Mathematics - 2023-02-16 (page 1 of 2)

Sine of the Time

00:08

everyone sleeping :(

Ted Shifrin

00:45

snore

Xander Henderson

00:58

@mick This has been asked and answered before on Math SE.

It essentially comes down to whether or not the denominator is a Fermat prime.

shintuku

01:12

i'm on the verge of a breakthrough. I've got $a_1z + a_0 = a_1(z-w_1)$, and $a_2z^2 + a_1z + a_0 = a_2(z-w_2)(z+w_2+\frac{a_1}{a_2})$

$w_1$ is a root for the degree 1 polynomial, $w_2$ is a root for the degree 2 polynomial

any clue how to show $-w_1 = w_2 + \frac{a_1}{a_2}$?

Ted Shifrin

What are you doing?

How do induction proofs work?

shintuku

i've got the induction proof down

but uh

i don't understand it

the base step is getting $a_1z + a_0 = a_1(z-w_1)$

the inductive step is

Ted Shifrin

Sure. Done.

Suppose I know that any polynomial of degree $n-1$ can be written in the desired form.

shintuku

Suppose it is true for any $n$ degree polynomial. Let $f$ be an $n+1$ degree polynomial. Then, from (1), it has at least one root, say $w$, and so from (2), it has a form $f(z) = (z-w)g(z)$, where $g$ is an $n$ degree polynomial. But by our inductive hypothesis, $g$ has the desired form, the form for $f$ we just mentioned is our desired result.

(1) is the fact that a polynomial with nonzero coefficients has at least one root

(2) is the fact that we can write a degree $n$ polynomial $f$ with at least one root $w$ as $f(z) = (z-w)g(z)$, where $g$ is degree $n-1$

but I don't understand where the coefficient is coming from

Ted Shifrin

Yes, call that root $w_{n+1}$ rather than $w$.

The coefficient is there just because the polynomial needn't have leading coefficient $1$. I told you that already.

shintuku

01:19

right but I don't know how to parse that logically into the proof

Ted Shifrin

So write $f(z)=(z-w_{n+1})g(z)$, and we know $g(z) = w_0(z-w_1)\dots (z-w_n)$. Combine.

shintuku

right, we get our coefficient from our hypothesis

Ted Shifrin

You do need to know that the leading coefficient is nonzero (or else the degree isn't right).

shintuku

now what's bothering me is, how is the coefficient for the leading term of $f$ coming from a lower degree polynomial $g$

so i'm trying to work it out for a degree 2 polynomial

i'm trying to understand why we aren't forcing the degree $n+1$ term and degree 1 term of $f$ to have the same coefficient, since the leading coefficient always comes from the lower degree polynomial

Ted Shifrin

I would factor out the leading term from the beginning and then work with leading coefficients of $1$ everywhere. But if you write $(z-\alpha)g(z)$, the leading coefficient is just the leading coefficient of $g$.

Better understanding is what I suggested. Write $f(z)=C F(z)$ where the leading coefficient of $F$ is $1$. Do induction with $F$ and you'll just get $F(z)=(z-w_1)(z-w_2)\dots (z-w_n)$. Then put the constant back. Anyhow, I'm outta here.

shintuku

01:25

hm i'll think through those thanks

Rithaniel

By the way, Ted. I was watching one of your lectures on YT, and I like how you write your j. Mine is way too similar to my i.

D.C. the III

I am working on your second deivative test for constrained extrema question @TedShifrin

I parameterized the constraint surface M locally by $\Phi$ with $\Phi(\bar{a}) = a$. Now using $f \circ \Phi$ and setting up the lagrangian I end up with:

$(Df(a) - \lambda Dg(a))D\Phi(\bar{a}) = 0$. Now you also gave me that the restriction of the hessian of $f-\lambda g$ to the tangent space is negative definite. But I'm stuck here because to show something is a local maximum with the hessian means to show the hessian is negative definite

Ted Shifrin

@Rithaniel you don’t hook on the bottom? My 9s confuse people, cuz I do them European style. America 9s can look lime lower case printed g.

katle

I have got a fatal error when i compile my thesis and i tried revoking all the changes i did, but the error still persists. Is there maybe a person out there, who could glean from the error log, what i have to change? T_T

Rithaniel

Yeah, but your hook is much more pronounced. Mine can be mistaken for an i that I just got a little sloppy while writing. I do a loop on my 2, though (also, I don't cross my 7 or my Z)

Ted Shifrin

01:34

Isolate the page it’s happening on and run a few lines at a time @katle

I cross 7 and z definitely for clarity. I’ve done that since I started teaching.

The point, DC, is that it’s just regular unconstrained stuff for $f\comp \Phi$. Now it’s careful chain rule.

D.C. the III

I think...well actually I know the issue is, I don't know where I'm going.

As in I don't know what the final result should look like once I do the chain rule

Ted Shifrin

Maybe wait until you”ve thought more about parametrizing (abstractly) submanifolds.

