Mathematics - 2018-10-04 (page 3 of 3)

Will Hunting

23:01

No excuse is needed. Just do it. Nike.

Impossible is nothing. Adidas.

Fargle

Eat your Wheaties. Wheaties.

Ted Shifrin

Oh, it's a @Fargle.

hi Jasper

Fargle

It is indeed. Heya @Ted.

Will Hunting

Hiiii.

Ted Shifrin

No math to discuss, @Fargle?

Fargle

23:09

I'm almost through section 3. Still gathering my thoughts.

Focus is not my strongest suit.

Ted Shifrin

Section 3 of whom?

Kasmir Khaan

btw Ted

to do ring theory

what does one need to know about group theory ?

just the iso theorems and some basic stuff yeah ?

Ted Shifrin

You certainly don't need Sylow theorems. But, as usual, you ask a very vague question.

Kasmir Khaan

yeah you basiclly read my mind ><

ring theory seems not very dependant on group theory

i mean ok a ring is an abelian group with +

and some other properties

Ted Shifrin

It all depends how advanced you get.

Kasmir Khaan

23:11

and then one does quotient rings

just basic so far =p

quotient rings will need some knowledge of the quotient process

and some iso thm , and then it is kinda indepandant

we gonna go thru PID UFC etc

Ted Shifrin

Note that I did all that basic ring stuff first in my book.

Kasmir Khaan

I know ! :)

and guess what kasmir will read now ?

Ted Shifrin

But as you get to more advanced stuff with modules, representation theory, etc., you start using a lot more group theory and linear algebra, of course.

Kasmir Khaan

okay, but am planning to do algebra properly

so gonna do full linear algebra course at home

the books am using are serge lang and steve roman books

first lang than roman

after that am gonna repeat group theory and ring theory properly

Ted Shifrin

I don't know Roman. Lang's is a standard university course.

Kasmir Khaan

23:15

when i took LA, we did not do anything usefull

litterly that is why i have been struggling

Ted Shifrin

I might disagree with you on the meaning of "useful."

Kasmir Khaan

so anyway :D kasmir got long hard year infront of him

okay ><

but really was not enough to take the next step

Lozansky

@KasmirKhaan You go to SU?

Ted Shifrin

We already discussed Artin.

Kasmir Khaan

@Lozansky uppsala and you ?

Lozansky

23:16

KTH

Kasmir Khaan

yes artin is very good book

@Lozansky nice what do you study ?

Ted Shifrin

If you know all the linear algebra in Artin, you're fine, @Kasmir.

Lozansky

Teknisk fysik ^^

Kasmir Khaan

coolt :)

Ted Shifrin

But Lozansky bugs us anyhow.

It's like 3 years now :P

Kasmir Khaan

23:17

@TedShifrin okay thanks Ted !

Ted Shifrin

Did you ever settle that mixed cylindrical/spherical coordinates problem, @Lozansky?

Lozansky

@TedShifrin Yeah mostly

Ted Shifrin

OK, good.

I really didn't want to spend 20 minutes trying to think about it.

Lozansky

I think the important thing to note was that $\hat{s}$ is $2\pi$ periodic in $\phi$ so the whole component vanished which is nice

Yeah it was a bit messy

Ted Shifrin

I mean: I like complicated computations, but ...

Lozansky

23:20

I prefer neat tricks :P

Ted Shifrin

As you know, I teach students to EXPLOIT SYMMETRY whenever possible :P

Lozansky

I'm gonna take a course in abstract algebra next semester, is that in your wheelhouse @TedShifrin?

Course book is Judson

Ted Shifrin

Never heard of that book.

I'm fonder of analysis/geometry, but the first book I wrote was an algebra textbook.

Lozansky

I might take a course in analysis as well :P

Ted Shifrin

There are lots of folks here you can talk to :P

Lozansky

23:24

Feel like I need to learn some real math

Leaky Nun

I'd prefer to learn some complex math

Ted Shifrin

I just answered a complex analysis question on the Poincaré half-plane, Leaky.

Leaky Nun

for real?

Ted Shifrin

I wasn't imagining it.

Leaky Nun

it would have been complex to imagine

Ted Shifrin

23:26

You're recycling now.

Leaky Nun

:(

today in one of the courses the professor said that if $f:\Omega\to\Bbb C$ is holomorphic at $p$ then $f'$ is also holomorphic at $p$

I thought for a while and came up with $f(z)=|z|^2$ and told the lecturer at the end of the lesson

Ted Shifrin

Depends on the definition.

