Mathematics - 2017-06-26 (page 4 of 6)

Akiva Weinberger

3:00 PM

Or, equivalently, use the Gamma function

SteamyRoot

Like, I'll believe it when someone with actual knowledge of RH reviews that paper...

Secret

@AkivaWeinberger ah, I see, so that's how one wrote that product in compact form

Simply Beautiful Art

@Secret en.wikipedia.org/wiki/…

s.harp

@Semiclassical do you mean the hilbert polya conjecture or osmething more specific?

Secret

The finite difference and antidifference made up calculus of finite difference just as $\int$ and $D$ made up calculus

I once tried to use that machinery to solve Waiting's Maclaurian formula zeta 7 integral

Semiclassical

3:02 PM

@s.harp Yeah, but specifically the story which Bender et al tell in the paper I linked.

Secret

@SimplyBeautifulArt Yeah I was aware of the rational case, but I never knew one can go to irrationals

Semiclassical

For criticism on it, see here: math.stackexchange.com/q/2211278/137524

Akiva Weinberger

@Secret Presumably it follows by continuity

Well, uniform continuity?

Secret

btw, there's also multiplicative calculus:

Multiplicative calculus

In mathematics, a multiplicative calculus is a system with two multiplicative operators, called a "multiplicative derivative" and a "multiplicative integral", which are inversely related in a manner analogous to the inverse relationship between the derivative and integral in the classical calculus of Newton and Leibniz. The multiplicative calculi provide alternatives to the classical calculus, which has an additive derivative and an additive integral. There are infinitely many non-Newtonian multiplicative calculi, including the geometric calculus and the bigeometric calculus discussed below. These...

Semiclassical

The authors have since written a response to the critical comment which Bellisard put up on arxiv

Akiva Weinberger

3:03 PM

Though I suppose with something like $(1+D)^h$, it's gonna end up being a finite sum anyway, so you just need continuity of each of the terms

s.harp

Yes I remember these things

people also wondered why this was published in PRL if it is mainly interesting for mathematicians

Semiclassical

Well, it's part and parcel of the whole pt-symmetry thing

Though I think that's been oversold as well.

Akiva Weinberger

Huh, I just realized, $D\Delta^{-1}$ is uniquely defined

$\Delta^{-1}$ introduces an arbitrary constant, but the $D$ gets rid of it.

Semiclassical

@AkivaWeinberger In their case I think they get rid of it via a particular choice of boundary condition.

Akiva Weinberger

It's bijective, too, since its inverse $\Delta D^{-1}$ is uniquely defined as well

Semiclassical

3:07 PM

(they want to have eigenfunctions of $\Delta^{-1}$)

Akiva Weinberger

(and it keeps the same degree)

@Semiclassical Ah, I see. I know that $2^x$ is an eigenfunction of $\Delta$ (analogous to $e^x$ for $D$). What would they get for $\Delta^{-1}$?

Semiclassical

If I'm remembering right, it's basically the Hurwitz zeta function.

Akiva Weinberger

…huh

Secret

Meanwhile the eignefunction for $\int$ is $e^{ax}$

Semiclassical

But again, the paper (and the MSE post to a lesser extent) has the details.

Akiva Weinberger

3:09 PM

Oh, wait, $n^x$ is an eigenfunction for $\Delta$ for all $n$

@Secret Not of $\int_0^x$.

Like, not if you want your antiderivative to be zero at zero.

Semiclassical

@akiva More explicitly, one has $\displaystyle \frac{x}{1-e^{-x}}=\sum_{k=0}^\infty \frac{B_n}{n!}(-x)^n$

Akiva Weinberger

Right, yes.

Semiclassical

Where $B_n$ are the Bernoulli numbers

So at least formally one can replace $x$ with $D$.

Akiva Weinberger

And applying that to $D$ gives us the aforementioned $D\Delta^{-1}$.

Semiclassical

Right.

Secret

3:11 PM

ah the (insert word) of definite integrals and their boundaries...
That means, I kinda have to think more about this...

$$\int_0^x f(ax) dx= a(f(ax)-f(a0))$$
hmm... set $f(0)=0$ gives..

Semiclassical

The critical comment from Bellisard is here and the response from Bender et al is here.

