Mathematics - 2023-10-01

shintuku

12:40 AM

"and nobody will know what that person does or why they are paid so much" hilarious

i wonder whether there's some sort of high-level view or diagram that makes sense of the bureacratic apparatus of a university

cool article reference

EE18

2:25 AM

If anyone is able to comment on my last question in the chat above I would greatly appreciate it :)

leslie townes

3:03 AM

EE18: i'm not sure i understand the scrollback. there is the definition of multiplication in formal polynomial rings which is as you note a matter of definition. there is also just a formula for multiplying "polynomials in an element of an arbitrary ring." it sounds like you're looking to prove the latter and if so, yes, induction is one way to do that. lots of choices in how to structure the argument.

e.g. fix one of the polynomials and prove by induction on the degree of the other one, or induct on the total or maximum of the degrees of the pairs of factors, or something

if you get stuck you can probably google it. there will be a lot of results relating to infinite series but it wouldn't surprise me if it is also just an algebra book exercise.

MathCrackExchange

3:27 AM

@LukasHeger how much are you looking to make for 1 hour of tutoring?

EE18

5:40 AM

Yes I am looking for the latter @Jakobian :)

you would induct on the degree of one polynomial? I ask because I've been unsuccessful finding this online

leslie townes

there are all sorts of ways to structure the induction. i wouldn't let my recommendation be 'the way.'

you'll need to be clear about what your definitions are, though. you'll probably have, i dunno, a set of formulas for coefficients that is itself defined inductively. somewhere there is a definition having properties that are verified by induction.

i'd wonder slightly why you are bothering with this, if this is self study.

less important than the abstract "any definition that uses sigma notation can be thought of as an inductive definition" concept is being able to work with sigma notation, e.g. being able to re-index two different things written in sigma notation so that they can be clearly added together without the usual thing calculus books do, where they write out a few representative terms and do " . . . "

the best practice for this kind of thing might still be in calculus books, however, and stuff involving manipulation of series.

or maybe the right kind of discrete probability theory textbook.

look at the hash that the questioner and various answerers make of math.stackexchange.com/questions/1937543/… for example. although the green box (interpreted as suggested by the commenter) is a good example of something that it would make sense to base a definition around, and prove by induction. in that example the inner sum on k in the RHS is itself defined inductively.

math.stackexchange.com/questions/4778584/… is a cool question. someone asks how google sheets fills in 'smoothed' curves to fit a finite set of points, and someone posts various plausible guesses that come visually close but do not quite get there.

a good experimental question.

i'm assuming that google does not simply document how it does that somewhere, in which case it is a less interesting question.

leslie townes

6:33 AM

10

Q: Lower bounding $|1+z+\cdots + z^{n-1}|$ when $z\approx 1$

I am trying to lower bound $|1+z+\cdots + z^{n-1}|$ when $z$ is a complex number close to $1$ (and $n$ is sufficiently large). My main concern occurs in the case $z = 1 + it$, where $t$ is small. In these cases, the terms of the geometric series spiral around the origin, but because $|z| > 1$ t...

cv.complex-variables

also cool

Soumik Mukherjee

6:45 AM

A Co-contributor of the site is an ntributor

Thomas Finley

7:20 AM

0

Q: Solve the system of equations $x_1+10x_2-x_3=3,2x_1+3x_2+20x_3=7,10x_1-x_2+2x_3=4$ using the Gauss-Elimination with partial pivoting.

Solve the system of equations $$x_1+10x_2-x_3=3,$$$$2x_1+3x_2+20x_3=7,$$$$10x_1-x_2+2x_3=4$$ using the Gauss-Elimination with partial pivoting. I tried solving the problem as follows: We have the augmented matrix as $(A,b)=\begin{pmatrix} 1 && 10 && -1 && 3\\ 2 && 3 && 20 && 7\\ 10 && -1 && 2 && ...

numerical-methods systems-of-equations gaussian-elimination

Need some help with this.

Shashaank

9:30 AM

integral from 0 to pi tanx dx is divergent right?

since integral tanxdx from 0 to pi/2 is divergent!

right?

leslie townes

yes

Shashaank

Thank you :-)

Lukas Heger

9:56 AM

@MathCrackExchange 15$/h

sunny

11:43 AM

I'm working with a non-zero sequence characterized by increasing by a factor of $2$, i.e. $$x_n\geq 2\cdot x_{n-1}.$$ Does this sequence diverge?

shintuku

@LukasHeger do you do comm algebra/algebraic geo tutoring?

sunny

@sunny My educated guess is yes, since $x_n\geq 2x_{n-1}\geq 2^2x_{n-2}\geq\ldots\geq 2^{k-1}x_1$, and so as $k\to\infty$, the RHS of the inequality tends to $\infty$.

