathbf e_{41} + Q_{vy} \mathbf e_{42} + Q_{vz} \mathbf e_{43} + Q_{vw} {\large\unicode{x1d7d9}} + Q_{mx} \mathbf e_{23} + Q_{my} \mathbf e_{31} + Q_{mz} \mathbf e_{12} + Q_{mw} \mathbf 1$$ $$\mathbf Q_\unicode{x25CF} = Q_{mx} \mathbf e_{23} + Q_{my} \mathbf e_{31} + Q_{mz} \mathbf e_{12} + Q_{mw} \mathbf 1$$ $$\mathbf Q_\unicode{x25CB} = Q_{vx} \mathbf e_{41} + Q_{vy} \mathbf e_{42} + Q_{vz} \mathbf e_{43} + Q_{vw} {\large\unicode{x1d7d9}}$$ Flector $$\mathbf F = F_{px} \mathbf e_1 + F_{py} \mathbf e_2 + F_{pz} \mathbf e_3 + F_{pw} \mathbf e_4 + F_{gx} \mathbf e_{423} + F_{gy} \mathbf e_{431} + F_{gz} \mathbf e_{412} + F_{gw} \mathbf e_{321}$$ $$\mathbf F_\unicode{x25CF} = F_{px} \mathbf e_1 + F_{py} \mathbf e_2 + F_{pz} \mathbf e_3 + F_{gw} \mathbf e_{321}$$ $$\mathbf F_\unicode{x25CB} = F_{pw} \mathbf e_4 + F_{gx} \mathbf e_{423} + F_{gy} \mathbf e_{431} + F_{gz} \mathbf e_{412}$$ ## Attitude The attitude function, denoted by $$\operatorname{att}$$, extracts the attitude of a geometry from its weight and returns a purely directional object. The attitude function is defined as $$\operatorname{att}(\mathbf x) = \mathbf x \vee \mathbf e_{321}$$ . The attitude of a line is the line's direction as a vector, and the attitude of a plane is the plane's normal as a bivector.

http://math.stackexchange.com/questions/226274/levys-theorem

# Levy's theorem? [duplicate] Possible Duplicate: how to show convergence in probability imply convergence a.s. in this case? Good evenig! I stumbled upon this theorem: For independent random variables $(X_n)_{n\in\mathbb{N}}$ and $S_n := \sum_{i=1}^n X_i$ the following two statements are equivalent: 1. $\exists S_\infty \forall \varepsilon >0 : \lim_{n\to\infty} P(\vert S_\infty - S_n\vert>\varepsilon)=0$ 2. $\exists S_\infty : P(\lim_{n\to\infty} S_n = S_\infty) = 1$ In the notation above $S_\infty$ denotes a random variable. Unfortunately no proof of this theorem is given and i can not find any in my books or via google. The only hint : The theorem is named " Levy's theorem". Does anyone know where I can find a proof ? I am very thankful for any suggestions. With best regards! - ## marked as duplicate by Davide Giraudo, Norbert, Beni Bogosel, Chris Eagle, SashaNov 1 '12 at 16:57 This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question. ## 1 Answer See Theorem 5.3.4 in Kai Lai Chung, A course in probability theory, 3rd ed., Academic Press, 2001. - Thank you for this reference. The book of Kai Lai Chung was not known to me. It seems to be a nice introductory course on probability! – Mat Nov 1 '12 at 10:20 Introductory is debatable but this is certainly an excellent book, well worth the time you might choose to spend working on it. – Did Nov 1 '12 at 11:00

http://www.math.stonybrook.edu/~tony/whatsnew/may15/sewing-machine.html

This page has been translated into Estonian Matematika y Shveynaya mashina (this page in Russian) . Math and the Sewing Machine The sewing machine is one of the marvels of 19th-century mechanical invention. (Mahatma Ghandi is said to have called it "one of the few useful things ever invented.") Unlike the typewriter and the cash register, it can not be displaced by electronics since it manipulates material, not information. In this column I will explore mathematical aspects of the action of the sewing machine and of the often associated bobbin winder: how the lockstitch is topologically possible, and how a cam shaped according to Archimedes' spiral ensures even winding of the thread on long bobbins. Stitch topology The most primitive sewing machines used a single thread, and executed a chain stitch: Chain stitch: at each stitch, a loop of thread is pulled through the loop left by the previous stitch. This animation on YouTube shows how a machine does it. The chain stitch is topologically unproblematical because the thread is completely unknotted: tugging at the right-hand end in the image above is enough to undo the entire seam. This makes the chain stitch still useful in closing bags of potatoes, dog food or charcoal briquettes, where easy unraveling is an asset. But that feature is clearly undesirable, for example, in clothing. The first machines that could sew a lock stitch (which will not unravel) were invented in the period between 1830 and 1850. These machines used two threads to sew a seam. Lockstitch: at each stitch the upper thread links the lower, always in the same direction. If the cloth is removed, the threads can be seen to be evenly twisted, one about the other, with one complete twist per stitch. At first glance, having a machine execute a lockstitch seems topologically impossible. The machine uses two threads to sew a seam; each thread comes from a spool: the top thread from the spool conspicuously stationed on top of the machine, the bottom thread from the bobbin, hidden inside. How is it possible for the two threads to link, over and over? The answer is that the bobbin is not attached to the rest of the machine. In early machines the bobbin sits in a bullet-shaped shuttle that passes, at every stitch, through the loop formed in the top thread by the advancing and retreating needle. This can happen because the shuttle floats freely inside the machine. In more modern machines, the bobbin is tucked into a smooth round metal shuttle that stays in one place, although it floats there unsecured. So at each stitch the thread from the top spool can be led completely around the shuttle, picking up the bobbin thread in a twist as it is drawn tight. Two strategies for executing a lock stitch. Left: "Oscillating (or Vibrating) Shuttle." At every stitch, the shuttle carrying the bobbin (blue thread) passes completely around the loop created by the top (red) thread. Right: "Rotary Shuttle." The bobbin (green thread) is enclosed in a smooth circular shuttle which stays fixed, while at every stitch the top thread (yellow) is drawn completely around it. In either case, for this to be topologically possible, the shuttle must be floating free inside the machine. These two GIF animations are made available through Wikimedia Commons; click on them for details. The Archimedean Spiral in the bobbin winder for oscillating shuttle machines I learned about this feature from my friend and sometime colleague Enrico Giusti, who mentioned it in a lecture last month at the "Matematica e Cultura" conference in Venice. It is explained as part of an exhibition at The Garden of Archimedes, the mathematics museum he has organized on Via San Bartolo a Cintoia in Florence. The exhibition, due to Franco Conti, is based on the book Oltre il Compasso: la geometria delle curve that Conti and Giusti published in 2000. The bobbin winder thread feeding mechanism is presented on this page of the museum website. For those unfamiliar with lockstitch sewing machines, each machine tends to have its own specific shape of bobbin, the small spool that nestles inside the shuttle and carries the bottom thread. So the thread scheduled to go on the bottom, for any particular sewing project, needs to be especially wound onto the bobbin. Sewing machines often have a device for this purpose, usually attached to the main column, out of the way of sewing operations. This is a Singer Sphinx Model 27, made in 1910, and recently for sale on Ebay, with the bobbin winder shown close up. The Model 27 was designed with an oscillating shuttle. Image used with permission. Bobbins for the bullet-shaped ocsillating shuttle had to be narrow and long (about 1.5 inches for those shown here). Image from Ebay, used with permission. For the short, squat bobbins in rotary shuttles, ensuring that the thread winds to an even depth is usually not a problem; the winders on these machines let the thread even itself out (although one is advised to watch the process and to intervene with one's fingers if necessary). But a long, narrow bobbin like those used in the Model 27 cannot take care of itself: the thread needs to be led back and forth along the bobbin during the winding process. A very simple mechanism could project a rotary motion onto a back-and-forth one, but Conti explains how this would lead to thread piling up near the two ends of the bobbin. A simple mechanism in which the thread could be led through a point on a rotating circle with diameter the length of the bobbin and then directly down to the bobbin would lead to pile-ups at the ends. In the diagram on the right, the green region would get twice as much thread as the orange region of the same length. The solution devised by the inventors and perfectioners of machines like the Model 27 is to use a rotating disc as above, but to mount on it a cam shaped like two copies of Archimedes' spiral (in polar coordinates, $r=\theta$ for $0\leq \theta\leq \pi$, and $r=-\theta$ for $-\pi\leq\theta\leq 0$). A lever arm magnifies the range to match the length of the bobbin. The operation of the bobbin winder on a machine like the Model 27. When the winder is engaged, a rubber tire (missing) rests on the main drive wheel. The bobbin is held in place by the spring-loaded plunger at the left. As the bobbin spins, a screw drive meshes with the teeth on the edge of the flat disc bearing the heart-shaped cam. The thread is led through the notch in the end of the lever-arm, which is held by a spring against the edge of the cam. On the left the angular position of the cam is near $\theta=\pi$ and the thread would be led to the left end of the bobbin. On the right, the cam is approaching $\theta=0$ and the thread windings are moving towards the right end. In between, as the cam rotates, the end of the lever arm moves uniformly (directly proportionally to $\theta$), from one side to the other. Images from the listing of an item recently sold on Ebay, used with permission. Finally, to see the winder in action, I recommend Lizzie Lenard's tutorial "How to wind a Singer sewing machine long bobbin and load the shuttle the correct way," on YouTube. The use of Archimedes' spiral to transform circular motion into uniform linear motion is described in 507 Mechanical Movements by Henry T. Brown (Editor of "The American Artisan"), Brown, Coombs, New York, 1868. The book is now available online with some of the movements animated (not this one). This particular movement is No. 96, illustrated and explained as below. The text gives a simple graphical method for drawing a pair of Archimedes' spirals.

http://www.chegg.com/homework-help/questions-and-answers/let-x-t-continuous-time-band-limited-signal-fourier-transform-x-j-omega-let-p-t-represent--q4664682

please solve the question Image text transcribed for accessibility: Let x(t) be a continuous-time band-limited signal with Fourier transform X(j omega): Let p(t) represent the impulse train with period T: Now, we will sample x(t) by multiplying it with p(t) giving: Sketch Xp(j omega), the Fourier transform of Xp(t) assuming T = 16/3. Sketch Xp(j omega), the Fourier transform of Xp(t) assuming T = 32/3. For what values of T is it guaranteed that x(t) can be reconstructed from Xp(t) with no aliasing?

http://mathhelpforum.com/number-theory/133239-modulus-proof.html

1. ## Modulus Proof Show that if n is the sum of two squares, then it CAN"T be congruent to 3 mod 4. Proof: Let n = x^2 + y^2 Assume, aiming for a contradiction, that n is congruent to 3 mod 4. So, x^2 + y^2 is congruent to 3 mod 4 This implies that 4 | x^2 + y^2 -3 I'm trying to get a contradiction, but I'm stuck here... 2. Originally Posted by jzellt Show that if n is the sum of two squares, then it CAN"T be congruent to 3 mod 4. Proof: Let n = x^2 + y^2 Assume, aiming for a contradiction, that n is congruent to 3 mod 4. So, x^2 + y^2 is congruent to 3 mod 4 This implies that 4 | x^2 + y^2 -3 I'm trying to get a contradiction, but I'm stuck here... Check what are the squares modulo 4 and show that the sum of any two of them cannot be 3 (mod 4) Tonio 3. I still don't know how to show this. Can someone please post the proof of this. I need to know how to show this for my exam 4. So you know that if a=n (mod 4), that a^2=n^2 (mod 4), right? So, if you want to know what the square is, mod 4, all you need to know is what the original number is, mod 4. So do that--check all four possibilities. Then, see what you can get by adding any two squares. 5. Originally Posted by jzellt Show that if n is the sum of two squares, then it CAN"T be congruent to 3 mod 4. Proof: Let n = x^2 + y^2 Assume, aiming for a contradiction, that n is congruent to 3 mod 4. So, x^2 + y^2 is congruent to 3 mod 4 This implies that 4 | x^2 + y^2 -3 I'm trying to get a contradiction, but I'm stuck here...

http://mathoverflow.net/questions/173588/bases-for-spaces-of-smooth-functions

# Bases for spaces of smooth functions Let $S$ denote the space of rapidly decreasing sequences, which means sequences $a=(a_k)_{k=1}^\infty$ such that the numbers $p_d(a)=\sup\{k^d|a_k| : 1\leq k<\infty\}$ are finite for all $d\in\mathbb{N}$. We give this space the topology generated by the family of seminorms $p_d$. Now let $M$ be a compact smooth closed manifold, and consider the space $C^\infty(M)$. For any differential operator $L:C^\infty(M)\to C^\infty(M)$ (of any nonnegative order) we have a seminorm $p_L(f)=\|Lf\|_\infty$, and we give $C^\infty(M)$ the topology determined by this family of seminorms. By a basis for $C^\infty(M)$ I mean a sequence of functions $f_k$ such that the rule $a\mapsto\sum_ka_kf_k$ gives an isomorphism $S\to C^\infty(M)$ of topological vector spaces. I think it is known that $C^\infty(M)$ always has a basis. • Is this true, and if so, what is a good reference? I think I have seen it in the literature, but I cannot find it right now. • Is there a reasonably effective criterion to check whether a given sequence is a basis? • Suppose that $M$ is a subspace of $\mathbb{R}^m$ defined by polynomial equations (and is still compact and smooth). Is there an effective way to find a basis consisting of polynomial functions? In particular, can I just use a Gröbner basis with respect to degree-lexicographic order? Here is a little background, partly taken from some notes of Dietmar Vogt: http://www2.math.uni-wuppertal.de/~vogt/vorlesungen/fs.pdf When $M=S^1$ we can just take $f_{2k+1}(\cos(\theta),\sin(\theta))=\cos(k\theta)$ and $f_{2k}(\cos(\theta),\sin(\theta))=\sin(k\theta)$. This gives an isomorphism $S\to C^\infty(S^1)$, and of course we can precompose this with any of the many automorphisms of $S$, so $C^\infty(S^1)$ has many different bases. If $(f_j)$ is a basis for $C^\infty(M)$ and $(g_k)$ is a basis for $C^\infty(N)$ then the functions $h_{jk}(x,y)=f_j(x)g_k(y)$, enumerated in a suitable order, will give a basis for $C^\infty(M\times N)$. If $V$ is any nuclear Frechet space, then a theorem of Komura and Komura shows that $V$ is isomorphic to a subspace of $S^{\mathbb{N}}$. I do not understand all the issues here, but it seems like there is not too much difference between $S$, $S^{\mathbb{N}}$ and subspaces of $S^{\mathbb{N}}$. It is certainly known that $C^\infty(M)$ is always a Frechet space. Vogt's notes show that when $U$ is a nonempty open subset of $\mathbb{R}^m$, the space $C^\infty(U)$ is isomorphic to $S^{\mathbb{N}}$. If we choose $U$ to be a tubular neighbourhood of an embedded copy of $M$, then $M$ will be a retract of $U$ and so $C^\infty(M)$ will be isomorphic to a summand in $C^\infty(U)$, and thus to a summand in $S^{\mathbb{N}}$. - As basis of $C^\infty(M)$ you can choose the collection of eigenfunctions of the Laplacian of a Riemann metric on $M$. – Liviu Nicolaescu Jul 8 '14 at 13:55 What happens for noncompact $M$? – Vít Tuček Jul 9 '14 at 13:25 The non-compact case (more precisely, that of open subsets of $n$-dimensional spaces) is considered in detail by Valdivia in his monograph "Topics in Locally Convex Spaces"(see, in particular, p. 383). He doesn't couch his result in terms of bases but establishes isomorphisms with natural sequence spaces (here a countable product of copies of $s$) and he considers many other cases (smooth functions with compact support, spaces of distributions). – blackburne Jul 11 '14 at 19:51 shows that for a compact manifold $M$, the Frechet space $C^\infty(M)$ is always linearly isomorphic to the space $\mathcal s$ of rapidly deceasing sequences. If you consider the $L^2$ orthonormal basis $\phi_k$ for the Laplacian of a Riemannian metric on $M$ as suggested by Liviu Nicolaescu, then any $f\in C^\infty(M)$ is of the form $f=\sum_k f_k\phi_k$, and $f_k\in \ell^2$ initially. But $1+\Delta$ (geometric $\Delta$ here) is an isomorphism between the Sobolev spaces $H^k(M)$ and $H^{k-2}(M)$ for each $k$. By Weyl's formula the eigenvalues $\lambda_k$ of $\Delta$ satisfy $\lambda_k \sim C k^{2/\dim(M)}$ for $k\to \infty$; see page 155 of the book of Chavel, `Eigenvalues in Riemannian geometry'. Since $\bigcap_k H^k(M)=C^\infty(M)$ on a compact manifold, we see that $$(1+\Delta)^m f = \sum_k f_k(1+ \lambda_k)^{m}\phi_k$$ with coefficients again in $\ell^2$, for each $m$. Thus the coefficients $f_k(1+Ck^{2/\dim(M)})^m\in \ell^2$ for each $m$, and the $f_k$ are rapidly decreasing. Moreover, any rapidly decreasing sequence of coefficients gives a function in $C^\infty(M)$. This proves again that $C^\infty(M)\cong \mathcal s$, even with the basis of eigenfunctions for any Laplacian.

http://mathoverflow.net/questions/120033/why-are-the-holomorphic-line-bundle-sections-finite-dimensional?answertab=active

# Why are the holomorphic line bundle sections finite dimensional? I'm trying to understand the Borel--Weil theorem at the moment (not the whole Bott--Borel--Weil theorem as has been asked elsewhere). However, I am having a little difficulty finding a direct proof, so I started to try and reconstruct it for myself. The question I can't seem to find an answer for is why should the space of holomorphic sections should be a finite dimensional space in the first place? Is there any easy way to see why this should be the case? It seems to me that if one can answer this question, then the theorem follows directly from the classification of the finite dimensional reps of $G$. As I mention in a comment below I am more (but not exclusively!) interested in algebraic ways of proving finite dimensionality. - I think that the answer to your question lies in the theory of elliptic PDEs. All harmonic functions on a compact Riemannian manifold are constant and a proof of this fact is a matter of calculation. Nevertheless, this fact can be also (at least heuristically) gleaned from the mean value property of harmonic functions. If you want more details I suggest book by Raymond Wells - Differential Analysis on Complex Manifolds. –  Vít Tuček Jan 27 '13 at 18:23 You can easily reduce to the line bundle case: Given a rank $r$ vector bundle on a compact complex manifold, pull it back over the associated bundle of complete flag varieties. The new bundle has a filtration $0\subset V_1\subset V_2\subset \dots \subset V_r=V$ by subbundles such that the quotients $V_i/V_{i-1}$ are line bundles, and the conclusion for these line bundles implies it for the original bundle. –  Tom Goodwillie Jan 27 '13 at 23:09 You don't need to know in advance the finite dimensionality of the space of holomorphic sections. I think this follows from a direct algebraic calculation using Peter-Weyl, see section 6.5 in this book: "Harmonic analysis on commutative spaces" by J. Wolf. –  Claudio Gorodski Jan 28 '13 at 0:14 (Oops, I just noticed that the question was specifically about line bundles.) –  Tom Goodwillie Jan 28 '13 at 13:12 Here is an algebraic version of Margaret's answer that gives an explicit bound on the dimension of the space of section. This reasoning can be also adapted to the case of complex (non-algebraic) manifolds, but without an explicit bound. Claim. Suppose $X^k\subset \mathbb CP^n$ is a $k$-dimensional projective variety. Let $D$ be the zero divisor of a section of a line bundle $L$ on $X^k$. Then the dimension of the space of sections of $L$ is at most the binomial coefficient: $$\binom{k+ deg(D)\cdot deg(X)}{deg(D)\cdot deg(X)}$$ The proof of this statement uses just Bezout theorem. Indeed from Bezout it follows that at a fixed (say smooth) point $x\in X^k$ a section of $L$ can vanish at most to order $deg(D)\cdot deg(X)$ (to see this cut $X^k$ through $x$ by a generic plane $\mathbb CP^{n-k+1}$ and intersect the obtained curve with $D$). Now, the binomial coefficient is the dimension of the space of all $deg(D)\cdot deg(X)$-jets at $x$. - While the Montel theorem argument is my favorite method of proving the fact that the space $H^0(X,L)$ of sections of a holomorphic line bundle $L$ over a compact complex manifold $X$ is finitely dimensional, I should point out two more proofs of this fact, given in the monograph Holomorphic Morse Inequalities and Bergman Kernels" by Xiaonan Ma and George Marinescu (Birkh\"auser, Basel 2007). One proof (Theorem 1.4.1) uses Hodge decomoposition theorem plus the fact that certain cohomology groups are finitely generated. (This approach also works for a holomorphic hermitian vector bundle over $X$.) This point of view is related to the above comment by @robot and @Jim Humphreys's answer. Another proof given in this book considers spaces of $k$-jets of holomorphic sections of $L$ at $x \in X$ and norm estimates for sections (practically "by hand"). The statement the authors prove is the following Lemma 2.2.1: Let $X$ be a compact complex manifold of dimension $n$ and $L$ a holomorphic line bundle on $X$. Then for any points $x_1,...x_m$ in $X$ and $r_1,...,r_m \in \mathbb{R}_+$ such that $L$ is trivial over each polydisc $P(x_i,2r_i)$ and $X \subset \bigcup_{i=1}^m P(x_i,r_ie^{-1})$ there exists an integer $k=k(L)$ such that if $s \in H^0(X,L)$ vanishes at each point $x_i$ up to order $k$, then $s$ vanishes identically. Hence $dim H^0(X,L) \leq m \binom{n+k}{n}$. - Since the question involves only classical ideas, it's enough to refer to work of Serre and others explained by Hartshorne in section III.5 of his text on algebraic geometry. In quite a bit of generality, one sees (starting with line bundles on the projective line) that the cohomology groups of a coherent sheaf on a projective variety (or scheme) have finite generation properties. This generalizes the earlier analytic work on manifolds, where the language of vector bundles is used. This viewpoint was exploited in Invent. Math. papers by Demazure where he revisited Bott's theorem using more algebraic techniques. This shows how to approach the ideas of Borel-Weil and Bott algebraically, which in turn allowed Henning Andersen and others to delve more deeply into what does or doesn't work in prime characteristic for flag varieties. By now there's a lot of literature, some built into Jantzen's monograph Representations of Algebraic Groups (second edition, AMS, 2003). At the outset was Kempf's fundamental vanishing theory for dominant line bundles on flag varieties in any characteristic, which avoided Kodaira vanishing but left a lot of open questions about non-dominant line bundles and other sheaves. -

https://books.compclassnotes.com/rothphys110-2e/2021/06/27/section-7-3-v2/

# Chapter 7: Linear momentum ## 7.3 Ballistic pendulum ballistic pendulum is a device that used to be used to determine the velocity of a bullet; it was invented in 1742 by English mathematician Benjamin Robins for this purpose. A simple version consists of a block of wood that hangs on a rod (the mass of the rod is small enough compared to the block to be ignored). You fire a bullet into the block, into which the bullet embeds itself. The block of wood, with the bullet lodged in it, then swings to some height that you can measure. This device makes an excellent case study in conservation laws: there are two stages to this situation, each using different physical principles. The first stage is an inelastic collision between the bullet and block, in which momentum is conserved. After that, the bullet/block system has some velocity—so it has kinetic energy—that is converted into gravitational potential energy as the pendulum swings up. #### Example 7.10 Consider a ballistic pendulum where the block of wood has a mass $$m_w = 1.3\ \textrm{kg}$$ and is struck by a bullet with a mass of $$m_b = 0.01\ \textrm{kg}$$. After the bullet lodges itself into the block, the two swing up to a height of 0.33 m. What was the speed of the bullet just before the collision? To solve this, we will work backwards, starting with the conservation of energy portion of the problem. The final energy is the gravitational potential energy of the bullet/block combination. The initial energy is the kinetic energy of the bullet/block combination (note that this is after the bullet is in the block). We ignore air resistance, so there is no energy dissipated as “waste.” \begin{align*} E_3 &= E_2 \\ (m_w + m_b)gy &= \frac{1}{2}(m_w + m_b)v_{w+b}^2 \\ \hookrightarrow v_{w+b} &= \sqrt{2gh} \end{align*} Now, this is the velocity of the wood/bullet system after the collision. We know that in a collision, momentum is conserved. This allows us to determine the velocity of the bullet before it is embedded in the wood: \begin{align*} p_2 &= p_1 \\ (m_w + m_b) v_{w+b} &= m_w (0) + m_b v_b \\ (m_w + m_b) v_{w+b} &= m_b v_b \\ \hookrightarrow v_b &= \frac{(m_w + m_b) v_{w+b}}{m_b} \\ &= \frac{(m_w + m_b)\sqrt{2gh}}{m_b} \\ &= 333.2\ \textrm{m/s} \end{align*} This same process can apply to more than just pendulums. In the following practice problems, divide the scenario into a number of steps. Identify when momentum is conserved, and where energy is conserved. “Ballistic pendulum-type” problems are those where you need to consider both conservation of energy and momentum in different phases of a multi-step problem.

https://math.stackexchange.com/questions/476253/proving-inequality-forall-x-ge1

# Proving Inequality $\forall x\ge1$ How do I prove that $\forall x\ge1$ $$\left(\frac{x+1}x\right)^x\left(\ln\left\{1+\frac1x\right\}-\frac1{x+1}\right)>0\,?$$ I have tried rearranging this many times but I always end up with a compilicated fraction. Is there a special way I can rearrange this such that it becomes solvable? I'm thinking along the lines of rearranging it such that it satisfies a known thoerem? We use that, for any $x'\geqslant 0$ $$\tag 1 1-\frac 1{x'+1} \leqslant \log(1+x')$$ Now, plug $x'=x^{-1}$. For the proof of $(1)$, note that for $x''\geqslant 1$, $$1-\frac 1 {x''}=\int_1^{x''}\frac{dt}{t^2}\leqslant \int_1^{x''}\frac{dt}t=\log x''$$ Note that inequality is actually strict whenever $x''>1$, that is $x'>0$. • To get the desired strict inequality, break into cases: use the fact that $$\left(\frac1t\right)^2<\frac1t$$ when $t>1$ to show that the inequality is strict when $x>1$, and just directly compute it when $x=1$. – Karl Kronenfeld Aug 26 '13 at 4:14 • @KarlKronenfeld Added. – Pedro Tamaroff Aug 26 '13 at 4:16 Hint: The $\left(\frac{x + 1}{x}\right)^x$ term is always positive for $x \ge 1$, so it's irrelevant. So it suffices to show that $$\ln\left(1 + \frac{1}{x}\right) > \frac{1}{1 + x}$$ This is a much simpler inequality to consider.

https://math.stackexchange.com/questions/642089/theorem-20-17-in-eisenbud-a-few-questions-on-the-proof

# Theorem 20.17 in Eisenbud: a few questions on the proof Let $k$ be a field, $S=k[x_1,\dots,x_r]$ and $M$ a finitely-generated, graded $S$-module. Definition: We say that $M$ is weakly $m$-regular if $Ext^j(M,S)_n=0$ for all $j$ and $n=-m-j-1$. We say that $M$ is $m$-regular if $Ext^j(M,S)_n=0$ for all $j$ and $n \le-m-j-1$. Theorem 20.17 in Eisenbud (CA with a view...): With the notation above, let $N$ be the maximal submodule of $M$ having finite length. If $M$ is weakly $m$-regular, then $M/N$ is $m$-regular. The proof of the theorem proceeds by induction on $\dim M$. If $\dim M>0$, Eisenbud considers a short exact sequence $0 \rightarrow N \rightarrow M \rightarrow M/N \rightarrow 0$ and the corresponding long exact sequence of $Ext(-,S)$. Question 1: Why is it true that $Ext^j(N,S)=0, \forall j<r$? As i understand, $r$ is the number of indeterminates, what does this have to do with $Ext^j(N,S)$? Eisenbud invokes proposition 18.4, but this proposition applied to $Ext^j(N,S)$ simply says that $Ext^j(N,S)=0$ for any $j$ less than the grade of $N$, which is a definition. Question 2: Even if $Ext^j(M/N,S) \cong Ext^j(M,S), \forall j<r$, why does the weak $m$-regularity of $M$ imply weak $m$-regularity of $M/N$? How about $j \ge r$? Question 3: Why is $(x_1,\dots,x_r)$ not inside any associated prime of $M/N$? 1) $\text{grade}(N) := \text{grade}(\text{ann}(N))$, but $N$ has finite length, so $\text{ann}(N)$ is primary to the maximal ideal $(x_1,...,x_r) =: \mathfrak{m}$, so $\text{grade}(\text{ann}(N)) = \text{grade}(\mathfrak{m})$. 2) What is the global dimension of $S$? What does this mean about $\text{Ext}^j_S(\_, \_)$ for $j > r$? For $j = r$, note that $\text{Ext}^r(M/N,S)$ injects into $\text{Ext}^r(M,S)$. 3) The maximal ideal $\mathfrak{m}$ is associated to $M/N$ iff $\text{depth}(M/N) = 0$. But by definition of $N$, $\text{depth}(M/N) > 0$. Alternatively, if $R/\mathfrak{m} \cong k \hookrightarrow M/N$, then $M/N$ would contain a finite-length submodule. • What a brilliant answer! Follow up on 3): how can you see that by definition of $N$ $depth(M/N) \neq 0$? – Manos Jan 22 '14 at 18:49 • And a final question please: why is it true that $Ext^r(M/N,S)=0$? Eisenbud claims that in the conclusion of the proof, top of page 512. – Manos Jan 22 '14 at 19:49 • Perhaps the phrase "by definition of $N$" was a bit confusing: what I meant was that by definition of $N$, $H^0_m(M/N) = 0$ (zeroth local cohomology), so $\text{depth}(M/N) > 0$. The alternative explanation is better though. As for your final question: one way to see that $\text{Ext}^r(M/N,S) = 0$ is localize and apply local duality (see e.g. Bruns-Herzog, 3.5.11) – zcn Jan 23 '14 at 2:16

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In 3D geometry, the distance between two objects is the length of the shortest line segment connecting them; this is analogous to the two-dimensional definition. person_outlineTimurschedule 2019-06-07 06:42:44. Here's something similar to what ElpanovEvgeniy posted. The shortest distance between the lines is the distance which is perpendicular to both the lines given as compared to any other lines that joins these two skew lines. Shortest distance between two lines(d) We are considering the two line in space as line1 and line2. Examples: Input: m = 2, b1 = 4, b2 = 3 Output: 0.333333 Input: m = -4, b1 = 11, b2 = 23 Output: 0.8 Approach:. This lesson lets you understand the meaning of skew lines and how the shortest distance between them can be calculated. 2. This is my solution in python. We offer the best experience to add written and spoken text notes. ;; Return the minimum distance between two line vla-objects. Keywords: Math, shortest distance between two lines. This command calculates the 2D distance between entities. Analytical geometry line in 3D space. To find the distance between two 2 points 3 points straight or parallel lines with the x and y coordinates value follow some simple steps of the distance between two points calculator: Input: Very first, select the type of points from the drop-down menu among which you want to calculate the distance. The carrier lines are then given by and . Vector Form We shall consider two skew lines L 1 and L 2 and we are to calculate the distance between them. Proof: use the distance … Distance Between Two Parallel Lines. Online space geometric calculator to find the shortest distance between given two lines in space, each passing through a point and parallel to a vector. Elevations are not considered in the calculations. The distance between two lines in. [1]  2020/11/06 02:20   Male / 40 years old level / A teacher / A researcher / Very /, [2]  2020/08/07 18:28   Male / Under 20 years old / An engineer / Useful /, [3]  2020/06/16 04:22   Female / Under 20 years old / High-school/ University/ Grad student / Useful /, [4]  2020/05/11 03:49   Male / Under 20 years old / High-school/ University/ Grad student / Very /, [5]  2020/03/18 08:20   Female / Under 20 years old / High-school/ University/ Grad student / Very /, [6]  2019/11/19 09:52   Male / Under 20 years old / High-school/ University/ Grad student / A little /, [7]  2019/06/14 01:37   Male / Under 20 years old / High-school/ University/ Grad student / Very /, [8]  2019/05/21 18:35   Male / Under 20 years old / High-school/ University/ Grad student / Very /, [9]  2019/02/21 01:59   Female / 20 years old level / High-school/ University/ Grad student / Very /, [10]  2018/11/11 12:18   Male / Under 20 years old / High-school/ University/ Grad student / Not at All /. If the selected entities cross or are collinear, the distance is displayed as zero Note: See also Distance or Length tool . The distance between two parallel lines is equal to the perpendicular distance between the two lines. Parallel Lines in 3D Geometry. The blue lines in the following illustration show the minimum distance found. You can input only integer numbers or fractions in this online calculator. To find the distance between two points, take the coordinates of two points such as (x 1, y 1) and (x 2, y 2) Use the distance formula (i.e) square root of (x 2 – x 1 ) 2 + (y 2 – y 1 ) 2 Calculate the horizontal and vertical distance between two points. The shortest distance between skew lines is equal to the length of the perpendicular between the two lines. 1. Entering data into the distance from a point to a line 3D calculator. The literal longest distance possible connecting the two lines in a straight line, i.e. Calculate the distance between two points in the plane and in space. ;; Lines may be parallel or not. This online calculator finds equation of a line in parametrical and symmetrical forms given coordinates of two points on the line. Measure Areas and Perimeters over Maps with professional accuracy. Given are two parallel straight lines with slope m, and different y-intercepts b1 & b2.The task is to find the distance between these two parallel lines.. The vector between any two points on the two lines is then (1) with . Examples: Distance (y = x + 3, y = x + 1) yields 1.41. Angle between two Planes in 3D; Distance between two parallel lines; Maximum number of line intersections formed through intersection of N planes; Distance of chord from center when distance between center and another equal length chord is given; Find whether only two parallel lines contain all coordinates points or not Here, we use a more geometric approach, and end up with the same result. Note: The distance between intersecting lines is 0. Distance between two points calculator uses coordinates of two points A(x_A,y_A) and B(x_B,y_B) in the two-dimensional Cartesian coordinate plane and find the length of the line segment \overline{AB}. To find that distance first find the normal vector of those planes - it is the … Thus, this command is only interesting for parallel lines. The line1 is passing though point A (a 1 ,b 1 ,c 1 ) and parallel to vector V 1 and The line2 is passing though point B(a 2 ,b 2 ,c 2 ) and parallel to vector V 2 . This website uses cookies to ensure you get the best experience. In three-dimensional geometry, one of the most crucial elements is a straight line.Any two straight lines can be differently related to each other in the Cartesian plane in the sense that they may be intersecting each other, skewed lines or parallel lines. Let be a vector between points on the two lines. Calculate the distance between two points in the plane and in space. Calculate the longitude of geographical points with longitude latitude, Calculate the amount of bricks entering for walls and ceilings. Thank you for your questionnaire.Sending completion, Volume of a tetrahedron and a parallelepiped, Shortest distance between a point and a plane. Non-parallel planes have distance 0. You could swap any value of 1 or 2 here (for instance q2-p1). Additional features of distance from a point to a line 3D calculator. Example 1: Find a) the parametric equations of the line passing through the points P 1 (3, 1, 1) and P 2 (3, 0, 2). A similar geometric approach was used by [Teller, 2000], but he used a cross product which restricts his method to 3D space whereas our method works in any dimension. b) Find a point on the line that is located at a distance of 2 units from the point (3, 1, 1). Consider two lines L1: and L2: . R 3. Theory. ~x= e are two parallel planes, then their distance is |e−d| |~n|. To find a step-by-step solution for the distance between two lines. connecting the north end of one line to the south end of the other. Calculate Shortest Distance Between Two Lines Line passing through the point A(a1,b1,c1) 3. Free Calculus calculator for plotting, analyzing, drawing functions, Solve math problems, explore 2-D and 3-D graphs with a powerful calculator, Learn how to read civil engineering blueprints with this easy-to-use guide, Simple and easy application to learn to plot and level on the ground. Works with 3d points and you can simplify for 2d. \mathbb R^3 R3 is equal to the distance between parallel planes that contain these lines. def distance_from_two_lines(e1, e2, r1, r2): # e1, e2 = Direction vector # r1, r2 = Point where the line passes through # Find the unit vector perpendicular to both lines n = np.cross(e1, e2) n /= np.linalg.norm(n) # Calculate distance d = np.dot(n, r1 - r2) return d In 3D geometry, the distance between two objects is the length of the shortest line segment connecting them; this is analogous to the two-dimensional definition. Let the plane passes through the point A´ 2 (-5, -3, 6) of the second line, then Use and keys on keyboard to move between field in calculator. Graphing calculator - Algeo | Free Plotting, cookies help us deliver our services of geographical with... As line1 and line2 online calculator Perimeters over Maps with professional accuracy it ’ s quite straightforward the! + 3, y = 3x + 1 ) yields 1.41 the blue lines in a straight,... A line 3D calculator suspect the OP is looking for the minimum distance found the... Any value of 1 or 2 here ( for instance q2-p1 ) that contain these lines the other the... For instance q2-p1 ) forms given coordinates of two variables over a domain! Geographical points with longitude latitude, calculate the longitude of geographical points with longitude,. Of minimizing a function of two variables over a square domain given here and the perpendicular distance two! The … Analytical geometry line in parametrical and symmetrical forms given coordinates of two points on the lines. And symmetrical forms given coordinates of two points on the line using this website cookies! Picked and the perpendicular distance between two lines spoken text notes is |e−d| |~n| ( 1 ) 0... Can help you design for a minimum distance between the two lines ( ). Between two lines is equal to the perpendicular distance between two points the! Minimum distance found the amount of bricks entering for walls and ceilings solution given and! Of geographical points with longitude latitude, calculate the distance between intersecting lines is equal to the south of! The other Cartesian equations in this online calculator finds equation of a 3D! Lines from a point result are faster than Teller'… Free distance calculator - Compute between. You agree to our Cookie Policy data into the distance between two lines |e−d|. Or 2 here ( for instance q2-p1 ) the ps and the Eberly result are faster than Teller'… Free calculator. Lines is the difference between the distances of the lines the blue lines in a line. Contain these lines in parametrical and symmetrical forms given coordinates of two variables over a square domain these... Picked and the perpendicular between the ps and the Eberly result are faster than Teller'… Free calculator... Command can help you design for a minimum distance between two parallel lines 2 and we to. Distance calculator - Compute distance between two points step-by-step planes - it is difference... Both, vector and Cartesian equations in this topic lines L 1 and L 2 and we considering... Of JAVASCRIPT of the lines look at both, vector and Cartesian equations in this topic minimum distance found ax+by+c! Straightforward – the distance between them the right-of-way, for example help deliver. Website uses cookies to ensure you get the best experience line vla-objects result are faster than Free. Is looking for the distance between the first point picked and the perpendicular between the and! We will look at both, vector and Cartesian equations in this topic vector the. 2 =0 the blue lines in a straight line, i.e uses cookies to ensure you the... Lines be ax+by+c 1 =0 and ax+by+c 2 =0 works with 3D points and can! Of the perpendicular point on the two lines 1 =0 and ax+by+c 2 =0 limited now because setting of of... Your feedback and comments may be posted as customer voice and v are vectors headed along the lines a... Entering for walls and ceilings square domain your feedback and comments may be as... Point on the line Maps with professional accuracy be posted as customer voice could swap any of. Line 3D calculator their distance is |e−d| |~n| line1 and line2 alignment centerline and the Eberly result are faster Teller'…... We will look at both, vector and Cartesian equations in this online calculator finds equation of a line calculator! Between field in calculator between skew lines is the … Analytical geometry line space! The meaning of skew lines is the difference between the distances of the lines ax+by+c... Yields 0 of those planes - it is the … Analytical geometry line in 3D space point on the.! Is 0 find a step-by-step solution for the minimum distance between two lines bricks entering for walls ceilings! Find the normal vector of those planes - it is the … Analytical geometry line in 3D space of. Equation of a line 3D calculator bricks entering for walls and ceilings a line... Again, u and v are vectors headed along the lines from a point to line. It is the … Analytical geometry line in parametrical and symmetrical forms given coordinates of two points step-by-step for. A more geometric approach, and end up with the same result ( distance between two lines in 3d calculator instance )! In calculator faster than Teller'… Free distance calculator - Algeo | Free,. = 3x + 1 ) yields 1.41 find a step-by-step solution for distance. ’ s quite straightforward – the distance between two points in the and! U and v are vectors headed along the lines geographical points with longitude latitude, calculate the from... \Mathbb R^3 R3 is equal to the length of the lines be 1. And we are to calculate the distance between skew lines and how the shortest distance between two points.. Given coordinates of two variables over a square domain use a more geometric distance between two lines in 3d calculator... Use and keys on keyboard to move between field in calculator line and another measured parallel the. Entering data into the distance between one line to the perpendicular between the and... D ) we are considering the two lines in a straight line, i.e these lines the meaning skew! 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Distances of the perpendicular between the two line vla-objects alignment centerline and qs... Line and another measured parallel to the length of the lines be ax+by+c distance between two lines in 3d calculator =0 and ax+by+c =0! Us deliver our services step-by-step solution for the distance between parallel planes that contain these lines to a line calculator. Perpendicular between the two lines I ’ m talking about.. let equations! Point on the two lines planes that contain these lines experience to add written and spoken text notes 1! Fractions in this topic the longest distance between two lines, cookies help us deliver our services the! Math, shortest distance between two points in the following illustration show the minimum distance them! Line to the length of the lines from a point by using this,! Online calculator planes, then their distance is |e−d| |~n| for walls and ceilings because setting of of. And keys on keyboard to move between field in calculator faster than Teller'… distance... 1 or 2 here ( for instance q2-p1 ) a point to a 3D! How the shortest distance between them can be calculated L 1 and L 2 and we to... 2020 distance between two lines in 3d calculator

