Mathematics - 2012-05-30 (page 1 of 4)

hhh

00:31

I cannot see immediately why it is 0 if is othogonal i.e. $m\not = n$.

well $\int_0^{2\pi}e^{2inx} dx=\int_0^{2\pi}e^{ix*2n} dx$

where $\cos(a x ) +i \sin(a x)$ where $a=2n$.

anon

It is basically $\oint_\gamma zdz$, where $\gamma$ is a contour around the circle $n$ times (orientation according to $n$'s sign). This integral is $0$ by symmetry: multiplying it by any $w$ (for which $|w|=1$) leaves the integral unchanged..

leo

@PeterTamaroff hey

Daniel Montealegre

hi!

leo

:-)

hhh

@anon By symmetry? I cannot see...

Is there some geometric explanation?

Peter Tamaroff

00:42

@leo Yello

leo

@PeterTamaroff, I did write the full solution and notice the homework tag when push "Post your answer"

so I have edited

the post

anon

@hhh If $|w|=1$, the map $z\mapsto wz$ leaves $\gamma$ unchanged. Therefore multiplying by $w$ ... Anyway the geometric interpretation is that averaging the numbers on the unit circle can't privilege any direction over any other direction - just look at it! - so it must be zero.

hhh

Hey wait, I can now remember if this is the same topic: if you have a singular point like $\frac{1}{z}$ then you get $2\pi$ otherwise zero -- this is about the same thing?

leo

good night all!

anon

...no, not really. The latter has to do with multivaluedness/branching of a primitive/antiderivative around a singular point.

hhh

00:47

I am now stubborn, cannot see...can I draw it?

anon

I hope you can draw a circle. Drawing the branches of a complex-valued function is harder...

2

(relevant to the latter question)

hhh

Can I see the basic case from circle about the orthogonal thing?

Peter Tamaroff

You can start with $$\int_0^{2 \pi} \cos^m(x)\sin^n(x) dx$$ maybe?

hhh

...now I am thinking how I can express the original thing with contour integrals i.e. to express the orthogonality $\oint_\gamma z\bar{z}dz$?

anon

Nope, definitely don't do that.

Because that turns out to be $2\pi$.

(Although that shows normality ... but anyway)

hhh

00:59

?

Nope to Peter or to mine point?

anon

Let me put it this way. Up to a scale factor, $\oint zdz$ is the average of the numbers on the unit circle. Multiplying by $a$ (where $|a|=1$) simply rotates the circle, thus doesn't change the numbers being averaged, and therefore the average is the same even if you multiply the integral by $a$. But then we have $a\square=\square\implies(a-1)\square=0\implies \square=0$ if $a\ne1$.

Nope to your comment. I don't see what Peter's getting at either though.

Also, $\langle e^{inx},e^{imx}\rangle=\int_0^{2\pi} e^{ikx}dx=\oint_\gamma zdz$, where $n\ne m$ and $k=n-m$.

And $\gamma$ goes around the circle $k$ times, orientation and sign and yadda yadda.

Actually wait, your idea was valid for $n+m=0$, because you end up with $\oint_\gamma (z\bar{z})^ndz=0$...

(Assuming $n,m\ne0$ of course.)

Peter Tamaroff

@anon I wrote it wrongly.

@anon How would one prove $$\mathop {\lim }\limits_{n \to \infty } \int\limits_{ - \pi }^\pi {\frac{{\sin nx}}{{1 + {x^2}}}dx} = 0$$ without Lebesgue Riemann?

I'm working on it. Do you have any hints?

Wait.

The integrand is always odd.

anon

heh

Peter Tamaroff

Why would Spiegel put that problem

@MarianoSuárezAlvarez Are you around?

hhh

01:24

To remind me what I am trying to understand:

$$\langle e_n, e_m\rangle = \frac{1}{2\pi} \int_{0}^{2\pi} e_n(x)\,\overline{e_m(x)}\,dx = \delta_{mn}$$

where $\bar{\square}$ is a complex conjugate, $\gamma_{mn}$ is the knecker-delta (if m=n, then 1 otherwise 0), $e_n=e^{inx}$.

I can see this as $e^{i\beta}$ where $\beta$ is an angle i.e. like a circle.

