Mathematics

12:42 AM

@leo Yello

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

so I have edited

the post

@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.

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?

good night all!

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

12:47 AM

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

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

2

(relevant to the latter question)

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

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

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

Nope, definitely don't do that.

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

(Although that shows normality ... but anyway)

12:59 AM

?

Nope to Peter or to mine point?

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.)

@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.

heh

Why would Spiegel put that problem

@MarianoSuárezAlvarez Are you around?

1:24 AM

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...

1:48 AM

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

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

@leo Teorema del Estabilizador de Orbita?

:-)

The problem is Stabilizer do not stabilize orbits

:-|

It should be something like "Teorema Estabilizador-Orbita"

más bien órbita

Teorema del Estabilizador-Orbita?

hi

2:01 AM

@PeterTamaroff I think so. Something like that

@DanielMontealegre hi

hey could you lend me a hand with galois theory?

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

im going to post it in the forum

@PeterTamaroff It will be Teorema 2

interesting

2:43 AM

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

@PeterTamaroff That would be pretty useful

@PeterTamaroff this can like you

3:12 AM

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?

@Eugene Hey.

I'm working on this thingy leo suggested.

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

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

Should that work?

bounded is not enough

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

3:23 AM

yep

@leo I guess we can assume the function is friendly

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

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

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

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

3:32 AM

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

@leo Oh.

I couldn't understand that effing symbol.

what effing symbol?

@leo I mean, how it was written

I realized it was something of the sort.

Oh.

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

I don't follow you

Rajesh D

Hi guys

3:36 AM

hi

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

2

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

I mean that the way Wikipedia wrote it confused me.

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

Oh. I just catch it. :-D

@PeterTamaroff but now you understand?

@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

3:39 AM

so f*~%!

Maybe you can complete the proof.

Hi

can i ask a quick question

@DanielMontealegre Sure.

@anon

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

3:50 AM

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

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

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 $H_1\cap H_2=G(L/K_1(K_2))$

You forgot the last $

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.

nevermind

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

David Wheeler

5:28 AM

hello?

6:12 AM

@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.