Mathematics - 2013-03-08 (page 4 of 4)

skullpatrol

21:00

(removed)

Gustavo Bandeira

Rly?

Ok. Put the fun in ma hand, then.

skullpatrol

Warning: Money may result in excessive greed!

Charlie

@skullpatrol κρανίον

skullpatrol

@Charlie patrol

Charlie

@skullpatrol Yes!!!

skullpatrol

21:05

:-)

Jonas Teuwen

I'm on FIRE.

Gustavo Bandeira

Why?

Charlie

@skullpatrol comes from greek, Kranion, in portuguese, it is Crânio :)

@JonasTeuwen your spine?

Jonas Teuwen

No, my head. It will explode - I'm out for a drink 8-).

Charlie

Oh!

Arkamis

21:07

Today is Friday.

skullpatrol

TGIF

Arkamis

I need to decide whether to do poker with the guys, or drinks with the ladies.

skullpatrol

@Arkamis Why not both?

Gustavo Bandeira

►

Charlie

NOOOOOOOOOOOOOOOO

Argon

21:08

@JacobBlack chat.stackexchange.com/transcript/message/8434484#8434484

Arkamis

@skullpatrol I'm old; only energy for one or the other. I'll probably do poker, and let my fiancee talk wedding things with her friend.

Charlie

@skullpatrol κρανίο περιπολία

@Arkamis poker sounds nicer

skullpatrol

@Charlie Thanks :-D

Gustavo Bandeira

@Arkamis But uh...

Oh, nvm.

Charlie

@skullpatrol I can read greek

:D

skullpatrol

21:12

@Charlie Cool.

@Charlie I have a "new" username.

Charlie

@skullpatrol hehe which?

skullpatrol

@Charlie κρανίοπεριπολία

Charlie

@skullpatrol doesn't it sound nice???

Kranionperitolia

Gustavo Bandeira

Perineum in greek, nice.

skullpatrol

@Charlie Yes, I like it.

Charlie

21:20

@κρανίοπεριπολία GREAT!

κρανίοπεριπολία

@Charlie Your "black swan" looks very nice :-D

Charlie

@κρανίοπεριπολία Thank you!

Arkamis

I still haven't seen Black Swan...

Argon

@robjohn Why can the contour integration method not be used to find the value of $\int_0^\infty \frac{x^2}{e^x-1}\, dx$?

user58512

im bored :(

Arkamis

21:25

@Argon the singularity on the origin.

Argon

@Arkamis Removable, no?

Charlie

@user58512 and your group theory :(?

Arkamis

I don't know offhand -- what's the Laurent expansion look like?

κρανίοπεριπολία

@Arkamis It's the same as White Swan, just darker ;-)

Arkamis

@skullpatrol With more Mila Kunis...

Argon

21:27

@Arkamis No need: $$\lim_{x \to 0} \frac{x^2}{e^x-1} = 0$$

Charlie

@Arkamis she doesn't appear much

Arkamis

@Argon Good point -- I forgot about L'Hopitals rule, and it looked like too much work to compute otherwise on a friday afternoon ;)

Argon

hehe Fridays

user58512

why are there no fun questions for me to answer on the site

Argon

What is a fun question you would like?

Arkamis

21:29

@Argon Well then, if that singularity is removable, then the Laurent expansion has a zero coefficient on the -1 term

So the residue is zero

Argon

Right, so we can "ignore" it

user58512

i would only know if I see it :(

Arkamis

So the integral around any contour you draw will be zero.

Argon

Avoiding the poles at $2 \pi i n$ though

user2041143

Hello guys.

user58512

21:32

hi

κρανίοπεριπολία

hi

Charlie

hi

Arkamis

@Argon Yes but e^*x* only equals 1 at x=0

user2041143

I would like to ask a question, if you were part of the chat yesterday, I asked a similar question.

Argon

@Arkamis Hm? What do you mean? $e^{2 \pi i } = 1$

Arkamis

21:33

Don't forget that when using a contour integral to evaluate a real integral, we want to cast the problem as a contour integral, but ultimately show that the other segments are irrelevant

user2041143

If we have three numbers, 4.00, 3.20 and 1.95. Their sum is 9.15 in total.