So what is the q then two circles -multiplication...

So it means there are two choices but what are they in terms of circles?

Probably $\pm$, Cartesian production $<\pm, \pm>$ has two choices with commutative so something along the lines but I need still some more direct way of deducing this...I am now more like guessing...

hhh

01:48

Perhaps I need to understand inner-product more deeply, investigating...

leo

@MarianoSuárezAlvarez Hi! How do you translate Orbit Stabilizer Theorem?

Peter Tamaroff

@leo Teorema del Estabilizador de Orbita?

leo

:-)

The problem is Stabilizer do not stabilize orbits

:-|

It should be something like "Teorema Estabilizador-Orbita"

más bien órbita

Peter Tamaroff

Teorema del Estabilizador-Orbita?

Daniel Montealegre

hi

leo

02:01

@PeterTamaroff I think so. Something like that

@DanielMontealegre hi

Daniel Montealegre

hey could you lend me a hand with galois theory?

leo

@DanielMontealegre I don't know Galois Theory. But ask your question, I'm pretty sure someone here can help you :-)

Daniel Montealegre

im going to post it in the forum

leo

@PeterTamaroff It will be Teorema 2

interesting

Peter Tamaroff

02:43

@leo I'd ask first for the option to star answers apart from questions.

leo

@PeterTamaroff That would be pretty useful

@PeterTamaroff this can like you

leo

03:12

It is seems that it's assumed that $f$ Riemann integrable in $[a,b]$ implies $f$ of bounded variation on $[a,b]$. Is that true?

Peter Tamaroff

@Eugene Hey.

I'm working on this thingy leo suggested.

leo

@PeterTamaroff is there a reason to assume that the $f$ in the post indeed is of VB?

Peter Tamaroff

@leo If the function is Riemann integrable then it is bounded.

Should that work?

leo

bounded is not enough

Peter Tamaroff

@leo Right. If it oscilates infinitely then the variation is not bounded right?

leo

03:23

yep

Peter Tamaroff

@leo I guess we can assume the function is friendly

leo

it is pretty easy to see that $V_f[a,b]\leq \int_a^b |g|$

Peter Tamaroff

I can't grasp what it means for $\sup$ to "run over the partitions".

I mean

I have a minimal clue. But I can't even put it into words

It's just there in my head.

I wrote this so far for a help: Just reading the definition in Wikipedia, I can give you some ideas:

$$V_a^b\left( f \right) = \mathop {\sup }\limits_P \sum\limits_{i = 0}^{{n_P} - 1} {\left| {f\left( {{x_{i + 1}}} \right) - f\left( {{x_i}} \right)} \right|} $$

where $\sup$ runs over the set of all partitions $P$ of $[a,b]$,

$$\mathcal P=\{P=\{x_0,\dots, x_{n_P}\}:P\text{ is a partition of }[a,b]\}$$

If $f$ differentiable in $[a,b]$ then it is continuous and we can apply the MVT, which means

leo

In the case when $g$ is continuous you have proved what is requested

Peter Tamaroff

@leo I need to understand how$\sup\limits_P$ operates on the sum. What does it retrieve?

leo

03:32

@PeterTamaroff $$V_a^b (f)=\sup\{\sum_{k=0}^n |f(t_k)-f(t_{k-1})|:a=t_0\lt t_1\lt\ldots\lt t_n=b\}$$

Peter Tamaroff

@leo Oh.

I couldn't understand that effing symbol.

leo

what effing symbol?

Peter Tamaroff

@leo I mean, how it was written

I realized it was something of the sort.

Oh.

"effing" is used to avoid "f*~%ing"

leo

I don't follow you

Rajesh D

Hi guys

leo

03:36

hi

anon

effing = "eff"-ing = F-ing = sex

2

Peter Tamaroff

@anon Could you illustrate leo? I can't XD

I mean that the way Wikipedia wrote it confused me.

anon

You want me to draw leo? Like one of my French girls presumably?

leo

Oh. I just catch it. :-D

@PeterTamaroff but now you understand?

Peter Tamaroff

@leo Yes yes.

Maybe all it is left is to show the Riemannish sum is increasing, so that the $\sup$ is the integral. I'm lost

leo

03:39

so f*~%!

