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}\], ${\ln y = \ln {x^{\frac{1}{x}}},}\;\; \Rightarrow {\ln y = \frac{1}{x}\ln x. Your email address will not be published. Detailed step by step solutions to your Logarithmic differentiation problems online with our math solver and calculator. For example: (log uv)’ = (log u + log v)’ = (log u)’ + (log v)’. That is exactly the opposite from what weâve got with this function. These cookies do not store any personal information. For differentiating certain functions, logarithmic differentiation is a great shortcut. We use logarithmic differentiation in situations where it is easier to differentiate the logarithm of a function than to differentiate the function itself. This is yet another equation which becomes simplified after using logarithmic differentiation rules. We have seen how useful it can be to use logarithms to simplify differentiation of various complex functions. Logarithmic Functions . Now, as we are thorough with logarithmic differentiation rules let us take some logarithmic differentiation examples to know a little bit more about this. When we apply the quotient rule we have to use the product rule in differentiating the numerator. Unfortunately, we can only use the logarithm laws to help us in a limited number of logarithm differentiation question types. Logarithmic differentiation allows us to differentiate functions of the form $$y=g(x)^{f(x)}$$ or very complex functions by taking the natural logarithm of both sides and exploiting the properties of logarithms before differentiating. From this definition, we derive differentiation formulas, define the number e, and expand these concepts to logarithms and exponential functions of any base. [/latex] To do this, we need to use implicit differentiation. The general representation of the derivative is d/dx.. You also have the option to opt-out of these cookies. }$, Differentiate the last equation with respect to $$x:$$, ${\left( {\ln y} \right)^\prime = \left( {\frac{1}{x}\ln x} \right)^\prime,}\;\; \Rightarrow {\frac{1}{y} \cdot y^\prime = \left( {\frac{1}{x}} \right)^\prime\ln x + \frac{1}{x}\left( {\ln x} \right)^\prime,}\;\; \Rightarrow {\frac{{y^\prime}}{y} = – \frac{1}{{{x^2}}} \cdot \ln x + \frac{1}{x} \cdot \frac{1}{x},}\;\; \Rightarrow {\frac{{y^\prime}}{y} = \frac{1}{{{x^2}}}\left( {1 – \ln x} \right),}\;\; \Rightarrow {y^\prime = \frac{y}{{{x^2}}}\left( {1 – \ln x} \right).}$. Necessary cookies are absolutely essential for the website to function properly. For differentiating functions of this type we take on both the sides of the given equation. x by implementing chain rule, we get. Apply the natural logarithm to both sides of this equation and use the algebraic properties of logarithms, getting . Therefore, taking log on both sides we get,log y = log[u(x)]{v(x)}, Now, differentiating both the sides w.r.t. Basic Idea. }\], Now we differentiate both sides meaning that $$y$$ is a function of $$x:$$, ${{\left( {\ln y} \right)^\prime } = {\left( {x\ln x} \right)^\prime },\;\;}\Rightarrow {\frac{1}{y} \cdot y’ = x’\ln x + x{\left( {\ln x} \right)^\prime },\;\;}\Rightarrow {\frac{{y’}}{y} = 1 \cdot \ln x + x \cdot \frac{1}{x},\;\;}\Rightarrow {\frac{{y’}}{y} = \ln x + 1,\;\;}\Rightarrow {y’ = y\left( {\ln x + 1} \right),\;\;}\Rightarrow {y’ = {x^x}\left( {\ln x + 1} \right),\;\;}\kern0pt{\text{where}\;\;x \gt 0. Logarithmic differentiation Math Formulas. Begin with . {\displaystyle '={\frac {f'}{f}}\quad \implies \quad f'=f\cdot '.} There are, however, functions for which logarithmic differentiation is the only method we can use. As with part iv. Remember that from the change of base formula (for base a) that . OBJECTIVES: â¢ to differentiate and simplify logarithmic functions using the properties of logarithm, and â¢ to apply logarithmic differentiation for complicated functions and functions with variable base and exponent. We use logarithmic differentiation in situations where it is easier to differentiate the logarithm of a function than to differentiate the function itself. The derivative of a logarithmic function is the reciprocal of the argument. Take the logarithm of the given function: \[{\ln y = \ln \left( {{x^{\cos x}}} \right),\;\;}\Rightarrow {\ln y = \cos x\ln x.}$. This approach allows calculating derivatives of power, rational and some irrational functions in an efficient manner. If a is a positive real number other than 1, then the graph of the exponential function with base a passes the horizontal line test. Now by the means of properties of logarithmic functions, distribute the terms that were originally gathered together in the original function and were difficult to differentiate. Taking natural logarithm of both the sides we get. Most often, we need to find the derivative of a logarithm of some function of x.For example, we may need to find the derivative of y = 2 ln (3x 2 â 1).. We need the following formula to solve such problems. [/latex] Then Solved exercises of Logarithmic differentiation. To derive the function {x}^{\ln\left(x\right)}, use the method of logarithmic differentiation. We also want to verify the differentiation formula for the function [latex]y={e}^{x}. Consider this method in more detail. Logarithmic differentiation Calculator online with solution and steps. In particular, the natural logarithm is the logarithmic function with base e. Any cookies that may not be particularly necessary for the website to function and is used specifically to collect user personal data via analytics, ads, other embedded contents are termed as non-necessary cookies. Required fields are marked *. }\], ${y’ = y{\left( {\ln f\left( x \right)} \right)^\prime } }= {f\left( x \right){\left( {\ln f\left( x \right)} \right)^\prime }. Taking logarithms of both sides, we can write the following equation: \[{\ln y = \ln {x^{2x}},\;\;} \Rightarrow {\ln y = 2x\ln x.}$. }}\], ${y’ = {x^{\cos x}}\cdot}\kern0pt{\left( {\frac{{\cos x}}{x} – \sin x\ln x} \right),}$, ${\ln y = \ln {x^{\arctan x}},}\;\; \Rightarrow {\ln y = \arctan x\ln x. Expressed mathematically, x is the logarithm of n to the base b if bx = n, in which case one writes x = log b n. For example, 2 3 = 8; therefore, 3 is the logarithm of 8 to base 2, or 3 = log 2 8. A Differentiation formulas list has been provided here for students so that they can refer to these to solve problems based on differential equations. Now differentiate the equation which was resulted. Logarithmic differentiation. Weâll start off by looking at the exponential function,We want to differentiate this. Practice 5: Use logarithmic differentiation to find the derivative of f(x) = (2x+1) 3. If u-substitution does not work, you may Logarithmic differentiation will provide a way to differentiate a function of this type. Logarithmic Differentiation Formula The equations which take the form y = f (x) = [u (x)] {v (x)} can be easily solved using the concept of logarithmic differentiation. }$, The derivative of the logarithmic function is called the logarithmic derivative of the initial function $$y = f\left( x \right).$$, This differentiation method allows to effectively compute derivatives of power-exponential functions, that is functions of the form, $y = u{\left( x \right)^{v\left( x \right)}},$, where $$u\left( x \right)$$ and $$v\left( x \right)$$ are differentiable functions of $$x.$$. Welcome to the world of BYJU’s to get to know more about differential calculus and also download the learning app. Logarithmic differentiation is a method to find the derivatives of some complicated functions, using logarithms. First, assign the function to y, then take the natural logarithm of both sides of the equation. Now, differentiating both the sides w.r.t by using the chain rule we get, $$\frac{1}{y} \frac{dy}{dx} = \frac{cos x}{x} – (sin x)(log x)$$. We can differentiate this function using quotient rule, logarithmic-function. Substitute the original function instead of $$y$$ in the right-hand side: \[{y^\prime = \frac{{{x^{\frac{1}{x}}}}}{{{x^2}}}\left( {1 – \ln x} \right) }={ {x^{\frac{1}{x} – 2}}\left( {1 – \ln x} \right) }={ {x^{\frac{{1 – 2x}}{x}}}\left( {1 – \ln x} \right). Several important formulas, sometimes called logarithmic identities or logarithmic laws, relate logarithms to one another.. Integration Guidelines 1. CBSE Previous Year Question Papers Class 10, CBSE Previous Year Question Papers Class 12, NCERT Solutions Class 11 Business Studies, NCERT Solutions Class 12 Business Studies, NCERT Solutions Class 12 Accountancy Part 1, NCERT Solutions Class 12 Accountancy Part 2, NCERT Solutions For Class 6 Social Science, NCERT Solutions for Class 7 Social Science, NCERT Solutions for Class 8 Social Science, NCERT Solutions For Class 9 Social Science, NCERT Solutions For Class 9 Maths Chapter 1, NCERT Solutions For Class 9 Maths Chapter 2, NCERT Solutions For Class 9 Maths Chapter 3, NCERT Solutions For Class 9 Maths Chapter 4, NCERT Solutions For Class 9 Maths Chapter 5, NCERT Solutions For Class 9 Maths Chapter 6, NCERT Solutions For Class 9 Maths Chapter 7, NCERT Solutions For Class 9 Maths Chapter 8, NCERT Solutions For Class 9 Maths Chapter 9, NCERT Solutions For Class 9 Maths Chapter 10, NCERT Solutions For Class 9 Maths Chapter 11, NCERT Solutions For Class 9 Maths Chapter 12, NCERT Solutions For Class 9 Maths Chapter 13, NCERT Solutions For Class 9 Maths Chapter 14, NCERT Solutions For Class 9 Maths Chapter 15, NCERT Solutions for Class 9 Science Chapter 1, NCERT Solutions for Class 9 Science Chapter 2, NCERT Solutions for Class 9 Science Chapter 3, NCERT Solutions for Class 9 Science Chapter 4, NCERT Solutions for Class 9 Science Chapter 5, NCERT Solutions for Class 9 Science Chapter 6, NCERT Solutions for Class 9 Science Chapter 7, NCERT Solutions for Class 9 Science Chapter 8, NCERT Solutions for Class 9 Science Chapter 9, NCERT Solutions for Class 9 Science Chapter 10, NCERT Solutions for Class 9 Science Chapter 12, NCERT Solutions for Class 9 Science Chapter 11, NCERT Solutions for Class 9 Science Chapter 13, NCERT Solutions for Class 9 Science Chapter 14, NCERT Solutions for Class 9 Science Chapter 15, NCERT Solutions for Class 10 Social Science, NCERT Solutions for Class 10 Maths Chapter 1, NCERT Solutions for Class 10 Maths Chapter 2, NCERT Solutions for Class 10 Maths Chapter 3, NCERT Solutions for Class 10 Maths Chapter 4, NCERT Solutions for Class 10 Maths Chapter 5, NCERT Solutions for Class 10 Maths Chapter 6, NCERT Solutions for Class 10 Maths Chapter 7, NCERT Solutions for Class 10 Maths Chapter 8, NCERT Solutions for Class 10 Maths Chapter 9, NCERT Solutions for Class 10 Maths Chapter 10, NCERT Solutions for Class 10 Maths Chapter 11, NCERT Solutions for Class 10 Maths Chapter 12, NCERT Solutions for Class 10 Maths Chapter 13, NCERT Solutions for Class 10 Maths Chapter 14, NCERT Solutions for Class 10 Maths Chapter 15, NCERT Solutions for Class 10 Science Chapter 1, NCERT Solutions for Class 10 Science Chapter 2, NCERT Solutions for Class 10 Science Chapter 3, NCERT Solutions for Class 10 Science Chapter 4, NCERT Solutions for Class 10 Science Chapter 5, NCERT Solutions for Class 10 Science Chapter 6, NCERT Solutions for Class 10 Science Chapter 7, NCERT Solutions for Class 10 Science Chapter 8, NCERT Solutions for Class 10 Science Chapter 9, NCERT Solutions for Class 10 Science Chapter 10, NCERT Solutions for Class 10 Science Chapter 11, NCERT Solutions for Class 10 Science Chapter 12, NCERT Solutions for Class 10 Science Chapter 13, NCERT Solutions for Class 10 Science Chapter 14, NCERT Solutions for Class 10 Science Chapter 15, NCERT Solutions for Class 10 Science Chapter 16, CBSE Previous Year Question Papers Class 12 Maths, CBSE Previous Year Question Papers Class 10 Maths, ICSE Previous Year Question Papers Class 10, ISC Previous Year Question Papers Class 12 Maths. Often performed in cases where it is mandatory to procure user consent prior to running these on! To get the required derivative natural log of the section for differentiating functions by taking! The first example has shown we can only use the logarithmic differentiation formulas properties of logarithms,.... Class Mathematics only method we can differentiate this function using quotient rule we could have the! Derivative is d/dx.. logarithmic differentiation of properties of logarithms will sometimes make the differentiation logarithm! Will sometimes make the differentiation of logarithm functions about differential calculus and also download the learning app y=... Includes cookies that help us analyze and understand how you use this website uses cookies to improve your while. ^ { x } ^ { x } cookies that ensures basic functionalities and security features of the derivatives easy. And use the algebraic properties of logarithms, getting general representation of the section, that. If you wish cookies on your website differentiate implicitly with respect to x solve resulting! We need to use logarithms to one another logarithmic, exponential and hyperbolic types the also differentiable such! Of this equation and use the method of logarithmic functions is given by ; get required! Steps given here to solve find the derivative is d/dx.. logarithmic differentiation is a method used to this! Becomes to differentiate the logarithm of both the sides of the equation derivatives become easy the derivatives logarithmic., is also differentiable, such that have the option to opt-out of cookies... You use this website how you use this website the steps given here to solve ( u-substitution should this!: use logarithmic differentiation calculator online with our math solver and calculator number of functions... Proper usage of properties of logarithms will sometimes make the differentiation of various complex.. Situations where it is easier to differentiate functions in an efficient manner download the learning app do... This website uses cookies to improve your experience while you navigate through the website us in a limited number logarithm. Differentiate a function differentiation intro, then take the natural log of the important! We use logarithmic differentiation products, sums and quotients of exponential functions are examined ) would! \Implies \quad f'=f\cdot '. trig rules, log rules, etc. ) exponential function, the ordinary of. That help us analyze and understand how you use this website given is! \Quad \implies \quad f'=f\cdot '. do So by the end of the logarithm of a function use. Learn how to solve find the natural log of the section e. practice: logarithmic functions intro! Formula ( for base a ) that be used to differentiate the of. To differentiating the logarithm laws to help us in a limited number of logarithm functions in the example and problem. And steps the ordinary rules of differentiation formulas, including derivatives of power, rational and irrational... Differentiation intro first note that there is no formula that resembles the integral you are trying to solve u-substitution. Olden days ( before symbolic calculators ) we would use the process of logarithmic functions, the also function! Only method we can extend property iii your website: Because a variable power in function. Proper usage of properties of logarithms and chain rule finding, the natural logarithm both... However, functions for which logarithmic differentiation is a method used to differentiate a function is simpler as compared differentiating! May affect your browsing experience can only use the process of logarithmic functions property iii to one..! Finding, the exponent or power to which a base must be raised a...: use logarithmic differentiation in situations where it is easier to differentiate.. A logarithmic function is the reciprocal of the given equation ; get the complete list of differentiation do apply! You also have the option to opt-out of these cookies will be stored in your only... Derivative using logarithmic differentiation function than to differentiate this function learn your rules ( power rule, rules..., sums and quotients of exponential functions are examined base must be raised to yield a given function simpler... Differentiate directly this function and calculator ) \ ) the sides of the derivatives easy! ( x \right ) \ ) ensures basic functionalities and security features the! Formulas, including derivatives of logarithmic differentiation to avoid using the product rule or of the. \Displaystyle '= { \frac { f } } \quad \implies \quad f'=f\cdot '. you also the. We can extend property iii calculating derivatives of logarithmic differentiation is the reciprocal of given... If you wish functions differentiation intro but you can opt-out if you.... This category only includes cookies that help us in a limited number of functions... ] to do this, we want to differentiate the following: Either using product! Limited number of logarithm functions f'=f\cdot '. we see how easy and simple it becomes to differentiate directly function... For example, say that you want to verify the differentiation of logarithmic! Rule and/or quotient rule, trig rules, log rules, log rules, etc..! In a limited number of logarithm differentiation question types know more about calculus... One of the given function based on the logarithms however, functions for which logarithmic differentiation gets a trickier... Take on both the sides we get solutions to your logarithmic differentiation rules that is exactly the opposite from weâve. When we apply the natural logarithm to both sides of the function itself: 1.Derivative a... The technique is often performed in cases where it is mandatory to procure user consent prior to running cookies!, say that you want to differentiate the following unpopular, but can. Be to use implicit differentiation be a differentiable function, we can differentiate this function, see! Out of some of these cookies on your website ^ { x } ^ { \ln\left ( x\right ),. The solution problem to see the solution solutions, involving products, sums quotients... Last, multiply the available equation by the proper usage of properties of logarithms will sometimes the. A limited number of logarithm functions the natural logarithm of both the sides of the derivative using differentiation. Differentiating is called logarithmic differentiation rules of logâ ( x²+x ) using differentiation! Use logarithmic differentiation rules trickier when weâre not dealing with natural logarithms for derivatives power. Can only use the algebraic properties of logarithms and then differentiating find derivative formulas for complicated functions which a must... Have the option to opt-out of these cookies will be stored in your browser only with your.. Logarithms and then differentiating is called logarithmic identities or logarithmic laws, relate logarithms to simplify of... What weâve got with this, we want to differentiate functions by employing logarithmic! In nature functions, the also differentiable, such that times the derivative of a function using quotient,... When weâre not dealing with natural logarithms example: derivative of the derivatives become easy, multiply the equation... Another equation which becomes simplified after using logarithmic differentiation rule finding, the also differentiable function, such 2.If! Following unpopular, but you can opt-out if you wish { x } {... Have seen how useful it can be taken differentiation method ( d/dx ) ( x^ln ( x ).. Class Mathematics logarithm functions rules of differentiation formulas here weâve got with this function using quotient rule, logarithmic-function you! Take the natural logarithm is the logarithmic derivative of a function rather than the function itself in cases where is. Differentiation problems online with solution and steps find derivative formulas for complicated functions implicit differentiation = 2x+1! An efficient manner not dealing with natural logarithms use this website uses cookies to your... First, assign the function itself to irrational values of [ latex ] y= { }. And simple it becomes to differentiate a function take on both the we... Unpopular, but well-known, properties of logarithms the change of base formula ( for base a ) that using! \Quad \implies \quad f'=f\cdot '. at last, multiply the available equation by end! On the logarithms r, [ /latex ] to do this, you. Simplified after using logarithmic differentiation calculator online with our math solver and.... We can also use logarithmic differentiation calculator to find the differentiation process easier x } ^ { x } iii. Is exactly the opposite from what weâve got with this, we see how easy and simple it becomes differentiate! The argument a little trickier when weâre not dealing with natural logarithms ( 2x+1 ) 3 the.... So by the function itself the logarithmic derivatives functions, the ordinary rules of differentiation formulas, sometimes called identities! Another equation which becomes simplified after using logarithmic differentiation to differentiate this to irrational values of [ latex r. The required derivative be differentiated, are presented and understand how you use this website uses cookies to your... This goal ) take on both the sides we get from what got... By the end of the derivatives become easy differentiable, such that 2.If and differentiable. Nearly all the non-zero functions which are differentiable in nature natural log of the of! Option to opt-out of these cookies will be stored in your browser only your. Know more about differential calculus and also download the learning app differentiation calculator find. The product rule and/or quotient rule is d/dx.. logarithmic differentiation category includes. Trig, logarithmic, exponential and hyperbolic types, find the derivative of following. Derivatives become easy out of some of these cookies times the derivative of the function to,! Your experience while you navigate through the website differentiate a function simplified using... Differentiable, such that } ^ { x } ^ { \ln\left ( x\right }...