Arkamis

Yes but you wrote $e^x-1$; I am assuming you're trying to evaluate a real-valued integral

user2041143

Is it possible to find a multipler for each number separately, whose sum of multipliers doesn't exceeded the result of any of the results by multiplication of the numbers

Argon

@Arkamis Yes, but contours are in the complex plane, so $x$ is complex

Arkamis

no no no

The contour integral technique for real valued integrals is basically this:

κρανίοπεριπολία

21:35

@user2041143 Why?

Dominic Michaelis

huhu how are yo u

Arkamis

The real part is a non-closed segment, and the integral along that segment has some value. We want to treat this segment as part of a closed curve in the complex plane, so we can use contour integral techniques, like the residue theorem

user2041143

I need to figure this out.

Math problem

user58512

@user2041143 math is boring

Arkamis

The key here is that when we close the curve, we can select a curve such that all the parts that we introduce in the complex plane cancel out in the contour integral

user2041143

21:35

No, it's not..

Arkamis

Leaving only the contribution of the purely real segment

Argon

@Arkamis I know all this, I have done complex integrals before. But poles are not only real.

user58512

it sure looks it from the thing you posted

user2041143

Everything is boring, if you're not doing it with love. :-)

Argon

$2 \pi i $ is a pole of $$\frac{x^2}{e^x-1}$$

Arkamis

21:36

Yes, but there is only one pole that you care about for the real integral

Argon

Right. I was replying to this: chat.stackexchange.com/transcript/message/8434852#8434852

Arkamis

And $2\pi i$ is not a pole of $x^2/(e^x+1)$ because $x \neq 2\pi i$ for any $x \in \Bbb R$

Argon

It is not always zero, it depends on the poles within the contour

Charlie

@DominicMichaelis good

Argon

@Arkamis Wolfram disagrees: wolframalpha.com/input/?i=poles+x%5E2%2F%28e%5Ex-1%29

Arkamis

21:37

Argon

user58512

it's a pole of the function when considered C -> C, but not when R -> R

Arkamis

the only point in the reals that sends your function to infinity is x=0

You're integrating over the reals. You can introduce curves in the complex plane, but only in such a way that they ultimately don't affect your computation of the real valued integral

Argon

The singularity in the reals is removable. The complex poles are not, and these must be considered.

Arkamis

Listen, please. You asked for the explanation

Argon

You are confusing me. Contours depend on the poles within it, not just real ones

Arkamis

21:39

Yes, of course, but you're missing a crucial part of contour integration

A really important one -- let me explain

$\int_{D \subset \Bbb R} f(x) dx = \int_\Gamma f(z) dz = \int_{D \subset \Bbb R} f(z) dz + \int_\gamma f(z) dz$ if $\int_\gamma f(z) dz = 0$

Argon

Go ahead

Arkamis

So if you introduce some curve such that $\int_\gamma f(z) dz \neq 0$, then you're not actually computing $\int_{D \subset \Bbb R} f(x) dx$

user58512

ok so your are saying: sum of residues = integral_{complex contour} = integral_{real line} + integral_{complex rest of path}

so integral_{real line} = residues - integral_{complex rest of path}?

Arkamis

Yes, basically. But that's not really how we want to use it

user58512

but unless the complex path integral part goes to zero or something (by R -> infinity) we are basically going to find this just as hard

Arkamis

21:42

What we really want is to select the "rest of path" denoted $\gamma$ such that the integral goes to 0

user58512

ok

Arkamis

So, Argon, your poles on the imaginary axis will contribute to the contour integral, but only by way of the curve $\gamma$ that you introduce, and hence are not part of the real-valued integral

@Argon That's different -- you can't close the curve without including those poles

user58512

but if that's true then there will be no residues?

Arkamis

But from 0 to infinity you can draw a bunch of closed curves that completely avoid the poles

Argon

A real valued integral is not a contour integral.

