There's all sorts of problems with just writing integral signs and not making things precise. He's replacing $y$ with $t$, but what he's written makes no sense.

okay. Is my integration correct?

I can't follow any of this;

$$\phi =\int Q dy - \int Pdx+h(z).$$

@TedShifrin :(

Hi @Ted

Watching a youtube video where a guy tackles $lim_{n\rightarrow \infty}(ln(n^n))^{1/n}$, and the first step he takes is to just rename the $n$ to $x$, which confuses me, at it strikes me as entirely unnecessary.

@TedShifrin Can you give hints. How do I find $U$?

@Mathgeek: Are you trying to do the algorithm for a specific example or write a proof?

hi @Alessandro

Hi @Ted @Alessandro

@TedShifrin proof

Hi @Mathei how is it going?

Pretty well, thanks. And yourself?

but I don't know how to start

Do anyone have artin book of abstract algebra that want to sell ?

@MatheinBoulomenos @TedShifrin Hello !

hi @Mathein

hi @Jacksoja

I just learned that the ANT3 exam is going to be a lot more doable than I thought

@MatheinBoulomenos $\mathcal O_X$-modules and quasicoherent sheaves are killing me

But pretty well apart from that

@TedShifrin Can you give hint?

@Mathgeek: I suggest you start by writing definite integrals with limits, and keeping track of variables very carefully. What that person never does (but you should) is differentiate under the integral sign and use the curl = 0 condition after you do so.

@AlessandroCodenotti ah, I think your AG class is going faster than the one here

@Mathgeek: The proof I prefer to do is to write down the integral from the origin to a point $(x,y,z)$ along the line segment. Parametrize, write it down directly. Then show that the partial derivatives of that function are the right things by using the curl = 0 condition.

hi @MatheinBoulomenos

I think it's going too fast for me actually

Hi @LeakyNun

@TedShifrin okay. I will try

so handsome ppl , do anyone have a solution for my question ?

hi @Kasmir

copying it here

@MatheinBoulomenos mathein ! :D

Hey everybody

long time no see ! how is everyting ?

@JakeRose Hi

Pretty well, thanks

I should really be finishing my BA soon

@KasmirKhaan Kasmir should stop calling people "handsome"

and yourself?

@MatheinBoulomenos it is going good thanks ! =p

@MatheinBoulomenos what subject did you write on ?

or gonna present

number theory

neat , I would have guessed rep theory or lie groups

rep theory is involved

:D

I knew it !

if I do the convolution of $|\omega|$ with a constant (let’s say for simplicity it’s 1). How do I notate this? $\int_{-\infty}^{\infty} |\omega - y| dy$?

@KasmirKhaan Any reason you can't just use row operations?

@Rithaniel that is not what I really need =p

i want to find a map from the upper triangular matrices in GL (3,F) to some other group s.t the kernel of the map is identity matrix exept the entry 1,3 can be anything

am not sure if this is a good approch however

@TedShifrin @MatheinBoulomenos please take a look at my Q if you have some time

@Jake: $\omega$ is a function of what? You should put the variable in explicitly and then rewrite your convolution.

@TedShifrin so it's not a geometric reasoning?

@Kasmir: You're not sure it's a good approach? To what?

i want to study the quotient group

@Lucas: The geometry comes from my comment about the unit vectors being the same.

H = upper 3x3 matrices in GL_3

@Kasmir: What quotient group? That's the whole group.

K = matrices that look like identity exept that the entry 1,3 can be any number

H/K

Oh, and you want $H/K$?

yes :)

Is $K$ a normal subgroup?

i thought about a map for long time but could not fidn any

yes it is

if there is a map, dont tell me yet, just want to be sure, that such problems can be solved by finding amap

because using first iso theorem , will help me understand the quotient group by understandign the image

I do not want to think of infinite cosets

@TedShifrin I have to go for 20 mins to buy something, brb ! :)

@TedShifrin you mean by applying stokes theorem? by defining $U(x,y,z)=\int_{\gamma} Pdx+Qdy+Rdz$ . $\gamma$ is a polygonal line. from$(0,0,0)\to (x,0,0)\to (x,y,0)\to (x,y,z)$

No Stokes's Theorem in what I suggested.

You can do that path, but I suggested a different one. But that should work too. I said straight line segment from $(0,0,0)$ to $(x,y,z)$.

How do I parametrize then? $X(t)=xt,Y(t)=yt,Z(t)=zt,t\in [0,1]$

For mine, yeah. Your way might actually come out a little easier. Did you write it out?

