Mathematics - 2021-05-19 (page 2 of 2)

copper.hat

6:23 PM

@TedShifrin Hope all is good.

Ted Shifrin

Oh, you're back. Now I'm in trouble :)

I don't believe baristas question or your comment.

copper.hat

I will check again.

Ted Shifrin

When we do usual bounds, I don’t see why they have to be tight on a given circle. Maybe I'm being dumb.

copper.hat

I could easily have made a mistake. I frequently do.

Ted Shifrin

Well, even the problem hint seems wrong to me. I haven't thought about the exponential yet.

copper.hat

6:29 PM

I need a few mins, stuck in a work thing and I am a strict monotasker :-(.

Ted Shifrin

No problem.

copper.hat

I deleted my comments, I think you are correct, thanks for spotting.

barista

quite sure I need to apply modulus theorem for my problem

Ted Shifrin

The hint seems to send you in the wrong direction. Why does the real part actually give you those bounds?

I am not sure I even believe it.

What is the link to a valid solution? Don't make us search, because I won't.

barista

0

Q: Cauchy's integral formula or Bounds for analytic functions

The problem is here. Consider the polynomial $P(z)=a_0+a_1z+...+a_{n-1}z^{n-1}+z^n $. Show that, for all sufficiently large values of $r$, there must exist points $z_0, z_1$ on the circle $|z|=r$ such that $|e^{P(z_0)}|=e^{(1/2)r^n}, |e^{P(z_1)}|=e^{-(1/2)r^n} $. I tried to show that there e...

I actually didn't read the accepted answer seriously

Ted Shifrin

6:41 PM

The $r$ is definitely going to be different from the one in the hint.

barista

What you mean the bound of real part? $|P(z)|$ in the statement?

Ted Shifrin

Modulus and real part are related but not strongly related.

Real part is bounded by bounds on modulus, but usually never strictly achieves the bounds.

barista

Yes so I tried to apply modulus theorem for $e^z$

Ted Shifrin

How is that relevant?

The image of $f$ on large circles is never a circle unless you have a monomial.

barista

Your talking apply the theorem to $f$ but apply to $e^z$ with $z = f(w)$ gives relation

Ted Shifrin

6:47 PM

I don't see it.

Maybe try your proof for a general polynomial of degree $1$ and convince me.

dc3rd

What is more computationally efficient for computing the inverse of a matrix?

1) Using the gaussian reduction algorithm?

2) Using the characteristic polynomial?

Both have a mechanically labor intensive process. So I'm curious what has been found to be the case when we are dealing with larger values. I'm guessing characteristic because it doesn't involve as much "division" through the process.

leslie townes

is the characteristic polynomial known? if not, how do you compute it?

and don't say determinants :)

dc3rd

Oooh... good point. Because in my text it is defined as................(well I can't say how now because you banned it :p

$\det(A-tI)$ :)

Ted Shifrin

No, Gaussian elimination wins.

leslie townes

you don't wanna take any determinants

dc3rd

6:57 PM

I'm guessing because of how crazy calculating determinants can get

I've noticed that.....there is good reason why texts only go up to $n=3$ with determinants and for fun throw in one $n=4$ so you know why it is crazy

copper.hat

@dc3rd are you asking a question about practical numerical inversion of arbitrary matrices? if so, LU with partial pivoting is good.

Thorgott

how do you calculate the inverse using the characteristic polynomial

leslie townes

if you have integer or rational coefficients and want to do exact arithmetic that is some overhead but a computer doesn't really care if it's multiplying or dividing in that case.

thorgott, i'm guessing cayley hamilton theorem, coefficient of I will be nonzero, solve for A [polynomial in A] = I

copper.hat

@Thorgott the multiplier of $I$ in $\chi(A)$ is non zero so you can factor $A$.

leslie townes

if you somehow already know the characteristic polynomial, it's not a bad approach

Ted Shifrin

7:00 PM

Makes a good standard exercise in every LA book.

dc3rd

@Thorgott a question I just solved applying the Cayley Hamilton theorem:

$A^{-1} = \frac{-1}{a_{0}}[(-1)^{n}A^{n-1}+a_{n-1}A^{n-2}+\dots+a_{1}I_{n}]$

or in the "condense" PhD talk of copper and Leslie.....how they said it

but getting that characteristic polynomial is where the problem lies isn't it @leslietownes ?