Arkamis

21:44

You're missing the point

user58512

help

Dominic Michaelis

it is one but a trivial one

user58512

@Arkamis, if you make the complex contour so that its zero there wil be no poles ??

so why bother

Arkamis

@user58512 That's the point -- that's why you can't use contour integration to do this integral

user58512

I dont get it I thought the whole point was that there were interesting poles which evaluate to your integral

Arkamis

21:45

If we went from $-\infty$ to $\infty$ then we include poles

Infinitely many, in fact

And we can get an infinite sum

And that's something we know how to handle

Argon

@Arkamis So why can I use it to integrate $\int_0^\infty \frac{x}{e^x-1}\, dx$?

I did it yesterday

user58512

@Argon,

Argon

Yes?

user58512

$$\int_\gamma \frac{z}{e^z-1}\, dz$$

with $\gamma$ the rectangle $[0, R, R + i \pi, i \pi]$ was it?

Argon

That was it, and I let $R \to \infty$

I ended up with $$\int_0^\infty \frac{x}{e^x-1}\, dx + \int_0^\infty \frac{x}{e^x+1}\, dx = \zeta(2) +\frac{\zeta(2)}{2}= \frac{\pi^2}{4}$$

Arkamis

21:50

Over 4?

user58512

$$\int_0^R \frac{x}{e^x-1}\, dx + \int_0^{\pi} \frac{R + it}{e^R e^{it}-1}\, dt - \int_0^R \frac{x + i\pi}{e^{i \pi }e^x-1}\, dx - \int_0^{\pi} \frac{it}{e^{it}-1}\, dt$$

Arkamis

Wait, exp(x)+1 or minus 1

Argon

Both appeared.

user58512

see the 4 integrals?

Argon

Yes, that was it

user58512

21:51

@Argon, can we throw away the second one, becaue e^R is huge

Arkamis

Ok I see that's a + not an equals sign

Argon

@user58512 Yup, and leave the $i \pi$ in #3 because imaginary parts not needed

user58512

what about the 4th one? can we evaluate it explicitly

Argon

@user58512 I took the real part, then it was easy

user58512

what is the real part of the 4th integral?

Argon

21:53

@user58512 Note that #4 has no $i$ in the numerator, I think you wrote it wrong

@user58512 $\frac{x}{2}$

user58512

apparently it's $\pi \log(2)$

I think it should have i

are you sure it shouldn't?

Argon

@user58512 It squares: $$\int_a^b f(g(x))g'(x)\, dx$$

$g'(x) = i$

user58512

what :(

Argon

Sorry, I am rushing things here

Arkamis

the basic formula for a contour integral ;)

Argon

21:55

That part of the contour is parametrized by $iz$

user58512

let $\delta$ be the contour $[i\pi,0]$ then I think $$\int_\delta \frac{z}{e^z - 1} dz = - \int_0^\pi \frac{it}{e^{it} - 1} dt$$ is that wrong?

Argon

Square the $i$

user58512

can you just say that im wrong

Argon

Methods of contour integration

In the mathematical field of complex analysis, contour integration is a method of evaluating certain integrals along paths in the complex plane. Contour integration is closely related to the calculus of residues, a methodology of complex analysis. One use for contour integrals is the evaluation of integrals along the real line that are not readily found by using only real variable methods. Contour integration methods include * direct integration of a complex-valued function along a curve in the complex plane (a contour) * application of the Cauchy integral formula * application of th...

Arkamis

Are we talking about $x^2$ or $x$?

user2041143

21:56

If we have THREE numbers
a=4 b=3.2 c=1.95
Their sum is 9.15
Can we find THREE multipliers for those numbers m1,m2,m3 whose sum m1+m2+m3 is equal or smaller by multiplication of a*m1, b*m2 or c*m3
Note: you can't multiplicate a*m3.

user58512

this isn't working....

Argon

@Arkamis Yesterday I did $x$, and I don't know why it isn't working for $x^2$

Arkamis

Right, and I'm trying to work out why you got a result for $x$ with a curve that includes no poles ;)

Argon

@user58512 Do you know how to turn a contour integral into a regular one?

user58512

1 min ago, by user58512

let $\delta$ be the contour $[i\pi,0]$ then I think $$\int_\delta \frac{z}{e^z - 1} dz = - \int_0^\pi \frac{it}{e^{it} - 1} dt$$ is that wrong?

