Mathematics - 2023-11-22

Obliv

03:00

How can I understand Euler's formula better

How come it follows that a wave can be expressed as $\cos(kx-\omega t)$ or $e^{i(kx-\omega t)}$?

(without simply stating the formula I guess)

$e^{i\theta}$ in general is cyclic, which doesn't seem intuitive

Since we're raising $2.718...^{\sqrt{-1}x}$ where $x\in \mathbb{R}$

without $i$, it'd obviously not be cyclic..

Thorgott

03:16

think about the exponential in terms of the differential equation it solves, $(e^x)^{\prime}=e^x$

it follows that $(e^{ix})^{\prime}=ie^{ix}$, so the velocity of this curve is always (counterclockwise) perpendicular to the current position

it's intuitive to see geometrically/physically that the motion that describes is circular

copper.hat

03:35

Or, equivalent to @Thorgott's suggestion the 2d real equations $x'=y, y'=-x$. no need for fancy exponents.

looking at the time derivative of $x^2+y^2$ shows the radius is constant.

it helps to think of numbers as stretching & rotating.

Ted Shifrin

03:53

Complex numbers.

Cotton Headed Ninnymuggins

What's an easy way to find if $2$ is a square $\mod 2009$? Euler's Criterion doesn't apply since $2009$ is composite. In general, what do I do to easily determine if $a$ is a square $\mod b$ when $b$ isn't prime?

Obliv

@Thorgott I appreciate you trying to make it more physically intuitive, but what do you mean velocity of this curve is always (counterclockwise) perpendicular to the current position? Are we taking $e^{ix}$ as the position? I guess I should be thinking of this in the complex plane?

@copper.hat $\frac{d}{dt}(x^2+y^2)=2xx'+2yy'=-2yy'-2xx'$?

copper.hat

go one more little step

you made a mistake somewhere

Obliv

ah i did it in my head one sec.

I just subbed in the $x'=y,y'=-x$ into that

copper.hat

and you got...

Obliv

04:05

was I supposed to cancel to get $\frac{d}{dt}(x^2+y^2)=0$?

copper.hat

jeez, just substitute the values for the derivative

use paper

you have made no substitutions above, and one odd arithmetic mistake.

Obliv

I..I did. The ordinary time derivative is $x\dot{x}+\dot{x}x+y\dot{y}+\dot{y}y=2x\dot{x}+2y\dot{y}$ $\dot{x}=y$ and $\dot{y}=-x$ so $2(-\dot{y})(y)+2(\dot{x})(-x)$

copper.hat

look again. come on.

Obliv

so $-2(\dot{y}y+\dot{x}x)$

ok ok

copper.hat

you did not do any substitution.

Obliv

04:09

Waiit.. do I only substitute the derivatives lol

copper.hat

look, i really hate repeating myself unless i am whining about something.

Obliv

I was substituting the $x$s and $y$s too

copper.hat

replace the derivatives by their rhs

Obliv

Mmk so we have $2xy+2y(-x)=0$

There we go :p

copper.hat

whew

so the solutions have constant radius.

it is a Lyapunov function, in case you care.

anyhow, from this you can show that it is periodic, and define $\pi$.

Obliv

04:12

Seems oddly relevant to a lot of things that I've been covering in physics..

copper.hat

Rudin does this nicely in the first few pages of real & complex analysis

Obliv

it reminds me of the rotation matrix, also cross product definition of angular momentum and torque

copper.hat

the matrix exponential is fundamental to a lot of physics

Obliv

$x'=y$ and $y'=-x$ then $x''=-y$ and $y''=x$?

and so on?

wait that looks awful

I'm just thinking of it as $y=ie^{ix}$ and $x=e^{ix}$ so $y' = -e^{ix}$

and $x'=ie^{ix}$

but then $x'' = -e^{ix} = y' = -x$

copper.hat

no, its not awful. it leads to the taylor expansions of $\cos, \sin$.

Obliv

04:17

but wouldn't $x'' = -y$ mean $x'' = -ie^{ix}$

copper.hat

decide if you want to work in $\mathbb{R}^2$ with matrices, or in $\mathbb{C}$ with scalars.