Some people say that holomorphic means complex differentiable on an open set ...

Leaky Nun

aha

interesting

Ted Shifrin

Different books have different conventions.

Leaky Nun

well he didn't say that

Ted Shifrin

23:29

But you're right. Your $f$ is complex diff. only at $0$.

To prove what he wants you need an open set to get a Taylor expansion ...

Leaky Nun

right

Ted Shifrin

Or the Cauchy integral formula to differentiate.

Leaky Nun

the proof is really beautiful

when I told my friend the proof he was like wow

Ted Shifrin

Regardless, either way, you need a little disk.

Leaky Nun

and then when I told him the proof for the annulus he was like double wow

Ted Shifrin

23:30

Complex analysis is one of my favorite courses ...

Wait ... what proof for the annulus?

Leaky Nun

and then when I showed that the formula for the coefficients is the same he was like triple wow

Ted Shifrin

you talking about Laurent expansions now?

Leaky Nun

right

if $f:\{z ~:~ r<|z|<R\} \to \Bbb C$ is holomorphic then it has a Laurent expansion at $0$

Ted Shifrin

Make sure you can show your friend explicit examples of how you do it. :)

Leaky Nun

I did

Ted Shifrin

23:31

OK

Leaky Nun

I made it a corollary :P

wait

I misunderstood

Ted Shifrin

I've had to explain concrete computations to several people in here

Leaky Nun

maybe I'll do that next time

Ted Shifrin

You have to use geometric series expansions, dividing through by different things.

Leaky Nun

oh and do you have a function that cannot be continued beyond the unit disc?

or should I think about it more?

Ted Shifrin

23:32

The standard example is $\sum z^{n!}$.

Leaky Nun

hmm

Ted Shifrin

You mean analytically continuable NOWHERE on the boundary, right?

Leaky Nun

right

Ted Shifrin

Yeah, the one I just gave you is the standard example. Figure out why.

Leaky Nun

I feel like I only need to consider $z=\exp(i\pi\theta)$ for rational $\theta$

Ted Shifrin

23:34

Presumably that's relevant.

Leaky Nun

there the series eventually becomes 1

so it can't converge

Ted Shifrin

Be careful.

Leaky Nun

that the limit of the series is not the series of the limit

Ted Shifrin

$\sum z^n$ fails to converge at every $z$ with $|z|=1$, but the function is still analytically continuable almost everywhere.

This is one of the really subtle things in complex analysis.

I've trapped students with that on many a complex analysis qualifying exam. :P

Leaky Nun

I think I'll want to travel on the line $z=r\exp(i\pi\theta)$, $0 \le r \le 1$ for rational $\theta$

hmm...

and reduce to the case to $\theta = 0$

because after a finite number of terms, $(r\exp(i\pi\theta))^{n!} = r^{n!}$

so I need to prove that $\lim_{r\to1^-} \sum r^{n!} = \infty$

Ted Shifrin

23:41

And then you need to write a proof why that shows there's no analytic continuation across.

Leaky Nun

because the rationals are dense in the reals

Ted Shifrin

You have more to say than this.

But I'm leaving for now, so you can contemplate.

Secret

Last night dream there are two things:

There are two family of sets consists of disjoint union of intervals. One of these is:

Leaky Nun

if there's continuation in the direction of $\theta$, then we can use the fact that $[0,\exp(i\pi\theta)]$ is compact to deduce that the function is bounded there, and then some test should be able to tell you that the function converges uniformly there

and that's the Weierstrass M-test

aha I can use the same test

since the function $\sum r^{n!}$ is increasing, if it doesn't go to $\infty$ then it must go to some $L$

so we have a continuous function $[0,1] \to \Bbb R$

so we can exchange limit with series

@TedShifrin how does this sound?

Secret

$\bigcup_{n<\infty} [n,2n+1]$ and another are $S_m = \bigcup_{n<\infty}[n,f_m(n)]$ where $f_m$ is a sequence of functions: $\{n^{m}|m\in \Bbb{Z}\}$. This eventually this union will cover $\bigcup_{n<\infty} [n,2n+1]$

Then there is a paragraph following these two families saying the second family has a higher abruptness than the first, with abruptness defines by:

$\text{abrupt}(r)=\frac{1+f(r)}{f(r)}$ where the $f(r)$ is the probability that the given function will take different values within any interval of size $|r-a|$

A plot of an abrupt function is then shown (to be posted)

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$Mathematics$ Mathematics