Mathematics - 2017-06-22 (page 4 of 7)

Semiclassical

12:24

Observation: D(e^{n 2\pi i x})=n 2\pi i (e^{n 2\pi i x}$ and $E(e^{n 2\pi i x})=e^{n 2\pi i (x+1)}=e^{n 2\pi i x})$ if $n$ is an integer

So that's a simultaneous eigenfunction of $D,E.$

In fact, it's also an eigenfunction of $E^{1/2}.$

SteamyRoot

12:39

@LeakyNun Right, I was bored enough to actually think about that puzzle.
$$\sum_{r=0}^m(-1)^r \begin{pmatrix}m\\ r\end{pmatrix}(x+r)^n = (-1)^m \sum_{r=0}^m(-1)^{m-r} \begin{pmatrix}m\\ r\end{pmatrix}(x+r)^n = (-1)^m \Delta^m x^n = (-1)^m \Delta^{m-n} n! = 0$$

Liad

12:51

i have $f: S \ ^1 \to \Bbb R $ continuous i need to prove that there exists $x \in S \ ^ 1 $ s.t $f(x) = f(-x)$.
So, suppose there isn't such a point. so define $g(x) = f(x) - f(-x) $ , $g \ne 0$ so we can assume w.l.o.g that $g \gt 0$ . so if we define $h(x) = \dfrac{g(x)}{||g(x)||}$ we have a map $h : S \ ^ 1 \to S \ ^ 1$. how can i get to a contradiction from here? or maybe im not in the right way?

(it is a question in topology course, we learned about retracts maybe i need to use it somehow.)

Akiva Weinberger

We also know that $g(-x)=-g(x)$

and $h(-x)=-h(x)$

SteamyRoot

What is $-x$ for $x \in S^1$ ?

Akiva Weinberger

Opposite point

SteamyRoot

Antipodal?

Akiva Weinberger

Yeah

Liad

12:53

i thought $-x = -(x_1,x_2) = (-x_1,-x_2)$ doesn't it?

Akiva Weinberger

Yeah, that's the antipodal point

SteamyRoot

It depends how you define $S^1$...

Liad

huh. ok

Sha Vuklia

@Semi you have time?

Leaky Nun

@SteamyRoot $S^n = \left\{ x \in \mathbb{R}^{n+1} : \left\| x \right\| = r \right\}$ from Wikipedia

SteamyRoot

12:54

A common definition is a subset of $\mathbb{C}$, where $-x$ would have a different meaning.

Leaky Nun

@SteamyRoot not really

Liad

alright, didn't know that :)

SteamyRoot

@LeakyNun How so?

Liad

@AkivaWeinberger so $h$ must get $0$

Leaky Nun

@SteamyRoot because $-x$ would still be the opposite point.

SteamyRoot

12:55

Oh, right.

In that sense

I was thinking of $e^{i\theta}$ where you consider $\theta \in \mathbb{R}/2\pi \mathbb{Z} \cong S^1$

Liad

@AkivaWeinberger if $g(-x) = -g(x)$ and $g$ is continuous then $g$ must get a $0$, is that what you meant?

SteamyRoot

It's a question that is commonly asked as "prove there exist two antipodal points on the equator having the same temperature".

Balarka Sen

$h$ is a map $S^1 \to S^0$ though, is it not?

Akiva Weinberger

@Liad Wait ignore all that

I got confused

Sorry sorry sorry

SteamyRoot

It tends to be easier here if you work with a map $T: [0,2\pi] \to \mathbb{R}$ for the "temperature"

Leaky Nun

12:59

@BalarkaSen what is $S^0$? Two points?

Liad

everything is ok :P

Balarka Sen

Yes.

Akiva Weinberger

@Liad You're right

I thought $g$ was from $S^1$ to $\Bbb R^2$ for some reason

or that $h$ was from $S^1$ to $S^1$

But yeah, $g$ is from $S^1$ to $\Bbb R$

Balarka Sen

Well, Liad wrote it wrong.