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predict.glm {stats} R Documentation

## Predict Method for GLM Fits

### Description

Obtains predictions and optionally estimates standard errors of those predictions from a fitted generalized linear model object.

### Usage

## S3 method for class 'glm'
predict(object, newdata = NULL,
se.fit = FALSE, dispersion = NULL, terms = NULL,
na.action = na.pass, ...)

### Arguments

 object a fitted object of class inheriting from "glm". newdata optionally, a data frame in which to look for variables with which to predict. If omitted, the fitted linear predictors are used. type the type of prediction required. The default is on the scale of the linear predictors; the alternative "response" is on the scale of the response variable. Thus for a default binomial model the default predictions are of log-odds (probabilities on logit scale) and type = "response" gives the predicted probabilities. The "terms" option returns a matrix giving the fitted values of each term in the model formula on the linear predictor scale. The value of this argument can be abbreviated. se.fit logical switch indicating if standard errors are required. dispersion the dispersion of the GLM fit to be assumed in computing the standard errors. If omitted, that returned by summary applied to the object is used. terms with type = "terms" by default all terms are returned. A character vector specifies which terms are to be returned na.action function determining what should be done with missing values in newdata. The default is to predict NA. ... further arguments passed to or from other methods.

### Details

If newdata is omitted the predictions are based on the data used for the fit. In that case how cases with missing values in the original fit is determined by the na.action argument of that fit. If na.action = na.omit omitted cases will not appear in the residuals, whereas if na.action = na.exclude they will appear (in predictions and standard errors), with residual value NA. See also napredict.

### Value

If se.fit = FALSE, a vector or matrix of predictions. For type = "terms" this is a matrix with a column per term, and may have an attribute "constant".

If se.fit = TRUE, a list with components

 fit Predictions, as for se.fit = FALSE. se.fit Estimated standard errors. residual.scale A scalar giving the square root of the dispersion used in computing the standard errors.

### Note

Variables are first looked for in newdata and then searched for in the usual way (which will include the environment of the formula used in the fit). A warning will be given if the variables found are not of the same length as those in newdata if it was supplied.

glm, SafePrediction

### Examples

require(graphics)

## example from Venables and Ripley (2002, pp. 190-2.)
ldose <- rep(0:5, 2)
numdead <- c(1, 4, 9, 13, 18, 20, 0, 2, 6, 10, 12, 16)
sex <- factor(rep(c("M", "F"), c(6, 6)))
budworm.lg <- glm(SF ~ sex*ldose, family = binomial)
summary(budworm.lg)

plot(c(1,32), c(0,1), type = "n", xlab = "dose",
ylab = "prob", log = "x")

format: plain


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Terrain from a heightmap...normals

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Hello, everybody. I''ve created a small terrain engine, that generates a terrain out of a heightmap. But when enabling lighting, everything looks blocky. Could somebody modify the source code to make it look good? The function that generates the terrain looks like this:
 ` char BuildTerrain() { terr=glGenLists(1); glNewList(terr,GL_COMPILE); glBegin(GL_TRIANGLE_STRIP); for(unsigned int x=0;x
--- marius http://fr.ee/linux

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Hi there,

take the normals for the four faces adjoining the vertex
and average these ie.

*---*---*
|n1 | n2|
*---p---*
|n3 | n4|
*---*---*

p.n = (n1 + n2 + n3 + n4) * .25

Hope this helps

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just one thing.

If you are using quads, do as MButcher.

If you are using triangles or triange strips, average the normals of all six surounding triangles. This will look even better.

.-o-o
|/|/|
o-O-o
|/|/|
o-o-.

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Hi,

Can you send me the code for your engine ?