Not able to reduce with this parametrization

$U(x,y,z)=\int_{\gamma} Pdx+Qdy+Rdz$

Go ahead and do yours. Write $$U(x,y,z) = \int_0^x P(t,0,0)dt + \int_0^y Q(x,t,0)dt + \int_0^z R(x,y,t)dt.$$

@TedShifrin But there is a problem. for the well definition of the function. integral must not depend on the path.

Here how do I guarantee that?

That follows from Stokes's Theorem, but if you define $U$ by a specific path and check that its gradient is correct, you don't need to worry about it.

By fundamental theorem of calculus I am getting $U_x=P(x,0,0)$, $U_y=Q(x,y,0)$ and $U_z=R(x,y,z)$

No you're not.

Only the $z$ one is correct.

There are $x$'s in both the second integrals.

There is a $y$ in the $z$ integral.

I am not getting. Can you do the first one?

I can imitate the second one.

Does anybody use the symmetric convention of fourier transform?

@TedShifrin Ted :D

am back !

@TedShifrin sorry about leaving before I got distracted

@Mathgeek: When you take the $y$ partial, you get $Q(x,y,0) + \int_0^z \frac{\partial R}{\partial y}(x,y,t)dt$.

Now proceed.

@Kasmir: I have no idea how to do your problem.

@TedShifrin Thank you. :)

@TedShifrin okay thank you anyway ! you are helping others also so you did not think about it much :D

@TedShifrin $U_y=Q(x,y,0) + \int_0^z Q_t(x,y,t)dt=Q(x,y,0)+Q(x,y,z)-Q(x,y,0)$ similarly $U_x=P(x,y,z)$. Thank you very much :)

There you go. That's a valid proof. That answer was just mostly nonsense.

mmm

@TedShifrin do you know anything about other conventions for FT?

Huh?

Ted! :D

Can you please latex that question so i can post it on main ?

sadly i dont know how to do it

No. Seriously. Learn.

I would type it in words but do you think that is allowed? did not post anything there since multivar course

okay :(

As in $F = \frac{1}{\sqrt{2\pi}} \int_{-\infty}{\infty} (...)dt$

essntially you just have a root 2pi for both forward and backwards transformation

Well, different people take different conventions there.

But that's usual, yes.

@KasmirKhaan LaTeX is possibly the easiest thing you could ever learn. Google the table of syntax’s and you can learn as you go

@JakeRose It involves matrices ><

3x3

i think if i explain it in words it could also work

but Ted will be angry at me for posting such thing on main

@Jake: For example, when I taught an applied math course out of Strang's book, he did not have the $\sqrt{2\pi}$ in the transform and had $2\pi$ in the inverse transform.

not just Ted many other ppl :D

when it comes to convolution, I think there should be a $\frac{1}{\sqrt{2\pi}} $ too

yeah I agree @TedShifrin

btw Ted , did you work with strang ?

Convolution has no factors. $f*g(x) = \int f(y)g(x-y)\,dy$.

sorry I phrased that horrendoulsy

when it comes to the FT of a product (or a FT of a convolution)

@Kasmir: I know him well. He also taught a recitation section when I lectured multivariable calculus for 350 students one year. That was cool :)

@Jake: I don't think it matters. If the FT has the factor, it has the factor with the convolution. If it doesn't, it doesn't.

@KasmirKhaan again. Go learn. You can do it fairly easily. And even I would get annoyed at you for doing that and I literally never use the main site for maths

@TedShifrin neat ! =p i watched his lectures on LA , they werent so involved like the ones you did but i think his courses was meant for non math ppl

@TedShifrin

ive finally got photo privelsges in chat

exciting times ahead

I actually find his lectures frustrating and his boardwork disappointing :P

LOL @Jake

@JakeRose where is the latex site?

that i can find the codes?

@TedShifrin if you watch the determinant lectures, it was pretty good for me

@KasmirKhaan just google latex symbols

@TedShifrin if you see here, the red follows the convention you spoke about, and the blue is the addition of symmetry convention. Resulting in an extra root 2pi

Aha, so that shows a good reason NOT to have the $1/\sqrt{2\pi}$ in the definition to start with, @Jake.

yes I’m afraid it does

$\Psi $

hmm

did not work

It did

$$\Psi $$

@KasmirKhaan you need to have a program that can read latex

I do not see it

aha

You need to use the LaTeX in chat bookmark, silly.

and how do I do that ._.'

It’s literally in the top right of your screen

plesse don’t spam the chat trying to learn

:D

if you go on the main site and type your question you can learn adequately there with the preview section beneath it

no, he's right

LOL

okay Kasmir shall return later :)

see you yall ! thanks for help !