Lukas Heger

@shintuku yes. I'm more familiar with scheme theory than classical AG though

shintuku

where could I contact you?

Lukas Heger

do you have Discord?

shintuku

11:52 AM

yeah

Sha Vuklia

12:55 PM

@F.White I've made progress. Using (a) I can show that $\mathbb Z[1/2+1/2\sqrt{-3}]$ is a PID, since (a) tells us that taking the Euclidean norm is a norm function

so now what's left to do is argue (b) using that $\mathbb Z[1/2+1/2\sqrt{-3}]$ is a PID. yesterday I thought I had an argument, but it was faulty, so I'm revising it now

Jakobian

@EE18 you meant leslie?

user123234

can someone help me here?

1

Q: How can I show $\|XMY\|=\|M\|$?

Let $M\in \operatorname{Mat}_{d\times d}(\Bbb{C})$ and define $\|M\|:=\sqrt{\sum_{i,j} |M_{i,j}|^2}$. Let $X,Y\in \operatorname{GL_{d\times d}}(\Bbb{C})$ s.t. $X^HX=I, Y^HY=I$. I want to show that $\|XMY\|=\|M\|$ My Idea was the following. First we know that $\langle AX,Y\rangle=\langle X, A^H...

real-analysis matrices functional-analysis analysis inner-products

EE18

2:13 PM

Ah yes I did mean Leslie, my apologies @Jakobian I got the icons mixed up!

Sha Vuklia

2:37 PM

@LukasHeger does you offer still holds that I can ask alg numb thy questions? (I did some reading last year, but this year I'm actually taking the course)

EE18

@leslietownes To your point, I'm just trying to make rigorous the argument for reindexing the sum and what not but I don't know why I can't seem to find the argument online. Seems like none of the polynomial related wikipedia articles have it

Lukas Heger

2:57 PM

@ShaVuklia sure! I love to answer alg num thy questions :)

user123234

I need to find $M\in M_2(\Bbb{R})$ s.t. $e^M=\left(\begin{matrix}
-1 & 0 \\
0 & -1
\end{matrix}\right)$. They wrote that the eigenvalues should satisfy $e^\lambda=-1$ but I mean then I don't get a matrix with real coeff. Is there an error? @LukasHeger

EE18

I see how we have $c_k = \sum_{i,j,\, i+j = k} a_ib_j$ in polynomial multiplication, but how we get that to our classic $c_k = \sum_{i = 0}^k a_ib_{k-i}$ I can't quite see rigorously

Lukas Heger

3:22 PM

@user123234 do you know about how you can embed complex numbers into real 2x2-matrices?

you can use that

you know that $e^{\pi i}=-1$

now $\pi i$ corresponds to the matrix $\begin{pmatrix} 0 & \pi \\ -\pi & 0\end{pmatrix}$

@user123234 your mistake is that a real matrix can have complex eigenvalues

Sha Vuklia

@LukasHeger great! I've also asked the question on the main site (with no response). shall I forward the link or shall I rephrase it here?

Lukas Heger

just linking is fine

Sha Vuklia

1

Q: Fractional $\mathbb Z[\sqrt{-3}]$-ideals

I want to show that each fractional ideal of $\mathbb Z[\sqrt{-3}]$ is of the form $\mathbb Z[\sqrt{-3}]a$ or $\mathbb Z[1/2+1/2\sqrt{-3}]a$ with $a\in\mathbb Q(\sqrt{-3})$. Initial considerations: Write $R=\mathbb Z[\sqrt{-3}]$ and $S=\mathbb Z[1/2+1/2\sqrt{-3}]$. Given a fractional ideal $I$ of...

number-theory algebraic-number-theory ideals

Lukas Heger

it's enough to show that any $R$-fractional ideal is a $S$-fractional ideal, right?

Sha Vuklia

but that's not true right?