The comment from Bellisard basically being "this approach doesn't give a rigorous proof" and the response being "we never claimed that this was a proof."

Akiva Weinberger

I doubt it has eigenfunctions if you want it to give zero at zero. @Secret

@Semiclassical What exactly were they trying to prove?

Well, "prove"

Balarka Sen

What's new today?

s.harp

Jonathan Keating gave a talk at my uni a few weeks ago

he said that he originally published his stuff about hilbert Polya in some NATO proceedings with Berry in the hopes that nobody would ever look at it

Astyx

I'm definitely better at English than I am at French, considering my oral exams today

Secret

3:16 PM

Well, since that is an indefinite integral, we can knock it out with the 1st fundamental theorem of calculus to give:
$$af(ax)=a^2f'(ax)$$

Akiva Weinberger

@BalarkaSen Discussing stuff related to the puzzle I gave Secret on linear operators over the polynomials

Semiclassical

I think the idea was: "This won't serve as a proof using the usual self-adjoint methods, but maybe the entire pt-symmetric program can be made rigorous."

s.harp

and that he sometimes gets mails from people saying "look we did the things you did too and it got us this far which looks super close!", his usual way of looking at it is "i got just as far as you did 20 years ago!"

i think this part was supposed to be a dig at the bender müller paper, but i only heard about it after the talk^

Semiclassical

@s.harp lol

I can buy that.

Secret

if $a=0$ it is trivial, thus assume $a\neq 0$, this gives the ODE:
$$f(ax)=af'(ax)$$

Akiva Weinberger

3:17 PM

@BalarkaSen Also, $\int_0^x$ (the antiderivative that has zero at zero) has no eigenfunctions, right?

Semiclassical

I sorta wish they'd taken that angle tbh

Akiva Weinberger

The only possible candidates are $e^{cx}$, and they all fail the boundary condition

s.harp

which angle?

Akiva Weinberger

@Semiclassical What was it they were trying to prove?

Semiclassical

Namely: If you take it seriously, this seems to get you awfully close to the Riemann hypothesis...but it doesn't get you there, and this is why.

@AkivaWeinberger That's a good question :/

Balarka Sen

3:18 PM

$\int_0^x f(t) dt = \lambda f(x)$ means $f'(x) =\lambda f(x)$ with $f(0) = 0$ isn't it?

Akiva Weinberger

Oh, they were trying to prove the Riemann hypothesis.

Semiclassical

Well, they don't claim that explicitly.

They more claim it as an approach which could possibly be made rigorous in the future.

Balarka Sen

So yeah, there isn't such a thing.

SteamyRoot

Anything that "seems to get close" to RH using techniques that anyone on this chat can understand, are extremely likely to get nowhere at all.

Secret

if $f(ax)=0$, which is trivial, then we are done and the result is trivial. Otherwise we can divide $f(ax)$ bot sides and then integrate to get:

$$1=a\frac{f'(ax)}{f(ax)}\implies x+C=a\ln f(ax)$$

That's... a constraint..., not a solution

Leaky Nun

3:19 PM

@AkivaWeinberger of course there is

Balarka Sen

Oh, what about $f(x) = 0$.

SteamyRoot

Uhh, yeah

Akiva Weinberger

I think it's techniquely grammatically incorrect to use "at all" in a positive clause or something

Semiclassical

lol.

Felix.C

hey I just did some quick calculations with CR-equation and I'm ashamed to ask this but I was quite unsure about chain rule again -.-
If I want to calculate ${\partial u} / {\partial r} = {\partial u} / {\partial x} * {\partial x} / {\partial r} + {\partial u} / {\partial y} * {\partial y} / {\partial r}$
I don't really understand anymore why we add the x/y partial derivative terms, I tried to explain it to myself that this follows from inner product when we replace $s := (x,y)$ and write $u(s)$ instead of $u(x,y)$ but I feel very unsecure, any easier way to see this?

Akiva Weinberger

3:21 PM

Kind of like using "for any $\epsilon>0$" in situations that aren't upwards closed

Leaky Nun

@AkivaWeinberger not if it modifies "nowhere"

Semiclassical

I'm not too sure that's true, though. For an example of a story which gets close and which is relatively understandable (including as to why it fails), there's the Hejhal-Haas story

That's what I linked to Paul Garrett's notes in the question.

