Mathematics

I think any conventional proof invokes the primitive element theorem

The primitive element theorem is also important in its own right, so maybe you should read up on that instead

1:15 AM

In other news, I just found an inversion-free proof of the injectivity of the Fourier transform that only requires tools from complex analysis and integration theory and it makes me very happy

Constant real part --> a holomorphic function is constant, right?

I think it is just Liouville.

That's just Cauchy-Riemann

That's true, it forces the imaginary part to be constant right away.

Bounded would need something more like Liouville affords.

Or maybe you can do it with CR.

That's essentially equivalent to Liouville

Since bounded implies bounded real part

1:33 AM

@Thorgott do you enjoy complex analysis?

I do, but I don't understand it

Do you loath applying the residue theorem to find improper integrals

lol I just had to do that in a tutorial today

A lot of applications of the residue theorem are pretty slick, but quite a few are also just plain ugly

I hate contour integrals with a passion and I have to do two tonight for an assignment.

Did you see the residue question Ted helped me with yesterday?

its.caltech.edu/~ivrii/practice-exam.pdf

#5

You seem much better at complex than me so you probably can do it without hints, but if you scroll back in the chat history you can see Ted help me work it out.

1:50 AM

Looks like a variant of the argument principle

2:09 AM

Hmm, I think I'm not seeing something

I feel like that integral should add up the zeros of $f(z)=a$ (with multiplicity) and substract the poles of $f$ (with multiplicity), but it seems I'm mistaken about that latter term

Two example calculations seem to agree with this

Agree with the mistaken part?

Agree with the poles turning up in the result

Hmmm, I didn't check about the poles.

Take $f(z)=\frac{1}{z-\frac{1}{2}}$, $a=0$ and the unit circle. Clearly, $\frac{1}{z-\frac{1}{2}}=0$ has no solutions. Now , $\frac{1}{2\pi i}\int_{|z|=1}\frac{-z\frac{1}{(z-1/2)^2}}{\frac{1}{z-1/2}}dz=-\frac{1}{2\pi i}\int_{|z|=1}\frac{z}{z-\frac{1}{2}}dz=-\frac{1}{2}$, which is minus the only pole of $f$ (counted with its multiplicity 1).

Akiva Weinberger

4:01 AM

Rithaniel

4:24 AM

What's that, Akiva?

Semiclassical

4:35 AM

looks fractal-ish

user76284

What's a good proposal distribution to sample from a multi-modal distribution with support in $[0,1]^n$?

The peaks of the distribution are very narrow, so I have to use importance sampling.

I was thinking something along the lines of a mixture of products of beta distributions, but I don't know the number of components to use, since I don't know the number of modes.

5:27 AM

@Thorgott just sat down and checked, and you are right about the poles.

Same proof effectively as the roots

Akiva Weinberger

$user image$

Cute @AkivaWeinberger

Akiva Weinberger

@Rithaniel Fractal I found somewhere

Not sure how it's made

Ajay Mishra

6:08 AM

(b) Prove that if A and B be two square invertible matrices, then AB and BA have same charac-
teristic roots.

Any hint on this question ^?

manooooh

7:18 AM

►

@TedShifrin ^^^^^^^ that was an amazing explanation. Thank you for letting upload your classes to YouTube, they are amazing!!

8:13 AM

Morning

8:57 AM

@EdwardEvans guten Morgen

9:07 AM

Grüß Gott

maths student

10:55 AM

0

Q: Set theory and relation

Let $\mathcal{C}$ be a collection of subsets of $[n]$ with the property that if $A, B \in \mathcal{C},$ then $A \cap B \neq \varnothing .$ (For example, $\mathcal{C}=\{\{1,2\},\{1,3\},\{2,3\},\{1,2,3\}\}$ has this property.) What is the largest that $\mathcal{C}$ can be? Hint: You will have to pr...

11:47 AM

Question: $A$ is a Dedekind ring, $\mathfrak{p}$ a non-zero prime ideal in $A$ and suppose $\mathfrak{a} \subset A_\mathfrak{p}$ is a non-zero ideal in the localisation. Since Dedekind rings are stable under localisation we have the factorisation $\mathfrak{a} = (\mathfrak{p}A_\mathfrak{p})^n$ for some $n \in \Bbb N$.

Take an $a \in (\mathfrak{p}A_\mathfrak{p})^n \setminus (\mathfrak{p}A_\mathfrak{p})^{n+1}$. Why does it follow that $(a) = (\mathfrak{p}A_\mathfrak{p})^n$

(that factorisation exists because Dedekind rings have prime ideal factorisation and that guy is the only prime of the localisation)

jeea

Does $E[X|Y=y] = E[X]$ mean X, Y are independent random variables??

No, take $X=Y\sim\delta_1$

Obviously $(a) \subseteq (\mathfrak{p}A_\mathfrak{p})^n$ but what's stopping smth like $(a) = (\mathfrak{p}A_\mathfrak{p})^{n-1}$

jeea

How we see mathjax in mobile @Thorgott

See here: math.ucla.edu/~robjohn/math/mathjax.html

All I'm saying is to take X and Y to be equal and dirac-distributed

12:05 PM

durr lower powers of primes contain the larger powers, not the other way

no wait

yeah

that's right hahaha

ccorn

12:24 PM

Looking for a nice geometry problem? There is one. It already has an answer, but finding a (mostly) coordinate-free reasoning would be great.

12:44 PM

@EdwardEvans is that even true? I'm looking at things like $(2) = (2,1+\sqrt{-5})^2$ in $\Bbb Z[\sqrt{-5}]$

oh what am I thinking about

yeah maybe just use the $\mathfrak p$-adic valuation

oh wait you're looking at the localization

then isn't it trivial

the localization is a DVR

@Leaky the proof is to show that $A_\mathfrak{p}$ is a DVR lol

if you still want a proof, let $a = \varpi^n u$; for any $\varpi^m v \in (\mathfrak p A_\mathfrak p)^n$, $m \ge n$, then $\varpi^m v = (\varpi^n u) (\varpi^{m-n}) (vu^{-1}) \in (a)$

aha

It's fine I got it hehe

thanks tho

ok