Obliv

i think I wrote that wrong

Ohh

Yeah I'm mix matching

copper.hat

alternatively, you could define the exponential as $e^z = \lim_n (1+ {z \over n})^n$.

user 726941

08:05

::crickets::

🦗🦗🦗

or

🏏🇦🇺🥇🏆🏏

🇮🇳😭🇮🇳

Soumik Mukherjee

The set of even degree monomials is dense in $C[0,1]$ but not in $C[-1,1]$

So domain matters

leslie townes

the linear span of the set of even degree monomials? :)

user 726941

but, a monomial can be defined as just a constant,

Bumblebee

08:21

@Koro How does a banana with a hole illustrate the fixed point theorem?

user 726941

a variable, or a product of a constant and one or more variables :)

where a polynomial is a monomial or sum of monomials.

Soumik Mukherjee

08:39

@leslietownes yes, I should've said spannend by

Mad

08:55

52

Q: When is the image of a linear operator closed?

Let $X$, $Y$ be Banach spaces. Let $T \colon X \to Y$ be a bounded linear operator. Under what circumstances is the image of $T$ closed in $Y$ (except finite-dimensional image). In particular, I wonder under which assumptions $T \colon X \to T(X)$ is a bounded linear bijection between Banach spac...

in this proof, he uses the fact, that the operator is Continiosu. Why is this true?

one potato two potato

bounded linear operator in a Banach space is continuous

Mad

but it is not bounded

It is bounded from below

math.stackexchange.com/questions/2686125/… this answer even says its not true

i refer to the comment of Jonas Meyer. to prove the statement, one needs to assume boundedness, IE continuoty.
I got my answer ) i didnt catch that part

one potato two potato

09:26

I've been to a conference once and my impression is that it's more like a social meeting than I thought.

there are people first than math

Lucky Chouhan

11:03

What is your Erdos Number?

one potato two potato

The field was not quite related to Erdos so there wasn't such conversation.

Lucky Chouhan

@onepotatotwopotato Now my Erdos number is infinity

Koro

11:29

@SoumikMukherjee yes, you can almost see this using Weierstrass's approximation. Generalized version of it tells you that there is no reason to believe denseness of even degree polynomials in C[-1,1].

Jakobian

Your Erdos number is imaginary

@Koro there's an easier way to see it. Why would an even function approximate a function that isn't even?

Koro

nvm, I misunderstood. I didn't know what I was talking about when I said something about Erdos number. I said so thoughtlessly hence I deleted the two messages.

@Jakobian yes. Such things often don't come to mind in timed settings say some exam. So naturally, one would think of WA when asked about this.

WA which is just an application of Fejer's Kernel shows the density in C[0,1].

Jakobian

WA can definitely be generalized to much wider settings, which makes it more than an application of Fourier theory

Stone-Weierstrass theorem

psie

It is said that if $f:G\to\mathbb C$ is analytic and $a\in G$ satisfies $f(a)=0$, then $a$ is a zero of multiplicity $m\geq1$ if there is an analytic $g:G\to\mathbb C$ such that $f(z)=(z-a)^mg(z)$ where $g(a)\neq 0$. By definition $(z-a)^m=e^{m\log(z-a)}$, which isn't defined at $z=a$, so how can $f(a)=0$?

Koro

I realised much later that Rudin's pf. of SW develops Fejer kernel and uses it as when I studied it the first time, I didn't know Fejer's kernel.

Having learnt Fejer kernel, the proof seems intuitive. 😊

@psie m is an integer.

Jakobian

11:41

@psie you got it wrong. Here $m$ is a natural number, its just powers of natural numbers

@Koro I doubt he proves Stone-Weierstrass theorem

psie

hmm, but it's the complex power function...which is defined in terms of the complex logarithm. Even for $m$ being a (positive) integer, $z^m$ is not defined at $z=0$, right?

Koro

@Jakobian well, he does.

Jakobian

Which book

Koro

pma

(7.26 onwards)

Soumik Mukherjee

@Koro yes, and on $C[a,b]$ with $a>0$ we have the Müntz–Szász theorem to show denseness

Jakobian

11:47

You can just use Stone-Weierstrass directly

@Koro not for his Stone-Weierstrass theorem proof though. For his Weierstrass approximation theorem

Soumik Mukherjee

@Jakobian Directly how? We don't have all the polynomials

Koro

@Jakobian he does in 7.32.