Akiva Weinberger

@BalarkaSen $0$ doesn't count as an eigenvector, does it?

SteamyRoot

Well, it doesn't hold for any hypothesis of course

Balarka Sen

Yeah probably

Semiclassical

3:22 PM

Specifically here: www-users.math.umn.edu/~garrett/m/v/pseudo-cuspforms.pdf

SteamyRoot

There's also this guy: quantamagazine.org/…

Semiclassical

Basically, it gives an approach which would prove RH if only certain functions were genuine eigenvalues of a certain spectral problem.

Akiva Weinberger

@LeakyNun I forget the exact condition, but there's a word for it

Secret

e both sides gives
$$Ke^x=f(ax)^a$$
and take $a$ root gives
$$(Ke^x)^{\frac{1}{a}}=f(ax)$$

Plug $f(0)=0$ gives
$$(Ke^0)^{\frac{1}{a}}=f(0)$$
$$(K)^{\frac{1}{a}}=0$$
Thus $K=0$

So $f(ax)=0$ thus zero function is the only eigenfunction of $\int_0^x$ (update, nope, 0 is not an eigenvector, thus no eigenfunctions)

Semiclassical

Alas, they aren't and so RH doesn't follow.

But it comes really close, in that the solutions are in a $+1-\epsilon$ Sobolev space (it'd need to be just plain $+1$ to work).

That to me is a pretty precise statement as to how close that approach comes to fulfilling the dream

Akiva Weinberger

3:26 PM

@LeakyNun If you care about the linguistics, I found an article

(Asterisks mark ungrammatical sentences)

Secret

$\Delta=E-I$

$\Delta^h f =(\sum_{k=0}^{\infty}\binom hk (E-I)^k) f$

uh what, this looks really scary...

Akiva Weinberger

@LeakyNun I heard the thing about downward entailing environments, but apparently it doesn't work 100%

Secret

ok, I see I can expand that $E-I$ and the polynomial $f$ will bubble out to the left for each term, the problem is how to convert the resulting expression back to $E^h$

Leaky Nun

@AkivaWeinberger your link addresses nothing at all

Akiva Weinberger

@LeakyNun I, uh… yes, I suppose it does

Leaky Nun

3:33 PM

@AkivaWeinberger "nowhere at all" is grammatically correct

I thought you would have noticed that I used the same grammar

Akiva Weinberger

I would probably call that a negative environment

s.harp

Cant say that it does everything at all

Akiva Weinberger

Wut

s.harp

its not correct to say "everything at all"

Akiva Weinberger

Also, it claims that "Exactly three people have ever been on the moon" is grammatical, but to me it feels somewhat off

(Also, isn't it more like twelve?)

Leaky Nun

3:39 PM

so which sentence are you complaining about?

Akiva Weinberger

Steamy accidentally said "extremely likely to get anywhere at all" before he fixed it

Semiclassical

Can we argue about math rather than grammar?

Leaky Nun

@AkivaWeinberger oh

Akiva Weinberger

I wasn't arguing about grammar, I just thought it was a fun fact

and an interesting topic

I like linguistics

Semiclassical

linguistics are a human construction

and like all human constructions are weird and strangely defined

s.harp

3:42 PM

unlike mathematics?

Akiva Weinberger

I didn't actually notice that Steamy fixed it to "nowhere". So when you started talking about "nowhere" I had no idea where that came from @LeakyNun

Semiclassical

Depends on what part of math we're talking about.

s.harp

which has been given to us by xeno

Semiclassical

xenu

s.harp

aka hilbert

Akiva Weinberger

3:42 PM

I thought it was just a random proposed counterexample you came up with @LeakyNun

Balarka Sen

"Stonewall Willingdone is an old maxy montrumeny"

This is a perfectly beautiful sentence.

Akiva Weinberger

@Semiclassical It's really a part of psychology, though

Secret

Ok, I have proved $\Delta^h f = f \Delta^h$ above by using power series and $[E,f]=0$ to bubble out $f$, now to figure out how to get the result back into the form of $E^h f$

Balarka Sen

Or "We nowhere she lives but you mussna tell annaone for the lamp of Jig-a-Lanthern!"