Thorgott

ah, that's what you mean, ok

yeah, that's not computationally efficient

copper.hat

@dc3rd in practice the matrix may have additional structure (symmetric, etc) or is evolving (rank one updates, etc) which can be exploited.

barista

@TedShifrin right.. maybe you're right. Continuity won't help?

copper.hat

7:22 PM

@barista my thinking is a little clouded this morning (well, more that just this morning :-)), but i think we have $\lim_{r \to \infty}{ |e^{f(r)}| \over e^{{1 \over 2}r^n}} \to \infty$ and $\lim_{r \to \infty}{ |e^{f(\alpha r)}| \over e^{-{1 \over 2}r^n}} \to 0$ (with $\alpha = e^{i { \pi \over n}}$), which you can use along with the IVT to get the desired result.

Onir

I wrote a solution about a matrix but I don't think it's rigorous enough

0

A: Find rank of very big matrix.

We have an integer matrix, we can do stuff related to its rank by reducing it $\bmod p$, the rank of the original matrix is larger than the rank of the reduced matrix. If we reduce it $\bmod 2$ we get the symmetric matrix in which $1$ is outside the diagonal and $0$ in it. Finding the rank of thi...

What I want to use is the following

we have an integer matrix and we know that if we reduce it mod_p then the rank is k (in the field mod p)

we want to say the rank of the original matrix is at least k

barista

7:38 PM

maybe I need some sleep it's 4:30am

shintuku

8:34 PM

So, in terms of the logic of the proof that, for any $A, B \subset \mathbb{R}$, we have $\sup (A+B) = \sup A + \sup B$.
If we produce the following steps in the proof:

1. Let $a,b$ respectively be elements of any subsets $A,B$ of $\mathbb{R}$.
2. Then, $a+b \leq \sup(A+B)$.
3. But, $a+b \leq \sup A + \sup B$.

Is the legitimate next step, in terms of the logic of the proof, $\sup(A+B) \leq \sup A + \sup B$ or $\sup A + \sup B \leq \sup(A+B)$?

leslie townes

in general, from $x \leq y$ and $x \leq z$ it's hard to deduce an order relation between $y$ and $z$, so you have to dig a little deeper and have a goal in mind.

the fact that 3 holds for arbitrary a in A and b in B shows you that $\sup A + \sup B$ is an upper bound of $A+B$. that tells you something about the relation between $\sup A + \sup B$ and $\sup(A+B)$, doesn't it?

shintuku

Right, it tells us $\sup(A+B) \leq \sup A + \sup B$, since $\sup(A+B)$ is the least upper bound of $A+B$ and up to that point $\sup A + \sup B$ is only some upper bound

leslie townes

8:55 PM

yeah. i guess that's the easier direction. makes sense.

shintuku

oh wait you were suggesting the contrary?

see, this is what is confusing me about the proof

leslie townes

i was not suggesting the contrary. i was hinting that 3. for all a and b was half of the problem.

i pretended not to know which half because i am fundamentally a dishonest individual.

shintuku

hahaha, well thanks it helped me make a step forward

leslie townes

for the other direction, lots of ways to do it. interposing sup(A + sup(B)) into it is one way. the more i think about it, i kinda like it.

i also like the limit-like argument.

copper.hat

@shintuku I think you are making more work than necessary. Since $a+b \le \sup A + \sup B$ you get $\sup (A+B) \le \sup A + \sup B$. since $a+b \le \sup (A+B)$, you get $\sup A + b \le \sup (A+B)$ and then $\sup A + \sup B \le \sup (A+B)$.

shintuku

9:06 PM

i keep feeling like I'm encountering possibilities for arguments that slip from me because I can't name and distinguish them properly

@copper.hat added to the repertory

thanks!

Ted Shifrin

@Onir You're right. Do the contrapositive.

Thorgott

@leslietownes hey, I think that's pretty much the argument I gave earlier except yours looks cooler

Onir

Thanks, yeah I went through the motions and it seemed to work out

I posted a question on it also

2

Q: rank of integer matrix reduced $\bmod p$ is smaller than that of the original matrix.

I am using a result in a problem but I am not sure how to write or support it correctly. We have an integer matrix $A$ over the reals. We can define a matrix $A'$ over $\mathbb F_p$ by just taking each entry $\bmod p$. I want to say that the rank of $A'$ is less than or equal to the rank of $A$. ...

linear-algebra

I've seen a couple of olympiad style problems where the reduction turns the problem into something easier

Ted Shifrin

Change the title to “at most” instead of “smaller than.”

Onir

done

Onir

9:27 PM

Oh if we notice the rank is the size of the largest non $0$ minor it becomes crystal clear

shintuku

what do you people use to do latex on your computer? is there anything simple to use like mathjax?