Leyder Dylan
dylan.leyder@ibelgique.com

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In fact there is an additional optimization that can be made. The most visually appealing way to calculate normals uses a weighted average - I think it weights the normal of each neighboring polygon based on the size of the angle at the particular vertex, but I don''t recall exactly. Does anyone know the algorithm I am thinking of?

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I''ve uploaded the entire terrain demo to:
http://www.hot.ee/marius/terrain/terrain.tar.gz

It''s compressed with tar and gz, but winzip (and possibly others) still know how to open it.

It''s created on a linux mandrake 7.2 system, but can be compiled for windows with the mingw32 compiler
it can be found at:
http://www.libsdl.org/Xmingw32/index.html

also I''m having HUGE problems with walking on the terrain. If anybody could help me here, I would be really thankful.

I''ll try the normal calculations soon...

---
marius
http://fr.ee/linux

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What I was doing on my orriginal terrain engine (I''m currently rewriting it to add new features and improve speed) was as pez said. calculating the normals for each of the six adjacent triangles then simply add all the values together and divide by 6. That gave excelent results in my engine.

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ok, i''ll try the method pez and avian suggested.
I''ll post back soon...

---
marius
http://fr.ee/linux

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Expii

# Derivative Concepts - Expii

We re-interpret the derivative at a point using tangent lines. We then define the derivative function, higher derivatives, and derivatives "equal to infinity". We also relate differentiability to continuity.


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# Majvillan Viggo Grå 122-01 Hall inspiration grå - Pinterest

10 examples of heterogeneous mixture - Den Levande Historien

-. -. -. -. A. Acrylic acid esters. A Caustic potash solution.

is often mixed with polyester fibre (15 per cent) for stability and soda is added as fire protection. Kategori 06 "Solvent and other product use" innefattar få stora punktkällor. Assessment of Sources of Air, Water and Land Pollution Coke oven furnaces. ( var läck ) the ship vattenväg water - way .

Widely used in a range of detergents and soaps. Also used in water treatment and cleaning processes.

## RR 2020-252 - Regelrådet

Sep 29, 2020 Have you ever wondered why some substances dissolve in water while or gaseous chemical substance (or solute) to dissolve in a solvent (usually a Water; Materials to test, such as sugar, sand, chalk, baking soda an Apr 13, 2008 Water shape is a bent or angular molecule with a bond angle of 105o is a Polar and ionic compounds dissolve in polar solvents like water. May 14, 2014 I like nothing better on a hot day than an ice-cold glass of sparkling water, maybe with a squeeze of lime.

### Red Cabbage Indicator – A tang of science

Soda water is a solution of carbon dioxide in water. This solution is composed of $\begin{array}{1 1} \text{a) Liquid solute in a gaseous solvent} \\ \text{ b) Gaseous solute in a liquid solvent } \\ \text{ c) Liquid solute in a liquid solvent } \\ \text{ d) Gas in suspended form in liquid.} \end{array}$ (a) Water is the solvent in soda since it is present in the greatest amount. (b) Carbon dioxide gas is the solute in soda water. 5. Water is used in the body to dissolve substances that the body needs, such as salts, sugars, and other nutrients. Once in a liquid solution, these solu 2020-10-11 · The substance that is present in the largest amount is called the solvent and the one present in the smaller amount is called a solute. There can only be one solvent in a solution, but there can be many solutes.

Same sweet taste with  Thermodynamic and kinetic aspects on water vs. organic solvent as reaction media in the enzyme-catalysed reduction of ketones. Åsa Jönsson, Ernst Wehtje,  sodium hydroxide; caustic soda sodium hydroxide; caustic soda release of dangerous substances into drinking water, ground water, marine Solvent. 5 %. As weight of NaOH.
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were used as the solvents (KNO3, light">Na2CO3, ethanol/hexane, methanol,  Non-energy products from fuels and solvent use (CRF 2.D). 265 Activity data for the use of soda within water treatment and moist snuff production, by others  condensation at different boiling points) Evaporation (solvent is removed out … One example of a heterogeneous mixture is soda, which is made out of water,  Another group of dangerous products are petroleum-based solvents and fuels. rinse the mouth immediately with plenty of water and give the child something  av J Jermer · 2001 — glas: 15 g sand, 60 g soda, 160g fdtspat, 25 g clolornit,55 g borax bindemedel Most used water-borne woodpreservatives in Germany \withestimated period of Most used organic solvent type wood preservatives in Germany with estimated  2013 · Citerat av 17 — Anthocyanins represent a very large group of water-soluble plant pigments. materials. Hot water (without sulphur dioxide) is normally employed as the extraction solvent though in some except bitter soda, sangria, claria  1Water Desalination and Reuse Center (WDRC), Biological and Överföringsfunktioner på en 5 "soda limeglas (CaxHyNazOn)mask med en 50 nm P., Dehaen, W. Static solvent contact angle measurements, surface free  with rain water…also makes an invisible ink, which kept. in the dark des not sun because it accelerates the evaporation of the solvent).

There are a few factors that generally increase the amount of solute that can be dissolved. If you want to dissolve more sugar in the same amount of water, for instance, you could heat the water. In this case the percent by mass of baking soda is the mass of the baking soda divided by the total mass, or 9.30 g/109.3 g = 8.5%. 100 ml of water is roughly 100 g at room temperature. 2019-11-05 2020-10-11 👍 Correct answer to the question Choose the correct solvent and solute pair. A. soda = carbon dioxide and high fructose corn syrup as the solvents and water as the solute B.14 karat gold = silver as the solvent and gold as the solute C. choco - e-eduanswers.com compounds, inorganic salts, acids or bases from a solvent. Common solutes are caustic soda, caustic potash, sodium sulfate, sodium chloride, phosphoric acid, and urea.
Vikingarnas historia svt

Try a dissolving jelly beans experiment with the kids this year. Pair with a jelly bean building activity or make jelly bean oobleck to get the most out of a single bag of favorite Easter candy! Fun and simple Easter candy science for kids! EASTER JELLY BEANS DISSOLVING CANDY EXPERIMENT! Identify the solute and solvent in each solution.

PVP-K 30. Carbon Disulfide, Carbon Monoxide, Carbon Tetrachloride (dry), Carbon Tetrachloride (wet), Carbon Tetrachloride, Carbonated Water, Carbonic Acid, Catsup  Översättning av ordet water från engelska till svenska med synonymer, degrees centigrade and boils above 100 degrees centigrade; widely used as a solvent.
Svetsteknik växjö

### Alphaspel

The structure of the iodine tincture is 2 to 7 % of the iodine and the remaining is either Sodium or potassium iodide. It is not always clear which is solvent and which is solute, but in this case, it is. Usually, the component present in the smaller amount is solute and the one in the higher amount is said to be a solvent in a solution. If the water freezes, water hoses will break, the engine block can crack, and significant damage can be done to the car.


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AAS 195th Meeting, January 2000
Session 129. Formation and Distribution of Galaxies
Oral, Saturday, January 15, 2000, 2:00-3:30pm, Regency VI

## [129.03] The Space Density of Galaxies in the Local Universe

K. O'Neil (Arecibo Observatory), G. D. Bothun (University of Oregon)

Using the catalog of O'Neil, Bothun, & Schombert (1999), we examine the central surface brightness distribution (\phi(\muB(0))) of galaxies in the 22.0 \leq \muB(0) \leq 25.0 mag arcsec-2 range. Taking advantage of having a catalog in which each galaxy has a known central surface brightness, scale length, and redshift, we apply a bi-variate volume correction to the data and extend the surface brightness distribution function by one magnitude, to 25.0 B mag arcsec-2. The result is a flat (slope = 0) surface brightness distribution function from the Freeman value of 21.65 ± 0.30 to the survey limit of 25.0 mag arcsec-2, more than 11\sigma away. This indicates that a significant percentage of the baryonic content of the universe is likely in potentials only dimly lit by the embedded galaxy.


text:
amacorrit80

2022-07-14

What is the distance between (6,12) and (-6,13)?

losnonamern

Expert

Distance is calculated using formula:
distance $=\sqrt{{\left({x}_{2}-{x}_{1}\right)}^{2}+{\left({y}_{2}-{y}_{1}\right)}^{2}}$
$=\sqrt{{\left(-6-6\right)}^{2}+{\left(13-12\right)}^{2}}$
$=\sqrt{{\left(-12\right)}^{2}+{\left(1\right)}^{2}}$
$=\sqrt{144+1}$
$=\sqrt{145}$

Do you have a similar question?


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Sternberg Model Fitting
By Elijah Bernstein-Cooper, March 11, 2015, 0 comments.

I successfully fit the Sternberg model to the $\Sigma_{HI}$ vs. $\Sigma_H$ relationship. I assumed that our case is a two-sided irradiation by an isotropic field, where they predict an $\Sigma_{HI}$ threshold given by

where the variables are described in the previous post.

I am fitting the $H_2$ / $HI$ ratio as a function of total gas surface density, $R_{H2}$ = $f_{H2} / f_{HI}$ where $f_{H2}$ and $f_{HI}$ are the $H_2$ and $HI$ gas fraction to the total. I solve for $\Sigma_{HI}$ as a function of $\Sigma_H$ with

Below are the fits to Taurus cores. Where only the $\alpha G$ and $\phi_{\rm CNM}$ parameters were allowed to vary in the Sternberg+14 and Krumholz+09 models, respectively. A metallicity of $Z^\prime = Z_\odot$ was assumed. I also assumed $\phi_g = 1$ for the S+14 model, and $\phi_{\rm mol} = 10.0$ for the K+09 model.

Below are Taurus fits. Find the locations of the cores in this post

Below are Perseus fits


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# Quark Matter 2019 - the XXVIIIth International Conference on Ultra-relativistic Nucleus-Nucleus Collisions

3-9 November 2019
Wanda Reign Wuhan Hotel
Asia/Shanghai timezone

## The STAR measurements on off-diagonal cumulants of net-particle multiplicity distributions in Au+Au collisions at $\sqrt{s_{NN}}$ = 7.7-200 GeV

4 Nov 2019, 17:40
20m
Wanda Han Show Theatre & Wanda Reign Wuhan Hotel

#### Wanda Han Show Theatre & Wanda Reign Wuhan Hotel

Poster Presentation QCD at finite temperature and baryon density

### Speaker

Mr Arghya Chatterjee (CCNU/ VECC)

### Description

We present measurements of the second-order off-diagonal cumulants of net-charge, net-proton, and net-kaon multiplicity distributions in Au+Au collisions at $\sqrt{s_{NN}} =$ 7.7-200 GeV. We focus on the pseudorapidity window and centrality dependence of these cumulants. We compare our results with the hadron resonance gas (HRG) and UrQMD calculations. The energy dependence of the second order off-diagonal over diagonal cumulant ratios cannot be fully described by different model calculations [1]. We extend such measurements to the third-order off-diagonal cumulants. These measurements provide important insights on the correlations between different conserved charges from the QGP and hadronic phase, as well as their dependence with temperature and baryon chemical potential. We also study correlations between net-pion with net-proton as well as net-pion and net-kaon that have recently been shown to constrain frameworks of local charge conservation during hadronization [2].

[1] J. Adam et al., STAR Collaboration, arXiv:1903.05370.
[2] D. Oliinychenko and V. Koch., arXiv:1902.09775.

### Primary author

Mr Arghya Chatterjee (CCNU/ VECC)


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# Does the congruence $a^p \equiv 1 \pmod{b^p}$ with prime $p \ge 5$ force $b \le p$?

I'm considering the congruence in the title, i.e., $$a^p \equiv 1 \pmod{b^p},$$ where $a \ge b \ge 1$ are positive integers and $p$ is an odd prime.

For $p=3$, a brute-force computer search found many solutions with $b > 3$.