Akiva Weinberger

It's made by all the everyone who speaks the language

Semiclassical

3:43 PM

I love how the Wiki page on Xenu includes a frame from the South Park "This is what Scientologists actually believe" bit.

Leaky Nun

"The old man the boat."

Akiva Weinberger

Yes they do :D

Secret

Btw, is the following true?

$$x^h=x^{\sum_{i=0}^{\infty}\frac{a_i}{b_i}}=\prod_{i=0}^{\infty}x^{\frac{a_i}{b_i}}$$?

where the sum converges to $h$

Akiva Weinberger

Sure, why not

Balarka Sen

Or "The extremes alone are stable as is stressed by the vibration to be observed when a pause occurs at some intermediate stage, no matter what its level and duration."

Secret

3:47 PM

$$E^h=\prod_{i=0}^{\infty}E^{\frac{a_i}{b_i}}$$

looks fishy...

Akiva Weinberger

Well, if $\Delta=E-I$, then $E=I+\Delta$

Secret

O yes, and then I can do $E^h=(I+\Delta)^h=\sum_{k=0}^{\infty}\binom h k (I+\Delta)^k$, that should work (Bonus, polynomials of the form $(I+A)$) are known to commute

Zee

How the hell can I start reading latex on this phone

Akiva Weinberger

You don't even need $[f,\Delta^h]=0$ for noninteger $h$

@Zee tinyurl.com/cfqcvpc

Scroll down

Semiclassical

Does that actually work? I had figured it'd be useless on mobile.

Fawad

3:51 PM

Hi

Zee

I still dont know what their talking about, might as well be topos theory

Akiva Weinberger

@Semiclassical I almost exclusively use the chat on mobile

Semiclassical

huh

Akiva Weinberger

I am currently on mobile with LaTeX on.

Fawad

What is $\Sigma_{r=1}^n\dfrac 1r$

Zee

3:52 PM

Where do I paste that script?

Simply Beautiful Art

@Fawad Harmonic numbers

Adeek

hi

Akiva Weinberger

What browser are you using? @Zee

Secret

@Fawad :

Harmonic number

In mathematics, the n-th harmonic number is the sum of the reciprocals of the first n natural numbers: H n = 1 + 1 2 + 1 3 + ⋯ + 1 n = ∑ k = 1 n ...

Akiva Weinberger

@Fawad You used $r$ as the index but $n$ in the sum

Zee

3:53 PM

Safari

Simply Beautiful Art

@Semiclassical Haven't you tried out the new glasses? They render MathJax on sight so you don't have to use any bookmarks and stuff.

Akiva Weinberger

Ah, yeah, I don't know how to do bookmarks on that

Felix.C

hey if we have a function $f:\mathbb{R}^n \to \mathbb{R}^n$ the partial derivative is a column vector right? While the total derivative of $g:\mathbb{R}^n \to \mathbb{R}$ is a row vektor right?

Zee

Well , would chrome work?

Fawad

@Zee yes

Semiclassical

3:54 PM

Oh damn, it works (and I use Safari on my phone)

Zee

How'd you do it?

Secret

@Felix.C $f$ from its definition should be a nxn matrix and $g$ is a row vector

Akiva Weinberger

@Secret Oh, you actually need to prove that $E^h=\sum\binom hk\Delta^k$… I mean, the formula works fine for actual numbers (within the radius of convergence), but how do you know it works on operators?

But, whatever, I think you're essentially done anyway

Semiclassical

Pretty much just the instructions there: Create a new bookmark, taking the link to be the text shown on robjohn's page, then open that bookmark when you're already in chat

Felix.C

@Secret You sure? I doubt so...I speak about the partial derivative not the total, but nice so we agree at least about the second one

Akiva Weinberger

3:56 PM

In any case, yeah, that's how you prove that if something commutes with $E$, it commutes with $E^h$. @Secret

Zee

It's saying JavaScript can't be used that way, oh well

Semiclassical

odd.

Akiva Weinberger

Works fine on Chrome, at least.