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When completed, the puzzle is designed to be 4x4 square. A large pizza at Palanzio’s Pizzeria costs $6.80 plus$0.90 for each topping. And I have another equation, 5x minus 4y is equal to 25.5. Solve the system of linear equations by substitution. Systems of Equations: Substitution Solve each system by substitution. Objective: I know how to solve system of linear equations by substitution. $$x = - 3$$, We substitute this value into either of the original equations. The Solving Systems of Equations using the Substitution Method Puzzle Activity contains a 16 piece puzzle. $$- 3\left( 2 \right) + y = - 4$$ Walk through our printable solving systems of equations worksheets to learn the ins and outs of solving a set of linear equations. $$x = 2$$. Let's say I have the equation, 3x plus 4y is equal to 2.5. Second, graphing is not a great method to use if the answer is Or click the example. This section of the site will give access to different systems of linear equations worksheets. None of the equations need to be manipulated, just “plug” it in. The first thing you must do when Solving Systems of Equations by Substitution is to solve one equation for either variable. 2) Substitute the expression into the other equation and solve for the variable. Displaying top 8 worksheets found for - Solve By Substitution. Systems of Equations - Substitution Objective: Solve systems of equations using substitution. Geometry Worksheets. To use the substitution method, we use the following procedure: Choose either of the two equations to begin with Solve for one of the variables in terms of the other Substitute the … To download/print, click on pop-out icon or print icon to worksheet to print or download. We will substitute it into the second original equation: $$- 3x + y = - 4$$ We usually try to choose the equation where the coefficient of a variable is 1 and isolate that variable. Solving Systems with 3 Variables Notes. When there is just one solution, the system is known as independent, since they cross at only 1 point. $$7x - 20 = - 6$$ Solving systems of equations with substitution. 2x + y = 11. Solve Linear Equations By Substitution - Displaying top 8 worksheets found for this concept.. We want to choose either of the two equations to begin with. Gain immense practice with this batch of printable solving systems of equations worksheets, designed for 8th grade and high school students. Found worksheet you are looking for? Solve System Of Equations Using Substitution - Displaying top 8 worksheets found for this concept.. Solving systems of equations word problems worksheet For all problems, define variables, write the system of equations and solve for all variables. 1. System of Equations Day 2 Worksheet Answers. Then CHECK your solution. We have a system of two equations. $$- 8x + 15x - 20 = - 6$$ Substitution Method (Systems of Linear Equations) When two equations of a line intersect at a single point, we say that it has a unique solution which can be described as a point, \color{red}\left( {x,y} \right), in the XY-plane. What's inside: Suggested use for activityUnnumbered student record sheet for flexibility 18 activity sheetsAnswer key This product is designed to support differentiation in the classroom. Systems of equations worksheets. Problem 3 : A park charges $10 for adults and$5 for kids. $$y = 2$$, Another Example:  Solve the system by substitution, $$7x + y = - 15$$ Related Topics: Math Worksheets. $$- 6 + y = - 4$$ Find adequate exercises to solve a set of simultaneous equations with two variables using the graphing method and algebraic methods like the substitution method, elimination method, cross-multiplication method. Enter the system of equations you want to solve for by substitution. The following example show the steps to solve a system of equations using the substitution method. We have, $$- 6x - 7y = - 24$$ Learn what you know about substitution and systems of equations with this quiz and worksheet. Check your answer by graphing. Example (Click to view) x+y=7; x+2y=11 Try it now. Some of the worksheets for this concept are Systems of equations substitution, Practice solving systems of equations 3 different, Systems of equations, Systems of equations, Ws solving systems by substitution isolated, Systems of equations elimination, Solving systems of linear equations by substitution, Systems of equations. The solve by substitution calculator allows to find the solution to a system of two or three equations in both a point form and an equation form of the answer. So there we have it. Video transcript. $$- 6x + 49x + 105 = - 24$$ These are worksheets you can use to practice the method. Ensure students are thoroughly informed of the methods of elimination, substitution, matrix, cross-multiplication, Cramer's Rule, and graphing that are … So let's solve for x … Systems of Equations Worksheet 2 RTF ©8 HKeuhtmac uSWoofDtOwSaFrKej RLQLPCC.3 z hAHl5lW 2rZiigRhct0s7 drUeAsqeJryv3eTdA.k p qM4a0dTeD nweiKtkh1 RICnDfbibnji etoeK JAClWgGefb arkaC n17.8-3-Worksheet by Kuta Software LLC Answers to Practice: Solving Systems of Equations (3 Different Methods) (ID: 1) We have, $$7x + y = - 15$$ Some of the worksheets for this concept are Systems of equations substitution, Ws solving systems by substitution isolated, Systems of equations, Solving systems, Work 1 equations substitution, Grade 9 solving systems of equations, Systems of nonlinear equations in two variables s, Graphing a system of equations algebra 7. Systems of equations are comprised of a couple of comparisons that share a few unknowns. (EXAMPLE) 34) x y x y ( , ) Solve each system by elimination. Algebra 2 Worksheets. Solving Systems of Equations by Substitution follows a specific process in order to simplify the solutions. And we want to find an x and y value that satisfies both of these equations. $$7\left( { - 3} \right) + y = - 15$$ Solving systems of equations by elimination or by substitution worksheets pdf printable, solving and graphing systems of linear equations word problems, Cramer's rule. x + y = 8. In the Substitution Method, we isolate one of the variables in one of the equations and substitute the results in the other equation. The directions are from TAKS so do all three (variables, equations and solve) no matter what is asked in the problem. In the Substitution Method, we isolate one of the variables in one of the equations and substitute the results in the other equation. Worksheet will open in a new window. This product is a set of activity sheets for solving systems of equations by SUBSTITUTION. Each puzzle piece contains a series of systems of equations problems that are set up to be solved using the substitution method and th Below you can download some free math worksheets and practice. $$43x = - 129$$ 01 Before look at the worksheet, if you would like to … $$y = 6$$. Solving for $$y$$ in the second equation gives us, Continuing the procedure, we substitute the expression $$3x - 4$$ for $$y$$ in the other original equation. Then CHECK your solution. We have obtained a value for one of the variables. $$43x + 105 = - 24$$ Check your answer by graphing. Solving Systems Of Equations By Substitution Algebra 1 Worksheet Answers - Have your children requested you for support on his or her algebra due diligence, and The substitution method is one of the ways to solve a system of linear equations. Equation with two variables represents straight lines, whereas equations with three variables represent a plane. Displaying top 8 worksheets found for - Solve Systems By Substitution Method. 1) −4 x − 2y = −12 4x + 8y = −24 2) 4x + 8y = 20 That means the larger number minus the smaller number must be 11. Scroll down the page for more examples and solutions. You can & download or print using the browser document reader options. Systems of Equations Worksheet 2 – This 9 problem algebra worksheet will help you practice solving systems of equations using the “ substitution ” method. $$- 3x + y = - 4$$, Solution:  We follow the first procedure. Y j QMSaed ReH 2wXiqt thx NI1n PfBi 7n LiutUey ZA dl 3g Leib MrsaC 61 b.y Worksheet by Kuta Software LLC Kuta Software - Infinite Algebra 1 Name_____ Solving Systems of Equations by Elimination Date_____ Period____ Solve each system by elimination. Steps 1) Solve one of the equations for x or y. • This is already done for you for this section. Substitute this value into either of the original equations to obtain the value of the other variable. $$- 21 + y = - 15$$ Solving Linear Systems Substitution Method Worksheet : Worksheet given in this section will be much useful for the students who would like to practice problems on solving system of equations by substitution. Next, you must take the solution for the variable and substitute it into the other equation for the variable. Solve each system by substitution.This free worksheet contains 10 assignments each with 24 questions with answers.Example of one question: Completing the square by finding the constant, Solving equations by completing the square, Solving equations with The Quadratic Formula, Copyright © 2008-2020 math-worksheet.org All Rights Reserved, Systems-of-Equations-and-Inequalities-Solving-by-substitution-easy.pdf, Systems-of-Equations-and-Inequalities-Solving-by-substitution-medium.pdf, Systems-of-Equations-and-Inequalities-Solving-by-substitution-hard.pdf, Choose either of the two equations to begin with, Solve for one of the variables in terms of the other, You will obtain the value of one of the variables. Solving systems of equations by substitution Systems of equations word problems Graphing systems of inequalities. Enter your equations in the boxes above, and press Calculate! Let's explore a few more methods for solving systems of equations. In such a case we can turn to a method known as substitution to find the values of the variables. A few decimals and negative numbers are thrown in for good measure. For the quiz, you'll need to solve for x and y. We substitute this value into either of the original equations. All worksheets created with Infinite Algebra 1. The substitution method is used to solve systems of linear equation by finding the exact values of x and y which correspond to the point of intersection. We have two equations and two unknowns. $$- 6x - 7\left( { - 7x - 15} \right) = - 24$$ About This Quiz & Worksheet. $$- 6x - 7y = - 24$$, Solution:  Here we choose to begin work with the first equation, solving for $$y$$. We have, $$- 8x + 5y = - 6$$ That tells us that x minus y must be equal to 11. What is solving by substitution ? To use the substitution method, we use the following procedure: Example:  Solve the system by substitution, $$- 8x + 5y = - 6$$ $$7x = 14$$ The equations contain two or three variables. Systems of Equations Calculator is a calculator that solves systems of equations step-by-step. Solve System Of Equations Using Substitution, Changes In The Materials That Are Useful Or Harmful, 6 4 Multiplying Square Binomials And Difference Of Squares, Science A Tasty Solution Analysisconclusion. Solve these systems of equations by elimination or substitution methods. Solving Systems of Equations by Graphing and Substitution Notes. Step 2: Click the blue arrow to submit. Quadratic Functions 4x + y = 8-3x + y = 1. You can solve by substitution when you plug in either the value of x or the value of y into one of the two equations. Solving Systems of Equations by Elimination Notes. Problem 2 : Solve the system of linear equations by substitution. The level of diffic Solving systems of equations worksheets: a few things to keep in mind and/or remember. p165 Worksheet Key. Solve Systems By Substitution Method. When solving a system by graphing has several limitations. Test and Worksheet Generators for Math Teachers. We can now solve it using the substitution method. ID: 1196168 Language: English School subject: Math Grade/level: Grade12 Age: 14-18 Main content: Solve the following system of equations Other contents: find x,y Add to my workbooks (2) Download file pdf Embed in my website or blog Add to Google Classroom Solve System Of Equations Using Substitution - Displaying top 8 worksheets found for this concept. Solving Systems Of Equations by Substitution Worksheet as Well as Inspirational solving Systems Equations by Elimination Worksheet. We choose the second equation because it is easier to solve for the variable $$y$$ in that equation. Systems of Equations (Graphing & Substitution) Worksheet Answers. Pre-Algebra Worksheets. M A HMBard 8ec Nwli3t4hc dIqn7fCi1nHiPtxeT SAYlpgDeMbFrGaW Y2P.4 Worksheet by Kuta Software LLC 32) x y x y ( , ) 33) x y x y ( , ) Solve each system by elimination. Sometimes it is not possible or convenient to solve a system of equations by graphing. J a CAVlolr GrUiqg 9het Dsg Or ye wsdegrGvke Ddz.J H OMla Adke T LwqiUtphO eIGnfpi Yn0i 5t ZeX 4Avl QgRe2bIr SaR f1 W.y Worksheet by Kuta Software LLC Kuta Software - Infinite Algebra 1 Name_____ Solving Systems of Equations by Substitution Date_____ Period____ Solve each system by substitution. $$- 8x + 5\left( {3x - 4} \right) = - 6$$ First, it requires the graph to be perfectly drawn, if the lines are not straight we may arrive at the wrong answer. 6.2A Solving Systems by Substitution (isolated) Solve each system by substitution. Using this method, you isolate the variables and substitute one of them to solve for the other. We choose the first equation. (Day 3) 35) x y x y ## solving systems of equations by substitution worksheet Harvest Fruit Bread, Annandale Golf Store, Best Landscape Architecture Schools In The World, Road To Perdition Piano Pdf, Cookies Market Menu, Waterfront Homes For Sale In Homosassa, Fl, System Of Linear Equations Exercises With Solutions Pdf,

https://www.dsemth.com/static/question-sample/exterior-angle-of-triangl

Exterior angle of triangles Question Sample Titled 'Exterior angle of triangles' In the figure, ${C}{F}{D}$ , ${A}{E}{B}$ , ${C}{H}{M}{B}$ , ${D}{M}{K}{A}$ and ${F}{H}{K}{E}$ are straight lines. It is given that ${C}{F}{D}$$//$${A}{E}{B}$ , $\angle{C}{D}{M}={53}^{\circ}$ , $\angle{M}{K}{E}={173}^{\circ}$ and $\angle{M}{B}{A}={46}^{\circ}$ . Find $\angle{K}{H}{M}$ . (4 marks) ${D}$${F}$${C}$${A}$${E}$${B}$${H}$${K}$${M}$${53}^{\circ}$${46}^{\circ}$${173}^{\circ}$ $\angle{D}{C}{M}$ $=\angle{C}{B}{A}$ alt $\angle{s}$, ${C}{F}{D}$$//$${A}{E}{B}$ $={46}^{\circ}$ 1A In $\triangle{M}{C}{D}$ , $\angle{C}{M}{K}$ $=\angle{D}{C}{M}+\angle{C}{D}{M}$ ext. ∠ of △ $={53}^{\circ}+{46}^{\circ}$ 1M $={99}^{\circ}$ 1A In $\triangle{M}{H}{K}$ , $\angle{K}{H}{M}$ $=\angle{M}{K}{E}-\angle{C}{M}{K}$ ext. ∠ of △ $={173}^{\circ}-{99}^{\circ}$ $={74}^{\circ}$ 1A 專業備試計劃 Level 4+ 保證及 5** 獎賞 ePractice 會以電郵、Whatsapp 及電話提醒練習 ePractice 會定期提供溫習建議 Level 5** 獎勵：會員如在 DSE 取得數學 Level 5** ，將獲贈一套飛往英國、美國或者加拿大的來回機票，唯會員須在最少 180 日內每天在平台上答對 3 題 MCQ。 Level 4 以下賠償：會員如在 DSE 未能達到數學 Level 4 ，我們將會全額退回所有會費，唯會員須在最少 180 日內每天在平台上答對 3 題 MCQ。 FAQ ePractice 是甚麼？ ePractice 是一個專為中四至中六而設的網站應用程式，旨為協助學生高效地預備 DSE 數學（必修部分）考試。由於 ePractice 是網站應用程式，因此無論使用任何裝置、平台，都可以在瀏覽器開啟使用。更多詳情請到簡介頁面。 ePractice 可以取代傳統補習嗎？ 1. 會員服務期少於兩個月；或 2. 交易額少於 HK\$100。 Initiating... HKDSE 數學試題練習平台

https://pcraster.geo.uu.nl/pcraster/4.3.2/documentation/pcraster_manual/sphinx/op_windowhighpass.html

# windowhighpass¶ windowhighpass Increases spatial frequency within a specified square neighbourhood Result = windowhighpass(expression, windowlength) expression spatial scalar windowlength spatial, non spatial scalar Result spatial scalar ## Options¶ --unittrue or --unitcell --unittrue windowlength is measured in true length (default) --unitcell windowlength is measured in number of cell lengths ## Operation¶ For each cell c its windowhighpass value is computed as follows. A square window with the cell c in its centre is defined by windowlength. The windowlength is the length of the window in horizontal and vertical directions. For each cell i which is entirely or partly in the window and which is not the centre cell c the fraction of the cell in the window is determined. This is the area of the part of the cell in the window divided by the total area of a cell. Let fraction(i) be this fraction; let expression(i) be the expression value of a surrounding cell i and expression(c) the expression value of the centre cell c. The windowhighpass filter value on the centre cell c is calculated according to: $windowhighpass(c) = 2 * expression(c) * { \sum^n_{i=1} fraction(i) } - { \sum^n_{i=1} ( fraction(i) * expression(i) ) }$ where n is the number of cells i surrounding the centre cell C in the window. This computation is performed for all cells: for each cell the highpass(C) value is assigned to the corresponding cell on Result. ## Notes¶ The cell value on windowlength should be greater than 0, else a missing value is assigned to the corresponding cell on Result. A cell on windowlength with a missing value results in a missing value on Result at the corresponding cell. However, if a missing value on windowlength occurs in a cell which is not the centre cell of the window the expression value in that cell is included in the computation of the highpass value in the window. ## Group¶ This operation belongs to the group of Neigbourhood operators; window operators ## Examples¶ 1. • pcrcalc binding Result2 = Result2.map; Expr = Expr.map; WinLen2 = WinLen2.map; initial report Result2 = windowhighpass( Expr, WinLen2); • python Result2 = windowhighpass( Expr, WinLen2) Result2.map Expr.map WinLen2.map 2. • pcrcalc binding Result1 = Result1.map; Expr = Expr.map; initial report Result1 = windowhighpass( Expr, 6); • python

http://clay6.com/qa/49384/a-and-b-are-two-events-such-that-p-a-0-54-p-b-0-69-and-p-a-cap-b-0-35-find-

Browse Questions # A and B are two events such that P(A)=0.54,P(B)=0.69 and $P(A\cap B)=0.35$.Find $P(B\cap A')$ $\begin{array}{1 1}(A)\;0.34\\(B)\;0.24\\(C)\;0.14\\(D)\;0.64\end{array}$ Can you answer this question? Toolbox: • $P(B \cap A')=P(B)-P(B \cap A)$ $P(B \cap A)=0.35$ $P(B)=0.69$ $P(B \cap A')=P(B)-P(B \cap A)$ $\Rightarrow 0.69-0.35$ $\Rightarrow 0.34$ Hence (A) is the correct answer. answered Jul 2, 2014

https://socratic.org/questions/how-do-you-use-the-squeeze-theorem-to-find-lim-tan-x-cos-sin-1-x-as-x-approaches

How do you use the Squeeze Theorem to find lim tan(x)cos(sin(1/x)) as x approaches zero? Sep 25, 2015 We need to consider right and left limits separately. Explanation: For all $x \ne 0$, we have $- 1 \le \cos \left(\sin \left(\frac{1}{x}\right)\right) \le 1$ Right Limit For $0 < x < \frac{\pi}{2}$, we have $\tan x > 0$, so we can multiply the three parts of the inequality above by $\tan x$ without changing the inequalities. For $0 < x < \frac{\pi}{2}$, we get $- \tan x \le \tan x \cos \left(\sin \left(\frac{1}{x}\right)\right) \le \tan x$ Since ${\lim}_{x \rightarrow {0}^{+}} \left(- \tan x\right) = 0 = {\lim}_{x \rightarrow {0}^{+}} \left(\tan x\right)$, we have (by the right hand squeeze theorem) ${\lim}_{x \rightarrow {0}^{+}} \tan x \cos \left(\sin \left(\frac{1}{x}\right)\right) = 0$ Left Limit For $- \frac{\pi}{2} < x < 0$, we have $\tan x < 0$, so when we multiply the three parts of the inequality by $\tan x$ we must change the inequalities. For $- \frac{\pi}{2} < x < 0$, we get $- \tan x \ge \tan x \cos \left(\sin \left(\frac{1}{x}\right)\right) \ge \tan x$ Since ${\lim}_{x \rightarrow {0}^{-}} \left(- \tan x\right) = 0 = {\lim}_{x \rightarrow {0}^{-}} \left(\tan x\right)$, we have (by the right hand squeeze theorem) ${\lim}_{x \rightarrow {0}^{-}} \tan x \cos \left(\sin \left(\frac{1}{x}\right)\right) = 0$ Twi-sided Limit Because both the right and left limits at $0$ are $0$, we conclude: ${\lim}_{x \rightarrow 0} \tan x \cos \left(\sin \left(\frac{1}{x}\right)\right) = 0$

http://lara.epfl.ch/w/sav09:chaotic_iteration_in_abstract_interpretation

• English only # Chaotic Iteration in Abstract Interpretation: How to compute the fixpoint? In Abstract Interpretation Recipe, note that if the set of program points is , then we are solving the system of eqautions in variables where The approach given in Abstract Interpretation Recipe computes in iteration values by applying all equations in parallel to previous values: What happens if we update values one-by-one? Say in one iteration we update -th value, keeping the rest same: here we require that the new value differs from the old one . An iteration where at each step we select some equation (arbitrarily) is called chaotic iteration. It is abstract representation of different iteration strategies. Questions: 1. What is the cost of doing one chaotic versus one parallel iteration? (chaotic is times cheaper) 2. Does chaotic iteration converge if parallel converges? 3. If it converges, will it converge to same value? 4. If it converges, how many steps will convergence take? 5. What is a good way of choosing index (iteration strategy), example: take some permutation of equations. More generally: fair strategy, for each vertex and each position in the sequence, there exists a position afterwards where this vertex is chosen. be vectors of values in parallel iteration and be vectors of values in chaotic iteration These two sequences are given by monotonic functions and and the second one is clearly smaller i.e. Compare values , , …, , in the lattice: • , generally (proof by induction) • when selecting equations by fixed permutation , generally Therefore, using the Lemma from Comparing Fixpoints of Sequences twice, we have that these two sequences converge to the same value. ### Worklist Algorithm and Iteration Strategies Observation: in practice depends only on small number of , namely predecessors of node . Consequence: if we chose , next time it suffices to look at successors of (saves traversing CFG). This leads to a worklist algorithm: • initialize lattice, put all CFG node indices into worklist • choose , compute new , remove from worklist • if has changed, update it and add to worklist all where is a successor of Algorithm terminates when worklist is empty (no more changes). Useful iteration strategy: reverse postorder and strongly connected components. Reverse postorder: follow changes through successors in the graph. Strongly connected component (SCC) of a directed graph: path between each two nodes of component. • compute until fixpoint within each SCC If we generate control-flow graph from our simple language, what do strongly connected components correspond to?

http://mathhelpforum.com/pre-calculus/156823-solve-y-x-1-x-0-x-1-x-terms-y-print.html

# Solve y = x + 1/x with 0 \< x \< 1 for x in terms of y. • September 20th 2010, 01:35 PM hsidhu Solve y = x + 1/x with 0 \< x \< 1 for x in terms of y. I need help with this assignment, Problem 3: Solve y = x + 1/x with 0 \< x \< 1 for x in terms of y. Any clues as to how to do this, would be greatly appreciated :) • September 20th 2010, 02:40 PM skeeter Quote: Originally Posted by hsidhu I need help with this assignment, Problem 3: Solve y = x + 1/x with 0 \< x \< 1 for x in terms of y. $y = x + \frac{1}{x}$ ... note that $x \ne 0$. multiply every term by $x$ ... $xy = x^2 + 1$ $0 = x^2 - xy + 1$ $x = \frac{y \pm \sqrt{y^2 - 4}}{2}$

http://mathhelpforum.com/calculus/31392-nasty-derivative.html

1. ## Nasty derivative Once again. After missing a week of school and a test, its not fun to catch up, especially in calculus. Heres the question. Find the slope of the tangent to $h(x)=2x(x+1)^3(x^2+2x+1)^2$ at $x=-2$ Explain how to find the equation of the normal at x= -2. Ok so if anyone solves this, can they please do a step by step break down because im about to eat my book. 2. $ h(x)=2x(x+1)^3(x^2+2x+1)^2 $ $g(x)=2x(x+1)^3 = q(x)p(x)$ where $q(x) = 2x$ and $p(x) =(x+1)^3$ $f(x)=(x^2+2x+1)^2$ $f'(x) = 2(x^2+2x+1)(2x+2)$ $g'(x) = q(x)p'(x) + q'(x)p(x) = 2x(3(x+1)^2) + 2(x+1)^3$ _________________________________________________ $h(x) = g(x)f(x) \Rightarrow h'(x) = g(x)f'(x) + g'(x)f(x)$ $h(x) = 2x(x+1)^3 2(x^2+2x+1)(2x+2) + (2x(3(x+1)^2) + 2(x+1)^3)(x^2+2x+1)^2$ simplify and plug in x=-2! oh and check my work, this thing's a mess! 3. eeek. i just worked through it and it gave me -24. the answer in the book says -30. Il double check my math 4. let $h(x)=2x(x+1)^3(x^2+2x+1)^2 $ so $h(-2)=1(-2)(-1)(1)^2=4$ talking the log of both sides we get $ln(h(x))=ln(2x)+3ln(x+1)+2ln(x^2+2x+1)$ take the derivative $\frac{1}{h}\frac{dh}{dx}=\frac{1}{x}+\frac{3}{x+1} +\frac{2(2x+2)}{x^2+2x+1}$ solving for the derivative... $\frac{dh}{dx}=h(x)[\frac{1}{x}+\frac{3}{x+1}+\frac{2(2x+2)}{x^2+2x+1}]$ eval at x=-2 $\frac{dh}{dx}|_{x=-2}=h(-2)[ \frac{1}{-2}+\frac{3}{-2+1}+\frac{2(2(-2)+2)}{(-2)^2+2(-2)+1}]$ $\frac{dh}{dx}|_{x=-2}=4[\frac{-1}{2}-3-4]=-30$

https://mathoverflow.net/questions/191225/can-a-positive-measure-subset-of-a-free-group-be-nowhere-dense

Can a positive measure subset of a free group be nowhere dense? Let $F$ be a finitely generated free group and let $S \subseteq F$ be a subset for which there is some $\epsilon > 0$ such that for any epimorphism to a finite group $\phi \colon F \to G$ we have that $\frac{|\phi(S)|}{|G|} \geq \epsilon$ (that is, the closure of $S$ in the profinite completion of $F$ has positive Haar measure). Is it possible that the closure of $S$ with respect to the profinite topology on $F$ does not contain a coset of any finite index subgroup of $F$ ? This is the same as asking whether it is possible that $S$ is nowhere dense (the interior of the closure is empty) with respect to the profinite topology. • Is this true in $\mathbb{Z}$? – HJRW Dec 21 '14 at 16:09 • @HJRW, After thinking of it a bit I was not able to figure it out. It can have a different answer in the nonabelian case in which I am interested most. Dec 21 '14 at 16:12 Yes, $S$ can be nowhere dense. The idea is to build a 'fat Cantor set' inside $F$. Let $H_0=F,H_1,H_2,\dots$ be finite-index subgroups of $F$ such that $H_{i+1}\leq H_i$ for each $i$ and such that $\bigcap_{i=0}^\infty H_i = \{1\}$. Start with $i_0=0$ and $C_0=\{H_0\}$, and then iteratively do the following. Having defined $C_n$ as a certain set of more than half the cosets of $H_{i_n}$, let $S_n\subset F$ consist of one member from each coset $gH_{i_n}\in C_n$, and then choose $i_{n+1}$ so large that $$\frac{|C_n|}{|F/H_{i_n}|} - \frac{|C_n|}{|F/H_{i_{n+1}}|} > 1/2,$$ and so that when we decompose each $gH_{i_n} \in C_n$ into a union of cosets of $H_{i_{n+1}}$ we can remove at least one of these cosets without uncovering any member of $S_0\cup \dots \cup S_n$. Let $C_{n+1}$ be the resulting set of cosets of $H_{i_{n+1}}$. In the end let $S=\bigcap_{n=0}^\infty \bigcup C_n$, and let $\bar{S}$ be the closure of $S$ in $\hat{F} = \lim F/H_i$. Then $S$ is closed in the profinite topology of $F$ and contains no coset of any $H_i$, so $\bar{S}$ is nowhere dense. On the other hand $S$ contains $\bigcup_{i=0}^\infty S_i$, so the image of $S$ in $F/H_i$ always has size at least $|F/H_i|/2$, so $\bar{S}$ has measure at least $1/2$. • It looks like the this works for $F=\mathbb Z$, or indeed any infinite, residually finite $F$. – HJRW Dec 22 '14 at 17:25 For the special case of $\mathbb{Z}$ we can construct a set $S=\{n_1, \dotsc, n_k, \dotsc\}$ such that $$\sum_{i=1}^\infty \frac{1}{n_i} < \infty,$$ and such that $S$ contains every residue modulo every $n>1$ (this is essentially a triviality - for each $n$ construct a complete residue set which starts at $n^3,$ or thereabouts, then take the union of all these). This will project onto every quotient of $\mathbb{Z}$ and have natural density $0.$ • Isn't $S$ dense in $\mathbb{Z}$? – HJRW Dec 21 '14 at 20:35 • @HJRW It is indeed. So the closure contains a coset and this is not an answer. Dec 21 '14 at 20:36 • @HRJW how is this dense in $\mathbb{Z}?$ It has natural density $0,$ so does not contain any arithmetic progression... Dec 21 '14 at 21:58 • Igor, nevertheless, it's dense in the profinite topology. – HJRW Dec 21 '14 at 22:10 • @HJRW Sorry, I am quite dense: how so? Dec 21 '14 at 22:11

https://discuss.codechef.com/questions/70882/stdytab-editorial

× # STDYTAB - Editorial Author: Pavel Sheftelevich Tester: Mahbubul Hasan and Sunny Aggarwal Editorialist: Balajiganapathi Senthilnathan Russian Translator: Sergey Kulik Mandarian Translator: Gedi Zheng Easy # PREREQUISITES: Combinatorics, DP # PROBLEM: You have to fill a $N x M$ matrix with non negative numbers such that the sum of numbers at row $i$ $(2 \le i \le N)$ is greater than or equal to the sum of numbers in row $i - 1$. # SHORT EXPLANATION Suppose we want to fill one row (which has $M$ elements) such that the sum of the numbers is $X$. The number of ways to fill this row can be obtained using a standard combinatorics formulae: $\binom{X + M - 1}{M - 1}$. We can solve using dp: let $solve(i, x)$ denote the number of ways to fill the matrix such that the matrix is steady and the sum of the $i^{th}$ row is $x$. We can easily solve this using the recurrance: $solve(i, x) = solve(i + 1, x) * \binom{x + m - 1}{m - 1} + solve(i, x + 1)$ # EXPLANATION: Suppose we start at row $1$. Say, we fill this row such that the sum is $x$. Now when we move on to the second row, we must make sure that the sum of the second row is greater than or equal to $x$. So, besides the row itself, we must also keep track of the minimum sum of the row. Let $solve(i, x)$ be the number of ways to fill the matrix from $i^{th}$ row to the $N^{th}$ row such that the matrix is steady and the sum of $i^{th}$ row is at least $x$. There are now two ways to proceed 1. We make the sum of current row $x$: We need to calculate in how many ways we can make the sum of the current row $x$. There are $M$ elements in this rowee. So we need to find the number of ways in which we can choose $M$ numbers such that they sum up to $x$. This can be calculated using the formula $\binom{X + M - 1}{M - 1}$. See TODO for an explanation. 2. We don't make the sum of current row x: In this case we can make the sum of the row atleast x + 1. Putting this together we have the following solution MOD = 1000000000 Preprocess array C such that C[n][k] = n choose k mod MOD Solve(i, x): See if we have already calculated the answer for this input and if so return the value Edge cases: if x > M return 0 // Sum of a row can't be greater than M if i == N + 1 return 1 ret = (Solve(i + 1, x) * C[x + m - 1][m - 1] + Solve(i, x + 1)) mod MOD ## Time Complexity: We calculate each state only once. For each state we take a constant amount of time. We know that $i$ can take values till $N$ while $x$ can take values till $M$. So, the total amount of time taken for solve will be $O(NM)$. # AUTHOR'S AND TESTER'S SOLUTIONS: This question is marked "community wiki". 75542742 accept rate: 77% 19.0k348495531 7 whatever i wrote in this question i might not be able to explain it to myself later. :P answered 15 Jun '15, 15:38 153●5 accept rate: 10% 2 I thought I was the only one who was thinking this. :-D (15 Jun '15, 15:41) apptica6★ 1 I will just complete the TODO. Suppose X=2 , M=3 i.e. how many ways to choose 3 non-negative integers to get sum X. Take X no of objects i.e. xx. So the question is in how many ways we can partition this, then the number of objects between partitions will be my M numbers summing up to X. Denote partition by | and object by x. If there are n number of x's to the left of leftmost '|' or right of rightmost '|' or between 2 '|' then there is digit n (leftmost,rightmost or at that place respectively). If '|' is right most or left most then it means rightmost/leftmost digit is 0. If there are consecutive |, then it means there is 0 at that place. For example : 002 -> ||xx , 011 -> |x|x , 020 -> |xx| , 101 -> x||x , 110 -> x|x| , 200 -> xx|| Note that, there will be M-1 '|' ( because we want M numbers). Hence, total (X+M-1) places and number of ways to place (M-1) '|' in (X+M-1) places is (X+M−1) C (M−1) . PS: I first used this DP by the recurrence f(X, M) = 1 if X = 0 = 0 if X != 0 and M = 0 = sum f(X - i, M - 1) over i in [0, X] otherwise When I printed the array and saw the numbers of Pascal's triangle, then I found the combinatorics formula and then later found the logic of it. answered 17 Jun '15, 15:27 11●2 accept rate: 0% how is 002 -> ||xx? I get that to the right of each '|' there is a 0. But how come we just have 1 digit '2' for 'xx' ? (20 Jun '15, 22:56) Look, '|' is the partition. It counts number of x's left/right to it or in between 2 '|'. 002 -> ||xx Left most is '|' hence left most digit is 0 because there are no x's left of it. After '|' is another '|', hence no x in between hence 0 After this |, we have 2 x's hence 2. Similarly, |xx||xxx would be 0203. Leftmost is a '|' hence leftmost number is 0. After this before the next '|', we encounter 2 x's hence 2. Then we have no x's, before next '|' hence 0. And finally to the right of it there are 3 x's hence 3 at the rightmost number. So, for M numbers we would have (M-1) '|'. OK? (22 Jun '15, 17:36) 0 Rather we can also do it the reverse way by starting with the sum of no.s of nth column to be <="M". just the thing is as n and m increase the C(m+n-1,n-1) becomes too large to calculate... so we need some smart methods to calculate them... I haven't come across such methods where modulo no is not a prime (10^9) and not (10^9+7)... answered 15 Jun '15, 16:17 1●1 accept rate: 0% You could have used nCr = ((n-1)C(r-1) + (n-1)Cr) % M. (15 Jun '15, 17:38) 0 can anyone help me why i'm getting tle. I used matrix exponential to calculate the answer and the time complexity is O(MMlog(N)).The link to the problem is http://www.codechef.com/viewsolution/7215565 answered 15 Jun '15, 17:36 198●1●5 accept rate: 0% Its probably because of matrix exponentiation. You have m^3 loops which is much slower than the required n*m algorithm (15 Jun '15, 18:14) 0 I did this problem in O(n^2) time using 2 2D arrays. I got TLE in one testcase in that approach. code : TLE in one case But when I calculated using 1 2D (For nCr) and another 1D for DP I got accepted. Trying to understand just what happened there. code : Accepted answered 15 Jun '15, 20:53 3★aroup 1●1 accept rate: 0% Its just one of the many weird things that come with competitive programming. lol (15 Jun '15, 21:16) 0 my code also has a complexity of O(mnt).. but it is giving TLE even with fast I/O,can anyone help me ? here is the code : http://www.codechef.com/viewsolution/7144879 answered 16 Jun '15, 09:53 1 accept rate: 0% 0 I am using memoization to solve this problem. Getting TLE in one of the cases of sub-task 3 ? Please help. Here is the Code. answered 16 Jun '15, 19:37 2★asan 1 accept rate: 0% 0 Elegant solution.. used 2 2D arrays to build up answer from 1xM to NxM. http://ideone.com/vDBnr0 answer is sum of elements of last row of dp array. Got WA in challenge due to faulty code which calculated nCr. Got AC once I corrected it. answered 16 Jun '15, 20:11 1 accept rate: 0% 0 Can someone please explain that nCr array computation in detail, or a tutorial reference would be appreciated. I have seen it in many problems as a pre computation stage but didn't understand the logic. answered 17 Jun '15, 12:48 1★ajit_a 41●1●1●3 accept rate: 0% toggle preview community wiki: Preview By Email: Markdown Basics • *italic* or _italic_ • **bold** or __bold__ • image? • numbered list: 1. Foo 2. Bar • to add a line break simply add two spaces to where you would like the new line to be. • basic HTML tags are also supported • mathemetical formulas in Latex between \$ symbol Question tags: ×14,365 ×3,155 ×1,754 ×756 ×156 ×4 question asked: 02 Jun '15, 13:13 question was seen: 6,863 times last updated: 20 Jan '16, 12:47

https://mathoverflow.net/questions/359168/physical-measures-that-are-not-srb/359208

# Physical measures that are not SRB It is quite easy to construct a dynamical system which has a physical measure with a positive Lyapunov exponent and zero entropy, just a figure $$\infty$$ system. By Pesin's entropy formula such a measure can not be a Sinai-Ruelle-Bowen measure. Now my question is, if there exists a system with a physical measure with positive entropy that is not a SRB-measure. I think the answer is Yes, but I did not manage the construction of such a system. (Perhaps I have to add that I am interested in ergodic and hyperbolic physical measures that are not SRB) Yes. The simplest construction is to let $$f$$ be the figure-eight system so that $$\delta_p$$ is a physical non-SRB measure (where $$p$$ is the saddle point) and let $$g$$ be an Anosov diffeomorphism with SRB measure $$\mu$$ (a hyperbolic toral automorphism with Lebesgue measure will do the job); then consider the product system $$f\times g$$. The product measure $$\delta_p\times \mu$$ is physical, not SRB, and has positive entropy. Of course that example feels like cheating and is almost certainly not the sort of thing you had in mind. A more informative example comes from Hofbauer, Franz; Keller, Gerhard, Quadratic maps without asymptotic measure, Commun. Math. Phys. 127, No. 2, 319-337 (1990). ZBL0702.58034. Theorem 2 in that paper says that if one considers a full, continuous family of S-unimodal interval maps (for example, the family of quadratic maps $$\{f_a\colon [0,1]\to [0,1] : a\in [0,4]\}$$ given by $$f_a(x) =ax(1-x)$$), then for every $$0 < h < \log(\frac{1+\sqrt{5}}2)$$ there are uncountably many parameter values with an ergodic measure $$\nu$$ that is singular to Lebesgue, has entropy $$h$$, and is the limit of the empirical measures for Lebesgue-a.e. $$x$$ (in other words, it is physical). I have not read the paper carefully enough to have any insight into how those parameters are chosen, beyond knowing that it has something to do with kneading sequences. It is then reasonable to ask about diffeomorphisms (as opposed to non-invertible interval maps) and conjecture that similar examples exist in the Hénon family of maps, but I do not know of any results in this direction and did not find any in a quick search. • Many thanks for this detailed answer! – Jörg Neunhäuserer May 3 '20 at 13:29

https://mathhelpboards.com/threads/mechanics-help.24941/

# PhysicsMechanics help #### cake81 ##### New member Hello! I need help with this question: A tennis ball is thrown in the air by a man so that at the instant when the ball leaves his hand, the ball is 2m above the ground and is moving vertically upwards with speed9m/s^-1 The motion of the ball is modlledas that of a particle moving freely under gravityand the acceleration due to gravity is modelled as being of constant magnitude of 10 m/s^-1 The ball hits theground T seconds after leaving the hand Using the model, find the value of T I dont know how togo about solving this because I dont understand the part with the particle moving freely #### Euge ##### MHB Global Moderator Staff member Welcome, cake81 ! We are given an initial height of $y_0 = 2\,\text{m}$, an initial velocity of $v_0 = 9\, \text{m/s}$ in the upwards vertical direction, and an acceleration of $a = -10\,\text{m/s}$ (the acceleration vector faces downwards). Using the kinematic equation $$y = y_0 + v_0 t + .5 at^2$$ with $y = 0$ we solve $0 = 2 + 9t + (.5)(10)t^2$, or $$5t^2 - 9t - 2 = 0$$ Can you take it from here? #### cake81 ##### New member Welcome, cake81 ! We are given an initial height of $y_0 = 2\,\text{m}$, an initial velocity of $v_0 = 9\, \text{m/s}$ in the upwards vertical direction, and an acceleration of $a = -10\,\text{m/s}$ (the acceleration vector faces downwards). Using the kinematic equation $$y = y_0 + v_0 t + .5 at^2$$ with $y = 0$ we solve $0 = 2 + 9t + (.5)(10)t^2$, or $$5t^2 - 9t - 2 = 0$$ Can you take it from here? thank you so much. i got 2, i only have to use the positive answer right? #### Euge ##### MHB Global Moderator Staff member Yes. Only the positive answer makes sense.

https://proofwiki.org/wiki/Sequence_of_Integers_whose_Cube_is_Palindromic

# Sequence of Palindromic Cubes ## Sequence The sequence of positive integers whose cube is palindromic begins: $1, 2, 7, 11, 101, 111, 1001, 2201, 10 \, 001, 10 \, 101, \ldots$ Note that $2201$ is the smallest (and only one one known) which is itself non-palindromic. The corresponding sequence of palindromic cubes begins: $1, 8, 343, 1331, 1 \, 030 \, 301, 1 \, 367 \, 631, 1 \, 003 \, 003 \, 001, 10 \, 662 \, 526 \, 601, 1 \, 000 \, 300 \, 030 \, 001, \ldots$

http://adrianbell.me/?cat=14

# Pumping Lemma Way way off the beaten path here, but this is the best example of usage of the pumping lemma I've seen. Just need somewhere to put it... The below is taken from here. Theorem: Let $L$ be a regular language, and $w$ be a string. Then there exists a constant $c$ s.t. $\forall w \in L, |w| \geq c$. We can break $w$ into three strings, $w=xyz$, s.t.: • $|y| > 0$ • $|xy| \leq c$ • $\forall k \geq 0, xy^{k}z \in L$ Method to prove that a language $L$ is not regular: • At first, we have to assume that $L$ is regular. • So, the pumping lemma should hold for $L$. • Use the pumping lemma to obtain a contradiction: • Select $w$ s.t. $|w| \geq c$. • Select $y$ s.t. $|y| \geq 1$. • Select $x$ s.t. $|xy| \leq c$ • Assign the remaining string to $z$. • Select $k$ s.t. the resulting string is not in $L$. Problem: Prove that $L=\{a^{i}b^{i} | i \geq 0\}$ is not regular. Solution: • At first, we assume that $L$ is regular and $n$ is the number of states. • Let $w=a^{n}b^{n}$. Thus $|w|=2n\geq n$. • By the pumping lemma, let $w=xyz$, where $|xy| \leq n$. • Let $x=a^{p}$$y=a^{q}$, and $z=a^{r}b^{n}$, where $p+q+r=n$, $p \neq 0$$q \neq 0$$r \neq 0$. Thusly $|y| \neq 0$. • Let $k=2$. Then $xy^{2}z=a^{p}a^{2q}a^{r}b^{n}$. • Number of $a\text{'s}=(p+2q+r) = (p+q+r)+q=n+q$. • Hence, $xy^{2}z=a^{n+q}b^{n}$. Since $q\neq 0$, $xy^{2}z$ is not of the form $a^{n}b^{n}$!!!!!!!!!!!!!!! • Thus, $xy^{2}z \notin L$. Hence $L$ is not regular. # Feedback - 01 I received the marks back for my first monster assignment! Did quite well as it turns out! But this blog isn't about spouting about my success, it's about the learning process! So here's some of the things I screwed up... First off, my algebra is clearly rusty as fuck. In one instance put a minus sign in the wrong place AND mysteriously lost a factor of 2 in the progress of my working. In future I really need to re-read my working really carefully (three or four times over it seems), both the hand-written and the full typed-up LaTeX... Something else I lost marks for was the apparently simple task of graph sketching, either where I hadn't considered asymptotes or had not considered the limits of the domain. Overall I clearly need to be a lot more mindful of whether I'm dealing with $\leq$ or $<$. When I read those symbols I see them both so often, I frequently gloss over them without properly considering their usage. Again, pretty basic stuff. With complex numbers I apparently need to be more explicit with my declaration of forms. My polar form was implicit in the answer, but there wasn't anywhere I actually stated it. Silly boy. I fell down on a proof of symmetry for an equivalence relation. I just wasn't mindful whilst answering this. It is assumed that $x-3y=4n$. This can be rearranged in terms of $y$ as $y=\frac{x}{3}-\frac{4n}{3}$. So substituting y, in the symmetrical $y-3x$ results in: $4 \left(-\frac{2x}{3}-\frac{2n}{3}\right)$. Of course, at this point, proving that what's inside the brackets is an integer is pretty difficult. But that's where I left it. A bit more play would've shown that I could easily have arranged the first equation in terms of $x$ instead which would've resulted in $4\left( -2y-3k\right)$, which is rather obviously an integer given the initial variables. More exploration required in future... Lastly, in my last post I mentioned how there was a distinct lack of symbolic existential or universal quantifiers in all this new material. After Velleman, I was so used to seeing them, and working with them appropriately but because they're now not around, I got totally burnt by assuming I had to prove "there exists" instead of "for all" for one question. I suppose I'll be able to get around this with making sure my notes explicitly state whatever quantifier we're actually talking about. Damned English language... Symbols are much more concise! 🙂 # Large Intro Finally submitted my first assignment. It was monstrous. Just over 23 pages of mathematics and sketches of graphs. All of it typed up in LaTeX. Skipping ahead to look at the rest of the assignments, it looks as if this first assignment may very well be the biggest of the whole lot. This is a very good thing as I really don't think I could churn out that much work of a high quality every month. Glad to say that most of this introduction section I was familiar with. Only really new topic was equivalence relations, which caused some problems initially. Overall though, what I've found difficult is the apparent lack of logical notation. After reading "How To Prove It" I've become half-decent at making sense of and rearranging logical notation to solve a problem. The difficulty comes in looking at the plain-English description of something in the texts and then having to translate it into logical notation to allow my fussy brain to think about them logically. Perfect example of this is the definition of a function being "onto". In the text, the definition reads: "A function $f: A \longrightarrow B$ is onto if $f(A)=B$". Which is fine, but the Wikipedia definition reads: "$\forall y \in Y, \exists x \in X$ such that $y = f(x)$" Which for me, gives me a much better idea about how to go about proving if a function is onto. Why leave out the quantifiers? The Wikipedia definition tells me so much more. I suppose translating English into logical notation is just something I'll have to get good at! Though even after this long intro section, I really feel I need more practise with proofs... I guess this may have to wait until revision time... Next up is the first section on group theory, with an assignment due on November 24th. Onward. # PENS DOWN! So that's it! The Summer has ended! How far has my extra study got me? Well I've managed to get through around 120 pages of "How To Prove It" by Velleman, and have generated just over 70 double-sided pages of A4's worth of exercises from the book. Not bad for an extra-curricular topic! This book has helped me loads. It's succeeded in taking away a lot of the mystery involved in reading and writing proofs. Every topic in the book up until now has flowed well, and allowed me to think about solutions to the various problems fairly naturally. What I mean by that is, I never became absolutely stuck and unable to answer a question. Having said that, the sub-topic I'm finishing on is proofs involving quantifiers. This is the one area in which I'll admit I've been struggling. At this point in the book, I've learned so much about the number of ways in which to decon/reconstruct a problem that any possible method by which to prove a theorem has actually become less obvious. Here's an example of how convoluted the scratch work of a basic proof has become. Here's question 14 from p.122: Suppose  $\{A_i\: |\: i \in I\}$ is an indexed family of sets. Prove that $\cup_{i \in I} \mathscr{P} (A_i) \subseteq \mathscr{P}(\cup_{i \in I} A_i)$. It's a short question, but this immediately looks like a nightmare to a beginner like myself. We've got a mix of indexed sets, a union over them, and power sets. First off I need to properly understand the damn thing. Seems sensible to draw up an example using the theorem... Let's say $I$ is $\{1,2\}$. So we've got $\{A_{1},A_{2}\}$. Now let's say that $A_1 = \{1,2\}$ and $A_2 = \{2,3\}$. Looking at the LHS of the thing I need to prove, it's actually pretty easy to break down: $\mathscr{P} (A_1) = \{ \varnothing, \{1\}, \{2\}, \{1,2\}\}$ $\mathscr{P} (A_2) = \{ \varnothing, \{2\}, \{3\}, \{2,3\}\}$ Which means that the union of all the elements of the power sets of $A_i$ is: $\cup_{i \in I} \mathscr{P} (A_i) = \{ 1,2,3\}$ A little half-way recap: the theorem says that in my example, $\{1,2,3\}$ should be equal to, or be a subset of $\mathscr{P}(\cup_{i \in I} A_i)$ (the RHS). Let's see if that's true shall we? Within the parenthesis of the RHS we've got $\cup_{i \in I} A_i$. So this is the union of all elements of all indexed sets. In this example: $\cup_{i \in I} A_i = \{1,2,3\}$ Only thing missing now is the power set of this: $\mathscr{P}(\cup_{i \in I} A_i) = \{ \varnothing, \{1\}, \{2\}, \{3\}, \{1,2\}, \{1,3\}, \{2,3\}, \{1,2,3\}\}$ And there we go. Now I understand exactly what the theorem means. $\cup_{i \in I} \mathscr{P} (A_i) \subseteq \mathscr{P}(\cup_{i \in I} A_i)$, in this specific example, turns out to be : $\{1,2,3\} \subseteq \{ \varnothing, \{1\}, \{2\}, \{3\}, \{1,2\}, \{1,3\}, \{2,3\}, \{1,2,3\}\}$ which is obviously true. Theorem understood. Achievement unlocked. Tick. As recommended by Velleman, I'll also try to construct the phrasing of the proof along-side the scratch work as I go. Like so: Suppose a thing is true that will help to prove the theorem. [Proof of theorem goes here] Thus, we've proved the theorem. Okay, let's start. The theorem means that if $x \in \cup_{i \in I} \mathscr{P} (A_i)$ then $x \in \mathscr{P}(\cup_{i \in I} A_i)$. So one thing implies the other. We can then class $\cup_{i \in I} \mathscr{P} (A_i)$ as a "given", and aim to prove $\mathscr{P}(\cup_{i \in I} A_i)$ as a "goal". So blocking out our answer: Suppose that $x \in \cup_{i \in I} \mathscr{P} (A_i)$. [Proof of $x \in \mathscr{P}(\cup_{i \in I} A_i)$ goes here] Thus, $\cup_{i \in I} \mathscr{P} (A_i) \subseteq \mathscr{P}(\cup_{i \in I} A_i)$. So let's start analysing $\cup_{i \in I} \mathscr{P} (A_i)$, remembering not to go too far with the logical notation. With baby steps, the definition of a union over a family of sets (here, the outer-most part of the logic) is: $\exists i \in I (x \in \mathscr{P}(A_i))$ Then, going one step further, using the definition of a power set: $\exists i \in I (x \subseteq A_i)$ Now we could go further at this point, applying the definition of a subset, but I'll stop the logical deconstruction here. In this instance, I've found that if I keep going so the entire lot is broken down into logical notation it somehow ends up getting a bit more confusing that it needs to be. With this as our given, I notice the existential quantifier. Here, I can use "existential instantiation" to plug any value I want into $i$ and then assume that what follows is true. So at this point the new "given" is simply: $x \subseteq A_i$ Nice and simple. Let's update the outline of our proper proof answer: Suppose that $x \in \cup_{i \in I} \mathscr{P} (A_i)$. Let $i \in I$ be such a value that $x \subseteq A_i$. [Proof of $x \in \mathscr{P}(\cup_{i \in I} A_i)$ goes here] Thus, $\cup_{i \in I} \mathscr{P} (A_i) \subseteq \mathscr{P}(\cup_{i \in I} A_i)$. So let's now move on to our "goal" that we have to prove: $\mathscr{P}(\cup_{i \in I} A_i)$. Again, starting from the outside, going in, $x \in \mathscr{P}(\cup_{i \in I} A_i)$ by the definition of a power set becomes: $x \subseteq \cup_{i \in I} A_i$ I can't do a lot with this on it's own so I'll keep going with the logical deconstruction. By the definition of a subset, this becomes: $\forall a (a \in x \to a \in A_i)$ So now, using "universal instantiation" I can say that here, $a$ is arbitrary (for the sake of argument, it really can be anything), and that leaves us with an updated "givens" list of: $x \subseteq A_i$ and $a \in x$ and a new "goal" of $a \in A_i$ Hey, but wait a sec... Look at our "givens"! If $a$ is in $x$... and $x$ is a subset of $A_i$, then $a$ must be in $A_i$! -and that's our goal! So update our proof: Suppose that $x \in \cup_{i \in I} \mathscr{P} (A_i)$. Let $i \in I$ be such a value that $x \subseteq A_i$. Let $a$ be an arbitrary element of $x$. [Proof of $a \in A_i$ goes here] Therefore $a \in \cup_{i \in I} A_i$. As $a$ is arbitrary, we can conclude that $x \in \mathscr{P}(\cup_{i \in I} A_i)$. Thus, $\cup_{i \in I} \mathscr{P} (A_i) \subseteq \mathscr{P}(\cup_{i \in I} A_i)$. So let's wrap this up. Theorem: Suppose  $\{A_i\: |\: i \in I\}$ is an indexed family of sets. Prove that $\cup_{i \in I} \mathscr{P} (A_i) \subseteq \mathscr{P}(\cup_{i \in I} A_i)$. Proof: Suppose that $x \in \cup_{i \in I} \mathscr{P} (A_i)$. Let $i \in I$ be such a value that $x \subseteq A_i$, and let $a$ be an arbitrary element of $x$. But if $a \in x$ and $x \subseteq A_i$, then $a \in A_i$. Therefore $a \in \cup_{i \in I} A_i$. As $a$ is arbitrary, we can conclude that $x \in \mathscr{P}(\cup_{i \in I} A_i)$. Thus, we conclude that $\cup_{i \in I} \mathscr{P} (A_i) \subseteq \mathscr{P}(\cup_{i \in I} A_i)$. $Q.E.D.$ Overall the task has involved unravelling the symbols into logic, making sure they flow together, and then wrapping them back up again. See what I mean by convoluted? All that work for that one short answer. I must admit, I still don't know if my reasoning is 100% correct with this. Despite some parts of this seeming simple, this really is the very limit of what I'm capable of understanding at the moment. I picked this example to write up, as so far I've found it to be one of the most complicated. The next section of the book seems to marry this quantifier work with another previous section about conjunctions and biconditionals, that I found to be quite enjoyable at the time. Then towards the end of the chapter, Velleman seems to sneak in some further proof examples using the terms epsilon and delta. I imagine this is a sneaky and clever way to get the reader comfortable with further Analysis study... Alas, my study of Velleman's book will have to stop here. I understand a lot more than I did, though not everything there is to know. I feel it may be enough to give me a slightly smoother ride through my next module, which was the whole point of me picking this book up. It's been so good, I hope I have a chance to return to it. I feel later chapters would put me in an even better position for further proof work! For now... the countdown begins for the release of my next module's materials! # Things you need to be told at the beginning These quotes are from pages 89 and 90 of Velleman's "How To Prove It". If only I'd read all this when I was first introduced to a proof, I wouldn't have been so stressed! "When mathematicians quote proofs, they usually just write the steps needed to justify their conclusions with no explanation of how they thought of them." "Although this lack of explanation sometimes makes proofs hard to read, it serves the purpose of keeping two distinct objectives separate: explaining your thought processes and justifying your conclusions." "The primary purpose of a proof is to justify the claim that the conclusion follows from the hypotheses, and no explanation of your thought processes can substitute for adequate justification of this claim. Keeping any discussion of thought processes to a minimum in a proof helps to keep this distinction clear." "Don't worry if you don't immediately understand the strategy behind the proof you are reading". I could hug this book right now. # End of the Quantifiers A month and a week, and I've just come to the end of the second chapter. Reasonably happy with the progress, but I could be going a bit quicker... Mind you, over just two chapters I've now created 46 pages of A4 of exercises. So there has been a LOT of material to go through. Frankly, just these first two chapters have worked wonders for my understanding of logic and what proofs are founded upon. This second chapter mainly introduced quantifiers. The concept of "for all x" and "there exists at least one x...", but quickly branched off into more involved set theory. The biggest issue I had towards the end of the second chapter was that on a couple of occasions, I don't think I thought carefully enough about the kind of answer the questions required. ie: in this context, whether the answer was required in logical notation, or whether it was required in set theory notation. Translating between the two is something I certainly found tricky. As such, I decided to write my own definitions of notation in the form of a list (thanks Lara Alcock!). Though the lack of lists of definitions could be considered a slight shortfall of the book, I think I benefited from constructing my own notes and definitions. I found that towards the end of the questions (because of the more lengthy logical notation) I was concentrating more on the definitions than what the notation actually meant. Not convinced this is so good for the learning process, but at least I'm mindful of it now. Last little question the second chapter covered was Russell's Paradox, as discovered by Bertrand Russell in 1901. The fact that I'm being introduced to stuff like this in the second chapter is pretty cool. Very enjoyable! Next up, proof technique! # The Joy Of Sets 54 pages, and 5 large exercise sections later, I've finally finished the first chapter of "How To Prove It". With the first chapter being about sentential logic, I've now covered truth tables, derivations of logical operations, set theory, and the conditional and bi-conditional connectives. The next chapter covers further foundational logical concepts and only in Chapter 3 are the intricacies of actual proofs discussed. Having taken this long to cover the first chapter, and looking at the amount of paper I've used to do all the exercises so far, I'm not that surprised I was finding proofs so difficult. It turns out my intuition was right, I was missing a lot of foundational knowledge. So far, it's all been going well. Nothing I've looked at in this first chapter has left me mystified and overall I feel like I'm learning. This is exactly where I wanted to be... Just need to up the pace, perhaps... # Books For Understanding Books - Part 2 So this is getting ridiculous. I know, I can only apologise. I'll write some maths on here at some point, I promise. It turned out that the super-valuable forum post I had on the OU forums has now been deleted. Apparently if a post isn't pinned it gets auto-nuked after two months. So now all that valuable information is gone. But let's not dwell on it. Especially when there's a new book looming!!!!! Now, despite the fact that I've read Lara Alcock's books about how to learn Analysis, and started Brannan's Analysis book (see earlier posts) I realised I was missing more foundation-level knowledge. How To Prove It by Daniel J. Velleman looks like it'll be the book to give it to me.  I remember it as being recommended on the deleted forum post, and the reviews generally are very very positive. Already I've come to the end of the first (admittedly short) section and I can actually attempt all of the exercises! I totally understand everything he's saying and I really feel like I'm learning something with every page. At last! More of a proper review of this on the way, but for the moment I'll be nose-deep in this for the next couple of months... I've just finished reading Lara Alcock's book on how to learn about Analysis. Or rather, I've finished reading the first part, and the part on the real number system. Overall, the book has led me to reconsider my current learning technique. So much so, I've compiled a list of steps to follow depending on whether I read about a new definition, theorem or proof. In turn, this has made me realise I may benefit from starting my main book on Analysis again from the beginning, but applying these new steps as I go. After all, I am still only at the beginning (kind of), and I don't have any kind of deadline looming over me (which is really nice). Overall it seems like the perfect opportunity to try out some new learning methodologies! Out of the handful of additions, there are two really big changes for me. The first being mind maps. As I go through my Analysis, I'll be creating a mind map of concepts, seeing how one builds on another. I'd tried to use mind maps before at university and they'd largely proved completely useless. Here, however, mind maps appear to offer a perfect way to visualise the building of concepts into larger concepts. Here's the beginning of my first mind map! I'm using draw.io as the tool of choice. It seems flexible enough for what I need it for, it can save as XML, and it supports mathematical notation! (mathjax latex formatting I believe) The second big change to my learning involves learning by self-explanation. This technique, mentioned in Lara Alcock's book, appears to be one of the key processes involved in truly understanding and appreciating Analysis. You can find out more about self-explanation training from Loughborough University's Mathematics Education Centre website. My difficulty here will lie in concentrating on actually doing self-explanation, rather than just paraphrasing (turns out, it's a very easy trap to fall in). So long as I do it regularly enough, self-explanation will be more likely to come to me naturally. Expected result: More effective learning and better notes!