Lukas Heger

3:28 PM

hmm, yeah

this would be easier if one can use the conductor of an order

Sha Vuklia

unfortunately I'm just at the beginning of the course, so we have very little to work with

Sha Vuklia

3:47 PM

My guess is that the non-invertible ideals are those of the form $S\cdot a$, since I also have to show that the Picard group is trivial

Lukas Heger

Let $I$ be a fractional ideal of $R$. Consider $\{x \in K:xI \subset I\}$. One can check that this is a subring of $K$

it contains $R$ and is contained in $S$ (because any element has to be integral)

There is no nontrivial intermediate ring between $R$ and $S$

so we have two possibilities: $\{x \in K:xI \subset I\} = S$, then $I$ is in fact a $S$-fractional ideal and hence of the form $aS$. Or $\{x \in K:xI \subset I\}=R$, then this implies that $I$ is invertible

Sha Vuklia

ah right, our notes call that the multiplier ring $(I:I)$

I have to convince myself why it is contained in $S$

Lukas Heger

do you know different characterizations for when an element is integral?

and do you know that $S$ is the set of all integral elements over $\Bbb Z$ in $K$?

Sha Vuklia

yes, $S$ is the ring of integers of $\mathbb Q(\sqrt{-3})$. I am aware of the characterizations, but I have to refresh them, so give me one moment

I suppose you're thinking of $s\in K$ is integral over $\mathbb Z$ iff $\mathbb Z[s]$ is a finite $\mathbb Z$-module?

Lukas Heger

4:02 PM

not quite

$s \in K$ is integral if there exists a faithful $\Bbb Z[s]$-module that is f.g. over $\Bbb Z$

Sha Vuklia

hm, I might have seen that in Atiyah-Macdonald a long time ago

ah yes I have, Prop 5.1 (iv)

Lukas Heger

but it's useful here

Sha Vuklia

ok, so my guess is that $I$ would be our fatihful $\mathbb Z[s]$ module

Lukas Heger

at this point, all that is left is to compute the Picard group of $R$, which you said you wanted to do anyway

@ShaVuklia correct

Sha Vuklia

I guess because we have a domain

it is faithful

and it's a module by the above

Lukas Heger

4:09 PM

yup

Sha Vuklia

ok, cool, let me parse the remainder of what you wrote

ok, I'm convinced there is no intermediate subring

@LukasHeger that is true, but the exercise I'm doing asks me to first prove that each fractional $R$-ideal is of the forms mentioned, and after that they ask to show that $R$ has trivial Picard group, so my intention was to use the description to show that the Picard group is trivial

my guess would be to show that an $R$-fractional ideal of the form $Sa$ is not invertible, from which we can deduce that the Picard group is trivial

Sha Vuklia

4:34 PM

hm, and I also have to see why $(I:I)=R$ implies that $I$ is invertible

I've shown in a previous exercise that if an ideal is invertible, then it is proper (i.e., $(I:I)=R$)

so the logical way would be to argue that if $(I:I)=R$ then $I=Ra$, and then with the exercise just mentioned I find that the Picard group is trivial

Sha Vuklia

5:19 PM

I've hit a wall on this one, so I'll leave it for today. I will probably ask my TA on Tuesday how that to show $(I:I)=R\implies Ra$. I'm really not comfortable with algnumb, so having solved the main part of the exercise (there were several subexercises) is already an achievement :')

MathCrackExchange

5:35 PM

@LukasHeger let's make it $20 / hr

Lukas Heger

okay :)

MathCrackExchange

So I can't start rn, but maybe later today

what time is it where ur at

Lukas Heger

19:36

MathCrackExchange

Thx, let's connect over email

fruitful approach (one word) at the mother g-ship

Email me!

Lukas Heger

email sent

MathCrackExchange

5:41 PM

K uploading the book now it's 80MB

K, so I'm confortable starting as far as convergent power series

We could use Discord, MS Teams, Google Meet, Zoom, MSE Chat rooms

Koro

H is a Hilbert space. e_n's are orthogonal basis vectors of H. T:H-->H is bounded linear operator. Is it possible that T(e_n) =e_1 for all n>=1?

Monty

Hi guys, if you have a function which is lipschitz and infinitely differentiable it is straight forward to show its derivative is bounded. Can anything be said about higher order derivatives?

Koro

Its solution says: clearly it is not possible.

Lukas Heger

@MathCrackExchange I've made an MSE chatroom

Koro

@Monty on a closed & bounded interval, yes.

