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# Finding/explaining equilibrium constant, chemical significance, and standard enthalpy

I tried doing these questions and just wanted to know if there are any that I should look back at and fix. Please don't edit this. This looks exactly how it looks on my paper and in the book I am using.

The link for the part that says for example "See equation 11.1" and "See equation 11.5" is:

I just want to add that the "See equation 11.1" is on page 299 and is called "(26.1)" for some reason. Also, for the "See equation 11.5" is on page 300 and is called "(26.5)".

1. The standard free energy change for the formation of two moles of $\ce{H2O(l)}$ in a strong acid-strong base neutralization reaction at $25\ \mathrm{^\circ C}$ is $-79.9\ \mathrm{kJ}$.

$$\ce{H3O+(aq) + OH- (aq) -> 2 H2O(l)}$$

$$\Delta G^\circ = -79.9\ \mathrm{kJ}$$

a. Calculate the equilibrium constant for the reaction. See equation 11.1.

$\Delta G^\circ = -RT\ln K$

$\Delta G^\circ = -79.9\ \mathrm{kJ}$

$R = 8.314 \times 10^{-3}\ \mathrm{kJ/mol\cdot K} = 0.008314\ \mathrm{kJ\cdot mol^{-1}\ K^{-1}}$

$T = 25\ \mathrm{^\circ C} = 298\ \mathrm K$

$-79.9\ \mathrm{kJ} = -\left(0.008314\ \mathrm{kJ\cdot mol^{-1}\ K^{-1}}\right)\left(298\ \mathrm K\right)\ln K$

$K = 1.013 \times 10^{14}$

The equilibrium constant calculated for the reaction is $K = 1.013 \times 10^{14}$

b. Explain the chemical significance of the calculated equilibrium constant for the neutralization reaction.

The explanation for the chemical significance of the calculated equilibrium constant for the neutralization reaction is that its the ratio of the concentrations of the products to the reactants at which the reaction reaches equilibrium. The significance of a larger number is that it has a relatively large amount product.

c. The standard enthalpy change, $\Delta H^\circ$, for a strong acid-strong base reaction is $-57.8\ \mathrm{kJ}$. Determine the standard entropy change, $\Delta S^\circ$, for the reaction at $25\ \mathrm{^\circ C}$. See equation 11.5.

$\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$

$\Delta G^\circ = -79.9\ \mathrm{kJ}$

$\Delta H^\circ = -57.8\ \mathrm{kJ}$

$T = 25\ \mathrm{^\circ C} = 298\ \mathrm K$

$-79.9\ \mathrm{kJ} = -57.8\ \mathrm{kJ}-298K\Delta S^\circ$

$\Delta S^\circ = 0.0742\ \mathrm{kJ}$

The standard entropy change, $\Delta S^\circ$, for the reaction at $25\ \mathrm{^\circ C}$ determined is $\Delta S^\circ$ = 0.0742 kJ

d. Explain the chemical significance of the calculated entropy change for the neutralization reaction.

The explanation for the chemical significance of the calculated entropy change for the neutralization reaction is that an increase in the number of microstates results in the calculated change in entropy to be positive.

• Well, just plug in the $\Delta\rm G^0$ and get the $\rm K$. Also, think of a more descriptive title, because you know, every single question on this site is a chemistry question. – Ivan Neretin Nov 23 '15 at 18:17
• This seems like a homework question. We have a policy which states that you should show your thoughts and/or efforts into solving the problem. It'll make us certain that we aren't doing your homework for you. Otherwise, this question may get closed. Additionally, everything you need to do the question is given to you, you just need to plug in the numbers. What exactly about this is giving you difficulty?? – bon Nov 23 '15 at 18:34
• This question basically boils down to what is the inverse function of lnx. – bon Nov 23 '15 at 22:01
• Your problem is not with thermodynamics, chemistry, or physics. It is with mathematics, algebra in particular. You need to review logarithms. – Chet Miller Nov 24 '15 at 2:19
• Your enthalpy should be in $\mathrm{kJ/K}$. :) – timaeus222 Nov 25 '15 at 7:59

Also for part D, having a positive $\Delta S$ only implies that there is an increase in the number of microstates in the system and tell you nothing about the spontaneity.
Also, the inverse of $\ln x$ is $\mathrm{e}^x$.

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# What is the sum of all 4 digit numbers with distinct digits?

I tried solving this problem and got into a little trouble.I had done questions similar to this one before where the digits(that I could use) were specified so I thought maybe I could solve it like them. so I started writing all 4 digit numbers with distinct digits using the digits: $${0,1,2,3}$$ and I found the number of occurrence of each digit in units place, then the 10's place ,100's place and finally the 1000's place and I got this as the sum of all numbers : $$4\times(0+1+2+3)\times(1+10+100) + 6 \times (1+2+3) \times 1000$$ and with some more thought into it I found out that I could Write the sum of the numbers in the form $$(A\times(0+1+...+9) \times 111)+(B\times (1+2+...+9)\times 1000)$$ now my question arises I'm able to find the value of B which is just $$9 \times9\times8\times7$$ devided by $$9$$ which is $$9\times8\times7$$ but I really can't find the Value of A because it seems to have no relations with the value of $$9 \times9\times8\times7$$ so I could use some help here; or maybe what I've done is not the best way of doing it. I will appreciate your answers.

• Should i be more elaborate in my answer? – Akshaj Bansal Sep 15 '19 at 14:46
• @AkshajBansal that would help a lot especially if you consider all digits in there! – Emad Sep 15 '19 at 14:50
• Do u wnt an answer when we consider $0,1,2,3,4,5,6,7,8,9$ as digits for the 4 digit numbers? – Akshaj Bansal Sep 15 '19 at 14:52
• @AkshajBansal yes just as the title suggests, I used 4 digits so that I could write down all of the digits myself and use it to find the formula for the general case. – Emad Sep 15 '19 at 14:53
• Repetition not allowed ,right? – Akshaj Bansal Sep 15 '19 at 14:54

Take the $$k$$th column place. For every digit $$a$$ in the $$k$$th columns there should be $$9*8*7$$ digits in the remaining three columns. $$a$$ appears in the $$k$$th column $$9*8*7$$ times. So if you add up all the terms in the $$k$$ths column you should get $$9*8*7(0+1+2+..... + 9) = 45*9*8*7$$ and so the $$k$$th column provides $$45*9*8*7*10^{k-1}$$ to the sum. So the total sum should be $$45*9*8*7(10^3 + 10^2 + 10 + 1) = 45*9*8*7*1111$$.

Except that adds up "four digit" numbers that have $$0$$ in the thousands column which should not be included.

So.... we subtract the sum of all strings that start with $$0$$.

Now by the same reasoning, if $$a$$ is in the $$k$$th column there are $$8*7$$ options for the remaining two columns (they can't be zero) so the sum of all those strings is $$8*7*(1+2+3+ ..... + 9)*(10^2 + 10 + 1)= 45*8*7*111$$.

So our sum is $$45*9*8*7*1111- 45*8*7*111 = 45*8*7(9999-111)=45*8*7*9888$$

• that is true and the answer also matches my book's.thank you! – Emad Sep 15 '19 at 16:43

Think like this For the sum of all digits at thousands place the expression would be

$$1000(1+2+3+4+5+6+7+8+9)×{9\choose3}×3!.$$ Explanation- You surely dont have $$0$$ therefore now only $$9$$ possibilities left for thousands place suppose you take $$1$$ at thousands place now you need to see how many times does it repeat i.e $$9\choose3$$ since you need to select $$3$$ digits out of $$9$$(0,2,3...9) possibilites also you need to see in how many ways those $$3$$ would be arranged that is $$3!$$

Now- For hundreds, tens, units place there would be two separate cases for $$0$$ coming at those places and $${1,2,3...9}$$ coming at those places. For $$0$$ we see it repeats $$9×8×7$$ times (but it doesnt affect the sum) For digits $${1,2,...9}$$ they repeat $$8×8×7$$ times ,now how does that come since you dont have $$0$$ at thousanda place so only $$8$$ possibilities and for other place $$8$$ possibilities again (since you can put $$0$$) now and $$7$$ possibilities for the last place .

$9×8×7(0)(111)+8×8×7(1+2+3+4+5+6+7+8+9)(111)$\$

$$\big(1000(1+2+3+4+5+6+7+8+9)×{9\choose3}×3!\big)+ \big(8×8×7(1+2+3+4+5+6+7+8+9)(111)\big)$$ $$=1000×9×8×7×45+8×8×7×45×111=24917760$$

$$A$$ is $$8×8×7$$ and $$B$$ is $$9×8×7$$
You've done well so far. You're getting stuck where you assign a digit to one of the places other than the $$1000$$'s place. Note that if you assign $$0$$, it contributes nothing to the sum, so you can ignore that possibility.
Once you've assigned a non-zero digit, you have to assign $$3$$ more, but you can't assign $$0$$ to the $$1000$$'s place. So, calculate the number of ways to assign the three digits, without regard to the restriction, and then subtract the number of ways to assign $$0$$ to the $$1000$$'s place.
There are $$9$$ ways to assign a non-zero digit to a place other than the $$1000$$'s place. That leaves you with $$9$$ digits to assign to $$3$$ places, which can be done in $$9\cdot8\cdot7$$ ways. We have to subtract the $$8\cdot7$$ ways where we assigned $$0$$ to the $$1000$$'s place, giving $$9\cdot8\cdot7-8\cdot7=8\cdot8\cdot7=448$$ ways.