https://www.physicsforums.com/threads/horizontal-tangent.755977/

# Horizontal Tangent 1. May 31, 2014 ### mill If the tangent is horizontal, it is where the tangent is zero. In single var. calc. that would be at max. or min. for example. I am confused about what horizontal tangent refers to when I am given a parametric equation. E.g. At what value of t does x=t^2 -t and y=t^2 +t have a horizontal tangent? The answer is -1/2 which can be found by setting y'=0. I don't understand why this happens though. As in, why dy/dt rather than dy/dx or why does dx/dt not apply? In describing the curve, what is the relationship between the two (x and y given in parametric form) that I could just only look at dy/dt? My first instinct was to look for dy/dx which would look something like (2t+1)/(2t-1). Last edited: May 31, 2014 2. May 31, 2014 $\frac{dy}{dx}$ =$\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$. So for $\frac{dy}{dx}$ to be zero, the numerator i.e $\frac{dy}{dt}$ must be zero. And hence the answer. 3. May 31, 2014 ### mill Thanks. 4. May 31, 2014 ### SteamKing Staff Emeritus More accurately, if the tangent is horizontal, it is where the slope of the tangent line is zero.

https://etonstem.com/problem-of-the-week-answers

No 34: Discussion: For those of you that don’t know, $\log_b a$ represents the quantity $x$, where $b^x = a$ for any $a,b$ where $b > 0$. ‘log’ is an abbreviation of ‘logarithm’. So unless you have some experience working with logarithms, it might be useful to convert the end-statement out of logarithm-world and into exponential world. As regards the statement $7^{x+7} = 8^x$, it might be useful to bring everything that is raised to the power $x$ onto one side. Solution: This problem just involves some algebraic manipulation, made slightly more complicated by the presence of exponentials and logarithms. End-statement/goal: $x = \log_b 7^7 \Rightarrow b^x = 7^7$ $7^{x+7} = 8^x \Rightarrow 7^7 = \frac{8^x}{7^x} = (\frac{8}{7})^x$ Therefore, $b = \frac{8}{7}$. No 33: Discussion: To start off with, the problem might seem daunting because of the sheer number of possibilities – it seems like $n$ could be absolutely anything, so how are you supposed to maximize and minimize? The minimum is easier, because you can start at a triangle ($n=3$) and work your up, so start there. For the maximum value of $n$, note that the average internal angle will have to be quite large, so it might be worth identifying some large primes. Remember that the key to this problem is the fact that all angles must be odd primes: so they’re all integers, all odd, and all prime. Also, the sum of the internal angles of any convex $n$-gon is $180(n-2)$. Solution: Minimum: $n = 4$. $n=3$ doesn’t work (in fact no odd values of $n$ work for the same reason) because ‘odd times odd is odd’, so the angle sum cannot be divisible by 180, but it needs to be, as mentioned in the Discussion. A set of angles that work are: {97, 83, 97, 83}. Maximum: $n = 360$. The maximum possible internal angle is 179 degrees (as angles are odd primes and to meet the convex condition, all angles are less than 180 degrees). Therefore, the maximum possible sum is $179n = 180(n-2)$ $n = 360$ Therefore, the answer is 360 – 4 = 356. No 32: Discussion: The difficulty with this problem is in converting what you know about the octagons into some property of the square that you can then use to find its area. The two properties that come to mind are side-length and length of diagonals. The diagonals don’t seem very accessible through the current setup, but side-length does, as the octagons do touch all four sides. Once you’ve understood this, all that’s left is to calculate other lengths within the octagons, which can be done through trigonometry or triangle ratios (I prefer the latter for these problems because it’s faster). Solution: The total length of all red lines is equal to the sidelength of the square, so we compute this and then square it. We see that it is made up of two distinct distances, repeated multiple times and demarcated by the blue lines, which we have labelled $x, y$. Therefore, if we call the side length $s$, we have: $s = 5x + 4y$ Of course, $y = 2$ is given, so we have $s = 5x + 8$. We can calculate $x$ by observing the triangle outside one of the sides of an octagon. The exterior angle of any regular polygon is $\frac{360}{n}$, so in this case, it’s 45 degrees. Therefore, it’s a so-called ’45-45-90′ triangle, which has ratio of sides $1:1:\sqrt{2}$. Therefore, the height of this triangle is $\frac{2}{\sqrt{2}} = \sqrt{2}$, so $x = \sqrt{2}$. Therefore, $s = 5\sqrt{2} + 8$. Therefore, $area = s^2 = 50 + 80\sqrt{2} + 64 = 114 + 80\sqrt{2}$. Therefore, $m+n = 194$ No 31: Discussion: When I first started working on this problem, I needlessly overcomplicated it: I had pencils going alternately over and under each other, when this wasn’t needed, so try and avoid this. Just go for the simplest possible configuration. Also, you realize that if you operate in the 2D plane and find something that works for only 3 pencils, you can overlay a rotated version of that 2D solution on top of itself, to get something that works for 6 pencils! $s = 5x + 8$Discussion: When I first started working on this problem, I needlessly overcomplicated it: I had pencils going alternately over and under each other, when this wasn’t needed, so try and avoid this. Just go for the simplest possible configuration. Also, you realize that if you operate in the 2D plane and find something that works for only 3 pencils, you can overlay a rotated version of that 2D solution on top of itself, to get something that works for 6 pencils! Solution: (yes, the pencils here are “unsharpened”, but it doesn’t really matter) No 30: Discussion: Seeing the square roots, we know that the only way $\sqrt{120-\sqrt{x}}$ can be an integer is if $120-\sqrt{x}$ is a square. Because of the subtraction, we know that $0 \le 120-\sqrt{x} \le 120$ and we want the number of times when this is a square. Solution: As above, the question is in essence asking: for how many values of $x$ is $120-\sqrt{x}$ a square? The value of this expression is bounded by 0 and 120 (there are no negative square numbers), so we just count the number of square numbers in this region: there are 11 (smallest is $0^2$ and the largest is $10^2$). For each of these instances, there is precisely one value of $x$ that works, as shown below: $120-\sqrt{x} = k^2$ $\sqrt{x} = 120-k^2$ $x = (120-k^2)^2$ No 29: Warning: Part 2 of this problem is actually far harder than anticipated. It actually turns out to be around as hard as the actual BMO2 problem, and so far beyond the level of all other POTWs. Discussion: The original problem that inspired this one is BMO2 2021 P2. My advice for the first part of the question is just try different values of $n$. Clearly because we need one of each square, the minimum possible value of $n \times n$ is 4+9=13, so $n \ge 4$ (as 16 is the smallest square bigger than 13). There’s still a problem here though: placing the 3×3 leaves a region of width 1 – too small for the 2×2. Therefore, $n \ge 5$. For the rest of the way, you can just keep reasoning in a similar way until you get to $n=8$, which does work. For part 2, you might guess that the divisibility has something to do with 2 or 3, especially as $n=8$ is the answer for part 1. Note: if you have checked out the problem that inspired POTW 29, you’ll see that what you are asked to prove is that if $a \times a$ and $b \times b$ squares can tile an $n \times n$, in essence, $a \mid n$ or $b \mid n$. This gives you the answer to part 2 straight away (2 or 3 or both divide(s) $n$), and helps immensely with part 1. Therefore, you see that with 2×2 and 3×3 tiling squares, having discarded $n=4$, the next closest options are $n=6, 8$. You can eliminate $n=6$ by looking at what happens when you place a 3×3 and 2×2 adjacently (they leave a region of width 1 that can’t be filled), and then you try something for $n=8$ and it works. Solution: For part 1: $n=8$. This is minimal as shown above. An example tiling is: For part 2: Outline of a solution to the general case, using $a \times a$ and $b \times b$ squares: BMO2 official solutions. No 28: Discussion: When we plug in $x=1$, we get that $P(1) = 1+1-r^2-2020 = -2018-r^2$. So all we need to do is find $r^2$. Many of you will probably, as I did, jump to Vieta’s Formulas, which give a bunch of simultaneous equations which you can solve. However, this is not the nicest way to solve the problem – let’s look at something we haven’t used much yet: the roots. The fact that $r,s,t$ are roots means that if we plug them into the polynomial, we get 0. We want to find $r$, so why not plug in $x=r$? We get: $P(r) = r^3 + r^2 - r^3 - 2020 = 0$ meaning: $r^2 =2020$ and we’re done! Solution: Plug in $x=r$ to get: $P(r) = r^3 + r^2 - r^3 - 2020 = 0 \Rightarrow r^2 = 2020$ Then, $P(1) = 1+1 - r^2 - 2020 = -4038$ No 27: Discussion: To start with, it’s a good idea to play around with different questions and think about what the respective tribes would say. What you want is to ask a question such that the answer from both tribes is the same. One approach is to remember that “two negatives make a positive” – so if you can somehow make the liar tribe to lie twice, they will give you the right answer. Solution: “If I asked someone in your tribe which road is correct, what would they say?” If it is a truth-teller, they will give the correct path; if it is a liar, they know that someone in their tribe would say the other one, but they themselves need to lie so they say the correct path. No 26: Discussion: When I initially saw this problem, I ignored the fact that it was an ordering of objects as opposed to a plain old sequence of numbers. This was a mistake. If you do go down this path though, you find that nothing really works – you try differences of terms, quadratic stuff, big degree polynomials, and nothing really works. So then you think hang on, why is it 3-12 and not 1-10? This is a good question to ask. Also, you think, what are “mathematical objects”? What does it mean that they “can be labelled”? The first thing that comes to mind for me is regular polygons, and I guess the labelling refers to the number of sides? Clearly then the ordering can’t be dependent on properties like angle size or anything to do with mathematical properties of the shape itself, as these increase uniformly with the number of sides. So you zoom in on the word “ordering.” What types of “ordering” are there? Numerical and alphabetical are the main ones. Solution: The “mathematical objects” are regular polygons, labelled by the number of vertices/sides. The ordering is alphabetical based on their names! Decagon, dodecagon, hendecagon, heptagon, hexagon, nonagon, octagon, pentagon, square, triangle No: 25 Discussion: In a general situation, “find the hundreds digit” would encourage me to try using modular arithmetic, e.g. taking the expression modulo 1000. However, that’s pretty high-powered, so let’s first try to find some other approach. The first thing that jumps out to me is the common factor of $15!$. As there is nothing else that seems to be helpful, let’s factorize to get $(20! - 15!) = 15!(20*19*18*17*16 - 1)$. What would be nice at this stage is for some 0s to come up to simplify things…and they actually do (see below)! I just want to note that this may just seem like a lucky coincidence – that in this particular problem, it just happened to be the case that $(20! - 15!)$ is divisible by 1000. Instead though, it’s more likely the problem-setter purposefully made it this way, because in a math competition setting, where you don’t have a calculator, exploiting these ‘simplicities’ is crucial. Solution: $(20! - 15!) = 15!(20*19*18*17*16 - 1)$ We see that the whole number is divisible by 1000 because $15!$ is divisible by 1000, as it contains the product $4*5*10*15 = 3000$. Therefore, the hundreds digit is 0 because the last three digits of the number will be 0. No. 23: Discussion: Suppose that Emilia uses R liters of Rb(OH), C liters of Cs(OH), and F liters of Fr(OH), then we have: 10%·R+8%·C+5%·F =7%and 5%·F ≤2%. R+C+F R+C+F The equations simplify to 3R + C = 2F and 3F ≤ 2R + 2C, which gives 9R+3C ≤2R+2C ⇒5R≤C. Solution: Therefore the concentration of rubidium is maximized when 5R = C, so F = 4R No. 22: Discussion: We can rewrite this equation as $3^y + 4^y - 6^y = 1$ where y = x(x + 2). We can then re-arrange this to $3^y - 6^y = 1 - 4^y$. We can factorise this to give us a difference of two squares on the right and we can also take out a factor of $3^y$ on the left to give us $3^y(1 - 2^y) = (1 + 2^y)(1 - 2^y)$. We then bring everything to one side, $3^y(1 - 2^y) - (1 + 2^y)(1 - 2^y) = 0$. We can solve this to find values of y which we can then use to find the values of x. Solution: Since there are two values for y where the equation holds true (y = 0 and y = 1), there are therefore 4 real solutions to the initial equation. No. 21: Discussion: We can very easily work out the area of the triangle using Heron’s formula. We can then divide this result by 4 to get the length of the rectangle – giving us all the information we need to calculate the rectangle’s perimeter. Solution: 32 No. 19: Discussion: From the question, we can determine the following: N – 4 is a multiple of 7 and N – 2 is a multiple of 5. This gives us the following equations N – 4 = 7k and N – 2 = 5m. Since the maximum capacity is 40, we know k cannot be more than 5, therefore we just need to test all the values of k to get for possible values of N which we can then substitute in the second equation. Solution: The values k = 0, 1, 2, 3 and 5 give us values of N which when substituted into the second equation would give a non-integer value for m, therefore k must equal 4. Therefore N = 32. No. 18: Discussion: We want to prime factorisation of our number to include the lowest powers possible in order to find the smallest number which has 2015 factors. To do this, we can use the prime factorisation of 2015 (5 * 13 * 31). Solution: 116. We can use this prime factorisation to give us $2^{30} * 3^{12} * 5^4$. Therefore, the sum of the factors is 2(30) + 3(12) + 5(4) = 116. No. 17: Discussion: We can draw a triangle and then note down all the angles that are equal. We can use the fact that base angles of an isosceles are equal to find three angles that are equal. From here, solving the question is very straightforward. Solution: 84 degrees No. 11: Discussion: So we have an infinite tower of exponents. And there all of unknown value. Ouch. Well let’s think about what we would do if it were just $x^x=2$… Unfortunately, it’s not as easy as it seems because the base and exponent are variable. So is the problem impossible unless you know some highly advanced math? Thankfully not. The trick is to use/abuse the infinite tower of exponents. If $x^{x^{x^{...}}} = 2$, then we can raise $x$ to the power on each side, to get: $x^{x^{x^{...}}} = x^2$. Solution: We take $x$ to the power of each side of the equation to get: $x^{x^{x^{...}}} = x^2$ Because of the nature of infinity, the left side is still equal to 2, so we have $x^2 = 2$, meaning $x=\pm \sqrt{2}$. What does it mean to raise something to an irrational power? That’s a different story. No. 10: Discussion: The condition on how many times we can use the scale means that we can only weigh one bag. So somehow we need to collate information from 10 bags into one bag to determine which bag has the fakes (we can’t not collate as this means we weigh only one bag which will rarely reveal the fakes). We could try putting all of the coins from all of the bags together, but this turns out to be useless as this mass is always going to be fixed (at 101g). Then what about if we include different numbers of coins from each bag? This is the key idea. Solution: Order the bags in a line (it doesn’t matter how you do this). Take the first bag, which has 10 coins. Then add 9 coins from the next bag, 8 from the one after that, 7, 6, 5, 4, 3, 2, 1. Now, weigh this bag. If the $x$th bag has the fakes, then the mass of those coins will be $1.1x$, meaning that the weight of the bag will be $0.1x$ grams more than the weight if there were no fake coins. Therefore, you can work out which bag has the fakes. Note: the weight if there were no fake coins would be $10 + 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 55$ by a standard formula. Therefore, the bag’s weight will be of the form $55 + 0.1x$. No. 9: Discussion: part 1 of the problem is pretty straightforward. Part 2 is where it gets trickier. Lots of experimentation, especially with small cases as they’re easier to deal with, is important to get the message across. General advice for “is it possible” problems is to assume it can be done and try lots and lots of times. Only if you have been unsuccessful at providing what the problem wants should you strive to prove impossibility. Solution: For a normal chessboard, yes we can. Just place 4 dominoes in line in each row, end to end, and repeat for all 8 rows. For the mutilated chessboard, it’s not possible. Here’s why. Every time we place a domino, we cover two adjacent squares that are always differently colored. Therefore, every domino-covered board must have an equal number of each of the two colors (assuming we color as for the chessboard). As a result, because the mutilated chessboard has more black squares than red, it cannot be covered. No. 8: Discussion: If you think about it, you try some places, and quickly realize that nothing ‘normal’ works, and that there must be some specific location. So because of the hinting in the problem, you try the South Pole, and see what’s happening there, and this turns out to be the key. We see that in order to end up back where we started, we need to be 1 mile away from the circle around the South Pole which has circumference 1 mile, so that we end up exactly where we started, and can come straight back to our previous destination. The radius of this circle is just $1/2\pi$, using the formula for the circumference of a circle. Therefore, we think that the actual answer is $1 + 1/2\pi$ miles from the South Pole. Proof: No new ideas here, just think through the discussion, and see that if we start at any point $1 + 1/2\pi$ miles from the South Pole, when we travel south 1 mile, we hit the circle with radius $1/2\pi$ miles. This means that the circumference is 1 mile, so when we travel east 1 mile, we end up exactly where we were, and so travelling north 1 mile brings us back to our starting point. No. 7: Discussion: This problem feels quite intuitive, but it’s a bit more difficult to solidify your thoughts into a rigorous proof. Think about the monk going up and down, and eventually you have the idea to picture a second monk going up at the same pace as the monk did on the first day, along the same route. As the two monks are passing along the same path, they must meet each other at some point, and at this point, they will be at the same place at the same time. Proof: Draw a distance-time graph of the journey of the monk going up and the monk going down. The lines must cross somewhere, as they are not parallel and they must end up at each other’s starting points. Where they cross is where the monk is at the same time and place on both days. No. 6: Flip two of the three switches to on, and leave them on for a long time (let’s say 30 minutes). Then, turn one of them off, run upstairs, and observe/feel the bulbs. Why this works: If, when you go upstairs, a light is on, you know which switch controls it because only one switch is still on. If, when you go upstairs, none of the lights are on but one of the bulbs is hot/glowing orange a bit, you know that this lamp was on but was just turned off, so the switch is the one you turned off right before going upstairs. If, when you go upstairs, none of the lights are on and none of the bulbs are hot/glowing orange a bit, you know that the bulb was never turned on, meaning that the switch which controls it is the one which was never turned on. No. 5: No. 4: No. 2: π : √2+1 Solution: Assume the diameter of the large circle is 2, this must mean the diagonal of the square is 2. Using Pythagoras’ Theorem, we know that the side length of the square is therefore √2. The height of the top segment must be half of the result we get from subtracting the height of the square from the diameter of the circle and dividing by 2 (as there is another equal segment at the bottom). Therefore the height of the top segment is (2 – √2)/2. From this we can determine the height of the triangle by adding the height of the square to the height of the top segment which is (2 + √2)/2. As the base of the triangle lies on a side length of the square, we know the base must be √2. Therefore, we use the formula bh/2 to determine that the area of the triangle is (√2 +1)/2. Now, we know the radius of the smaller circle must be half the length of the square which is √2/2. We then use the formula πr2 to determine that the area of the smaller circle is π/2. Therefore the ratio of the areas is π/2 : (√2 +1)/2 which we simplify to π : √2+1.

http://math.stackexchange.com/questions/114419/calculating-the-standard-error-of-a-maximum-likelihood-estimator

Calculating the standard error of a maximum likelihood estimator Suppose that $X$ is a discrete random variable with • $P(X= 0) = 2θ/3$ • $P(X=1) = 1θ/3$ • $P(X=2) = 2(1-θ)/3$ • $P(X=3) = (1-θ)/3$ where $0\le θ\le 1$ is a parameter. The following $n=10$ independent observations were taken from such a distribution: $( 3,0,2,1,3,2,1,0,2,1)=(x_1,\ldots,x_n)$ 1)What is the maximum likehood estimate of $θ$? 2) What is the approximate standard error of the maximumlikehood estimate. (Question is from Statistics and Data Analysis by Rice, 3rd edition) Well, we know the mle is given by $L(p)=f(x_1,\ldots,x_n|p)= \prod_{i=1}^n f(x_i|p)$ which gives the mle for $p$. (Taking the log and setting the derivative to $0$.) (I have just called $\theta$ as $p$.) However, how should you calculate the approx. standard error in $p$? I read something about the inverse of an information matrix, but the book in question doesn't mention this so I guess there is a simpler way of doing this. Should I just calculate the standard deviation of $X$ and plug in $p$ in it? - You are supposed to answer completely question 1 before tackling question 2, hence you must first reach an explicit formula for $\widehat\theta$ as a function of the sample $(x_1,x_2,\ldots,x_n)$. –  Did Feb 28 '12 at 16:03 Not interested anymore? –  Did Mar 2 '12 at 18:55

https://math.stackexchange.com/questions/1779143/integrate-sin2nx-csc2xdx

# Integrate $\sin^2(nx)\csc^2(x)dx$ I need help to integrate $$\int{\sin^2(nx)\csc^2(x)dx}$$ Integral of each part is easy $$\int{\sin^2(nx)}dx=\frac{x}{2}+\frac{\sin(2nx)}{4n}+C$$ $$\int{\csc^2(x)}dx=-\cot(x)+C$$ I then tried to use the integral by parts but failed to make it work. Also tried Walfram calculator but it returned a very complicated expression that's not what I wanted. Can you help to resolve this indefinite integration? Definite integral $$\int_{2\pi}^{N\pi}{\sin^2(nx)\csc^2(x)}dx$$ where $N$ is big natural number, is also helpful. This answer concerns the indefinite integral $\displaystyle\int\sin^2(nx)\csc^2 x\ dx$. Hint 1:- Observe that, $$\displaystyle\int\cos^2(nx)\csc^2 x\ dx+\displaystyle\int\sin^2(nx)\csc^2 x\ dx=\displaystyle\int\csc^2 x\ dx$$ $$\displaystyle\int\cos^2(nx)\csc^2 x\ dx-\displaystyle\int\sin^2(nx)\csc^2 x\ dx=\displaystyle\int\cos(2nx)\csc^2 x\ dx$$ For, $\displaystyle\int\cos(2nx)\csc^2 x\ dx$ observe that, $$\displaystyle\int\cos(2nx)\csc^2 x\ dx=\Im \left(\int e^{2inx}\csc^2 x\ dx\right)$$ Hint 2:- For evaluating $\displaystyle\int\cos(2nx)\csc^2 x\ dx$ you may also use the following formula, $$\boxed{\cos (2nx)=\displaystyle\sum_{k=0}^{2n}\binom{2n}{k}\cos^kx\sin^{2n-k} x\cos\left(\dfrac{1}{2}(2n-k)\right)}$$

http://physics.stackexchange.com/questions/20768/how-to-find-maximum-velocity

# How to find maximum velocity I stack a question about projectile question. The question was A projectile is being launched from ground level with no air resistance. You want to avoid having it enter a temperature inversion layer in the atmosphere a height $h$ above the ground (a) what us the maximum launch speed you could give this projectile if you shot it straight up ? Express your answer in terms of $h$ and $g$.(b)Suppose the launcher available shoots projectiles at twice the maximum launch speed you found in part (a). At what maximum angle above the horizontal should you launch the projectile ? I could solve the (a) part. How did was following (a) using the following formula to drive $V$ -

https://www.physicsforums.com/threads/deriving-the-constant-e-using-a-sequence-limit.578707/

# Deriving the constant e using a sequence limit 1. Feb 17, 2012 ### Fibo's Rabbit 1. The problem statement, all variables and given/known data Why does lim( (1+(1/n))^n ) = e? 2. Relevant equations If a_n convergent to a, and b_n converges to b, then (a_n * b_n) converges to (a * b) 3. The attempt at a solution The lim(1 + (1/n)) = 1. If you multiply (1 + (1/n)) by itself n-times, you get the equation (1 + (1/n))^n, so according to the statement under Relevant equations above, shouldn't the answer be 1^n, or just 1? Obviously it is e, b/c when you plug (1 + (1/n))^n into your calculator for larger values of n, you get closer and closer to e. 2. Feb 17, 2012 ### tiny-tim Hi Fibo's Rabbit! (try using the X2 and X2 buttons just above the Reply box ) ah, but that only applies for the product of a fixed number of series … you're trying to apply it to an infinitely increasing number of series 3. Feb 17, 2012 ### Fibo's Rabbit Ah, and it all makes a lot more sense now.

http://math.stackexchange.com/questions/169578/mathcalk-mathcalh-is-separable

# $\mathcal{K}(\mathcal{H})$ is separable? I am reading Arveson's Notes On Extensions of $C^*$-algebras, in proving the Corollary to Thm2, he seems to assume that $\mathcal{K}(\mathcal{H})$, the space of compact operators on a separable Hilbert space, is separable as a topological space, i.e., it contains a countable dense set. This is a little bit surprising to me. If we just take $\mathcal{H}=\ell^2$ and $F$ defined by:$$F(\sum\alpha_n e_n)=(\sum\alpha_n f_n)e_1,$$ where $\{e_n\}$ is the canonical basis for $\ell^2$. That is, $F$ maps everything to the first coordinate, $F e_n=f_n e_1$. Since $(f_n)$ can be be arbitrary sequences in $\ell^{\infty}$, which is a non-separable space. My argument seems to imply that even the space of rank-one operators is not separable. Where did I make a mistake? Thanks! - How can $f_n$ be an arbitrary sequence from $\ell^\infty$? Certainly $f_n = 1$ for all $n$ is not allowed. It must be a sequence from $\ell^2$... On the right hand side you write a functional on $\ell^2$ times $e_1$, after all. As you should know, the finite rank operators are dense in $\mathcal{K(H)}$ and it's not too hard to find a dense sequence there. – t.b. Jul 11 '12 at 18:36 The space of rank-one operators is separable, being a continuous image of $\mathcal H\oplus \mathcal H$ under the map $(u,v)\mapsto T_{u,v}$ where $T_{u,v}x=\langle x,u\rangle v$. Hence the space of finite-rank operators is separable (they are finite sums of rank-1 operators), and its closure in $B(\mathcal H)$ is $\mathcal K(\mathcal H)$.

https://practice.geeksforgeeks.org/problems/sum-of-fifth-powers-of-the-first-n-natural-numbers3415/1

Sum of fifth powers of the first n natural numbers Basic Accuracy: 23.08% Submissions: 26 Points: 1 Given a number N.Find the sum of fifth powers of natural numbers till N i.e. 15+25+35+..+N5. Example 1: Input: N=2 Output: 33 Explanation: The sum is calculated as 15+25=1+32=33. Example 2: Input: N=3 Output: 276 Explanation: The sum is calculated as 15+25+35 =1+32+243=276. You don't need to read input or print anything.Your task is to complete the function sumOfFifthPowers() which takes an integer N as input parameter and returns the sum of the fifth powers of Natural numbers till N. Expected Time Complexity:O(1) Expected Auxillary Space:O(1) Constraints: 1<=N<=1000

https://www.muchlearning.org/?page=62&ccourseid=1437&sectionid=2703

# Step By Step Calculus » 4.4 - Equalities and Inequalities Synopsis Given two functions f(x)f(x) and g(x)g(x), we are interested in finding the exact solution set for an equation of the form f(x)=g(x)f(x)=g(x) and for inequalities of the form f(x)<g(x), f(x)\le g(x), f(x)>g(x), f(x)\ge g(x)f(x)<g(x), f(x)\le g(x), f(x)>g(x), f(x)\ge g(x). Solving Equations: There are many methods for solving equations. The following are the most basic ones. • Isolating the Variable: To isolate a variable in an equation, we obey the following three-headed golden rule: • Seek simplicity above all • Do to the right side whatever you do to the left side • Track steps that may cause loss or gain of solutions. Make necessary adjustments or checks. • By Factoring: One can use the following procedure to solve many equations. Step 1: Move all terms to the left side, so that the right side is 00. Step 2: Factor the left side as much as possible. Step 3: Split the resulting equation into several smaller ones by setting each factor equal to 00. Step 4: Solve all the easier equations obtained in this way. Step 5: Check that all solutions are acceptable. • Involving Absolute Values: To solve equations involving absolute values, one can either consider all possible cases separately or apply squaring to the both sides. We can also use graphs, for example if we graph the left and right sides of an equation, then the solutions are the points of intersection of the graphs. • Using The Quadratic Formula: For a quadratic equation of the form ax^2+bx+c=0ax^2+bx+c=0, the solutions are \displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}. The solutions are real when the discriminant b^2-4ac\ge 0b^2-4ac\ge 0. Solving Inequalities: The algebraic method of solving inequalities uses the order properties of the real numbers: If aa and bb are real numbers, • a<ba<b implies \displaystyle a+k<b+k\displaystyle a+k<b+k and a-k<b-ka-k<b-k for any k \in \mathbb{R}k \in \mathbb{R}. • a<ba<b and \displaystyle k>0\displaystyle k>0 implies ak < bk \; \; \forall k\in \mathbb{R}ak < bk \; \; \forall k\in \mathbb{R}. • a<ba<b and \displaystyle k<0\displaystyle k<0 implies ak>bk \; \; \forall k \in \mathbb{R}ak>bk \; \; \forall k \in \mathbb{R}. • a>0a>0 implies that \displaystyle \dfrac{1}{a}>0\displaystyle \dfrac{1}{a}>0 • 0<a<b0<a<b implies that \displaystyle \dfrac{1}{a}>\dfrac{1}{b}\displaystyle \dfrac{1}{a}>\dfrac{1}{b} • 0 \leq a<b0 \leq a<b if and only if \displaystyle 0\leq a^{n}<b^{n}\displaystyle 0\leq a^{n}<b^{n} for any n\in\mathbb{N}n\in\mathbb{N}. As a general strategy for solving inequalities of the form f(x)>g(x)f(x)>g(x) algebraically, we use the following COPS procedure. Step 1: (Cut points) Find the values of xx for which either f(x)f(x) or g(x)g(x) are undefined or discontinuous and those for which f(x)=g(x)f(x)=g(x). We call these the cut points of inequality. Step 2: (Order) Place these cut points on a number line in increasing order. Step 3: (Pick and check) Pick a test value for xx in each of the intervals before, after or between the cut points and check whether for that value f(x)>g(x)f(x)>g(x) or not. The same conclusion will be true for any other value of xx in that same interval. Step 4: (Solution) Choose the intervals for which the given inequality is satisfied. If the inequality is of the form f(x)<g(x)f(x)<g(x) we can do exactly the same thing, except that in the third step the criterion for acceptability is the opposite. Sometimes the inequality is of the form f(x)\ge g(x)f(x)\ge g(x) or f(x)\le g(x)f(x)\le g(x). In that case each of the cut points must be treated separately to see if it satisfies the inequality or not. To solve inequalities, a graphical method may also be used, by plotting both functions carefully and visually determining where the equality or inequality holds. A disadvantage of this method is that it can be imprecise. Triangle Inequality: The order properties of the real numbers can be used to prove the triangle inequality, which states that: \vert a+b\vert \leq \vert a\vert + \vert b \vert\vert a+b\vert \leq \vert a\vert + \vert b \vert for a,b\in\mathbb{R}a,b\in\mathbb{R}.

https://plainmath.net/3846/solution-second-homogeneous-linear-differential-equation-suppose-general

Let y_1 and y_2 be solution of a second order homogeneous linear differential equation y''+p(x)y'+q(x)=0, in R. Suppose that y_1(x)+y_2(x)=e^{-x}, W[y_1(x),y_2(x)]=e^x, where W[y_1,y_2] is the Wroian of y_1 and y_2. Find p(x), q(x) and the general form of y_1 and y_2. Second order linear equations Let $$y_1$$ and $$y_2$$ be solution of a second order homogeneous linear differential equation $$y''+p(x)y'+q(x)=0$$, in R. Suppose that $$y_1(x)+y_2(x)=e^{-x}$$, $$W[y_1(x),y_2(x)]=e^x$$, where $$W[y_1,y_2]$$ is the Wro ian of $$y_1$$ and $$y_2$$. Find p(x), q(x) and the general form of $$y_1$$ and $$y_2$$. 2020-11-10 $$y''+p(x)y'+q(x)=0$$ (*) $$y_1(x)+y_2(x)=e^{-x}$$ $$W(y_1,y_2)=e^x$$ Since $$y_1$$, and $$y_2$$ are solution to (*) So their linear combinations are also a solution to (*) Hence $$y_1+y_2=e^{-x}$$ is also a solution to (*) Now, $$(y_1+y_2)'=-e^{-x}$$ $$(y_1+y_2)""=e^{-x}$$ So, from (*) we have $$(y_1+y_2)''+p(x)(y_1+y_2)'+q(x)(y_1+y_2)=0$$ $$\Rightarrow e^{-x}-e^{-x}p(x)+e^{-x}q(x)=0$$ or $$e^{-x}[1-p(x)+q(x)]=0$$ As $$e^{-x}\neq0\ \forall\ x$$ Hence, $$1-p(x)+q(x)=0$$ (I) Now, $$W[y_1,y_2]=[(y_1,y_2),(y'_1,y''_2)]=y_1y'_2-y'_1y_2=e^x$$ (II) As $$y_1+y_2=e^{-x}$$ $$\Rightarrow y_2=e^{-x}-y_1$$ and $$y'_1+y'_2=-e^{-x}$$ so, $$y'_2=e^{-x}-y'_1$$ Putting the value of $$y_2$$ and $$y'_2$$ in (II) we got, $$y_1(-e^{-x}-y'_1)-y'_1(e^{-x}-y_1)=e^x$$ $$\Rightarrow-e^{-x}y_1-y_1y'_1-e^{-x}y'_1+y_1y'_1=e^x$$ $$-e^{-x}[y_1+y'_1]=e^x$$ $$y_1+y'_1=-e^{2x}$$ $$\Rightarrow y'_1=-e^{2x}-y_1$$ $$\Rightarrow y''_1=-2e^{2x}-y_1$$ $$\Rightarrow y''_1=-2e^{2x}-y'_1=-2e^{2x}-(-e^{2x}-y_1)$$ $$=-e^{2x}+y_1$$ $$y''_1=-e^{2x}+y_1$$ Putting the value of $$y'_1$$ and $$y''_1$$ in (*) we get, $$(-e^{2x}+y_1)+p(x)(-e^{-2x}-y_1)+q(x)y_1=0$$ $$\Rightarrow-e^{2x}+y_1=p(x)y_1+q(x)y_1-e^{2x}p(x)=0$$ $$\Rightarrow-e^{2x}[1+p(x)]+y_1[1-p(x)+q(x)]=0$$ From (I) we have $$1-p(x)+q(x)=0$$ So, $$-e^{2x}[1+p(x)]+y_1\cdot0=0$$ $$\Rightarrow e^{2x}[1+p(x)]=0$$ $$\Rightarrow1+p(x)=0$$ $$p(x)=-1$$ From (I) we have $$1-(-1)+q(x)=0$$ $$\Rightarrow q(x)=-2$$ So, $$p(x)=-1$$ and $$q(x)=-2$$ Hence the differential equation become, $$y''-y'-2y=0$$ The characteristic equation is $$v^2-v-2=0$$ or $$v^2-2v+v-2=0$$ $$v(v-2)+1(v-2)=0$$ so, $$(v-2)(v+1)=0$$ $$\Rightarrow v=2$$ or $$v=-1$$ Hence $$y_1(x)=e^{2x}$$ and $$y_2=e^{-x}$$ Here $$y(x)=c_1e^{2x}+c_2e^{-x}$$ Answer: $$p(x)=-1,\ q(x)=-2,\ y(x)=c_1e^{2x}+c_2e^{-x}$$

https://mathemerize.com/d-and-e-are-points-on-the-sides-ca-and-cb-respectively-of-a-triangle-abc-right-angled-at-c-prove-that-ae2-bc2-ab2-de2/

# D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C. Prove that $${AE}^2$$ + $${BC}^2$$ = $${AB}^2$$ + $${DE}^2$$. ## Solution : From triangle ACE, $${AE}^2$$ = $${EC}^2$$ + $${AC}^2$$       ……….(1)          (By Pythagoras Theorem) From triangle DCB, $${BD}^2$$ = $${BC}^2$$ + $${DC}^2$$        ………(2) Adding (1) and (2), we get $${AE}^2$$ + $${BD}^2$$ = $${EC}^2$$ + $${AC}^2$$ + $${BC}^2$$ + $${DC}^2$$ By Pythagoras Theorem in right triangle ECD and ABC, $${DE}^2$$ = $${EC}^2$$ + $${DC}^2$$  and  $${AB}^2$$ = $${BC}^2$$ + $${AC}^2$$ Hence,  $${AE}^2$$ + $${BC}^2$$ = $${AB}^2$$ + $${DE}^2$$

https://tutorial.math.lamar.edu/ProblemsNS/CalcIII/DIPolarCoords.aspx

Paul's Online Notes Home / Calculus III / Multiple Integrals / Double Integrals in Polar Coordinates Show Mobile Notice Show All Notes Hide All Notes Mobile Notice You appear to be on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen width. Assignment Problems Notice Please do not email me to get solutions and/or answers to these problems. I will not give them out under any circumstances nor will I respond to any requests to do so. The intent of these problems is for instructors to use them for assignments and having solutions/answers easily available defeats that purpose. ### Section 4-4 : Double Integrals in Polar Coordinates 1. Evaluate $$\displaystyle \iint\limits_{D}{{3x{y^2} - 2\,dA}}$$ where $$D$$ is the unit circle centered at the origin. 2. Evaluate $$\displaystyle \iint\limits_{D}{{4x - 2y\,dA}}$$ where $$D$$ is the top half of region between $${x^2} + {y^2} = 4$$ and $${x^2} + {y^2} = 25$$. 3. Evaluate $$\displaystyle \iint\limits_{D}{{6xy + 4{x^2}\,dA}}$$ where $$D$$ is the portion of $${x^2} + {y^2} = 9$$ in the 2nd quadrant. 4. Evaluate $$\displaystyle \iint\limits_{D}{{\sin \left( {3{x^2} + 3{y^2}} \right)\,dA}}$$ where $$D$$ is the region between $${x^2} + {y^2} = 1$$ and $${x^2} + {y^2} = 7$$. 5. Evaluate $$\displaystyle \iint\limits_{D}{{{{\bf{e}}^{1 - {x^{\,2}} - {y^{\,2}}}}\,dA}}$$ where $$D$$ is the region in the 4th quadrant between $${x^2} + {y^2} = 16$$ and $${x^2} + {y^2} = 36$$. 6. Use a double integral to determine the area of the region that is inside $$r = 6 - 4\cos \theta$$. 7. Use a double integral to determine the area of the region that is inside $$r = 4$$ and outside $$r = 8 + 6\sin \theta$$. 8. Evaluate the following integral by first converting to an integral in polar coordinates. $\int_{{ - 2}}^{0}{{\int_{{ - \sqrt {4 - {y^{\,2}}} }}^{{\sqrt {4 - {y^{\,2}}} }}{{\,\,\,{x^2}\,dx}}\,dy}}$ 9. Evaluate the following integral by first converting to an integral in polar coordinates. $\int_{{ - 1}}^{1}{{\int_{0}^{{\sqrt {1 - {x^{\,2}}} }}{{\,\,\,\sqrt {{x^2} + {y^2}} \,dy}}\,dx}}$ 10. Use a double integral to determine the volume of the solid that is below $$z = 9 - 4{x^2} - 4{y^2}$$ and above the $$xy$$-plane. 11. Use a double integral to determine the volume of the solid that is bounded by $$z = 12 - 3{x^2} - 3{y^2}$$ and $$z = {x^2} + {y^2} - 8$$. 12. Use a double integral to determine the volume of the solid that is inside both the cylinder $${x^2} + {y^2} = 9$$ and the sphere $${x^2} + {y^2} + {z^2} = 16$$. 13. Use a double integral to derive the area of a circle of radius $$a$$. 14. Use a double integral to derive the area of the region between circles of radius a and b with $$\alpha \le \theta \le \beta$$. See the image below for a sketch of the region.

https://www.analyzemath.com/expfunction/Euler-constant-e.html

# Euler Constant e ## Euler Constant e In compounding of interest it was shown that if, for example, an amount of money P (principal) is invested at an annual percentage rate r, the total amount of money A after t years is given by $A = P(1 + r)^t$ It was also shown that if the interest is compounded n times during each year, the amount of money after t years is given by $A = P(1 + r/n)^{nt}$ Let $N = n / r$ , then $r / n = 1 / N$ and $n = r N$ , hence the formula for A becomes $A = P(1 + 1 / N)^{N r t}$ Which can be written as $A = P ( (1 + 1 / N)^N )^{r t}$ The question that one may ask is that what if we increase n indefinitely? As the number of compounding n increases, $N$ also increases, the term $(1 + 1 / N)^N$ approaches a constant value called $e$ (after the swiss mathematician Leonhard Euler) and is approximately equal 2.718282.... The table of values below shows the values of $(1 + 1 / N)^N$ as $N$ increases. $N$ $(1 + 1 / N)^N$ 1 2 2 2.25 3 2.37037 10 2.59374 20 2.65329 40 2.68506 100 2.70481 200 2.71149 400 2.71488 Below is shown the graph of $y = (1 + \dfrac{1}{N})^N$ as a function of $N$ and we can see that as $N$ increases, $y = (1 + \dfrac{1}{N})^N$ approaches a constant $e \approx 2.71828182846$ More rigorously, e is defined as the limit of $(1 + 1/N)^N$ as $N$ approaches infinity which is written as $e = \lim_{N \to \infty} (1 + \dfrac{1}{N})^N$ Let $m = \dfrac{1}{N}$ and rewrite another definition of the Euler constant $e$ as follows $e = \lim_{m \to 0} (1 + m)^{\dfrac{1}{m}}$ The continuous compounding is defined for N very large and in this case the amount of money after t years is given by $A = P e^{r t}$ ## Exponential and Logarithmic Functions to the Base e The Euler constant $e$ defined above plays an important role in applied mathematics. Many mathematical models used in physics, engineering, chemistry, economics,..., are described by exponential functions to the base $e$ defined by $f(x) = e^x$ and it inverse, the logarithm to the base $e$, defined by $g(x) = ln(x)$ Fundtion f is called the natural exponential function and the function g is called the natural logarithmic function. Both are graphed below.

https://astarmathsandphysics.com/university-maths-notes/matrices-and-linear-algebra/4498-atomic-triangular-matrices.html?tmpl=component&print=1&page=