SteamyRoot

And you define a secondary map $T_a: [0,2\pi] \to \mathbb{R}: \theta \mapsto T(\theta + \pi) - T(\theta)$

Leaky Nun

13:01

@BalarkaSen why would $h$ map $S^1$ to two points?

Akiva Weinberger

scrolls back up

Liad

yea, sorry $h: S \ ^ 1 \to \Bbb R -\{0\}$

Akiva Weinberger

Oh, so he did

Liad

:P

Balarka Sen

@LeakyNun $|h| = 1$

What are the norm 1 elements in R?

Leaky Nun

13:01

10 mins ago, by Liad

i have $f: S \ ^1 \to \Bbb R $ continuous i need to prove that there exists $x \in S \ ^ 1 $ s.t $f(x) = f(-x)$.
So, suppose there isn't such a point. so define $g(x) = f(x) - f(-x) $ , $g \ne 0$ so we can assume w.l.o.g that $g \gt 0$ . so if we define $h(x) = \dfrac{g(x)}{||g(x)||}$ we have a map $h : S \ ^ 1 \to S \ ^ 1$. how can i get to a contradiction from here? or maybe im not in the right way?

Akiva Weinberger

In any case, $g(x)$ is either positive or negative

and $g(-x)$ is either negative or positive

In either case, IVT.

Balarka Sen

@Leaky Huh? $g$ maps to $\Bbb R$.

Leaky Nun

@BalarkaSen sorry, I see it now

you are right. $h:S^1 \to S^0$.

Balarka Sen

Borsuk-Ulam in dimension 1 is basically IVT, yup.

Liad

@AkivaWeinberger yea, thanks.

Balarka Sen

13:05

@AkivaWeinberger We should write an expository on the Borsuk-Ulam theorem lol

it came up on a lot of stuff we've done recently

I actually don't even know all the various proofs of B-U.

Simply Beautiful Art

14:32

WolframAlpha, convert united states to king

Akiva Weinberger

14:49

Fun fact: If Quebec were to become independent, the US would become larger than Canada

@BalarkaSen Is Borsuk–Ulam still true if we replace the antipodal map with any degree -1 map?

Simply Beautiful Art

Balarka Sen

@AkivaWeinberger That's interesting. I think so.

Krijn

@Bala I have a question for you

Assuming you want to talk about alg geom

Balarka Sen

go ahead

Krijn

Say you have this friendly and nice curve $C$ over a finite field

And you want to map it to $\mathbb P^1$ over the same field

Take $d >> 0$ and $D$ an effective divisor of $C$ of degree $d$

Akiva Weinberger

14:58

@SimplyBeautifulArt $\operatorname{km}\$^2$ is my favorite unit of measurement

Krijn

Can we find a (degree $d$) map $f: C \to \mathbb P^1$ with fiber $D$ above $0$?

I've been googling this for the past half an hour and I assume it has been answered or at least looked at, but can' t find too much

Simply Beautiful Art

@AkivaWeinberger I think $\$^2$ is almost like saying $\text{radians}^2$

Balarka Sen

@Krijn What's an "effective" divisor?

Krijn

@BalarkaSen No negative coefficients

Balarka Sen

Ah, ok. I was actually thinking about the exceptional divisor in blowups, or something of that sort.

But of course we're working with curves here.

Akiva Weinberger

15:02

@SimplyBeautifulArt Radians squared is a legit unit of measurement, actually

Measures areas of parts of the sphere

Simply Beautiful Art

Hm, right... didn't think that through too well.

Akiva Weinberger

You'll see it for apparent size, I think

Simply Beautiful Art

So what does $\$^2$ mean?