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Zero cycles on surfaces with ${p}_{g}=0$
Compositio Mathematica, Volume 33 (1976) no. 2, p. 135-145
@article{CM_1976__33_2_135_0,
author = {Bloch, Spencer and Kas, Arnold and Lieberman, D. I.},
title = {Zero cycles on surfaces with $p\_g = 0$},
journal = {Compositio Mathematica},
publisher = {Noordhoff International Publishing},
volume = {33},
number = {2},
year = {1976},
pages = {135-145},
zbl = {0337.14006},
mrnumber = {435073},
language = {en},
url = {http://www.numdam.org/item/CM_1976__33_2_135_0}
}
Bloch, S.; Kas, A.; Lieberman, D. Zero cycles on surfaces with $p_g = 0$. Compositio Mathematica, Volume 33 (1976) no. 2, pp. 135-145. http://www.numdam.org/item/CM_1976__33_2_135_0/

[1] K. Kodaira: On Compact Analytic surfaces II, III. Ann. of Math. (2) 77 (1963) 563-626, 78 (1963) 1-40. | MR 184257 | Zbl 0171.19601

[2] K. Kodaira: On the structure of compact complex analytic surfaces IV. Amer. J. Math. 90 (1968) 1048-1066. | MR 239114 | Zbl 0193.37702

[3] D. Mumford: Rational equivalence of zero cycles on surfaces. J. Math. Kyoto 9 (1969) 195-204. | MR 249428 | Zbl 0184.46603

[4] A.A. Roitman: (a) Rational equivalence of zero cycles. Math. USSR Sbornik, 18 (1972) 571-588.(b) Ruledness of an algebraic surface with finite dimensional nonzero group of classes of zero dimensional cycles of degree zero modulo rational equivalence. Funct. Anal. i Pril. 8 (1974) 88-89. | MR 463165 | Zbl 0337.14027

[5] I. R ŠAFAREVIC, et al.: Algebraic surfaces. Proc. Steklov Inst. Math. 75 (1965). | MR 215850

[6] P. Samuel: Relations d'équivalence en géometrie algébrique. Proc. Int. Congress Math. 1958, 470-487. | MR 116010 | Zbl 0119.36901

[7] S. Bloch: K2 or artinian Q-algebras with application to algebraic cycles. Communications in Algebra. (To appear). | Zbl 0327.14002

[8] O. Zariski: Algebraic Surfaces, Appendix by D. Mumford to the second supplemented edition. Ergeb. du Math. 61 (1971). | MR 469915 | Zbl 0219.14020

[9] D. Mumford: Abelian Varieties. Oxford University Press (1970). | MR 282985 | Zbl 0223.14022


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# Sample Certificate Of Recognition For Speaker Cbeeddceadccb

Sample Certificate Of Recognition For Speaker Cbeeddceadccb

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# A brewery's filling machine is adjusted to fill quart bottles with a mean of 32.0 oz of ale and a variance of $0.003 .$ Periodically, a bottle is checked and the amount of ale is noted.a. Assuming the amount of fill is normally distributed, what is the probability that the next randomly checked bottle contains more than 32.02 oz?b. Let's say you buy 100 quart bottles of this ale for a party; how many bottles would you expect to find containing more than 32.02 oz of ale?

## a .3581; b. about 35-36 bottles

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### Video Transcript

okay for this problem. We are given some information about ah, a breweries filling machine of court bottles. So we want Teoh drawer are normal distribution. That said, that's Norman. Normally distributed with the mean of 32 ounces is what it's supposed to fill at. Well sketched that out there and we said, We have a variance of 0.0. Uh, there there are three So most problems have been giving us the standard deviation. So with the mean and the standard deviation of Newsom's Ys fours and then some either technology or a table Teoh figure out the the probability this was giving us a variance. But we just got to go back to our understanding off the variances. Ah, a value stocks about variability and the standard deviation soda step water was gonna take the square root of the variance to get the standard deviation. So the standard deviation which we use this symbol, this is just the square root of 0.3 So let's just get that out of the way. And then the next two steps will be binding our boundaries, looking at our direction and then working backwards. We're working using the technology to get the probabilities. So it's first things first. Let's take this number 0.3 square root of it and thats gonna tell us that corresponds to ah, standard deviation is 0.5 It was dinner TV show. You can see that's ounces. So 0.54055 ounces. So that means that we're going above 0.55 each of those steps up and then below. I'm not gonna sketch that all out, but maybe we'll get somebody of the boundaries. So let's go over here and look at A We want to find its probability of 32.2 ounces more than so second stuff, we're gonna write what we want using probability notation. So we want to the probability that, given this information, um, it's normally the field is normally distributed. So what? The probably next bottle contains more than 32 point zero to write that out some more than 32.2 So I'm gonna sketch that just like to look at the direction. So 32 eyes here 0.5 the whole standard deviation above the 32.2 He's just a little bit above the the meaning there, but not even a full standard deviation above, um, it says more than 32.2. So actually going to sketch that to the rate I d like to show my directions. And I bet your teacher I want you to think your show your direction. Um, your table A table three user, You can use a table to get that. Since we have a calculator up here, we're gonna use our calculator. And were you second distribution? We know it's a normal distribution. I almost think their frequencies here. So in this case, we're going to look. A boundary of 32.2 is our left boundary upper boundary is going to the right towards infinity typing up ninth. That many nines is gonna give us plenty of digits to evaluate that. And, ah, we got to use the information we're given for this problem. We were told that 32 ounces is the mean and we were told the variants, but we figured out the standard deviation is 0.558 typical distance from the mean. So let's look at this and it says the 35% chance of rain over filler bottles there. So, uh, right that answer down 350.358 Dan, it's above that threshold. Okay, um and then we'll look at part B and see what it asks us. So for power bi, they see by 100 court bottles of the sale for party. How many bottles would you expect to be containing more than 32.2 ounces of Ailes? So we want to know for part B is rephrase that. Our own words. We want to know if we have 100 bottles. How money you're gonna be overfilled defense we call overfilled. So our plan here is just to take that simple Crip probability, Um, and just multiply it, um, the three d printer to So let's just take our number here, draw A little euro se will take him. We've been we figured out from party that about 36% more precisely 30.3581 overfilled If we consider that being overfilled, you were the most play that by 100. And we know that this is going with the decimal over to places. So 34 5.81 bottles. Okay, so let's say, um, everybody, would you expect? We'll say about 35 Maybe could see 36. But since there's not a full bottle, say, about 35 bottles. So that's what we do probability to make predictions. All right, so that's our problem about the feeling of bottles.

City College of New York

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# A map between Banach spaces is continuous

I'm trying to prove this statement:

Let $$(X_0, \| \cdot \|_{X_0})$$ and $$(X_1, \|\cdot \|_{X_1})$$ be Banach spaces and $$(Y_0, \| \cdot \|_{Y_0})$$ and $$(Y_1, \|\cdot \|_{Y_1})$$ normed spaces so that $$X_0$$ is a vector subspace of $$Y_0$$ and $$X_1$$ is a vector subspace of $$Y_1$$.

Further assume that $$i_0: (X_0, \|\cdot\|_{X_0}) \rightarrow (Y_0, \|\cdot\|_{Y_0}),\; x \mapsto x$$ and $$i_1: (X_1, \|\cdot\|_{X_1}) \rightarrow (Y_1, \|\cdot\|_{Y_1}),\; x \mapsto x$$ are continuous.

If $$T \in L(Y_0, Y_1)$$ so that $$T(X_0) \subseteq X_1$$, define $$S: (X_0, \|\cdot\|_{X_0}) \rightarrow (X_1, \|\cdot\|_{X_1}), \;x \mapsto Tx$$ and show that $$S$$ is continous.

Any ideas how to prove it?

• What are $X,Y$? Jan 5 '20 at 15:14
• My bad. $T \in L (Y_0, Y_1)$ Jan 5 '20 at 15:17
• That seems false to me. Are you sure $X_1$ only has a continuous inclusion into $Y_1$ and not a bi-lipschitz inclusion? (or the norm on $X_1$ is the restriction of the norm on $Y_1$) Jan 5 '20 at 16:49
• I don't really understand what you mean. $i_0$ and $i_1$ are embeddings, $T$ is a linear and continuous operator from $Y_0$ onto $Y_1$. Jan 5 '20 at 17:05

It is true (which seems surprising), being a direct application of the closed graph theorem.

If we have a continuous linear map $$T:Y_0\to Y_1$$ so that $$T(X_0)\subseteq X_1$$ denote with $$T': X_0\to X_1$$ the induced map. It is clearly linear and is a map between Banach spaces. We verify that it is has a closed graph and then by the closed mapping theorem it is continuous.

The condition you need to check is if $$x_n\in X_0$$ and $$T'(x_n)\in X_1$$ both converge in $$X_0$$ and $$X_1$$ respectively, that then $$T'(\lim_n x_n) = \lim T'(x_n)$$, where the limits are in $$X_0$$ and $$X_1$$ respectively. For this we will use $$i_1\circ T' = T\circ i_0$$.

So suppose the two limits are not the same, ie $$T'(x_n)\overset{X_1}\to y$$ and $$x_n\overset{X_0}\to x$$ with $$T'(x)\neq y$$. Then it follows that

$$i_1(T'(x)) = (T\circ i_0)(x) = \lim_n (T\circ i_0)(x_n)\neq i_1(y) = i_1(\lim_n T'(x_n)) = \lim_n (i_1\circ T')(x_n)=\lim_n (T\circ i_0)(x_n)$$ which is a contradiction.

• Is $x_n \in X_0$ ? Jan 5 '20 at 19:29
• @s.harp Could you have a look at my follow-up question? math.stackexchange.com/questions/3498621/… I thought that I had a counterexample, but your proof looks very convincing and still I cannot find my mistake. Jan 5 '20 at 20:42
• @SeverinSchraven RhysSteele has already pointed out an error in your example. In trying to find counter-examples I made almost exactly the same error (and a whole bunch of other ones too). In fact there are 3 deleted answers to this post describing counter examples, so really you are not alone in believing this statement must be false, only to be frustrated by some detail not working as expected in a counter-example! Jan 5 '20 at 23:59
• @MathStudent yes, the sequence $x_n$ is in $X_0$, $i_0(x_n)$ is in $Y_0$, $T'(x_n)\in X_1$ and $i_1(T'(x_n))=T(i_0(x_n))\in Y_1$. That can be confusing, since I didn't write explicitly where the sequence must live. Jan 6 '20 at 0:00


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Enter the amount of coolant the vehicle can hold (L) and the percentage of concentrated coolant desired (%) into the Coolant Ratio Calculator. The calculator will evaluate and display the Coolant Ratio.

## Coolant Ratio Formula

The following formula is used to calculate the Coolant Ratio.

CLR = AC * C/100
• Where CLR is the Coolant Ratio (L)
• AC is the amount of coolant the vehicle can hold (L)
• C is the percentage of concentrated coolant desired (%)

To calculate the coolant ratio, multiply the amount of coolant the vehicle can hold by percent of coolant desired.

## How to Calculate Coolant Ratio?

The following example problems outline how to calculate Coolant Ratio.

Example Problem #1:

1. First, determine the amount of coolant the vehicle can hold (L).
• The amount of coolant the vehicle can hold (L) is given as: 12.
2. Next, determine the percentage of concentrated coolant desired (%).
• The percentage of concentrated coolant desired (%) is provided as: 50.
3. Finally, calculate the Coolant Ratio using the equation above:

CLR = AC * %C/100

The values given above are inserted into the equation below and the solution is calculated:

CLR = 12 *50/100 = 6.00 (L)

Example Problem #2:

For this problem, the variables required are provided below:

amount of coolant the vehicle can hold (L) = 6

percentage of concentrated coolant desired (%) = 30


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# Help needed in NT!