@TedShifrin I’m sorry if I have ever been like this

I’m just gonna assume I have

LOL

I'm still stuck on his algebra question, regardless.

something I could do?

or attempt?

I dunno. You're busy enough.

Hi chat

Hi @Astyx.

What's the question ? :)

back to FT question anyway. So I’ve just done a question which was the FT of a product. It only worked out if I ignored the factors for the individual FT and kept the one for the convolution

hey @Astyx

What's a homomorphism defined on the group of invertible upper-triangular $3\times 3$ matrices whose kernel consists of matrices $\begin{bmatrix} 1 & 0 & x \\ 0 & 1 & 0 \\ 0 &0 & 1\end{bmatrix}$?

Nope haven’t done that stuff yet

Well, you get the product by doing the inverse transform to the product of transforms, so the factors are gonna mess up with your blue convention.

And Wikipedia says that is the correct result which I agree with

this is so confusing

Fourier transform conventions are like that

I defer to @Semiclassic.

ugh

that's my considered opinion on Fourier transform conventions

@TedShifrin thanks for putting up with my ramblings

I sorta like the convention that you get as a limit of Fourier series expansions. That's how I motivated the whole thing when I taught it.

@Semiclassical so if I use the symmetric convention. When I take the individual FT I should include factor too right?

Can you write out the convention you're using for the forward FT?

That'll avoid some headache.

He put in the $1/\sqrt{2\pi}$.

^^

on both forward and inverse

@TedShifrin I will forever remember the moment you were asked to kick me

LOL ... you starred it; of course.

Of course, that is hilarious

that's not the only part of the convention, though. you also have the issue of whether it's $e^{-i \omega t}$ or $e^{-i 2\pi \nu t}$

I like to think I’m somewhat trustworthy around here these days

hence why I wanted to see it written out

Oh, @Semiclassic makes an excellent point.

@Semiclassical omega

I've never put in the $2\pi$ there !

apologies, I forgot that even was part of it

Nor I, frankly

But it's one of those dumb things

the real annoying part is that in my tripos questions if they say frequency they’re referring to $\omega$

Aha ... putting it there with the $1/\sqrt{2\pi}$ factor might save headaches.

ugh

angular vs. ordinary frequency, yeah

the usual tip-off is the units.

angular frequency is radians per second, ordinary is cycles per second = hertz

They’re maths exams so they put the units in a deep deep hole and never bring them up ever again

yeah, I figured as much

which is weird because 95% of the people who take this course aren physicists

Are*

So $$\tilde{f}(\omega)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty e^{-i \omega t}f(t)\,dt$$ and $$f(t)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty e^{i \omega t}\tilde{f}(\omega)\,d\omega$$

?

si

@TedShifrin the obvious map is the det map

but this does not work !

Mmkay. So the question is how the convolution theorem works in this case?

yes, but I agree with wikipedia

So do I, under this convention: you do get a factor of sqrt(2pi) overall

can't avoid that, since you need to get a factor of 1/sqrt(2pi) for each transform

would you mind if I post a question and see if you can spot what’s wrong

in this that is

so I could only get the solution when I ignored it in the FT

@Ted I think I have an idea about your question

@Kasmir: That's not the obvious map. It doesn't see anything but diagonals.

@Astyx: It's Kasmir's question, not mine.

sure

Ah right

if you mean you're posting it in here, that is

Well, I have an idea about Kasmi's question then

LOL, feel free to talk to Kasmir :P

One good check for this kind of problem, btw, is to consider the case $\omega=0$

This isn’t my neatest work too guys be kind

Do $\int_a^c f(x,\ y)\ dx = \int_b^c f(x,\ y)\ dx + \int_a^b f(x,\ y)\ dx$, $\int_a^c f(x,\ y)\ dS = \int_b^c f(x,\ y)\ dS + \int_a^b f(x,\ y)\ dS$, and $\int_a^c \vec f(x,\ y)\cdot d\vec r = \int_b^c \vec f(x,\ y)\cdot d\vec r + \int_a^b \vec f(x,\ y)\cdot d\vec r$ hold for line integrals just as for regular integrals?

@user10478: What in the world is the second stuff? And no, line integrals are on curves, not intervals.

Wait nvm

You do have additivity for curves if you take one curve followed by another (assuming one starts where the other ends), yes.

That doesn't work

@TedShifrin You mean the second identity? It's $\int_a^c f(x,\ y)\ dS = \int_b^c f(x,\ y)\ dS + \int_a^b f(x,\ y)\ dS$

what is $dS$?

That's usually a surface integral.