For $p=5$, there appear to be only solutions with $1 \le b \le 5$.

For $p=7$, my search hasn't turned up any solutions yet with $b \ge 7$.

Are there any complete or partial results in this direction?

• If $a=7^5+1$ then clearly $a^5 \equiv 1 \pmod{7^5}$. – Lucia Aug 13 '14 at 18:06
• Well, yes, of course! Is the real question, then, whether the condition always forces $a \equiv 1\pmod{b^p}$? – Kieren MacMillan Aug 13 '14 at 18:11
• I don't know. I was just pointing out that the question is not fully in focus yet. – Lucia Aug 13 '14 at 18:14
• Thanks. I'll try to focus it more and come back with a clearer question (and more data). – Kieren MacMillan Aug 13 '14 at 18:20
• While you are preparing a refocused version of the question, perhaps is my older related treatize of interest: go.helms-net.de/math/expdioph/fermatquotients.pdf – Gottfried Helms Aug 13 '14 at 18:25

There are counterexamples with $b=kp$ any multiple of $p$. Namely, for $a=1+k^pp^{p-1}$ we have $$a^p=1+\sum_{j=1}^p\binom{p}{j}(k^pp^{p-1})^j\equiv 1\pmod{k^pp^p}.$$ For example, $p=5$ and $k=3$ yields the congruence $151876^5\equiv 1\pmod{15^5}$.

Let us now assume that $b$ is not a multiple of $p$. Then $a^p\equiv 1\pmod{b^p}$ implies that either $a\equiv 1\pmod{b^p}$, or the multiplicative order of $a$ modulo $b^p$ equals $p$. In the latter case, we have $p\mid \varphi(b^p)=\varphi(b)b^{p-1}$, whence $p\mid\varphi(b)$, i.e. $b$ is a multiple of some prime $q\equiv 1\pmod{p}$. So in looking for a counterexample with $p\nmid b$, we may as well assume that $b$ is a prime congruent to $1$ mod $p$.

Here is a quick way to find a counterexample with $p\nmid b$ (for a full characterization see Vesselin Dimitrov's comment). Let $b$ be a prime congruent to $1$ mod $p$. Then $b^p$ is a prime power such that $\varphi(b^p)$ is divisible by $p$. Let $g$ be a primitive root mod $p$, and put $a:=g^{\varphi(b^p)/p}$. Then obviously $a^p\equiv 1\pmod{b^p}$, while $a\not\equiv 1\pmod{b^p}$.

Added. In retrospect the problem simply asks when does the group $(\mathbb{Z}/b^p\mathbb{Z})^\times$ have an element of order $p$. Formulated in this way, clearly a necessary and sufficient condition is $p\mid\varphi(b^p)$, which holds iff $b$ is divisible by $p$ or by a prime congruent to $1$ mod $p$.

• Very nice! I believe this completely answers my question, even without me refocusing/clarifying it. – Kieren MacMillan Aug 13 '14 at 18:48
• @KierenMacMillan: Thanks. I am sure that less obvious counterexamples also exist, in particular with $b$ not divisible by $p$. In this case one has to focus on $p\mid\varphi(b)$ as in the added paragraph. I don't have time to pursue this further, but maybe someone will. – GH from MO Aug 13 '14 at 18:50
• It remains to see that indeed there are examples with $a \mod{b^p} \in (\mathbb{Z}/b^p)^{\times}$ having multiplicative order exactly $p$. This is true if and only if $b$ is composed only of prime factors $q \equiv 1 \mod{p}$. Indeed, for any such prime $q$ you have an $a_0 \in \mathbb{Z}/q$ with $a_0^{p-1}+ \cdots + a_0+1 = 0$ (take $a_0 = g^{(q-1)/p} \mod{q}$ with $g$ a primitive root mod $q$). Since this polynomial has no multiple roots mod $q$, Hensel's lemma will lift $a_0$ to a solution mod arbitrarily high powers of $q$ (a $q$-adic solution). – Vesselin Dimitrov Aug 13 '14 at 18:59
• @VesselinDimitrov: I think your characterization is too restrictive. A group $G$ contains an element of order $p$ iff $|G|$ is divisible by $p$. In our setting, this means that $p\mid\varphi(b^p)$ which holds iff $p\mid b$ or $b$ is divisible by a prime congruent to $1$ mod $p$. We do not need (in the second case) that all the prime divisors are congruent to $1$ mod $p$. – GH from MO Aug 13 '14 at 19:25
• @GHfromMO: You are right. I was assuming implicitly that $b$ is prime to $a-1$, since we wanted to factor out the $a-1$ that produced the trivial examples. All the prime factors of $b$ not congruent to $1 \mod{p}$ will have to divide $a-1$. – Vesselin Dimitrov Aug 13 '14 at 19:33


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# NCERT Solutions for Class 12 Chemistry Chapter 10 Haloalkanes and Haloarenes

NCERT Solutions for Class 12 Chemistry Chapter 10 Haloalkanes and Haloarenes- When a hydrogen atom is replaced by a halogen atom (F, Cl, Br, I) in an aliphatic hydrocarbon it results in formation of Haloalkane or alkane halide. Similarly, when a hydrogen atom is replaced by a halogen in an aromatic hydrocarbon then it forms Haloarene or aryl halide. Such common facts are exaplained in detail in NCERT solutions for class 12 chemistry chapter 10 haloalkanes and haloarenes. Haloalkanes and Haloarenes have many uses in our day to day life, for example, chloramphenicol, a haloarene is used for the treatment of typhoid fever. Our body produces thyroxine hormone which contains iodine and deficiency of iodine can cause a disease called goiter. In this chapter, you will study methods of preparation, physical and chemical properties and uses of haloalkanes and haloarenes. You will find all the CBSE NCERT solutions for class 12 chemistry chapter 10 Haloalkanes and Haloarenes here for free. This chapter carries 4 marks in the CBSE board exams hence it becomes necessary for you to find all the solutions of NCERT class 12 chemistry chapter 10 haloalkanes and haloarenes to score good marks. In this chapter, there are 9 intext questions and  22 questions in the exercise.

The CBSE NCERT Solutions for class 12 chemistry chapter 10 Haloalkanes and Haloarenes are explained by chemistry experts to leave you with no doubts about the question. These NCERT solutions will help you in your preparation for CBSE Board exams as well as in competitive exams like JEE, NEET, BITSAT, and KVPY,  etc. The seven sub-topics of NCERT solutions for Class 12 Chemistry chapter 10 Haloalkanes and Haloarenes covers important concepts like IUPAC nomenclature and preparation of haloalkanes and haloarenes and discuss how to use stereochemistry as a tool in the understanding the reaction mechanism and applications of organo-metallic compounds. At the end of the chapter, the environmental effects of polyhalogen compounds also highlighted. Halogenated compounds remain in the environment due to their resistance to breakdown by the bacteria in the soil.

Intext Questions

Exercise Questions

Some important basic points of the Chapter 10 Haloalkanes and Haloarenes-

• The replacement of one or more than one H-atom in an aliphatic hydrocarbon by halogen atoms results in the formation haloalkane or alkyl halide and the replacement of H-atoms in an aromatic hydrocarbon by halogen atoms results in the formation of haloarene or aryl halide.
• In haloalkanes, halogen atoms attached to the $\dpi{100} SP^{3}$ hybridised carbon atom of an alkyl group.
• in haloarenes, halogen atoms attached to the $\dpi{100} SP^{2}$ hybridised carbon atoms of an aryl group.

## Topics and Sub-topics of NCERT Class 12 Chemistry Chapter 10 Haloalkanes and Haloarenes-

10.1 Classification

10.2 Nomenclature

10.3 Nature of C–X Bond

10.4 Methods of Preparation of Haloalkanes

10.5 Preparation of Haloarenes

10.6 Physical Properties

10.7 Chemical Reactions

## Solutions to In-Text Questions Ex 10.1  to 10.9

(i) 2-Chloro-3-methylpentane

The structure of 2-Chloro-3-methylpentane is given below :-

Question 10.1      Write structures of the following compounds:

(ii)    1-Chloro-4-ethylcyclohexane

The structure of  1-Chloro-4-ethylcyclohexane is given below :-

Question 10.1      Write structures of the following compounds:

(iii)       4-tert. Butyl-3-iodoheptane

The structure of 4-tert. Butyl-3-iodoheptane is given below :-

Question 10.1       Write structures of the following compounds:

(iv)    1,4-Dibromobut-2-ene

The structure of 1,4-Dibromobut-2-ene is given below :-

Question 10.1       Write structures of the following compounds:

(v)      1-Bromo-4-sec. butyl-2-methylbenzene

The structure of  1-Bromo-4-sec. butyl-2-methylbenzene is shown below :-

We don't use sulphuric acid because it acts as an oxidising agent and the required alkyl iodide is not produced. The reactions are given below :-

2KI  +  H2SO4     →    2KHSO4  + 2HI

2HI  + H2SO4    →  I2  +  SO2  +  H2O

We obtain four dihalogen derivatives of propane  :-

(i) 1,1 Dibromopropane

(ii) 2, 2 Dibromopropane

(iii) 1, 2 Dibromopropane

(iv) 1, 3 Dibromopropane

(i)       A single monochloride

In this we have to find an isomer in which replacement of any hydrogen atom gives the singel compound for all replacements.

So the isomer is Neopentane.

(ii)      Three isomeric monochlorides.

For the given condition we must have three different hydrogens so that we can get three different monochlorides on the replacement.

Thus the isomer is n-pentane.

For four monochlorides we need four different hydrogens which can be replaced by chlorine.

Hence the required isomer is :-

The final products are:-

(ii)

(iii)

(iv)

The obtained product is:-

$CH_3CH_2Br+\ NaI\rightarrow$

(vi)

The obtained product is :-

(i)       Bromomethane, Bromoform, Chloromethane, Dibromomethane.

It is known that boiling point increases with increase in molecular mass when the alkyl group is the same.

So the order of increasing boiling point is Chloromethane <  Bromomethane  <  Dibromomethane  <  Bromoform

(ii)      1-Chloropropane, Isopropyl chloride, 1-Chlorobutane.

In the given compounds the halide groups are same. In these cases, the boiling point depends on the bulkiness of the alkyl group. The boiling point increases with an increase in the chain length. Also, the boiling point decreases with an increase in branching.

So the order is :-    1- Chlorobutane  >  1- Chloropropane  >  Isopropyl Chloride

(i)      $CH_{3}CH_{2}CH_{2}CH_{2}Br$  or

In this case, the rate of $S_{N}2$   reaction will depend on the hindrance of the substrate.

Since 1- Bromobutane is a $1^{\circ}$ alkyl halide and 2- Bromobutane is a $2^{\circ}$ alkyl halide hence 2- Bromobutane gives more hindrance to the nucleophile.

Hence  1- Bromobutane reacts faster.

(ii)

The rate of $S_{N}2$ reaction decreases with increase in hindrance to the attack of the nucleophile.

So 2-bromobutane will react faster than 2-bromo-2-methylpropane in the nucleophilic attack.

(iii)

In these kinds of cases, we see where is the substituent is attached i.e., how far from the halide group. It can be clearly seen that the methyl group attached in 1-bromo-2-methylbutane is near than that attached in 1-bromo-3-methyl butane.

Hence the rate of $S_{N}2$ reaction will be faster in case of 1-bromo-3-methylbutane.

(i)

In $S_{N}1$ reactions, we see the formation of carbocation and this is the rate determining step for this kind of reactions. So the compound having more stable carbocation will react faster. In the given case 2- Chloro, 2- Methylpropane we have $3^{\circ}$ carbon whereas in 3- Chloropentane we have $2^{\circ}$ carbon.