Secret

@Felix.C The partial derivatives of $f$ will be given by the jacobian matrix if I recall

Zee

Am gonna try it on chrome, so I can start sharing my mathematical wisdom with you all

Secret

3:58 PM

but otherwise, yes, each partial derivative on $f$ is given by a column or row vector (depending on which jacobian convention you are using)

@AkivaWeinberger I like the derivative approach better. No need to worry about convergence issues which I am currently still so bad to handle them rigorously

but nice to see the other approach, will keep that in mind

Simply Beautiful Art

Use the following integral formulas to get a form of the harmonic numbers:
$$\frac1r=\int_0^1x^{r-1}~\mathrm dx=\int_0^\infty e^{-rx}~\mathrm dx$$

Fawad

@Zee you need to bookmark any site and replace it url by javascript. Then when you open math.se chat tab,click on bookmarks and run that edited url

Akiva Weinberger

@Secret To be complete, here is my entire solution (for the case of $E^{1/2}$):

Define $\Delta_h=E^h-I$. Note that $E=(\Delta_{1/2}+I)^2=\Delta_{1/2}^2+2\Delta_{1/2}+I$.

Fawad

@SimplyBeautifulArt not required As of now.

Akiva Weinberger

Which means that $\Delta=E-I={\Delta_{1/2}}^2+2\Delta_{1/2}$.

Fawad

4:02 PM

I am solving this

Felix.C

hm... I just thought the total derivatives of the $f_i$ would form the rows of our matrix (total differential of f, e.g. df) as with $f_i : \mathbb{R}^n \to \mathbb{R}$ and $f = (f_1,...,f_n)^T$ our original function and then the partial derivatives would form the column vectors of df.
Edit: typo
@TedShifrin You mind having a quick look on this please :)

Akiva Weinberger

We know that by hypothesis that it commutes with $E$, and this with $\Delta={\Delta_{1/2}}^2+2\Delta_{1/2}$. Right?

Simply Beautiful Art

@Fawad You don't need the harmonic numbers for that. Note the general term is given by:
$$a_n=1-\frac1{2^n}$$
And so it breaks down mostly to a geometric sum

Akiva Weinberger

Now, $\Delta_{1/2}=\sqrt{({\Delta_{1/2}}^2+2\Delta_{1/2})+I}$. Yeah? @Secret

Semiclassical

Plus, it's a multiple choice question.

So you could just test terms until only one possibility remains.

Simply Beautiful Art

4:04 PM

@Fawad It should be $\frac1{2^r}$ starting with $r=1$.

Ted Shifrin

@Felix.C: Yes, the ith row of $df$ is the derivative $df_i$.

Balarka Sen

Hi @Ted

Secret

yup, if it commutes with $E$, it also commutes with $E-I$

woa, you took the squareroot of this bulky thing...

Akiva Weinberger

Expanding out binomially, we get $\Delta_{1/2}=\sum\binom{1/2}n ({\Delta_{1/2}}^2+2\Delta_{1/2})^n$.

Semiclassical

So $n=1$ gives 1/2 (rules out a,d)

Ted Shifrin

4:05 PM

Hi @Balarka ... My last night in Croatia. To the airport at 5 AM.

Semiclassical

and $n=2$ gives 5/4 (rules out b)

So it'd better be c.

Balarka Sen

@TedShifrin Where next? :)

Akiva Weinberger

(This is legal because each coefficient is determined by a finite sum)

@Secret And thus, if it commutes with ${\Delta_{1/2}}^2+2\Delta_{1/2}$, it commutes with $\Delta_{1/2}$. QED.

Felix.C

@TedShifrin And the partials the columns so?

Semiclassical

Not the most substantive approach, but probably the fastest.

Ted Shifrin

4:06 PM

A fixed partial gives a column.

Simply Beautiful Art

@Semiclassical Have you checked if c is correct? :D Perhaps the question is wrong

Ted Shifrin

@Balarka: 21 hours home.

Balarka Sen

@Ted: Wee!

Safe flight.

Akiva Weinberger

@Secret Note that, in the ring $\Bbb R[[\Delta_{1/2}]]$, every element has at most two square roots. Since, in the above equation, one can check that the RHS squared is the LHS, we can be sure that the equation is correct.

Felix.C

@TedShifrin thanks a lot, have a good flight tho :)

Akiva Weinberger

4:07 PM

Well, up to a possible minus sign. Not that it matters.

Fawad

@SimplyBeautifulArt I wrote 1=2-1 ,3=4-1,7=8-1 etc. nth term will be $\dfrac{n-1}{n}$ so sum will be $n-\Sigma_{r=1}^n\dfrac 1n$ no?