Peter Tamaroff

Maybe you can complete the proof.

Daniel Montealegre

Hi

can i ask a quick question

Peter Tamaroff

@DanielMontealegre Sure.

@anon

Daniel Montealegre

Say we have that L/F is a finite galois extension, and consider L/K_1/F and L/K_2/F, and denote by H_i=G(K_i/F)

what does it mean for \tau to be in H_1\cap H_2

Rajesh D

yes @Daniel

Daniel Montealegre

03:50

math.ucla.edu/~chh/110c.1.12s/elman.pdf, I am looking at problem 49.11.6

anon

If you put $ signs around latex we can parse them in-tab using chatjax.

Daniel Montealegre

ok

so basically if you look at the above pdf, I am a little confused on what they are trying to say. They define $H_1=G(K_1/F)$ and $H_2=G(K_2/F)$

and they want us to show that $H_1\cap H_2=G(L/K_1(K_2))$

Peter Tamaroff

You forgot the last $

anon

Are not $H_1$ and $H_2$ subgroups of $\mathrm{Gal}(L/F)$?

(Disclaimer: I don't know Galois theory!)

Also, see here if you want to see latex in chat like the kool kids.

Daniel Montealegre

nevermind

there is a typo. I message someone in the class.

David Wheeler

05:28

hello?

leo

06:12

@robjohn hi! any insight in this

t.b.

@leo what's the problem? the identity is standard in Lebesgue theory and since for Riemann integrable functions Riemann and Lebesgue integrals coincide we have the equality.

leo

@tb yes you are right. There is an exercise in the Zygmund book asking for that. I'll probably elaborate on it in some hours

I was defined the norm of a partition as the minimum length of his intervals :-S

Matt N.

07:07

Good morning everyone.

Jonas Teuwen

07:28

@MattN Haaiii.

Matt N.

@JonasTeuwen Last night it was really hot here. : ( I was awake for hours, melting.

Jonas Teuwen

@MattN Hotttt? Here too.

Ilya

@JonasTeuwen oh, don't say that

Jonas Teuwen

Was this hhh doing some good old trolling this night?

Ilya

who is hhh?

@MattN: what about air conditioning/sleeping with windows opened?

Jonas Teuwen

07:42

@Ilya Some guy saying that he cannot compute the Fourier series of his function because although it is periodic, it is not periodic on $[-\pi, \pi]$...

Ilya

@Jonas: our wall of stars is again full of math! I knew some guys here who star all math advises given to them which they found to be useful

500?

2

Jonas Teuwen

@Ilya Cool.

Matt N.

@Ilya Yes, we sleep with open windows and we don't have airconditioning.

Jonas Teuwen

Mosquitos?

Matt N.

No. Not yet.

Ilya

07:52

@JonasTeuwen why? Matt didn't say they sleep with closed windows.

@MattN if your windows are open, you can always open them a $\delta$-bit. If they're closed, you can do nothing :)

Jonas Teuwen

Mosquitos are quite $\delta$s...

Matt N.

Actually, if they're closed you can open them : )

Ilya

@JonasTeuwen only in this part of the world :D you should ask Rajesh

or maybe also JM is aware of $\Delta$-mosquitoes

@Jonas: listen, is it possible to show that
$$
\langle \gamma, x\cdot\log |x|\rangle \geq \langle \gamma, x\rangle \langle \gamma, \log |x|\rangle
$$
where $\gamma$ is $1$-dim Gaussian measure?

Matt N.

@Ilya If they're not too big our ninja cats will sort them out.

Jonas Teuwen

@Ilya I go to the office now. I check there okay?

Matt N.

07:56

When the things get too big then the cats are scared too.

Rajesh D

Hi

Ilya

@JonasTeuwen sure. I thought, maybe convexity arguments

@MattN too big like...

Matt N.

There was this huge black spider last summer. Approximately 8cm diameter.

shudders

I guess I'm not manly enough. A proper man would've grabbed it by one of its legs and thrown it out of the window.

Ilya

@MattN not me, so Gigili's doubts are coming true :)

Matt N.

: )

Ilya

08:19

@Matt: here?

t.b.