@Astyx @TedShifrin you can talk to kasmir directly, I dont bite -.-

A tiny step with respect to arclength.

So in particular you should have $\tilde{s}(0)=\pi$

Along the curve

If you mean arclength integral along a curve, then same comment holds as for the line integral comment I made.

I thought you left @Kasmir

@Astyx mdr , pas de problem

Was the verdict that my statements of the identities was invalid?

@JakeRose What's the issue you're seeing? I seem to have missed it

@Semiclassical gimme 2 min just moves over to my phone and it's latex has broke

@Kasmir Well I think I know an answer

@Astyx am dying to know !

Not sure what you meant by "And no, line integrals are on curves, not intervals."

@user10478: I answered you. Go back and read.

Line integrals are on curves, not on intervals.

If you parametrize and write in the parametrization, then they are on intervals.

@JakeRose One thing I do notice: They say that the Fourier transform of $1/t^2$ is $-\pi |\omega|$

Oh okay, gotcha

@Kasmir First of all, do you know what multiplication by such a matrix does to a triangular one ?

On the left or on the right, @Astyx?

Both

The relevant entry of Wikipedia's Fourier transform table is 310

@Semiclassical I didn't include $\frac{1}{\sqrt{2\pi}}$ on the terms being convolved

@Jake: That's convolved :)

@TedShifrin thankyou

LOL

ah.

Donyou see what I mean?

but, what I was going to say: If you look at 310, in the case of n=2, in the second column

@Astyx am here, just need to answer something, but i read what you say

I might be gone soon

@Kasmir: He's waiting for you to answer him.

you get $-i\sqrt{\pi/2}(-i\omega)\text{sgn}(\omega)=-\pi |\omega|/\sqrt{2\pi}$

I'm so volatile

I could disintegrate at any moment

I knew you had a temper.

it adds

entry 1,3

To what?

a_1,3 + b _ 1,3

in other words, the result $f(t)=1/t^2\implies \tilde{f}(\omega)=-\pi |\omega|$ is not the correct result under that convention

will be added, in the product AB

So the correct way to state them would be $\int_{c+d} f(x,\ y)\ dx = \int_c f(x,\ y)\ dx + \int_d f(x,\ y)\ dx$, $\int_{c+d} f(x,\ y)\ dS = \int_c f(x,\ y)\ dS + \int_d f(x,\ y)\ dS$, and $\int_{c+d} \vec f(x,\ y)\cdot d\vec r = \int_c \vec f(x,\ y)\cdot d\vec r + \int_d \vec f(x,\ y)\cdot d\vec r$, right?

No.

of upper 3x3

the correct result has an extra factor of $1/\sqrt{2\pi}$ in the first place.

For the convention I was using?

Is the first one also a line integral, then? @user10478? I wouldn't use $c$ and $d$ for curves, but whatever.

What's A, what's B ? @Kasmir

Right. So my claim is that the hint they give you is not using the same convention as you were.

@TedShifrin Yeah they're all line integrals

it is, however, what you get in entry 310 in the third column

OK, I didn't know that. I would use $C$ for a curve, but whatever you are used to is fine. Yes, line integrals add, provided, as I said, that you have the second curve starting where the first curve ends.

@Semiclassical I agree, they use the convention where the full $\sqrt{2\pi}$ is in the inverse transform

i.e. under the convention where there's no factor for the forward transform but there's a factor of 1/2pi for the inverse transform

Okay, thanks for the answer

are they using distributions?

@JakeRose full $\sqrt{2\pi}$, or full $2\pi$?

Howdy @Mathein

My claim is it's the latter.

becaue I don't see how to make sense of the Fourier transform of $\frac{1}{x^2}$ else

hi @Ted

Wait did I not convert it??

i could have sworn I put the factor back in

hmm

yeah, you do

hmmm

I think we're getting mixed up on which convention is being used when.

I purposefully omitted it with the convolution because it gave the answer

But I agree I shouldn't have

I mean, the answer they give for $\hat{s}(\omega)$ is itself subject to the convention

Hey guys

Hi @Daminark

Yes but it should itself on differ by a factor

Hi Demonark.

hi @Daminark

Let's try to make life less confusing and stick with the convention the problem is using

which is apparently the non-symmetric one

@Semiclassic: Maybe a wall would help.

@kasmir How much group theory have you done ?