My doubt has been resolved by Satyajit Mohanty and Pi Han Goh.How would you go about solving this problem?:$10^n \equiv 2\pmod{19};n \in \mathbb{Z}^{+}\\ Smallest\ such\ n=?$Please help!

1 year, 1 month ago

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@Satyajit Mohanty @Pi Han Goh · 1 year, 1 month ago

The smallest $$n$$ is $$17$$. Study residue systems modulo $$n$$. · 1 year, 1 month ago

You need to check that 17 is the smallest, though. · 1 year, 1 month ago

Kindly tell the method! · 1 year, 1 month ago

Oh,wait I got it:did you do it this way:$10^{18}\equiv1\pmod{19}\\ from\ Euler's\ totient\ function\\ 10^{17}\equiv\dfrac{1}{10}\equiv\dfrac{2}{20}\equiv2\pmod{19}$? · 1 year, 1 month ago

You have only shown that n = 17 is a solution but you didn't show that it's the smallest solution. · 1 year, 1 month ago

Show that the residues of a prime is (prime - 1). So if you found anything less than a prime, then it's minimum. · 1 year, 1 month ago

Yup. · 1 year, 1 month ago


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# Derivative of $2\sqrt{1+\sin{t}}$

Using the standard method to find the derivative of $f(t)=2\sqrt{1+\sin{t}}$ yields $f'(t)=\sqrt{1-\sin{t}}$ but plotting both of the equations into a graph shows that this is not true.

The plot can be seen here: WolframAlpha

The thing to note is that at around $\pi$, $f(\pi)$ is decreasing, meaning that its derivative $f'(\pi)$ should be negative; but instead, $\sqrt{1-\sin{\pi}} = 1$. And the definite integral of $f'(t)$ from $0$ to $2\pi$ should not be $0$, but $f(2\pi)-f(0)=0$.

Is this because $f(t)$ is not continuous?

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What I think of as the standard method leads mechanically to $\dfrac{\cos t}{\sqrt{1+\sin t}}$.

If we now replace $\cos t$ by $\sqrt{1-\sin^2 t}$, which is presumably how you obtained your expression, we are sometimes making a mistake. For when $\cos t$ is negative, we have $\cos t=-\sqrt{1-\sin^2 t}$. That would account for the trouble at $\pi$.

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Wow, that was fast. Thanks! –  user803253 Dec 4 '12 at 6:22

$f(t)$ is, in fact, continuous since $1+\sin(t)\geq0$ for all $t$. The problem is the derivative you have. Here, you would have to use the chain rule with $g(t)=2\sqrt{t}$ and $h(t)=1+\sin(t)$. Then $f(t)=g\circ h(t)$, and when we take the derivative of $f$, we end up with $$f'(t)=g'(h(t))\cdot h'(t)=\frac{\cos(t)}{\sqrt{1+\sin(t)}}.$$ Hope this helps!

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Your answer is ultimately the same as mine. If you multiply your $f'(t)$ with $\frac{\sqrt{1-sin{t}}}{\sqrt{1-sin{t}}}$, you would get the one I've found. –  user803253 Dec 4 '12 at 6:21
As Andre Nicolas pointed out, though, the $\cos(t)$ maintains its positive and negative values, whereas when we eliminate as you have, we lose out on some of that information (since $\sqrt{t}\geq0$ for all positive real numbers $t$). –  Clayton Dec 4 '12 at 6:23

The derivative of this function is $f'(t) = \frac{\cos(t)}{\sqrt{1+\sin(t)}}$. Writing $\cos(t) = \sqrt{\cos^2(t)} = \sqrt{1 - \sin^2(t)}$ you get the simplification to your expression. However, this leads to a mistake, especially for $\frac{\pi}{2} < t < \pi$. Namely, for $t$ in this range, $\cos(t) <0$ which implies $f'(t)<0$, but the simplification absorbs the negative sign as a square inside the square root, hence the wrong answer. If you compute $f'(\pi) = \frac{\cos(\pi)}{\sqrt{1 + \sin(\pi)}} = -1$ you see that the function is decreasing, as expected. The function is not only continuous, it is differentiable for $t < \frac{3\pi}{2}$.

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# VotingClassifier in scikit-learn: building and optimizing an ensemble of classification models

As part of the implementation of a large task on Sentiment Analysis (analysis of reviews), I decided to devote some time to additional study of its separate element - using the VotingClassifier from sklearn.ensemble as a tool for building an ensemble of classification models and improving the final quality of predictions. Why is this important and what are the nuances?

It often happens that in the course of solving an applied problem of analyzing data, it is not immediately obvious (or not at all) what training model is best suited. One solution may be to choose the most popular and / or intuitively suitable model based on the nature of the data available. In this case, the parameters of the selected model are optimized (for example, via GridSearchCV) and it is used in the work. Another approach may be the use of an ensemble of models, when the results of several of them are involved in the formation of the final result. At once I will make a reservation that the purpose of the article is not to describe the advantages of using an ensemble of models or the principles of its construction (you can read about it here), and rather in one separately applied approach to solving the problem on a specific example and analysis of the nuances arising in the course of such a decision.

Setting a global problem is the following : there are only 100 reviews on mobile phones as a test sample, and we need a pre-trained model that, with these 100 reviews, will show the best result — that is, determine whether a review is positive or negative. An additional complexity, as follows from the conditions of the problem, is the absence of a training sample. To overcome this difficulty, 10,000 reviews of mobile phones and ratings from one of the Russian sites were successfully sparred with the help of the Beautiful Soup library.

Skipping the steps of parsing, data preprocessing and studying their original structure, we are transitioning to the moment when there are:

• a training sample of 10,000 phone reviews, each feedback is marked binary (positive or negative). The markup of receiving the definition of reviews with grades 1-3 as negative and grades 4-5 as positive.
• using Count Vectorizer, data is presented in a form that is suitable for training classifier models

How to decide which model will work best? We do not have the ability to manually iterate through the models, since A test sample of only 100 reviews creates a huge risk that some model will simply better fit this test sample, but if you use it on an additional sample hidden from us or in a “battle”, the result will be below average.

To solve this problem, in the Scikit-learn library there is a VotingClassifier module , which is an excellent tool for using several machine learning models that are not similar among themselves, and combining them into one classifier. This reduces the risk of retraining, as well as misinterpretation of the results of any one single model. The VotingClassifier module is imported by the following command :
`from sklearn.ensemble import VotingClassifier`

Practical details when working with this module:

1) The first and most important thing is how a separate taken prediction of a combined classifier is obtained after receiving predictions from each of its member models. Among the VotingClassifier parameters there is a voting parameter with two possible values: 'hard' and 'soft'.

1.1) In the first case, the final answer of the combined classifier will correspond to the “opinion” of the majority of its members. For example, your combined classifier uses data from three different models. Two of them on a specific observation predict a response “positive feedback”, the third - “negative feedback”. Thus, for this observation, the final prediction will be a “positive feedback”, since we have 2 - “for” and 1 “against”.

1.2) In the second case, i.e. when using the parameter 'soft', the voting parameter goes to full “voting” and weighting model predictions for each class, so the final answer of the combined classifier is argmax of the sum of the predicted probabilities. IMPORTANT!To be able to use this “voting” method, each classifier from within your ensemble must support the predict_proba () method in order to obtain a quantitative estimate of the probability of entering each of the classes. Please note that not all models of classifiers support this method and, accordingly, can be used within the VotingClassifier when using the method of weighted probability (Soft Voting).

We will understand an example: there are three classifiers and two classes of feedback: positive and negative. Each classifier through the method predict_proba will give a certain value of probability (p), with which a specific observation is assigned to it to class 1 and, accordingly, with probability (1-p) to class two. The combined classifier after receiving a response from each of the models weighs the estimates obtained and gives the final result, obtained as

$display max \left(w1 * p1 + w2 * p1 + w3 * p1, w1 * p2 + w2 * p2 + w3 * p3\right) display$

where w1, w2, w3 are the weights of your classifier that are included in the ensemble, have equal weights by default, and p1, p2 is the score for belonging to class 1 or class 2 of each of them. Note also that the weights of classifiers using Soft Vote can be changed using the weights parameter, so the module call should look like this:
`... = VotingClassifier(estimators=[('..', clf1), ('..', clf2), ('...', clf3)], voting='soft', weights=[*,*,*]) `where the asterisks can be specified the required weights for each model.

2) The ability to simultaneously use the module VotingClassifier and GridSearch to optimize the hyperparameters of each of the classifiers in the ensemble.

When you plan to use an ensemble and you want the models included in it to be optimized, you can use GridSearch already in the combined classifier. And the code below shows how you can work with the models included in it (Logistic regression, naive Bayes, stochastic gradient descent) while remaining within the combined classifier (VotingClassifier):

``````clf1 = LogisticRegression()
clf2 = MultinomialNB()
clf3 = SGDClassifier(max_iter=1000, loss='log')
eclf = VotingClassifier(estimators=[ ('lr', clf1), ('nb', clf2),('sgd', clf3)], voting='hard') # задаем метод голосования через большинство (hard voting), см. п. 1.1
<b>params = {'lr__C' : [0.5,1,1.5], 'lr__class_weight': [None,'balanced'],
'nb__alpha' : [0.1,1,2],
'sgd__penalty' : ['l2', 'l1'], 'sgd__alpha': [0.0001,0.001,0.01]} </b> # задаем сетку параметров для перебора и сравнения, важен синтаксис, чтобы оптимизировалась нужная модель
grid = GridSearchCV(estimator=eclf, param_grid=params, cv=5, scoring='accuracy', n_jobs=-1)
grid = grid.fit(data_messages_vectorized, df_texts['Binary_Rate']) # когда задали все условия, проводим обучение и оптимизацию на 5 фолдах на собранной обучающий выборке
``````

Thus, the params dictionary must be defined in such a way that when accessing it via GridSearch, it is possible to determine which of the models in the ensemble is the parameter whose value is to be optimized.