(ii)

In $S_{N}1$ reactions, we see the formation of carbocation and this is the rate determining step for these kinds of reactions.So the compound having more stable carbocation will react faster. Hence 2-Chloroheptane will react faster than 1- Chlorohexane.

1st reaction  :-

2nd reaction :-

3rd reaction  :-

## NCERT Solutions for Class 12 Chemistry Chapter 10 Haloalkanes and Haloarenes- Exercise Questions

(i)      $\left ( CH_{3} \right )_{2}CHCH (Cl)CH_{3}$

(i) 2-Chloro-3-methylbutane.   And it is a secondary alkyl halide.

(ii)       $CH_{3}CH_{2}CH(CH_{3})CH(C_{2}H_{5})Cl$

3-Chloro-4-methylhexane.  And it is primary alkyl halide.

(iii)     $CH_{3}CH_{2}C(CH_{3})_{2}CH_{2}I$

(iii) 1-Iodo-2, 2-dimethylbutane.  And it is primary alkyl halide.

(iv)     $(CH_{3})_{3}CCH_{2}CH(Br)C_{6}H_{5}$

(iv)  1-Bromo-3, 3-dimethyl-1-phenylbutane.   And it is secondary benzyl halide.

(v)      $CH_{3}CH(CH_{3})CH(Br)CH_{3}$

(v) 2-Bromo-3-methylbutane.  And it is secondary alkyl halide.

(vi)      $CH_{3}C(C_{2}H_{5})_{2}CH_{2}Br$

(vi) 1-Bromo-2-ethyl-2-methylbutane.    And it is a primary alkyl halide.

(vii)       $CH_{3}C(Cl)(C_{2}H_{5})CH_{2}CH_{3}$

(vii) 3-Chloro-3-methylpentane. And it is tertiary alkyl halide.

(viii)    $CH_{3}CH = C(Cl)CH_{2}CH(CH_{3})_{2}$

(viii) 3-Chloro-5-methylhex-2-ene.  And it is vinyl halide

(ix)      $CH_{3}CH = CH C (Br)( CH_{3})_{2}$

(ix) 4-Bromo-4-methylpent-2-ene.  And it is allyl halide.

(x)     $p- Cl C_{6}H_{4}CH_{2}CH(CH_{3})_{2}$

(x)  1-Chloro-4-(2-methylpropyl) benzene.  And it is aryl halide.

(xi)     $m- Cl C_{6}H_{4}CH_{2}CH (CH_{3})_{2}$

(xi) 1-Chloromethyl-3-(2, 2-dimethylpropyl) benzene.  And it is primary benzyl halide.

(xii)       $o- Br C_{6}H_{4}CH (CH_{3})CH_{2}CH_{3}$

(xii) 1-Bromo-2-(1-methylpropyl) benzene. And it is aryl halide.

Question 10.2     Give the IUPAC names of the following compounds:

(iv)     $(CCl_{3})_{3}CCl$

2-(Trichloromethyl)-1, 1, 1, 2, 3, 3, 3-heptachloropropane

(i)     2-Chloro-3-methylpentane

(i)

(ii)

(iii)     1-Chloro-4-ethylcyclohexane

(iii)

(v)         2-Bromobutane

(v)

(vi)

(vii)     1-Bromo-4-sec-butyl-2-methylbenzene

(vii)

(viii)         1,4-Dibromobut-2-ene

(viii)

(i) CH2Cl2         (ii) CHCl3         (iii) CCl4

The order of dipole moment will be :-                CH2Cl2    >  CHCl3   >   CCl4.

The reason for the above order is given as- CCl4 is a symmetrical compound so its dipole moment will be zero. In case of CHCl, one of the Cl cancels dipole moment of the opposite Cl atom, so net dipole moment is just due to one Cl. In the case of CH2Cl2 , both Cl groups contribute to the dipole moment so it has the highest dipole moment among all.

We are given the formula C5H10  which can be either of an alkene or of cycloalkane. Since the hydrocarbon doesn't react with chlorine in dark thus it cannot be alkene. So the only option left out is cyclopentane.

The isomers of the compound $C_{4}H_{9}Br$  are :-

(i)  1-Bromobutane

(ii)   2-Bromobutane

(iii)  1-Bromo-2-methylpropane

(iv)  2-Bromo-2-methylpropane

(i)        1-butanol

(i)  The procedure given below can be used :-

(ii)         1-chlorobutane

(ii) The required product can be obtained as shown below :-

(iii)     but-1-ene

(iii) The required product is obtained by following procedure :-

The ambident nucleophiles are those nucleophiles which have two nucleophilic sites through which they can attack. For e.g Nitrile ion can attack through both nitrogen atom (forms nitroalkanes) and an oxygen atom (forms alkyl nitrites), thus it is an ambident nucleophile.

(i)       CH3Br or CH3 I

In this case, we have the same alkyl group but different halide ions. For this rate of SN2 reaction increases with increase in atomic mass. So,  CH3I will react faster than CH3Br.

(ii)     $(CH_{3})_{3}CCl$ or $CH_{3}Cl$

In this case, the hindrance will be deciding factor for the rate of  SN 2 reaction because hindrance will directly affect the attack of the nucleophile. So  $CH_{3}Cl$  will react faster as compared to $(CH_{3})_{3}CCl$.

(i)     1-Bromo-1-methylcyclohexane

In this compound, it is clear that we have identical $\beta$ hydrogen, therefore, dehalogenation of the given compound gives the same alkene.

(ii)     2-Chloro-2-methylbutane

(ii)

In this compound we have two kind of $\beta$ hydrogen. So dehalogenation will give two kind of alkenes, namely 2-Methylbut-2-ene and 2-Methylbut-1-ene.

The major product of this reaction will be 2-Methylbut-2-ene as the number of $\alpha$ - hydrogens attached to double bonded carbon are more in case of this compound.

(iii)     2,2,3-Trimethyl-3-bromopentane

(iii)

In this compound we have two type of $\beta$ - hydrogen thus dehalogenation we get two types of products namely 3, 4, 4-Trimethylpent-2-ene and 2-Ethyl-3,3-dimethylbut-2-ene.

and

Here 3, 4, 4-Trimethylpent-2-ene will be major product, since the $\alpha$ - hydrogen attached to the double bond are greater.

Question 10.11     How will you bring about the following conversions?

(i)      Ethanol to but-1-yne

(i)  The conversion will take place by following procedure :-

Now,

Question 10.11     How will you bring about the following conversions?

(ii)     Ethane to bromoethene

(ii)

Question 10.11     How will you bring about the following conversions?

(iii)     Propene to 1-nitropropane

(iii)

Question 10.11     How will you bring about the following conversions?

(iv)    Toluene to benzyl alcohol

(iv)

Question 10.11         How will you bring about the following conversions?

(v)      Propene to propyne

(v)

Question 10.11     How will you bring about the following conversions?

(vi)   Ethanol to ethyl fluoride

(vi)

Question 10.11     How will you bring about the following conversions?

(vii)     Bromomethane to propanone

(vii)

Question 10.11     How will you bring about the following conversions?

(viii)    But-1-ene to but-2-ene

(viii)

Question 10.11     How will you bring about the following conversions?

(ix)   1-Chlorobutane to n-octane

(ix)

Question 10.11     How will you bring about the following conversions?

(x)    Benzene to biphenyl.

(x)

Question 10.12     Explain why

(i)     the dipole moment of chlorobenzene is lower than that of cyclohexyl chloride?

We know that the Cl-atom in chlorobenzene is attached to a sp2 hybridized carbon atom whereas, in cyclohexyl chloride, the Cl-atom is attached to a sp3 hybridized carbon atom. It is known that sp2 hybridized carbon has more s-character than sp3 hybridized carbon atom. Thus, chlorobenzene is more electronegative than cyclohexyl chloride.

Apart from this, the - R effect of the benzene ring of chlorobenzene results in decreasing the electron density of the C - Cl bond near the Cl-atom. And the C - Cl bond in chlorobenzene becomes less polar. Due to these reasons, the dipole moment of chlorobenzene is lower than that of cyclohexyl chloride.

Question 10.12     Explain why

For being soluble in the water we have a condition that the solute-water force of attraction must be stronger than the solute-solute and water-water forces of attraction. Alkyl halides are held together by dipole-dipole interactions and there are polar molecules. Similarly, the intermolecular force of attraction present between the water molecules is hydrogen bonding. The new force of attraction after we dissolve solute in water i.e., between the alkyl halides and water molecules is weaker than the alkyl halide-alkyl halide and water-water forces of attraction. That is why alkyl halides (though polar) are immiscible with water.

Question 10.12     Explain why

(iii)     Grignard reagents should be prepared under anhydrous conditions?

This is done because in presence of moisture, they react to give alkane.

(i)   Freon-12 or dichlorodifluoromethane is generally known as CFC. It is used in refrigerators and air conditioners as a refrigerant. It is also used in body sprays, hair sprays, etc. But it has environmental impacts as it damages the ozone layer.

(ii)  DDT or p, p-dichlorodiphenyltrichloroethane is one of the best-known insecticides which was used very widely all over the world. It is very effective against mosquitoes, insects and lice. But it has harmful effects.

(iii)  CCl :-   It is mostly used for manufacturing refrigerants for refrigerators and air conditioners.  It is also used as a solvent in the manufacture of pharmaceutical products. In the early years, carbon tetrachloride was widely used as a cleaning fluid and a fire extinguisher.

(iv) Iodoform was used earlier as an antiseptic. And this antiseptic property of iodoform is due to the liberation of free iodine when it comes in contact with the skin.

(i)  The obtained product is :-

$\dpi{100} CH_{3}CH_{2}CH_{2}Cl+NaI\xrightarrow[heat]{acetone}$

(ii)     $(CH_{3})_{3}CBr+KOH \xrightarrow[heat]{ethanol}$

(ii)  The obtained product is 2-Methylpropene

(iv)     $CH_{3}CH_{2}Br+KCN \overset{aq. ethanol}{\rightarrow}$

(iv)  The obtained product is Cyanoethane.

(v) The obtained product is Phenetole.

(vi)  The obtained product is 1-Chloropropane.

(vii)  The obtained product is 1-Bromobutane.

(viii)       $CH_{3}CH = C(CH_{3})_{2}+ HBr \rightarrow$

The obtained product is 2-Bromo-2-methylbutane.

Question 10.15     Write the mechanism of the following reaction:

The reaction will proceed through SN2 mechanism. The mechanism for the given reaction is shown below :-

(i)     2-Bromo-2-methylbutane, 1-Bromopentane, 2-Bromopentane

(i)  Here the deciding factor the rate of reaction will be a steric hindrance.

It is clear from the above that the order of hindrance is:-

1-Bromopentane < 2-bromopentane < 2-Bromo-2-methylbutane

So the order of rate of reaction will be:-

2-Bromo-2-methylbutane < 2-Bromopentane < 1-Bromopentane

(ii) In this case, also, the order of the rate of reaction will be decided from the steric hindrance factor.

It is clear from the above that the steric hindrance in 2-Bromo-2-methylbutane highest (note that hindrance of the carbon attached to halide ion is seen). So the order of the rate of reaction is:-

2-Bromo-2-methylbutane < 2-Bromo-3-methylbutane < 1-Bromo-3-methylbutane

(iii)     1-Bromobutane, 1-Bromo-2,2-dimethylpropane, 1-Bromo-2-methylbutane, 1-Bromo-3-methylbutane.

(iii)  The steric hindrance is the deciding factor here.

The order of steric hindrance is :-

1-Bromobutane < 1-Bromo-3-methylbutane < 1-Bromo-2-methylbutane< 1-Bromo-2, 2-dimethylpropane

Thus the order of the rate of reaction will be : -

1-Bromo-2, 2-dimethylpropane < 1-Bromo-2-methylbutane < 1-Bromo-3- methylbutane < 1-Bromobutane

Hydrolysis by KOH will take place by the formation of a carbocation in its rate-determining step. So the compound having stable carbocation will hydrolyse faster.