@TedShifrin a wall between time space and frequency space, eh

LOL

So, we have: $$-\pi |\omega|=\int_{-\infty}^\infty e^{-i\omega t}\frac{1}{t^2}\,dt$$

Yes

mmkay. And the FT transform of $1-\cos t=1-\frac12 e^{it}-\frac12 e^{-it}$ under this convention would be $\delta(\omega)-\frac12 \delta(\omega-1)-\frac12 \delta(\omega+1)$

@Astyx @TedShifrin okay I see

so, taking the FT of $\frac{1}{t^2}(1-\cos t)$ should yield... $$\frac{1}{2\pi}\int_{-\infty}^\infty (-\pi |\omega-\omega'|)\left(\delta(\omega')-\frac12 \delta(\omega'-1)-\frac12 \delta(\omega'+1)\right)\,d\omega'=\frac{1}{2\pi}(-\pi)(|\omega|-\frac12 |\omega-1|-\frac12|\omega+1|)$$

it turns out this is an an algebraic group

@Astyx I did a good bit group theory

Sure, any group of matrices is.

@Astyx you trying to conjugate?

I still don't see the answer :(

and you think kasmir has an answer? :D

i will keep trying Ted!

or just $-\frac12 |\omega|+\frac14 |\omega-1|+\frac14 |\omega+1|$

If $\omega>1$, that reduces to $-\frac12 \omega+\frac14(\omega-1)+\frac14(\omega+1) =0$

If $0<\omega <1$, it's $-\frac12 \omega+\frac14(1-\omega)+\frac14(\omega+1)=\frac12(1-\omega)$

What's the question?

Do you know about normal subgroups ?@Kasmir

...which differs from their reported answer by a factor of $2\pi$.

huh.

Something's fishy here

Is the 1/2pi the correct convention?

That's what wikipedia has in front of the convolution

But you're right that it's what is at stake

Mhm

K is normal in H

that is why we can talk about H/K as a group

Wait what is the question exactly ?

Hi. Consider a ring morphism $g:k\to A$ with $k$ a field. Suppose $u,v:A\to A\otimes _kA$ given by $a\mapsto 1\otimes a,a\mapsto a\otimes 1$ are equal. Why does this imply $g$ is surjective?

brb

Ok, let's check. Suppose $$H(\omega)=(F\star G)(\omega)=\int_{-\infty}^\infty F(\omega')G(\omega-\omega')\,d\omega'$$

@Arrow the condition implies that you have an epimorphism and since $k$ is a field $g$ is faithfully flat. faithfully flat epimorphisms are isomorphisms

Then $$h(t)=\frac{1}{2\pi}\int_{-\infty}^\infty e^{i \omega t}\int_{-\infty}^\infty F(\omega')G(\omega-\omega')\,d\omega'\,d\omega$$

assuming wlog that $A \neq 0$

Shifting $\omega$ by $\omega'$, we get $$h(t)=\frac{1}{2\pi}\int_{-\infty}^\infty e^{i\omega t}G(\omega)\,d\omega\int_{-\infty}^\infty e^{i\omega' t}F(\omega')\,d\omega'=2\pi g(t)f(t)$$

@MatheinBoulomenos thank you! Is there a more elementary proof, not involving the words "faithfully flat"? I ask because this answer seems to hint there might be a simpler argument (and this comment).

Which is exactly the $2\pi$ that Wikipedia has.

Howndidnyoy shift the G?

$\omega\mapsto \omega+\omega'$

And then note that $G(\omega)$ has no $\omega'$ dependence

(Strictly speaking, I'm really doing $\omega''=\omega-\omega'$, subbing that in for $\omega$, and then relabeling $\omega''$ to $\omega$)

oh, wait, I see my problem. the convolution is fine

Yeah I see

the issue is that $g(t)=1\implies G(\omega)=2\pi \delta(\omega)$ under this convention

that cancels out the overall factor of $1/2\pi$ and gives the stated answer

i.e. $\int_{-\infty}^\infty 1 e^{-i\omega t}\,dt = 2\pi \delta(\omega)$

So with that in play, the calculation with respect to this convention makes sense.

@Arrow it's just linear algebra. If $A$ has dimension greater than $1$, choose $x$ linearly independent of $1$, then $x \otimes 1 \neq 1 \otimes x$

@JakeRose So I think that, with respect to the non-symmetric convention that the problem assumes, the calculation works out as it should

@MatheinBoulomenos ah! thanks again!

So the question will be how things change in the symmetric convention

@MatheinBoulomenos I was trying to find a better proof than bases

Mhmm

I couldn't think of one

i guess you need to make use of the field assumption eventually

It should still have the same form when using thenother conventikn

just scaled

I want to mention that the general proof for "faithfully flat epimorphisms are isos" is also really easy

Would love to learn it

Right. An overall multiplicative factor is to be expected

yes