That's all you need to know to fully use the VotingClassifier tool as a way to build an ensemble of models and optimize it. Let's look at the results:

`````` print grid.best_params_
{'lr__class_weight': 'balanced', 'sgd__penalty': 'l1', 'nb__alpha': 1, 'lr__C': 1, 'sgd__alpha': 0.001}
``````

The optimal values of the parameters are found, it remains to compare the results for the ensemble of classifiers (VotingClassifier) with the optimal parameters, cross-validate the training set and compare the models with the optimal parameters and the ensemble consisting of them:

``````for clf, label in zip([clf1, clf2, clf3, eclf], ['Logistic Regression', 'Naive Bayes', 'SGD', 'Ensemble_HardVoting']):
scores = cross_val_score(clf, data_messages_vectorized, df_texts['Binary_Rate'], cv=3, scoring='accuracy')
print("Accuracy: %0.2f (+/- %0.2f) [%s]" % (scores.mean(), scores.std(), label))
``````

Final result:

Accuracy: 0.75 (± 0.02) [Logistic Regression]
Accuracy: 0.79 (± 0.02) [Naive Bayes]
Accuracy: 0.79 (± 0.02) [SGD]
Accuracy: 0.79 (± 0.02) [Ensemble_HardVoting]

As you can see, the models showed themselves in several different ways on the training set (with the standard parameters this difference was more noticeable). In this case, the total value (for the accuracy metric) of the ensemble does not have to exceed the best value of the models included in it, since The ensemble is more likely to be a more stable model that can show ± similar results on a test sample and in “combat”, and thus reduce the risk of retraining, fitting a training sample, and other related classifiers of problems. Good luck in solving applied problems and thank you for your attention!

PS Considering the specifics and rules of publication in the sandbox, I cannot provide a link to github and the source code for the analysis given in this article, as well as references to Kaggle, in the framework of the InClass competition which provided a test set and tools for testing models on it. I can only say that this ensemble significantly broke the baseline and took a worthy place on the leaderboard after checking on the test set. I hope in the following publications I can share.


text:
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text:
# What do I conclude if I found the eigenvalues of a matrix, then noticed that one of those eigenvalues resulted in a zero eigenvector?

By definition:

$$Ax = \lambda x$$

for $x \neq 0$

I was using this to calculate the eigenvectors for $A$:

$$A = \begin{bmatrix}2 & 0 & 0\\1 & 3 & 0\\ 2 & 3& 4\end{bmatrix}$$

I already found that the eigenvalues for this matrix are:

$$\lambda = 2,3,4$$

However, when I began evaluating the eigenvectors for these eigenvalues, I noticed that for $\lambda = 4$, the resulting eigenvector is:

$$x = \begin{bmatrix}0\\0\\0\end{bmatrix}$$

But how is this possible??

Please help me find my mistake.

To find the eigenvectors, we must solve the equation:

$$(\lambda I -A)x = 0$$

$$\begin{bmatrix} 4 & 0 & 0\\0 & 4& 0\\ 0 & 0 & 4 \end{bmatrix} - \begin{bmatrix}2 & 0 & 0\\1 & 3 & 0\\ 2 & 3& 4\end{bmatrix} = \begin{bmatrix}2 & 0 & 0\\-1 & 1 & 0\\ -2 & -3 & 0\end{bmatrix}$$

Row reducing, we see that

$$\left[ \begin{array}{ccc|c} 1&0&0&0\\ 0&-1&0&0\\ 0&3&0&0 \end{array} \right]$$

...which tells us that $x_1=0, x_2=0, 3x_2=0$.

• There is a nonzero eigenvector; you must have made a mistake somewhere. May 1, 2016 at 19:27
• Well, it is always true that $(A-\lambda I) 0 = 0$, but an eigenvector must be non zero. In this case, you can 'read off' an eigenvector corresponding to $\lambda =4$ as $(0,0,1)^T$. May 1, 2016 at 19:27
• The last line is wrong, $x_3$ is not $0$ from that equation system. May 1, 2016 at 20:00
• @whatwhatwhat Au contraire: $\;z\;$ doesn't appear at all in the homogeneous system and it thus can take any value whatsoever! May 1, 2016 at 20:05
• z is free choice because it has a 0 coefficient for each row. There is no equation that can restrict it if it is multiplied by 0 in the equations. May 1, 2016 at 20:05

## 3 Answers

The resulting homogeneous system for $\lambda=4$ ( i.e., $\;(\lambda I-A)\vec x=\vec0\;$) is:

$$\begin{cases}2x=0\\-x+y=0\\-2x-3y=0\end{cases}\implies\;\;x=y=0\implies \begin{pmatrix}0\\0\\1\end{pmatrix}$$

is an eigenvector for $\;\lambda=4\;$ and, in fact, it is a basis for the corresponding eigenspace

• The difference between my work and yours is the $x+y$. I got $-x+y$. I got a negative value for $x$ because when doing $\lambda I -A$ we see that the 1st column 2nd row of $\lambda I$ is $0$. Subtracting the corresponding value from $A$, we get $0-1=-1$. May 1, 2016 at 19:43
• So how did you get a positive value? May 1, 2016 at 19:43
• @whatwhatwhat Thank you, you're completely right: it must be $\;-x\;$ there and, in fact, also $\;-2x-3y=0\;$ in the last equation. I did it in my mind and, of course, it makes no difference in this case. I shall edit now. May 1, 2016 at 19:52

There seems to be some confusion here. The eigenvalues solve the characteristic equation $\det({\bf A}-\lambda {\bf I}) = 0$ , but there can be several eigenvalues for one eigenvector (multiple roots). Take for example the matrix $$A = \left[\begin{array}{ccc} 2&0&0\\1&2&0\\0&0&4 \end{array}\right] \hspace{1cm} A-2I = \left[\begin{array}{ccc} 0&0&0\\1&0&0\\0&0&2 \end{array}\right]$$

This only gives one vector ($[0,1,0]^T$), although there are two roots $\lambda = 2$.

You made a mistake, the (normed) eigenvector for the eigenvalue $4$ is $e_z$.

>> A
A =

2   0   0
1   3   0
2   3   4

>> [x,y]=eig(A)
x =

0.00000   0.00000   0.66667
0.00000   0.31623  -0.66667
1.00000  -0.94868   0.33333

y =

Diagonal Matrix

4   0   0
0   3   0
0   0   2

And this is easy to verify as $e_z = (0,0,1)^t$ picks the third column, $(0,0,4)^t$, of $A$ and this is $4 \cdot e_z$.

By hand you would solve $A_1 u = 0$ for $$A_1 = \begin{pmatrix} 2 - 4 & 0 & 0 \\ 1 & 3-4 & 0 \\ 2 & 3 & 4-4 \end{pmatrix} = \begin{pmatrix} -2 & 0 & 0 \\ 1 & -1 & 0 \\ 2 & 3 & 0 \end{pmatrix}$$ This has the solutions $u = (x,y,z) = (0,0,z)$ for arbitrary $z \in \mathbb{R}$. Here is the Gauss elimination, using augmented matrices:

$$A_1 u = 0 \iff \\ \left[ \begin{array}{rrr|c} -2 & 0 & 0 & 0 \\ 1 & -1 & 0 & 0 \\ 2 & 3 & 0 & 0 \end{array} \right] \to \left[ \begin{array}{rrr|c} 1 & 0 & 0 & 0 \\ 1 & -1 & 0 & 0 \\ 2 & 3 & 0 & 0 \end{array} \right] \to \left[ \begin{array}{rrr|c} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 3 & 0 & 0 \end{array} \right] \to \left[ \begin{array}{rrr|c} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 3 & 0 & 0 \end{array} \right] \to \left[ \begin{array}{rrr|c} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right]$$

Update:

...which tells us that x1=0,x2=0,x3=0.

No it does not. It just tells you $x_1$ and $x_2$ are zero.

• I don't understand what this is supposed to tell me. May 1, 2016 at 19:44
• $x$ is the matrix of eigenvectors, $y$ is the diagonal matrix with the corresponding eigenvalues. In other words I asked my machine for the results to compare with yours.
– mvw
May 1, 2016 at 19:45


text:
zbMATH — the first resource for mathematics

Complete classification of surfaces with a canonical principal direction in the Euclidean space $$\mathbb{E}^{3}$$. (English) Zbl 1222.53009
A constant angle surface is defined as a surface whose unit normal forms a constant angle with a fixed direction $$\vec{k}$$. The projection of its fixed direction $$\vec{k}$$ on the tangent plane (denoted by $$U$$) is a principal direction with vanishing corresponding principal curvature.
The study of constant angle surfaces in $$\mathbb{R}^3$$ can be generalized for surfaces whose angle function is no longer constant, provided that certain properties are preserved. For example, when $$U$$ is a principal direction with corresponding principal curvature different from $$0$$, these surfaces are called surfaces with a canonical principal direction. In the paper, the authors classify them, construct some examples and prove that the only minimal surface with a canonical principal direction in the Euclidean space $$\mathbb{R}^3$$ is the catenoid.

MSC:
 53A05 Surfaces in Euclidean and related spaces
Full Text:
References:
 [1] Blair D.E., On a generalization of the catenoid, Canad. J. Math., 1975, 27, 231-236 http://dx.doi.org/10.4153/CJM-1975-028-8 · Zbl 0307.53003 [2] do Carmo M.P., Dajczer M., Rotation hypersurfaces in spaces of constant curvature, Trans. Amer. Math. Soc, 1983, 277(2), 685-709 http://dx.doi.org/10.2307/1999231 · Zbl 0518.53059 [3] Cermelli P., Di Scala A.J., Constant-angle surfaces in liquid crystals, Philosophical Magazine, 2007, 87(12), 1871-1888 http://dx.doi.org/10.1080/14786430601110364 [4] Dillen F., Fastenakels J., Van der Veken J., Surfaces in $$\mathbb{S}$$ 2×ℝ with a canonical principal direction, Ann. Global Anal. Geom., 2009, 35(4), 381-396 http://dx.doi.org/10.1007/s10455-008-9140-x · Zbl 1176.53031 [5] Dillen F., Munteanu M.I., Nistor A.I., Canonical coordinates and principal directions for surfaces in $$\mathbb{H}$$ 2×ℝ, Taiwanese J. Math., 2011 (in press), preprint available at http://arxiv.org/abs/0910.2135 · Zbl 1241.53010 [6] Morita S., Geometry of Differential Forms, Transl. Math. Monogr., 201, American Mathematical Society, Providence, 2001 · Zbl 0987.58002 [7] Munteanu M.I., Nistor A.-I., A new approach on constant angle surfaces in $$\mathbb{H}$$ 3, Turkish J. Math., 2009, 33(2), 169-178 [8] O’Neill B., Elementary Differential Geometry, 2nd ed. revised, Academic Press, Amsterdam, 2006 [9] Tojeiro R., On a class of hypersurfaces in $$\mathbb{S}$$ n×ℝ and ℍn×ℝ, Illinois J. Math., 2011 (in press), preprint available at http://arxiv.org.abs/0909.2265
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:
# SuanShu, a Java numerical and statistical library

com.numericalmethod.suanshu.algebra.linear.matrix.doubles.linearsystem

## Class Kernel

• java.lang.Object
• com.numericalmethod.suanshu.algebra.linear.matrix.doubles.linearsystem.Kernel

• public class Kernel
extends Object
The kernel or null space (also nullspace) of a matrix A is the set of all vectors x for which Ax = 0. The kernel of a matrix with n columns is a linear subspace of n-dimensional Euclidean space. Also, the kernel of a matrix A is exactly the same as the kernel of the linear mapping defined by the matrix-vector multiplication, x → Ax. That is, the set of vectors that map to the zero vector. The rank-nullity theorem says that the rank of A + the dimension of the kernel of A = the number of columns in A.