Jakobian

@Koro he proves SW theorem, but not using Fejer kernels or similar analysis

@SoumikMukherjee SW theorem says that a subalgebra of $C(X)$ where $X$ is compact Hausdorff with non-zero constant functions is dense iff it separates points

Here if $X = [a, b]$ with $a\geq 0$ then $x^2 = y^2 \implies x = y$ so it separates points

Soumik Mukherjee

ohh okay

Jakobian

Similarly if $b\leq 0$. But if $a < 0 < b$ then it fails for $c = \min(|a|, |b|)$.

Koro

12:00

@Jakobian I should have said 'kernel' not necessarily Fejer's kernel. P_n(x) = f* Q_n(x)

Jakobian

That proof isn't of SW theorem

Koro

Q_n is a kernel a.k.a. approx. to the identity.

Jakobian

Its of WA theorem

Koro

I mixed up the abbreviations above -WA and SW. my bad.

@SoumikMukherjee you should also check the subalgebra of even deg. polynomials to be non -vanishing on [a,b], a>0, which is obvious but important. Then your result follows by SW.

Jakobian

Thats included in them containing a non-zero constant

Soumik Mukherjee

12:10

Okay, I was thinking about WA when I wrote the earlier comment:/

Jakobian

Also even degree polynomial $\neq$ even polynomial

Koro

Oh, Rudin uses the terminology -non-vanishing and I'm used to that.

Jakobian

@SoumikMukherjee You can prove this for $[0, 1]$ using WA like this: approximate $x\mapsto f(\sqrt{x})$ by polynomials.

Koro

typo: even polyn. not even degree poly.

There exist polynomials f_n(x)---> f(x) uniformly on [0,1] so f_n(x^2)--> f(x^2) and f_n(x^2) are even polynomials.

Soumik Mukherjee

12:27

@Jakobian Okay, so if $f(\sqrt{x})$ is approximated by $p(x)$ then $f(x)$ is approximated by $p(x^2)$, which is an even function.

Koro

yes.

Lucky Chouhan

15:24

@Jakobian what about you?

Jakobian

what about me

Lucky Chouhan

Once Brian Greene was told in 'World Science Festivals'" video that "I used to go to college with my wife" while talking about Paul Dirac with Leonardo Suskind and Author of 'The Strangest Man'.

@Jakobian Your Erdos number.

@XanderHenderson do you know any couple who are both teacher at university?

@Jakobian what if I email you?

Jakobian

I've already said that I don't like answering personal questions and I will refrain from doing so

Tony Pizza

In a book I was reading there was a commutative ring $A$ and a commutative algebra over it $B$. Then there are modules which are projective over $B$ call the $X_i$. Then there is statement "Modules X_i are projective, hence become free (of rank 1) over some ring $C$ containing $B$. I tried searching but couldn't find the proof of this result. If someone can lead me to it I will glad.

Lucky Chouhan

@Jakobian I am wondering, what type of person you're in-person. Imagine in campus someone ask you your name then "You'll be like : I don't like answering personal questions" that guy would be like "WTF".

Jakobian

15:30

Can you just respect my personal space, please?

Soumik Mukherjee

@LuckyChouhan Politeness is chasing you but you're always faster

3

Lucky Chouhan

@SoumikMukherjee oh bro, btw how are you doing?

@SoumikMukherjee this made me laugh so hard :))

Jakobian

15:55

@Thorgott Tony asked a question you might be able to answer

ChoMedit

Could someone define that one continuous curve is tangent to another continuous curve? Could tangency be extended to the $C^0$-regularity?

shintuku

16:19

@Thorgott I vaguely remember you talking about topos theory and Hegel, did you ever do any work on these?

Jakobian

16:42

Fun fact, is that the reason for our definition of a topological space is that people were seeking a definition that could be applied to subsets of $\mathbb{R}$. That is, subspace topology.

Today we take subspaces of topological spaces as a God given fact

Note: This is subject to me more or less twisting the story

Jakobian

17:17

-2 C hits differently

Thorgott

17:34

@TonyPizza This seems to miss context. For example, a module that is free is projective and remains free of the same rank when extended along any ring homomorphism $B\rightarrow C$, so the "of rank $1$" statement certainly means there are further assumptions on the $X_i$. I wonder also if there are further assumptions on $B$.