Simply Beautiful Art

@Fawad Nope. Try $n=3$

Ted Shifrin

Three flights with several customs clearances ....

Akiva Weinberger

(You can always have at most two square roots in rings without zero divisors.)

Simply Beautiful Art

@Fawad the denominators are powers of two.

Akiva Weinberger

4:08 PM

('Cause $A^2=B^2$ implies $(A-B)(A+B)=0$, so either $A=B$ or $A=-B$.)

Ted Shifrin

Not usually, DigAteMy!

Simply Beautiful Art

@TedShifrin Haha, I do hit that shift or caps lock instead of the 'a' key from time to time.

Akiva Weinberger

@Secret Without that last bit, all we could know is that $\sum\binom nk(\Delta_{1/2}{}^2+2\Delta_{1/2})=:A$ is just some operator that satisfies $A^2=E$. The last bit confirms that $A$ indeed equals $E^{1/2}$.

@TedShifrin In a ring without zero divisors, I said

Zee

Wow! Finally can read latex, it's like magic

Ted Shifrin

I thought matrices ...

Abdullah UYU

4:10 PM

Hi, i am taking care of this question at this time, i can talk with the guys also interested in, math.stackexchange.com/questions/2336500/…

Semiclassical

@SimplyBeautifulArt Ehhh. If the question is wrong, there's nothing to do in the first place. So might as well go with the only answer you haven't ruled out as absurd :P

Akiva Weinberger

@TedShifrin Well, these can be infinite matrices… The context is linear operators from the polynomials to themselves

Semiclassical

@Zee I'm kinda amazed it works on mobile myself.

Fawad

@SimplyBeautifulArt yes. I was wrong.

Simply Beautiful Art

@Zee xD

Akiva Weinberger

4:11 PM

@TedShifrin Fun fact: Define $E$ to be the shift operator defined by $Ef(x)=f(x+1)$. (And $D$ is the differentiation operator $Df(x)=f'(x)$, which is much easier to write in matrix form.) Then:

Secret

Ok... that's quite different from my usual thinking patterns. For me, I tend to try to use a given $A$ and tries to massage $B$ in terms of $A$ so I can bubble things out. Your approach is more like you express $B$ in terms of $A$ and since $A$ is true, then $B$ is true

I recall my linear algebra professor have pointed out this problem of mine before, and she and one of my abstract algeba maths friends tend to find my proofs go backwards

Ah, so there is indeed something I don't quite aware in your solution: Polynomial rings!

Semiclassical

That said, one check to run on c is what the difference between consecutive terms is.

Akiva Weinberger

@TedShifrin If two linear operators commute with $E$, they commute with each other

That is, shift-invariant linear operators commute

Semiclassical

the difference between the (n+1)th term and the nth term, for instance, would be $[2^{-n-1}+n]-[2^{-n}+n-1]=1+2^{-n}(1/2-1)=1-2^{-n}$

Akiva Weinberger

Also, commuting with $E$ is equivalent to commuting with $D$.

Ted Shifrin

4:12 PM

This isn't something I recognize offhand.

Semiclassical

And, hey, that's exactly the form expected.

Akiva Weinberger

(Just for things that go from polynomials to polynomials.)

Ted Shifrin

I flunk backwards proofs, @Secret.

Fawad

What is $\Sigma_{r=0}^n\dfrac{2r+1}{2^{r+1}}$?

Secret

what's even worse is that I often don't recognise my proofs are backwards, and I always thoguht I am proofing in the forward direction. Imagine a person who walks backward with the head facing forward, that's my issue in maths thinking in general

Akiva Weinberger

4:15 PM

Don't use \Sigma, use \sum @Fawad

Semiclassical

@Fawad Is that what you think that sum is? If so, it's not.

Zee

@TedShifrin may you have a marvelous and magical vacation

Simply Beautiful Art

@Fawad Nope, the numerator is equal to the denominator minus one.

Secret

@TedShifrin Here's an example of what it feels like:

3

Q: Why, in terms of the structure of proofs and proof strategy, is this proof of mine said to be backwards in logic?