Secret

but how can we be sure any function $f$ can be decomposed into a fourier transform, hence allowing us to exploit the linearity and that $D$ and $E^n$ have simultaneous eigenfunctions $e^{2\pi i x}$ thus allowing us to commute the $E^n$ and $D$ for each term in the expansion?

although I suspect for Akiva's question, I think we should be fine since the fourier transform of a polynomial always exists thus we might be able to use that to show the required commutation relations

although, I felt like when Akiva made that question, he might have a more elegant (read: simpler) way in proving that commutation relation, which probably requires theorems that I either don't know or I don't remember

Akiva Weinberger

I'm not actually sure if the proof is that simple

Secret

15:05

NB. I also noticed that $\frac{d}{dx}$ is actually a shift operator in disguise...

Akiva Weinberger

By the way, note that it's important that these be over polynomials.

$f(x)\mapsto f(x)+\cos(2\pi x)$ commutes with $E$ but not $E^{1/2}$, for example.

(Luckily, there are no nonzero periodic polynomials)

@Secret How so?

Secret

$$\frac{df(x)}{dx}=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to 0}\frac{E^hf(x)-f(x)}{h}=\lim_{h\to 0} \left(\frac{E^h-\textrm{id}}{h}f(x)\right)$$

Akiva Weinberger

Oh yeah

Krijn

@BalarkaSen Yeah, so we are in the easiest case

SteamyRoot

@AkivaWeinberger That's not a very linear operator though, is it?

Secret

15:12

While yes under such form, $\textrm{id}$ and $\frac{1}{h}$ will definitely commute with any operator, the issue is that we need to deal with $E^h L$ where $L$ is some linear operator. But that will require proving that $[E,L]\implies [E^n,L]=0,\forall n \in \Bbb{R}$ which is not true for the general case as the periodic counterexample showed, but it might work on polynomials for reasons I need to check out how...

One issue with doing polynomial proofs is I often have no idea what is happening when a polynomial $f(x)$ becomes $f(x+1)$ until I expand it

and that's where things get messy...

Akiva Weinberger

@SteamyRoot Oh. You're right

Woops

Is $f(x)\mapsto f(x)\cdot\cos(2\pi x)$ good?

Secret

Meanwhile, the action by the derivative on polynomials is easy to see when we understood that all one variable differential operators have the form $D=\sum_{i=1^n}a_i(x)\frac{d^i}{dx^i}$, thus considering each powers of $\frac{d}{dx}$, each integer application of $\frac{d}{dx}$ on the polynomial effectively shift the polynomial "to the left" if we use the standard basis $\{1,x,x^2,x^3,x^4,...,x^n\}$, and scale its coefficients in a linear manner

Akiva Weinberger

Don't use $D$ for something other than $\frac d{dx}$, that's confusing

Use like $A$ or something

Secret

Ok, will do that next time since I timed out

Meanwhile graphically, $E^n$ will translate the graph of the polynomial (or in fact, any function) to the right by $n$ units, however how the coefficients are affected I think I need to work out its matrix representation

Dattier

Hi, $$\textbf{ A classic revived : } \text{this series , is-it converged } \sum\limits_{k=2}^n \frac{1}{\sqrt{k}}\cos(\frac{k^2+1}{k-1}) \text{ ?}$$

SteamyRoot

15:28

@AkivaWeinberger That seems fine, yes.

Dattier

bye

Krijn

@Bala, I' ve just asked it in MSE here, if you' re interested in following it

It' s probably an easy question for someone with the right knowledge

Secret

in The h Bar, 1 min ago, by 0celouvskyopoulo7

@Secret $d/dx$ is unbounded

in The h Bar, 1 min ago, by 0celouvskyopoulo7

so "converge" is tricky

Semiclassical

Just saw someone in the engineering building wearing a t-shirt with a statement of the Collatz conjecture on the back

Secret

O crap, I guess I really need to have a solid background in functional calculus before I started trying to write formal expressions and manipulation on operators...

SteamyRoot

15:40

Did he claim to have proven it?

Dodsy

like this

cdn.shopify.com/s/files/1/0149/3544/products/…

?

SoumyoB

@Astyx sorry didn't get time yesterday to answer your question in more details

SteamyRoot

That font looks suspiciously like the xkcd font...

