Mathematics - 2014-04-13 (page 3 of 3)

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Kevin Driscoll

21:00

@Robjohn AH that makes sense. Now all we have are internet memes.

robjohn

@KevinDriscoll Monty Python is still available and still funny.

Kevin Driscoll

Its true, but they arent as big a part of nerd culture as they once were.

Daniel Fischer

Is it still a nerd culture at all then?

Kevin Driscoll

I guess that depends on if you allow your definition to 'float'

user127001

My question got re-opened after 17 days lol

Ted Shifrin

21:06

@Daniel: I find that so fulfilling that you are so offended :P

hi @robjohn, @Kevin

Gabriel R.

@KevinDriscoll this is exactly why I asked :P

Ted Shifrin

hi again @Gabriel and @user127001

What question, @user127001?

Kevin Driscoll

@Ted Howdy. Sorry if Ive already pestered you about it, but did you see that I thought of a way to compute my horrible integrals?

Gabriel R.

hey @TedShifrin

Ted Shifrin

No, I didn't go look for it. @Kevin

Kevin Driscoll

21:08

not rigorous at all @Ted

r9m

@robjohn nooooooooooooooo

Kevin Driscoll

The answers are not simple.....

Ted Shifrin

Not exactly my cup of tea, @Kevin ...

Kevin Driscoll

Haha, I imagined not. At least you were right though! There's a way to do it. @Ted

robjohn

@r9m I have the Luke Skywalker "Noooooo!" as a ringtone on my phone.

Ted Shifrin

21:10

Now that I see the question, I'm surprised to have been right.

Kevin Driscoll

Hahaha, fake it til you make it!

r9m

@robjohn lol ..

Chris's sis

A funny double inequality I met while learning some stuff in complex analysis $$\left|e^{z}-\sum_{k=0}^{n} \frac{z^k}{k!}\right|\le \left| e^{|z|}-\sum_{k=0}^{n} \frac{|z|^k}{k!}\right| \le |z|^{n+1} e^{|z|}$$

It seems a piece of cake to finish Fresnel integrals by complex analysis.

(A proper slice of pizza makes the job :-))

robjohn

@Chris'ssis the first inequality is pretty easy. The second one is not so obvious (or I am missing it).

Chris's sis

21:25

@robjohn Yeah, right! Well, maybe it's not that obvious, but it doesn't require any special manipulation.

Daniel Fischer

@robjohn The second one is very brutal, $$\sum_{k=n+1}^\infty \frac{\lvert z\rvert^k}{k!} = \lvert z\rvert^{n+1}\sum_{m=0}^\infty \frac{\lvert z\rvert^m}{(m+n+1)!} \leqslant \lvert z\rvert^{n+1}\sum_{m=0}^\infty \frac{\lvert z\rvert^m}{m!}.$$

robjohn

@DanielFischer yeah, I was just writing that out to check. Nice.

Chris's sis

That's the work!

r9m

Prestige

robjohn

@DanielFischer sometimes when the inequality is so overwhelming, it is hard to find a proof at first. :-)

Daniel Fischer

21:30

It's a waste, we could write $$\leqslant \frac{\lvert z\rvert^{n+1}}{(n+1)!}e^{\lvert z\rvert}.$$

Chris's sis

21:42

@DanielFischer Do you remember some particular cases involving the evaluation of some integrals by complex analysis where these inequalities are very useful?

@DanielFischer I noticed that often one needs to be pretty good at manipulating inequalities for nicely showing that some integrals tend to $0$.

Daniel Fischer

@Chris'ssis Let me quote Peter: "I don't remember. I don't recall. I have no memory of anything at all." No, I don't remember any such occurrence.

Chris's sis

@robjohn I previous had to show that $$\left| \int_0^{\pi/4} e^{-R^2} e^{2x i} R e^{x i} i \ dx\right| \rightarrow 0$$ as $R$ goes to $\infty$.