The kernel is the solution of a system of homogeneous linear equations. With the transformation matrix T, which turns A into the reduced row echelon form, it can solve also a system of non-homogeneous linear equations. Specifically, to find a particular solution for a non-homogeneous system of linear equations with a matrix A, i.e.,

Ax = b
we solve
T * A = U
and then
x = T * b
where x is a particular solution.
• ### Nested Class Summary

Nested Classes
Modifier and Type Class and Description
static class  Kernel.Method
These are the available methods to compute kernel basis.
• ### Constructor Summary

Constructors
Constructor and Description
Kernel(Matrix A)
Construct the kernel of a matrix.
Kernel(Matrix A, Kernel.Method method, double epsilon)
Construct the kernel of a matrix.
• ### Method Summary

All Methods
Modifier and Type Method and Description
List<Vector> basis()
Get the kernel basis.
Map<Integer,Vector> basisAndFreeVars()
Get the kernel basis and the associated free variables for each basis/column.
boolean isZero()
Check if the kernel has zero dimension, that is, if A has full rank.
int nullity()
Get the nullity of A.
int rank()
Get the rank of A.
Matrix T()
Get the transformation matrix, T, such that T * A = U.
Matrix U()
Get the upper triangular matrix U, such that T * A = U.
• ### Methods inherited from class java.lang.Object

clone, equals, finalize, getClass, hashCode, notify, notifyAll, toString, wait, wait, wait
• ### Constructor Detail

• #### Kernel

public Kernel(Matrix A,
Kernel.Method method,
double epsilon)
Construct the kernel of a matrix.
Parameters:
A - a matrix
method - the kernel computation method
epsilon - a precision parameter: when a number |x| ≤ ε, it is considered 0; the ε is used to determine the numerical rank of the linear space
• #### Kernel

public Kernel(Matrix A)
Construct the kernel of a matrix.
Parameters:
A - a matrix
• ### Method Detail

• #### nullity

public int nullity()
Get the nullity of A. That is, the dimension of the kernel.
Returns:
the nullity
• #### rank

public int rank()
Get the rank of A. That is, the number of linearly independent columns or rows in A.
Returns:
the rank
• #### basisAndFreeVars

public Map<Integer,Vector> basisAndFreeVars()
Get the kernel basis and the associated free variables for each basis/column. This method is only meaningful when the kernel is computed using Kernel.Method.GAUSSIAN_JORDAN_ELIMINATION. Using this process, a free variable corresponds to an entry '1' in a column. For the example in Wikipedia, x3 (the 3rd variable) is a free variable, and it corresponds to [3,-5,1,0,0,0]t.

If the kernel basis is computed using a different method, call basis() instead.

Returns:
a map of indices and kernel basis vectors
Wikipedia: Basis
• #### basis

public List<Vector> basis()
Get the kernel basis.
Returns:
the kernel basis
• #### isZero

public boolean isZero()
Check if the kernel has zero dimension, that is, if A has full rank.
Returns:
true if the kernel is null
• #### T

public Matrix T()
Get the transformation matrix, T, such that T * A = U. To find a particular solution for a non-homogeneous system of linear equations Ax = b, we do
x = T * b
where x is a particular solution.
Returns:
the transformation matrix T
• #### U

public Matrix U()
Get the upper triangular matrix U, such that T * A = U.
Returns:
the U matrix


text:
# The Crepant Transformation Conjecture and Fourier--Mukai Transforms

13 May 2014
14:00
Abstract

Suppose that X and Y are Kahler manifolds or orbifolds which are related by a crepant resolution or flop F.  It is expected that the Gromov--Witten potentials of X and Y should be related by analytic continuation in Kahler parameters combined with a linear symplectomorphism between Givental's symplectic spaces for X and Y.  This linear symplectomorphism is expected to coincide, in a precise sense which I will explain, with the Fourier--Mukai transform on K-theory induced by F.  In this talk I will prove these conjectures, as well as their torus-equivariant generalizations, in the case where X and Y are toric.
This is joint work with Hiroshi Iritani and Yunfeng Jian

• Algebraic and Symplectic Geometry Seminar


text:
fundamental theorem of calculus part 2 calculator

Sample Calculus Exam, Part 2. where is any antiderivative of . 5. b, 0. The Fundamental Theorem of Calculus Part 1, Creative Commons Attribution-ShareAlike 3.0 License. You can use the following applet to explore the Second Fundamental Theorem of Calculus. Here, the F'(x) is a derivative function of F(x). 26. And as discussed above, this mighty Fundamental Theorem of Calculus setting a relationship between differentiation and integration provides a simple technique to assess definite integrals without having to use calculating areas or Riemann sums. Fundamental Theorem of Calculus. About Pricing Login GET STARTED About Pricing Login. Integration is an important tool in calculus that can give an antiderivative or represent area under a curve. The Fundamental Theorem tells us how to compute the derivative of functions of the form R x a f(t) dt. The Fundamental Theorem of Calculus Part 2, \begin{align} g(a) = \int_a^a f(t) \: dt \\ g(a) = 0 \end{align}, \begin{align} F(b) - F(a) = [g(b) + C] - [g(a) + C] \\ = g(b) - g(a) \\ = g(b) - 0 \\ \end{align}, Unless otherwise stated, the content of this page is licensed under. Fundamental Theorem of Calculus, Part 1 . The Fundamental Theorem of Calculus formalizes this connection. You recognize that sin ‘t’  is an antiderivative of cos, so it is rational to anticipate that an antiderivative of  cos(π²t)  would include  sin(π²t). You da real mvps! Pro Lite, Vedantu 30. This applet has two functions you can choose from, one linear and one that is a curve. So, don't let words get in your way. Both are inter-related to each other, even though the former evokes the tangent problem while the latter from the area problem. See . 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The Fundamental Theorem of Calculus The Fundamental Theorem of Calculus shows that di erentiation and Integration are inverse processes. The Fundamental Theorem of Calculus (Part 2) FTC 2 relates a definite integral of a function to the net change in its antiderivative. The Fundamental Theorem of Calculus Part 2. Pick any function f(x) 1. f x = x 2. This means . Traditionally, the F.T.C. The Fundamental Theorem of Calculus The Fundamental Theorem of Calculus shows that di erentiation and Integration are inverse processes. Fundamental Theorem of Calculus Part 2; Within the theorem the second fundamental theorem of calculus, depicts the connection between the derivative and the integral— the two main concepts in calculus. Log InorSign Up. This math video tutorial provides a basic introduction into the fundamental theorem of calculus part 1. Step-by-step math courses covering Pre-Algebra through Calculus 3. Using calculus, astronomers could finally determine distances in space and map planetary orbits. 4. b = − 2. Find out what you can do. The first part of the theorem, sometimes called the first fundamental theorem of calculus, states that one of the antiderivatives, say F, of some function f may be obtained as the integral of f with a variable bound of integration. Fundamental Theorem of Calculus, Part 2: The Evaluation Theorem. Now moving on to Anie, you want to evaluate. The Fundamental Theorem of Calculus. Pick any function f(x) 1. f x = x 2. See pages that link to and include this page. 28. The Fundamental Theorem of Calculus, Part 2, is perhaps the most important theorem in calculus. F is any function that satisfies F’(x) = f(x). The Fundamental Theorem of Calculus Part 1. Everyday financial … After tireless efforts by mathematicians for approximately 500 years, new techniques emerged that provided scientists with the necessary tools to explain many phenomena. The second fundamental theorem of calculus holds for f a continuous function on an open interval I and a any point in I, and states that if F is defined by the integral (antiderivative) F(x)=int_a^xf(t)dt, then F^'(x)=f(x) at each point in I, where F^'(x) is the derivative of F(x). Deﬁnition: An antiderivative of a function f(x) is a function F(x) such that F0(x) = f(x). A ball is thrown straight up from the 5th floor of the building with a velocity v(t)=−32t+20ft/s, where t is calculated in seconds. The Fundamental Theorem of Calculus, Part 2 is a formula for evaluating a definite integral in terms of an antiderivative of its integrand. Answer: As per the fundamental theorem of calculus part 2 states that it holds for ∫a continuous function on an open interval Ι and a any point in I. The first part of the fundamental theorem stets that when solving indefinite integrals between two points a and b, just subtract the value of the integral at a from the value of the integral at b. $\displaystyle y = \int^{x^4}_0 \cos^2 \theta \,d\theta$ This outcome, while taught initially in primary calculus courses, is literally an intense outcome linking the purely algebraic indefinite integral and the purely evaluative geometric definite integral. \int_{ a }^{ b } f(x)d(x), is the area of that is bounded by the curve y = f(x) and the lines x = a, x =b and x – axis \int_{a}^{x} f(x)dx. The Fundamental Theorem of Calculus justifies this procedure. It looks like your problem is to calculate: d/dx { ∫ x −1 (4^t5−t)^22 dt }, with integration limits x and -1. We will now look at the second part to the Fundamental Theorem of Calculus which gives us a method for evaluating definite integrals without going through the tedium of evaluating limits. The fundamental theorem of calculus tells us-- let me write this down because this is a big deal. … 29. The Fundamental Theorem of Calculus, Part 2 (also known as the Evaluation Theorem) If is continuous on then . Recall that the The Fundamental Theorem of Calculus Part 1 essentially tells us that integration and differentiation are "inverse" operations. The integral of f(x) between the points a and b i.e. This FTC 2 can be written in a way that clearly shows the derivative and antiderivative relationship, as ∫ a b g ′ (x) d x = g (b) − g (a). The Fundamental theorem of calculus links these two branches. Something does not work as expected? If you're seeing this message, it means we're having trouble loading external resources on our website. Fundamental Theorem of Calculus, Part 2: The Evaluation Theorem. The Fundamental Theorem of Calculus, Part 1 shows the relationship between the derivative and the integral. There are really two versions of the fundamental theorem of calculus, and we go through the connection here. For now lets see an example of FTC Part 2 in action. is broken up into two part. It has two main branches – differential calculus and integral calculus. The Fundamental Theorem of Calculus denotes that differentiation and integration makes for inverse processes. Click here to edit contents of this page. It is essential, though. Type in any integral to get the solution, free steps and graph The Fundamental Theorem of Calculus Three Different Concepts The Fundamental Theorem of Calculus (Part 2) The Fundamental Theorem of Calculus (Part 1) More FTC 1 The Indefinite Integral and the Net Change Indefinite Integrals and Anti-derivatives A Table of Common Anti-derivatives The Net Change Theorem The NCT and Public Policy Substitution If you give me an x value that's between a and b, it'll tell you the area under lowercase f of t between a and x. Calculus also known as the infinitesimal calculus is a history of a mathematical regimen centralize towards functions, limits, derivatives, integrals, and infinite series. The second part of the theorem gives an indefinite integral of a function. Fundamental Theorem of Calculus says that differentiation and … As we learned in indefinite integrals, a primitive of a a function f(x) is another function whose derivative is f(x). As mentioned earlier, the Fundamental Theorem of Calculus is an extremely powerful theorem that establishes the relationship between differentiation and integration, and gives us a way to evaluate definite integrals without using Riemann sums or calculating areas. Question 5: State the fundamental theorem of calculus part 2? There are 2 primary subdivisions of calculus i.e. One of the largely significant is what is now known as the Fundamental Theorem of Calculus, which links derivatives to integrals. Problem Session 7. The Fundamental Theorem of Calculus (part 1) If then . If we know an anti-derivative, we can use it to find the value of the definite integral. $\int_{a}^{b} f(x) dx = F(x)|_{a}^{b} = F(b) - F(a)$. That was until Second Fundamental Theorem. Fundamental Theorem of Calculus (Part 2): If f is continuous on [ a, b], and F ′ (x) = f (x), then ∫ a b f (x) d x = F (b) − F (a). Check out how this page has evolved in the past. Popular German based mathematician of 17th century –Gottfried Wilhelm Leibniz is primarily accredited to have first discovered calculus in the mid-17th century. Concepts with geometry, but almost all the steps very intimidating name Calculus to find the derivative of functions the. Theorem tells us how we can use it to find the value of Theorem... … so all fair and good with velocity and interpret, ∫10v fundamental theorem of calculus part 2 calculator t dt. Each other, even though the former evokes the tangent problem while the latter the. 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text:
# Circuit with two resistances and inductance [closed]