@shintuku I probably mentioned that to make fun of it.

shintuku

argh

Ted Shifrin

18:03

Don’t let @Thor’s serious demeanor fool you!

Lucky Chouhan

Hello Ted.

Hello Copper-Hat

@copper.hat how did you decide to start a business and leaving academia?

copper.hat

@LuckyChouhan I was never in academia apart from graduate studies.

Lucky Chouhan

18:23

But you have Phd @copper.hat

@copper.hat were you fully sure that you don't wanna do research or teaching?

copper.hat

@LuckyChouhan Yes. I liked teaching & research, but I could not handle the politics.

Lucky Chouhan

Politics? What kind of politics in teaching?

@copper.hat What do you think how different Copper-hat who was in graduate school from where you're now sometimes when you sometimes remember

I typed 'sometimes' two times :( I am terrible at Math and English.

Jakobian

18:39

@LuckyChouhan don't email me with stuff like that

Ted Shifrin

19:15

Talk about chutzpah!

psie

19:28

Consider the complex power function $z^\mu$ with $\mu=a+ib$ fixed and the branch $0< \arg z <2\pi$. Is this computation correct? $$|z^\mu|=|e^{(a+ib)(\log|z|+i\arg z)}|=e^{a\log|z|-b\arg z}=|z|^{\mathrm{Re}(\mu)}e^{-\mathrm{Im}(\mu)\mathrm{Im}(\log(z))}.$$ My teacher claims the "growth" of this function only depends on $|z|^{\mathrm{Re}(\mu)}$. I guess they have a point, since $e^{-\mathrm{Im}(\mu)\mathrm{Im}(\log(z))}$ would be bounded for fixed $\mu$ and on the branch given.

Ted Shifrin

Yup.

psie

Cool, thanks for checking!

Ted Shifrin

It’s basically no different from $|e^z| = e^{\Re z}$.

psie

👍

user 726941

Edward Witten - How is Mathematics Truth and Beauty?

►

Jakobian

19:58

@psie You're sunny, right?

psie

@Jakobian indeed :) you can call me whatever, but psie should be pronounced as pee-zee (it actually refers to my name)

Jakobian

oh, my next question would be what psie means

psie

p is just the letter in my first name, sie the three letters in my last name, but it is an actual nickname ('pee-zee')

Ted Shifrin

As in the American English "eazy-peezy."

psie

exactly :)

Ted Shifrin

20:15

When will I learn not to bother with questions like this?

Ted Shifrin

20:29

Precisely. Are you agreeing or disagreeing? (The poor OP probably didn't get the guy's point.)

Jakobian

Both at the same time

user 726941

Neither, actually.

Ted Shifrin

So what was your point?

Jakobian

See, you avoid danger by stating a neutral opinion. I take it on by stating a controversial opinion no matter what option is correct.

Ted Shifrin

It's always helpful to confuse a confused OP even further.

shintuku

20:38

OP must atone for their sins

user 726941

Nice play on the word "tone."

shintuku

?????????????????? i was just saying that OP must suffer

user 726941

💯🥇

By being the recipient of a condescending tone.

shintuku

whatever makes OP suffer

it was an unmediated and unabated wish of ill-being to a complete stranger. but then again, OP is structurally guilty. and if they aren't yet, they will probably be soon enough

Jakobian

21:24

better be safe than sorry...

Ted Shifrin

@user726941 No, the condescending tone was to me.

user 726941

21:44

I apologize then for my misunderstanding @TedShifrin

mick

23:20

any ideas on this ?

0

Q: For all complex $|z| \neq 1$ : $\frac{z}{1+z+z^{2}+z^{3}+z^{4}} = \sum_{n=0}^{\infty} T_n \frac{z^n}{1+z^n+z^{2n}}$?

Inspired by this one For all complex $|z| \neq 1$ : $\sum_{n=0}^{\infty} w_n \frac{z^n}{1+z^n+z^{2n}+z^{3n}+z^{4n}} = \sum_{n=0}^{\infty} u_n \frac{z^n}{1+z^n+z^{2n}}$? It made sense to me, to take the terms apart and split the problem into 2 subproblems. Allow me to explain. if $\sum_n a_n x_n =...

sequences-and-series taylor-expansion rational-functions analytic-continuation cyclotomic-polynomials

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$Mathematics$ Mathematics