Monty

5:48 PM

hmmm

so on the whole space there is no such condition that guarantees uniformly bounded derivatives

robjohn

6:28 PM

@Monty Not if your whole space is not compact.

Monty

no yeh I meant $\mathbb{R}$.

robjohn

For example $f(x)=x^2$ on $\mathbb{R}$.

Monty

ok thanks:)

Jakobian

@Koro Try $x_n = \sum_{k=1}^n \frac{1}{n}e_n$

$x_n\to \sum_{k=1}^\infty \frac{1}{n}e_n$, but $\|T(x_n)\|\to\infty$

Gwyn

6:53 PM

Is this proof correct? Problem: show that if $a_n>0$ and $a_n \to +\infty$ then $\frac{a_n}{e^{a_n}} \to 0$. Proof: since $a_n>0$, we have $\frac{a_n}{e^{a_n}}=e^{\log a_n -a_n}=e^{-a_n \left(1-\frac{\log a_n}{a_n}\right)}$. Since we proved that $\frac{\log x}{x} \to 0$ when $x \to +\infty$, from the theorem that binds limits of functions with limits of sequences I $\frac{\log a_n}{a_n} \to 0$ as well because by hypothesis $a_n \to +\infty$

Hence $-a_n \left(1-\frac{\log a_n}{a_n}\right) \to -\infty$ and finally $e^{-a_n \left(1-\frac{\log a_n}{a_n}\right)}\to 0$.

In our course, we proved both the theorem I quoted and the fact that $\lim_{x \to +\infty} \frac{\log x}{x}=0$; just to give context. :D

Gwyn

7:09 PM

Re-reading this, I could shorten this by using that $x/e^x \to 0$ as $x \to +\infty$.

Ted Shifrin

@Monty Hardly likely without more info.

@Gwyn If you know that, yes.

Gwyn

Thanks for checking that, Ted! I suspect that a more elementary proof is required, I will think about that a little more. In the worst case scenario, I can always invoke this proof. :D

Ted Shifrin

This is equivalent to that statement about limiting growth of $e^x$.

Which is equivalent to the statement about limiting growth of log.

Gwyn

Yes, indeed initially I was trying to use the ratio test or trying to find an inequality. I used $e^{a_n} \ge 1+a_n$ without success. I want to think about this more. Math is so fun.

user123234

7:24 PM

I would like to show that $e^X=\lim_{n\rightarrow \infty}\left(I +\frac{1}{n}X\right)^n$ where $X\in M_d(\Bbb{C})$. We should use the log-function as a hint. But I only know that $\log(e^X)=X$ for $\|X\|<\log(2)$ and this additional assumption is not made in the exercise. Can someone give me a hint?

Ted Shifrin

@Gwyn Go one term more.

Gwyn

That's trueeee, adding one more term gives me the inequality with $\le$! This works! How could I not see that, I even knew that

Thanks again Ted, part of my graduation will be thanks to you :D

Maybe I need two more terms? Because I need $e^{a_n} \ge 1+a_n+a_n^2/2+a_n^3/6$.

Ted Shifrin

8:03 PM

Nah, the quadratic does it.

Gwyn

8:16 PM

You're right, I don't know why I thought the inequalities obtained from Taylor's expansion alternate for the exponential as well, but that's not true. So the quadratic is indeed enough. Thanks!

Ted Shifrin

8:31 PM

@user123234 Doesn’t it suffice to prove it for sufficiently small $X$?

user123234

@TedShifrin I'm not sure. I have now tried to show it for diagonalisable matrices and then used that the set of diagonalisable matrices dense is

but I have not used the logarithm for the diagonalisable part, I used the binomial theorem and the serie definition of $e^X$

Ted Shifrin

8:46 PM

Right. I would not have tried to use log. But if it holds for $X$, doesn’t it follow that it holds for $cX$?

user123234

It should hold for $cX$ but I don't see where you want to go

Ted Shifrin

Then we’re done if you prove it for all small $X$.

user123234

ah okey I see what you mean so I could then take the property that $\log(e^X)=X$ for $\|X\|<\log(2)$

Ted Shifrin

9:04 PM

Right, if you see how to prove what they claim.

robjohn

11:24 PM

@user123234 Can you assume that $\lim\limits_{n\to\infty}\left(1+\frac1n\right)^n=e$?

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$Mathematics$ Mathematics