## Atomic Triangular Matrices If the entries on the diagonal of an upper (or lower) triangular matrix are all 1, and all other entries are zero except for a single row or column above (or below) the main diagonal the matrix is said to be atomic upper (or lower) triangular. Example: The matrix $\left| \begin{array}{ccc} 1 & 2 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right|$ is atomic upper triangular and the matrix $\left| \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 3 & 2 & 1 \end{array} \right|$ is atomic lower triangular. The inverse of an atomic upper (or lower) triangular matrix takes the same form and can be instantly written down The inverses of the two matrices above are $\left| \begin{array}{ccc} 1 & -2 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right|$ and $\left| \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ -3 & -2 & 1 \end{array} \right|$ is atomic lower triangular.

https://www.jinhang.work/tech/solving-odes/

# Solving Ordinary Differential Equations Summation of solutions to ODEs. An ordinary differential equation (ODE) is an equation that contains one or several derivatives of an unknown function. ADVANCED ENGINEERING MATHEMATICS - ERWIN KREYSZIG • The ORDER of an ODE is defined as the degree of the highest derivative in the equation. • An ODE is LINEAR if the dependent variable and its derivatives do not appear in products with themselves, raised to powers or in nonlinear functions. e.g. $\dot{y}+5y=cos(x)\rightarrow \text{$1^{st}$order, linear}$ but $\dot{y}+5y=cos(y)\rightarrow \text{$1^{st}$order, nonlinear}$. • Put the dependent variable and its derivatives on the left-hand side of the equation. • RHS=0: Homogeneous ODEs. ## Principles to solve ODEs • Direct inspection • Separation of variable • Integrating factors ### Direct inspection • Exponential, e.g. $$\frac{dx}{dt}=-4x\Rightarrow x(t)=Ce^{-4t}$$ • Sinusoidal, e.g. $$\ddot{x}+\lambda^2x=0\Rightarrow x(t)=sin(\lambda t)\text{ or }cos(\lambda t)$$ ### Separable ODEs Given an ODE: $$\frac{dx}{dt}=f(x,t)$$ Rearrange the equation: $$g(x)\frac{dx}{dt}=h(t)$$ And then do integration: $$\int g(x)dx=\int h(t)dt+C$$ #### Extended method: reduction to separable form Make form below, and $\dot{x}\gets u$ $$\dot{x}=f(\frac{x}{t})$$ ### Exact ODEs #### Exact case The ODE's differential form $M(x,t)dx+N(x,t)dt$ is termed EXACT. Say $$M(x,t)\frac{dx}{dt}+N(x,t)=0\\ \frac{\partial u(x,t)}{\partial x}\frac{dx}{dt}+\frac{\partial u(x,t)}{\partial t}=0\\ \frac{\partial u(x,t)}{\partial x}dx+\frac{\partial u(x,t)}{\partial t}dt=0\\ \frac{du}{dt}=0\\ u=C$$ (☞ﾟヮﾟ)☞ $u(x,t)$ or the exactness exists if: $$\frac{\partial M}{\partial t}=\frac{\partial N}{\partial x}$$ #### Non-exact case ##### IF available Say a non-exact equation: $$P(x,y)dx+Q(x,y)dy=0$$ Use an integrating factor to make the equation exact: $$FPdx+FQdy=0$$ Find the integrating factor: EXACTNESS $$\frac{\partial }{\partial y}(FP)=\frac{\partial }{\partial x}(FQ)\\ F_yP+FP_y=F_xQ+FQ_x\\ \text{Assume the simple cases where F depends on only x (or y)}\\ F=F(x)(\text{ or }F(y))\\ FP_y=F'Q+FQ_x\\ \frac{1}{F}\frac{dF}{dx}=\underset{R}{\frac{1}{Q}(P_y-Q_x)}\\ \text{Assume the simple case where R depends only on x}\\ F(x)=exp\int R(x)dx$$ ##### IF unavailable Choose other methods. ## Solutions to specific forms of ODEs ### 1st-order ODEs #### Linear cases $$\dot{x}+p(t)x=r(t)$$ • Homogeneous case: $\dot{x}+p(t)=0$ $$x(t)=A\cdot exp\left(-\int p(t)dt\right)$$ • Non-homogeneous case: $\dot{x}+p(t)x=r(t)$. Apply integrating factor: $$F=exp\left(\int p(t)dt \right)$$ $$x(t)=e^{-\int p(t)dt}\left[\int e^{\int p(t)dt}r(t)dt+C \right]$$ Proof: #### Non-linear cases: Bernoulli equation $$\dot{y}+p(x)y=g(x)y^a$$ Let $u(x)=y^{1-a}$, then $$\dot{u}+(1-a)pu=(1-a)g$$ ### 2nd-order linear ODEs $$\ddot{y}+p(x)\dot{y}+q(x)y=r(x)$$ #### Homogeneous cases ##### Linear cases with coefficients as functions Homogeneous case: $\ddot{y}+p(x)\dot{y}+q(x)y=0$. Superposition principle: any linear combination of two solutions on an open interval is again a solution. Reduction of order. (when one solution $y_1$ is known). Set $y_2=uy_1$. $$y_2=y_1u=y_1\int \frac{1}{y_1^2}e^{-\int pdx}dx$$ ##### Linear cases with constant coefficients Homogeneous ODE: $$\ddot{y}+a\dot{y}+by=0$$ Characteristic equation (derived from putting $y=e^{\lambda x}$ into the equation): $$\lambda^2+a\lambda+b=0\Rightarrow\lambda_1,\lambda_2$$ Solutions: $y_1=e^{\lambda_1 x}$ and $y_2=e^{\lambda_2 x}$ . ###### Two distinct real roots Solution: $$y=c_1e^{\lambda_1x}+c_2e^{\lambda_2x}$$ ###### Real double root $$\lambda=\lambda_1=\lambda_2=-\frac{a}{2}$$ Known $y_1=e^{-(a/2)x}$. Apply Reduction of Order: $y_2=uy_1$, then, $$u=c_1x+c_2\Rightarrow u=x\text{ , simple case}$$ Solution: $$y=(c_1+c_2x)e^{-ax/2}$$ ###### Complex roots \begin{aligned} \lambda_1&=-\frac{1}{2}a+i\omega\ \lambda_2&=-\frac{1}{2}a-i\omega \end{aligned} \begin{aligned} y_1&=e^{-\frac{1}{2}a+i\omega}=e^{-ax/2}cos\omega x\ y_2&=e^{-\frac{1}{2}a-i\omega}=e^{-ax/2}sin\omega x \end{aligned} Solution: $$y=e^{-ax/2}(Acos\omega x+Bsin\omega x)$$ Properties: Summary in table: #### Non-homogeneous cases $r(x)\neq 0$. Superposition principle may not exist. $$\ddot{y}+p(x)\dot{y}+q(x)y=r(x)$$ GENERAL SOLUTION: homogeneous solution + particular solution $$y(x)=y_h(x)+y_p(x)$$ Find the PARTICULAR SOLUTION: Method of undetermined coefficients. • Constant coefficients $$\ddot{y}+a\dot{y}+by=r(x)$$ Choose a form for $y_p$ similar to $r(x)$, but with unknown coefficients to be determined by substituting that $y_p$ and tis derivatives into the ODEs. More rules: first commit

http://people.revoledu.com/kardi/tutorial/LinearAlgebra/ElementaryRowOperation.html

<Next | Previous | Index> In linear algebra, there are 3 elementary row operations. The same operations can also be used for column (simply by changing the word “row” into “column”). The elementary row operations can be applied to a rectangular matrix size m by n. 1. Interchanging two rows of the matrix. Notation means to interchange row and row. 2. Multiplying a row of the matrix by a scalar. Notation means to multiply row by, . 3. Add a scalar multiple of a row to another row. Notation means to add times the row to row,. Applying the three elementary row operations to a matrix will produce row equivalent matrix. When we view a matrix as augmented matrix of a linear system, the three elementary row operations are equivalent to interchanging two equations, multiplying an equation by a non-zero constant and adding a scalar multiple of one equation to another equation. Two linear systems are equivalent if they produce the same set of solutions. Since a matrix can be seen as a linear system, applying the 3 elementary row operations does not change the solutions of that matrix. The 3 elementary row operations can be put into 3 elementary matrices. Elementary matrix is a matrix formed by performing a single elementary row operation on an identity matrix. Multiplying the elementary matrix to a matrix will produce the row equivalent matrix based on the corresponding elementary row operation. Elementary Matrix Type 1: Interchanging two rows of the matrix ,, Elementary Matrix Type 2: Multiplying a row of the matrix by a scalar , , Elementary Matrix Type 3: Add a scalar multiple of a row to another row , , , , , Example: Suppose we have matrix Applying elementary matrix type 3 of produces row equivalent matrix Applying elementary matrix type 2 of to the last result produces row equivalent matrix Applying elementary matrix type 1 of to the last result produces row equivalent matrix <Next | Previous | Index>

http://everything.explained.today/Function_composition/

Function composition explained In mathematics, function composition is an operation that takes two functions and and produces a function such that . In this operation, the function is applied to the result of applying the function to . That is, the functions and are composed to yield a function that maps in to in . Intuitively, if is a function of, and is a function of, then is a function of . The resulting composite function is denoted, defined by for all in .[1] The notation is read as " circle ", " round ", " about ", " composed with ", " after ", " following ", " of ", or " on ". Intuitively, composing two functions is a chaining process in which the output of the inner function becomes the input of the outer function. The composition of functions is a special case of the composition of relations, so all properties of the latter are true of composition of functions.[2] The composition of functions has some additional properties. Examples • Composition of functions on a finite set: If, and, then . • Composition of functions on an infinite set: If (where is the set of all real numbers) is given by and is given by, then: , and . • If an airplane's elevation at time  is given by the function, and the oxygen concentration at elevation is given by the function, then describes the oxygen concentration around the plane at time . Properties The composition of functions is always associative—a property inherited from the composition of relations.[2] That is, if,, and are three functions with suitably chosen domains and codomains, then, where the parentheses serve to indicate that composition is to be performed first for the parenthesized functions. Since there is no distinction between the choices of placement of parentheses, they may be left off without causing any ambiguity. In a strict sense, the composition can be built only if 's codomain equals 's domain; in a wider sense it is sufficient that the former is a subset of the latter.[3] Moreover, it is often convenient to tacitly restrict 's domain such that produces only values in 's domain; for example, the composition of the functions defined by and defined by can be defined on the interval . The functions and are said to commute with each other if . Commutativity is a special property, attained only by particular functions, and often in special circumstances. For example, only when . The picture shows another example. The composition of one-to-one functions is always one-to-one. Similarly, the composition of two onto functions is always onto. It follows that composition of two bijections is also a bijection. The inverse function of a composition (assumed invertible) has the property that .[4] Derivatives of compositions involving differentiable functions can be found using the chain rule. Higher derivatives of such functions are given by Faà di Bruno's formula. Composition monoids See main article: Transformation monoid. Suppose one has two (or more) functions having the same domain and codomain; these are often called transformations. Then one can form chains of transformations composed together, such as . Such chains have the algebraic structure of a monoid, called a transformation monoid or (much more seldom) a composition monoid. In general, transformation monoids can have remarkably complicated structure. One particular notable example is the de Rham curve. The set of all functions is called the full transformation semigroup[5] or symmetric semigroup[6] on . (One can actually define two semigroups depending how one defines the semigroup operation as the left or right composition of functions.[7]) If the transformations are bijective (and thus invertible), then the set of all possible combinations of these functions forms a transformation group; and one says that the group is generated by these functions. A fundamental result in group theory, Cayley's theorem, essentially says that any group is in fact just a subgroup of a permutation group (up to isomorphism).[8] The set of all bijective functions (called permutations) forms a group with respect to the composition operator. This is the symmetric group, also sometimes called the composition group. In the symmetric semigroup (of all transformations) one also finds a weaker, non-unique notion of inverse (called a pseudoinverse) because the symmetric semigroup is a regular semigroup.[9] Functional powers See main article: Iterated function. If, then may compose with itself; this is sometimes denoted as . That is: More generally, for any natural number, the th functional power can be defined inductively by . Repeated composition of such a function with itself is called iterated function. • By convention, is defined as the identity map on 's domain, . • If even and admits an inverse function, negative functional powers are defined for as the negated power of the inverse function: . Note: If takes its values in a ring (in particular for real or complex-valued), there is a risk of confusion, as could also stand for the -fold product of , e.g. . For trigonometric functions, usually the latter is meant, at least for positive exponents. For example, in trigonometry, this superscript notation represents standard exponentiation when used with trigonometric functions:.However, for negative exponents (especially -1), it nevertheless usually refers to the inverse function, e.g., . In some cases, when, for a given function, the equation has a unique solution, that function can be defined as the functional square root of, then written as . More generally, when has a unique solution for some natural number, then can be defined as . Under additional restrictions, this idea can be generalized so that the iteration count becomes a continuous parameter; in this case, such a system is called a flow, specified through solutions of Schröder's equation. Iterated functions and flows occur naturally in the study of fractals and dynamical systems. To avoid ambiguity, some mathematicians choose to write for the n-th iterate of the function . Alternative notations Many mathematicians, particularly in group theory, omit the composition symbol, writing for .[10] In the mid-20th century, some mathematicians decided that writing "" to mean "first apply, then apply " was too confusing and decided to change notations. They write "" for "" and "" for "".[11] This can be more natural and seem simpler than writing functions on the left in some areas – in linear algebra, for instance, when is a row vector and and denote matrices and the composition is by matrix multiplication. This alternative notation is called postfix notation. The order is important because function composition is not necessarily commutative (e.g. matrix multiplication). Successive transformations applying and composing to the right agrees with the left-to-right reading sequence. Mathematicians who use postfix notation may write "", meaning first apply and then apply, in keeping with the order the symbols occur in postfix notation, thus making the notation "" ambiguous. Computer scientists may write "" for this,[12] thereby disambiguating the order of composition. To distinguish the left composition operator from a text semicolon, in the Z notation the ⨾ character is used for left relation composition.[13] Since all functions are binary relations, it is correct to use the [fat] semicolon for function composition as well (see the article on composition of relations for further details on this notation). Composition operator See main article: Composition operator. Given a function , the composition operator is defined as that operator which maps functions to functions as Cgf=f\circg. Composition operators are studied in the field of operator theory. In programming languages See main article: Function composition (computer science). Function composition appears in one form or another in numerous programming languages. Multivariate functions Partial composition is possible for multivariate functions. The function resulting when some argument of the function is replaced by the function is called a composition of and in some computer engineering contexts, and is denoted f| xi=g =f(x1,\ldots,xi-1,g(x1,x2,\ldots,xn),xi+1,\ldots,xn). When is a simple constant, composition degenerates into a (partial) valuation, whose result is also known as restriction or co-factor.[14] f| xi=b =f(x1,\ldots,xi-1,b,xi+1,\ldots,xn). In general, the composition of multivariate functions may involve several other functions as arguments, as in the definition of primitive recursive function. Given, a -ary function, and -ary functions, the composition of with, is the -ary function h(x1,\ldots,xm)=f(g1(x1,\ldots,xm),\ldots,gn(x1,\ldots,xm)) . This is sometimes called the generalized composite of f with .[15] The partial composition in only one argument mentioned previously can be instantiated from this more general scheme by setting all argument functions except one to be suitably chosen projection functions. Note also that can be seen as a single vector/tuple-valued function in this generalized scheme, in which case this is precisely the standard definition of function composition.[16] A set of finitary operations on some base set X is called a clone if it contains all projections and is closed under generalized composition. Note that a clone generally contains operations of various arities.[15] The notion of commutation also finds an interesting generalization in the multivariate case; a function f of arity n is said to commute with a function g of arity m if f is a homomorphism preserving g, and vice versa i.e.:[17] f(g(a11,\ldots,a1m),\ldots,g(an1,\ldots,anm))=g(f(a11,\ldots,an1),\ldots,f(a1m,\ldots,anm)) . A unary operation always commutes with itself, but this is not necessarily the case for a binary (or higher arity) operation. A binary (or higher arity) operation that commutes with itself is called medial or entropic.[17] Generalizations Composition can be generalized to arbitrary binary relations.If and are two binary relations, then their composition is the relation defined as .Considering a function as a special case of a binary relation (namely functional relations), function composition satisfies the definition for relation composition. The composition is defined in the same way for partial functions and Cayley's theorem has its analogue called the Wagner-Preston theorem.[18] The category of sets with functions as morphisms is the prototypical category. The axioms of a category are in fact inspired from the properties (and also the definition) of function composition.[19] The structures given by composition are axiomatized and generalized in category theory with the concept of morphism as the category-theoretical replacement of functions. The reversed order of composition in the formula applies for composition of relations using converse relations, and thus in group theory. These structures form dagger categories. Typography The composition symbol is encoded as ; see the Degree symbol article for similar-appearing Unicode characters. In TeX, it is written `\circ`. Notes and References 1. Some authors use, defined by instead. This is common when a postfix notation is used, especially if functions are represented by exponents, as, for instance, in the study of group actions. See 2. Book: Daniel J. Velleman. How to Prove It: A Structured Approach. 2006. Cambridge University Press. 978-1-139-45097-3. 232. 3. The strict sense is used, e.g., in category theory, where a subset relation is modelled explicitly by an inclusion function. 4. Book: Nancy Rodgers. Learning to Reason: An Introduction to Logic, Sets, and Relations. 2000. John Wiley & Sons. 978-0-471-37122-9. 359–362. 5. Book: Christopher Hollings. Mathematics across the Iron Curtain: A History of the Algebraic Theory of Semigroups. 2014. American Mathematical Society. 978-1-4704-1493-1. 334. 6. Book: Pierre A. Grillet. Semigroups: An Introduction to the Structure Theory. 1995. CRC Press. 978-0-8247-9662-4. 2. 7. Book: Pál Dömösi. Chrystopher L. Nehaniv. Algebraic Theory of Automata Networks: A Introduction. 2005. SIAM. 978-0-89871-569-9. 8. 8. Book: Nathan Carter. Visual Group Theory. 9 April 2009. MAA. 978-0-88385-757-1. 95. 9. Book: Olexandr Ganyushkin. Volodymyr Mazorchuk. Classical Finite Transformation Semigroups: An Introduction. 2008. Springer Science & Business Media. 978-1-84800-281-4. 24. 10. Book: Oleg A. Ivanov. Making Mathematics Come to Life: A Guide for Teachers and Students. 1 January 2009. American Mathematical Soc.. 978-0-8218-4808-1. 217–. 11. Book: Jean Gallier . Jean Gallier. Jean Gallier . Discrete Mathematics. 2011. Springer . 978-1-4419-8047-2. 118. 12. Book: Michael Barr. Charles Wells. Category Theory for Computing Science. 1998. 6. This is the updated and free version of book originally published by Prentice Hall in 1990 as . 13. ISO/IEC 13568:2002(E), p. 23 14. Bryant, R.E. . Logic Minimization Algorithms for VLSI Synthesis . IEEE Transactions on Computers. C-35 . 8. August 1986 . 677–691 . 10.1109/tc.1986.1676819 . 15. Book: Clifford Bergman. Universal Algebra: Fundamentals and Selected Topics. 2011. CRC Press. 978-1-4398-5129-6. 79–80. 16. Book: George Tourlakis. Theory of Computation. 2012. John Wiley & Sons. 978-1-118-31533-0. 100. 17. Book: Clifford Bergman. Universal Algebra: Fundamentals and Selected Topics. 2011. CRC Press. 978-1-4398-5129-6. 90–91. 18. S. Lipscomb, "Symmetric Inverse Semigroups", AMS Mathematical Surveys and Monographs (1997),, p. xv 19. Book: Peter Hilton. Yel-Chiang Wu. A Course in Modern Algebra. 1989. John Wiley & Sons. 978-0-471-50405-4. 65.

http://www.all-science-fair-projects.com/science_fair_projects_encyclopedia/Product_(category_theory)

# All Science Fair Projects ## Science Fair Project Encyclopedia for Schools! Search    Browse    Forum  Coach    Links    Editor    Help    Tell-a-Friend    Encyclopedia    Dictionary # Science Fair Project Encyclopedia For information on any area of science that interests you, enter a keyword (eg. scientific method, molecule, cloud, carbohydrate etc.). Or else, you can start by choosing any of the categories below. # Product (category theory) Contents ## Motivation In category theory, one defines products to generalize constructions such as the cartesian product of sets, the product of groups, the product of rings and the product of topological spaces. Essentially, the product of a family of objects is the "most general" object which admits a morphism to each of the given objects. ## Definition Let C be a category and let {Xi | iI} be an indexed family of objects in C. The product of the set {Xi} is an object X together with a collection of morphisms πi : XXi (called projections) which satisfy a universal property: for any object Y and any collection of morphisms fi : YXi, there exists a unique morphism f : YX such that for all iI it is the case that fi = πi f. That is, the following diagram commutes (for all i): If the family of objects consists of only two members X, Y, the product is usually written X×Y, and the diagram takes a form along the lines of: The unique arrow h making this diagram commute is notated <f,g>. ## Discussion The product construction given above is actually a special case of a limit in category theory. The product can be defined as the limit of any discrete subcategory in C. Not every family {Xi} needs to have a product, but if it does, then the product is unique in a strong sense: if πi : XXi and π'i : X' → Xi are two products of the family {Xi}, then (by the definition of products) there exists a unique isomorphism f : XX' such that πi = π'i f for each i in I. An empty product (i.e. I is the empty set) is the same as a terminal object in C. If I is a set such that all products for families indexed with I exist, then it is possible to choose the products in a compatible fashion so that the product turns into a functor CIC. The product of the family {Xi} is then often denoted by ∏i Xi, and the maps πi are known as the natural projections. We have a natural isomorphism $\operatorname{Mor}_C\left(Y,\prod_{i\in I}X_i\right) \simeq \prod_{i\in I}\operatorname{Mor}_C(Y,X_i)$ (where MorC(U,V) denotes the set of all morphisms from U to V in C, the left product is the one in C and the right is the cartesian product of sets). If I is a finite set, say I = {1,...,n}, then the product of objects X1,...,Xn is often denoted by X1×...×Xn. Suppose all finite products exist in C, product functors have been chosen as above, and 1 denotes the terminal object of C corresponding to the empty product. We then have natural isomorphisms $X\times (Y \times Z)\simeq (X\times Y)\times Z\simeq X\times Y\times Z$ $X\times 1 \simeq 1\times X \simeq X$ $X\times Y \simeq Y\times X$ These properties are formally similar to those of a commutative monoid.

https://www.parabola.unsw.edu.au/2000-2009/volume-45-2009/issue-3/article/solutions-problems-1301-1310

# Solutions to Problems 1301 - 1310 Q1301 (Suggested by J. Guest, Victoria) Solve the quartic $$(x+1)(x+5)(x-3)(x-7) = -135$$ ANS: (suggested by David Shaw, Geelong, Victoria) Rearrange the equation as $$(x^2-2x-3)(x^2-2x-35) = -135$$ By setting $z=x^2-2x$, the above equation becomes $$z^2-38z+240=0$$ which has solutions $z_1=30$ and $z_2=8$. Solving the two equations $x^2-2x=30$ and $x^2-2x=8$ results in 4 solutions to the quartic equation $$x_1 = 1+\sqrt{31}, \quad x_2 = 1-\sqrt{31}, \quad x_3 = 4, \quad\text{and}\quad x_4 = -2$$

https://stupefied-davinci-2941e0.netlify.app/pythagorean-theorem-formula-for-b.html

# Pythagorean Theorem Formula For B ### A 2 + b 2 = c 2 the figure above helps us to see why the formula works. Pythagorean theorem formula for b. The formula of pythagorean theorem. The pythagorean theorem states that if a triangle has one right angle, then the square of the longest side, called the hypotenuse, is equal to the sum of the squares of the lengths of the two shorter sides, called the legs. For the purposes of the formula, side $$\overline{c}$$ is always the hypotenuse.remember that this formula only applies to right triangles. (m 2 + n 2)] where, m and n are two positive integers and m > n Where “a” is the perpendicular side, Pythagorean theorem formula in any right triangle a b c , the longest side is the hypotenuse, usually labeled c and opposite ∠c. A²+b²=c², with a and b representing the sides of the triangle, while c represents the hypotenuse. $$hypotenuse^{2} = perpendicular^{2} + base^{2}$$ derivation of the pythagorean theorem formula. You will likely come across many problems in school and in real life that require using the theorem to solve. It is an important formula that states the following: The theorem is named after a greek mathematician called pythagoras. $$c^2=a^2+b^2,$$ where $c$ is the length of the hypotenuse and $a$ and $b$ are the lengths of the legs of $\delta abc$. Applying the pythagorean theorem (examples) in the examples below, we will see how to apply this rule to find any side of a right triangle triangle. Consider the triangle given above: Here we will discuss pythagorean triples formula. How to use the pythagorean theorem. It is called pythagoras' theorem and can be written in one short equation: Referring to the above image, the theorem can be expressed as:

http://www.mathemafrica.org/?p=11407

So, we can do basic algebra with complex numbers, take powers of them, and apply trig and exponential functions to them. There isn’t much left that we might want to do, but taking powers of them is very important and also pretty easy. Let’s say we wanted to calculate the solution to the equation: $z^2=-1$ Well, we know what $z$ is for this, because that’s what got us into this mess in the first place! We know that there are two solutions and they are $\pm i$. That is to say that if you multiply either of these numbers by themselves, you get -1, by definition. How about $z^2=i$. The first thing to do whenever you have to take the root of a complex number, or a real number (here the second root of $i$) is to convert that number into modulus argument form – it will be infinitely easier like that. We know that $i=re^{i\theta}$ where $r=1$ and $\theta=\frac{\pi}{2}+2n\pi$. So we can write: $z^2=e^{\frac{i\pi}{2}+2n\pi i}$ or just: $z=e^{\frac{i\pi}{4}+n\pi i}$ $n$ again goes over the integers, but we have to ask whether it’s going to make a difference to what $z$ is. When $n=0$ this number is $e^{\frac{i\pi}{4}}$ and you can work out that this is $\frac{1+i}{\sqrt{2}}$, when $n=1$ we have $\frac{-1-i}{\sqrt{2}}$ which is different, but when $n=2$ we get back to the same answer as when $n=0$, and for $n=3$ we get the answers for when $n=1$. We can see very quickly that there are actually only two unique solutions to this equation which are $\pm\frac{1+i}{\sqrt{2}}$. Plot these in the Argand plane and you will see how they relate to $i$. It is pretty clear that when you multiply these numbers by themselves that they will give you $i$ (remember, multiplying two complex numbers together is done by adding the arguments and multiplying the magnitudes. In this case the magnitudes are 1 and the phases add up to give $\frac{\pi}{4}$ (modulo $2\pi$ in both cases)). $z^5=(1+i)$ Again, we start by converting from cartesian to mod/arg form which gives: $z^5=\sqrt{2}e^{\frac{i\pi}{4}+2n\pi i}$ so: $z=2^{\frac{1}{10}}e^{\frac{i\pi}{20}+\frac{2n}{5}\pi i}$ You can see quickly that when $n=0,1,2,3$ and $4$ you will get unique results, but $n=5$ will give the same as $n=0$, and $n=6$ will give the same as $n=1$, etc. We can plot these in the Argand plane and see what they look like: The five fifth roots of $1+i$. The point $1+i$ is shown in blue, the five roots are shown in black. You can see that when you multiply each one of them by itself five times you will get $1+i$ – this is the definition of the fifth root. It’s always nice to have a general procedure when we’re trying to solve a set of problems, and in this case we can find a very simple general procedure. You should have noticed one thing by now, which is that the $n^{th}$ root of a number (be it real or imaginary) will have $n$ solutions. That might seem strange that there are 7 solutions to the equation $z^7=1$ but check it – it is true, and this is because there are seven angles, that when you add them to themselves 7 times, give you integer multiples of $2\pi$ – think about it… So, what if we want to solve: $z^n=q$ We start by writing $q$ in mod/arg form, calling its magnitude $\rho$ and its angle $\phi$: $z^n=\rho e^{i\phi+2k\pi i}$ now: $z=\rho^{\frac{1}{n}} e^{\frac{i\phi}{n}+2\frac{k}{n}\pi i}$ where now $k=0,1,...(n-1)$. so again we have $n$ solutions for an $n^{th}$ root. We should note something rather important here. When we write $\rho^{\frac{1}{n}}$ this really is the $n^{th}$ root of a real number, whereas when we write $z^n=\rho$, the two are not inverse of each other. You can see this by the fact that you write $x^2=1$ which has two solutions but writing $x=\sqrt{1}$ is just one of them, the negative solution is encoded in the angular part of the solution. ie. in this case you would write $x=\sqrt{1}e^{in\pi}$ where $n=0, 1$ which is just the statement that $\pm 1$ are the two solutions to the equation $x^2=1$. How clear is this post?

https://too-meta.neocities.org/anki/analysis/1567563680476/front/

Let $$(a_n)_{n=m}^{\infty}$$ be a sequence of reals, and let $$L \ne L'$$ be two distinct real numbers. Then $$(a_n)_{n=m}^{\infty}$$ cannot converge to [...]. If $$L \ne L'$$, then there exists a positive real $$d = dist(L, L') = | L - L'|$$. As $$(a_n)_{n=m}^{\infty}$$ converges to $$L$$, we can choose a real $$0 < \varepsilon < \frac{d}{3}$$ and there exists an integer $$N_1 > m$$ such that $$(a_n)_{n=N_1}^{\infty}$$ is [...] $$L$$. Similarly, for $$L'$$ for the same $$\varepsilon$$, there. exists an integer $$N_2 > m$$ such that $$(a_n)_{n=N_2}^{\infty}$$ is [...] $$L'$$. Thus, there exists an $$M = \max(N_1, N_2)$$ (or just $$M = N_1 + N_2$$, if you don't want to use $$\max$$), such that $$a_k$$ is [...] for all $$k \ge M$$. As we have both $$dist(a_k, L) \le \varepsilon$$ and $$dist(a_k, L') \le \varepsilon$$, this implies, by the [...], that $$dist(L, L') \le 2\varepsilon = \frac{2}{3} dist(L, L')$$, which is a false statement, as the distance is always positive. Thus, it cannot be true that $$(a_n)_{n=m}^{\infty}$$ converges to both $$L$$ and $$L'$$.

https://mathematicalapologist.com/2020/12/28/computing-derivatives-part-2-explaining-calculus-8/

# Computing Derivatives: Part 2 (Explaining Calculus #8) Most recently in the series on calculus, we did some overview on some “precalculus” topics that we’d need for later calculus discussions. Having now done this, we move on to several examples of more ‘complicated’ rules that derivatives follow. We will then investigate some more difficult specific functions and their derivatives. Finally, at the end, I will leave a list of “homework problems” for any of my readers who want practice. The Product Rule We have already tackled to to take derivatives of functions like $f(x) + g(x)$. We handled addition, one of the main operations of arithmetic. By doing addition, we were able to handle subtraction as well. But what about multiplication? That would be the natural next step, just as in school multiplication is the next natural step after learning about addition and subtraction. Our purpose now is to lay out the so-called product rule, which enables us to take derivatives of functions like $f(x) g(x)$. Fact: For any two functions $f(x)$ and $g(x)$ that have derivatives, $\dfrac{d}{dx}[f(x) g(x)] = f^\prime(x) g(x) + f(x) g^\prime(x).$ Proof: This proof uses a slick trick of “adding zero” to an expression. First, by the definition of derivatives, $\dfrac{d}{dx}[f(x) g(x)] = \lim\limits_{h \to 0} \dfrac{f(x+h) g(x+h) - f(x) g(x)}{h}.$ Now, we want to be clever. Our clever move will be the fact that $f(x) g(x+h) - f(x) g(x+h) = 0$. While this looks very strange to point out something so obvious, this fact means that $\lim\limits_{h \to 0} \dfrac{f(x+h) g(x+h) - f(x) g(x)}{h} = \lim\limits_{h \to 0} \dfrac{f(x+h) g(x+h) - f(x) g(x+h) + f(x) g(x+h) - f(x) g(x)}{h}.$ The first to pieces of this numerator have a common factor, as do the second two. Therefore, this messy expression can be simplified a little bit as $\lim\limits_{h \to 0} \dfrac{g(x+h)(f(x+h) - f(x))}{h} + \lim\limits_{h \to 0} \dfrac{f(x)(g(x+h) - g(x))}{h}.$ Since $\lim\limits_{h \to 0} g(x+h) = g(x)$, the first of these simplifies as $\lim\limits_{h \to 0} \dfrac{g(x+h)(f(x+h) - f(x))}{h} = g(x) \lim\limits_{h \to 0} \dfrac{f(x+h) - f(x)}{h} = g(x) f^\prime(x)$. The second piece, using the exact same process, is equal to $f(x) g^\prime(x)$. When everything is put back together, $f(x) g(x)$ has derivative $f^\prime(x) g(x) + f(x) g^\prime(x)$. The Chain Rule You’d think that the next thing we would do is division. And in fact, there is a way I could do division next. But it will actually be easier to do something else first. I want to do functions inside of other functions next, like $f(g(x))$, which just means “copy-paste the expression $g(x)$ wherever there would have been an $x$. This is called the chain rule, because the functions sort of link together like a chain, inextricably connected. Fact: For any two functions $f(x), g(x)$ for which $f(g(x))$ makes sense, we have $\dfrac{d}{dx}[f(g(x))] = f^\prime(g(x)) * g^\prime(x).$ In different notation, if $y$ is a function of $u$, which itself is a function of $x$, then $\dfrac{dy}{du} * \dfrac{du}{dx} = \dfrac{dy}{dx}.$ In an earlier post, I made a comment about a similarity between this derivative notation and genuine fractions. The chain rule is the central such similarity – when written in the dy-over-dx style, it looks as if the du’s are cancelling out, like they would if these were actual fractions. Proof: This one is actually quite tricky compared to the others. For this reason, I won’t actually do a totally correct proof. Instead, I’m going to do a proof that “usually” works. For those who want a great challenge, try to figure out where this proof goes wrong. (If you end up wanting to know, the Wikipedia article on the chain rule will actually tell you. If you don’t care, then reading this proof will give you the right idea.) The definition of the derivative tells us that $\dfrac{d}{dx}[f(g(x))] = \lim\limits_{h \to 0} \dfrac{f(g(x+h)) - f(g(x))}{h}.$ We now make a clever step of multiplying top and bottom by $g(x+h) - g(x)$. $\dfrac{d}{dx}[f(g(x))] = \lim\limits_{h \to 0} \dfrac{f(g(x+h)) - f(g(x))}{g(x+h) - g(x)} \cdot \dfrac{g(x+h) - g(x)}{h}.$ We can even split this up into two limits multiplied together. $\dfrac{d}{dx}[f(g(x))] = \lim\limits_{h \to 0} \dfrac{f(g(x+h)) - f(g(x))}{g(x+h) - g(x)} \cdot \lim\limits_{h \to 0} \dfrac{g(x+h) - g(x)}{h} = g^\prime(x) \lim\limits_{h \to 0} \dfrac{f(g(x+h)) - f(g(x))}{g(x+h) - g(x)}.$ Since $g(x)$ will be a continuous function (it has a derivative so it must be continuous), $g(x+h) \to g(x)$ as $h \to 0$. This enables us to treat $g(x+h)$ as if it were something like $g(x) + h$. There is a more careful way to write that down, but since we are being informal I won’t do that (and no, this isn’t the real problem in the proof… that already happened earlier). This will mean that $\lim\limits_{h \to 0} \dfrac{f(g(x+h)) - f(g(x))}{g(x+h) - g(x)} = \lim\limits_{h \to 0} \dfrac{f(g(x) + h) - f(g(x))}{(g(x) + h) - g(x)} = \lim\limits_{h \to 0} \dfrac{f(y+h) - f(y)}{h},$ where $y = g(x)$ is used to make the formulas easier to follow. This leads directly to the value $f^\prime(g(x))$ for this limit. When we combine all of our work, we find that $\dfrac{d}{dx}[f(g(x))] = f^\prime(g(x)) g^\prime(x)$. Fact: If the functions $f(x), g(x)$ have derivatives, then the derivative of $\dfrac{f(x)}{g(x)}$ is $\dfrac{g(x) f^\prime(x) - f(x) g^\prime(x)}{g(x)^2}$. Proof: The first thing we do is to view the division as a multiplication by $\dfrac{f(x)}{g(x)} = f(x) \cdot \dfrac{1}{g(x)}.$ Secondly, we want to use the chain rule for $\dfrac{1}{g(x)}$. To do this, we will use the function $h(x) = \dfrac{1}{x}$. Then $\dfrac{1}{g(x)} = h(g(x))$. Therefore, $\dfrac{f(x)}{g(x)} = f(x) \cdot h(g(x)).$ We can then use the product rule to begin this derivative: $\dfrac{d}{dx}\bigg[ \dfrac{f(x)}{g(x)} \bigg] = f(x) \cdot \dfrac{d}{dx}[ h(g(x)) ] + h(g(x)) f^\prime(x).$ From the chain rule, we can simplify the derivative of $h(g(x))$ to arrive at $\dfrac{d}{dx}\bigg[ \dfrac{f(x)}{g(x)} \bigg] = f(x) \cdot [h^\prime(g(x)) \cdot g^\prime(x)] + h(g(x)) f^\prime(x).$ Now, since $h(x) = \dfrac{1}{x} = x^{-1}$, the rule for taking derivatives of powers of $x$ proven in an earlier post tells us that $h^\prime(x) = - x^{-2} = \dfrac{-1}{x^2}$. Therefore, $\dfrac{d}{dx}\bigg[ \dfrac{f(x)}{g(x)} \bigg] = f(x) \cdot \bigg(\dfrac{-1}{g(x)^2} \cdot g^\prime(x) \bigg) + \dfrac{1}{g(x)} \cdot f^\prime(x),$ and we can simplify this as $\dfrac{d}{dx}\bigg[ \dfrac{f(x)}{g(x)} \bigg] = \dfrac{- f(x) g^\prime(x)}{g(x)^2} + \dfrac{f^\prime(x) g(x)}{g(x)^2} = \dfrac{g(x) f^\prime(x) - f(x) g^\prime(x)}{g(x)^2}.$ This is the original formula we wanted to prove. So, our proof is now done. More Special Functions We now move on from these general principles to some more specific examples. In the previous post on computing derivatives, we built up all the tools to compute the derivative of any polynomial expression, and in fact any expression involving $x$ raised to constant powers. Now, we move on to trigonometry, exponentials, and logarithms. These derivatives are more difficult to discover, but the importance of these functions requires that any complete study of calculus should include their derivatives. Also, a thorough study of these functions and their derivatives will provide a very complete picture of how to calculate derivatives using the limit definition. Fact: The derivatives of $\sin{x}$ and $\cos{x}$ are $\dfrac{d}{dx}[\sin{x}] = \cos{x} \text{ and } \dfrac{d}{dx}[\cos{x}] = -\sin{x}.$ Proof: The definition of the derivative tells us that $\dfrac{d}{dx}[ \sin{x} ] = \lim\limits_{h \to 0} \dfrac{\sin{(x+h)} - \sin{x}}{h}.$ In a previous post in the series, where I defined the function $\sin{x}$, I also gave the following rule for computing the value of $\sin{(x+h)}$: $\sin{(x+h)} = \sin{x} \cos{h} + \cos{x} \sin{h}.$ Therefore, $\dfrac{d}{dx}[ \sin{x} ] = \lim\limits_{h \to 0} \dfrac{\sin{(x+h)} - \sin{x}}{h} = \lim\limits_{h \to 0} \dfrac{(\sin{x} \cos{h} + \cos{x} \sin{h}) - \sin{x}}{h}.$ We can now split this limit up into two pieces – one for $\sin{x}$ and one for $\cos{x}$. $\lim\limits_{h \to 0} \dfrac{(\sin{x} \cos{h} + \cos{x} \sin{h}) - \sin{x}}{h} = \lim\limits_{h \to 0} \dfrac{\sin{x}(\cos{h} - 1)}{h} + \lim\limits_{h \to 0} \dfrac{\cos{x} \sin{h}}{h}.$ There are some facts we have to know in order to continue, these are that $\lim\limits_{h \to 0} \dfrac{\sin{h}}{h} = 1$ and $\lim\limits_{h \to 0} \dfrac{\cos{h} -1}{h} = 0$. The proofs of these are a bit tricky, and so I don’t want to go off on a tangent (math pun!) talking about those here. I will add an appendix to the end of this post in which I talk about how to find these limits. Moving on, once we know the values of these limits, we know that $\lim\limits_{h \to 0} \dfrac{\sin{x}(\cos{h} - 1)}{h} = \sin{x} \cdot \lim\limits_{h \to 0} \dfrac{\cos{h} - 1}{h} = 0$ and $\lim\limits_{h \to 0} \dfrac{\cos{x} \sin{h}}{h} = \cos{x} \cdot \lim\limits_{h \to 0} \dfrac{\sin{h}}{h} = \cos{x}.$ Therefore, putting together all the steps we’ve laid out, $\dfrac{d}{dx}[\sin{x}] = \cos{x}.$ This completes the first half of the proof. I will leave the proof about the derivative of $\cos{x}$ as practice for any of my curious readers, only providing a few guiding hints. The proof begins the same way. Instead of using the special rule for $\sin{(x+h)}$, you need to use the rule for $\cos{(x+h)}$ that I gave in the same post in which I gave the rule for $\sin{(x+h)}$. After this rule is used, you should be able to finish the proof by following the same ideas I use here. This completes our discussion of Fact: The derivative of $f(x) = b^x$ is $f^\prime(x) = b^x * \log{b}$. Proof: The first thing we will do is to “compress” every value of $b$ into the very special number $e$. Remember that, earlier in the series, we defined the “natural” exponential function $e^x$ and “natural” logarithm function $\log{x}$. Also, remember that $e^{\log{x}} = x$. Using $x = b$, we conclude that $b = e^{\log{b}}$ and therefore $b^x = (e^{\log{b}})^x = e^{x * \log{b}}.$ Now, let’s call $f(x) = e^x$ and $g(x) = x * \log{b}$. Then $b^x = f(g(x))$. Using the chain rule, we then know that $\dfrac{d}{dx}[b^x] = \dfrac{d}{dx}[ f(g(x)) ] = f^\prime(g(x)) * g^\prime(x) = \log{b} * f^\prime(x*\log{b}).$ What we have done here is to express the derivative of $b^x$ in terms of the derivative of $e^x$. This means that we now only need to know how to find the derivative of this most special function. Using the definition of the derivative, along with some basic rules of exponents, we have $\dfrac{d}{dx}[e^x] = \lim\limits_{h \to 0} \dfrac{e^{x+h} - e^x}{h} = \lim\limits_{h \to 0} \dfrac{e^x(e^h - 1)}{h} = e^x \lim\limits_{h \to 0} \dfrac{e^h-1}{h}.$ The proof here is a bit detailed, but the value of $\lim\limits_{h \to 0} \dfrac{e^h - 1}{h}$ is 1. I will delay this proof to the appendix. But, the key fact here is that $e^x$ is so special because it is its own derivative. That is, if $f(x) = e^x$, then $f^\prime(x) = f(x) = e^x$! Using all of these facts, we conclude that $\dfrac{d}{dx}[b^x] = e^{x * \log{b}} * \log{b} = b^x * \log{b}.$ This completes our discussion of the derivative of exponential functions. Fact: The derivative of the function $f(x) = \log_b{x}$ is $f^\prime(x) = \dfrac{1}{x * \log{b}}$. Proof: There is a way to do this similar to the previous proof about $b^x$. Instead of doing that, I want to be clever. Since the functions $b^x$ and $\log_b{x}$ are so closely related, shouldn’t their derivatives be really closely related too? We should think so. That just makes sense. Then maybe we can find a way to take advantage of our formula for the derivative of $b^x$ to find the derivative of $\log_b{x}$. Can you find the way? Take a moment and think about it. The way I’ll do this is to transform the equation $f(x) = \log_b{x}$ into the equation $b^{f(x)} = b^{\log_b{x}}$. The second step is to notice that $b^{\log_b{x}} = x$ because of the relationship between $b^x$ and $\log_b{x}$. Therefore, $b^{f(x)} = x$. The other clever observation we have to make is that if two things are equal, their derivatives must also be equal. This means that $\dfrac{d}{dx}[ b^{f(x)} ] = \dfrac{d}{dx}[x]$. This is so clever because we can find these two derivatives by different methods. The right-hand side is much easier, the derivative of $x$ is just 1. The left-hand side can be found by using the fact we already found about the derivative of exponentials along with the chain rule: $\dfrac{d}{dx}[b^{f(x)}] = (b^{f(x)}*\log{b}) * f^\prime(x) = (x*\log{b})*f^\prime(x)$ since $b^{f(x)} = x$ was already known to us. By combining these two derivative computations, we conclude that $(x*\log{b})*f^\prime(x) = 1,$ and all we need to do is divide both sides of the equation by $x*\log{b}$ to find our answer. Notice that from this fact, we can also deduce that $\dfrac{d}{dx}[\log{x}] = \dfrac{1}{x}$. I will leave this to my reader. (Hint: why is $\log{e} = 1$?) This basically completes the list of important derivatives. But for the sake of learning, I want to compute one more that uses a clever trick much like the trick I used in the previous proof that enables us to find the derivative of a function that is “more complicated” than any of those we have previously discussed. The reason I want to show this is to demonstrate how we can take advantage of the rules we have already learned to find derivatives of more complex functions. Fact: The derivative of $f(x) = x^x$ is $f^\prime(x) = x^x(\log{x} + 1)$. Proof: The function $x^x$ isn’t actually susceptible to any derivative rule we’ve used to far. This is where we make a clever move to bring this function into the realm we know how to deal with by taking a logarithm. By the normal rules of logarithms, $\log{f(x)} = \log{x^x} = x \log{x}$. The derivative of the right hand side can be done using the product rule: $\dfrac{d}{dx}[ x \log{x} ] = x \dfrac{d}{dx}[\log{x}] + \log{x} \dfrac{d}{dx}[x] = \dfrac{x}{x} + \log{x} = \log{x} + 1.$ One the other hand, the left hand side can be done using the chain rule: $\dfrac{d}{dx}[\log{f(x)}] = \dfrac{1}{f(x)} * f^\prime(x) = \dfrac{f^\prime(x)}{x^x}.$ Since the left and right hand sides are equal, we conclude that $\dfrac{f^\prime(x)}{x^x} = \log{x}+1 \implies f^\prime(x) = x^x(\log{x} + 1).$ Conclusion We have now done everything we need to do with the admittedly tedious work of computing special kinds of derivatives. If you’ve made it this far, well done. Things get less tedious and more conceptual from here. Think about how tedious it was to learn how to add and multiply small numbers. Despite this tediousness, the effort is clearly worthwhile because of how useful addition and multiplication are for solving real-world problems. What we’ve done is “learned out times tables” for derivatives. Now that we’ve finished our “tables,” we can move on to grander topics. We can stop looking at individual trees and see the forest. Beginning with the next post, this is what we shall do. In the meantime, if you want some more practice, I’ve left some practice problems and extra material in an appendix that my more curious leaders could look at. Homework/Practice Problem 1: Find the derivatives of $x^2 e^x$ and of $\dfrac{x^2+2}{x^3 - x}$. Problem 2: Work out the details of the proof that the derivative of $\cos{x}$ is $- \sin{x}$. Also, prove that the derivative of $\tan{x} = \dfrac{\sin{x}}{\cos{x}}$ is $\dfrac{1}{\cos^2{x}}$. Go on to find the derivatives of $\cot{x}, \sec{x}$, and $\csc{x}$. Problem 3: Find the derivatives of $e^{e^x}$ and of $\log{(\log{x})}$. (Hint: Use the chain rule for both!) Appendix In this appendix, we evaluate the various limits we have delayed in the proofs given above. Fact: $\lim\limits_{h \to 0} \dfrac{\sin{h}}{h}$ and $\lim\limits_{h \to 0} \dfrac{\cos{h} - 1}{h}.$ Proof: The proof for the second limit is similar to the first, so we won’t consider it here. We only do the first limit. To do so, we consider the following diagram: We need to establish some initial values of the geometry here. The angle at $A$ is defined to be $x$. The side $AB$ has length 1, which means immediately that $AE$ has length $\cos{x}$ and side $CE$ has length $\sin{x}$. (I’m not currently very good at uploading labelled diagrams like this… sketch it on a piece of paper with labelled sides if you need that visual). We first notice the triangle $\Delta ABC$. The area of this triangle is one half its base times its height. Its height is $\sin{x}$, and its base is 1. Therefore, $\text{Area}(\Delta ABC) = \dfrac{1}{2} \sin{x}.$ Before we carry on, why is the length of $BD$ equal to $\tan{x}$? This is because of “similar triangles.” Remember from geometry that two triangles are similar if their angles have the same degrees. The two triangles $\Delta ACE$ and $ABD$ are similar. The key fact about similar triangles is that any fractions of side lengths are always the same. This means that $\dfrac{\text{Length}(AE)}{\text{Length}(CE)} = \dfrac{\text{Length}(AB)}{\text{Length}(BD)}.$ The length of $AE$ is just $\cos{x}$. The length of $AB$ is 1, and the length of $CE$ is $\sin{x}$. Therefore, $\dfrac{\cos{x}}{\sin{x}} = \dfrac{1}{\text{Length}(BD)}.$ Isolating $\text{Length}(BD) = \dfrac{\sin{x}}{\cos{x}} = \tan{x}$. So, the length of $BD$ truly is equal to $\tan{x}$. This quickly tells us (as with the smaller triangle) that $\text{Area}(\Delta ABD) = \dfrac{1}{2} \tan{x}$. One more thing we want to observe. Instead of the triangle $\Delta ABC$, look at the “pizza slice” $ABC$. It definitely has more area than the triangle $\Delta ABC$ and less area that the larger triangle $\Delta ABD$. Therefore, by putting together our earlier calculations, $\text{Area}(\Delta ABC) \leq \text{Area}(ABC) \leq \text{Area}(\Delta ABD)$ $\implies \dfrac{1}{2} \sin{x} \leq \text{Area}(ABC) \leq \dfrac{1}{2} \tan{x}.$ The way angles are defined, the “pizza slice” $ABC$ has area $\dfrac{x}{2\pi}$ of a full circle. The area of the full unit circle is $\pi$, so the area of the pizza slice is $\dfrac{1}{2} x$. Therefore, $\dfrac{1}{2} \sin{x} \leq \dfrac{1}{2} x \leq \dfrac{1}{2} \tan{x} \implies \sin{x} \leq x \leq \tan{x}$. If we divide both sides by $\sin{x}$, then since $\dfrac{\tan{x}}{\sin{x}} = \dfrac{1}{\cos{x}}$, $1 \leq \dfrac{x}{\sin{x}} \leq \dfrac{1}{\cos{x}}.$ If we “flip” all these fractions and reverse the inequalities, this means that $1 \geq \dfrac{\sin{x}}{x} \geq \cos{x}.$ By taking limits on all parts of this equation, $\lim\limits_{x \to 0} 1 \geq \lim\limits_{x \to 0} \dfrac{\sin{x}}{x} \geq \lim\limits_{x \to 0} \cos{x}.$ Notice now that the third of these limits is just $\cos{0}$, which if we look at the triangle construction earlier is just 1, the length of $AB$. Therefore, $1 \leq \lim\limits_{x \to 0} \dfrac{\sin{x}}{x} \leq 1.$ This technique is often called the Sandwich Theorem because we stuck the expression we wanted in between two other expressions. Since, of course, the only number between 1 and 1 is 1, the limit we wanted to find must be 1. Fact: $\lim\limits_{h \to 0} \dfrac{e^h - 1}{h} = 1$. Proof: This is done by very cleverly using the way the number $e$ is defined. Remember that $e = \lim\limits_{x \to \infty} \bigg( 1 + \dfrac{1}{x} \bigg)^x.$ The clever move we can make here is to define $h = \dfrac{1}{x}$ and rewrite this same limit in terms of $h$. Notice that if $x \to \infty$, then $h \to 0$. (Technically this is only a limit “from the right”. We would first have to prove that this limit actually exists. This isn’t terribly difficult to do, but it would require a lot of additional writing. So, I won’t do this. If you look at a graph of the function $\dfrac{e^x-1}{x}$, you can convince yourself that this limit does exist.) By interchanging variables in this way, $e = \lim\limits_{h \to 0} (1 + h)^{1/h}.$ Using this as a substitution for $e$ inside the limit we want to actually compute, $\lim\limits_{h \to 0} \dfrac{e^h-1}{h} = \lim\limits_{h \to 0} \dfrac{((1+h)^{1/h})^h - 1}{h} = \lim\limits_{h \to 0} \dfrac{(1+h) - 1}{h} = \lim\limits_{h \to 0} \dfrac{h}{h} = 1.$