Matt will be awfully surprised...

Morning all

Jonas Teuwen

@Ilya Let's see.

Ilya

morning, Theo

@tb: how did you sleep?

t.b.

Okay.

Ilya

@tb not only Matt 0_o Did did something surprising

@Jonas: I've asked it here, so there is a precise formulation

Jonas Teuwen

08:23

@Ilya So what is $\langle \cdot, \cdot \rangle$ and is $\gamma(x) = \pi^{-1/2} e^{-|x|^2}$?

@Ilya Aha, looks like my thing 8-).

Ilya

my thing looks certainly different :-)

it doesn't contain parallel lines | |

Jonas Teuwen

Mwah, we just kick out all the ugly constants, they hardly matter. All you need is that the measure of the space is $1$.

Yes, that is in $d$-dim.

Ilya

hah, I knew!\

@Jonas: so $\pi^{-1/2}$ is not ugly?

Jonas Teuwen

@Ilya Hmm, I put $d = 1$ there.

Hybrid formula!

t.b.

yes, it is ugly. Modern people call it $\sqrt{2/\tau}$

Jonas Teuwen

08:25

I am not modern!

Ilya

@tb: who is $\tau$?

t.b.

@Ilya the correct value of $\pi$, namely $2\pi$. At least according to some people :)

Ilya

oh, that is too abstract for me ;)

Jonas Teuwen

@Ilya Markus's talk is today.

Ilya

Supertree adventures: "My bag!" "Somebody, help!"

@JonasTeuwen cool, 4pm? I'm coming

Jonas Teuwen

08:29

@Ilya Yes.

Matt N.

@Ilya Hey, sorry was in the shower. What's up?

Ilya

@MattN again?!!

Matt N.

@Ilya Don't be so horrified. I shower every day.

Jonas Teuwen

Matt likes showers...

Matt N.

@tb Nice : )

@JonasTeuwen ...

Ilya

08:33

@JonasTeuwen I like Matt. I like showers. Local transitivity

t.b.

So, showers like Matt?

Matt N.

@Ilya What did I do?

t.b.

Oh, I forgot. To like is highly non-symmetric :/

Ilya

not you did, Did did

Matt N.

.

Jonas Teuwen

08:34

@Ilya Are you sure that is possible? Maybe for some $\mu$ and $\sigma$ can make the LHS $0$.

Ilya

@tb look who's talking

Matt N.

@Ilya Ah, sorry, didn't read properly.

Ilya

@JonasTeuwen that's the inequality we discussed yesterday and proven to be true

Jonas Teuwen

@Ilya Excuse me?

Ilya

@MattN Please do not feel sorry, my previous hint was misleading hence I am the one who should apologize. See revised version:
"Didier did something surprising"

Matt N.

08:37

Highly is an exaggeration. I like Ilya too. So it's not that non-symmetric.

Ilya

@tb why did you add "I" there? I miss the meaning now

@JonasTeuwen well, yesterday we wanted to show that $\mathsf E\log|\mu+\sigma X|$ whici is symmetric in $\mu$, increases whenever $|\mu|$ increases

@Jonas so we asked Mathematica to compute that and obtained the hypergeometric form for this function

t.b.

I'm, like, whatever...

Ilya

hm... you just puzzled me more

Matt N.

You're not alone.

t.b.

allusion to noise that tortured our ears every day a few years ago

Rajesh D

08:41

anon goin great \looking at the starred messages ------------------>

Ilya

@Jonas: on the other hand, we have $\mathsf E\log|\mu+\sigma X| = \mathsf E \log|Y|$ where $Y$ now is $\mathscr N(\mu,\sigma^2)$

that's why we have this integrals, where it is easy to find the derivative, namely the derivative is

Jonas Teuwen

@Ilya Need to wake up 8-).

Ilya

In short words,
$$
f(x,y) = \frac{1}{\sqrt{2\pi}}\int\limits_\mathbb R \log|x+yt|\mathrm e^{-t^2/2}\mathrm dt
$$

t.b.

so, I see that @Ilya is not actually ignored.

Ilya

@tb by whom?

$$
=\frac{1}{y\sqrt{2\pi}}\int\limits_\mathbb R \log|t|\exp\left(-\frac{(t-x)^2}{2y^2}\right)\mathrm dt
$$

Matt N.