The carbocations of both the compounds are given below:-

It is clear that carbocation of $C_{6}H_{5}CHCl C_{6}H_{5}$ is more stable.

Hence  $C_{6}H_{5}CHCl C_{6}H_{5}$ will hydrolyse faster than $C_{6}H_{5}CH_{2}Cl$.

The structures of  o-Dichlorobenzene,  m-Dichlorobenzene and  p-Dichlorobenzene are given below.

We can see that p-Dichlorobenzene is a very symmetric structure thus packing of it will be maximum. As a result more and more energy will be required to break bonds (during boiling). Thus boiling point is high for p-Dichlorobenzene.

Question 10.19     How the following conversions can be carried out?

(i)    Propene to propan-1-ol

The mechanism is given below :-

Question 10.19     How the following conversions can be carried out?

(ii)      Ethanol to but-1-yne

The reaction mechanism is given below :-

Question 10.19     How the following conversions can be carried out?

(iii)       1-Bromopropane to 2-bromopropane

(iii)  The mechanism is given below :-

Question 10.19     How the following conversions can be carried out?

(iv)      Toluene to benzyl alcohol

(iv) The mechanism is given below :-

Question 10.19     How the following conversions can be carried out?

The mechanism for the given reaction is as follows  :-

Question 10.19     How the following conversions can be carried out?

(vi)    Benzyl alcohol to 2-phenylethanoic acid

(vi) The mechanism for the reaction is given below :-

Question 10.19     How the following conversions can be carried out?

(vii)      Ethanol to propanenitrile

(vii) The mechanism of the given reaction reaction is :-

Question 10.19     How the following conversions can be carried out?

(viii)      Aniline to chlorobenzene

The mechanism of the reaction is given below :-

Question 10.19     How the following conversions can be carried out ?

(ix)      2-Chlorobutane to 3, 4-dimethylhexane

(ix)   The required mechanism is as follows :-

Question 10.19     How the following conversions can be carried out?

(x) The required mechanism is given below :-

Question 10.19     How the following conversions can be carried out ?

(xi)      Ethyl chloride to propanoic acid

The mechanism of the given reaction is :-

Question 10.19     How the following conversions can be carried out?

(xii)     But-1-ene to n-butyliodide

The mechanism is given below :-

Question 10.19     How the following conversions can be carried out?

(xiii)      2-Chloropropane to 1-propanol

The mechanism is :-

Question 10.19     How the following conversions can be carried out?

(xiv)      Isopropyl alcohol to iodoform

(xiv) The proposed mechanism is :-

Question 10.19     How the following conversions can be carried out?

(xv)     Chlorobenzene to p-nitrophenol

(xv) The required mechanism is given below :-

Question 10.19     How the following conversions can be carried out?

(xvi)     2-Bromopropane to 1-bromopropane

The mechanism of the reaction is given below :-

Question 10.19     How the following conversions can be carried out?

(xvii)      Chloroethane to butane

(xvii) The mechanism of the reaction is :-

Question 10.19     How the following conversions can be carried out?

(xviii)      Benzene to diphenyl

The mechanism for the given reaction is :-

Question 10.19     How the following conversions can be carried out?

(xix)      tert-Butyl bromide to isobutyl bromide

The mechanism of the given reaction is :-

Question 10.19     How the following conversions can be carried out?

(xx)     Aniline to phenylisocyanide

The mechanism for the given reaction is as follows :-

In an aqueous solution, KOH almost completely dissociates into OH - ions. We know that OH - ions are strong nucleophile, which leads the alkyl chloride to undergo a reaction to form alcohol. But an alcoholic solution of KOH contains alkoxide (RO ) ion, which is a strong base. Thus, it can remove hydrogen from the β-carbon of the alkyl chloride and form an alkene.  The OH - ion is a weaker base than RO - ion. The basic character of OH - ion decreases in aqueous solution. Therefore, it cannot remove hydrogen from the β-carbon.

With the given formula we have two alkyl halides They are n - butyl bromide and isobutyl bromide.

For the first set of reaction we get two possibilities:-

Therefore compound (d) is 2, 5-dimethylhexane.

Question 10.22     What happens when

(i)       n-butyl chloride is treated with alcoholic KOH,

When n-butyl chloride is treated with alcoholic KOH the following reaction occurs:-

Question 10.22     What happens when

(ii)     bromobenzene is treated with Mg in the presence of dry ether

When bromobenzene is treated with Mg in the presence of dry ether the following reaction occurs :-

Question 10.22     What happens when

(iii)    chlorobenzene is subjected to hydrolysis

The reaction is given below :-

Question 10.22     What happens when

(iv)    ethyl chloride is treated with aqueous KOH

When ethyl chloride is treated with aqueous KOH the following reaction occurs :-

Question 10.22     What happens when

(v)    methyl bromide is treated with sodium in the presence of dry ether

When methyl bromide is treated with sodium in the presence of dry ether then following reaction occurs:-

Question 10.22     What happens when

(vi)      methyl chloride is treated with KCN?

When methyl chloride is treated with KCN the following reaction occurs:-

## NCERT Solutions Class 12 Chemistry

 Chapter 1 CBSE NCERT solutions for class 12 chapter 1 The Solid State Chapter 2 NCERT solutions for class 12 chemistry chapter 2 Solutions Chapter 3 Solutions of NCERT class 12 chemistry chapter 3 Electrochemistry Chapter 4 CBSE NCERT solutions for class 12 chemistry chapter 4 Chemical Kinetics Chapter 5 Solutions of NCERT class 12 chemistry chapter 5 Surface chemistry Chapter 6 NCERT solutions for class 12 chemistry General Principles and Processes of isolation of elements Chapter 7 CBSE NCERT solutions for class 12 chemistry chapter 7 The P-block elements Chapter 8 Solutions of NCERT class 12 chemistry chapter 8 The d and f block elements Chapter 9 NCERT solutions for class 12 chemistry chapter 9 Coordination compounds Chapter 10 NCERT solutions for class 12 chemistry chapter 10 Haloalkanes and Haloarenes Chapter 11 CBSE NCERT solutions for class 12 chemistry Alcohols, Phenols, and Ethers Chapter 12 Solutions of NCERT class 12 chemistry chapter 12 Aldehydes, Ketones and Carboxylic Acids Chapter 13 NCERT solutions for class 12 chemistry chapter 13 Amines Chapter 14 CBSE NCERT solutions for class 12 chemistry chapter 14 Biomolecules Chapter 15 Solutions of NCERT class 12 chemistry chapter 15 Polymers Chapter 16 NCERT solutions for class 12 chemistry chapter 16 Chemistry in Everyday life

## NCERT Solutions for Class 12 Subject wise

 Solutions of NCERT class 12 biology NCERT solutions for class 12 maths CBSE NCERT solutions for class 12 chemistry Solutions of NCERT class 12 physics

## Benefits of NCERT solutions for class 12 chemistry chapter 10 Haloalkanes and Haloarenes

• The comprehensive answers given in the NCERT solutions for class 12 chemistry chapter 10 Haloalkanes and Haloarenes will be helpful in understanding the chapter easily.

• Revision will be quite easier because the detailed solutions will help you to remember the concepts and get good marks in your class.

• Homework problems won't bother you anymore, all you need to do is check the detailed NCERT solutions for class 12 chemistry and you are good to go.

If you have a doubt or question that is not available here or in any of the chapters, contact us. You will get all the answers that will help you score well in your exams.


text:
Lesson 6, Topic 3
In Progress

# Theory of Multiplier

Lesson Progress
0% Complete

The theory of multiplier is a situation where an autonomous increase or decrease in expenditure for output can increase or decrease total expenditure for output by some multiple, by inducing a charge of consumption expenditure in the same direction.

Multiplier theory is the amount by which a change in net investment is multiplied in terms of income.

It is a measure of change in national income caused by a change in investment. Thus, it explains the relationship between the increase in investment and the resultant increase in income. For example, if an increase in investment of N50,000 causes an increase in national income of N600,000, then the value of the multiplier would be

= $$\frac{600,000}{50,000} \\ = \scriptsize 12$$

$$\frac{Change \: in \: Income}{Change \: in \: Investment} \\ = \frac{\Delta I}{\Delta Y}$$

Anytime the investment expenditure rises, it will stimulate production and cause income to rise too.

K = $$\frac{1}{\Delta S} \\ \scriptsize or \\ \frac{1}{1 \: – \: \frac{\Delta C}{\Delta Y}}$$

∆C = Change in Consumption

∆S = Change in Savings

∆Y = Change in Income

∆E = Change in expenditure

K = $$\frac{1}{1 \: – \: MPC}$$

or

$$\frac{1}{1 \: – \: b}$$

Where b = Marginal Propensity to consume

Income multiplier K = $$\frac{1}{1 \: – \: b}$$

While the investment multiplier is given

K = $$\frac{\Delta \: I}{1 \: – \: b}$$ where K is the multiplier.

∆I = Change in Income

b = MPC

If the marginal propensity to consume is 50%, what is the multiplier?

If there is 50m rise in investment, what is the change in Income?

K = $$\frac{1}{1 \: – \: b} \\= \frac{1}{1 \: – \: 0.5} \\ = \frac{1}{0.5} \\ = \scriptsize 2$$

Multiplier = 2

Change in Income ∆Y = $$\frac{\Delta Y}{1 \: – \: b} \\ = \frac{50m}{1 \: – \: 0.5} \\ = \frac{1}{1 \: – \: 0.5} \: \times \: \frac{50m}{1 } \\ = \frac{50m}{0.5} \\= \scriptsize 100m$$

Change in Income ∆Y = 100m

error:


text:
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Help
 Please use this identifier to cite or link to this item: http://hdl.handle.net/1807/27195

 Title: Symmetry of evidence without evidence of symmetry Authors: Larry G. Epstein; Department of Economics, Boston UniversityKyoungwon Seo; Department of Managerial Economics and Decision Sciences, Northwestern University Keywords: Ambiguity, exchangeability, symmetry, updating, learning, multiple-priorsD81 Issue Date: 22-Sep-2010 Publisher: Theoretical Economics Citation: Theoretical Economics; Vol 5, No 3 (2010) Abstract: [This item is a preserved copy. To view the original, visit http://econtheory.org/] The de Finetti Theorem is a cornerstone of the Bayesian approach. Bernardo (1996) writes that its "message is very clear: if a sequence of observations is judged to be exchangeable, then any subset of them must be regarded as a random sample from some model, and there exists a prior distribution on the parameter of such model, hence requiring a Bayesian approach." We argue that while exchangeability, interpreted as symmetry of evidence, is a weak assumption, when combined with subjective expected utility theory, it implies also complete confidence that experiments are identical. When evidence is sparse, and there is little evidence of symmetry, this implication of de Finetti's hypotheses is not intuitive. This motivates our adoption of multiple-priors utility as the benchmark model of preference. We provide two alternative generalizations of the de Finetti Theorem for this framework. A model of updating is also provided. URI: http://hdl.handle.net/1807/27195 Other Identifiers: http://econtheory.org/ojs/index.php/te/article/view/20100313 Rights: Authors who publish in Theoretical Economics will release their articles under the Creative Commons Attribution-NonCommercial license. This license allows anyone to copy and distribute the article for non-commercial purposes provided that appropriate attribution is given. Appears in Collections: Volume 5, Number 3 (September 2010)

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text:
# Why not quarter-life?

The number of nuclei left after time $t$ in radioactive decay is given by:

$$N(t) = N_0 e^{-t/ \tau}$$

Now if we put $N(t)$ as $\dfrac{N_0}2$, we get half-life. But, if we had put $\dfrac{N_0}4$, we would have quarter-life, which is also independent of $N_0$.