Semiclassical

yeah, similar to that. except in red and with a big tree on the back as well.

same font, though.

Alessandro Codenotti

Dodsy

15:43

like this

cdn.shopify.com/s/files/1/0149/3544/products/…

Alessandro Codenotti

I don't understand how the existence of a real eigenvalue follows in this passage

SoumyoB

@Astyx could you point me to a more complete calculation of the probabilities you were suggesting for the random walk using that exponent method? It might refresh my memories and I might be able to better answer your question

Dodsy

yeah okay, I'll give @Fargle a star.

Semiclassical

right. I'm surprised I can't seem to find the version of it I saw, though.

Dodsy

what are those shirts called where the ape stands up

and is then human

Semiclassical

15:49

my guess is it's something like "the evolution of man"

Dodsy

haha, yes.

Semiclassical

it's an old image, though, so I wonder where it first comes from.

Dodsy

I guess it would be offensive to some.

Semiclassical

Ahah: en.wikipedia.org/wiki/March_of_Progress

Dodsy

sniped so hard..

Semiclassical

15:52

lol

Secret

Linear algebra question. Given a vector $v \in \Bbb{R}^n$, I want to find the matrix such that $\forall v \mapsto (1,1,1,1,...,1)$. How should I approach this problem?

Dodsy

did you ever solve that problem semi

SteamyRoot

Ummm.

Semiclassical

Not yet. I'll be returning to it at some point.

SteamyRoot

You claim to be asking a linear algebra question but the function you describe is everything but linear.

Dodsy

15:55

ah I see.

Semiclassical

@SteamyRoot I think it'd be linear algebra if you asked for the component of $v$ in the 111...1 direction, though.

Secret

o wait, yes... I forgot the statement of the problem will imply $\mathbf{0}\mapsto (1,1,1,1,...,1)$ which cannot be linear

nvm then...

SteamyRoot

@Semiclassical In that case, yes.

Semiclassical

But that's not what was asked, so yeah.

SteamyRoot

The closest to an answer of the original question would be to embed $\mathbb{R}^n$ in $\mathbb{R}^{n+1}$ by $i(x_1,\dots,x_n) = (x_1, \dots, x_n, 1)$

Dodsy

15:57

@Semiclassical So the lady said that if I do part time I would be considered by the Ontario government as a full-time student and get full-time funding, and I'd be able to live on campus. She said the part-time application costs like 58 bucks. It turns out it costs 145 bucks, so now she's looking into it to see if I can get it for 58 bucks.

Hey zee.

SteamyRoot

And then consider the matrix given by all zeros except the last column, consisting of only ones.

Semiclassical

@Dodsy Good. Keep plugging away at that.

Zee

Hey dod

SteamyRoot

Because that matches the affine map $x \mapsto 0 x + 1$ on $\mathbb{R}^n$...

Zee

Well it sounds to me like you got accepted where you want

Yet you ain't happy...

Semiclassical

15:59

bureaucratic limbo is never fun.

Dodsy

Thanks semi.

Great to have some support!

Semiclassical

even if it can lead somewhere good, it's still not a pleasant place to be.

Zee

It sure is fun

Bureaucratic limbo is where you can hack the system so to speak

Semiclassical

Right up until the system hacks you.

Zee

The systems is always hacking you so it don't matter

Dodsy

16:00

Richard Muller: I Was wrong on Climate Change

►

Zee

Oh god

Dodsy

SMACK ONNNNNNN

Semiclassical

That's right, I'd forgotten that he was previously a skeptic.

Reminds me, there was something I got in my campus mail slot lately

Dodsy

if he was a schitzophrenic, we would've nailed all three of the people you need to convince:

Convince yourself
Convince your friend
Convince a skeptic

Semiclassical

And apparently a bunch of other physics people (e.g. grad students)

Dodsy

16:03

What was it

Semiclassical

which was an anti-global warming booklet from the Heartland Institute

Dodsy

hm interesting

Semiclassical

I tossed it. Anyone who is going to try to convince you that way is doing propaganda.