I have this circuit and I want to know $i_1$ and $i_2$ (currents in $R_1$ and $R_2$). $\varepsilon$ is electromotive force and L is inductance.

I'd like to know if the following system of equations is correct and if I can get $i_1$ and $i_2$. Thanks a lot

\begin{cases} \varepsilon=R_1(i_1+i_2)+R_2i_2 \\ \varepsilon-L\frac{d}{dt}i_1=R_1(i_1+i_2) \end{cases}

-
According to your drawing (and the equations you wrote), $i_1$ is the current in $L$, and not in $R_1$ ($i_2$ is still the current in $R_2$ like you wrote). – Joe Jan 21 at 15:30
Also, according to the drawing you have is a battery that supplies the power to the circuit, and not an EMF like you wrote. – Joe Jan 21 at 15:31
The equations seem correct to me. – Joe Jan 21 at 15:32
@Joe can you tell me how can I get currents in R1 and R2? Thanks a lot – sunrise Jan 21 at 16:09
Please see our homework policy. We expect homework problems to have some effort put into them, and deal with conceptual issues. If you edit your question to explain (1) What you have tried, (2) the concept you have trouble with, and (3) your level of understanding, I'll be happy to reopen this. (Flag this message for ♦ attention with a custom message, or reply to me in the comments with @Manishearth to notify me) – Manishearth Jan 21 at 16:30


text:
## 7.2 Fitting simple models

### 7.2.1 The Question (2)

We are interested in modelling the change in life expectancy for different countries over the past 60 years.

### 7.2.2 Get the data

library(tidyverse)
library(gapminder) # dataset
library(finalfit)
library(broom)

theme_set(theme_bw())
gapdata <- gapminder

### 7.2.3 Check the data

Always check a new dataset, as described in Section 6.3.

glimpse(gapdata) # each variable as line, variable type, first values
missing_glimpse(gapdata) # missing data for each variable
ff_glimpse(gapdata) # summary statistics for each variable

### 7.2.4 Plot the data

Let’s plot the life expectancies in European countries over the past 60 years, focussing on the UK and Turkey. We can add in simple best fit lines using geom_smooth().

gapdata %>%
filter(continent == "Europe") %>%    # Europe only
ggplot(aes(x = year, y = lifeExp)) + # lifeExp~year
geom_point() +                       # plot points
facet_wrap(~ country) +              # facet by country
scale_x_continuous(
breaks = c(1960, 2000)) +          # adjust x-axis
geom_smooth(method = "lm")           # add regression lines
## geom_smooth() using formula 'y ~ x'

### 7.2.5 Simple linear regression

As you can see, ggplot() is very happy to run and plot linear regression models for us. While this is convenient for a quick look, we usually want to build, run, and explore these models ourselves. We can then investigate the intercepts and the slope coefficients (linear increase per year):

First let’s plot two countries to compare, Turkey and United Kingdom:

gapdata %>%
filter(country %in% c("Turkey", "United Kingdom")) %>%
ggplot(aes(x = year, y = lifeExp, colour = country)) +
geom_point()

The two non-parallel lines may make you think of what has been discussed above (Figure 7.6).

First, let’s model the two countries separately.

# United Kingdom
fit_uk <- gapdata %>%
filter(country == "United Kingdom") %>%
lm(lifeExp~year, data = .)

fit_uk %>%
summary()
##
## Call:
## lm(formula = lifeExp ~ year, data = .)
##
## Residuals:
##      Min       1Q   Median       3Q      Max
## -0.69767 -0.31962  0.06642  0.36601  0.68165
##
## Coefficients:
##               Estimate Std. Error t value Pr(>|t|)
## (Intercept) -2.942e+02  1.464e+01  -20.10 2.05e-09 ***
## year         1.860e-01  7.394e-03   25.15 2.26e-10 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.4421 on 10 degrees of freedom
## Multiple R-squared:  0.9844, Adjusted R-squared:  0.9829
## F-statistic: 632.5 on 1 and 10 DF,  p-value: 2.262e-10
# Turkey
fit_turkey <- gapdata %>%
filter(country == "Turkey") %>%
lm(lifeExp~year, data = .)

fit_turkey %>%
summary()
##
## Call:
## lm(formula = lifeExp ~ year, data = .)
##
## Residuals:
##     Min      1Q  Median      3Q     Max
## -2.4373 -0.3457  0.1653  0.9008  1.1033
##
## Coefficients:
##               Estimate Std. Error t value Pr(>|t|)
## (Intercept) -924.58989   37.97715  -24.35 3.12e-10 ***
## year           0.49724    0.01918   25.92 1.68e-10 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 1.147 on 10 degrees of freedom
## Multiple R-squared:  0.9853, Adjusted R-squared:  0.9839
## F-statistic: 671.8 on 1 and 10 DF,  p-value: 1.681e-10

Accessing the coefficients of linear regression

A simple linear regression model will return two coefficients - the intercept and the slope (the second returned value). Compare these coefficients to the summary() output above to see where these numbers came from.

fit_uk$coefficients ## (Intercept) year ## -294.1965876 0.1859657 fit_turkey$coefficients
##  (Intercept)         year
## -924.5898865    0.4972399

The slopes make sense - the results of the linear regression say that in the UK, life expectancy increases by 0.186 every year, whereas in Turkey the change is 0.497 per year. The reason the intercepts are negative, however, may be less obvious.

In this example, the intercept is telling us that life expectancy at year 0 in the United Kingdom (some 2000 years ago) was -294 years. While this is mathematically correct (based on the data we have), it obviously makes no sense in practice. It is important to think about the range over which you can extend your model predictions, and where they just become unrealistic.

To make the intercepts meaningful, we will add in a new column called year_from1952 and re-run fit_uk and fit_turkey using year_from1952 instead of year.

gapdata <- gapdata %>%
mutate(year_from1952 = year - 1952)

fit_uk <- gapdata %>%
filter(country == "United Kingdom") %>%
lm(lifeExp ~ year_from1952, data = .)

fit_turkey <- gapdata %>%
filter(country == "Turkey") %>%
lm(lifeExp ~ year_from1952, data = .)
fit_uk$coefficients ## (Intercept) year_from1952 ## 68.8085256 0.1859657 fit_turkey$coefficients
##   (Intercept) year_from1952
##    46.0223205     0.4972399

Now, the updated results tell us that in year 1952, the life expectancy in the United Kingdom was 69 years. Note that the slopes do not change. There was nothing wrong with the original model and the results were correct, the intercept was just not meaningful.

Accessing all model information tidy() and glance()

In the fit_uk and fit_turkey examples above, we were using fit_uk %>% summary() to get R to print out a summary of the model. This summary is not, however, in a rectangular shape so we can’t easily access the values or put them in a table/use as information on plot labels.

We use the tidy() function from library(broom) to get the variable(s) and specific values in a nice tibble:

fit_uk %>% tidy()
## # A tibble: 2 x 5
##   term          estimate std.error statistic  p.value
##   <chr>            <dbl>     <dbl>     <dbl>    <dbl>
## 1 (Intercept)     68.8     0.240       287.  6.58e-21
## 2 year_from1952    0.186   0.00739      25.1 2.26e-10

In the tidy() output, the column estimate includes both the intercepts and slopes.