In a textbook is given a way that I think it can be improved.

@DanielFischer OK

previously

Chris's sis

22:06

@N3buchadnezzar you should probably know how to nicely prove that integral converges to $0$. It seems to me that you're involved in learning complex analysis these days. :D

N3buchadnezzar

Yeah =)

Chris's sis

N3buchadnezzar

I just proved Jordans lemma and corollary

Chris's sis

Good.

robjohn

@Chris'ssis is that to show that the integral along a big arc goes to $0$?

Chris's sis

22:12

@robjohn Exactly.

Mike

@Daniel It turns out I misunderstood what my homework was, and my professor forgot that the spectral theorem was nontrivial.

robjohn

@Chris'ssis usually noting that on $[0,\pi/2]$, $\frac{2x}{\pi}\le\sin(x)\le x$ often helps

Chris's sis

@robjohn Yeah! :-)

Daniel Fischer

@Mike And what aws the homework meant to be?

Mike

@Daniel "Prove the spectral theorem for compact self- adjoint operators." No fancy machinery intended.

He didn't realize it wasn't an easy proof.

(on a Hilbert space)

Daniel Fischer

22:25

I should look at the proof of the spectral theorem for compact operators an a Banach space in Rudin, and see what can be simplified and how much for compact normal operators on a Hilbert space. I guess some things get substantially easier, but the entire thing probably remains somewhat nontrivial.

Chris's sis

@robjohn btw, I think I'll try the summation by parts to that series (just thought of it while working on some different questions)

N3buchadnezzar

@Chris'ssis It is enough to prove jordans lemma then, who cares about each special case that follows?

robjohn

@Chris'ssis to which series?

Chris's sis

@robjohn the one involving the convergence you proved it diverges.

robjohn

@Chris'ssis I thought I used summation by parts.

Chris's sis

22:35

@robjohn I didn't note that. (honestly)

robjohn

@Chris'ssis second equation to $(2)$.

Chris's sis

@robjohn The summation by parts may be applied in various way though, depending on the way you choose the sequences.

robjohn

@Chris'ssis okay. I will be interested to see how that goes.

Alec Teal

Continuity from the right $\forall\epsilon>0\exists\delta>0:0\le x-c<\delta\implies|f(x)-f(c)|<\epsilon$ is okay right? If the limit from the right exists then $f(c)$ is that limit, so I took that definition and strapped the $x=c$ case on to it. I am confident this is fine, is it?

robjohn

@AlecTeal it looks okay to me.

Alec Teal

22:46

Thanks.

Chris's sis

@robjohn in my proof I apply the summation by parts from the beginning ... then I saw that I reached the same expressions as yours ...

@robjohn I'll modify many things there to be different.

@robjohn the idea is to get a different proof, not the same proof a bit changed.

@robjohn from the beginning I used $$ a_k=\frac{1}{k}, \space b_k=\frac{\sin(H_k)}{H_k}$$

then applied the summation by parts

robjohn

@Chris'ssis and $\frac1k=H_k-H_{k-1}$

Chris's sis

@robjohn OK, I'll check another way.

@robjohn It's unbelievable it cannot be done by other means... Some professors I talk to failed to finish it too.

robjohn

@Chris'ssis I haven't tried anything else since that approach is so strongly suggested.

Chris's sis

@robjohn I agree that it's a natural thing to think of the summation by parts.

@robjohn If I posted this question on MSE, then I'd also add a 500 points bounty since I'm curious if someone might come up with a better proof than what you found. :-)

@robjohn At the moment I really doubt this is possible ....

@robjohn the proof is right in the heart of the question ... it's a deadly proof

robjohn

23:13

@Chris'ssis :-p

Chris's sis

@robjohn I think I should read some more on convergence series topic.

Now I'm out for a while. :-)

robjohn

Ugh... I was just serially upvoted. Luckily, only 3 votes counted, the others were past the cap.

@Chris'ssis have fun