D.C. the III

Fair enough I do feel the key is in the parameterization and I kind of just took it as is and tried to shoehorn it in symbol pushin-wise

Well that will come after I do Chapter 7, which is now on deck.....maybe with a quick refresh of integrals in Spivak first...

Prateek Mourya

02:11

@TedShifrin i tried the same but not sure how using those boundary values cancels out the term as we have no clue about wave function being odd or even

Ted Shifrin

No, no. Boundary terms are $0$ because we always know the functions decay at $\infty$. That is the usual assumption.

It is not a consequence of $L^2$; it is an additional assumption. Physics is unfortunately sloppy about being explicit.

@Rithaniel I’m glad your observations were purely orthographic!

Rithaniel

@TedShifrin Lol, yeah, no math was learned

Ted Shifrin

Nor pedagogy. 🤷‍♂️🤷‍♂️

Rithaniel

It was an undergraduate linear algebra course. It's fun to see a new style on teaching it

You were definitely encouraging, too. I took note of that

leslie townes

writing nines the european way in america is an affectation. just wear a beret, professeur.

Ted Shifrin

02:24

The course moved very quickly, and more abstractly, by comparison with my typical LA courses.

Ugh. Chat malfunctioning.

shintuku

it is no secret i am a doofus

but i still don't understand the induction proof

is there not a complete exposition of it somewhere? also, does it have anything to do with the factor theorem (or the remainder theorem)?

there must exist an exposition of it somewhere

Ted Shifrin

03:18

You’re using the root-factor theorem, doofus.

shintuku

on towards google

copper.hat

those continental europeans and their funny numbers.

Koro

03:41

if $p^m|o(G)$, o(G)= o(G/K). o(K). If $p^m$ does not divide o(K), then why does $p^m$ divide o(G/K)?

this statement actually feels false to me: take p=2, m=3, o(G/K) =2, o(K)=2^2.

Then why is it used in Sylow theorem's proof?

nvm, I think I got it.

Ted Shifrin

That’s an incorrect statement.

Koro

yes, it is. I'm adding context where it is true.

Ted Shifrin

03:58

Yeah, $o(K)=p^m$.

Koro

Let me write the proof here now. The following is variation of Sylow's theorem. I claim the following: Suppose that p is prime. If $p^m$ divides |G|, then G has a subgroup of order $p^m$.
Proof: It is true for |G|=1, 2. If m=1, then the result follows by Cauchy's theorem so assume that m>1. We'll use induction on |G|. So suppose the statement to be true for all groups with order < |G|. Now, if G has a proper subgroup H s.t. $p^m$ divides |H|, then by the induction hypothesis, H has a subgroup K of order $p^m$, hence $K\le G$.

Ted Shifrin

Then $p$ does not divide $o(G/K)$.

Koro

So we are done in this case. So suppose that G has no proper subgroup whose order is divisible by $p^m$. We write the class equation of G: $|G|= o(Z(G))+\sum_{a\notin Z(G)} |G|/|stab(a)$. Here a is not in Z(G), so stab(a) is a proper subgroup of G, hence $p^m$ does not divide |stab(a)|. $p^m||G|/|stab(a)| . |stab(a)|\implies p| |G|/|stab(a)|.$

Hence, p||Z(G)|. By Cauchy's theorem, there exists x in Z(G), |x|=p. (x) =:K is normal in G. We consider the quotient G/K. o(G/K)< o(G). Now $p^m| |G/K|. |K|\implies p^{m-1}||G/K|$

This is where the statement is used. It makes sense because |K|=p.

By induction hypothesis, there is a subgroup A/K of G/K, $|A/K|=p^{m-1}$, where $A\le G$, whence $|A|=p^m$. Q.E.D.

'where $A\le K$ ' holds by correspondence theorem.

i.e., every group contains subgroups of every prime power that divides the order of the group.

Node.JS

04:48

I am wondering can someone please help with this question: math.stackexchange.com/questions/4640147/…

leslie townes

much of this is, i don't want to say explained exactly, but worked out in some detail on wikipedia, en.wikipedia.org/wiki/Preorder

i don't see an exact dupe of the question on MSE, but you should maybe try google for something that talks more about it

shintuku

where can I find a proof that $A$ is nonsingular iff $\forall b \in \Bbb R^n$, $A\vec x = \vec b$ has a unique solution?

Ted Shifrin

Every linear algebra text.

leslie townes

this is going to sound sarcastic, but it isn't. almost any linear algebra book?

ah, ted beat me to it.

shintuku

nice, let's all give a round of applause to every linear algebra book

Koro

05:52

Strong convergence of a sequence of f_n (bounded linear functionals) and weak* convergence of the f_n seems to be the same concept.

I’m confused.

PNDas

06:03

@Koro Why do you think so?

leslie townes

yeah, if you think they're the same notion, first thing to do is try to prove that they're the same.

your book or notes may have established relations between those concepts that stop short of establishing that they are exactly the same.

reviewing those relations might give an indication of where to begin looking for examples that they are not the same, if indeed they are not.

Koro