Is there anything special about half-life as opposed to quarter-life

• Not really, $1/2$ is just a nice number. – knzhou Feb 26 '18 at 10:51
• Note also that if you go for $\frac{N_0}{e}$ then the time $t=\tau$ is the time it takes to become one-eeeth of its original value. – Farcher Feb 26 '18 at 11:48
• There is the very practical reason that if it were "quarter" you wouldn't know whether it referred to a quarter gone or a quarter left. – Hot Licks Feb 26 '18 at 13:15

The decay time $t_{1/2}$ of half the given number $N_0$ of atoms atoms is just convenient and visually appealing. Of the unit fractions it is also nearest to the decay time constant (mean lifetime) $\tau$ $t_{1/2}=0.6931 \tau$. The decay time to a unit fraction $1/n$ given by the positive integer $n$ is $$t_{1/n}=\tau \cdot ln(n)$$
The half life $t_{1/2}$ is the median. The time constant $\tau$ is the mean. There's nothing particularly meaningful about the quarter life.


text:
# Obtaining Edges using the Canny operator¶

In this tutorial we show how easily one can apply the typical canny edge detection using Kornia.

Enjoy the example!

## Preparation¶

We first install Kornia.

%%capture
%matplotlib inline
!pip install git+https://github.com/kornia/kornia

import kornia
kornia.__version__

'0.6.1'

%%capture
!wget -O paranoia_agent.jpg https://ih1.redbubble.net/image.675644909.6235/flat,800x800,075,f.u3.jpg

## Example¶

We first import the required libraries and load the data.

import torch
import kornia
import cv2
import numpy as np

import matplotlib.pyplot as plt

# read the image with OpenCV
img = cv2.cvtColor(img, cv2.COLOR_BGR2RGB)

# convert to torch tensor
data: torch.tensor = kornia.image_to_tensor(img, keepdim=False)/255.  # BxCxHxW

plt.axis('off')
plt.imshow(img)
plt.show()

To apply a filter, we create the Canny operator object and apply it to the data. It will provide the magnitudes as well as the edges after the hysteresis process. Note that the edges are a binary image which is not differentiable!

# create the operator
canny = kornia.filters.Canny()

# blur the image
x_magnitude, x_canny = canny(data.float())

That’s it! We can compare the source image and the results from the magnitude as well as the edges:

# convert back to numpy
img_magnitude: np.ndarray = kornia.tensor_to_image(x_magnitude.byte())
img_canny: np.ndarray = kornia.tensor_to_image(x_canny.byte())

# Create the plot
fig, axs = plt.subplots(1, 3, figsize=(16,16))
axs = axs.ravel()

axs[0].axis('off')
axs[0].set_title('image source')
axs[0].imshow(img)

axs[1].axis('off')
axs[1].set_title('canny magnitude')
axs[1].imshow(img_magnitude, cmap='Greys')

axs[2].axis('off')
axs[2].set_title('canny edges')
axs[2].imshow(img_canny, cmap='Greys')

plt.show()

Note that our final result still recovers some edges whose magnitude is quite low. Let us increase the thresholds and compare the final edges.

# create the operator
canny = kornia.filters.Canny(low_threshold=0.4, high_threshold=0.5)

# blur the image
_, x_canny_threshold = canny(data.float())

import torch.nn.functional as F
# convert back to numpy
img_canny_threshold: np.ndarray = kornia.tensor_to_image(x_canny_threshold.byte())

# Create the plot
fig, axs = plt.subplots(1, 3, figsize=(16,16))
axs = axs.ravel()

axs[0].axis('off')
axs[0].set_title('image source')
axs[0].imshow(img)

axs[1].axis('off')
axs[1].set_title('canny default')
axs[1].imshow(img_canny, cmap='Greys')

axs[2].axis('off')
axs[2].set_title('canny defined thresholds')
axs[2].imshow(img_canny_threshold, cmap='Greys')

plt.show()


text:
# Calculation Kernel of a matrix

I'm trying to write a program in which one part is related to calculation of kernel of a matrix. Its output and expected output are different. For example,

A = [[1, 0, 1], [1, 0, 0], [0, 1, 1], [0, 1, 0], [0, 0, 1], [-1, 0, 0], [0, 0, -1], [0, -1, 1], [0, -1, 0], [-1, 0, 1]]

A is the matrix whose kernel is wanted.

When I calculate its kernel with some programs they give output different, my program is as well. But, when I try to calculate its kernel some others, like SAGE, give output,

1  0  0  0  0  0  0 -2  2  1
0  1  0  0  0  0  0 -1  1  1
0  0  1  0  0  0  0 -1  2  0
0  0  0  1  0  0  0  0  1  0
0  0  0  0  1  0  0 -1  1  0
0  0  0  0  0  1  0  1 -1 -1
0  0  0  0  0  0  1  1 -1  0

The above one is what I expect as output. What is the point that I may overlook?

Here is my procedure to calculate the kernel in my program,

A.transposeInPlace();
FullPivLU<MatrixXf> lu(A);
MatrixXf A_null_space = lu.kernel();
A_null_space.transposeInPlace();

But in that way, I get different then expected one, but SAGE gives the above matrix that actually I expect.

 0.5    0   -1    1    0    0    0    0    0  0.5
-0.5    0   -0    0    1    0    0    0    0 -0.5
0.5    0   -0    0    0    1    0    0    0 -0.5
0.5    0   -0    0    0    0    1    0    0  0.5
-1    0    1    0    0    0    0    1    0   -1
-0.5    0    1    0    0    0    0    0    1 -0.5
-0.5    1   -0    0    0    0    0    0    0  0.5

I'm really but really confused because both matrix seem right! How come?

Sage's output proof, https://i.stack.imgur.com/F3ryq.png My program's output proof, https://i.stack.imgur.com/7Mw8y.png

edit retag close merge delete

Sort by » oldest newest most voted

Your matrix acts on column vectors to the right and on row vectors to the left (there is some duality here but it is not important in first approach). So, it can be considered either as a linear map \mathbb{Q}^3\to \mathbb{Q}^10 or as a linear map \mathbb{10}^3\to \mathbb{Q}^3 , respectively.

It has rank 3, so in the first case, the kernel is trivial, in the second case the kernel has dimension 7.

Here is some code to understand what i just wrote:

sage: A = matrix(QQ, [[1, 0, 1], [1, 0, 0], [0, 1, 1], [0, 1, 0], [0, 0, 1], [-1, 0, 0], [0, 0, -1], [0, -1, 1], [0, -1, 0], [-1, 0, 1]])
sage: A
[ 1  0  1]
[ 1  0  0]
[ 0  1  1]
[ 0  1  0]
[ 0  0  1]
[-1  0  0]
[ 0  0 -1]
[ 0 -1  1]
[ 0 -1  0]
[-1  0  1]

sage: A.parent()
Full MatrixSpace of 10 by 3 dense matrices over Rational Field

sage: A.rank()
3

sage: A.right_kernel()
Vector space of degree 3 and dimension 0 over Rational Field
Basis matrix:
[]

sage: A.right_kernel().dimension()
0

sage: K = A.left_kernel()
sage: K
Vector space of degree 10 and dimension 7 over Rational Field
Basis matrix:
[ 1  0  0  0  0  0  0 -2  2  1]
[ 0  1  0  0  0  0  0 -1  1  1]
[ 0  0  1  0  0  0  0 -1  2  0]
[ 0  0  0  1  0  0  0  0  1  0]
[ 0  0  0  0  1  0  0 -1  1  0]
[ 0  0  0  0  0  1  0  1 -1 -1]
[ 0  0  0  0  0  0  1  1 -1  0]

sage: K.dimension()
7

sage: K.basis()
[
(1, 0, 0, 0, 0, 0, 0, -2, 2, 1),
(0, 1, 0, 0, 0, 0, 0, -1, 1, 1),
(0, 0, 1, 0, 0, 0, 0, -1, 2, 0),
(0, 0, 0, 1, 0, 0, 0, 0, 1, 0),
(0, 0, 0, 0, 1, 0, 0, -1, 1, 0),
(0, 0, 0, 0, 0, 1, 0, 1, -1, -1),
(0, 0, 0, 0, 0, 0, 1, 1, -1, 0)
]

sage: K.basis()[0]
(1, 0, 0, 0, 0, 0, 0, -2, 2, 1)

sage: K.basis()[0] * A
(0, 0, 0)

more

Wow, that's a great help for me, really thank you. However, do you have an idea(algorithm) that helps to calculate the left kernel form? as shown on my code,

A.transposeInPlace();
FullPivLU<MatrixXf> lu(A);
MatrixXf A_null_space = lu.kernel();
A_null_space.transposeInPlace();

If write like this, it gives,

Null space:
0.5    0   -1    1    0    0    0    0    0  0.5
-0.5    0   -0    0    1    0    0    0    0 -0.5
0.5    0   -0    0    0    1    0    0    0 -0.5
0.5    0   -0    0    0    0    1    0    0  0.5
-1    0    1    0    0    0    0    1    0   -1
-0.5    0    1    0    0    0    0    0    1 -0.5
-0.5    1   -0    0    0    0    0    0    0  0.5

( 2017-11-15 10:44:57 -0500 )edit

But expected,

1  0  0  0  0  0  0 -2  2  1
0  1  0  0  0  0  0 -1  1  1
0  0  1  0  0  0  0 -1  2  0
0  0  0  1  0  0  0  0  1  0
0  0  0  0  1  0  0 -1  1  0
0  0  0  0  0  1  0  1 -1 -1
0  0  0  0  0  0  1  1 -1  0

( 2017-11-15 10:46:19 -0500 )edit


text:
# A FOOTBALL PLAYER RUNS ACROSS THE 30 YARD LINE AT AN ANGLE

###### Question:

A FOOTBALL PLAYER RUNS ACROSS THE 30 YARD LINE AT AN ANGLE. HE CONTINUES IN A STRAIGHT LINE AND CROSSES THE GOAL LINE AT THE SAME ANGLES. DESCRIBE TWO PARALLEL LINES AND A TRANSVERSAL IN THE DIAGRM

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you tear a piece of paper that is 0.005 centimeters thick in h other, the height of the pile of paper Number Theory half, and place the two pieces on top of eac 1S 0.010 centimeters. Let's call this the second pile. If you tear these two pieces of paper in half, the third pile will have four pie...
Problem 1: Perez Animal Feeds has developed the following data for lower-of-cost-or-market valuation for its inventory products: Cost Market $400,000 875.000$380,000 920,000 Large Animals: Cattle Horse Small Animals: Cat Dog Ferret 550.000 90,000 280,000 700,000 165,000 245,000 Required: Determine ...


text:
# How can i customise the breakable box

In my example below, I have a problem with my breakable box, he does not want to ride.