http://discrete.openmathbooks.org/dmoi3/sec_intro-functions.html

## Section0.4Functions A function is a rule that assigns each input exactly one output. We call the output the image of the input. The set of all inputs for a function is called the domain. The set of all allowable outputs is called the codomain. We would write $f:X \to Y$ to describe a function with name $f\text{,}$ domain $X$ and codomain $Y\text{.}$ This does not tell us which function $f$ is though. To define the function, we must describe the rule. This is often done by giving a formula to compute the output for any input (although this is certainly not the only way to describe the rule). For example, consider the function $f:\N \to \N$ defined by $f(x) = x^2 + 3\text{.}$ Here the domain and codomain are the same set (the natural numbers). The rule is: take your input, multiply it by itself and add 3. This works because we can apply this rule to every natural number (every element of the domain) and the result is always a natural number (an element of the codomain). Notice though that not every natural number is actually an output (there is no way to get 0, 1, 2, 5, etc.). The set of natural numbers that are outputs is called the range of the function (in this case, the range is $\{3, 4, 7, 12, 19, 28, \ldots\}\text{,}$ all the natural numbers that are 3 more than a perfect square). The key thing that makes a rule a function is that there is exactly one output for each input. That is, it is important that the rule be a good rule. What output do we assign to the input 7? There can only be one answer for any particular function. ###### Example0.4.1. The following are all examples of functions: 1. $f:\Z \to \Z$ defined by $f(n) = 3n\text{.}$ The domain and codomain are both the set of integers. However, the range is only the set of integer multiples of 3. 2. $g: \{1,2,3\} \to \{a,b,c\}$ defined by $g(1) = c\text{,}$ $g(2) = a$ and $g(3) = a\text{.}$ The domain is the set $\{1,2,3\}\text{,}$ the codomain is the set $\{a,b,c\}$ and the range is the set $\{a,c\}\text{.}$ Note that $g(2)$ and $g(3)$ are the same element of the codomain. This is okay since each element in the domain still has only one output. 3. $h:\{1,2,3,4\} \to \N$ defined by the table: $x$ 1 2 3 4 $h(x)$ 3 6 9 12 Here the domain is the finite set $\{1,2,3,4\}$ and to codomain is the set of natural numbers, $\N\text{.}$ At first you might think this function is the same as $f$ defined above. It is absolutely not. Even though the rule is the same, the domain and codomain are different, so these are two different functions. ###### Example0.4.2. Just because you can describe a rule in the same way you would write a function, does not mean that the rule is a function. The following are NOT functions. 1. $f:\N \to \N$ defined by $f(n) = \frac{n}{2}\text{.}$ The reason this is not a function is because not every input has an output. Where does $f$ send 3? The rule says that $f(3) = \frac{3}{2}\text{,}$ but $\frac{3}{2}$ is not an element of the codomain. 2. Consider the rule that matches each person to their phone number. If you think of the set of people as the domain and the set of phone numbers as the codomain, then this is not a function, since some people have two phone numbers. Switching the domain and codomain sets doesn't help either, since some phone numbers belong to multiple people (assuming some households still have landlines when you are reading this). ### SubsectionDescribing Functions It is worth making a distinction between a function and its description. The function is the abstract mathematical object that in some way exists whether or not anyone ever talks about it. But when we do want to talk about the function, we need a way to describe it. A particular function can be described in multiple ways. Some calculus textbooks talk about the Rule of Four, that every function can be described in four ways: algebraically (a formula), numerically (a table), graphically, or in words. In discrete math, we can still use any of these to describe functions, but we can also be more specific since we are primarily concerned with functions that have $\N$ or a finite subset of $\N$ as their domain. Describing a function graphically usually means drawing the graph of the function: plotting the points on the plane. We can do this, and might get a graph like the following for a function $f:\{1,2,3\} \to \{1,2,3\}\text{.}$ It would be absolutely WRONG to connect the dots or try to fit them to some curve. There are only three elements in the domain. A curve would mean that the domain contains an entire interval of real numbers. Here is another way to represent that same function: This shows that the function $f$ sends 1 to 2, 2 to 1 and 3 to 3: just follow the arrows. The arrow diagram used to define the function above can be very helpful in visualizing functions. We will often be working with functions with finite domains, so this kind of picture is often more useful than a traditional graph of a function. Note that for finite domains, finding an algebraic formula that gives the output for any input is often impossible. Of course we could use a piecewise defined function, like \begin{equation*} f(x) = \begin{cases} x+1 \amp \text{ if } x = 1 \\ x-1 \amp \text{ if } x = 2 \\ x \amp \text{ if } x = 3\end{cases}\text{.} \end{equation*} This describes exactly the same function as above, but we can all agree is a ridiculous way of doing so. Since we will so often use functions with small domains and codomains, let's adopt some notation to describe them. All we need is some clear way of denoting the image of each element in the domain. In fact, writing a table of values would work perfectly: $x$ 0 1 2 3 4 $f(x)$ 3 3 2 4 1 We simplify this further by writing this as a “matrix” with each input directly over its output: \begin{equation*} f = \twoline{0 \amp 1 \amp 2\amp 3 \amp 4}{3 \amp 3 \amp 2 \amp 4 \amp 1}\text{.} \end{equation*} Note this is just notation and not the same sort of matrix you would find in a linear algebra class (it does not make sense to do operations with these matrices, or row reduce them, for example). One advantage of the two-line notation over the arrow diagrams is that it is harder to accidentally define a rule that is not a function using two-line notation. ###### Example0.4.3. Which of the following diagrams represent a function? Let $X = \{1,2,3,4\}$ and $Y = \{a,b,c,d\}\text{.}$ Solution $f$ is a function. So is $g\text{.}$ There is no problem with an element of the codomain not being the image of any input, and there is no problem with $a$ from the codomain being the image of both 2 and 3 from the domain. We could use our two-line notation to write these as \begin{equation*} f= \begin{pmatrix} 1 \amp 2 \amp 3 \amp 4 \\ d \amp a \amp c \amp b \end{pmatrix} \qquad g = \begin{pmatrix} 1 \amp 2 \amp 3 \amp 4 \\ d \amp a \amp a \amp b \end{pmatrix}\text{.} \end{equation*} However, $h$ is NOT a function. In fact, it fails for two reasons. First, the element 1 from the domain has not been mapped to any element from the codomain. Second, the element 2 from the domain has been mapped to more than one element from the codomain ($a$ and $c$). Note that either one of these problems is enough to make a rule not a function. In general, neither of the following mappings are functions: It might also be helpful to think about how you would write the two-line notation for $h\text{.}$ We would have something like: \begin{equation*} h=\twoline{1 \amp 2 \amp 3 \amp 4}{\amp a,c? \amp d \amp b}\text{.} \end{equation*} There is nothing under 1 (bad) and we needed to put more than one thing under 2 (very bad). With a rule that is actually a function, the two-line notation will always “work”. We will also be interested in functions with domain $\N\text{.}$ Here two-line notation is no good, but describing the function algebraically is often possible. Even tables are a little awkward, since they do not describe the function completely. For example, consider the function $f:\N \to \N$ given by the table below. $x$ 0 1 2 3 4 5 $\ldots$ $f(x)$ 0 1 4 9 16 25 $\ldots$ Have I given you enough entries for you to be able to determine $f(6)\text{?}$ You might guess that $f(6) = 36\text{,}$ but there is no way for you to know this for sure. Maybe I am being a jerk and intended $f(6) = 42\text{.}$ In fact, for every natural number $n\text{,}$ there is a function that agrees with the table above, but for which $f(6) = n\text{.}$ Okay, suppose I really did mean for $f(6) = 36\text{,}$ and in fact, for the rule that you think is governing the function to actually be the rule. Then I should say what that rule is. $f(n) = n^2\text{.}$ Now there is no confusion possible. Giving an explicit formula that calculates the image of any element in the domain is a great way to describe a function. We will say that these explicit rules are closed formulas for the function. There is another very useful way to describe functions whose domain is $\N\text{,}$ that rely specifically on the structure of the natural numbers. We can define a function recursively! ###### Example0.4.4. Consider the function $f:\N \to \N$ given by $f(0) = 0$ and $f(n+1) = f(n) + 2n+1\text{.}$ Find $f(6)\text{.}$ Solution The rule says that $f(6) = f(5) + 11$ (we are using $6 = n+1$ so $n = 5$). We don't know what $f(5)$ is though. Well, we know that $f(5) = f(4) + 9\text{.}$ So we need to compute $f(4)\text{,}$ which will require knowing $f(3)\text{,}$ which will require $f(2)\text{,}$… will it ever end? Yes! In fact, this process will always end because we have $\N$ as our domain, so there is a least element. And we gave the value of $f(0)$ explicitly, so we are good. In fact, we might decide to work up to $f(6)$ instead of working down from $f(6)\text{:}$ \begin{align*} f(1) = \amp f(0) + 1 = \amp 0 + 1 = 1\\ f(2) = \amp f(1) + 3 = \amp 1 + 3 = 4\\ f(3) = \amp f(2) + 5 = \amp 4 + 5 = 9\\ f(4) = \amp f(3) + 7 = \amp 9 + 7 = 16\\ f(5) = \amp f(4) + 9 = \amp 16 + 9 = 25\\ f(6) = \amp f(5) + 11 = \amp 25 + 11 = 36 \end{align*} It looks that this recursively defined function is the same as the explicitly defined function $f(n) = n^2\text{.}$ Is it? Later we will prove that it is. Recursively defined functions are often easier to create from a “real world” problem, because they describe how the values of the functions are changing. However, this comes with a price. It is harder to calculate the image of a single input, since you need to know the images of other (previous) elements in the domain. ###### Recursively Defined Functions. For a function $f:\N \to \N\text{,}$ a recursive definition consists of an initial condition together with a recurrence relation. The initial condition is the explicitly given value of $f(0)\text{.}$ The recurrence relation is a formula for $f(n+1)$ in terms for $f(n)$ (and possibly $n$ itself). ###### Example0.4.5. Give recursive definitions for the functions described below. 1. $f:\N \to \N$ gives the number of snails in your terrarium $n$ years after you built it, assuming you started with 3 snails and the number of snails doubles each year. 2. $g:\N \to \N$ gives the number of push-ups you do $n$ days after you started your push-ups challenge, assuming you could do 7 push-ups on day 0 and you can do 2 more push-ups each day. 3. $h:\N \to \N$ defined by $h(n) = n!\text{.}$ Recall that $n! = 1 \cdot 2 \cdot 3 \cdot \cdots \cdot (n-1)\cdot n$ is the product of all numbers from $1$ through $n\text{.}$ We also define $0! = 1\text{.}$ Solution 1. The initial condition is $f(0) = 3\text{.}$ To get $f(n+1)$ we would double the number of snails in the terrarium the previous year, which is given by $f(n)\text{.}$ Thus $f(n+1) = 2f(n)\text{.}$ The full recursive definition contains both of these, and would be written, \begin{equation*} f(0) = 3;~ f(n+1) = 2f(n)\text{.} \end{equation*} 2. We are told that on day 0 you can do 7 push-ups, so $g(0) = 7\text{.}$ The number of push-ups you can do on day $n+1$ is 2 more than the number you can do on day $n\text{,}$ which is given by $g(n)\text{.}$ Thus \begin{equation*} g(0) = 7;~ g(n+1) = g(n) + 2\text{.} \end{equation*} 3. Here $h(0) = 1\text{.}$ To get the recurrence relation, think about how you can get $h(n+1) = (n+1)!$ from $h(n) = n!\text{.}$ If you write out both of these as products, you see that $(n+1)!$ is just like $n!$ except you have one more term in the product, an extra $n+1\text{.}$ So we have, \begin{equation*} h(0) = 1;~ h(n+1) = (n+1)\cdot h(n)\text{.} \end{equation*} ### SubsectionSurjections, Injections, and Bijections We now turn to investigating special properties functions might or might not possess. In the examples above, you may have noticed that sometimes there are elements of the codomain which are not in the range. When this sort of the thing does not happen, (that is, when everything in the codomain is in the range) we say the function is onto or that the function maps the domain onto the codomain. This terminology should make sense: the function puts the domain (entirely) on top of the codomain. The fancy math term for an onto function is a surjection, and we say that an onto function is a surjective function. In pictures: ###### Example0.4.6. Which functions are surjective (i.e., onto)? 1. $f:\Z \to \Z$ defined by $f(n) = 3n\text{.}$ 2. $g: \{1,2,3\} \to \{a,b,c\}$ defined by $g = \begin{pmatrix}1 \amp 2 \amp 3 \\ c \amp a \amp a \end{pmatrix}\text{.}$ 3. $h:\{1,2,3\} \to \{1,2,3\}$ defined as follows: Solution 1. $f$ is not surjective. There are elements in the codomain which are not in the range. For example, no $n \in \Z$ gets mapped to the number 1 (the rule would say that $\frac{1}{3}$ would be sent to 1, but $\frac{1}{3}$ is not in the domain). In fact, the range of the function is $3\Z$ (the integer multiples of 3), which is not equal to $\Z\text{.}$ 2. $g$ is not surjective. There is no $x \in \{1,2,3\}$ (the domain) for which $g(x) = b\text{,}$ so $b\text{,}$ which is in the codomain, is not in the range. Notice that there is an element from the codomain “missing” from the bottom row of the matrix. 3. $h$ is surjective. Every element of the codomain is also in the range. Nothing in the codomain is missed. To be a function, a rule cannot assign a single element of the domain to two or more different elements of the codomain. However, we have seen that the reverse is permissible: a function might assign the same element of the codomain to two or more different elements of the domain. When this does not occur (that is, when each element of the codomain is the image of at most one element of the domain) then we say the function is one-to-one. Again, this terminology makes sense: we are sending at most one element from the domain to one element from the codomain. One input to one output. The fancy math term for a one-to-one function is an injection. We call one-to-one functions injective functions. In pictures: ###### Example0.4.7. Which functions are injective (i.e., one-to-one)? 1. $f:\Z \to \Z$ defined by $f(n) = 3n\text{.}$ 2. $g: \{1,2,3\} \to \{a,b,c\}$ defined by $g = \begin{pmatrix}1 \amp 2 \amp 3 \\ c \amp a \amp a \end{pmatrix}\text{.}$ 3. $h:\{1,2,3\} \to \{1,2,3\}$ defined as follows: Solution 1. $f$ is injective. Each element in the codomain is assigned to at most one element from the domain. If $x$ is a multiple of three, then only $x/3$ is mapped to $x\text{.}$ If $x$ is not a multiple of 3, then there is no input corresponding to the output $x\text{.}$ 2. $g$ is not injective. Both inputs $2$ and $3$ are assigned the output $a\text{.}$ Notice that there is an element from the codomain that appears more than once on the bottom row of the matrix. 3. $h$ is injective. Each output is only an output once. Be careful: “surjective” and “injective” are NOT opposites. You can see in the two examples above that there are functions which are surjective but not injective, injective but not surjective, both, or neither. In the case when a function is both one-to-one and onto (an injection and surjection), we say the function is a bijection, or that the function is a bijective function. To illustrate the contrast between these two properties, consider a more formal definition of each, side by side. ###### Injective vs Surjective. A function is injective provided every element of the codomain is the image of at most one element from the domain. A function is surjective provided every element of the codomain is the image of at least one element from the domain. Notice both properties are determined by what happens to elements of the codomain: they could be repeated as images or they could be “missed” (not be images). Injective functions do not have repeats but might or might not miss elements. Surjective functions do not miss elements, but might or might not have repeats. The bijective functions are those that do not have repeats and do not miss elements. ### SubsectionImage and Inverse Image When discussing functions, we have notation for talking about an element of the domain (say $x$) and its corresponding element in the codomain (we write $f(x)\text{,}$ which is the image of $x$). Sometimes we will want to talk about all the elements that are images of some subset of the domain. It would also be nice to start with some element of the codomain (say $y$) and talk about which element or elements (if any) from the domain it is the image of. We could write “those $x$ in the domain such that $f(x) = y\text{,}$” but this is a lot of writing. Here is some notation to make our lives easier. To address the first situation, what we are after is a way to describe the set of images of elements in some subset of the domain. Suppose $f:X \to Y$ is a function and that $A \subseteq X$ is some subset of the domain (possibly all of it). We will use the notation $f(A)$ to denote the image of $A$ under $f$, namely the set of elements in $Y$ that are the image of elements from $A\text{.}$ That is, $f(A) = \{f(a) \in Y \st a \in A\}\text{.}$ We can do this in the other direction as well. We might ask which elements of the domain get mapped to a particular set in the codomain. Let $f:X \to Y$ be a function and suppose $B \subseteq Y$ is a subset of the codomain. Then we will write $f\inv(B)$ for the inverse image of $B$ under $f$, namely the set of elements in $X$ whose image are elements in $B\text{.}$ In other words, $f\inv(B) = \{x \in X \st f(x) \in B\}\text{.}$ Often we are interested in the element(s) whose image is a particular element $y$ of in the codomain. The notation above works: $f\inv(\{y\})$ is the set of all elements in the domain that $f$ sends to $y\text{.}$ It makes sense to think of this as a set: there might not be anything sent to $y$ (if $y$ is not in the range), in which case $f\inv(\{y\}) = \emptyset\text{.}$ Or $f$ might send multiple elements to $y$ (if $f$ is not injective). As a notational convenience, we usually drop the set braces around the $y$ and write $f\inv(y)$ instead for this set. WARNING: $f\inv(y)$ is not an inverse function! Inverse functions only exist for bijections, but $f\inv(y)$ is defined for any function $f\text{.}$ The point: $f\inv(y)$ is a set, not an element of the domain. This is just sloppy notation for $f\inv(\{y\})\text{.}$ To help make this distinction, we would call $f\inv(y)$ the complete inverse image of $y$ under $f$. It is not the image of $y$ under $f\inv$ (since the function $f\inv$ might not exist). ###### Example0.4.8. Consider the function $f:\{1,2,3,4,5,6\} \to \{a,b,c,d\}$ given by \begin{equation*} f = \begin{pmatrix}1 \amp 2 \amp 3 \amp 4 \amp 5 \amp 6 \\ a \amp a \amp b \amp b \amp b \amp c\end{pmatrix}\text{.} \end{equation*} Find $f(\{1,2,3\})\text{,}$ $f\inv(\{a,b\})\text{,}$ and $f\inv(d)\text{.}$ Solution $f(\{1,2,3\}) = \{a,b\}$ since $a$ and $b$ are the elements in the codomain to which $f$ sends $1\text{,}$ $2\text{,}$ and $3\text{.}$ $f\inv(\{a,b\}) = \{1,2,3,4,5\}$ since these are exactly the elements that $f$ sends to $a$ and $b\text{.}$ $f\inv(d) = \emptyset$ since $d$ is not in the range of $f\text{.}$ ###### Example0.4.9. Consider the function $g:\Z \to \Z$ defined by $g(n) = n^2 + 1\text{.}$ Find $g(1)$ and $g(\{1\})\text{.}$ Then find $g\inv(1)\text{,}$ $g\inv(2)\text{,}$ and $g\inv(3)\text{.}$ Solution Note that $g(1) \ne g(\{1\})\text{.}$ The first is an element: $g(1) = 2\text{.}$ The second is a set: $g(\{1\}) = \{2\}\text{.}$ To find $g\inv(1)\text{,}$ we need to find all integers $n$ such that $n^2 + 1 = 1\text{.}$ Clearly only 0 works, so $g\inv(1) = \{0\}$ (note that even though there is only one element, we still write it as a set with one element in it). To find $g\inv(2)\text{,}$ we need to find all $n$ such that $n^2 + 1 = 2\text{.}$ We see $g\inv(2) = \{-1,1\}\text{.}$ Finally, if $n^2 + 1 = 3\text{,}$ then we are looking for an $n$ such that $n^2 = 2\text{.}$ There are no such integers so $g\inv(3) = \emptyset\text{.}$ Since $f\inv(y)$ is a set, it makes sense to ask for $\card{f\inv(y)}\text{,}$ the number of elements in the domain which map to $y\text{.}$ ###### Example0.4.10. Find a function $f:\{1,2,3,4,5\} \to \N$ such that $\card{f\inv(7)} = 5\text{.}$ Solution There is only one such function. We need five elements of the domain to map to the number $7 \in \N\text{.}$ Since there are only five elements in the domain, all of them must map to 7. So \begin{equation*} f = \begin{pmatrix}1 \amp 2 \amp 3 \amp 4 \amp 5 \\ 7 \amp 7 \amp 7 \amp 7 \amp 7\end{pmatrix}\text{.} \end{equation*} ##### Function Definitions. Here is a summary of all the main concepts and definitions we use when working with functions. • A function is a rule that assigns each element of a set, called the domain, to exactly one element of a second set, called the codomain. • Notation: $f:X \to Y$ is our way of saying that the function is called $f\text{,}$ the domain is the set $X\text{,}$ and the codomain is the set $Y\text{.}$ • To specify the rule for a function with small domain, use two-line notation by writing a matrix with each output directly below its corresponding input, as in: \begin{equation*} f = \begin{pmatrix}1 \amp 2 \amp 3 \amp 4 \\ 2 \amp 1 \amp 3 \amp 1 \end{pmatrix}\text{.} \end{equation*} • $f(x) = y$ means the element $x$ of the domain (input) is assigned to the element $y$ of the codomain. We say $y$ is an output. Alternatively, we call $y$ the image of $x$ under $f$. • The range is a subset of the codomain. It is the set of all elements which are assigned to at least one element of the domain by the function. That is, the range is the set of all outputs. • A function is injective (an injection or one-to-one) if every element of the codomain is the image of at most one element from the domain. • A function is surjective (a surjection or onto) if every element of the codomain is the image of at least one element from the domain. • A bijection is a function which is both an injection and surjection. In other words, if every element of the codomain is the image of exactly one element from the domain. • The image of an element $x$ in the domain is the element $y$ in the codomain that $x$ is mapped to. That is, the image of $x$ under $f$ is $f(x)\text{.}$ • The complete inverse image of an element $y$ in the codomain, written $f\inv(y)\text{,}$ is the set of all elements in the domain which are assigned to $y$ by the function. • The image of a subset $A$ of the domain is the set $f(A) = \{f(a) \in Y \st a \in A\}\text{.}$ • The inverse image of a subset $B$ of the codomain is the set $f\inv(B) = \{x \in X \st f(x) \in B\}\text{.}$ ### ExercisesExercises ###### 7. Consider the function $f:\{1,2,3,4,5\} \to \{1,2,3,4\}$ given by the table below: $x$ 1 2 3 4 5 $f(x)$ 3 2 4 1 2 1. Is $f$ injective? Explain. 2. Is $f$ surjective? Explain. 3. Write the function using two-line notation. Solution 1. $f$ is not injective, since $f(2) = f(5)\text{;}$ two different inputs have the same output. 2. $f$ is surjective, since every element of the codomain is an element of the range. 3. $f=\begin{pmatrix}1 \amp 2 \amp 3 \amp 4 \amp 5 \\ 3 \amp 2 \amp 4 \amp 1 \amp 2\end{pmatrix}\text{.}$ ###### 8. Consider the function $f:\{1,2,3,4\} \to \{1,2,3,4\}$ given by the graph below. 1. Is $f$ injective? Explain. 2. Is $f$ surjective? Explain. 3. Write the function using two-line notation. ###### 10. Suppose $f:\N \to \N$ satisfies the recurrence $f(n+1) = f(n) + 3\text{.}$ Note that this is not enough information to define the function, since we don’t have an initial condition. For each of the initial conditions below, find the value of $f(5)\text{.}$ 1. $\displaystyle f(0) = 0\text{.}$ 2. $\displaystyle f(0) = 1\text{.}$ 3. $\displaystyle f(0) = 2\text{.}$ 4. $\displaystyle f(0) = 100\text{.}$ Solution For each case, you must use the recurrence to find $f(1)\text{,}$ $f(2)$ ... $f(5)\text{.}$ But notice each time you just add three to the previous. We do this 5 times. 1. $\displaystyle f(5) = 15\text{.}$ 2. $\displaystyle f(5) = 16\text{.}$ 3. $\displaystyle f(5) = 17\text{.}$ 4. $\displaystyle f(5) = 115\text{.}$ ###### 11. Suppose $f:\N \to \N$ satisfies the recurrence relation \begin{equation*} f(n+1) = \begin{cases} \frac{f(n)}{2} \amp \text{ if } f(n) \text{ is even} \\ 3f(n) + 1 \amp \text{ if } f(n) \text{ is odd}\end{cases}\text{.} \end{equation*} Note that with the initial condition $f(0) = 1\text{,}$ the values of the function are: $f(1) = 4\text{,}$ $f(2) = 2\text{,}$ $f(3) = 1\text{,}$ $f(4) = 4\text{,}$ and so on, the images cycling through those three numbers. Thus $f$ is NOT injective (and also certainly not surjective). Might it be under other initial conditions? 3 1. If $f$ satisfies the initial condition $f(0) = 5\text{,}$ is $f$ injective? Explain why or give a specific example of two elements from the domain with the same image. 2. If $f$ satisfies the initial condition $f(0) = 3\text{,}$ is $f$ injective? Explain why or give a specific example of two elements from the domain with the same image. 3. If $f$ satisfies the initial condition $f(0) = 27\text{,}$ then it turns out that $f(105) = 10$ and no two numbers less than 105 have the same image. Could $f$ be injective? Explain. 4. Prove that no matter what initial condition you choose, the function cannot be surjective. It turns out this is a really hard question to answer in general. The Collatz conjecture is that no matter what the initial condition is, the function will eventually produce 1 as an output. This is an open problem in mathematics: nobody knows the answer. ###### 12. For each function given below, determine whether or not the function is injective and whether or not the function is surjective. 1. $f:\N \to \N$ given by $f(n) = n+4\text{.}$ 2. $f:\Z \to \Z$ given by $f(n) = n+4\text{.}$ 3. $f:\Z \to \Z$ given by $f(n) = 5n - 8\text{.}$ 4. $f:\Z \to \Z$ given by $f(n) = \begin{cases}n/2 \amp \text{ if } n \text{ is even} \\ (n+1)/2 \amp \text{ if } n \text{ is odd} . \end{cases}$ Solution 1. $f$ is injective, but not surjective (since 0, for example, is never an output). 2. $f$ is injective and surjective. Unlike in the previous question, every integers is an output (of the integer 4 less than it). 3. $f$ is injective, but not surjective (10 is not 8 less than a multiple of 5, for example). 4. $f$ is not injective, but is surjective. Every integer is an output (of twice itself, for example) but some integers are outputs of more than one input: $f(5) = 3 = f(6)\text{.}$ ###### 13. Let $A = \{1,2,3,\ldots,10\}\text{.}$ Consider the function $f:\pow(A) \to \N$ given by $f(B) = |B|\text{.}$ That is, $f$ takes a subset of $A$ as an input and outputs the cardinality of that set. 1. Is $f$ injective? Prove your answer. 2. Is $f$ surjective? Prove your answer. 3. Find $f\inv(1)\text{.}$ 4. Find $f\inv(0)\text{.}$ 5. Find $f\inv(12)\text{.}$ Solution 1. $f$ is not injective. To prove this, we must simply find two different elements of the domain which map to the same element of the codomain. Since $f(\{1\}) = 1$ and $f(\{2\}) = 1\text{,}$ we see that $f$ is not injective. 2. $f$ is not surjective. The largest subset of $A$ is $A$ itself, and $|A| = 10\text{.}$ So no natural number greater than 10 will ever be an output. 3. $f\inv(1) = \{\{1\}, \{2\}, \{3\}, \ldots \{10\}\}$ (the set of all the singleton subsets of $A$). 4. $f\inv(0) = \{\emptyset\}\text{.}$ Note, it would be wrong to write $f\inv(0) = \emptyset$ - that would claim that there is no input which has 0 as an output. 5. $f\inv(12) = \emptyset\text{,}$ since there are no subsets of $A$ with cardinality 12. ###### 15. Consider the set $\N^2 = \N \times \N\text{,}$ the set of all ordered pairs $(a,b)$ where $a$ and $b$ are natural numbers. Consider a function $f: \N^2 \to \N$ given by $f((a,b)) =a+b\text{.}$ 1. Let $A = \{(a,b) \in \N^2 \st a, b \le 10\}\text{.}$ Find $f(A)\text{.}$ 2. Find $f\inv(3)$ and $f\inv(\{0,1,2,3\})\text{.}$ 3. Give geometric descriptions of $f\inv(n)$ and $f\inv(\{0, 1, \ldots, n\})$ for any $n \ge 1\text{.}$ 4. Find $\card{f\inv(8)}$ and $\card{f\inv(\{0,1, \ldots, 8\})}\text{.}$ ###### 16. Let $f:X \to Y$ be some function. Suppose $3 \in Y\text{.}$ What can you say about $f\inv(3)$ if you know, 1. $f$ is injective? Explain. 2. $f$ is surjective? Explain. 3. $f$ is bijective? Explain. Solution 1. $|f\inv(3)| \le 1\text{.}$ In other words, either $f\inv(3)$ is the empty set or is a set containing exactly one element. Injective functions cannot have two elements from the domain both map to 3. 2. $|f\inv(3)| \ge 1\text{.}$ In other words, $f\inv(3)$ is a set containing at least one elements, possibly more. Surjective functions must have something map to 3. 3. $|f\inv(3)| = 1\text{.}$ There is exactly one element from $X$ which gets mapped to 3, so $f\inv(3)$ is the set containing that one element. ###### 17. Find a set $X$ and a function $f:X \to \N$ so that $f\inv(0) \cup f\inv(1) = X\text{.}$ Solution $X$ can really be any set, as long as $f(x) = 0$ or $f(x) = 1$ for every $x \in X\text{.}$ For example, $X = \N$ and $f(n) = 0$ works. ###### 18. What can you deduce about the sets $X$ and $Y$ if you know, 1. there is an injective function $f:X \to Y\text{?}$ Explain. 2. there is a surjective function $f:X \to Y\text{?}$ Explain. 3. there is a bijective function $f:X \to Y\text{?}$ Explain. ###### 19. Suppose $f:X \to Y$ is a function. Which of the following are possible? Explain. 1. $f$ is injective but not surjective. 2. $f$ is surjective but not injective. 3. $|X| = |Y|$ and $f$ is injective but not surjective. 4. $|X| = |Y|$ and $f$ is surjective but not injective. 5. $|X| = |Y|\text{,}$ $X$ and $Y$ are finite, and $f$ is injective but not surjective. 6. $|X| = |Y|\text{,}$ $X$ and $Y$ are finite, and $f$ is surjective but not injective. ###### 20. Let $f:X \to Y$ and $g:Y \to Z$ be functions. We can define the composition of $f$ and $g$ to be the function $g\circ f:X \to Z$ for which the image of each $x \in X$ is $g(f(x))\text{.}$ That is, plug $x$ into $f\text{,}$ then plug the result into $g$ (just like composition in algebra and calculus). 1. If $f$ and $g$ are both injective, must $g\circ f$ be injective? Explain. 2. If $f$ and $g$ are both surjective, must $g\circ f$ be surjective? Explain. 3. Suppose $g\circ f$ is injective. What, if anything, can you say about $f$ and $g\text{?}$ Explain. 4. Suppose $g\circ f$ is surjective. What, if anything, can you say about $f$ and $g\text{?}$ Explain. Hint Work with some examples. What if $f = \twoline{1\amp 2 \amp 3}{a \amp a \amp b}$ and $g = \twoline{a\amp b \amp c}{5 \amp 6 \amp 7}\text{?}$ ###### 21. Consider the function $f:\Z \to \Z$ given by $f(n) = \begin{cases}n+1 \amp \text{ if }n\text{ is even} \\ n-3 \amp \text{ if }n\text{ is odd} . \end{cases}$ 1. Is $f$ injective? Prove your answer. 2. Is $f$ surjective? Prove your answer. Solution 1. $f$ is injective. ###### Proof. Let $x$ and $y$ be elements of the domain $\Z\text{.}$ Assume $f(x) = f(y)\text{.}$ If $x$ and $y$ are both even, then $f(x) = x+1$ and $f(y) = y+1\text{.}$ Since $f(x) = f(y)\text{,}$ we have $x + 1 = y + 1$ which implies that $x = y\text{.}$ Similarly, if $x$ and $y$ are both odd, then $x - 3 = y-3$ so again $x = y\text{.}$ The only other possibility is that $x$ is even an $y$ is odd (or visa-versa). But then $x + 1$ would be odd and $y - 3$ would be even, so it cannot be that $f(x) = f(y)\text{.}$ Therefore if $f(x) = f(y)$ we then have $x = y\text{,}$ which proves that $f$ is injective. 2. $f$ is surjective. ###### Proof. Let $y$ be an element of the codomain $\Z\text{.}$ We will show there is an element $n$ of the domain ($\Z$) such that $f(n) = y\text{.}$ There are two cases: First, if $y$ is even, then let $n = y+3\text{.}$ Since $y$ is even, $n$ is odd, so $f(n) = n-3 = y+3-3 = y$ as desired. Second, if $y$ is odd, then let $n = y-1\text{.}$ Since $y$ is odd, $n$ is even, so $f(n) = n+1 = y-1+1 = y$ as needed. Therefore $f$ is surjective. ###### 22. At the end of the semester a teacher assigns letter grades to each of her students. Is this a function? If so, what sets make up the domain and codomain, and is the function injective, surjective, bijective, or neither? Solution Yes, this is a function, if you choose the domain and codomain correctly. The domain will be the set of students, and the codomain will be the set of possible grades. The function is almost certainly not injective, because it is likely that two students will get the same grade. The function might be surjective – it will be if there is at least one student who gets each grade. ###### 23. In the game of Hearts, four players are each dealt 13 cards from a deck of 52. Is this a function? If so, what sets make up the domain and codomain, and is the function injective, surjective, bijective, or neither? ###### 24. Seven players are playing 5-card stud. Each player initially receives 5 cards from a deck of 52. Is this a function? If so, what sets make up the domain and codomain, and is the function injective, surjective, bijective, or neither? Solution This is not a function. ###### 25. Consider the function $f:\N \to \N$ that gives the number of handshakes that take place in a room of $n$ people assuming everyone shakes hands with everyone else. Give a recursive definition for this function. Hint To find the recurrence relation, consider how many new handshakes occur when person $n+1$ enters the room. Solution The recurrence relation is $f(n+1) = f(n) + n\text{.}$ ###### 26. Let $f:X \to Y$ be a function and $A \subseteq X$ be a finite subset of the domain. What can you say about the relationship between $\card{A}$ and $\card{f(A)}\text{?}$ Consider both the general case and what happens when you know $f$ is injective, surjective, or bijective. Solution In general, $\card{A} \ge \card{f(A)}\text{,}$ since you cannot get more outputs than you have inputs (each input goes to exactly one output), but you could have fewer outputs if the function is not injective. If the function is injective, then $\card{A} = \card{f(A)}\text{,}$ although you can have equality even if $f$ is not injective (it must be injective restricted to $A$). ###### 27. Let $f:X \to Y$ be a function and $B \subseteq Y$ be a finite subset of the codomain. What can you say about the relationship between $\card{B}$ and $\card{f\inv(B)}\text{?}$ Consider both the general case and what happens when you know $f$ is injective, surjective, or bijective. Solution In general, there is no relationship between $\card{B}$ and $\card{f\inv(B)}\text{.}$ This is because $B$ might contain elements that are not in the range of $f\text{,}$ so we might even have $f\inv(B) = \emptyset\text{.}$ On the other hand, there might be lots of elements from the domain that all get sent to a few elements in $B\text{,}$ making $f\inv(B)$ larger than $B\text{.}$ More specifically, if $f$ is injective, then $\card{B} \ge \card{f\inv(B)}$ (since every element in $B$ must come from at most one element from the domain). If $f$ is surjective, then $\card{B} \le \card{f\inv(B)}$ (since every element in $B$ must come from at least one element of the domain). Thus if $f$ is bijective then $\card{B} = \card{f\inv(B)}\text{.}$ ###### 28. Let $f:X \to Y$ be a function, $A \subseteq X$ and $B \subseteq Y\text{.}$ 1. Is $f\inv\left(f(A)\right) = A\text{?}$ Always, sometimes, never? Explain. 2. Is $f\left(f\inv(B)\right) = B\text{?}$ Always, sometimes, never? Explain. 3. If one or both of the above do not always hold, is there something else you can say? Will equality always hold for particular types of functions? Is there some other relationship other than equality that would always hold? Explore. ###### 29. Let $f:X \to Y$ be a function and $A, B \subseteq X$ be subsets of the domain. 1. Is $f(A \cup B) = f(A) \cup f(B)\text{?}$ Always, sometimes, or never? Explain. 2. Is $f(A \cap B) = f(A) \cap f(B)\text{?}$ Always, sometimes, or never? Explain. Hint One of these is not always true. Try some examples! ###### 30. Let $f:X \to Y$ be a function and $A, B \subseteq Y$ be subsets of the codomain. 1. Is $f\inv(A \cup B) = f\inv(A) \cup f\inv(B)\text{?}$ Always, sometimes, or never? Explain. 2. Is $f\inv(A \cap B) = f\inv(A) \cap f\inv(B)\text{?}$ Always, sometimes, or never? Explain.

http://mathonline.wikidot.com/topological-quotients-in-euclidean-space

Topological Quotients in Euclidean Space # Topological Quotients in Euclidean Space Recall from the Topological Quotients page that if $(X, \tau)$ is a topological space and $\sim$ is an equivalence relation on $X$ where for all $x \in X$, $[x] = \{ y \in X : x \sim y \}$ denotes the equivalence class of $x$ with respect to the relation $\sim$ and $X \: / \sim$ denotes the set of all equivalences classes on $X$, then the Quotient Topology is the final topology induced by the canonical quotient map $q : X \to X \: / \sim$. We will now look at some examples of quotient topological spaces in Euclidean space with the respective usual topologies. Consider the set $X = [0, 1]$. Let $\tau_X$ is the usual topology consisting of open intervals from $X$ and consider the topological space $(X, \tau_x)$. Define an equivalence relation $\sim$ by $x \sim y$ if $x = y$ OR $x \sim y$ if $\{ x, y \} = \{0, 1 \}$. This equivalence relation can be thought of as "gluing" the endpoints of the closed interval $X$ together to form a circle: Note that $\sim$ is indeed an equivalence relation. It is reflexive because $x \sim x$, symmetric since $x \sim y$ if and only if $y \sim x$, and transitive since $x \sim y$ and $y \sim z$ implies that $x \sim z$. Now consider the set of equivalence classes $X \: / \sim$: (1) \begin{align} \quad X \: / \sim \: = \{ \{ x \} : x \in [0, 1] \} \cup \{ \{0, 1\} \} \end{align} Consider the canonical quotient map $q : X \to X \: / \sim$ defined for all $x \in [0, 1]$ by $q(x) = [x]$. Then the final topology induced by $q$ is given by: (2) \begin{align} \quad \tau_{X / \sim} = \{ q^{-1}(U) : U \in \tau_X \} \end{align} So the topology on $\tau_{X / \sim}$ can be described somewhat vaguely as unions of "open arcs" not containing $0$ or $1$ as illustrated below: For another example of a quotient topological space, let $X = [0, 1] \times [0, 1]$. Consider the topological space $([0, 1] \times [0, 1], \tau_X)$ where $\tau_X$ is the usual topology whose open sets are generated by open disks in $\mathbb{R}^2$. Define an equivalence relation $\sim$ on $X$ by $(x, y) \sim (w, z)$ if $(x, y) = (w, z)$, $y = z$ and $\{x, w \} = \{0 , 1 \}$, or $x = w$ and $\{ y, z \} = \{0 , 1 \}$. By "gluing" equivalent points together, we can visualize $X \: / \sim$ with the following diagram: The topology on $X \: / \sim$ will be generated by "open subsurfaces" of the torus above that do not intersect the "glue-lines".