And we use the glance() function to get overall model statistics (mostly the r.squared).

fit_uk %>% glance()
## # A tibble: 1 x 12
##   r.squared adj.r.squared sigma statistic  p.value    df logLik   AIC   BIC
##       <dbl>         <dbl> <dbl>     <dbl>    <dbl> <dbl>  <dbl> <dbl> <dbl>
## 1     0.984         0.983 0.442      633. 2.26e-10     1  -6.14  18.3  19.7
## # … with 3 more variables: deviance <dbl>, df.residual <int>, nobs <int>

### 7.2.6 Multivariable linear regression

Multivariable linear regression includes more than one explanatory variable. There are a few ways to include more variables, depending on whether they should share the intercept and how they interact:

Simple linear regression (exactly one predictor variable):

myfit = lm(lifeExp ~ year, data = gapdata)

myfit = lm(lifeExp ~ year + country, data = gapdata)

Multivariable linear regression (interaction):

myfit = lm(lifeExp ~ year * country, data = gapdata)

This equivalent to: myfit = lm(lifeExp ~ year + country + year:country, data = gapdata)

These examples of multivariable regression include two variables: year and country, but we could include more by adding them with +, it does not just have to be two.

We will now create three different linear regression models to further illustrate the difference between a simple model, additive model, and multiplicative model.

Model 1: year only

# UK and Turkey dataset
gapdata_UK_T <- gapdata %>%
filter(country %in% c("Turkey", "United Kingdom"))

fit_both1 <- gapdata_UK_T %>%
lm(lifeExp ~ year_from1952, data = .)
fit_both1
##
## Call:
## lm(formula = lifeExp ~ year_from1952, data = .)
##
## Coefficients:
##   (Intercept)  year_from1952
##       57.4154         0.3416
gapdata_UK_T %>%
mutate(pred_lifeExp = predict(fit_both1)) %>%
ggplot() +
geom_point(aes(x = year, y = lifeExp, colour = country)) +
geom_line(aes(x = year, y = pred_lifeExp))

By fitting to year only (lifeExp ~ year_from1952), the model ignores country. This gives us a fitted line which is the average of life expectancy in the UK and Turkey. This may be desirable, depending on the question. But here we want to best describe the data.

How we made the plot and what does predict() do? Previously, we were using geom_smooth(method = "lm") to conveniently add linear regression lines on a scatter plot. When a scatter plot includes categorical value (e.g., the points are coloured by a variable), the regression lines geom_smooth() draws are multiplicative. That is great, and almost always exactly what we want. Here, however, to illustrate the difference between the different models, we will have to use the predict() model and geom_line() to have full control over the plotted regression lines.

gapdata_UK_T %>%
mutate(pred_lifeExp = predict(fit_both1)) %>%
select(country, year, lifeExp, pred_lifeExp) %>%
group_by(country) %>%
slice(1, 6, 12)
## # A tibble: 6 x 4
## # Groups:   country [2]
##   country         year lifeExp pred_lifeExp
##   <fct>          <int>   <dbl>        <dbl>
## 1 Turkey          1952    43.6         57.4
## 2 Turkey          1977    59.5         66.0
## 3 Turkey          2007    71.8         76.2
## 4 United Kingdom  1952    69.2         57.4
## 5 United Kingdom  1977    72.8         66.0
## 6 United Kingdom  2007    79.4         76.2

Note how the slice() function recognises group_by() and in this case shows us the 1st, 6th, and 12th observation within each group.

Model 2: year + country

fit_both2 <- gapdata_UK_T %>%
lm(lifeExp ~ year_from1952 + country, data = .)
fit_both2
##
## Call:
## lm(formula = lifeExp ~ year_from1952 + country, data = .)
##
## Coefficients:
##           (Intercept)          year_from1952  countryUnited Kingdom
##               50.3023                 0.3416                14.2262
gapdata_UK_T %>%
mutate(pred_lifeExp = predict(fit_both2)) %>%
ggplot() +
geom_point(aes(x = year, y = lifeExp, colour = country)) +
geom_line(aes(x = year, y = pred_lifeExp, colour = country))

This is better, by including country in the model, we now have fitted lines more closely representing the data. However, the lines are constrained to be parallel. This is the additive model that was discussed above. We need to include an interaction term to allow the effect of year on life expectancy to vary by country in a non-additive manner.

Model 3: year * country

fit_both3 <- gapdata_UK_T %>%
lm(lifeExp ~ year_from1952 * country, data = .)
fit_both3
##
## Call:
## lm(formula = lifeExp ~ year_from1952 * country, data = .)
##
## Coefficients:
##                         (Intercept)                        year_from1952
##                             46.0223                               0.4972
##               countryUnited Kingdom  year_from1952:countryUnited Kingdom
##                             22.7862                              -0.3113
gapdata_UK_T %>%
mutate(pred_lifeExp = predict(fit_both3)) %>%
ggplot() +
geom_point(aes(x = year, y = lifeExp, colour = country)) +
geom_line(aes(x = year, y = pred_lifeExp, colour = country))

This fits the data much better than the previous two models. You can check the R-squared using summary(fit_both3).

Advanced tip: we can apply a function on multiple objects at once by putting them in a list() and using a map_() function from the purrr package. library(purrr) gets installed and loaded with library(tidyverse), but it is outside the scope of this book. Do look it up once you get a little more comfortable with using R, and realise that you are starting to do similar things over and over again. For example, this code:

mod_stats1 <- glance(fit_both1)
mod_stats2 <- glance(fit_both2)
mod_stats3 <- glance(fit_both3)

bind_rows(mod_stats1, mod_stats2, mod_stats3)

returns the exact same thing as:

list(fit_both1, fit_both2, fit_both3) %>%
map_df(glance)
## # A tibble: 3 x 12
##   r.squared adj.r.squared sigma statistic  p.value    df logLik   AIC   BIC
##       <dbl>         <dbl> <dbl>     <dbl>    <dbl> <dbl>  <dbl> <dbl> <dbl>
## 1     0.373         0.344 7.98       13.1 1.53e- 3     1  -82.9 172.  175.
## 2     0.916         0.908 2.99      114.  5.18e-12     2  -58.8 126.  130.
## 3     0.993         0.992 0.869     980.  7.30e-22     3  -28.5  67.0  72.9
## # … with 3 more variables: deviance <dbl>, df.residual <int>, nobs <int>

What happens here is that map_df() applies a function on each object in the list it gets passed, and returns a df (data frame). In this case, the function is glance() (note that once inside map_df(), glance does not have its own brackets.

### 7.2.7 Check assumptions

The assumptions of linear regression can be checked with diagnostic plots, either by passing the fitted object (lm() output) to base R plot(), or by using the more convenient function below.

library(ggfortify)
autoplot(fit_both3)
## Warning: arrange_() is deprecated as of dplyr 0.7.0.
## Please use arrange() instead.
## See vignette('programming') for more help
## This warning is displayed once every 8 hours.
## Call lifecycle::last_warnings() to see where this warning was generated.


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The five circles making up this archery target have diameters of length and. What is the total red area?

Simplify your answer as much as you can. You can use pi in your answer if necessary (for example, if the answer were, you could enter "3pi" or "3*pi" or "$3\pi$").

Apr 7, 2019

#1
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You didn't specify the diameters but.....if memory serves from answering this problem before....I believe that the diameters were 2, 4, 6, 8 and 10

So....the radiuses are 1, 2, 3, 4 and 5

The inner red area  is easy  =  pi (1)^2  =  1pi

The   next red area  =   area of 3rd circle - area of second circle  =  pi (3^2 - 2^2) = pi (9 - 4)  = 5pi

The outer red area is =   area of 5th circle - area of 4th circle  = pi (5^2 - 4^2) = pi (25 - 16) = 9pi

So....the total of the red areas  = [ 1 + 5 + 9 ]  =  15 pi   [units^2 ]

Apr 7, 2019
#2
+64
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thanks!

donkey  Apr 13, 2019


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## African Diaspora Journal of Mathematics

### Entropy Solution for Some $p(x)$-Quasilinear Problem with Right-Hand Side Measure

#### Abstract

In this paper we study the existence of entropy solution for the following $p(x)$-quasilinear elliptic problem $$\mbox{div}(a(x,u,\nabla u))+ g(x,u,\nabla u) = \mu$$ where the right-hand side $\mu$ is a measure, which admits a decomposition in $L^{1}(\Omega)+W^{-1,p'(x)}(\Omega)$ and $g(x,s,\xi)$ is a nonlinear term which has a growth condition with respect to $\xi$ and has no growth with respect to $s$ while satisfying a sign condition on $s$.

#### Article information

Source
Afr. Diaspora J. Math. (N.S.), Volume 13, Number 2 (2012), 23-44.

Dates
First available in Project Euclid: 2 November 2012

Azroul, E.; Benboubker, M. B.; Rhoudaf, M. Entropy Solution for Some $p(x)$-Quasilinear Problem with Right-Hand Side Measure. Afr. Diaspora J. Math. (N.S.) 13 (2012), no. 2, 23--44. https://projecteuclid.org/euclid.adjm/1351864731


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# zbMATH — the first resource for mathematics

## Strang, William Gilbert

Compute Distance To:
 Author ID: strang.gilbert Published as: Strang, G.; Strang, Gilbert; Strang, W. G.; Strang, W. Gilbert Homepage: http://www-math.mit.edu/~gs/ External Links: MGP · Math-Net.Ru · Wikidata · ORCID · ResearchGate · dblp · GND
 Documents Indexed: 190 Publications since 1957, including 27 Books Biographic References: 2 Publications
all top 5

#### Co-Authors

 98 single-authored 11 Kohn, Robert Vita 5 Fix, George J. 5 Shen, Jianhong 4 MacNamara, Shev F. 4 Nguyen, Truong Quang 4 Temam, Roger Meyer 4 Zhou, Dingxuan 3 Edelman, Alan Stuart 3 Strela, Vasily 2 Berger, Alan E. 2 Borre, Kai 2 Flaschka, Hermann 2 Gao, Xiqi Q. 2 Gao, Yang 2 Grinfeld, Pavel 2 Heil, Christopher E. 2 Iserles, Arieh 2 Liu, Xiangwei 2 Matthies, Hermann Georg 2 Persson, Per-Olof 1 Abdel Naby, Mokhtar A. 1 Albert, Chase 1 Bathe, Klaus-Jürgen 1 Boyd, Geoff 1 Chan, Raymond Hon-Fu 1 Crandall, S. H. Stephen H. 1 Dharmaraja, Sohan 1 Ehrlich, Louis W. 1 Eisenberg, Bennett 1 Engelman, Michael S. 1 Fujita, Hiroshi 1 Gao, Xiqi 1 Gartling, David K. 1 Genton, Marc G. 1 Gorsich, David J. 1 Green, William H. 1 Hager, William W. 1 He, Li 1 Hoffmann, Roald 1 Horn, Berthold Klaus Paul 1 Klarbring, Anders 1 Lax, Peter David 1 Lee, Steven L. 1 Li, Chi-Kwong 1 Lindner, Marko 1 Lippert, Ross A. 1 Martins, João A. C. 1 McCarthy, Charles E. 1 Micchelli, Charles A. 1 Miller, John James Henry 1 Moreau, Jean Jacques 1 Mosco, Umberto 1 Movassagh, Ramis 1 Nguyen, Tri-Thanh 1 Nickell, Robert E. 1 Olshevsky, Vadim 1 Ott, Susan 1 Panagiotopoulos, Panagiotis D. 1 Parlett, Beresford Neill 1 Philpott, Andy B. 1 Riley, James D. 1 Rota, Gian-Carlo 1 Schneider, Hans 1 Scott, Larkin Ridgway 1 Serre, Denis 1 Speth, Raymond L. 1 Stengle, Gilbert 1 Storch, Joel A. 1 Troesch, B. Andreas 1 Tsuji, Yuta 1 Van Dooren, Paul M. 1 Wang, Liang 1 Wang, Yinghui 1 Wathen, Andrew John 1 Yu, Gexin 1 Zhlobich, Pavel
all top 5