08:45

@tb That's before my time. Or I'm not cool enough.

t.b.

@Ilya by did. did did write something on your question

on that happy note, I wish you a pleasant day. Ice cubes will be falling from the skies very soon!

Rajesh D

@Ilya Hey

Matt N.

@Gigili Anyway. If you do then you could just send him an email and ask him out. That's the easiest way : )

Ilya

@tb have a nice day as well, enjoy a good weather

@Gigili: hi

Jonas Teuwen

@Ilya We can pick $\sigma = 1$ by scaling :-).

Now let's see about $\mu$...

Matt N.

08:48

@tb Bye.

Ilya

@JonasTeuwen yes

Matt N.

"ice cubes will be falling from the skies very soon"? What is going on?

Rajesh D

@Ilya : Hey

Ilya

@RajeshD sorry, hi!

Rajesh D

iHi

Matt N.

08:51

Whatever.

Ilya

@MattN that's the famous Swiss forecast

Rajesh D

Man o' man I can't believe it!

Jonas Teuwen

@Ilya You know if it depends on $\mu$?

Ilya

@JonasTeuwen what should depend on $\mu$?

Jonas Teuwen

@Ilya If we scale $\mu$ does the inequality change?

@Ilya Hmm, if $x \to -x$, then we have $=$?

Ilya

08:56

@JonasTeuwen it shouldn't, it only depends on the sign of $\mu$

@tb nice choice of music :)

Jonas Teuwen

@Ilya But do the substitution $x \to -x$, then we get a minus sign on both sides?

Oh, wait.

@Ilya Need to wake up.

@Ilya Actually you only want the RHS in the thing in your comment to be positive right?

hhh

Jonas Teuwen

09:11

Please troll no more.

I have trouble to believe that you try to do these things but have such... basic questions.

hhh

<--- how can I deduce this formally using $L^2$ inner product and orthogonaly of $e_n(x)=e^{inx}$?

Jonas Teuwen

Holy cow, check out the starred comment by @tb.

hhh

$\langle e_n, e_m\rangle = \frac{1}{2\pi} \int_{0}^{2\pi} e_n(x)\,\overline{e_m(x)}\,dx = \delta_{mn}$, yes, I was given this and now I try to connect dots...

Jonas Teuwen

Try harder, everything is there.

hhh

$\langle e_n, e_m\rangle = \frac{1}{2\pi} \int_{0}^{2\pi} e_n(x)\,\overline{e_m(x)}\,dx = \delta_{mn}$ <--- only about exponential functions while the $f(x)$ can be anything...or wait have to check $f(x)$'s def

mixedmath

09:18

Good morning all

my avatar hasn't shown up yet on my screen at right ---------->

oh - there it is

hhh

Gigili

jj

Hello.

Huh, it works.

mixedmath

it works?

hhh

Expressing the sum somehow with the integrand, thinking?

mixedmath

@JonasTeuwen Just wondering, is this a recurring thing? Not in the recent chat, it seems?

hhh

09:28

(well I fall asleep last time...)

Jonas Teuwen

@mixedmath What is? :-).

@hhh Fill in the bloody $f$ into the bloody integral.

mixedmath

The chat line I replied to, in particular. About a troll?

Jonas Teuwen

@mixedmath Oh, yes. I think so... Maybe I'm too sensitive, but some people come with quite "advanced" questions and then some time later ask really elementary things.

hhh

$c_n = \frac{1}{2\pi} \int_{-\pi}^{\pi} \left( \sum_{n=-\infty}^{\infty}c_n \right) dx$

Jonas Teuwen

@hhh You have to verify the formula. So try to get $c_n$ out of it, not... what the.

hhh

09:33

If $c_n$ not depended on $x$, then something absurd $c_n = \sum_{n=-\infty}^{\infty} c_n$

David Wheeler

well, i'm an idiot, but it looks like linear algebra to me....