Argon

21:57

@Arkamis You mean why it does not equal 0?

Arkamis

Because for $z(t) = t(R+i\pi)$, $\frac{z}{e^z-1}$ has no poles.

$e^z+1$ would induce a pole with that curve.

Argon

@user58512 Yes :). Let $g(z) = iz$ be the parametrization of the contour and $f(x) = \frac{x}{e^x-1}$. Then $f(g(x))g'(x) = \frac{i^2 x}{e^{ix}-1}$

@Arkamis But the segment from $\pi i$ to $0$ contributes the $\frac{\pi^2}{4}$.

Arkamis

Yes, but shouldn't it be canceled out by the segment at $\Re z = R$? I don't know offhand -- haven't done the math, just expecting it to by symmetry.

Argon

The one from $R$ to $R + i \pi$ disappears, the other one does not.

Arkamis

Ok, hang on a second. I see what you're doing

Let me get this right

user58512

22:02

$$\int_\gamma \frac{z}{e^z-1}\, dz = \int_0^R \frac{x}{e^x-1}\, dx + \int_0^{\pi} \frac{iR - t}{e^R e^{it}-1}\, dt - \int_0^R \frac{x + i\pi}{e^{i \pi }e^x-1}\, dx + \int_0^{\pi} \frac{t}{e^{it}-1}\, dt$$

Arkamis

Earlier, you were computing $\int_0^\infty \left[\frac{x}{e^x-1} + \frac{x}{e^x+1}\right]\ dx$, right?

Argon

@user58512 Looks better

@Arkamis Noting that $\int_0^\infty \frac{x}{e^x-1}\, dx = \zeta(2)$ and $\int_0^\infty \frac{x}{e^x+1}\, dx = \frac{1}{2}\zeta(2)$, I used this method to find $\zeta(2) = \frac{\pi^2}{6}$

Arkamis

Right right

But here's the important thing

user58512

$$\Re \left(\int_0^{\pi} \frac{t}{e^{it}-1}\, dt \right) = -\frac{\pi^2}{4}$$

so $$\int_\gamma \frac{z}{e^z-1}\, dz = \int_0^R \frac{x}{e^x-1}\, dx + \int_0^R \frac{x}{e^x + 1}\, dx - \frac{\pi^2}{4}$$

and this is zero because all the singularities are removable?

Arkamis

Think of $f(x)$ as the sum of both of those parts, taken as a whole -- then, when you pop the contour into the complex plane, you get something that basically cancels out to leave just an integral along the real part. But that contour, applied individually to each term, doesn't necessarily disappear for each term. So you have to be careful.

user58512

22:06

that is wrong

Arkamis

Or rather, the same parts might not disappear

The point of doing contour integration to evaluate the real integral is to pop a curve into the complex plane, and then say, "well these complex components go away, leaving only the real segment"

user58512

what is the sum of residues of the integral?

Argon

@Arkamis Or, if you can turn it into a sum of integrals and/or residues that can be evaluated, you can solve for the integral you want.

user19161

@argon It's Friday for you!

Argon

@JacobBlack YAY! I pinged you

user58512

22:08

help :(

Arkamis

@Argon That's functionally the same thing ;)

But the point is -- if your curve doesn't go away, then you're not computing the real integral

Argon

@Arkamis Or sometimes, as I love, you can have a contour that gives the integral you want a bunch of times with different coefficients and solve for the integral. This is how I use contours to find $\int_{-\infty}^\infty$

user58512

@Argon, acan you help me

Arkamis

And then we also want to have path independence -- the contour integral depends only on its poles, not the curve

Argon

@Arkamis But why wouldn't they disappear when $x$ is squared, like before?

@user58512 With what?

Arkamis

22:10

So whenever we have a situation where you could pick two different curves, and get different results, that's bad.

user58512

4 mins ago, by user58512

and this is zero because all the singularities are removable?

What is the value of $\displaystyle\int_\gamma \frac{z}{e^z-1}\, dz $?