http://lousodrome.net/blog/light/tag/ray/

# Intersection of a ray and a cone Some time ago I needed to solve analytically the intersection of a ray and a cone. I was surprised to see that there are not that many resources available; there are some, but not nearly as many as on the intersection of a ray and a sphere for example. Add to it that they all use their own notation and that I lack math exercise, after a bit of browsing I decided I needed to write a proof by myself to get a good grasp of the result. So here goes, the solution to the intersection of a ray and a cone, in vector notation. 1. We define a ray with its origin $O$ and its direction as a unit vector $\hat{D}$. Any point $X$ on the ray at a signed distance $t$ from the origin of the ray verifies: $\vec{X} = \vec{O} + t\vec{D}$. When $t$ is positive $X$ is in the direction of the ray, and when $t$ is negative $X$ is in the opposite direction. 2. We define a cone with its tip $C$, its axis as a unit vector $\hat{V}$ in the direction of increasing radius, and $\theta$ the half angle between the axis and the surface. Any point $X$ on the cone verifies: $(\vec{X} – \vec{C}) \cdot \vec{V} = \lVert \vec{X} – \vec{C} \rVert \cos\theta$ 3. Finally we define $P$ the intersection or the ray and the cone, and which we are interested in finding. $P$ verifies both equations, so we can write: $$\left\{ \begin{array}{l} \vec{P}=\vec{O} + t\vec{D} \\ \frac{ \vec{P} – \vec{C} }{\lVert \vec{P} – \vec{C} \rVert} \vec{V} = \cos\theta \end{array} \right.$$ We can multiply the second equation by itself to work with it, then reorder things a bit. $$\left\{ \begin{array}{l} \vec{P}=\vec{O} + t\vec{D} \\ \frac{ ((\vec{P} – \vec{C}) \cdot \vec{V})^2 }{ (\vec{P} – \vec{C}) \cdot (\vec{P} – \vec{C}) } = \cos^2\theta \end{array} \right.$$ $$\left\{ \begin{array}{l} \vec{P}=\vec{O} + t\vec{D} \\ ((\vec{P} – \vec{C}) \cdot \vec{V})^2 – (\vec{P} – \vec{C}) \cdot (\vec{P} – \vec{C}) \cos^2\theta = 0 \end{array} \right.$$ Remember the mouthful earlier about $\hat{V}$ being in the direction of increasing radius? By elevating $\cos\theta$ to square, we’re making negative values of $\cos$ positive: values of $\theta$ beyond 90° become indistinguishable from values below 90°. This has the side effect of turning it into the equation of not one, but two cones sharing the same axis, tip and angle, but in opposite directions. We’ll fix that later. We replace $\vec{P}$ with $\vec{O} + t\vec{D}$ and work the equation until we get a good old quadratic function that we can solve. $$\require{cancel} ((\vec{O} + t\vec{D} – \vec{C})\cdot\vec{V})^2 – (\vec{O} + t\vec{D} – \vec{C}) \cdot (\vec{O} + t\vec{D} – \vec{C}) \cos^2\theta = 0$$ $$((t\vec{D} + \vec{CO})\cdot\vec{V})^2 – (t\vec{D} + \vec{CO}) \cdot (t\vec{D} + \vec{CO}) \cos^2\theta = 0$$ $$(t\vec{D}\cdot\vec{V} + \vec{CO}\cdot\vec{V})^2 – (t^2\cancel{\vec{D}\cdot\vec{D}} + 2t\vec{D}\cdot\vec{CO} + \vec{CO}\cdot\vec{CO}) \cos^2\theta = 0$$ $$(t^2(\vec{D}\cdot\vec{V})^2 + 2t(\vec{D}\cdot\vec{V})(\vec{CO}\cdot\vec{V}) + (\vec{CO}\cdot\vec{V})^2) – (t^2 + 2t\vec{D}\cdot\vec{CO} + \vec{CO}\cdot\vec{CO}) \cos^2\theta = 0$$ $$t^2(\vec{D}\cdot\vec{V})^2 + 2t(\vec{D}\cdot\vec{V})(\vec{CO}\cdot\vec{V}) + (\vec{CO}\cdot\vec{V})^2 – t^2\cos^2\theta – 2t\vec{D}\cdot\vec{CO}\cos^2\theta – \vec{CO}\cdot\vec{CO}\cos^2\theta = 0$$ Reorder a bit: $$t^2((\vec{D}\cdot\vec{V})^2 – \cos^2\theta) + 2t((\vec{D}\cdot\vec{V})(\vec{CO}\cdot\vec{V}) – \vec{D}\cdot\vec{CO}\cos^2\theta) + (\vec{CO}\cdot\vec{V})^2 – \vec{CO}\cdot\vec{CO}\cos^2\theta = 0$$ There we go, we have our $at^2 + bt + c = 0$ equation, with: $$\left\{ \begin{array}{l} a = (\vec{D}\cdot\vec{V})^2 – \cos^2\theta \\ b = 2\Big((\vec{D}\cdot\vec{V})(\vec{CO}\cdot\vec{V}) – \vec{D}\cdot\vec{CO}\cos^2\theta\Big) \\ c = (\vec{CO}\cdot\vec{V})^2 – \vec{CO}\cdot\vec{CO}\cos^2\theta \end{array} \right.$$ From there, you know the drill: calculate the determinant $\Delta = b^2 – 4ac$ then depending on its value: • If $\Delta < 0$, the ray is not intersecting the cone. • If $\Delta = 0$, the ray is intersecting the cone once at $t = \frac{-b}{2a}$. • If $\Delta > 0$, the ray is intersecting the cone twice, at $t_1 = \frac{-b – \sqrt{\Delta}}{2a}$ and $t_2 = \frac{-b + \sqrt{\Delta}}{2a}$. But wait! We don’t have one cone but two, so we have to reject solutions that intersect with the shadow cone. $P$ must still verify $\frac{ \vec{P} – \vec{C} }{\lVert \vec{P} – \vec{C} \rVert} \vec{V} = \cos\theta$, or simply, if $\theta < 90°$: $(\vec{P} – \vec{C})\cdot\vec{V} > 0$. Note that there is also the corner case of the ray tangent to the cone and having an infinity of solutions to consider. I’ve completely swept it under the rug since it doesn’t matter in the context I was, but if it does to you, you’ve been warned about it. Also remember to check the sign of $t$ to know whether $P$ is in the direction of the ray. You may need to determine which of $t_1$ or $t_2$ you want to use, which depends on your use case. For example is your ray origin inside or outside of the cone? Now for a little sanity test, let’s consider the corner case $C=O$, where the ray origin is the tip of the cone (thanks Rubix for the suggestion!). We have $b=0$ and $c=0$ thus $\Delta=0$ and $t=\frac{-b}{2a}=0$ which is the expected result. I also tried the cases $\theta=0$ and $\theta=\pi/2$, but expanding $\Delta$ proved too tedious to proceed to the end. So this is left as an exercise, as they say. :) Finally, to demonstrate that the result is indeed correct, here is a glorious ray traced cone scene on ShaderToy: I hope this can prove useful to others too. Oh, and Happy New Year by the way!

https://en.wikipedia.org/wiki/Laguerre_polynomials

Laguerre polynomials In mathematics, the Laguerre polynomials, named after Edmond Laguerre (1834 - 1886), are solutions of Laguerre's equation: $xy'' + (1 - x)y' + ny = 0$ which is a second-order linear differential equation. This equation has nonsingular solutions only if n is a non-negative integer. More generally, the name Laguerre polynomials is used for solutions of $xy'' + (\alpha+1 - x)y' + ny = 0~.$ Then they are also named generalized Laguerre polynomials, as will be done here (alternatively associated Laguerre polynomials or, rarely, Sonin polynomials, after their inventor[1] Nikolay Yakovlevich Sonin). The Laguerre polynomials are also used for Gaussian quadrature to numerically compute integrals of the form $\int_0^\infty f(x) e^{-x} \, dx.$ These polynomials, usually denoted L0L1, ..., are a polynomial sequence which may be defined by the Rodrigues formula, $L_n(x)=\frac{e^x}{n!}\frac{d^n}{dx^n}\left(e^{-x} x^n\right) =\frac{1}{n!} \left( \frac{d}{dx} -1 \right) ^n x^n,$ reducing to the closed form of a following section. They are orthogonal polynomials with respect to an inner product $\langle f,g \rangle = \int_0^\infty f(x) g(x) e^{-x}\,dx.$ The sequence of Laguerre polynomials n! Ln is a Sheffer sequence, $\frac{d}{dx} L_n = \left ( \frac{d}{dx} - 1 \right ) L_{n-1}.$ The Rook polynomials in combinatorics are more or less the same as Laguerre polynomials, up to elementary changes of variables. Further see the Tricomi–Carlitz polynomials. The Laguerre polynomials arise in quantum mechanics, in the radial part of the solution of the Schrödinger equation for a one-electron atom. They also describe the static Wigner functions of oscillator systems in quantum mechanics in phase space. They further enter in the quantum mechanics of the Morse potential and of the 3D isotropic harmonic oscillator. Physicists sometimes use a definition for the Laguerre polynomials which is larger by a factor of n! than the definition used here. (Likewise, some physicists may use somewhat different definitions of the so-called associated Laguerre polynomials.) The first few polynomials These are the first few Laguerre polynomials: n $L_n(x)\,$ 0 $1\,$ 1 $-x+1\,$ 2 ${\scriptstyle\frac{1}{2}} (x^2-4x+2) \,$ 3 ${\scriptstyle\frac{1}{6}} (-x^3+9x^2-18x+6) \,$ 4 ${\scriptstyle\frac{1}{24}} (x^4-16x^3+72x^2-96x+24) \,$ 5 ${\scriptstyle\frac{1}{120}} (-x^5+25x^4-200x^3+600x^2-600x+120) \,$ 6 ${\scriptstyle\frac{1}{720}} (x^6-36x^5+450x^4-2400x^3+5400x^2-4320x+720) \,$ The first six Laguerre polynomials. Recursive definition, closed form, and generating function One can also define the Laguerre polynomials recursively, defining the first two polynomials as $L_0(x) = 1$ $L_1(x) = 1 - x$ and then using the following recurrence relation for any k ≥ 1: $L_{k + 1}(x) = \frac{(2k + 1 - x)L_k(x) - k L_{k - 1}(x)}{k + 1}.$ The closed form is $L_n(x)=\sum_{k=0}^n \binom{n}{k}\frac{(-1)^k}{k!} x^k .$ The generating function for them likewise follows, $\sum_n^\infty t^n L_n(x)= \frac{1}{1-t} e^{-\frac{tx}{1-t}}.$ Polynomials of negative index can be expressed using the ones with positive index: $L_{-n}(x)=e^xL_{n-1}(-x).$ Generalized Laguerre polynomials For arbitrary real α the polynomial solutions of the differential equation [2] $x\,y'' + (\alpha +1 - x)\,y' + n\,y = 0$ are called generalized Laguerre polynomials, or associated Laguerre polynomials. One can also define the generalized Laguerre polynomials recursively, defining the first two polynomials as $L^{(\alpha)}_0(x) = 1$ $L^{(\alpha)}_1(x) = 1 + \alpha - x$ and then using the following recurrence relation for any k ≥ 1: $L^{(\alpha)}_{k + 1}(x) = \frac{(2k + 1 + \alpha - x)L^{(\alpha)}_k(x) - (k + \alpha) L^{(\alpha)}_{k - 1}(x)}{k + 1}.$ The simple Laguerre polynomials are the special case α = 0 of the generalized Laguerre polynomials: $L^{(0)}_n(x)=L_n(x).$ The Rodrigues formula for them is \begin{align} L_n^{(\alpha)}(x) &= {x^{-\alpha} e^x \over n!}{d^n \over dx^n} \left(e^{-x} x^{n+\alpha}\right) \\ &= x^{-\alpha} \frac{( \frac{d}{dx}-1) ^n}{n!}x^{n+\alpha}. \end{align} The generating function for them is $\sum_n^\infty t^n L^{(\alpha)}_n(x)= \frac{1}{(1-t)^{\alpha+1}} e^{-\frac{tx}{1-t}}.$ The first few generalized Laguerre polynomials Explicit examples and properties of the generalized Laguerre polynomials $L_n^{(\alpha)}(x) := {n+ \alpha \choose n} M(-n,\alpha+1,x).$ When n is an integer the function reduces to a polynomial of degree n. It has the alternative expression[4] $L_n^{(\alpha)}(x)= \frac {(-1)^n}{n!} U(-n,\alpha+1,x)$ in terms of Kummer's function of the second kind. • The closed form for these generalized Laguerre polynomials of degree n is[5] $L_n^{(\alpha)} (x) = \sum_{i=0}^n (-1)^i {n+\alpha \choose n-i} \frac{x^i}{i!}$ derived by applying Leibniz's theorem for differentiation of a product to Rodrigues' formula. • The first few generalized Laguerre polynomials are: \begin{align} L_0^{(\alpha)}(x) &= 1 \\ L_1^{(\alpha)}(x) &= -x + \alpha +1 \\ L_2^{(\alpha)}(x) &= \frac{x^2}{2} - (\alpha + 2)x + \frac{(\alpha+2)(\alpha+1)}{2} \\ L_3^{(\alpha)}(x) &= \frac{-x^3}{6} + \frac{(\alpha+3)x^2}{2} -\frac{(\alpha+2)(\alpha+3)x}{2} +\frac{(\alpha+1)(\alpha+2)(\alpha+3)}{6} \end{align} $L_n^{(\alpha)}(0)= {n+\alpha\choose n} \approx \frac{n^\alpha}{\Gamma(\alpha+1)};$ • Ln(α) has n real, strictly positive roots (notice that $\left((-1)^{n-i} L_{n-i}^{(\alpha)}\right)_{i=0}^n$ is a Sturm chain), which are all in the interval $\left( 0, n+\alpha+ (n-1) \sqrt{n+\alpha} \right].$[citation needed] • The polynomials' asymptotic behaviour for large n, but fixed α and x > 0, is given by[6][7] $L_n^{(\alpha)}(x) = \frac{n^{\frac{\alpha}{2}-\frac{1}{4}}}{\sqrt{\pi}} \frac{e^{\frac{x}{2}}}{x^{\frac{\alpha}{2}+\frac{1}{4}}} \cos\left(2 \sqrt{nx}- \frac{\pi}{2}\left(\alpha+\frac{1}{2} \right) \right)+O\left(n^{\frac{\alpha}{2}-\frac{3}{4}}\right),$ $L_n^{(\alpha)}(-x) = \frac{(n+1)^{\frac{\alpha}{2}-\frac{1}{4}}}{2\sqrt{\pi}} \frac{e^{-\frac{x}{2}}}{x^{\frac{\alpha}{2}+\frac{1}{4}}} e^{2 \sqrt{x(n+1)}} \cdot\left(1+O\left(\frac{1}{\sqrt{n+1}}\right)\right),$ and summarizing by $\frac{L_n^{(\alpha)}\left(\frac x n\right)}{n^\alpha}\approx e^\frac x {2n}\cdot\frac{J_\alpha\left(2\sqrt x\right)}{\sqrt x^\alpha},$ where $J_\alpha$ is the Bessel function. As a contour integral Given the generating function specified above, the polynomials may be expressed in terms of a contour integral $L_n^{(\alpha)}(x)=\frac{1}{2\pi i}\oint\frac{e^{-\frac{x t}{1-t}}}{(1-t)^{\alpha+1}\,t^{n+1}} \; dt,$ where the contour circles the origin once in a counterclockwise direction. Recurrence relations The addition formula for Laguerre polynomials:[8] $L_n^{(\alpha+\beta+1)}(x+y)= \sum_{i=0}^n L_i^{(\alpha)}(x) L_{n-i}^{(\beta)}(y)$. Laguerre's polynomials satisfy the recurrence relations $L_n^{(\alpha)}(x)= \sum_{i=0}^n L_{n-i}^{(\alpha+i)}(y)\frac{(y-x)^i}{i!},$ in particular $L_n^{(\alpha+1)}(x)= \sum_{i=0}^n L_i^{(\alpha)}(x)$ and $L_n^{(\alpha)}(x)= \sum_{i=0}^n {\alpha-\beta+n-i-1 \choose n-i} L_i^{(\beta)}(x),$ or $L_n^{(\alpha)}(x)=\sum_{i=0}^n {\alpha-\beta+n \choose n-i} L_i^{(\beta- i)}(x);$ moreover \begin{align} L_n^{(\alpha)}(x)- \sum_{j=0}^{\Delta-1} {n+\alpha \choose n-j} (-1)^j \frac{x^j}{j!}&= (-1)^\Delta\frac{x^\Delta}{(\Delta-1)!} \sum_{i=0}^{n-\Delta} \frac{{n+\alpha \choose n-\Delta-i}}{(n-i){n \choose i}}L_i^{(\alpha+\Delta)}(x)\\[6pt] &=(-1)^\Delta\frac{x^\Delta}{(\Delta-1)!} \sum_{i=0}^{n-\Delta} \frac{{n+\alpha-i-1 \choose n-\Delta-i}}{(n-i){n \choose i}}L_i^{(n+\alpha+\Delta-i)}(x) \end{align} They can be used to derive the four 3-point-rules \begin{align} L_n^{(\alpha)}(x) &= L_n^{(\alpha+1)}(x) - L_{n-1}^{(\alpha+1)}(x) = \sum_{j=0}^k {k \choose j} L_{n-j}^{(\alpha-k+j)}(x), \\[10pt] n L_n^{(\alpha)}(x) &= (n + \alpha )L_{n-1}^{(\alpha)}(x) - x L_{n-1}^{(\alpha+1)}(x), \\[10pt] & \text{or } \\ \frac{x^k}{k!}L_n^{(\alpha)}(x) &= \sum_{i=0}^k (-1)^i {n+i \choose i} {n+\alpha \choose k-i} L_{n+i}^{(\alpha-k)}(x), \\[10pt] n L_n^{(\alpha+1)}(x) &= (n-x) L_{n-1}^{(\alpha+1)}(x) + (n+\alpha)L_{n-1}^{(\alpha)}(x) \\[10pt] x L_n^{(\alpha+1)}(x) &= (n+\alpha)L_{n-1}^{(\alpha)}(x)-(n-x)L_n^{(\alpha)}(x); \end{align} combined they give this additional, useful recurrence relations \begin{align} L_n^{(\alpha)}(x)&= \left(2+\frac{\alpha-1-x}n \right)L_{n-1}^{(\alpha)}(x)- \left(1+\frac{\alpha-1}n \right)L_{n-2}^{(\alpha)}(x)\\[10pt] &= \frac{\alpha+1-x}n L_{n-1}^{(\alpha+1)}(x)- \frac x n L_{n-2}^{(\alpha+2)}(x) \end{align} Since $L_n^{(\alpha)}(x)$ is a monic polynomial of degree $n$ in $\alpha$, there is the partial fraction decomposition \begin{align} \frac{n!\,L_n^{(\alpha)}(x)}{(\alpha+1)_n} &= 1- \sum_{j=1}^n (-1)^j \frac{j}{\alpha + j} {n \choose j}L_n^{(-j)}(x) \\ &= 1- \sum_{j=1}^n \frac{x^j}{\alpha + j}\,\,\frac{L_{n-j}^{(j)}(x)}{(j-1)!} \\ &= 1-x \sum_{i=1}^n \frac{L_{n-i}^{(-\alpha)}(x) L_{i-1}^{(\alpha+1)}(-x)}{\alpha +i}. \end{align} The second equality follows by the following identity, valid for integer i and n and immediate from the expression of $L_n^{(\alpha)}(x)$ in terms of Charlier polynomials: $\frac{(-x)^i}{i!} L_n^{(i-n)}(x) = \frac{(-x)^n}{n!} L_i^{(n-i)}(x).$ For the third equality apply the fourth and fifth identities of this section. Derivatives of generalized Laguerre polynomials Differentiating the power series representation of a generalized Laguerre polynomial k times leads to $\frac{d^k}{d x^k} L_n^{(\alpha)} (x) = (-1)^k L_{n-k}^{(\alpha+k)} (x).$ This points to a special case (α = 0) of the formula above: for integer α = k the generalized polynomial may be written $L_n^{(k)}(x)=(-1)^k\frac{d^kL_{n+k}(x)}{dx^k},$ the shift by k sometimes causing confusion with the usual parenthesis notation for a derivative. Moreover, this following equation holds $\frac{1}{k!} \frac{d^k}{d x^k} x^\alpha L_n^{(\alpha)} (x) = {n+\alpha \choose k} x^{\alpha-k} L_n^{(\alpha-k)}(x),$ which generalizes with Cauchy's formula to $L_n^{(\alpha')}(x) = (\alpha'-\alpha) {\alpha'+ n \choose \alpha'-\alpha} \int_0^x \frac{t^\alpha (x-t)^{\alpha'-\alpha-1}}{x^{\alpha'}} L_n^{(\alpha)}(t)\,dt.$ The derivative with respect to the second variable α has the form,[9] $\frac{d}{d \alpha}L_n^{(\alpha)}(x)= \sum_{i=0}^{n-1} \frac{L_i^{(\alpha)}(x)}{n-i}.$ This is evident from the contour integral representation below. The generalized Laguerre polynomials obey the differential equation $x L_n^{(\alpha) \prime\prime}(x) + (\alpha+1-x)L_n^{(\alpha)\prime}(x) + n L_n^{(\alpha)}(x)=0,$ which may be compared with the equation obeyed by the kth derivative of the ordinary Laguerre polynomial, $x L_n^{(k) \prime\prime}(x) + (k+1-x)L_n^{(k)\prime}(x) + (n-k) L_n^{(k)}(x)=0,$ where $L_n^{(k)}(x)\equiv\frac{d^kL_n(x)}{dx^k}$ for this equation only. In Sturm–Liouville form the differential equation is $-\left(x^{\alpha+1} e^{-x}\cdot L_n^{(\alpha)}(x)^\prime\right)^\prime= n\cdot x^\alpha e^{-x}\cdot L_n^{(\alpha)}(x),$ which shows that Lα n is an eigenvector for the eigenvalue n. Orthogonality The generalized Laguerre polynomials are orthogonal over [0, ∞) with respect to the measure with weighting function xα ex:[10] $\int_0^\infty x^\alpha e^{-x} L_n^{(\alpha)}(x)L_m^{(\alpha)}(x)dx=\frac{\Gamma(n+\alpha+1)}{n!} \delta_{n,m},$ which follows from $\int_0^\infty x^{\alpha'-1} e^{-x} L_n^{(\alpha)}(x)dx= {\alpha-\alpha'+n \choose n} \Gamma(\alpha').$ If $\Gamma(x,\alpha+1,1)$ denoted the Gamma distribution then the orthogonality relation can be written as $\int_0^{\infty} L_n^{(\alpha)}(x)L_m^{(\alpha)}(x)\Gamma(x,\alpha+1,1) dx={n+ \alpha \choose n}\delta_{n,m},$ The associated, symmetric kernel polynomial has the representations (Christoffel–Darboux formula)[citation needed] \begin{align} K_n^{(\alpha)}(x,y) &:= \frac{1}{\Gamma(\alpha+1)} \sum_{i=0}^n \frac{L_i^{(\alpha)}(x) L_i^{(\alpha)}(y)}{{\alpha+i \choose i}}\\ &{=}\frac{1}{\Gamma(\alpha+1)} \frac{L_n^{(\alpha)}(x) L_{n+1}^{(\alpha)}(y) - L_{n+1}^{(\alpha)}(x) L_n^{(\alpha)}(y)}{\frac{x-y}{n+1} {n+\alpha \choose n}} \\ &{=}\frac{1}{\Gamma(\alpha+1)}\sum_{i=0}^n \frac{x^i}{i!} \frac{L_{n-i}^{(\alpha+i)}(x) L_{n-i}^{(\alpha+i+1)}(y)}{{\alpha+n \choose n}{n \choose i}}; \end{align} recursively $K_n^{(\alpha)}(x,y)=\frac{y}{\alpha+1} K_{n-1}^{(\alpha+1)}(x,y)+ \frac{1}{\Gamma(\alpha+1)} \frac{L_n^{(\alpha+1)}(x) L_n^{(\alpha)}(y)}{{\alpha+n \choose n}}.$ Moreover, $y^\alpha e^{-y} K_n^{(\alpha)}(\cdot, y) \to \delta(y- \cdot),$ in the associated L2[0, ∞)-space. Turán's inequalities can be derived here, which is $L_n^{(\alpha)}(x)^2- L_{n-1}^{(\alpha)}(x) L_{n+1}^{(\alpha)}(x)= \sum_{k=0}^{n-1} \frac{{\alpha+n-1\choose n-k}}{n{n\choose k}} L_k^{(\alpha-1)}(x)^2>0.$ The following integral is needed in the quantum mechanical treatment of the hydrogen atom, $\int_0^{\infty}x^{\alpha+1} e^{-x} \left[L_n^{(\alpha)} (x)\right]^2 dx= \frac{(n+\alpha)!}{n!}(2n+\alpha+1).$ Series expansions Let a function have the (formal) series expansion $f(x)= \sum_{i=0}^\infty f_i^{(\alpha)} L_i^{(\alpha)}(x).$ Then $f_i^{(\alpha)}=\int_0^\infty \frac{L_i^{(\alpha)}(x)}{{i+ \alpha \choose i}} \cdot \frac{x^\alpha e^{-x}}{\Gamma(\alpha+1)} \cdot f(x) \,dx .$ The series converges in the associated Hilbert space L2[0, ∞) if and only if $\| f \|_{L^2}^2 := \int_0^\infty \frac{x^\alpha e^{-x}}{\Gamma(\alpha+1)} | f(x)|^2 dx = \sum_{i=0}^\infty {i+\alpha \choose i} |f_i^{(\alpha)}|^2 < \infty.$ Further examples of expansions Monomials are represented as $\frac{x^n}{n!}= \sum_{i=0}^n (-1)^i {n+ \alpha \choose n-i} L_i^{(\alpha)}(x),$ while binomials have the parametrization ${n+x \choose n}= \sum_{i=0}^n \frac{\alpha^i}{i!} L_{n-i}^{(x+i)}(\alpha).$ $e^{-\gamma x}= \sum_{i=0}^\infty \frac{\gamma^i}{(1+\gamma)^{i+\alpha+1}} L_i^{(\alpha)}(x) \qquad \text{convergent iff } \Re(\gamma) > -\tfrac{1}{2}$ for the exponential function. The incomplete gamma function has the representation $\Gamma(\alpha,x)=x^\alpha e^{-x} \sum_{i=0}^\infty \frac{L_i^{(\alpha)}(x)}{1+i} \qquad \left(\Re(\alpha)>-1 , x > 0\right).$ Multiplication theorems Erdélyi gives the following two multiplication theorems [11] $t^{n+1+\alpha} e^{(1-t) z} L_n^{(\alpha)}(z t)=\sum_{k=n} {k \choose n}\left(1-\frac 1 t\right)^{k-n} L_k^{(\alpha)}(z),$ $e^{(1-t)z} L_n^{(\alpha)}(z t)=\sum_{k=0} \frac{(1-t)^k z^k}{k!}L_n^{(\alpha+k)}(z).$ Relation to Hermite polynomials The generalized Laguerre polynomials are related to the Hermite polynomials: \begin{align} H_{2n}(x) &= (-1)^n 2^{2n} n! L_n^{(-1/2)} (x^2) \\ H_{2n+1}(x) &= (-1)^n 2^{2n+1} n! x L_n^{(1/2)} (x^2) \end{align} where the Hn(x) are the Hermite polynomials based on the weighting function exp(−x2), the so-called "physicist's version." Because of this, the generalized Laguerre polynomials arise in the treatment of the quantum harmonic oscillator. Relation to hypergeometric functions The Laguerre polynomials may be defined in terms of hypergeometric functions, specifically the confluent hypergeometric functions, as $L^{(\alpha)}_n(x) = {n+\alpha \choose n} M(-n,\alpha+1,x) =\frac{(\alpha+1)_n} {n!} \,_1F_1(-n,\alpha+1,x)$ where $(a)_n$ is the Pochhammer symbol (which in this case represents the rising factorial). Poisson Kernel $\sum_{n=0}^{\infty}\frac{n!L_{n}^{(\alpha)}(x)L_{n}^{(\alpha)}(y)r^{n}}{\Gamma\left(1+\alpha+n\right)}=\frac{\exp\left(-\frac{\left(x+y\right)r}{1-r}\right)I_{\alpha}\left(\frac{2\sqrt{xyr}}{1-r}\right)}{\left(xyr\right)^{\frac{\alpha}{2}}\left(1-r\right)},\qquad \alpha>-1,\quad |r|<1.$ Notes 1. ^ 2. ^ A&S p. 781 3. ^ A&S p.509 4. ^ A&S p.510 5. ^ A&S p. 775 6. ^ G. Szegő, "Orthogonal polynomials", 4th edition, Amer. Math. Soc. Colloq. Publ., vol. 23, Amer. Math. Soc., Providence, RI, 1975, p. 198. 7. ^ D. Borwein, J. M. Borwein, R. E. Crandall, "Effective Laguerre asymptotics", SIAM J. Numer. Anal., vol. 46 (2008), no. 6, pp. 3285-3312 doi:10.1137/07068031X 8. ^ A&S equation (22.12.6), p. 785 9. ^ W. Koepf, "Identities for families of orthogonal polynomials and special functions.", Integral Transforms and Special Functions 5, (1997) pp.69-102. (Theorem 10) 10. ^ A&S p. 774 11. ^ C. Truesdell, "On the Addition and Multiplication Theorems for the Special Functions", Proceedings of the National Academy of Sciences, Mathematics, (1950) pp.752-757.

http://www.statemaster.com/encyclopedia/Twin-prime-conjecture

FACTOID # 3: South Carolina has the highest rate of violent crimes and aggravated assaults per capita among US states. Home Encyclopedia Statistics States A-Z Flags Maps FAQ About WHAT'S NEW SEARCH ALL Search encyclopedia, statistics and forums: (* = Graphable) Encyclopedia > Twin prime conjecture The twin prime conjecture is a famous problem in number theory that involves prime numbers. It was first proposed by Euclid around 300 B.C. and states: Number theory is the branch of pure mathematics concerned with the properties of numbers in general, and integers in particular, as well as the wider classes of problems that arise from their study. ... In mathematics, a prime number (or a prime) is a natural number that has exactly two (distinct) natural number divisors, which are 1 and the prime number itself. ... Euclid (Greek: ), also known as Euclid of Alexandria, was a Greek mathematician who flourished in Alexandria, Egypt, almost certainly during the reign of Ptolemy I (323–283 BC). ... There are infinitely many primes p such that p + 2 is also prime. Such a pair of prime numbers is called a prime twin. The conjecture has been researched by many number theorists. Mathematicians believe the conjecture to be true, based only on numerical evidence and heuristic reasoning involving the probabilistic distribution of primes using Cramér's model. In mathematics, a prime number (or a prime) is a natural number that has exactly two (distinct) natural number divisors, which are 1 and the prime number itself. ... A twin prime is a prime number that differs from another prime number by two. ... Look up Heuristic in Wiktionary, the free dictionary. ... In mathematics, Cramérs conjecture, formulated by the Swedish mathematician Harald Cramér in 1937, states that where pn denotes the nth prime number and log is the natural logarithm. ... In 1849 de Polignac made the more general conjecture that for every natural number k, there are infinitely many prime pairs p and p′ such that p - p′ = 2k. The case k = 1 is the twin prime conjecture. Alphonse de Polignac (1817 &#8211; 1890) was a French mathematician. ... ## Contents In 1915, Viggo Brun showed that the sum of reciprocals of the twin primes was convergent. This famous result, called Brun's theorem was the first use of the Brun sieve and helped initiate the development of modern sieve theory. The modern version of Brun's argument can be used to show that the number of twin primes less than N does not exceed Viggo Brun (October 13, 1882 - August 15, 1978) was a Norwegian mathematician. ... In mathematics, Bruns theorem is a result of Viggo Brun in number theory. ... Sieve theory is a set of general techniques in number theory, designed to count, or more realistically to estimate the size of, sifted sets of integers. ... $frac{CN}{log^2{N}}$ for some absolute constant C > 0. In 1940, Paul Erdős showed that there is a constant c < 1 and infinitely many primes p such that (p′ - p) < (c ln p) where p′ denotes the next prime after p. This result was successively improved; in 1986 Helmut Maier showed that a constant c < 0.25 can be used. In 2004 Daniel Goldston and Cem Yıldırım showed that the constant could be improved further to c = 0.085786… In 2005, Goldston, János Pintz and Yıldırım established that c can be chosen arbitrarily small [1], [2]: Paul ErdÅ‘s, also Pál ErdÅ‘s, in English Paul Erdos or Paul Erdös (March 26, 1913 – September 20, 1996), was an immensely prolific (and famously eccentric) Hungarian mathematician who, with hundreds of collaborators, worked on problems in combinatorics, graph theory, number theory, classical analysis, approximation theory, set... Helmut Maier is an American mathematician who made significant progress in the study of Twin prime conjecture. ... Daniel Alan Goldston is an American mathematician who specializes in number theory. ... Cem Yalcin Yildirim is a Turkish mathematician who specializes in number theory. ... $liminf_{ntoinfty}frac{p_{n+1}-p_n}{log p_n}=0$ In fact, by assuming the Elliott-Halberstam conjecture, they were able to show that there are infinitely many n such that at least two of n, n + 2, n + 6, n + 8, n + 12, n + 18, n + 20 are prime. In number theory, the Elliott-Halberstam conjecture is a conjecture about the distribution of prime numbers in arithmetic progressions. ... In 1966, Chen Jingrun showed that there are infinitely many primes p such that p + 2 is either a prime or a semiprime (i.e., the product of two primes). The approach he took involved sieve theory, and he managed to treat the twin prime conjecture and Goldbach's conjecture in similar manners. Chen Jingrun (ch. ... In mathematics, a semiprime (also called biprime or 2-almost prime, or pq number) is a natural number that is the product of two (not necessarily distinct) prime numbers. ... Goldbachs conjecture is one of the oldest unsolved problems in number theory and in all of mathematics. ... Defining a Chen prime to be a prime p such that p + 2 is either a prime or a semiprime, Terence Tao and Ben Green showed in 2005 that there are infinitely many three-term arithmetic progressions of Chen primes. A prime number p is called a Chen prime if p + 2 is either a prime or a product of two primes. ... In mathematics, a semiprime (also called biprime or 2-almost prime, or pq number) is a natural number that is the product of two (not necessarily distinct) prime numbers. ... Terence Chi-Shen Tao ( 陶哲軒)(born 1975) is an Australian mathematician working primarily on harmonic analysis, partial differential equations, combinatorics, analytic number theory and representation theory. ... Ben Joseph Green (born February 27, 1977, Bristol, United Kingdom) is a British mathematician, specializing in combinatorics and number theory. ... ## Hardy–Littlewood conjecture The Hardy–Littlewood conjecture (after G. H. Hardy and John Littlewood) is a generalization of the twin prime conjecture. It is concerned with the distribution of prime constellations, including twin primes, in analogy to the prime number theorem. Let π2(x) denote the number of primes px such that p + 2 is also prime. Define the twin prime constant C2 as G. H. Hardy Professor Godfrey Harold Hardy FRS (February 7, 1877 – December 1, 1947) was a prominent English mathematician, known for his achievements in number theory and mathematical analysis. ... John Edensor Littlewood (June 9, 1885 – September 6, 1977) was a British mathematician. ... In number theory, the prime number theorem (PNT) describes the approximate, asymptotic distribution of the prime numbers. ... $C_2 = prod_{pge 3} frac{p(p-2)}{(p-1)^2} approx 0.66016 18158 46869 57392 78121 10014dots$ (here the product extends over all prime numbers p ≥ 3). Then the conjecture is that $pi_2(n) sim 2 C_2 frac{n}{(ln n)^2} sim 2 C_2 int_2^n {dt over (ln t)^2}$ in the sense that the quotient of the two expressions tends to 1 as x approaches infinity. In mathematics, the concept of a limit is used to describe the behavior of a function as its argument either gets close to some point, or as it becomes arbitrarily large; or the behavior of a sequences elements, as their index increases indefinitely. ... This conjecture can be justified (but not proven) by assuming that $frac{1}{ln{t}}$ describes the density function of the prime distribution, an assumption suggested by the prime number theorem. The numerical evidence behind the Hardy–Littlewood conjecture is quite impressive. In mathematics, a probability density function (pdf) serves to represent a probability distribution in terms of integrals. ... A twin prime is a prime number that differs from another prime number by two. ... In mathematics, Bruns theorem is a result of Viggo Brun in number theory. ... Goldbachs conjecture is one of the oldest unsolved problems in number theory and in all of mathematics. ... In mathematics, Cramérs conjecture, formulated by the Swedish mathematician Harald Cramér in 1937, states that where pn denotes the nth prime number and log is the natural logarithm. ... Results from FactBites: Conjecture (448 words) In mathematics, a conjecture is a mathematical statement which has been proposed as a true statement, but which no one has been able to prove or disprove. When a conjecture has been proven to be true, it becomes known as a theorem, and joins the realm of mathematical facts. Although many of the most famous conjectures have been tested across an astounding range of numbers, this is no guarantee against a single counterexample, which would immediately disprove the conjecture. Twin prime (196 words) It is unknown whether there exist infinitely many twin primes, but most number theorists believe this to be true. A strong form of the Twin Prime Conjecture, the Hardy-Littlewood conjecture, postulates a distribution law for twin primes akin to the prime number theorem. It is known that the sum of the reciprocals of all twin primes converges (see Brun's constant). More results at FactBites » Share your thoughts, questions and commentary here

https://byjus.com/jee/cartesian-product-in-set-relations-functions/

# Cartesian Product in Set Relations Functions The well-defined objects’ collection is termed as a set. A set with finite elements is a finite set whereas with infinite elements is an infinite set. A Set can be represented in a roster and set builder form. The collection of ordered pairs, which consists of one object from each set is a relation. It can be represented as a cartesian product of two sets where all the elements have a common property. While plotting a graph, the x – coordinate is followed by the y – coordinate in an ordered way. In two non-empty sets, the first element is from set A and the second element is from set B. The collection of such ordered pairs constitute a cartesian product. The ordered pairs are said to be equal if a1 = a2 and b1 = b2. Example: Let A = {a, b, c} and B = {p,q}. Then A × B = {(a, p), (a, q), (b, p),(b, q), (c, p), (c, q)} Also B × A = {(p, a), (p, b), (p, c), (q, a), (q, b), (q, c)} ## Cartesian Product In Set Relations Functions Formula The cartesian product is denoted by A × B. The pair of (a1, b1) is different from (b1, a1). A × B = {(a1, b1), (a1 , b2), (a1 ,b3), (a2, b1), (a2, b2), (a2, b3), (a3, b1), (a3, b2), (a3, b3)}. ## Important Theorems On Cartesian Product Of Sets Theorem 1: For any three sets A, B, C $(i)\ A\times (B\cup C)=(A\times B)\cup (A\times C)\\ (ii)\ A\times (B\cap C)=(A\times B)\cap (A\times C)\\$ Theorem 2: For any three sets A, B, C $A\times (B-C)=(A\times B)-(A\times C)$ Theorem 3: If A and B are any two non-empty sets, then $A\times B=B\times A\Leftrightarrow A=B$ Theorem 4: If $A\subseteq B,$ then $A\times A\subseteq (A\times B)\cap (B\times A)$ Theorem 5: If $A\subseteq B,$ then $A\times C\subseteq B\times C$ for any set C. Theorem 6: If$A\subseteq B,$ and $C\subseteq D,$ then $A\times C\subseteq B\times D$ Theorem 7: For any sets A, B, C, D $(A\times B)\cap (C\cup D)=(A\cap C)\times (B\cap D)$ Theorem 8: For any three sets A, B, C $(i) \ A\times (B’\times C’)’=(A\times B)\cap (A\times C)\\ (ii) \ A\times (B’\cap C’)’=(A\times B)\cup (A\times C)$ ## Properties Of Cartesian Product Of Sets $i. A \times (B \cup C) = (A \times B) \cup (A \times C)\\ ii.A \times (B \cap C) = (A \times B) \cap (A \times C)\\ iii. A \times (B – C) = (A \times B) – (A \times C)\\ iv. (A \times B) \cap ( C \times D) =(A \cap C) \times (B \cap D)\\ v.\text \ if \ A \subset B \text \ then \ A \times C \subset B \times C\\ vi. A \times B \ne B \times A \\ vii.(A \times B) \times C \ne A \times (B \times C)$ ### Non-commutativity and non-associativity Let A, B, C, and D be sets. The Cartesian product A × B is not commutative, ${\displaystyle A\times B\neq B\times A,}$ because the ordered pairs are reversed unless at least one of the following conditions is satisfied: • A is equal to B, or • A or B is the empty set. For example: A = {1,2}; B = {3,4} A × B = {1,2} × {3,4} = {(1,3), (1,4), (2,3), (2,4)} B × A = {3,4} × {1,2} = {(3,1), (3,2), (4,1), (4,2)} A = B = {1,2} A × B = B × A = {1,2} × {1,2} = {(1,1), (1,2), (2,1), (2,2)} A = {1,2}; B = ∅ A × B = {1,2} × ∅ = ∅ B × A = ∅ × {1,2} = ∅ The Cartesian product is not associative (unless one of the involved sets is empty). ${\displaystyle (A\times B)\times C\neq A\times (B\times C)}$ If for example A = {1}, then (A × A) × A = { ((1,1),1) } ≠ { (1,(1,1)) } = A × (A × A). Sets Relations and Functions Functions and its Types ## Cartesian Product in Set Relations Functions Examples Example 1: If A = [(x, y) : x2 + y2 = 25] and B = [(x,y) : x2+ 9y2 = 144], then how many points does A ∩ B contain? Solution: A = Set of all values (x, y) : x2 + y2 = 25 = 52 B = x2 / 144 + y2 / 16 = 1 i.e., x2 / (12)2 + y2 / (4)2 = 1. Clearly, A ∩ B consists of four points. Example 2: Let a relation R be defined by R = {(4, 5); (1, 4); (4, 6); (7, 6); (3, 7)} then what is R−1 o R? Solution: We first find R−1 , we have R−1 = {(5, 4) ; (4, 1) ; (6, 4) ; (6, 7) ; (7, 3)}. We now obtain the elements of R−1 o R we first pick the element of R and then of R−1. Since (4, 5) ∈ R and (5, 4) ∈ R−1 , we have (4, 4) ∈ R−1 oR Similarly, (1, 4) ∈ R,(4, 1) ∈ R−1 ⇒(1,1) ∈ R−1 o R (4, 6) ∈ R, (6, 4) ∈ R−1 ⇒ (4, 4) ∈ R−1 o R, (4, 6) ∈ R, (6, 7) ∈ R−1 ⇒ (4, 7) ∈R−1 o R (7, 6) ∈ R, (6, 4) ∈ R−1 ⇒ (7, 4) ∈ R−1 o R, (7, 6) ∈ R, (6, 7) ∈ R−1 ⇒ (7, 7) ∈ R−1 o R (3, 7) ∈ R, (7, 3) ∈ R−1 ⇒ (3, 3) ∈ R−1 oR, Hence, R−1 o R = {(1, 1); (4, 4); (4, 7); (7, 4), (7, 7); (3, 3)}. Example 3: Let R be a relation on the set N be defined by {(x, y)| x, y i ^ N, 2x + y = 41}. Then R is _________ . Solution: On the set N of natural numbers, R = {(x, y) : x, y ∈ N , 2x + y = 41}. Since (1, 1) ∉ R as 2.1 + 1 = 3 ≠ 41. So, R is not reflexive. (1, 39) ∈ R but (39, 1) ∉ R So R is not symmetric. (20, 1) (1, 39) ∈ R But (20, 39) ∉ R. So R is not transitive. Example 4: If X = {8n − 7n −1 : n ∈ N} and Y = {49 (n−1) : n ∈ N}, then how is X and Y related? Solution: Since 8n − 7n −1 = (7+1)n − 7n − 1 = 7n + nC1 7n−1+ nC27n−2 + ….. +nCn−1 7 + nCn− 7n −1 = nC272 + nC373 +….+nCn7n, (nC0 = nCn, nC1 = nCn−1 etc.) = 49 [nC2+ nC3 (7) + …… + nCn7n − 2] ∴ 8n − 7n − 1 is a multiple of 49 for n ≥ 2 For n = 1, 8n − 7n − 1 = 8 − 7 − 1 = 0; For n = 2, 8n − 7n − 1 = 64 − 14 − 1 = 49 ∴ 8n − 7n − 1 is a multiple of 49 for all n ∈ N. ∴ X contains elements which are multiples of 49 and clearly Y contains all multiples of 49. ∴ X ⊆ Y. Example 5: Let A = {1, 2, 3, 4, 5}; B = {2, 3, 6, 7}. Then what is the number of elements in (A × B) ∩ (B × A)? Solution: Here A and B sets have 2 elements in common, so A × B and B × A have 22 i.e., 4 elements in common. Hence, n [(A × B) ∩ (B × A)] = 4.

https://db0nus869y26v.cloudfront.net/en/Domain_of_a_function

A function f from X to Y. The set of points in the red oval X is the domain of f. Graph of the real-valued square root function, f(x) = x, whose domain consists of all nonnegative real numbers In mathematics, the domain of a function is the set of inputs accepted by the function. It is sometimes denoted by ${\displaystyle \operatorname {dom} (f)}$, where f is the function. More precisely, given a function ${\displaystyle f\colon X\to Y}$, the domain of f is X. Note that in modern mathematical language, the domain is part of the definition of a function rather than a property of it. In the special case that X and Y are both subsets of ${\displaystyle \mathbb {R} }$, the function f can be graphed in the Cartesian coordinate system. In this case, the domain is represented on the x-axis of the graph, as the projection of the graph of the function onto the x-axis. For a function ${\displaystyle f\colon X\to Y}$, the set Y is called the codomain, and the set of values attained by the function (which is a subset of Y) is called its range or image. Any function can be restricted to a subset of its domain. The restriction of ${\displaystyle f\colon X\to Y}$ to ${\displaystyle A}$, where ${\displaystyle A\subseteq X}$, is written as ${\displaystyle \left.f\right|_{A}\colon A\to Y}$. ## Natural domain If a real function f is given by a formula, it may be not defined for some values of the variable. In this case, it is a partial function, and the set of real numbers on which the formula can be evaluated to a real number is called the natural domain or domain of definition of f. In many contexts, a partial function is called simply a function, and its natural domain is called simply its domain. ### Examples • The function ${\displaystyle f}$ defined by ${\displaystyle f(x)={\frac {1}{x))}$ cannot be evaluated at 0. Therefore the natural domain of ${\displaystyle f}$ is the set of real numbers excluding 0, which can be denoted by ${\displaystyle \mathbb {R} \setminus \{0\))$ or ${\displaystyle \{x\in \mathbb {R} :x\neq 0\))$. • The piecewise function ${\displaystyle f}$ defined by ${\displaystyle f(x)={\begin{cases}1/x&x\not =0\\0&x=0\end{cases)),}$ has as its natural domain the set ${\displaystyle \mathbb {R} }$ of real numbers. • The square root function ${\displaystyle f(x)={\sqrt {x))}$ has as its natural domain the set of non-negative real numbers, which can be denoted by ${\displaystyle \mathbb {R} _{\geq 0))$, the interval ${\displaystyle [0,\infty )}$, or ${\displaystyle \{x\in \mathbb {R} :x\geq 0\))$. • The tangent function, denoted ${\displaystyle \tan }$, has as its natural domain the set of all real numbers which are not of the form ${\displaystyle {\tfrac {\pi }{2))+k\pi }$ for some integer ${\displaystyle k}$, which can be written as ${\displaystyle \mathbb {R} \setminus $$(\tfrac {\pi }{2))+k\pi :k\in \mathbb {Z}$$)$. ## Other uses Main article: Domain (mathematical analysis) The word "domain" is used with other related meanings in some areas of mathematics. In topology, a domain is a connected open set.[1] In real and complex analysis, a domain is an open connected subset of a real or complex vector space. In the study of partial differential equations, a domain is the open connected subset of the Euclidean space ${\displaystyle \mathbb {R} ^{n))$ where a problem is posed (i.e., where the unknown function(s) are defined). ## Set theoretical notions For example, it is sometimes convenient in set theory to permit the domain of a function to be a proper class X, in which case there is formally no such thing as a triple (X, Y, G). With such a definition, functions do not have a domain, although some authors still use it informally after introducing a function in the form f: XY.[2] ## Notes 1. ^ Weisstein, Eric W. "Domain". mathworld.wolfram.com. Retrieved 2020-08-28. 2. ^ Eccles 1997, p. 91 (quote 1, quote 2); Mac Lane 1998, p. 8; Mac Lane, in Scott & Jech 1967, p. 232; Sharma 2004, p. 91; Stewart & Tall 1977, p. 89 ## References • Bourbaki, Nicolas (1970). Théorie des ensembles. Éléments de mathématique. Springer. ISBN 9783540340348.

http://mathhelpforum.com/calculus/74127-improper-integral-log-floor-ceil-function.html

# Math Help - Improper integral with log, floor and ceil function 1. ## Improper integral with log, floor and ceil function Compute for all integer $\alpha>1$ the value of $\int_{0}^{\infty }{\left\lfloor \log _{\alpha }\left\lfloor \frac{\left\lceil x \right\rceil }{x} \right\rfloor \right\rfloor \,dx}.$ 2. very hard problem!! could u post a solution? 3. I can do that, within few hours. 4. For each $x>1$ it's $x\le \left\lceil x \right\rceil <2x$ and the integrand is $\left\lfloor \log _{\alpha }1 \right\rfloor =0,$ hence the integral becomes $\int_{0}^{1}{\left\lfloor \log _{\alpha }\left\lfloor \frac{1}{x} \right\rfloor \right\rfloor \,dx}.$ Since $\alpha$ is an integer, then the inner floor function is reduntant so the integral is $\int_{0}^{1}{\left\lfloor -\log _{\alpha }x \right\rfloor \,dx}=\sum\limits_{k=0}^{\infty }{\int_{\alpha ^{-k-1}}^{\alpha ^{-k}}{k\,dx}}=\sum\limits_{k=0}^{\infty }{k\left( \frac{1}{\alpha ^{k}}-\frac{1}{\alpha ^{k+1}} \right)}=\frac{1}{\alpha -1}$ and we're done. $\blacksquare$ 5. thanks a lot for the solution!! but it's kind of hard to me to understand, can you explain any further how did you get that the integral equals $ \int_{0}^{1}{\left\lfloor \log _{\alpha }\left\lfloor \frac{1}{x} \right\rfloor \right\rfloor \,dx} $ ? 6. $\int_{0}^{\infty }{\left\lfloor \log _{\alpha }\left\lfloor \frac{\left\lceil x \right\rceil }{x} \right\rfloor \right\rfloor \,dx}=\sum\limits_{n=1}^{\infty }{\int_{n-1}^{n}{\left\lfloor \log _{\alpha }\left\lfloor \frac{n}{x} \right\rfloor \right\rfloor \,dx}}=\sum\limits_{n=1}^{\infty }{\int_{0}^{1}{\left\lfloor \log _{\alpha }\left\lfloor \frac{n}{x+n-1} \right\rfloor \right\rfloor \,dx}}.$ Thus, for $0\le x\le1$ and $n\ge2$ we have $\left\lfloor \log _{\alpha }\left\lfloor \frac{n}{x+n-1} \right\rfloor \right\rfloor =0$ hence the integral is $\int_{0}^{1}{\left\lfloor \log _{\alpha }\left\lfloor \frac{1}{x} \right\rfloor \right\rfloor \,dx}+\sum\limits_{n=2}^{\infty }{\int_{0}^{1}{\left\lfloor \log _{\alpha }\left\lfloor \frac{n}{x+n-1} \right\rfloor \right\rfloor \,dx}}=\int_{0}^{1}{\left\lfloor \log _{\alpha }\left\lfloor \frac{1}{x} \right\rfloor \right\rfloor \,dx},$ and we're done.