#### Serials

 9 American Mathematical Monthly 6 SIAM Review 5 International Journal for Numerical Methods in Engineering 5 Linear Algebra and its Applications 4 Archive for Rational Mechanics and Analysis 4 Numerische Mathematik 4 Proceedings of the American Mathematical Society 4 SIAM Journal on Numerical Analysis 4 Studies in Applied Mathematics 4 IEEE Transactions on Signal Processing 3 Computer Methods in Applied Mechanics and Engineering 3 Communications on Pure and Applied Mathematics 3 Journal of Differential Equations 3 Transactions of the American Mathematical Society 3 Bulletin of the American Mathematical Society 2 Journal of Mathematical Analysis and Applications 2 Advances in Mathematics 2 SIAM Journal on Discrete Mathematics 2 Bulletin of the American Mathematical Society. New Series 2 Journal of Mathematics and Physics 1 Computers & Mathematics with Applications 1 IMA Journal of Numerical Analysis 1 Journal d’Analyse Mathématique 1 Journal of Applied Mechanics 1 Journal of the Franklin Institute 1 Journal of Mathematical Physics 1 Mitteilungen der Deutschen Mathematiker-Vereinigung (DMV) 1 Zeitschrift für Angewandte Mathematik und Mechanik (ZAMM) 1 Mathematics of Computation 1 Mathematics Magazine 1 The Mathematical Intelligencer 1 Applied Mathematics and Optimization 1 Canadian Mathematical Bulletin 1 Duke Mathematical Journal 1 Journal of Mathematics of Kyoto University 1 Journal of Multivariate Analysis 1 Mathematical Programming Study 1 Mathematical Programming 1 Mathematica Scandinavica 1 Proceedings of the Edinburgh Mathematical Society. Series II 1 Quarterly of Applied Mathematics 1 SIAM Journal on Scientific and Statistical Computing 1 Journal of Xi’an Jiaotong University 1 Acta Applicandae Mathematicae 1 Physica D 1 The Annals of Applied Probability 1 Journal of Global Optimization 1 Proceedings of the National Academy of Sciences of the United States of America 1 Proceedings of the Royal Society of Edinburgh. Section A. Mathematics 1 Notices of the American Mathematical Society 1 Applied and Computational Harmonic Analysis 1 Computational and Applied Mathematics 1 Advances in Computational Mathematics 1 Obozrenie Prikladnoĭ i Promyshlennoĭ Matematiki 1 The Journal of Fourier Analysis and Applications 1 ELA. The Electronic Journal of Linear Algebra 1 M2AN. Mathematical Modelling and Numerical Analysis. ESAIM, European Series in Applied and Industrial Mathematics 1 Foundations of Computational Mathematics 1 HERMIS-$\mu\pi$. Hellenic European Research on Mathematics and Informatics Science 1 Comptes Rendus Hebdomadaires des Séances de l’Académie des Sciences, Série A 1 Journal of Mathematics and Mechanics 1 Journal of the Society for Industrial & Applied Mathematics 1 Journal de Mécanique 1 North-Holland Mathematics Studies 1 Inverse Problems and Imaging 1 Nederlandse Akademie van Wetenschappen. Proceedings. Series A. Indagationes Mathematicae 1 Journal of the Society for Industrial and Applied Mathematics. Series B, Numerical Analysis
all top 5

#### Fields

 76 Numerical analysis (65-XX) 42 Linear and multilinear algebra; matrix theory (15-XX) 29 Mechanics of deformable solids (74-XX) 28 Harmonic analysis on Euclidean spaces (42-XX) 23 Partial differential equations (35-XX) 20 Calculus of variations and optimal control; optimization (49-XX) 16 Information and communication, circuits (94-XX) 15 Operations research, mathematical programming (90-XX) 12 Approximation and expansions (41-XX) 9 General mathematics (00-XX) 7 Systems theory; control (93-XX) 6 Real functions (26-XX) 4 Combinatorics (05-XX) 4 Convex and discrete geometry (52-XX) 4 Fluid mechanics (76-XX) 3 Biology and other natural sciences (92-XX) 3 Mathematics education (97-XX) 2 Ordinary differential equations (34-XX) 2 Functional analysis (46-XX) 2 Probability theory and stochastic processes (60-XX) 2 Geophysics (86-XX) 1 History and biography (01-XX) 1 Number theory (11-XX) 1 Group theory and generalizations (20-XX) 1 Special functions (33-XX) 1 Integral transforms, operational calculus (44-XX) 1 Geometry (51-XX) 1 Differential geometry (53-XX) 1 General topology (54-XX) 1 Manifolds and cell complexes (57-XX) 1 Statistics (62-XX) 1 Computer science (68-XX) 1 Mechanics of particles and systems (70-XX) 1 Quantum Theory (81-XX) 1 Statistical mechanics, structure of matter (82-XX)


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# Dehn’s theorem

We all know the elementary formula to compute the area of a triangle: basis times height divided by two. This formula can be justified with a scissor type argument: one divides the triangle into smaller polygons and rearranges these polygons to obtain a rectangle which should have the same area.

Can we use the same argument to compute the volume of a pyramid? This is the third Hilbert’s problem. Quite surprisingly the answer is negative, as states the theorem below. This means that the formulae to compute the volume of polyhedra cannot be proved without a limiting process (for example using integrals).

###### Definition 1.

We say that two polyhedra $P$ and $Q$ are scissor-equivalent if there exists a finite number $P_{1},\ldots,P_{N}$ of polyhedra and $\theta_{1},\ldots,\theta_{N}$ isometries such that

1. 1.

$P=\bigcup_{k=1}^{N}P_{k}$ and $Q=\bigcup_{k=1}^{N}\theta_{k}(P_{k})$;

2. 2.

$P_{j}\cap P_{k}$ and $\theta_{j}(P_{j})\cap\theta_{k}(P_{k})$ have empty interior for every $k\neq j$

The properties given above assure that two scissor-equivalent polyhedra must have the same volume. It is also simple to prove that the scissor-equivalence is indeed an equivalence relation.

###### Theorem 1.

The regular tetrahedron is not scissor-equivalent to any parallelepiped.

Title Dehn’s theorem DehnsTheorem 2013-03-22 16:18:04 2013-03-22 16:18:04 paolini (1187) paolini (1187) 11 paolini (1187) Theorem msc 51M04 msc 52B45 BanachTarskiParadox HilbertsProblems RegularTetrahedron3 scissor-equivalent


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# Post-edited

O

Post-edited Formulas for the limiting distribution of traces of Frobenius

Auteurs : Lachaud, Gilles (Auteur de la Conférence)
CIRM (Editeur )

 Loading the player... symplectic group integration formula of Weyl equidistribution of traces Katz-Sarnak theory Sato-Tate conjecture distribution of the trace curves of genus 2 Legendre function Meijer G-function Legendre elliptic integrals Viète's way Viète map Girard's formula the second trace formula for g=3

Résumé : We discuss the distribution of the trace of a random matrix in the compact Lie group USp2g, with the normalized Haar measure. According to the generalized Sato-Tate conjecture, if A is an abelian variety of dimension g defined over the rationals, the sequence of traces of Frobenius in the successive reductions of A modulo primes appears to be equidistributed with respect to this distribution. If g = 2, we provide expressions for the characteristic function, the density, and the repartition function of this distribution in terms of higher transcendental functions, namely Legendre and Meijer functions.

Codes MSC :
11G05 - Elliptic curves over global fields
11G10 - Abelian varieties of dimension > 1, See also {14Kxx}
14G10 - Zeta-functions and related questions
37C30 - Zeta functions, (Ruelle-Frobenius) transfer operators, and other functional analytic techniques in dynamical systems

 Informations sur la Vidéo Réalisateur : Hennenfent, Guillaume Langue : Anglais Date de publication : 28/04/14 Date de captation : 11/02/14 Collection : Research talks Format : QuickTime (.mov) Durée : 00:48:58 Domaine : Lie Theory and Generalizations ; Number Theory Audience : Chercheurs ; Doctorants , Post - Doctorants Download : http://videos.cirm-math.fr/2014-02-24_Lachaud.mp4 Informations sur la rencontre Nom du congrès : Jean-Morlet Chair - Doctoral school : Frobenius distribution on curves / Chaire Jean-Morlet - Ecole doctorale : distribution de Frobenius sur des courbesOrganisteurs Congrès : Kohel, David ; Ritzenthaler, Christophe ; Shparlinski, IgorDates : 17/02/14 - 28/02/14 Année de la rencontre : 2014 URL Congrès : http://shparlinskikohel.weebly.com/sched...

Bibliographie

Z


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Ubuntu – Secure shared memory

ramwebserver

In my Ubuntu deployment script I have written a function to secure shared memory. For a web server it's important to secure the shared memory to tighten up security. Below you can find the code:

echo "tmpfs       /dev/shm     tmpfs   tdefaults,noexec,nosuid        0 0" >> /etc/fstab

When I reboot the web server it gets stuck. I can't see where it goes wrong. Do you have an idea, please let me know.

Unless theres's a typo in your question, I think you're options are off. First, don't specify 'defaults' (or tdefaults, as you have it.. Here's the output of my shm directory taken from cat /proc/mounts:
none /run/shm tmpfs rw,nosuid,nodev,relatime 0 0

echo "tmpfs /dev/shm tmpfs rw,nosuid,nodev,relatime 0 0" >> /etc/fstab