Secret

Ok guys, I got the matrix representation of $\frac{d}{dx}$ in polynomial space. Not sure how much that will help on the proof:

In $\textrm{P}^n(\Bbb{R})$. using the standard basis
$$\frac{d}{dx}=\begin{pmatrix}
0 & 0 & 0 & 0 & \cdots & 0 \\
n & 0 & 0 & 0 & \cdots & 0 \\
0 & n-1 & 0 & 0 & \cdots & 0 \\
0 & 0 & n-2 & 0 & \cdots & 0 \\
0 & 0 & 0 & n-3 & \cdots & 0 \\
\vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & 0 & \cdots & 0 \\
\end{pmatrix}$$

Semiclassical

yeah, this: pbs.org/wgbh/frontline/article/…

Secret

16:05

I am currently computing the matrix reprsentation for the shift operator also, though this one is indeed messy because going from $x^n$ to $(x+1)^n$ suddenly you have a pascal triangle of coefficients need to add to the polynomials for each power of x

Semiclassical

One connection worth noting re: E and D

If we do a Taylor series on $Ef(x)=f(x+1)$, we get

$$Ef(x) = f(x+1)=f(1)+xf'(1)+\frac{x^2}2 f''(1)+\cdots$$

hang on. I need to say this differently to get the conclusion I want.

Let me write $E_y f(x)=f(x+y)$.

Akiva Weinberger

$E^y$, @Semiclassical

Semiclassical

oh, sure.

$$E^y f(x) = f(x+y)=f(y)+x f'(y)+\frac{x^2}2 f''(y)+\cdots$$

Akiva Weinberger

By the way, @Secret, re: convergence

Say I have the following infinite polynomial in $D$: $a_0+a_1D+a_2D^2+\dotsb$

Semiclassical

which I can rewrite in operator form as $$f(x+y)=\left(1+x \dfrac{d}{dy}+\frac{x^2}{2}\dfrac{d^2}{dy^2}+\cdots\right)f(y)$$

Akiva Weinberger

16:12

(which could be something horrible like $a_n=n!$ or something, doesn't matter)

That always converges when applied to a polynomial, because it turns into a finite sum.

High enough powers of $D$ would just send our polynomial to 0.

($D=\frac d{dx}$)

Semiclassical

Therefore $E^y f(x) = f(x+y)=e^{x (d/dy)}f(y)$.

Secret

right, so polynomial space formed a finite support in functional space at least for the differential operator

Akiva Weinberger

Yup. And (with $x=1$) $~E=e^D$!

Semiclassical

blarg. I should've expanded in powers of $y$.

Akiva Weinberger

(And with $x=h$, $~E^h=e^{hD}$)

Semiclassical

16:13

so that $E^y f(x) = f(x+y)=e^{y (d/dx)}f(x)$.

i.e. $E^y = e^{y(d/dx)}$.

With, yeah, the suggestive implication being $E=e^{d/dx}$.

Akiva Weinberger

Another way to get the same result, by the way (and I don't know if you know this, Semi)

Semiclassical

you use this in QM a lot, but written in terms of the momentum operator

Akiva Weinberger

but earlier Secret pointed out that$$D=\lim_{h\to0}\frac{E^h-I}h.$$

Semiclassical

in which case you'd prove it via the commutation relation

Oh, nice.

Akiva Weinberger

And then use L'Hôpital.

That gives you $D=\ln E$.

Which implies $E=e^D$ as well ^_^

Semiclassical

16:16

I know the reverse of it: $\left(1+\frac{1}{n}D\right)^n\to e^D=E$ as $n\to \infty$.

Akiva Weinberger

Yeah that's Newton's something or something

Semiclassical

yeah.