Argon

@user58512 $0$, because the removable singularities have $0$ as the coefficient of the $z^{-1}$ term of the series.

i.e. the residue there is $0$

user58512

when ever e^{z} = 1

so thats z = 0, 2pi, ..... oh...

so there's only one pole

in our rectangle

ok that is EXCELLENT!!!!!!!!

Argon

@user58512 $0$ is the only removable one, the others aren't. But the rectangle avoids all :)

user58512

This is by all means the fastest solution of the Basel problem I have ever seen

user19161

22:13

@user58512 WOW, so many !

Argon

@user58512 There is anther cool complex way noting that $$\sum_{n=-\infty}^\infty f(n) = \sum \text{Residues of }\pi \cot (\pi z)f(z) \text{ at poles of }f$$

Arkamis

@Argon What happens to the numerator in the neighborhood of $z = 2\pi i k$?

When you have $x^2$

Argon

@Arkamis Removable at $z=0$ but poles elsewhere

Arkamis

Right

So let's say you have a curve that includes those poles

What's the residue there?

Argon

Right

They are simple poles

So the easiest way is to differentiate the bottom

$(2 \pi i k)^2$ I think

Arkamis

22:18

Right, so you get $z^2/1$

Yes, which is real valued.

Argon

Right, yes

Arkamis

So what happens if you take a contour that includes one of the poles?

Argon

@Arkamis Then the whole integral equals $2 \pi i $ times that residue

Arkamis

Right

Now what happens when you take a contour that includes two of the poles?

(I'm getting somewhere here, I promise)

Carpediem

Hello

Arkamis

22:22

You get a different result

Carpediem

I have a question. Can an eigenvector be equal to zero ?

Argon

@Arkamis $2 \pi i \left(\sum \text{Residues in contour}\right)$

Arkamis

@Argon I know that

But we know that the residue is $-(2\pi k)^2$

So if our curve includes the pole at $k=1$, then our sum of residues is $-(2\pi)^2$.

Argon

$(2 \pi i k)^2$, no?

@Arkamis Yes

Arkamis

Then, if we take a curve that includes both poles at $k=1$ and $k=2$, then our sum of residues becomes

$-(2\pi)^2-(4\pi)^2$

And so on and so on.

So different curves lead to different values. HOWEVER

Carpediem

22:25

Anyone ?

Arkamis

Let's look at your previous function: $$f(x) = \frac{x}{e^x-1}+\frac{x}{e^x+1}$$

user58512

hi

robjohn

@Argon why do you say it can't be done?

Arkamis

There are poles at $2\pi i k$ just like before

user58512

@Carpediem, only if matrix is zero

Arkamis

22:26

But, when you compute the residues, then the first and second terms cancel

Argon

@robjohn I seem to be computing it wrongly, because I end up with some weird value with $\pi$ and $\log 2$

Carpediem

@user58512 I mean in the case of an endomorphism

user58512

???

Arkamis

@Argon I might be completely wrong, but it seems like in the case of $x^2$ you get a divergent series

robjohn

@Argon hang on

Arkamis

22:27

But in the case of the previous, you get things that are not divergent.

Argon

@Arkamis But I didn't include any poles in my contour. Like before, I just get other integrals

Arkamis

Well if you get no poles in your contour, it should be zero!

Carpediem

@user58512 Let us consider an endomorphism $T \in L(V)$

Argon

Right. I am not sure I am getting what you are saying exactly. What is summed here?

@robjohn Thanks

robjohn

@Argon You know it is $\Gamma(3)\zeta(3)$?

Arkamis

22:28

What's the residue at $2\pi i$ of $\frac{x}{e^x-1}+\frac{x}{e^x+1}$?

Argon

@robjohn Yes, but my computations gave me something elementary... I must have done something wrong

Carpediem

An eigenvalue of T is a scalar $\lambda$ for which there exists a nonzero vector $x\in V$ such that $T(x)=\lambda x$

Argon

@Arkamis Same as before, no?

Carpediem

And an eigenspace is the vector subspace of the eigenvectors

@user58512

Arkamis

Shouldn't be. You don't have the squared term involved!