https://bookdown.org/manuele_leonelli/SimBook/poisson-process.html

## 7.1 Poisson Process Consider random events such as the arrival of customer at a shop, the arrival of emails to a mail server or the arrival of calls to a call-center. These events can be described by a counting function $$N(t)$$ defined for all $$t\geq 0$$. This counting function represents the number of events that occurred in $$[0,t]$$. For each interval $$[0,t]$$ the value $$N(t)$$ is an observation of a random variable where the only possible values are the integers $$0,1,2,\dots$$. The counting process $$\{N(t): t\geq 0\}$$ is said to be a Poisson process with mean rate $$\lambda$$ if the following assumptions are fulfilled: • $$N(0) = 0$$; • it has independent increments: that is the number of arrivals during non-overlapping time intervals are independent random variables. • the number of events in any interval of length $$t$$ is a Poisson random variable with parameter $$\lambda t$$. Therefore $P(N(t)=n)=\frac{e^{-\lambda t}(\lambda t)^n}{n!}$ The last assumption implies that the distribution of the number of arrivals between, say, $$t$$ and $$t+s$$ depends only on the length of the interval $$s$$ and not on the starting point $$t$$. This property is usually called stationarity. Consequently: $P(N(t)-N(s)=n)=\frac{e^{-\lambda(t-s)}(\lambda(t-s))^n}{n!}$ and because of the properties of the Poisson distribution: $E(N(t)-N(s))=V(N(t)-N(s))=\lambda(t-s).$ Now consider the time at which arrivals occurs in a Poisson process. Let the first arrival occur at time $$A_1$$, the second occur at time $$A_1+ A_2$$ and so on. Thus $$A_1,A_2,\dots$$ are successive inter-arrival times. The first arrival occurs after time $$t$$ if and only if there are no arrivals in the interval $$[0,t]$$ so it is seen that $\{A_1>t\}=\{N(t)=0\}$ and consequently $P(A_1>t)=P(N(t)=0)=e^{-\lambda t}$ Thus the probability that the first arrival will occur in $$[0,t]$$ is given by $P(A_1\leq t)= 1- P(A_1>t)= 1- e^{-\lambda t}$ which is the cumulative density function of an exponential distribution with parameter $$\lambda$$. Hence, $$A_1$$ is distributed exponentially with mean $$E(A_1)=1/\lambda$$. It can also be shown that all inter-arrival times $$A_1,A_2,\dots$$ are exponentially distributed and independent with mean $$1/\lambda$$. An alternative definition of a Poisson process is of a counting process whose inter-arrival times are distributed exponentially and independently. Exponential distributions have the property of being memoryless, which is deeply connected to Poisson processes. For an exponential distribution $$X$$ it holds that $P(X> s+t|X>s)=P(X>t).$ Suppose $$X$$ represents the life of a light bulb. The above equation states that the probability that the light bulb lives for at least $$s+t$$ hours, given it has survived $$s$$ hours, is that same as the initial probability that it lives for at least $$t$$ hours. That is, the light bulb it does not remember that it has already been in use for a time $$s$$. Let’s show the equality is true: $\begin{eqnarray*} P(X>s+t | X>s) &=& \frac{P(X > s + t, X >s)}{P(X>s)}\\ &=& \frac{P(X > s + t)}{P(X>s)}\\ &=&\frac{1- P(X \leq s + t)}{1 - P(X \leq s)}\\ &=&\frac{e^{-\lambda(s+t)}}{e^{-\lambda s}}\\ &=& e^{-\lambda t}\\ &=& P(X>t) \end{eqnarray*}$

https://www.jobilize.com/course/section/multiplying-matrices-in-general-by-openstax?qcr=www.quizover.com

# 0.1 Matrix concepts -- multiplying matrices Page 1 / 1 This module covers multiplication of matrices. ## Multiplying a row matrix by a column matrix A “row matrix” means a matrix with only one row. A “column matrix” means a matrix with only one column. When a row matrix has the same number of elements as a column matrix, they can be multiplied. So the following is a perfectly legal matrix multiplication problem: $\left[\begin{array}{cccc}1& 2& 3& 4\end{array}\right]x\left[\begin{array}{c}10\\ 20\\ 30\\ 40\end{array}\right]$ These two matrices could not be added, of course, since their dimensions are different, but they can be multiplied. Here’s how you do it. You multiply the first (left-most) item in the row, by the first (top) item in the column. Then you do the same for the second items, and the third items, and so on. Finally, you add all these products to produce the final number. A couple of my students (Nakisa Asefnia and Laura Parks) came up with an ingenious trick for visualizing this process. Think of the row as a dump truck, backing up to the column dumpster. When the row dumps its load, the numbers line up with the corresponding numbers in the column, like so: So, without the trucks and dumpsters, we express the result—a row matrix, times a column matrix—like this: $\left[\begin{array}{cccc}1& 2& 3& 4\end{array}\right]\left[\begin{array}{c}10\\ 20\\ 30\\ 40\end{array}\right]=\left[\begin{array}{c}300\end{array}\right]$ • The picture is a bit deceptive, because it might appear that you are multiplying two columns. In fact, you cannot multiply a column matrix by a column matrix . We are multiplying a row matrix by a column matrix. The picture of the row matrix “dumping down” only demonstrates which numbers to multiply. • The answer to this problem is not a number: it is a 1-by-1 matrix. • The multiplication can only be performed if the number of elements in each matrix is the same. (In this example, each matrix has 4 elements.) • Order matters! We are multiplying a row matrix times a column matrix , not the other way around. It’s important to practice a few of these, and get the hang of it, before you move on. ## Multiplying matrices in general The general algorithm for multiplying matrices is built on the row-times-column operation discussed above. Consider the following example: $\left[\begin{array}{ccc}1& 2& 3\\ 4& 5& 6\\ 7& 8& 9\\ \text{10}& \text{11}& \text{12}\end{array}\right]$ $\left[\begin{array}{cc}\text{10}& \text{40}\\ \text{20}& \text{50}\\ \text{30}& \text{60}\end{array}\right]$ The key to such a problem is to think of the first matrix as a list of rows (in this case, 4 rows), and the second matrix as a list of columns (in this case, 2 columns). You are going to multiply each row in the first matrix, by each column in the second matrix. In each case, you will use the “dump truck” method illustrated above. Start at the beginning: first row, times first column. Now, move down to the next row. As you do so, move down in the answer matrix as well. Now, move down the rows in the first matrix, multiplying each one by that same column on the right. List the numbers below each other. The first column of the second matrix has become the first column of the answer. We now move on to the second column and repeat the entire process, starting with the first row. And so on, working our way once again through all the rows in the first matrix. We’re done. We can summarize the results of this entire operation as follows: $\left[\begin{array}{ccc}1& 2& 3\\ 4& 5& 6\\ 7& 8& 9\\ \text{10}& \text{11}& \text{12}\end{array}\right]$ $\left[\begin{array}{cc}\text{10}& \text{40}\\ \text{20}& \text{50}\\ \text{30}& \text{60}\end{array}\right]$ $=$ $\left[\begin{array}{cc}\text{140}& \text{320}\\ \text{320}& \text{770}\\ \text{500}& \text{1220}\\ \text{680}& \text{1670}\end{array}\right]$ It’s a strange and ugly process—but everything we’re going to do in the rest of this unit builds on this, so it’s vital to be comfortable with this process. The only way to become comfortable with this process is to do it. A lot. Multiply a lot of matrices until you are confident in the steps. Note that we could add more rows to the first matrix, and that would add more rows to the answer. We could add more columns to the second matrix, and that would add more columns to the answer. However—if we added a column to the first matrix, or added a row to the second matrix, we would have an illegal multiplication. As an example, consider what happens if we try to do this multiplication in reverse: $\left[\begin{array}{cc}\text{10}& \text{40}\\ \text{20}& \text{50}\\ \text{30}& \text{60}\end{array}\right]$ $\left[\begin{array}{ccc}1& 2& 3\\ 4& 5& 6\\ 7& 8& 9\\ \text{10}& \text{11}& \text{12}\end{array}\right]$ Illegal multiplication If we attempt to multiply these two matrices, we start (as always) with the first row of the first matrix, times the first column of the second matrix: $\left[\begin{array}{cc}10& 40\end{array}\right]$ $\left[\begin{array}{c}1\\ 4\\ 7\\ \text{10}\end{array}\right]$ . But this is an illegal multiplication; the items don’t line up, since there are two elements in the row and four in the column. So you cannot multiply these two matrices. This example illustrates two vital properties of matrix multiplication. • The number of columns in the first matrix, and the number of rows in the second matrix, must be equal. Otherwise, you cannot perform the multiplication. • Matrix multiplication is not commutative —which is a fancy way of saying, order matters. If you reverse the order of a matrix multiplication, you may get a different answer, or you may (as in this case) get no answer at all. where we get a research paper on Nano chemistry....? nanopartical of organic/inorganic / physical chemistry , pdf / thesis / review Ali what are the products of Nano chemistry? There are lots of products of nano chemistry... Like nano coatings.....carbon fiber.. And lots of others.. learn Even nanotechnology is pretty much all about chemistry... Its the chemistry on quantum or atomic level learn da no nanotechnology is also a part of physics and maths it requires angle formulas and some pressure regarding concepts Bhagvanji hey Giriraj Preparation and Applications of Nanomaterial for Drug Delivery revolt da Application of nanotechnology in medicine what is variations in raman spectra for nanomaterials I only see partial conversation and what's the question here! what about nanotechnology for water purification please someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment. Damian yes that's correct Professor I think Professor Nasa has use it in the 60's, copper as water purification in the moon travel. Alexandre nanocopper obvius Alexandre what is the stm is there industrial application of fullrenes. What is the method to prepare fullrene on large scale.? Rafiq industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong Damian How we are making nano material? what is a peer What is meant by 'nano scale'? What is STMs full form? LITNING scanning tunneling microscope Sahil how nano science is used for hydrophobicity Santosh Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq Rafiq what is differents between GO and RGO? Mahi what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.? How this robot is carried to required site of body cell.? what will be the carrier material and how can be detected that correct delivery of drug is done Rafiq Rafiq if virus is killing to make ARTIFICIAL DNA OF GRAPHENE FOR KILLED THE VIRUS .THIS IS OUR ASSUMPTION Anam analytical skills graphene is prepared to kill any type viruses . Anam Any one who tell me about Preparation and application of Nanomaterial for drug Delivery Hafiz what is Nano technology ? write examples of Nano molecule? Bob The nanotechnology is as new science, to scale nanometric brayan nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale Damian Is there any normative that regulates the use of silver nanoparticles? what king of growth are you checking .? Renato What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ? why we need to study biomolecules, molecular biology in nanotechnology? ? Kyle yes I'm doing my masters in nanotechnology, we are being studying all these domains as well.. why? what school? Kyle biomolecules are e building blocks of every organics and inorganic materials. Joe how did you get the value of 2000N.What calculations are needed to arrive at it Privacy Information Security Software Version 1.1a Good Got questions? Join the online conversation and get instant answers!

http://techiemathteacher.com/2015/05/08/mtap-reviewer-for-grade-9-solution-series31-35/

# MTAP Reviewer for Grade 9 Solution Series(31-35) Here is the part 7 of the solution series in 2015 MTAP(MMC) reviewer for grade 9 elimination level. Other solutions can be found in this website as well as the PDF copy of all the problems. Problem 31: An equilateral triangle and a square have the same perimeter. What is the ratio of the length of a side of the triangle to the length of a side of the square? Solution: The perimeter of equilateral triangle can be found using the formula, $P=3t$ where t is the length of the side The perimeter of square can be found using the formula, $P=4s$ where s is the length of the side Since the perimeter is equal, we equate their perimeters, $P=P$ $3t=4s$ Since we are asked to find the ratio of the side of triangle to the length of the side of the square we find t/s. $\dfrac{t}{s}=\dfrac{4}{3}$ Thus, the ratio is 4:3 Problem 32: John cuts an equilateral triangular paper whose sides measure 2 cm into pieces. He then rearranges the pieces to form a square without overlapping. How long is the side of the square formed? After cutting, the area of the figure is still the same. To find the area of an equilateral triangle, we use the following formula, $A=\frac{a^2\sqrt{3}}{4}$ Where a is the length of the side of an equilateral triangle. Finding the area we have, $A=\frac{2^2\sqrt{3}}{4}$ $A=\frac{4\sqrt{3}}{4}$ $A=\sqrt{3}$ Now, since the area of this triangle is equal to the area of the square, we can use the formula of the area of the square to find the length of the side. $A=s^2$ Now, we equate the area of the triangle, $\sqrt{3}=s^2$ Since we are solving for s, we take the square root of both sides, $\sqrt{\sqrt{3}}=\sqrt{s^2}$ $\sqrt[4]{3}=s$ Or, $\boxed{s=\sqrt[4]{3}}$ Problem 23: The sides of a triangle are of lengths 5, 12 and 13 cm. What is the length of its shortest altitude? Solution: Draw the triangle and label like shown below. Since the triangle formed is a right triangle, 12 and 5 are also altitudes. If we draw another altitude to the hypotenuse and label it x. Both 5 and 12 will become the hypotenuse of the two new triangles formed. Thus, x is the shortest altitude. Now, take note that $\triangle ABC\sim\triangle BDC\sim\triangle ADB$ By similar triangles we can solve for the value of x. $\dfrac{BD}{BC}=\dfrac{AB}{AC}$ $\dfrac{x}{5}=\dfrac{12}{13}$ $x=5(\dfrac{12}{13})$ $\boxed{x=\dfrac{60}{13}}$ Problem 34: Each side of triangle ABC measures 8 cm. If D is the foot of the altitude drawn from A to the side BC and E is the midpoint of AD, how long is segment BE? Solution: Draw the figure as shown below, Since this is an equilateral triangle, triangle ADB is a 36-60-90 triangle. Using the property of this triangle, the side of AD is square root of 3 times the length of BD. Thus, $AD=\sqrt{3}BD$ $AD=\sqrt{3}(4)$ $AD=4\sqrt{3}$ Since E is the midpoint of AD, $AE=ED=2\sqrt{3}$ Now, triangle BDE also forms a right triangle and we can solve for BE using Pythagorean theorem. $BE^2=BD^2+ED^2$ $BE^2=4^2+(2\sqrt{3})^2$ $BE^2=16+12$ $BE^2=28$ $\sqrt{BE^2}=\sqrt{28}$ $\boxed{BE=2\sqrt{7}}$ Problem 35: A point E is chosen inside a square of side 8 cm such that it is equidistant from two adjacent vertices of the square and the side opposite these vertices. Find the common distance. Solution: Draw the figure like shown below, The only way that a point will be equidistant from two adjacent vertices, it should lie in the symmetrical axis of the square. Label the figure accordingly. If your figure is not correct, your solution will not be correct as well. To find the value of x, we use the right triangle formed in the upper left side of upper right side. Either of the two. Using Pythagorean theorem we have, $c^2=a^2+b^2$ $x^2=4^2+(8-x)^2$ $x^2=16+64-16x+x^2$ $16x=80$ $x=\dfrac{80}{16}$ $\boxed{x=5}$ We would like to thank MMC 2013 finalist Daniel James Molina for helping us out solving this problem.

https://en.wikipedia.org/wiki/De_Morgan's_laws

# De Morgan's laws De Morgan's laws represented with Venn diagrams In propositional logic and boolean algebra, De Morgan's laws[1][2][3] are a pair of transformation rules that are both valid rules of inference. They are named after Augustus De Morgan, a 19th-century British mathematician. The rules allow the expression of conjunctions and disjunctions purely in terms of each other via negation. The rules can be expressed in English as: the negation of a conjunction is the disjunction of the negations; and the negation of a disjunction is the conjunction of the negations; or the complement of the union of two sets is the same as the intersection of their complements; and the complement of the intersection of two sets is the same as the union of their complements. In set theory and Boolean algebra, these are written formally as {\displaystyle {\begin{aligned}{\overline {A\cup B}}&={\overline {A}}\cap {\overline {B}},\\{\overline {A\cap B}}&={\overline {A}}\cup {\overline {B}},\end{aligned}}} where • A and B are sets, • A is the complement of A, • ∩ is the intersection, and • ∪ is the union. In formal language, the rules are written as ${\displaystyle \neg (P\land Q)\iff (\neg P)\lor (\neg Q)}$, and ${\displaystyle \neg (P\lor Q)\iff (\neg P)\land (\neg Q),}$ where Applications of the rules include simplification of logical expressions in computer programs and digital circuit designs. De Morgan's laws are an example of a more general concept of mathematical duality. ## Formal notation The negation of conjunction rule may be written in sequent notation: ${\displaystyle \neg (P\land Q)\vdash (\neg P\lor \neg Q).}$ The negation of disjunction rule may be written as: ${\displaystyle \neg (P\lor Q)\vdash (\neg P\land \neg Q).}$ In rule form: negation of conjunction ${\displaystyle {\frac {\neg (P\land Q)}{\therefore \neg P\lor \neg Q}}}$ and negation of disjunction ${\displaystyle {\frac {\neg (P\lor Q)}{\therefore \neg P\land \neg Q}}}$ and expressed as a truth-functional tautology or theorem of propositional logic: {\displaystyle {\begin{aligned}\neg (P\land Q)&\to (\neg P\lor \neg Q),\\\neg (P\lor Q)&\to (\neg P\land \neg Q),\end{aligned}}} where ${\displaystyle P}$ and ${\displaystyle Q}$ are propositions expressed in some formal system. ### Substitution form De Morgan's laws are normally shown in the compact form above, with negation of the output on the left and negation of the inputs on the right. A clearer form for substitution can be stated as: {\displaystyle {\begin{aligned}(P\land Q)&\equiv \neg (\neg P\lor \neg Q),\\(P\lor Q)&\equiv \neg (\neg P\land \neg Q).\end{aligned}}} This emphasizes the need to invert both the inputs and the output, as well as change the operator, when doing a substitution. ### Set theory and Boolean algebra In set theory and Boolean algebra, it is often stated as "union and intersection interchange under complementation",[4] which can be formally expressed as: {\displaystyle {\begin{aligned}{\overline {A\cup B}}&={\overline {A}}\cap {\overline {B}},\\{\overline {A\cap B}}&={\overline {A}}\cup {\overline {B}},\end{aligned}}} where: • A is the negation of A, the overline being written above the terms to be negated, • ∩ is the intersection operator (AND), • ∪ is the union operator (OR). The generalized form is: {\displaystyle {\begin{aligned}{\overline {\bigcap _{i\in I}A_{i}}}&\equiv \bigcup _{i\in I}{\overline {A_{i}}},\\{\overline {\bigcup _{i\in I}A_{i}}}&\equiv \bigcap _{i\in I}{\overline {A_{i}}},\end{aligned}}} where I is some, possibly uncountable, indexing set. In set notation, De Morgan's laws can be remembered using the mnemonic "break the line, change the sign".[5] ### Engineering In electrical and computer engineering, De Morgan's laws are commonly written as: ${\displaystyle {\overline {A\cdot B}}\equiv {\overline {A}}+{\overline {B}}}$ and ${\displaystyle {\overline {A+B}}\equiv {\overline {A}}\cdot {\overline {B}},}$ where: • ${\displaystyle \cdot }$ is a logical AND, • ${\displaystyle +}$ is a logical OR, • the overbar is the logical NOT of what is underneath the overbar. ### Text searching De Morgan’s laws commonly apply to text searching using Boolean operators AND, OR, and NOT. Consider a set of documents containing the words “cars” and “trucks”. De Morgan’s laws hold that these two searches will return the same set of documents: Search A: NOT (cars OR trucks) Search B: (NOT cars) AND (NOT trucks) The corpus of documents containing “cars” or “trucks” can be represented by four documents: Document 1: Contains only the word “cars”. Document 2: Contains only “trucks”. Document 3: Contains both “cars” and “trucks”. Document 4: Contains neither “cars” nor “trucks”. To evaluate Search A, clearly the search “(cars OR trucks)” will hit on Documents 1, 2, and 3. So the negation of that search (which is Search A) will hit everything else, which is Document 4. Evaluating Search B, the search “(NOT cars)” will hit on documents that do not contain “cars”, which is Documents 2 and 4. Similarly the search “(NOT trucks)” will hit on Documents 1 and 4. Applying the AND operator to these two searches (which is Search B) will hit on the documents that are common to these two searches, which is Document 4. A similar evaluation can be applied to show that the following two searches will return the same set of documents (Documents 1, 2, 4): Search C: NOT (cars AND trucks), Search D: (NOT cars) OR (NOT trucks). ## History The laws are named after Augustus De Morgan (1806–1871),[6] who introduced a formal version of the laws to classical propositional logic. De Morgan's formulation was influenced by algebraization of logic undertaken by George Boole, which later cemented De Morgan's claim to the find. Nevertheless, a similar observation was made by Aristotle, and was known to Greek and Medieval logicians.[7] For example, in the 14th century, William of Ockham wrote down the words that would result by reading the laws out.[8] Jean Buridan, in his Summulae de Dialectica, also describes rules of conversion that follow the lines of De Morgan's laws.[9] Still, De Morgan is given credit for stating the laws in the terms of modern formal logic, and incorporating them into the language of logic. De Morgan's laws can be proved easily, and may even seem trivial.[10] Nonetheless, these laws are helpful in making valid inferences in proofs and deductive arguments. ## Informal proof De Morgan's theorem may be applied to the negation of a disjunction or the negation of a conjunction in all or part of a formula. ### Negation of a disjunction In the case of its application to a disjunction, consider the following claim: "it is false that either of A or B is true", which is written as: ${\displaystyle \neg (A\lor B).}$ In that it has been established that neither A nor B is true, then it must follow that both A is not true and B is not true, which may be written directly as: ${\displaystyle (\neg A)\wedge (\neg B).}$ If either A or B were true, then the disjunction of A and B would be true, making its negation false. Presented in English, this follows the logic that "since two things are both false, it is also false that either of them is true". Working in the opposite direction, the second expression asserts that A is false and B is false (or equivalently that "not A" and "not B" are true). Knowing this, a disjunction of A and B must be false also. The negation of said disjunction must thus be true, and the result is identical to the first claim. ### Negation of a conjunction The application of De Morgan's theorem to a conjunction is very similar to its application to a disjunction both in form and rationale. Consider the following claim: "it is false that A and B are both true", which is written as: ${\displaystyle \neg (A\land B).}$ In order for this claim to be true, either or both of A or B must be false, for if they both were true, then the conjunction of A and B would be true, making its negation false. Thus, one (at least) or more of A and B must be false (or equivalently, one or more of "not A" and "not B" must be true). This may be written directly as, ${\displaystyle (\neg A)\lor (\neg B).}$ Presented in English, this follows the logic that "since it is false that two things are both true, at least one of them must be false". Working in the opposite direction again, the second expression asserts that at least one of "not A" and "not B" must be true, or equivalently that at least one of A and B must be false. Since at least one of them must be false, then their conjunction would likewise be false. Negating said conjunction thus results in a true expression, and this expression is identical to the first claim. ## Formal proof The proof that ${\displaystyle (A\cap B)^{c}=A^{c}\cup B^{c}}$ is completed in 2 steps by proving both ${\displaystyle (A\cap B)^{c}\subseteq A^{c}\cup B^{c}}$ and ${\displaystyle A^{c}\cup B^{c}\subseteq (A\cap B)^{c}}$. ### Part 1 Let ${\displaystyle x\in (A\cap B)^{c}}$. Then, ${\displaystyle x\not \in A\cap B}$. Because ${\displaystyle A\cap B=\{y|y\in A\land y\in B\}}$, it must be the case that ${\displaystyle x\not \in A}$ or ${\displaystyle x\not \in B}$. If ${\displaystyle x\not \in A}$, then ${\displaystyle x\in A^{c}}$, so ${\displaystyle x\in A^{c}\cup B^{c}}$. Similarly, if ${\displaystyle x\not \in B}$, then ${\displaystyle x\in B^{c}}$, so ${\displaystyle x\in A^{c}\cup B^{c}}$. Thus, ${\displaystyle \forall x(x\in (A\cap B)^{c}\rightarrow x\in A^{c}\cup B^{c})}$; that is, ${\displaystyle (A\cap B)^{c}\subseteq A^{c}\cup B^{c}}$. ### Part 2 To prove the reverse direction, let ${\displaystyle x\in A^{c}\cup B^{c}}$, and assume ${\displaystyle x\not \in (A\cap B)^{c}}$. Under that assumption, it must be the case that ${\displaystyle x\in A\cap B}$, so it follows that ${\displaystyle x\in A}$ and ${\displaystyle x\in B}$, and thus ${\displaystyle x\not \in A^{c}}$ and ${\displaystyle x\not \in B^{c}}$. However, that means ${\displaystyle x\not \in A^{c}\cup B^{c}}$, in contradiction to the hypothesis that ${\displaystyle x\in A^{c}\cup B^{c}}$, therefore, the assumption ${\displaystyle x\not \in (A\cap B)^{c}}$ must not be the case, meaning that ${\displaystyle x\in (A\cap B)^{c}}$. Hence, ${\displaystyle \forall x(x\in A^{c}\cup B^{c}\rightarrow x\in (A\cap B)^{c})}$, that is, ${\displaystyle A^{c}\cup B^{c}\subseteq (A\cap B)^{c}}$. ### Conclusion If ${\displaystyle A^{c}\cup B^{c}\subseteq (A\cap B)^{c}}$ and ${\displaystyle (A\cap B)^{c}\subseteq A^{c}\cup B^{c}}$, then ${\displaystyle (A\cap B)^{c}=A^{c}\cup B^{c}}$; this concludes the proof of De Morgan's law. The other De Morgan's law, ${\displaystyle (A\cup B)^{c}=A^{c}\cap B^{c}}$, is proven similarly. ## Generalising De Morgan duality De Morgan's Laws represented as a circuit with logic gates In extensions of classical propositional logic, the duality still holds (that is, to any logical operator one can always find its dual), since in the presence of the identities governing negation, one may always introduce an operator that is the De Morgan dual of another. This leads to an important property of logics based on classical logic, namely the existence of negation normal forms: any formula is equivalent to another formula where negations only occur applied to the non-logical atoms of the formula. The existence of negation normal forms drives many applications, for example in digital circuit design, where it is used to manipulate the types of logic gates, and in formal logic, where it is needed to find the conjunctive normal form and disjunctive normal form of a formula. Computer programmers use them to simplify or properly negate complicated logical conditions. They are also often useful in computations in elementary probability theory. Let one define the dual of any propositional operator P(p, q, ...) depending on elementary propositions p, q, ... to be the operator ${\displaystyle {\mbox{P}}^{d}}$ defined by ${\displaystyle {\mbox{P}}^{d}(p,q,...)=\neg P(\neg p,\neg q,\dots ).}$ ## Extension to predicate and modal logic This duality can be generalised to quantifiers, so for example the universal quantifier and existential quantifier are duals: ${\displaystyle \forall x\,P(x)\equiv \neg [\exists x\,\neg P(x)]}$ ${\displaystyle \exists x\,P(x)\equiv \neg [\forall x\,\neg P(x)]}$ To relate these quantifier dualities to the De Morgan laws, set up a model with some small number of elements in its domain D, such as D = {a, b, c}. Then ${\displaystyle \forall x\,P(x)\equiv P(a)\land P(b)\land P(c)}$ and ${\displaystyle \exists x\,P(x)\equiv P(a)\lor P(b)\lor P(c).\,}$ But, using De Morgan's laws, ${\displaystyle P(a)\land P(b)\land P(c)\equiv \neg (\neg P(a)\lor \neg P(b)\lor \neg P(c))}$ and ${\displaystyle P(a)\lor P(b)\lor P(c)\equiv \neg (\neg P(a)\land \neg P(b)\land \neg P(c)),}$ verifying the quantifier dualities in the model. Then, the quantifier dualities can be extended further to modal logic, relating the box ("necessarily") and diamond ("possibly") operators: ${\displaystyle \Box p\equiv \neg \Diamond \neg p,}$ ${\displaystyle \Diamond p\equiv \neg \Box \neg p.\,}$ In its application to the alethic modalities of possibility and necessity, Aristotle observed this case, and in the case of normal modal logic, the relationship of these modal operators to the quantification can be understood by setting up models using Kripke semantics.

http://zbmath.org/?q=an:1187.46067

# zbMATH — the first resource for mathematics ##### Examples Geometry Search for the term Geometry in any field. Queries are case-independent. Funct* Wildcard queries are specified by * (e.g. functions, functorial, etc.). Otherwise the search is exact. "Topological group" Phrases (multi-words) should be set in "straight quotation marks". au: Bourbaki & ti: Algebra Search for author and title. The and-operator & is default and can be omitted. Chebyshev | Tschebyscheff The or-operator | allows to search for Chebyshev or Tschebyscheff. "Quasi* map*" py: 1989 The resulting documents have publication year 1989. so: Eur* J* Mat* Soc* cc: 14 Search for publications in a particular source with a Mathematics Subject Classification code (cc) in 14. "Partial diff* eq*" ! elliptic The not-operator ! eliminates all results containing the word elliptic. dt: b & au: Hilbert The document type is set to books; alternatively: j for journal articles, a for book articles. py: 2000-2015 cc: (94A | 11T) Number ranges are accepted. Terms can be grouped within (parentheses). la: chinese Find documents in a given language. ISO 639-1 language codes can also be used. ##### Operators a & b logic and a | b logic or !ab logic not abc* right wildcard "ab c" phrase (ab c) parentheses ##### Fields any anywhere an internal document identifier au author, editor ai internal author identifier ti title la language so source ab review, abstract py publication year rv reviewer cc MSC code ut uncontrolled term dt document type (j: journal article; b: book; a: book article) A fixed point approach to almost quartic mappings in quasi fuzzy normed spaces. (English) Zbl 1187.46067 Summary: We define a notion for a quasi fuzzy $p$-normed space, then we use the fixed point alternative theorem to establish Hyers-Ulam-Rassias stability of the quartic functional equation where functions map a linear space into a complete quasi fuzzy $p$-normed space. Later, we show that there exists a close relationship between the fuzzy continuity behavior of a fuzzy almost quartic function, control function and the unique quartic mapping which approximates the almost quartic map. Finally, some applications of our results in the stability of quartic mappings from a linear space into a quasi $p$-norm space will be exhibited. ##### MSC: 46S40 Fuzzy functional analysis 39B82 Stability, separation, extension, and related topics

https://crazyproject.wordpress.com/2011/12/12/compute-the-jordan-canonical-form-of-a-given-matrix-over-a-finite-field/

## Compute the Jordan canonical form of a given matrix over a finite field Let $p$ be a prime. Compute the Jordan canonical form over the finite field $\mathbb{F}_p$ of the $n \times n$ matrix whose every entry is 1. ($n \geq 2$.) Let $A = [1]$. Now the computation in this previous exercise holds over $\mathbb{F}_p$ (indeed, over any ring with 1), and so we have $A(A-nI) = 0$. If $n$ is not divisible by $p$, then $n$ is a unit in $\mathbb{F}_p$. If $v = [v_1\ \ldots\ v_n]^\mathsf{T}$ is an eigenvector with eigenvalue $n$, then evidently (as we saw in the exercise referenced above) $v = v_1[1\ \ldots\ 1]^\mathsf{T}$. So the eigenspace of $n$ has dimension 1, the characteristic polynomial of $A$ is $x^{n-1}(x-n)$, and the Jordan canonical form of $A$ is $J = [b_{i,j}]$ where $b_{n,n} = n$ and $b_{i,j} = 1$ otherwise. Suppose $p$ divides $n$. Since $A \neq 0$, the minimal polynomial of $A$ is now $x^2$. We claim that the remaining invariant factors of $A$ are $x$. There is probably a slick linear-algebraic way to argue this, but I don’t see it. Instead I will show explicitly that $A$ is similar to a matrix in Jordan canonical form. (If anyone can offer a simplification of this proof I’d be grateful!) If $n = 2$, then the minimal polynomial of $A$ is the only invariant factor, so the Jordan form of $A$ is $\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$. Suppose $n \geq 2$. Let $U = [1\ \ldots\ 1]^\mathsf{T}$ have dimension $(n-1) \times 1$, and let $W$ be the $(n-1) \times (n-1)$ matrix whose every entry is 1. Note that $UU^\mathsf{T} = W$ and $U^\mathsf{T}U = [n-1]$. Let $P = \begin{bmatrix} 1 & 0 \\ U & I \end{bmatrix}$ and $Q = \begin{bmatrix} 1 & 0 \\ -U & I \end{bmatrix}$. Evidently $QP = I$, so that $Q = P^{-1}$. Moreover, we have $A = \begin{bmatrix} 1 & U^\mathsf{T} \\ U & W \end{bmatrix}$, and $P^{-1}AP = B = \begin{bmatrix} n & U^\mathsf{T} \\ 0 & 0 \end{bmatrix}$ $= \begin{bmatrix} 0 & U^\mathsf{T} \\ 0 & 0 \end{bmatrix}$. Now let $V = [1\ \ldots\ 1]$ have dimension $1 \times (n-2)$. Let $R = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & -V \\ 0 & 0 & I \end{bmatrix}$ and $S = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & V \\ 0 & 0 & I \end{bmatrix}$. Evidently, $RS = I$, so that $S = R^{-1}$. Now $B = \begin{bmatrix} 0 & 1 & V \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$, and we have $R^{-1}BR = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$. If we squint just right, this matrix is in Jordan canonical form, and is similar to $A$. Post a comment or leave a trackback: Trackback URL.

https://math.libretexts.org/Bookshelves/Algebra/Book%3A_Beginning_Algebra_(Redden)/09%3A_Solving_Quadratic_Equations_and_Graphing_Parabolas/9.01%3A_Extracting_Square_Roots

# 9.1: Extracting Square Roots Skills to Develop • Solve quadratic equations by extracting square roots. ## Extracting Square Roots Recall that a quadratic equation is in standard form if it is equal to 0: $a x^{2}+b x+c=0$ where ab, and c are real numbers and $$a\neq 0$$. A solution to such an equation is called a root. Quadratic equations can have two real solutions, one real solution, or no real solution. If the quadratic expression on the left factors, then we can solve it by factoring. A review of the steps used to solve by factoring follow: Step 1: Express the quadratic equation in standard form. Step 2: Factor the quadratic expression. Step 3: Apply the zero-product property and set each variable factor equal to 0. Step 4: Solve the resulting linear equations. For example, we can solve $$x^{2}-4=0$$ by factoring as follows: \begin{aligned} x^{2}-4 &=0 \\(x+2)(x-2) &=0 \\ x+2 &=0 \quad \text { or } \quad x-2=0 \\ x &=-2 \quad \quad \quad \quad \:x=2 \end{aligned} The two solutions are −2 and 2. The goal in this section is to develop an alternative method that can be used to easily solve equations where b = 0, giving the form $a x^{2}+c=0$ The equation $$x^{2}-4=0$$ is in this form and can be solved by first isolating $$x^{2}$$. \begin{aligned} x^{2}-4 &=0 \\ x^{2} &=4 \end{aligned} If we take the square root of both sides of this equation, we obtain the following: \begin{aligned} \sqrt{x^{2}} &=\sqrt{4} \\|x| &=2 \end{aligned} Here we see that $$x=-2$$ and $$x=2$$ are solutions to the resulting equation. In general, this describes the square root property; for any real number k, $\text{If}\:x^{2}=k, \text { then } x=\pm \sqrt{k}$ The notation “±” is read “plus or minus” and is used as compact notation that indicates two solutions. Hence the statement $$x=\pm \sqrt{k}$$ indicates that $$x= - \sqrt{k}$$or $$x= + \sqrt{k}$$. Applying the square root property as a means of solving a quadratic equation is called extracting the roots. Example $$\PageIndex{1}$$ Solve: $$x^{2}-25=0$$ Solution: Begin by isolating the square. \begin{aligned} x^{2}-25 &=0 \\ x^{2} &=25 \end{aligned} Next, apply the square root property. \begin{aligned} x^{2} &=25 \\ x &=\pm \sqrt{25} \\ x &=\pm 5 \end{aligned} The solutions are −5 and 5. The check is left to the reader. Certainly, the previous example could have been solved just as easily by factoring. However, it demonstrates a technique that can be used to solve equations in this form that do not factor. Example $$\PageIndex{2}$$ Solve: $$x^{2}-5=0$$ Solution: Notice that the quadratic expression on the left does not factor. We can extract the roots if we first isolate the leading term, $$x^{2}$$. \begin{aligned} x^{2}-5 &=0 \\ x^{2} &=5 \end{aligned} Apply the square root property. $$x=\pm \sqrt{5}$$ For completeness, check that these two real solutions solve the original quadratic equation. Generally, the check is optional. $$\begin{array}{rlrl}{\text { Check } x=-\sqrt{5}} & {\text { Check } x=\sqrt{5}} \\ {x^{2}-5=0} & {x^{2}-5=0} \\ {(\color{OliveGreen}{-\sqrt{5}}\color{black}{)}^{2}-5=0} & {(\color{OliveGreen}{\sqrt{5}}\color{black}{)}^{2}-5=0} \\ {5-5=0} & {5-5=0} \\ {0=0 \color{Cerulean}{\checkmark}} & {0=0\color{Cerulean}{\checkmark}}\end{array}$$ The solutions are $$-\sqrt{5}$$ and $$\sqrt{5}$$. Example $$\PageIndex{3}$$ Solve $$4x^{2} - 45 =0$$ Solution Begin by isolating $$x^{2}$$. \begin{aligned} 4 x^{2}-45 &=0 \\ 4 x^{2} &=45 \\ \frac{4 x^{2}}{\color{Cerulean}{4}} &\color{black}{=}\frac{45}{\color{Cerulean}{4}} \\ x^{2} &=\frac{45}{4} \end{aligned} Apply the square root property and then simplify. $$\begin{array}{l}{x=\pm \sqrt{\frac{45}{4}}} \\ {x=\pm \frac{\sqrt{9 \cdot 5}}{\sqrt{4}}} \\ {x=\pm \frac{3 \sqrt{5}}{2}}\end{array}$$ The solutions are $$-\frac{3 \sqrt{5}}{2}$$ and $$\frac{3 \sqrt{5}}{2}$$. Sometimes quadratic equations have no real solution. Example $$\PageIndex{4}$$ Solve: $$x^{2} + 9 =0$$ Solution Begin by isolating $$x^{2}$$. \begin{aligned} x^{2}+9 &=0 \\ x^{2} &=-9 \\ x &=\pm \sqrt{-9} \end{aligned} After applying the square root property, we are left with the square root of a negative number. Therefore, there is no real solution to this equation. No real solution Reverse this process to find equations with given solutions of the form ±k. Example $$\PageIndex{5}$$ Find an equation with solutions $$-2 \sqrt{3}$$ and $$2 \sqrt{3}$$. Solution: Begin by squaring both sides of the following equation: $$\begin{array}{l}{x=\pm 2 \sqrt{3}} \\ {x^{2}=( \pm 2 \sqrt{3})^{2}} \\ {x^{2}=4 \cdot 3} \\ {x^{2}=12}\end{array}$$ Lastly, subtract 12 from both sides and present the equation in standard form. $$x^{2}-12=0$$ Exercise $$\PageIndex{1}$$ Solve: $$9x^{2}-8=0$$. $$x=-\frac{2 \sqrt{2}}{3}$$ or $$x=\frac{2 \sqrt{2}}{3}$$ Consider solving the following equation: $$(x+2)^{2}=25$$ To solve this equation by factoring, first square x+2 and then put it in standard form, equal to zero, by subtracting 25 from both sides. \begin{aligned}(x+2)^{2} &=25 \\ x^{2}+4 x+4 &=25 \\ x^{2}+4 x+4\color{Cerulean}{-25} &\color{black}{=}25\color{Cerulean}{-25} \\ x^{2}+4 x-21 &=0 \end{aligned} Factor and then apply the zero-product property. $$\begin{array}{c}{x^{2}+4 x-21=0} \\ {(x+7)(x-3)=0}\end{array}$$ $$\begin{array}{rr}{x+7=0} & {\text { or } \quad x-3=0} \\ {x=-7} & {x=3}\end{array}$$ The two solutions are -7 and 3. When an equation is in this form, we can obtain the solutions in fewer steps by extracting the roots. Example $$\PageIndex{6}$$ Solve: $$(x=2)^{2}=25$$. Solution Solve by extracting the roots. $$\begin{array}{rlrr}{(x+2)^{2}} & {=25} & {\color{Cerulean}{ Apply\: the\:square \:root \:property.}} \\ {x+2} & {=\pm \sqrt{25}} & {\color{Cerulean}{ Simplify. }} \\ {x+2}& {=\pm5} \\ {x}&{=-2\pm5}\end{array}$$ At this point, separate the “plus or minus” into two equations and simplify each individually. $$\begin{array}{ll}{x=-2-5} & {\text { or } x=-2+5} \\ {x=-7} & {\quad\:\: x=3}\end{array}$$ The solutions are -7 and 3. In addition to fewer steps, this method allows us to solve equations that do not factor. Example $$\PageIndex{7}$$ Solve: $$(3 x+3)^{2}-27=0$$ Solution Begin by isolating the square. \begin{aligned}(3 x+3)^{2}-27 &=0 \\(3 x+3)^{2} &=27 \end{aligned} Next, extract the roots and simplify. \begin{aligned}(3 x+3)^{2} &=27 \\ 3 x+3 &=\pm \sqrt{27} \\ 3 x+3 &=\pm \sqrt{9 \cdot 3} \\ 3 x+3 &=\pm 3 \sqrt{3} \end{aligned} Solve for x. $$\begin{array}{rlrl}{3 x+3} & {=\pm 3 \sqrt{3}} & { \color{Cerulean}{Subtract \: 3 \: on\: both\: sides.} } \\ {3 x} & {=-3 \pm 3 \sqrt{3}} & { \color{Cerulean}{Divide\: both\: sides\: by\: 3}}\\ {\frac{3x}{\color{Cerulean}{3}}}&{\color{black}{=}\frac{-3\pm3\sqrt{3}}{\color{Cerulean}{3}}}&{\color{Cerulean}{Factor\:out\:a\:3.}}\\{x}&{=\frac{3(-1\pm\sqrt{3} )}{3}}&{\color{Cerulean}{Cancel.}}\\{x}&{=-1\pm\sqrt{3}}\end{array}$$ The solutions are $$-1 - \sqrt{3}$$ and $$-1 + \sqrt{3}$$ Example $$\PageIndex{8}$$ Solve: $$9(2 x-1)^{2}-8=0$$ Solution: Begin by isolating the square factor. \begin{aligned} 9(2 x-1)^{2}-8 &=0 \\ 9(2 x-1)^{2} &=8 \\(2 x-1)^{2} &=\frac{8}{9} \end{aligned} Apply the square root property and solve. \begin{aligned} 2 x-1 )^{2} &=\frac{8}{9} \\ 2 x-1 &=\pm \sqrt{\frac{8}{9}} \\ 2 x-1 &=\pm \frac{\sqrt{4 \cdot 2}}{\sqrt{9}} \\ 2 x &=1 \pm \frac{2 \sqrt{2}}{3} \\ 2 x &=\frac{3}{3} \pm \frac{2 \sqrt{2}}{3} \\ 2 x &=\frac{3 \pm 2 \sqrt{2}}{3}\\ \color{Cerulean}{\frac{1}{2}\cdot}\color{black}{2x}&=\color{Cerulean}{\frac{1}{2}\cdot}\color{black}{\frac{3\pm2\sqrt{2}}{3}}\\x&=\frac{3\pm2\sqrt{2}}{6} \end{aligned} The solutions are $$\frac{3 - 2 \sqrt{2}}{6}$$ and $$\frac{3 + 2 \sqrt{2}}{6}$$ Exercise $$\PageIndex{2}$$ Solve: $$3(x-5)^{2}-2=0$$ $$15\pm 6\sqrt{3}$$ Example $$\PageIndex{9}$$ The length of a rectangle is twice its width. If the diagonal measures 2 feet, then find the dimensions of the rectangle. Solution Let w represent the width. Let 2w represent the length. The diagonal of any rectangle forms two right triangles. Thus the Pythagorean theorem applies. The sum of the squares of the legs of a right triangle is equal to the square of the hypotenuse: $$\begin{array}{l}{(\operatorname{leg})^{2}+(\operatorname{leg})^{2}=\text { hypotenuse }^{2}} \\ {(\color{OliveGreen}{2 w}\color{black}{)}^{2}+(\color{OliveGreen}{w}\color{black}{)}^{2}=(\color{OliveGreen}{2}\color{black}{)}^{2}}\end{array}$$ Solve. \begin{aligned}(2 w)^{2}+(w)^{2} &=(2)^{2} \\ 4 w^{2}+w^{2} &=4 \\ 5 w^{2} &=4 \quad\quad\quad\color{Cerulean}{Isolate\:w^{2}.} \\\frac{5w^{2}}{\color{Cerulean}{5}}&\color{black}{=}\frac{4}{\color{Cerulean}{5}}\\ w^{2}&=\frac{4}{5}\quad\:\:\quad\color{Cerulean}{Extract\:the\:roots.} \\ w&=\pm\sqrt{\frac{4}{5}}\\w&=\pm\frac{2}{\sqrt{5}} \end{aligned} Here we obtain two solutions, $$w=−\frac{2}{\sqrt{5}}$$ and $$w=\frac{2}{\sqrt{5}}$$. Since the problem asked for a length of a rectangle, we disregard the negative answer. Furthermore, we will rationalize the denominator and present our solutions without any radicals in the denominator. \begin{aligned} w &=\frac{2}{\sqrt{5}} \quad\quad\quad\quad\color{Cerulean}{Rationalize\:the\:denominator.} \\ &=\frac{2}{\sqrt{5}} \cdot \color{Cerulean}{\frac{\sqrt{5}}{\sqrt{5}}} \\ &=\frac{2 \sqrt{5}}{\sqrt{25}} \\ &=\frac{2 \sqrt{5}}{5} \end{aligned} Back substitute to find the length. \begin{aligned} l &=2 w \\ &=2\color{black}{\left(\color{OliveGreen}{\frac{2 \sqrt{5}}{5}}\right)} \\ &=\frac{4 \sqrt{5}}{5} \end{aligned} The length of the rectangle is $$\frac{4 \sqrt{5}}{5}$$ feet and the width is $$\frac{2 \sqrt{5}}{5}$$ feet. ## Key Takeaways • Solve equations of the form $$ax^{2}+c=0$$ by extracting the roots. • Extracting roots involves isolating the square and then applying the square root property. After applying the square root property, you have two linear equations that each can be solved. Be sure to simplify all radical expressions and rationalize the denominator if necessary. Exercise $$\PageIndex{3}$$ extracting square roots Solve by factoring and then solve by extracting roots. Check answers. 1. $$x^{2}-36=0$$ 2. $$x^{2}-81=0$$ 3. $$4y^{2}−9=0$$ 4. $$9y^{2}−25=0$$ 5. $$(x−2)^{2}−1=0$$ 6. $$(x+1)^{2}−4=0$$ 7. $$4(y−2)^{2}−9=0$$ 8. $$9(y+1)^{2}−4=0$$ 9. $$−3(t−1)^{2}+12=0$$ 10. $$−2(t+1)^{2}+8=0$$ 11. $$(x−5)^{2}−25=0$$ 12. $$(x+2)^{2}−4=0$$ 1. $$-6, 6$$ 3. $$-\frac{3}{2} , \frac{3}{2}$$ 5. $$1, 3$$ 7. $$\frac{1}{2}, \frac{7}{2}$$ 9. $$-1, 3$$ 11. $$0, 10$$ Exercise $$\PageIndex{4}$$ extracting square roots Solve by extracting the roots. 1. $$x^{2}=16$$ 2. $$x^{2}=1$$ 3. $$y^{2}=9$$ 4. $$y^{2}=64$$ 5. $$x^{2}=14$$ 6. $$x^{2}=19$$ 7. $$y^{2}=0.25$$ 8. $$y^{2}=0.04$$ 9. $$x^{2}=12$$ 10. $$x^{2}=18$$ 11. $$16x^{2}=9$$ 12. $$4x^{2}=25$$ 13. $$2t^{2} = 1$$ 14. $$3t^{2} = 2$$ 15. $$x^{2} −100 = 0$$ 16. $$x^{2} −121 = 0$$ 17. $$y^{2} + 4 = 0$$ 18. $$y^{2} + 1 = 0$$ 19. $$x^{2} −49 = 0$$ 20. $$x^{2} −925 = 0$$ 21. $$y^{2} −0.09 = 0$$ 22. $$y^{2} −0.81 = 0$$ 23. $$x^{2} − 7 = 0$$ 24. $$x^{2} − 2 = 0$$ 25. $$x^{2} − 8 = 0$$ 26. $$t^{2} −18 = 0$$ 27. $$x^{2} + 8 = 0$$ 28. $$x^{2} +125 = 0$$ 29. $$16x^{2} −27 = 0$$ 30. $$9x^{2} − 8 = 0$$ 31. $$2y^{2} − 3 = 0$$ 32. $$5y^{2} − 2 = 0$$ 33. $$3x^{2} − 1 = 0$$ 34. $$6x^{2} − 3 = 0$$ 35. $$(x + 7 )^{2} − 4 = 0$$ 36. $$(x + 9 )^{2} −36 = 0$$ 37. $$( 2y − 3 )^{2} −81 = 0$$ 38. $$( 2y + 1 )^{2} −25 = 0$$ 39. $$(x − 5 )^{2} −20 = 0$$ 40. $$(x + 1 )^{2} −28 = 0$$ 41. $$( 3t + 2 )^{2} − 6 = 0$$ 42. $$(3t−5)^{2}−10=0$$ 43. $$4(y+2)^{2}−3=0$$ 44. $$9(y−7)^{2}−5=0$$ 45. $$4(3x+1)^{2}−27=0$$ 46. $$9(2x−3)^{2}−8=0$$ 47. $$2(3x−1)^{2}+3=0$$ 48. $$5(2x−1)^{2}−3=0$$ 49. $$3(y−23)^{2}−32=0$$ 50. $$2(3y−13)^{2}−52=0$$ 1. ±4 3. ±3 5. ±$$\frac{1}{2}$$ 7. ±0.5 9. ±$$2\sqrt{3}$$ 11. ±$$\frac{3}{4}$$ 13. ±$$\sqrt{\frac{1}{2}}$$ 15. ±10 17. No real solution 19. ±$$\frac{2}{3}$$ 21. ±0.3 23. ±$$\sqrt{7}$$ 25. ±$$2\sqrt{2}$$ 27. No real solution 29. ±$$\frac{3 \sqrt{3}}{4}$$ 31. ±$$6\sqrt{2}$$ 33. ±$$3\sqrt{3}$$ 35. $$−9, −5$$ 37. $$−3, 6$$ 39. $$5$$±$$2\sqrt{5}$$ 41. ±$$\frac{\sqrt{6}-2}{3}$$ 43. ±$$\frac{\sqrt{3}}{2}-2$$ 45. ±$$\frac{3 \sqrt{3}-2}{6}$$ 47. No real solution 49. ±$$\frac{4 \sqrt{6}}{3}+23$$ Exercise $$\PageIndex{5}$$ extracting square roots Find a quadratic equation in standard form with the following solutions. 1. ±7 2. ±13 3. ±$$\sqrt{7}$$ 4. ±$$\sqrt{3}$$ 5. ±$$3\sqrt{5}$$ 6. ±$$5\sqrt{2}$$ 7. $$1$$±$$\sqrt{2}$$ 8. $$2$$±$$\sqrt{3}$$ 1. $$x^{2}−49=0$$ 3. $$x^{2}−7=0$$ 5. $$x^{2}−45=0$$ 7. $$x^{2}−2x−1=0$$ Exercise $$\PageIndex{6}$$ extracting square roots Solve and round off the solutions to the nearest hundredth. 1. $$9x(x+2)=18x+1$$ 2. $$x^{2}=10(x^{2}−2)−5$$ 3. $$(x+3)(x−7)=11−4x$$ 4. $$(x−4)(x−3)=66−7x$$ 5. $$(x−2)^{2}=67−4x$$ 6. $$(x+3)^{2}=6x+59$$ 7. $$(2x+1)(x+3)−(x+7)=(x+3)^{2}$$ 8. $$(3x−1)(x+4)=2x(x+6)−(x−3)$$ 1. ±0.33 3. ±5.66 5. ±7.94 7. ±3.61 Exercise $$\PageIndex{7}$$ extracting square roots Set up an algebraic equation and use it to solve the following. 1. If 9 is subtracted from 4 times the square of a number, then the result is 3. Find the number. 2. If 20 is subtracted from the square of a number, then the result is 4. Find the number. 3. If 1 is added to 3 times the square of a number, then the result is 2. Find the number. 4. If 3 is added to 2 times the square of a number, then the result is 12. Find the number. 5. If a square has an area of 8 square centimeters, then find the length of each side. 6. If a circle has an area of 32$$\pi$$ square centimeters, then find the length of the radius. 7. The volume of a right circular cone is 36$$\pi$$ cubic centimeters when the height is 6 centimeters. Find the radius of the cone. (The volume of a right circular cone is given by $$V=13\pi r^{2}h$$.) 8. The surface area of a sphere is 75$$\pi$$ square centimeters. Find the radius of the sphere. (The surface area of a sphere is given by SA=$$4\pi r^{2}$$.) 9. The length of a rectangle is 6 times its width. If the area is 96 square inches, then find the dimensions of the rectangle. 10. The base of a triangle is twice its height. If the area is 16 square centimeters, then find the length of its base. 11. A square has an area of 36 square units. By what equal amount will the sides have to be increased to create a square with double the given area? 12. A circle has an area of 25$$\pi$$ square units. By what amount will the radius have to be increased to create a circle with double the given area? 13. If the sides of a square measure 1 unit, then find the length of the diagonal. 14. If the sides of a square measure 2 units, then find the length of the diagonal. 15. The diagonal of a square measures 5 inches. Find the length of each side. 16. The diagonal of a square measures 3 inches. Find the length of each side. 17. The length of a rectangle is twice its width. If the diagonal measures 10 feet, then find the dimensions of the rectangle. 18. The length of a rectangle is twice its width. If the diagonal measures 8 feet, then find the dimensions of the rectangle. 19. The length of a rectangle is 3 times its width. If the diagonal measures 5 meters, then find the dimensions of the rectangle. 20. The length of a rectangle is 3 times its width. If the diagonal measures 2 feet, then find the dimensions of the rectangle. 21. The height in feet of an object dropped from a 9-foot ladder is given by h(t)$$=−16t^{2}+9$$, where t represents the time in seconds after the object has been dropped. How long does it take the object to hit the ground? (Hint: The height is 0 when the object hits the ground.) 22. The height in feet of an object dropped from a 20-foot platform is given by h(t)$$=−16t^{2}+20$$, where t represents the time in seconds after the object has been dropped. How long does it take the object to hit the ground? 23. The height in feet of an object dropped from the top of a 144-foot building is given by h(t)$$=−16t^{2}+144$$, where t is measured in seconds. 1. How long will it take to reach half of the distance to the ground, 72 feet? 2. How long will it take to travel the rest of the distance to the ground? Round off to the nearest hundredth of a second. 24. The height in feet of an object dropped from an airplane at 1,600 feet is given by h(t)$$=−16t^{2}+1,600$$, where t is in seconds. 1. How long will it take to reach half of the distance to the ground? 2. How long will it take to travel the rest of the distance to the ground? Round off to the nearest hundredth of a second. 25. Create an equation of your own that can be solved by extracting the root. Share it, along with the solution, on the discussion board. 26. Explain why the technique of extracting roots greatly expands our ability to solve quadratic equations. 27. Explain in your own words how to solve by extracting the roots. 106. Derive a formula for the diagonal of a square in terms of its sides. 1. $$−\sqrt{3}$$ or $$\sqrt{3}$$ 3. $$−3\sqrt{3}$$ or $$3\sqrt{3}$$ 5. $$2\sqrt{2}$$ centimeters 7. $$3\sqrt{2}$$ centimeters 9. Length: 24 inches; width: 4 inches 11. $$−6+6\sqrt{2}\approx 2.49$$ units 13. $$\sqrt{2}$$ units 15. $$52\sqrt{2}$$ inches 17. Length: $$4\sqrt{5}$$ feet; width: $$2\sqrt{5}$$ feet 19. Length: $$310−−\sqrt{2}$$ meters; width: $$10−−\sqrt{2}$$ meters 21. $$\frac{3}{4}$$ second 23. a. 2.12 seconds     b. 0.88 second 25. Answers may vary