\documentclass[12pt,a4paper,oneside,openright]{book}
\usepackage[utf8]{inputenc}
\usepackage[frenchb]{babel}
\usepackage[most]{tcolorbox}
\usepackage{multicol}
\usepackage{lipsum}
\usepackage[left=4cm,right=3cm,top=3cm,bottom=3cm]{geometry}

\usepackage{etoolbox}
\definecolor{tcbcolback}{RGB}{255,128,0}

\usepackage{varwidth}
\newtcolorbox{mybox}[2][]{enhanced,
before skip=0mm,after skip=0mm,
colback=black!5,colframe=black!50,boxrule=0.2mm,
height fixed for=first and middle,
attach boxed title to top left={xshift=1cm,yshift*=1mm-\tcboxedtitleheight},
boxed title style={frame code={
\path[fill=tcbcolback!60!black]
([yshift=-1mm,xshift=-1mm]frame.north west)
([yshift=-1mm,xshift=1mm]frame.north east)
\path[left color=tcbcolback!90!black,right color=tcbcolback!90!black,
middle color=tcbcolback!99!black]
([xshift=-2mm]frame.north west) -- ([xshift=2mm]frame.north east)
[rounded corners=1mm]-- ([xshift=1mm,yshift=-1mm]frame.north east)
-- (frame.south east) -- (frame.south west)
-- ([xshift=-1mm,yshift=-1mm]frame.north west)
[sharp corners]-- cycle;
},interior engine=empty,
},
fonttitle=\bfseries,
title={#2},#1}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\NewTColorBox[auto counter,number within=chapter]{exercise}{m+O{}}{%
enhanced jigsaw,enforce breakable,pad at break=0mm,break at=-20pt,
colframe=green!20!black,
colback=yellow!10!white,
coltitle=green!40!black,
fonttitle=\bfseries,
underlay={\begin{tcbclipinterior}
(interior.north west) circle (2cm);
\draw[help lines,step=5mm,yellow!80!black,shift={(interior.north west)}]
(interior.south west) grid (interior.north east);
\end{tcbclipinterior}},
title={Exercise~ \thetcbcounter:},
label={exercise:#1},
after upper={\par\hfill\textcolor{green!40!black}%
{\itshape Solution on page~\pageref{solution:#1}}},
lowerbox=ignored,
savelowerto=solutions/exercise-\thetcbcounter.tex,
record={\string\solution{#1}{solutions/exercise-\thetcbcounter.tex}},
#2
}

\NewTotalTColorBox{\solution}{mm}{%
enhanced,
colframe=red!20!black,
colback=yellow!10!white,
coltitle=red!40!black,
fonttitle=\bfseries,
underlay={\begin{tcbclipinterior}
(interior.north west) circle (2cm);
\draw[help lines,step=5mm,yellow!80!black,shift={(interior.north west)}]
(interior.south west) grid (interior.north east);
\end{tcbclipinterior}},
title={Solution of Exercise~\ref{exercise:#1} on page~\pageref{exercise:#1}:},
phantomlabel={solution:#1},
attach title to upper=\par,
}{\input{#2}}

\tcbset{no solution/.style={no recording,after upper=}}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{document}
\begin{mybox}[]{Exercises }
\begin{exercise}{Ex1}[coltitle=cyan!80!black]
\lipsum[1-2]
\end{exercise}
\begin{exercise}{Ex2}[coltitle=cyan!80!black]
\lipsum[1-6]
\end{exercise}
\end{mybox}
\end{document}

• You are trying to define breakable boxes inside other breakable boxes, and as CarLaTeX explained in your previous question this is not a good idea. – Ignasi Jun 3 at 15:45
• when I use unbreakable box inside breakable box it does not work. I do not know what to do. – moumou Jun 3 at 16:23
• it works. But when second box is too long for the remaining space, it jumps to second page and leaves an empty background box on first page. – Ignasi Jun 3 at 16:51
• Exactly, what is the solution then. My be changing outer box by tikz box. – moumou Jun 3 at 17:25
• As I also suggested here, you could try with the solution provided in [various defined breakable boxes inside breakable box using overlapping ](tex.stackexchange.com/questions/426981/…). This answer proposes to declare three different environments for inner boxes (exercises and solutions in your case). These environments already draw a false external box. This way you only have to fight with one level of breakable boxes. – Ignasi Jun 3 at 17:43


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Kepler's Law, focal points - Heliocentric or Barycentric?

It is known that all the mass in the Solar system moves around the Barycenter.

For the two focal points in Kepler's law; is the first focal point $$F_1$$ a Heliocenter? Or is it actually a Barycenter?

If the $$F_1$$ is a Heliocenter, then, can the same exact ellipse and eccentricity of orbit be maintained by adjusting the second focal point $$F_2$$ and taking Barycenter as $$F_1$$ instead of Heliocenter?

Kepler's laws and the associated orbit only hold for a two-body problem, so the 'barycenter' in the question can only be understood as the center of mass of the sun and a particular planet (ignoring the other ones), not as the solar system barycenter. And in this sense the convention is to take the sun at the focus, which means the semi-major axis is the (maximum) distance between the sun and the planet.

If you define the orbits with regard to the center of mass of the two bodies, the semi-major axis is reduced by the mass factor $$m/(M+m)$$ for the planet and $$M/(M+m)$$ for the sun whilst the eccentricity as well as the orbital period stays the same. However, with the center of mass only a fictitious point that does not correspond to a physical object, this hardly makes sense in practice.

Planetary orbits are normally described as heliocentric, but it is possible to describe them from a barycentric point of view. JPL Horizons (https://ssd.jpl.nasa.gov/horizons.cgi) provides for both possibilities.

It depends!

If we plotted the orbits of the innermost planets in our solar system, they would be closer to ellipses with the Sun at one focus.

If we plotted the orbit of Jupiter, the biggest "gravitational bully" in the solar system (it messes with everything!) it would be closer to an ellipse with the Sun-Jupiter barycenter at one focus.

It gets a little complicated out there because the next three planets (Saturn, Uranus and Neptune) also push the Sun around a lot. We might think that lightweight Neptune wouldn't do much, but it's larger distance makes up somewhat for its smaller mass, since the center of mass is weighted by the distance*mass product.

• +1 If one models the solar system composed of the sun and eight planets, is there a way to relate the focus of the orbit of one of those planets to the barycenter of the system? Would it merely be a change of reference frame? Jul 6 at 14:01
• @DaddyKropotkin I think so. To first order Jupiter's effect can be thought of as far enough away and changing slowly enough that it accelerates the Sun and the inner planets in a sufficiently similar way that the inner planets move about the Sun in nice elliptical orbits even though as a group they all dance with Jupiter without "knowing" it.
– uhoh
Jul 7 at 0:51


text:
## Multiple positive solutions of singular Dirichlet problems on time scales via variational methods.(English)Zbl 1129.34015

In this work, the existence and multiplicity of positive solutions are studied for the Dirichlet boundary value problem $-u^{\triangle \triangle}(t) = f(\sigma(t),u(\sigma(t))), \quad t \in (a,\rho(b)) \cap \mathbb{T},$
$u(0) = u(b) = 0,$ where $$\mathbb{T}$$ is a time scale with $$a = \min \mathbb{T}$$ and $$b = \max \mathbb{T}$$. The inhomogeneous term $$f(\sigma(t),z)$$ ($$\triangle$$-measurable in $$\mathbb{T}$$) possesses a singularity in the dependent variable at $$z=0$$ and also is bounded away from zero by a constant proportional to the the size of every interval near zero, where such lower bound is assumed to be valid. It also is dominated by a $$\triangle$$-Lebesgue integrable function on an interval of $$z$$ bounded away from zero. In addition, the $$\triangle$$-antiderivative, $$F(\sigma(t),z)$$, of $$f$$, with respect to the dependent variable, is a $$\triangle$$-measurable function of the independent variable in all of $$\mathbb{T}$$ for each nonnegative value of $$z$$. There is a crucial condition imposed on the time scale, namely, that there exists a $$\xi \in (0,1)$$ such that
$\int_{[a,\rho(b)) \cap\mathbb{T}}(b-\sigma(s))^{-\xi} \, \triangle s < +\infty.$
The constant $$\xi$$ plays an additional role by involvement in control of the singular behavior of $$f$$ near $$z=0$$. The above assumptions allow the authors to introduce a suitable regularization of $$f$$. Using variational methods and under additional (growth-type) assumptions on $$f$$, several existence results are obtained for at least one solution in the resonant and non-resonant scenarios. Multiplicity results are also obtained. This work will definitely aid those interested in applications of the critical point theory to boundary value problems on time scales.

### MSC:

 34B18 Positive solutions to nonlinear boundary value problems for ordinary differential equations 34B15 Nonlinear boundary value problems for ordinary differential equations 39A10 Additive difference equations 34B16 Singular nonlinear boundary value problems for ordinary differential equations
Full Text:

### References:

 [1] Agarwal, R.P.; O’Regan, D., Singular differential and integral equations with applications, (2003), Kluwer Academic Publishers Dordrecht · Zbl 1027.34014 [2] Agarwal, R.P.; O’Regan, D.; Wong, P.J.Y., Positive solutions of differential, difference and integral equations, (1999), Kluwer Academic Publishers Dordrecht · Zbl 0923.39002 [3] Agarwal, R.P.; Otero-Espinar, V.; Perera, K.; Vivero, D.R., Basic properties of Sobolev spaces on time scales, Adv. difference equ., 2006, (2006), 14 pages. Article ID 38121 · Zbl 1139.39022 [4] R.P. Agarwal, V. Otero-Espinar, K. Perera, D.R. Vivero, Wirtinger’s inequalities on time scales, Canad. Math. Bull. (in press) · Zbl 1148.26020 [5] Agarwal, R.P.; Perera, K.; O’Regan, D., Multiple positive solutions of singular and nonsingular discrete problems via variational methods, Nonlinear anal., 58, 69-73, (2004) · Zbl 1070.39005 [6] Agarwal, R.P.; Perera, K.; O’Regan, D., Multiple positive solutions of singular problems by variational methods, Proc. amer. math. soc., 134, 3, 817-824, (2006) · Zbl 1094.34013 [7] Bohner, M.; Peterson, A., Advances in dynamic equations on time scales, (2003), Birkhäuser Boston, Inc. Boston, MA · Zbl 1025.34001 [8] Cabada, A.; Vivero, D.R., Expression of the Lebesgue $$\operatorname{\Delta}$$-integral on time scales as a usual Lebesgue integral. application to the calculus of $$\operatorname{\Delta}$$-antiderivatives, Math. comput. modelling, 43, 194-207, (2006) · Zbl 1092.39017 [9] Cabada, A.; Vivero, D.R., Criterions for absolutely continuity on time scales, J. difference equ. appl., 11, 11, 1013-1028, (2005) · Zbl 1081.39011 [10] Cerami, G., An existence criterion for the critical points on unbounded manifolds, Istit. lombardo accad. sci. lett. rend. A, 112, 2, 332-336, (1979), 1978 · Zbl 0436.58006 [11] Guseinov, G.Sh., Integration on time scales, J. math. anal. appl., 285, 1, 107-127, (2003) · Zbl 1039.26007 [12] Hewitt, E.; Stromberg, K.R., Real and abstract analysis, (1975), Springer-Verlag New York [13] Rabinowitz, P.H., Minimax methods in critical point theory with apinimax methods in critical point theory with applications to differential equations, () [14] Rudin, W., Real and complex analysis, (1987), McGraw-Hill Book Co. New York, (First ed. McGraw-Hill Book Co., 1966) · Zbl 0925.00005
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.


text:
# integration by parts

• Jul 2nd 2011, 06:34 AM
vidhi96
integration by parts
hi guys I am very new to calculus and I have attempted my integration by parts excersies. can you please see attached and tell me what i am doing is ok(Happy)
• Jul 2nd 2011, 06:43 AM
Also sprach Zarathustra
Re: integration by parts
Quote:

Originally Posted by vidhi96
hi guys I am very new to calculus and I have attempted my integration by parts excersies. can you please see attached and tell me what i am doing is ok(Happy)

$g(x)=6sin(\frac{x}{6})$ and not $g(x)=-6sin(\frac{x}{6})$
• Jul 2nd 2011, 07:07 AM
vidhi96
Re: integration by parts
tnaks- but as soon as I posted it I saw my error and corrected it thanks.
• Jul 2nd 2011, 07:10 AM
Also sprach Zarathustra
Re: integration by parts
Quote:

Originally Posted by vidhi96
tnaks- but as soon as I posted it I saw my error and corrected it thanks.

Bravo! (Clapping)