Ilya

@JonasTeuwen yes

David Wheeler

the integrals are just there to fool people

Ilya

@Gigili what does work?

mixedmath

@hhh This line is not correct

hhh

09:38

$c_n = \frac{1}{2\pi} \int_{-\pi}^{\pi} e^{-inx} \left( \sum_{m=-\infty}^{\infty}c_m e^{imx} \right) dx$

I mixed different vars with same letter $n$?

Jonas Teuwen

@DavidWheeler That's true :-).

David Wheeler

it's like, you have a vector space made out of periodic functions, with an orthonormal basis of complex exponentials (let's not kid ourselves, these are really trig functions)

Jonas Teuwen

Holy cow man, your integral is just $\langle f, e_n \rangle$. Then write $f$ in the same basis and pull it out.

@DavidWheeler It is much more instructive (to me) to see them as complex exponentials. Stuff that runs around the circle.

David Wheeler

and the $c_n$ are just the coordinates of our vectors/functions

Jonas Teuwen

As, that is what you do, you consider functions on the circle, not really periodic ones. Okay, that's the same, but...

David Wheeler

09:40

so what you want to do is take the projection of f in each "basis direction" and add them all up

integrals are linear functionals, so with some monkeying around, you can make inner products out of them

evaluating the integrals themselves is kind of tedious, and it doesn't interest me

hhh

(this so far corect here? I just plugged the two things together)

David Wheeler

@Jonas well, yes, i think "circular functions" is a better name than trig functions. it expresses the linkage between the ups and downs and side to sides better

mixedmath

@hhh Keep going

Ilya

@Gigili: are you here?

Jonas Teuwen

@Ilya Why would you wake a sleeping dog? :-(.

Ilya

09:48

@Jonas: who is a sleeping dog here?

Jonas Teuwen

@Ilya <-

Ilya

you?

or I?

Jonas Teuwen

No.

Ilya

no?

Jonas Teuwen

Gig.

Ilya

09:49

why is she dog?

David Wheeler

i had hoped to find an algebraist here, but apparently only analysts have been hanging out here as of late

Jonas Teuwen

@Ilya Hmm, dog is the wrong word. But the screen will be infested with thingies.

@DavidWheeler You can only conclude one thing now: Switch to analysis.

Plus Ilya is no analyst, but way cooler.

Ilya

@JonasTeuwen WHAT???

Jonas Teuwen

@Ilya 8-))))).

Oh man, I should just shut up.

Ilya

is it a challenge who puzzles whom? Then the challenge is accepted. Prepare to die :D

Zhen Lin

09:52

Hmmm, an "infinitary group" with an associative infinite multiplication must be trivial. :(

David Wheeler

i like algebra. it's so clean. analysis is messy.

Jonas Teuwen

@DavidWheeler Uh,... analysis messy, what?

Ilya

@JonasTeuwen Duel with him? There are two of us

Jonas Teuwen

Can be very messy, but algebra too.

@Ilya Yes.

David Wheeler

yes, algebra can get messy too. i don't like those parts.

Ilya

09:54

@JonasTeuwen If he kills me, you still have a chance for the revenge

Jonas Teuwen

Operator theory? Functional calculi?

Abstract harmonic analysis?

David Wheeler

there are just soooo many functions. and most of them behave rather poorly. so then we make up all sorts of names which mean: functions which are nice.

Jonas Teuwen

You probably don't know much about it. Usually analysts only work with very nice functions and the rest is just an extension using limits, you would hardly do anything with the strange things directly.

Jordan

It is the second day of my calculus 2 class and I think I need to drop out already

David Wheeler

i know integration is harder than it looks. and which functions we can integrate has a lot to do which which integral we use (apparently one size doesn't fit all)

Jonas Teuwen

09:58

Huh?

In analysis usually always we use the Lebesgue integral wrt to some measure. That is sufficient for almost everything which does not have an infinite dimensional range.

Jordan

I woke up at 4am to do homework before class at 8am lol

David Wheeler

at some point, apparently people stop even thinking about the functions them selves, and just about "good sets" and bad sets"

Jonas Teuwen

Have you ever seen what is in operator theory book?

Good sets and bad sets is something of real harmonic analysis.

David Wheeler

like with hilbert and banach spaces?

Jonas Teuwen

Which is actually quite elegant if you think about it, topologists speak about pants.

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$Mathematics$ Mathematics