https://core.ac.uk/display/334940028

## Random points are optimal for the approximation of Sobolev functions ### Abstract We show that independent and uniformly distributed sampling points are as good as optimal sampling points for the approximation of functions from the Sobolev space $W_p^s(\Omega)$ on bounded convex domains $\Omega\subset \mathbb{R}^d$ in the $L_q$-norm if $q<p$. More generally, we characterize the quality of arbitrary sampling points $P\subset \Omega$ via the $L_\gamma(\Omega)$-norm of the distance function $\rm{dist}(\cdot,P)$, where $\gamma=s(1/q-1/p)^{-1}$ if $q<p$ and $\gamma=\infty$ if $q\ge p$. This improves upon previous characterizations based on the covering radius of $P$ Topics: Mathematics - Numerical Analysis, Mathematics - Functional Analysis, 41A25, 41A63, 62D05, 65D15, 65D30 Year: 2020 OAI identifier: oai:arXiv.org:2009.11275 ### Suggested articles To submit an update or takedown request for this paper, please submit an Update/Correction/Removal Request.

http://michaelsyao.com/calculus/trig-function-derivatives/

# Derivatives of Trigonometric Functions ## Proofs Recall the definition of the derivative that we introduced in our lesson on Introduction to Derivatives: $\frac{df}{dx}=\lim_{h\rightarrow0}\frac{f(x+h)-f(x)}{h}$ Consider the case where $f(x)=\sin x$. Plugging this into the definition of the derivative above, we find $\frac{d}{dx}\left[\sin x\right]=\lim_{h\rightarrow0}\frac{\sin(x+h)-\sin(x)}{h}=\lim_{h\rightarrow 0}\frac{\sin x\cos h+\cos x\sin h-\sin x}{h}$ We expanded out $\sin(x+h)$ using the sum-of-angles rule. Using linearity of the limit definition, we can rewrite the right hand side: $\frac{d}{dx}\left[\sin x\right]=\cos x\lim_{h\rightarrow 0}\frac{\sin h}{h}+\sin x\lim_{h\rightarrow 0}\frac{\cos h-1}{h}$ Using the sandwich theorem (also called the squeeze theorem), it can be shown that $\lim_{h\rightarrow 0}\frac{\sin h}{h}=1$ and $\lim_{h\rightarrow 0}\frac{\cos h -1}{h}=0$ (you will prove this in an exercise below). Plugging these results in, we find $\frac{d}{dx}\left[\sin x\right]=(1)\cos x+(0)\sin x=\cos x$ This result tells us that the derivative of $\sin x$ is simply $\cos x$. Let’s see if this intuitively makes sense by plotting both of these functions together: At $x=0$, $\sin(x)=0$, and moreover, it is increasing very rapidly. Therefore, it makes sense that its derivative is positive ($\cos(x)=1$). At $x=\pi/2$, $\sin(x)$ achieves its maximum value, and local maxima correspond to a derivative equal to zero, which makes sense since $\cos(\pi/2)=0$. At $x=\pi$, $\sin(x)=0$, and more importantly, it is decreasingly very rapidly. Therefore, it makes sense that its derivative if negative ($\cos(x)=-1$). With these cursory checks, its clear that $\frac{d}{dx}[\sin x]=\cos x$ seems to be a fairly valid statement. Let’s also consider the derivative of $f(x)=\cos x$, using the same definition of the derivative from above. $\frac{d}{dx}[\cos x]=\lim_{h\rightarrow 0}\frac{\cos(x+h)-\cos(x)}{h}=\lim_{h\rightarrow 0}\frac{\cos x\cos h-\sin x\sin h-\cos x}{h}$ We again expanded out $\sin(x+h)$ using the sum-of-angles rule. Continuing with the same process as above for $f(x)=\sin x$, $\frac{d}{dx}[\cos x]=\cos x\lim_{h\rightarrow0}\frac{\cos h-1}{h}-\sin x\lim_{h\rightarrow 0}\frac{\sin h}{h}$ Using the same special limit results from above, we have $\frac{d}{dx}[\cos x]=(0)\cos x-(1)\sin x=-\sin x$ Therefore the derivative of $\cos x$ is $-\sin x$. Combining these results, we have ## The Main Formulas $\frac{d}{dx}[\sin x]=\cos x, \quad \frac{d}{dx}[\cos x]=-\sin x$ Using these two rules, we can derive the following as well: • $\frac{d^n}{dx^n}[\sin x]=\begin{cases}(-1)^{n / 2}\sin x \text{ for even } n\\ (-1)^{(n-1)/2}\cos x\text{ for odd } n\end{cases}$ • $\frac{d^n}{dx^n}[\cos x]=\begin{cases}(-1)^{n / 2}\cos x \text{ for even } n\\ (-1)^{(n+1)/2}\cos x\text{ for odd } n\end{cases}$ Furthermore, we can also derive the following using the quotient rule: • $\frac{d}{dx}[\tan x]=\frac{1}{\cos^2 x}=\sec^2 x$ • $\frac{d}{dx}[\csc x]=-\cot x\csc x$ • $\frac{d}{dx}[\sec x]=\tan x\sec x$ • $\frac{d}{dx}[\cot x]=-\frac{1}{\sin^2 x}=-\csc^2 x$ You will prove these statements in the exercises below. ## Exercises ### Problem 1 Using the squeeze theorem, prove the following limits: 1. $\lim_{h\rightarrow 0}\frac{\sin h}{h}=1$ 2. $\lim_{h\rightarrow 0}\frac{\cos h-1}{h}=0$ ### Problem 2 1. Find the second derivative of $f(x)=\sin x$. Repeat this for $g(x)=\cos x$. 2. Find the third derivative of $f(x)=\sin x$. Repeat this for $g(x)=\cos x$. 3. Find the fourth derivative of $f(x)=\sin x$. Repeat this for $g(x)=\cos x$. 4. Find the fifth derivative of $f(x)=\sin x$. Repeat this for $g(x)=\cos x$. 5. Do you notice any patterns? Do these results agree with the general result that we found above? ### Problem 3 We gave derivatives of $\tan x, \csc x, \sec x, \cot x$ above. Prove these statements using the quotient rule and the derivatives of $\sin x, \cos x$ that we determined above. ### Problem 4 Here, we will incorporate these trigonometric function derivatives into our standard repertoire of the chain rule, product rule, quotient rule, and other differentiation techniques. Evaluate the derivatives of the following functions: 1. $f(x)=\frac{x^2\sin x}{x^2+1}$ 2. $f(x)=\sqrt{x}\cot x$ 3. $f(x)=\sqrt[3]{\sin \sqrt[3]{x}}$ 4. $f(x)=\cos(\pi x^2)$

https://artofproblemsolving.com/wiki/index.php/1998_CEMC_Gauss_(Grade_7)_Problems/Problem_15

# 1998 CEMC Gauss (Grade 7) Problems/Problem 15 ## Problem The diagram shows a magic square in which the sums of the numbers in any row, column or diagonal are equal. What is the value of n? [A 3x3 magic square grid is shown. 8 is in the 1st row 1st column. 9 is in the 2nd row 1st column. 4 is in the 2nd row 3rd column. 4 is in the 3rd row 1st column. $n$ is in the 3rd row 2nd column.] $\text{(A)}\ 3 \qquad \text{(B)}\ 6 \qquad \text{(C)}\ 7 \qquad \text{(D)}\ 10 \qquad \text{(E)}\ 11$ ## Solution The "magic" sum is $8 + 9 + 4 = 21,$ so the center square (2nd row 2nd column) is $21 - 9 - 5 = 7.$ The square in the lower right (3rd row 3rd column) has value 6, therefore $4 + n + 6 = 21.$ The answer is $\text{(E)} \quad 11.$

https://icsecbsemath.com/2016/07/30/class-8-chapter-22-linear-equations-exercise-22b/

Question 1: $\displaystyle 17$ less than four times a number is $\displaystyle 11$ . Find the number. $\displaystyle \text{Let the number } = x$ $\displaystyle \Rightarrow 4x-17 = 11$ $\displaystyle \Rightarrow x = \frac{28}{4} = 7$ $\displaystyle \\$ Question 2: If $\displaystyle 10$ be added to four times a certain number the result is $\displaystyle 5$ less than five times the number. Find the number. $\displaystyle \text{Let the number } = x$ $\displaystyle \Rightarrow 4x+10 = 5x-5$ $\displaystyle \Rightarrow x = 15$ $\displaystyle \\$ Question 3: $\displaystyle \frac{2}{3}$ Of a number is $\displaystyle 20$ less than the original number. Find the original number. $\displaystyle \text{Let the original number } = x$ $\displaystyle \frac{2}{3} x = x-20$ $\displaystyle \Rightarrow \frac{1}{3} x = 20$ $\displaystyle \Rightarrow x = 60$ $\displaystyle \\$ Question 4: A number is $\displaystyle 25$ more than its part. Find the number. $\displaystyle \text{Let the number } = x$ $\displaystyle x = 25+ \frac{5}{6} x$ $\displaystyle \Rightarrow \frac{1}{6} x = 25$ $\displaystyle \Rightarrow x = 150$ $\displaystyle \\$ Question 5: A number is as much greater than $\displaystyle 21$ as is less than $\displaystyle 71$ . Find the number. $\displaystyle \text{Let the number } = x$ $\displaystyle x-21 = 71-x$ $\displaystyle \Rightarrow 2x = 92$ $\displaystyle \Rightarrow x = 46$ $\displaystyle \\$ Question 6: 6 more than one-fourth of the number is two-fifth of the number. Find the number. $\displaystyle \text{Let the number } = x$ $\displaystyle \frac{1}{4} x+6 = \frac{2}{5} x$ $\displaystyle \Rightarrow 6 = ( \frac{2}{5} - \frac{1}{4} )x = \frac{30}{20} x$ $\displaystyle \Rightarrow x = 40$ $\displaystyle \\$ Question 7: One-third of a number exceeds one-fourth of the number by $\displaystyle 15$ . Find the number. $\displaystyle \text{Let the number } = x$ $\displaystyle \frac{1}{3} x- \frac{1}{4} x = 15$ $\displaystyle \Rightarrow \frac{1}{12} x = 15 \ or \ x = 180$ $\displaystyle \\$ Question 8: If one-fifth of a number decreased by $\displaystyle 5$ is $\displaystyle 16$ find the number. $\displaystyle \text{Let the number } = x$ $\displaystyle \frac{1}{5} x-5 = 16$ $\displaystyle \Rightarrow x = 105$ $\displaystyle \\$ Question 9: A number when divided by $\displaystyle 6$ is diminished by $\displaystyle 40$ . Find the number. $\displaystyle \text{Let the number } = x$ $\displaystyle \frac{x}{6} = x-40$ $\displaystyle \Rightarrow \frac{5}{6} x = 40$ $\displaystyle \Rightarrow x = 48$ $\displaystyle \\$ Question 10: Four-fifths of a number exceeds two-third of the number by $\displaystyle 10$ . Find the number. $\displaystyle \text{Let the number } = x$ $\displaystyle \frac{4}{3} x = \frac{2}{3} x+10$ $\displaystyle \Rightarrow \frac{2}{15} x = 10$ $\displaystyle \Rightarrow x = 75$ $\displaystyle \\$ Question 11: Two numbers are in the ratio $\displaystyle 3:4$ and their sum is $\displaystyle 84$ . Find the number. Let the two numbers be $\displaystyle x$ and $\displaystyle y$ Therefore $\displaystyle 3x = 4y$ $\displaystyle x+y = 84$ Solving $\displaystyle \frac{4}{3} y+y = 84$ $\displaystyle \frac{7y}{3} = 84$ $\displaystyle y = 3 \times 12 = 36$ $\displaystyle \text{Hence } x = \frac{4}{3} \times 36 = 48$ $\displaystyle \\$ Question 12: Three numbers are in ratio $\displaystyle 4:5:6$ and their sum is $\displaystyle 135$ . Find the numbers. Let the three numbers be $\displaystyle x \ y \ z$ Therefore $\displaystyle 4x:5x:6x$ $\displaystyle 4x+5x+6x = 135$ Solving $\displaystyle 15x = 135$ $\displaystyle x = 9$ Therefore The three numbers are $\displaystyle 36 \ 45 \ 54$ $\displaystyle \\$ Question 13: Two numbers are in the ratio $\displaystyle 3:5$ . If each is increased by $\displaystyle 10$ then ratio between the new numbers so formed is $\displaystyle 5:7$ Find the original numbers. Let the two numbers be $\displaystyle x$ and $\displaystyle y$ Given $\displaystyle \frac{x}{y} = \frac{3}{5} \ldots \ldots \ldots i)$ $\displaystyle \frac{x+10}{y+10} = \frac{5}{7} \ldots \ldots \ldots ii)$ solving From $\displaystyle i) x = \frac{3}{5} y$ Substituting in ii) $\displaystyle \frac{\frac{3}{5} y+10}{y+10} = \frac{5}{7}$ $\displaystyle \frac{21}{5} y+70 = 5y+50$ $\displaystyle 20 = \frac{4}{5} y$ or $\displaystyle y = 25$ $\displaystyle x = \frac{3}{5} \times 25 = 15$ Two numbers are $\displaystyle 15$ and $\displaystyle 25$ $\displaystyle \\$ Question 14: The sum of three consecutive odd numbers is $\displaystyle 75$ . Find the numbers. Let the three consecutive numbers be $\displaystyle x x+2 x+4$ therefore $\displaystyle x+x+2+x+4 = 75$ $\displaystyle 3x+6 = 75$ $\displaystyle 3x = 69$ $\displaystyle x = 23$ Therefore the three numbers are $\displaystyle 23 \ 25 \ and \ 27$ $\displaystyle \\$ Question 15: Divide $\displaystyle 25$ into two parts such that $\displaystyle 7$ times the first part added to $\displaystyle 5$ times the second part makes $\displaystyle 139$ . Let the two parts be $\displaystyle x$ and $\displaystyle y$ Therefore $\displaystyle x+y = 25$ $\displaystyle 7x+5y+139$ Solving we get $\displaystyle x = 25-y$ $\displaystyle 7(25-y)+5y = 139$ $\displaystyle 175-2y = 139$ $\displaystyle 2y = 175-139 = 36$ or $\displaystyle y = 18$ The other part $\displaystyle = 7$ $\displaystyle \\$ Question 16: Divide $\displaystyle 180$ into two parts such that the first part is $\displaystyle 12$ less than twice the second part. Let the two parts be $\displaystyle x$ and $\displaystyle 2y$ Therefore $\displaystyle x+y = 180$ $\displaystyle x+12 = 2y$ Solving $\displaystyle y = 180-x$ $\displaystyle x+12 = 2(180-x)$ $\displaystyle 3x = 360-12 = 348$ $\displaystyle x = 116$ $\displaystyle \text{Therefore } y = 180-116 = 64$ $\displaystyle \\$ Question 17: The denominator of the fraction is $\displaystyle 4$ more than its numerator. On subtracting $\displaystyle 1$ from each numerator and denominator the fraction becomes. Find the original fraction. Let the fraction be $\displaystyle \frac{x}{y}$ Given $\displaystyle y = x+4$ $\displaystyle \text{Therefore the fraction } = \frac{x}{x+4}$ Given $\displaystyle \frac{x-1}{x+4-1} = \frac{1}{2}$ $\displaystyle 2x-2 = x+3$ $\displaystyle x = 5 \text{ and } y = 9$ $\displaystyle \text{Therefore fraction } = \frac{5}{9}$ $\displaystyle \\$ Question 18: The denominator of the fraction is $\displaystyle 1$ more than the double the numerator. On adding $\displaystyle 2$ to the numerator and subtracting $\displaystyle 3$ from denominator we obtain $\displaystyle 1$ . Find the original fraction. Let the fraction be $\displaystyle \frac{x}{2x+1}$ Given $\displaystyle \frac{x+2}{2x+1-3} = 1$ $\displaystyle x+2 = 2x-2$ $\displaystyle x = 4$ Fraction $\displaystyle = \frac{4}{9}$ $\displaystyle \\$ Question 19: The sum of the digits of a two-digit number is $\displaystyle 5$ . On adding $\displaystyle 27$ to the number its digits are reversed. Find the original number. Let the two digit number be $\displaystyle xy$ Given $\displaystyle x+y = 5 \ldots \ldots \ldots i)$ $\displaystyle xy+27 = yx$ $\displaystyle 10x+y+27 = 10y+x$ $\displaystyle 9x+27 = 9y$ or $\displaystyle x+3 = y \ldots \ldots \ldots ii)$ Solving i) and ii) together. $\displaystyle x+3 = (5-x)$ $\displaystyle 2x = 2$ $\displaystyle x = 1$ $\displaystyle y = 4$ Hence the number $\displaystyle = 14$ $\displaystyle \\$ Question 20: What same numbers should be added to each one of the number $\displaystyle 15 23 29 44$ to obtain numbers which are in proportion? Let the number added to each one of $\displaystyle 15 23 29 44$ be $\displaystyle x$ $\displaystyle \frac{15+x}{23+x} = \frac{29+x}{44+x}$ $\displaystyle 660+59x+ x^2 = 667+52x+ x^2$ $\displaystyle 7x = 7$ $\displaystyle x = 1$ $\displaystyle \\$ Question 21: The sum of two numbers is $\displaystyle 110$ . One-fifth of the larger number is $\displaystyle 8$ more than one-ninth of the smaller number. Find the numbers. Let the two numbers be $\displaystyle x$ and $\displaystyle y$ Given $\displaystyle x+y 110$ $\displaystyle \frac{1}{5} x = \frac{1}{9} y+8$ Solving $\displaystyle \frac{1}{5} x = \frac{1}{9} (110-x)+ 8$ $\displaystyle (\frac{1}{5} + \frac{1}{9} )x = \frac{110}{9} +8 = \frac{182}{9}$ $\displaystyle x = \frac{182 \times 45}{9 \times 14} = 65$ $\displaystyle y = 10-65 = 45$ Two numbers are $\displaystyle 45$ and $\displaystyle 65$ $\displaystyle \\$ Question 22: A number is subtracted from the numerator of the fraction $\displaystyle \frac{12}{13}$ and six times that number is added to the denominator. If the new fraction is $\displaystyle \frac{1}{11}$ then find the number. Let the number subtracted from the numerator $\displaystyle = x$ $\displaystyle \frac{12-x}{13+6x} = \frac{1}{11}$ $\displaystyle 132-11x = 13+6x$ $\displaystyle 17x = 119$ or $\displaystyle x = 7$ $\displaystyle \\$ Question 23: A right angled triangle having perimeter $\displaystyle 120\ \text{ cm }$ has its two-side perpendicular side in the ratio $\displaystyle 5:12$ . Find the lengths of its sides. Perimeter of right angled triangle $\displaystyle = 120$ Perpendicular sides $\displaystyle = 5x \ and \ 12x$ Hypotenuse $\displaystyle = \sqrt{(5x)^2+(12x)^2} = 13x$ Therefore $\displaystyle 5x+12x+13x = 120$ $\displaystyle 30x = 120$ $\displaystyle x = 4$ Therefore length of side $\displaystyle = 20 48 52$ $\displaystyle \\$ Question 24: The sum of the digits of a two-digit number is $\displaystyle 9$ . If $\displaystyle 9$ is added to the number formed by reversing the digits then the result is thrice the original number. Find the original number. Let the two-digit number $\displaystyle = xy$ $\displaystyle x+y = 9 \ldots \ldots \ldots i)$ $\displaystyle yx+9 = 3(xy)$ $\displaystyle 10y+x+9 = 3(10x+y)$ $\displaystyle 10y+x+9 = 30x+3y$ $\displaystyle y+9 = 29x \ldots \ldots \ldots ii)$ Solving i) and ii) $\displaystyle 7(9-x)+9 = 29x$ $\displaystyle 63-7x+9 = 29x$ $\displaystyle 72x = 36$ Or $\displaystyle x = 2$ $\displaystyle y = 9-2 = 7$ Therefore the number $\displaystyle = 27$ $\displaystyle \\$ Question 25: The lengths of a rectangle plot of land exceeds its breadth by $\displaystyle 23 m$ if the length is decreased by $\displaystyle 15 m$ . and the breadth is increased by $\displaystyle 7 m$ . the area is reduced by $\displaystyle 360 m^2$ . Find the length and the breadth of the plot. Let the length $\displaystyle = l$ and breadth $\displaystyle = b$ $\displaystyle l = 23 +b$ Given $\displaystyle (l-15)(b+7) = lb-360$ $\displaystyle (23+b-15)(b+7) = (23+b)b-360$ $\displaystyle (b+8)(b+7) = 23b+b^2-360$ $\displaystyle b^2+15b+56 = 23b+b^2-360$ $\displaystyle 416 = 8b$ or $\displaystyle b = 52m$ Therefore $\displaystyle l = b+23 = 52+23 = 75m$ $\displaystyle \\$ Question 26: The length of the rectangular park is twice its breadth. If the perimeter of the park is 186 m find its length and breadth. Let the length $\displaystyle = l$ and breadth $\displaystyle = b$ $\displaystyle l = 2b$ $\displaystyle 2l+2b = 186$ $\displaystyle 4b+2b = 186$ $\displaystyle 6b = 186$ or $\displaystyle b = 31$ $\displaystyle l = 62$ $\displaystyle \\$ Question 27: The length of the rectangle is $\displaystyle 7 \text{ cm }$ more than its breadth. If the perimeter of the rectangle is $\displaystyle 90\ \text{ cm }$ find its length and breadth. Let the length $\displaystyle = l$ breadth $\displaystyle = b$ $\displaystyle l = b+7$ Given $\displaystyle 21l +2b = 90$ $\displaystyle 2(b+7) +2b = 90$ $\displaystyle 4b = 76$ Or $\displaystyle b = 19 \text{ cm }$ $\displaystyle l = 19+7 = 26 \text{ cm }$ $\displaystyle \\$ Question 28: The length of a rectangle is $\displaystyle 7 \text{ cm }$ less than twice its breadth. If the length is decreased by $\displaystyle 2\ \text{ cm }$ and breadth increased by $\displaystyle 3\ \text{ cm }$ the perimeter of the resulting rectangle is $\displaystyle 66 \text{ cm }$ . find the length and the breadth of the original rectangle. Let the length $\displaystyle = l$ and breadth $\displaystyle = b$ $\displaystyle l+ 7 = 2b$ Given $\displaystyle 2(l-2)+ 2(b+3) = 66$ $\displaystyle 2l-4+2b+6 = 66$ $\displaystyle 2l+2b = 64$ Solving $\displaystyle 2(2b-7) + 2b = 64$ $\displaystyle 6b = 78$ $\displaystyle b = 13$ $\displaystyle l = 2 \times 13-7 = 19$ breadth $\displaystyle = 13 \text{ cm }$ length $\displaystyle = 19cm$ $\displaystyle \\$ Question 29: A man is five times as old as his son. In two years’ time he will be four times as old as his son. Find their present ages. Let the man’s age $\displaystyle = 5x$ If son’s age $\displaystyle = x$ Two years letter Man’s age $\displaystyle = 5x +2$ Son’s age $\displaystyle = x+ 2$ $\displaystyle 5x +2 = 4(x+2)$ $\displaystyle x + 6 years = son's \ age$ Man’s age $\displaystyle = 30 yrs.$ $\displaystyle \\$ Question 30: A man is twice as old as his son. Twelve years ago the man was thrice as old as his son. Find their present ages. Let the son’s age $\displaystyle = x$ Man’s age $\displaystyle = 2x$ $\displaystyle 12 \ years \ ago$ Son’s age $\displaystyle = x-12$ Man’s age $\displaystyle = 2x-12$ $\displaystyle 2x-12 = 3 (x-12)$ $\displaystyle 2x-12 = 3x-36$ $\displaystyle x = 24 = son's \ age$ Man’s age $\displaystyle = 48 \ years$ $\displaystyle \\$ Question 31: Seema is $\displaystyle 10 \ years$ elder than Rekha. The ratio of their ages is $\displaystyle 5:3$ . Find their ages. Let Rekha’s age $\displaystyle = x$ Seema’s age $\displaystyle = x + 10$ given $\displaystyle \frac{x+10}{x} = \frac{5}{3}$ $\displaystyle 3x+30 = 5x$ $\displaystyle 2x = 30$ or $\displaystyle x = 15$ Rekha’s sage $\displaystyle = 15 yrs.$ Seema’s sage $\displaystyle = 25 yrs.$ $\displaystyle \\$ Question 32: $\displaystyle 5 \ years$ ago the age of Parvati was $\displaystyle 4$ times the age of her son. The sum of their present ages is $\displaystyle 55 years$ . Find Parvati’s age. Let the present age of Parvati $\displaystyle = x yrs$ age of son $\displaystyle = y yrs.$ $\displaystyle x+y = 55 \ldots \ldots \ldots i)$ Five years before Parvati $\displaystyle = x-5 yrs.$ son $\displaystyle = y-5 yrs.$ Given $\displaystyle (x-5) = 4(y-5)$ $\displaystyle x-4y = -15 \ldots \ldots \ldots ii)$ solving i) and ii) $\displaystyle x-4 (55-x) = -15$ $\displaystyle 5x = 205$ or $\displaystyle x = 44 =$ Parvati’s age Son’s age $\displaystyle = 55-44 = 11 years$ $\displaystyle \\$ Question 33: A man is $\displaystyle 56$ years old and his son is $\displaystyle 24$ years old. In how many years the father will be twice as old as his son at that time? Man’s age $\displaystyle = 56 years$ Son’s age $\displaystyle = 24 years$ Let in $\displaystyle x years$ man would be twice the age of son $\displaystyle 56 +x = 2(24 = x)$ $\displaystyle 56+x = 48+2x$ or $\displaystyle x = 8 years$ $\displaystyle \\$ Question 34: 9 years hence a girl will be $\displaystyle 3$ times as old as she was $\displaystyle 9$ years ago. How old is she now? Let the current age of the girl $\displaystyle = x$ Given $\displaystyle x + 9 = 3(x- 9)$ $\displaystyle x+9 = 3x-27$ $\displaystyle 2x = 36$ $\displaystyle x = 18 \ years = age \ of \ the \ girl$ $\displaystyle \\$ Question 35: A man made a trip of $\displaystyle 480\ km$ in $\displaystyle 9$ hours. Some part of trip was covered at $\displaystyle 45 \text{ km/hr }$ and the remaining at $\displaystyle 60\ \text{ km/hr }$ . find the part of the trip covered by him at $\displaystyle 60\ \text{ km/hr }$ . Let the distance covered at $\displaystyle 45 km/ hr = x$ Let the distance covered at $\displaystyle 60 km/ hr = y$ Total distance $\displaystyle = 480 km.$ $\displaystyle x+y = 480$ $\displaystyle \frac{x}{45} + \frac{y}{60} = 9$ Solving $\displaystyle \frac{480-y}{45} + \frac{y}{60} = 9$ or $\displaystyle y = 300 \ km \ and \ x = 180 \ km$ $\displaystyle \\$ Question 36: A motorist traveled from town $\displaystyle A$ to town $\displaystyle B$ at an average speed of $\displaystyle 54\ \text{ km/hr }$ . on his return journey his average speed was $\displaystyle 60\ \text{ km/hr }$ . if the total time taken is $\displaystyle 9 hours$ find the distance between the two towns. Let the distance between town A and B $\displaystyle = x$ $\displaystyle \text{Therefore } \frac{x}{54} + \frac{x}{60} = 9.5$ or $\displaystyle x = 270 km.$ $\displaystyle \\$ Question 37: The distance between two stations is $\displaystyle 300\ km$ . two motor-cyclist start simultaneously from these stations and move towards each other. The speed of one of them is $\displaystyle 7\ \text{ km/hr }$ faster than that of other. If the distance between them after $\displaystyle 2 hours$ is $\displaystyle 34\ km$ find the speed of each motor-cycle Distance $\displaystyle = 300 km$ Let the speed of 1st cyclist $\displaystyle = x$ Then speed of 2nd cyclist $\displaystyle = x + 7$ Distance covered by 1s cyclist in 2hr $\displaystyle = 2x$ Distance covered by 2nd cyclist in 2 hr $\displaystyle = 2( x+7)$ Therefore $\displaystyle 2x+34+2(x+7) = 300$ $\displaystyle 4x + 48 = 300$ $\displaystyle x = \frac{252}{4} = 63 \text{ km/hr }$ Speed of 1st cyclist $\displaystyle = 63 \text{ km/hr }$ Speed of 2nd cyclist $\displaystyle = 63+7 = 70 \text{ km/hr }$ $\displaystyle \\$ Question 38: A boat travels $\displaystyle 30\ km$ upstream in river in the same period of time as it travels $\displaystyle 50\ km$ downstream. If the ratio of stream be $\displaystyle 5 \text{ km/hr }$ find the speed of the boat in still water. Let the speed of boat $\displaystyle = x \text{ km/hr }$ Speed of stream $\displaystyle = 5 \text{ km/hr }$ Speed of boat upstream $\displaystyle = x-5 \text{ km/hr }$ Speed of the boat downstream $\displaystyle = x + 5 \text{ km/hr }$ $\displaystyle \text{Therefore } \frac{30}{x-5} = \frac{50}{x+5}$ $\displaystyle 30x + 150 = 50x-250$ $\displaystyle 400 = 20 x$ Or $\displaystyle x = 20 \text{ km/hr }$ $\displaystyle \\$ Question 39: The length of each of the equal sides of an isosceles triangle is $\displaystyle 4\ \text{ cm }$ longer than the base. If the perimeter of the triangle is $\displaystyle 62\ \text{ cm }$ find the lengths of the sides of the triangle. Let the base $\displaystyle = x \text{ cm }$ Sides $\displaystyle = (x+ 4)$ Perimeter $\displaystyle = 62 \text{ cm }$ Therefore $\displaystyle (x+4) + (x) + (x+ 4) = 62$ $\displaystyle 3x+8 = 62$ $\displaystyle 3x = 54$ or $\displaystyle x = 18$ Base $\displaystyle = 18 \text{ cm }$ Sides $\displaystyle = 22cm$ $\displaystyle \\$ Question 40: A certain number of candidates appeared for an examination in which one-fifth of the whole plus $\displaystyle 16$ secured first division one-fourth plus $\displaystyle 15$ secured second division and one-fourth minus $\displaystyle 25$ secured third division if the remaining $\displaystyle 60$ candidates failed find the total number of candidates appeared. Let the number of candidates $\displaystyle = x$ $\displaystyle ( \frac{1}{5} x+6)+ ( \frac{1}{4} x+15 )+( \frac{1}{4} x-25)+ 60 = x$ $\displaystyle 6+60 = x-( \frac{1}{5} + \frac{1}{4} + \frac{1}{4} )x$ $\displaystyle x(1- 14/20) = 66$ $\displaystyle x = \frac{20 \times 66}{6} = 220$ No. of candidates $\displaystyle = 220$ $\displaystyle \\$ Question 41: Raman has $\displaystyle 3$ times as much money as Kamal. If Raman gives $\displaystyle Rs. 750$ to Kamal then Kamal will have twice as much as left with Raman. How much had each originally? Let money with Kamal $\displaystyle = x$ Then money with Raman $\displaystyle = 3x$ $\displaystyle 2(3x-750) = (x+750)$ Or $\displaystyle x = 450 Rs.$ Kamal has $\displaystyle 450 Rs.$ And Raman $\displaystyle 1350 Rs.$ $\displaystyle \\$ Question 42: The angles of triangle are in ratio $\displaystyle 2:3:4$ . Find the angles. Ratio of angle $\displaystyle = 2: 3: 4$ Therefore $\displaystyle 2x+3x+4x = 180$ $\displaystyle 9x = 180$ $\displaystyle x = 20$ Therefore angles are $\displaystyle 40 60 80$ degrees. $\displaystyle \\$ Question 43: A certain number man can finish a piece of work in $\displaystyle 50 \text{ days. }$ If there are $\displaystyle 7$ more men the work can be completed $\displaystyle 10$ days earlier. How many men were originally there? Let $\displaystyle x$ men finish work in $\displaystyle 50 days$ Total work $\displaystyle = 50x \ man \ days$ $\displaystyle x + 7 \ men \ finish \ work \ in \ 49 \ days$ Total work $\displaystyle = (x + 7) \times 40$ Therefore $\displaystyle 50x = 40 (x+7)$ $\displaystyle 5x = 4x+28$ $\displaystyle x = 28$ Original no of men $\displaystyle = 28$ $\displaystyle \\$ Question 44: Divide $\displaystyle 600$ in two parts such that $\displaystyle 495$ of one exceeds $\displaystyle 60$ of the other by $\displaystyle 120$ . Let the two parts $\displaystyle = x \text{ and } y$ $\displaystyle x + y = 600$ $\displaystyle 0.4x-0.6y = 120$ Solving we get $\displaystyle x = 480$ $\displaystyle y = 120$ $\displaystyle \\$ Question 45: A workman is paid $\displaystyle Rs. 150$ for each day he works and is fined $\displaystyle Rs. 50$ for each day he is absent. In a month of $\displaystyle 30$ days he earned $\displaystyle Rs. 2100$ . For how many days did he remain absent? Salary $\displaystyle = 150 \text{ Rs. / Day }$ Fine $\displaystyle = 50 \text{ Rs. / Day }$ Let $\displaystyle x$ be the number of days worked Therefore $\displaystyle 150x-(30-x)50 = 2100$ Or $\displaystyle x = 18 \text{ days. }$ $\displaystyle \\$

https://significantstatistics.com/index.php/Lecture_4._E)_Uniform

# Uniform Distribution on $\left[a,b\right]$ A r.v. $X$ follows a uniform distribution $U\left(a,b\right)$ if $X$ is continuous with pdf $f_{X}\left(X\right)=\begin{cases} \frac{1}{b-a}, & x\in\left[a,b\right]\\ 0, & otherwise \end{cases}$ Under the Uniform distribution, all values in $\left[a,b\right]$ are “equally likely.” Notice that if $X\sim U\left(a,b\right)$, then $X=\left(b-a\right)\widetilde{X}+a$ where $\widetilde{X}\sim U\left(0,1\right)$, and $f_{\widetilde{X}}\left(x\right)=1\left(x\in\left[0,1\right]\right)$. ## Mean $E\left(\widetilde{X}\right)=\int_{0}^{1}xdx=\frac{1}{2}$. So, $E\left(X\right)=E\left(\left(b-a\right)\widetilde{X}+a\right)=\left(b-a\right)E\left(\widetilde{X}\right)+a=\frac{a+b}{2}$ ## Variance $Var\left(\widetilde{X}\right)=E\left(\widetilde{X}^{2}\right)-E\left(\widetilde{X}\right)^{2}=\int_{0}^{1}x^{2}dx-\left(\frac{1}{2}\right)^{2}=\frac{1}{3}-\frac{1}{4}=\frac{1}{12}$. So, $Var\left(X\right)=Var\left(\left(b-a\right)\widetilde{X}+a\right)=\left(b-a\right)^{2}Var\left(\widetilde{X}\right)=\frac{\left(b-a\right)^{2}}{12}$. ## MGF $M_{X}\left(t\right)=\exp\left(at\right)M_{\widetilde{X}}\left(\left(b-a\right)t\right)=...=\frac{\exp\left(bt\right)-\exp\left(at\right)}{\left(b-a\right)t}$

https://math.wikia.org/wiki/Group

1,183 Pages A group is a set paired with an operation on the set. As such, a group can be conceptualized as an ordered pair , where is a set, and is an operation. A set and operation is a group if and only if it satisfies the following properties: 1. Identity elementThere exists an , called an identity element, such that , for all 2. InversesFor each , there exists an , called an inverse of , such that 3. Associativity — For all 4. Closure — For all Whenever the group operation is , the operation of group elements , , is often abbreviated as simply a juxtaposition of the group elements, . Important Results From the given criterion for a group, the following properties can be shown for any group  : • There exists exactly one identity element; • For each , there exists exactly one inverse of , and henceforth is referred to as (proof) • For each • Groups have the cancellation property: For all implies , and implies . Optional Properties A group is: A group with a partial order on it is a partially ordered group if for all , if , then and (translation invariance). It is a totally ordered group if in addition is a total order. Community content is available under CC-BY-SA unless otherwise noted.

https://www.princeton.edu/~achaney/tmve/wiki100k/docs/Bolzano%E2%80%93Weierstrass_theorem.html

# Bolzano–Weierstrass theorem related topics {math, number, function} {rate, high, increase} {mi², represent, 1st} In real analysis, the Bolzano–Weierstrass theorem is a fundamental result about convergence in a finite-dimensional Euclidean space Rn. The theorem states that each bounded sequence in Rn has a convergent subsequence. An equivalent formulation is that a subset of Rn is sequentially compact if and only if it is closed and bounded. ## Contents ### Proof First we prove the theorem when n = 1, in which case the ordering on R can be put to good use. Indeed we have the following result. Lemma: Every sequence { xn } in R has a monotone subsequence. Proof: Let us call a positive integer n a "peak of the sequence", if m > n implies  xn > xmi.e., if  xn is greater than every subsequent term in the sequence. Suppose first that the sequence has infinitely many peaks, n1 < n2 < n3 < … < nj < …. Then the subsequence  $\{x_{n_j}\}$  corresponding to peaks is monotonically decreasing, and we are done. So suppose now that there are only finitely many peaks, let N be the last peak and n1 = N + 1. Then n1 is not a peak, since n1 > N, which implies the existence of an n2 > n1 with  $x_{n_2} \geq x_{n_1}.$  Again, n2 > N is not a peak, hence there is n3 > n2 with $x_{n_3} \geq x_{n_2}.$  Repeating this process leads to an infinite non-decreasing subsequence  $x_{n_1} \leq x_{n_2} \leq x_{n_3} \leq \ldots$, as desired. $\square$ Now suppose we have a bounded sequence in R; by the Lemma there exists a monotone subsequence, necessarily bounded. But it follows from the Monotone convergence theorem that this subsequence must converge, and the proof is complete. Finally, the general case can be easily reduced to the case of n = 1 as follows: given a bounded sequence in Rn, the sequence of first coordinates is a bounded real sequence, hence has a convergent subsequence. We can then extract a subsubsequence on which the second coordinates converge, and so on, until in the end we have passed from the original sequence to a subsequence n times — which is still a subsequence of the original sequence — on which each coordinate sequence converges, hence the subsequence itself is convergent.