Secret

As 0celo have noted, that limit form will probably be an issue if we are working outside the polynomials (to be checked by some functional analysis people), but for polynomials, I think we will be fine because the differential operator is bounded in polynomial space

Akiva Weinberger

(In fact, since $E=e^D$ essentially is equivalent to Taylor's theorem, and the above limit is the definition of the derivative, that manipulation doubles as like the shortest proof of Taylor's theorem ever)

@Secret Right. And as I noted, the problem is false outside of the polynomials

$f(x)\mapsto f(x)\cos(2\pi x)$ being the example of something that commutes with $E^1$ but not $E^{1/2}$

Semiclassical

Typically in physics you'd instead restrict it by only considering periodic $f(x)$.

I have in mind Bloch's theorem like I said elsewhere.

0celouvskyopoulo7

16:19

@AkivaWeinberger Taylor's theorem?

Akiva Weinberger

Hi, 0celoubskyopoulo7!

0celouvskyopoulo7

Hi

Akiva Weinberger

(Is that Greek?)

0celouvskyopoulo7

Ask @BalarkaSen

Akiva Weinberger

But yeah, write that as $E^h=e^{hD}$, apply it to a function $f$ at $x$, and then do a substitution in the variables I think

I can post a fuller version later

Semiclassical

16:20

What's nice, as I said earlier, is that if $f(x)$ is periodic then $f(x)$ can be written as a sum of complex exponentials $e^{2\pi i n x}$.

Secret

i.e. fourier series expansion

Semiclassical

Right.

0celouvskyopoulo7

Not to be too pedantic, but with Fourier series one has to be careful with the words "can be written as."

Semiclassical

in which case $E e^{2\pi i n x}=e^{2\pi i n}e^{2\pi i n x}= e^{2\pi i n x}$.

Mary Star

Hello!! Do you maybe know which function has the following graph: https://lp.uni-goettingen.de/get/image/5182 ?
Is it maybe a cosine function?

0celouvskyopoulo7

16:22

I can write anything as a series of something, that doesn't mean the series has to converge or converge nicely.

Semiclassical

Yeah. In physics I'd also want it to be square-integrable

since wavefunction and all that.

0celouvskyopoulo7

At least that. But just square integrable doesn't mean the derivative has any meaning.

Semiclassical

Sure. You'd also want it to be smooth (so long as you're not doing some kind of delta-function potential).

$De^{2\pi i n x} = 2\pi i n e^{2\pi i n x}$.

Just win baby

@0celouvskyopoulo7 get out of here you pedant :P

Semiclassical

So $E$ acts as the identity operator on these eigenfunctions and $D$ just multiplies by $2\pi i n$.

Secret

16:25

I am ok with pedantics to ensure us not fall into wrong places

Akiva Weinberger

It's physics, I'm sure all functions are continuous and infinitely differentiable and all that jazz

Secret

yeah, and we do sometimes deal with distributions loosely...

Semiclassical

ya

Secret

e.g. in h bar we talked about a horror paper where you have power series expansions on delta functions

Semiclassical

Additionally, we have that $e^{2\pi i x }e^{2\pi i n x} = e^{2\pi i (n+1)x}$

Akiva Weinberger

16:27

There's a variation on classical logic in which the excluded middle fails and all functions are continuous

Semiclassical

So multiplying by $e^{2\pi i x}$ shifts us by one index.

Akiva Weinberger

$\begin{cases}1,&x=0\\0,&x\ne0\end{cases}$ isn't well-defined because it doesn't talk about things that are neither zero nor not zero :P

@Secret ??

Like, $\sum a_n\delta^n$?

0celouvskyopoulo7

@AkivaWeinberger Good

I reject classical logic

Secret

in The h Bar, 2 days ago, by ACuriousMind

Mar 16 at 18:29, by 0celo7

He's Taylor expanding a Dirac delta holy shit

Semiclassical

So something like $\delta(x+y)=\delta(x)+\delta'(x)y+\cdots$.

Akiva Weinberger

16:29

Oh god wow

0celouvskyopoulo7

$5 if you can make that precise.

Just win baby

How can you reject Aristotle?