Onir

vim

and pdflatex terminal command

shintuku

i'm a windows and i don't want to spend time debugging anything ever

last time i tried linux

well, it's fun but maybe later

leslie townes

you can use a text editor of your choosing, even in windows.

i wrote a lot of my dissertation in Wordpad because i was housesitting for someone and didn't want to install new stuff on their computer.

shintuku

oh sweet I'll try that

copper.hat

9:38 PM

@shintuku i would have said the same but with windows replaced by linux :-).

leslie townes

miktex is a pretty good windows distribution.

shintuku

getting it right now

Onir

Linux has a learning curve, windows doesn't have a learning curve because you're not supposed to learn anything.

shintuku

i'm gonna get into linux for the privacy part soon enough, just, i need to write down these proofs

copper.hat

i have found that later versions of windows have even less than before

Onir

9:41 PM

you just click where you're supposed to click and if there isn't a place to click for what you want ur pretty much out of luck

copper.hat

i used to (pointlessly) write to usoft to get my license fees returned because i was not using windows.

unfortunately most large it orgs are centred around windows. lcd.

shintuku

i want to become a good enough algebraic geometer to be able to draw pikachu with polynomials in latex

copper.hat

i suspect you could skip the first step there...

just need something to compute polynomial interpolation.

shintuku

no numerical methods

leslie townes

... you realize that pikachu isn't, like, an elliptic curve, right? :)

shintuku

9:49 PM

maybe pikachu isn't an elliptic curve. maybe, even, pikachu is not a conic section. but deep down i believe pikachu is an algebraic curve

copper.hat

how is latex going to help plotting that? you will prob need tikz or similar.

you could try tex.SE

Onir

you can draw pikachu with straight lines

can you make a decent semicircle with a piecewise degree $2$ polynomial?

shintuku

i can prove $\sup(A+B) = \sup A + \sup B$

copper.hat

:-)

prove $\sup -A = - \inf A$.

robjohn

10:36 PM

@leslietownes yes, elliptic curves are not yellow.

copper.hat

but they do have fits.

pika pika

hey chat

hey chatter

I have an exercise in a group theory problem set that is getting a bit boring

copper.hat

10:41 PM

so you want to share the boredom?

robjohn

Oh, no. I hope that Pikachat does not become a mascot here. Pika... Pika... Pikachat!

copper.hat

be forewarned, i am a 'wirey personality'.

Lucas Henrique

I have to find all the permutations from $S_8$ that commute with $(1234)(5678)$. Counting how many there are is easy using group actions and so. But I can't find a way to briefly describe the group of $32$ elements that commute with this guy.

Onir

10:57 PM

conmuting by $\sigma$ will give you permutation $(\sigma(1),\sigma(2),\sigma(3),\sigma(4))(\sigma(5),\sigma(6),\sigma(7),\sigma(8))$ So if you decide what $\sigma(1)$ is all of the $\sigma(i)$ with $i\in \{2,3,4\}$ are determined, and then you have $4$ options for $\sigma(5)$ and the rest are determined again.

Oh nvm there's a better one I think

leslie townes

thinking more constructively, the elements (1234)^a (5678)^b for 0 <= a, b < 3 will commute with that and be distinct elements. that gives you 16 elements. i think the involution sending 1 <-> 5, 2 <-> 6, 3 <-> 7, 4 <-> 8 or something similar to that might also commute. and maybe those generate everything.

Onir

They are generated by $\sigma$ and $\theta$ and $\gamma$, where $\sigma$ is the first cycle, $\theta$ is the second cycle, and $\gamma$ is the involution that transposes $I$ and $4+i$.

leslie townes

what onir just said.

Cornman

Hello, I wanted to ask the fellow math students, and preferably researchers on how a "mathematical day" should look like. I have always the feeling that I can not find a good balance between solving exercises, studying lecture notes, studying from books and so on. Especially when I attempt a problem and can not figure it out, I often feel like I would have been better off just attempting easier exercises or study more from the books, to advance.

Onir

11:12 PM

how is the rank of a matrix over a ring defined?

the size of the largest non-zero minor?

or is it left undefined?

oh you use module theory and have row rank and column rank and then they can be different

leslie townes

11:45 PM

cornman it varies a lot from person to person. everybody has different strengths and weaknesses. i think it is common to want to study theorems and theory maybe a little too much at the expense of understanding specific problems, key examples, and so on.

with the caveat that particularly at the undergraduate level and particularly with the internet being a firehose of mushy nonsense, it is sometimes difficult to find good sources of problems and sometimes easy to confuse yourself with problems that presuppose different definitions or a different set of working tools that the solver has access to.

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