Argon

22:29

But it does not matter, because my contour does not even touch that

Arkamis

You have $x(e^x+1)+x(e^x-1)$ in the numerator; evaluated at $2\pi i$, you should get $2\pi i - 2\pi i = 0$.

robjohn

@Arkamis $2\pi i$

Argon

@Arkamis Oh, I hadn't noticed. $2 \pi i$

Arkamis

Wait

Argon

@Arkamis $e^{2 \pi i }+1 = 2$

Arkamis

22:30

I'm an idiot

Yeah, 1+1 = 2, not 1

Argon

hehhehehe

Arkamis

Disregard the last 45 minutes of my life.

Argon

:)

Arkamis

I'm going to go walk off a cliff now.

Carpediem

@user58512 So in this case, can an eigenvector be equal to zero?

Arkamis

22:31

Sorry for confusing you -- I had a misunderstanding from the beginning!

Argon

@Arkamis No worries, thanks anyways :)

user58512

@Carpediem, i dont know

Event Horizon

@Charlie assimilation is just part of the game

Arkamis

@Argon But my point stands -- when you asked about doing it from $-\infty$ to $\infty$, you basically include infinite poles, the sum of the residues at which diverges, so $\int_{-\infty}^\infty \frac{x^2}{e^x-1}\ dx$ does not converge

Argon

@Arkamis That is assuming that the contour dissapears around $Re^{i\theta}$ so that the real integral can be found by summing the poles. But this doesn't work here

robjohn

22:36

@Argon what contour are you using?

Argon

@robjohn Same as last: $0, R, R+i\pi, i\pi$

robjohn

@Argon okay

Arkamis

@Argon Right, so, now that I understand what you're asking, and that I've confirmed that 1+1 = 2, let me see if I can brain a little better.

user19161

@Arkamis 1+1=2 is my favourite equation!

Arkamis

The rightmost vertical segment should vanish in the limit, for one

e^R grows faster than R^2

Argon

22:45

Yep

user19161

I think @argon will win the Fields medal for computing some hard integrals...

Argon

@JacobBlack Haha I don't think they give Fields medals for that :)

Arkamis

@Argon it should just break down into some integrals that you've computed before

user19161

@Argon The integral happens to solve RH!

Argon

@Arkamis This is what I got too, just with a $\log 2$ in there. But the integral should be nonelementary!

Arkamis

22:50

Since you're not enclosing any poles, you just have the real integral equal to the top, left, and right segments

Argon

@JacobBlack HAHAHAHAHAHAHA

Arkamis

But the right segment disappears in the limit

And then effectively you just have to compute $\int_0^a \frac{x^2}{e^x-1} dx$ a couple times, and $\int_0^a \frac{x}{e^x-1} dx$ a couple times, if I'm doing my mental math right.

along with some constants thrown in

I'm abusing a bit of notation here, by $\int_0^a$ I mean the integral along some segment on the curve

Argon

This is what all got, but all the integrals seem to have elementary closed forms

user19161

Hey @amzoti!

user58512

no one has proved that zeta(3) isn't elemetary...

:D

Amzoti

22:55

Hi @JacobBlack

maybe someone can answer a question

Argon

@user58512 Haha this is true. Perhaps this is its closed form! }:)

Amzoti

a new person posted a question and part of the question was hidden from the actual view (you had to click edit to see it)

user58512

link please @Amzoti

Amzoti

of course, this confused everyone and it was closed (about three seconds before I posted an answer)

I cannot see it

it dealt with eigenvectors

the poor guy was new and he messed up mathjax

also had a language barrier

I figured out his issue

was going to post answer and fix bad mathjax on question

however, i can no longer find the thing

I though they are tagged with closed

I wanted to reopen and fix

user58512

@Amzoti was it @Carpediem?

Amzoti

22:59

Not sure

I will try and look for it again as I am not sure why I can no longer see it

Nope - sorry I cannot find any trace of the thing!

Is there a way to see a list of closed questions

@JacobBlack: can I steal your quote "I am confused about your confusion" :-)

user58512

23:25

Hey

How do you get continued fraction for pi?

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