Akiva Weinberger

What, so $\delta(0)=\delta(-1)+\delta'(-1)+\dotsb$??

Semiclassical

Probably one way to do it is to Fourier transform in $x$, expand in powers of $y$ there, and then inverse Fourier transform back.

Akiva Weinberger

That's just infinity equals zero

@Secret On the train ride home I found a hole in my solution

but then I found a way to fix it

0celouvskyopoulo7

16:31

@AkivaWeinberger $\delta'(1)$ is pretty badly defined.

Semiclassical

I mean, the Fourier transform of $\delta(x-y)$ in $x$ is just $e^{i k y}$ I think.

modulo an overall constant etc.

0celo7

@Semiclassical Correct

Semiclassical

So you'd be doing $1+i k y-\frac{k^2}{2}y^2+\cdots$

...hrm. And then you'd be asking about the inverse Fourier transforms of $k^p$.

0celo7

Those are not Schwartz functions, good luck.

Semiclassical

Yeeah.

0celo7

16:33

(Neither is $1$, which is why the delta is not a function.)

Semiclassical

What makes me think there could still be some logic to it is what it would seem to mean physically.

Namely: Suppose I start with a point charge at $x=y$

And then move the charge just a little.

How does the charge distribution change?

...but, that seems kinda silly now that I think of it.

i mean, it sorta makes sense that the difference in charge distribution would be something like a dipole.

(which is what $\delta'$ more-or-less is as I remember)

But that's just the linear term.

0celo7

yeah it's a limit of dipole looking things

Semiclassical

I think this does have some interpretation in terms of multipole moments or some such

But uh

0celo7

so the idea would be that one dip cancels the original delta and the spike is the new one

Semiclassical

Yeah.

As usual with the dirac delta, it's probably best to think in terms its effect on an actual integral

e.g. $f(x)=\int_{\mathbb{R}} f(y)\delta(x-y)\,dy$

0celo7

16:38

how about setting $y=0$ for simplicity

Secret

$-f'(x)=\int_{\Bbb{R}} f(y)\delta'(x-y)dy$ if I recall...

Semiclassical

$y=0$ or $x=0$?

In the way I've written it just now I'd say $x=0$ since i'm integrating over $y$.

But I've also managed to reverse how I said it above, blah

So I'll say $f(y)=\int_{\mathbb{R}} f(x)\delta(y-x)\,dx$.

Secret

NB this is how engineers use $\delta'$

Unit doublet

In mathematics, the unit doublet is the derivative of the Dirac delta function. It can be used to differentiate signals in electrical engineering: If u1 is the unit doublet, then ( x ∗ u 1 ) ( t ) = d x ( t ) d t {\displaystyle (x*u_{1})(t)={\frac {dx(t...

Semiclassical

@Secret yeah.

Hence dipole kind of thing

I imagine, in fact, that it really is a dipole thing once you go to a point charge in 3-space and take $\nabla$ of the 3D dirac delta

0celo7

@Semiclassical ...hmm

one of us got switched

Semiclassical

16:43

Yeah, probably me.

If I want to say "the point charge is at $x=y$"

then I'd better have $y$ as a free variable not a dummy one

anyways. If I formally expand $\delta(x-y)=\sum_k \delta_k(x) (-)^k y^k$ and expand the LHS as $f(y)=\sum_k f^{(k)}(0) y^k$

then I can identify each side term-by-term in $y$

0celo7

that's pretty painful lol

Semiclassical

Not going to argue with that.

and get $f^{(k)}(0)=(-1)^k \int_{\mathbb{R}} \delta_k (x) f(x)\,dx.$

0celo7

@Semiclassical I won't argue that $\delta^{(k)}$ doesn't make sense...in fact what you're doing is just deriving the action of the $k$-th distributional derivative of $\delta$.

Semiclassical

Right.

There's a certain kind of logic to this, but I won't pretend that what I'm doing is terribly logical on the whole

If I do some vigorous hand-waving as regards integration by parts, that basically does give $\delta_k(x) = \delta^{(k)}(x)$.

0celo7

But the problem is that if $f$ has Taylor series like that, you have $f\in C^\omega$. But $C^\omega\cap \mathscr D=\{0\}$. So you need to work in $\mathscr S\cap C^\omega$. I don't know what topology that has.

(Besides the subspace topology.)

Semiclassical

16:48

Formally, this all has a certain coherence.

Rigorously....oh, look, a convenient distraction! flees

Just win baby

lol

0celo7

@Semiclassical Another thought. $\delta$ is a limit of gaussians in $\mathscr D'$. Those are $C^\omega$. Maybe the limit of their Taylor series has a meaning as a Taylor series for $\delta$.

Semiclassical

maaaybe.

It usually does help to think in terms of approximations rather than the dirac delta itself.

0celo7

@Semiclassical wait, wasn't the integral over $y$?

$x$ is fixed...

Secret

(Reading all of this makes me wish I had a solid functional analysis background, so I can participate in the investigation...)

Semiclassical

16:53

@0celo7 I swapped it, remember?

13 mins ago, by Semiclassical

So I'll say $f(y)=\int_{\mathbb{R}} f(x)\delta(y-x)\,dx$.

0celo7

Oops. Well still, a Taylor series coefficient does not depend on the dynamical variable

So it should be $\delta_k(0)$

Semiclassical

Yeah.

Probably wasn't worth worrying about in any case.

0celo7

It isn't.

But there was a paper that assumed $\delta$ was an analytic function.

Semiclassical

In fairness, as you say, you can pick an approximation of $\delta$ which is analytic

and do the arguments on that.

but uh, coulda-woulda-shoulda

0celo7

Yeah. The paper made about 0 sense unless you read the conclusion. They were trying to derive formal expressions that could be approximated numerically with Gaussians.

Semiclassical

16:56

ahhh

0celo7

@Semiclassical In particular, they needed a Taylor expansion of $\delta(x)$ so they could define $\delta(-i\partial_x)$.

Semiclassical

:S

Secret

Guys, here's the shift operator's matrix representation in the standard polynomial basis (and this is why I so hate to expand polynomials) :

Interestingly, its upper antitriangular

In $\textrm{P}^n(\Bbb{R})$. using the standard basis. Let $\begin{pmatrix}x \\ y\end{pmatrix}={}^xC_y$
$$E^k=\begin{pmatrix}
{}^nC_0k^0 & {}^{n-1}C_0k^0 & {}^{n-2}C_0k^0 & {}^{n-3}C_0k^0 & \cdots & {}^{0}C_0k^0 \\
{}^nC_1k^1 & {}^{n-1}C_1k^1 & {}^{n-2}C_1k^1 & {}^{n-3}C_1k^1 & \cdots & 0 \\
{}^nC_2k^2 & {}^{n-1}C_2k^2 & {}^{n-2}C_2k^2 & {}^{n-3}C_2k^2 & \cdots & 0 \\
{}^nC_3k^3 & {}^{n-1}C_3k^3 & {}^{n-2}C_3k^3 & {}^{n-3}C_3k^3 & \cdots & 0 \\
{}^nC_4k^4 & {}^{n-1}C_4k^4 & {}^{n-2}C_4k^4 & {}^{n-3}C_4k^4 & \cdots & 0 \\

0celo7

@Semiclassical Their formula for the Fourier transform:

$$\tilde g(y)=\sqrt{2\pi}\mathrm e^{\mathrm ixy}\delta(\mathrm i\partial_x-y)g(x).$$

Semiclassical

@0celo7 friends don't let friends drink and derive.

4

is my reaction to that

0celo7

16:59

$$\int_\Bbb Rg\,\mathrm dx=2\pi \delta\mathopen{}\left(\mathrm i \frac{\mathrm d}{\mathrm dt}\right) \mathclose{}g(t).$$

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$Mathematics$ Mathematics