Last year when I was doing the linear algebra and proof writing course, I was often said by my friends and my professors that the logical flow of my proofs are weird or even backwards. Recently, I accidentally stumbled upon this site. After analysing it and comparing with the standard approaches...

> At this point I realise that you've been writing 0=0 in all kinds of obscure and complicated ways.

jokingly speaking, its as if my mind is being placed into a klein bottle before I born

Akiva Weinberger

@Secret Originally my proof was in the same direction as yours. But then I had a subway ride in which I realized, to my horror, that I hadn't actually proven it commutes with $E^{1/2}$; I had proven it commutes with some $A$ that satisfies $A^2=E$

Ted Shifrin

4:17 PM

@Zee: I am at the end.

Akiva Weinberger

But then I realized everything in the proof lives in $\Bbb R[[\Delta_{1/2}]]$, where unique square roots exist. So all is good

and I rewrote my proof from that perspective.

Ted Shifrin

We'll discuss more when I'm home in a few days, Secret..

Bye all.

Semiclassical

There's a quotation which I can't remember regarding a proof that A=B amounts to proving $A\geq B$ and $A\leq B$.

Fawad

Alright. It is $n-\sum_{r=1}^n\dfrac {1}{2^r}$ so what is $\sum_{r=1}^n\dfrac {1}{2^r}$?

Semiclassical

@Fawad it's a geometric series.

you should be able to do that much yourself.

Secret

4:19 PM

Generally speaking, based on all the user feedbacks so far, I think I tend to do maths like a constructivist (and somehow it goes backwards for unknown reasons)

That is, given a problem, I analyse the given conditions available, treat those as resources, and then use them to build my way to the solution required by the problem

Fawad

The cringe is real now.

Secret

Most mathematicians, however work in the opposite way: They keep an eye on what is required, and then based their proofs by thinking back from the desired solution

Simply Beautiful Art

@Fawad Hint:
$$\frac12=1-\frac12\\\frac12+\frac14 =1-\frac14\\\frac12+\frac14+\frac18=1-\frac18$$

Secret

O and to top that, number theory still look like magic

and I still procrastinating on my chemistry literature review because maths is currently too interesting

Akiva Weinberger

@Secret Same

Secret

4:24 PM

Joke: Meanwhile, it took me $\omega$ steps to crawl from 1/3/2017 to 6/3/2017, talk about very slow growing function

M/D/Y or D/M/Y

D/M/Y

Oh :(

This period has a lot of Waiting integrals and some algebra and topology, hence I frequently get distracted an talk about them

Akiva Weinberger

(The best calendar system is YYYY.YYY…, of course)

(Wait, no, ssssssssss)

In all seriousness, though, it's YYYY-MM-DD.

Secret

4:29 PM

I like research, but I really don't like the dreadful feeling of reading something from scratch (past literature and so on), though like reading novels, once you start, you cannot really stop. the issue is how to minimise that so called "warm up period"

Akiva Weinberger

Think about it. The first digit is millenia; the next digit is centuries, the next is decades, years, tens of months, months, tens of days, days

No other calendar system goes all in one direction like that.

You can sort it easily, for example

Waiting

@Hippalectryon o/

Secret

2024.12.03.16.02.02.0999

@Waiting Weird question: How many functional identities do you know/derived/etc., rough estimate is ok?

and what proportions roughly are series identities?

Waiting

@Secret A few compared to all the questions I created, but most of them are not known in the mathematical literature.

On this realm I have something very special (but as I said, not many compared to all my created questions).

@Secret Surprisingly (if one can say that), some involve both integrals and series.

@Secret They (or better say, some of them) will be available soon in a publishing form or other.

Zee

I should do more math, am slacking

@Waiting I highly doubt that

Secret

4:39 PM

Nice, I am guessing you have quite a rich understanding to a largely uncharted region of what I called the space of all functional identities (which via mine and Leaky Lun's discussion, it is countably infinite).

Recently, as part of my personal investigation on integral symmetries, I had been discussing with various users on what makes integral in general hard and will eventually came to a consensus that nearly everything about closed forms hinge on pathways in this space

Your knowledge base kinda confirms my suspicion on this hypothesis

Semiclassical

Ugh, I can't track down this source.

I know I've seen it elsewhere, but I can't find it.

Secret

This is an illustration on what my personal research involving integrals are like

Here, I am mostly in a scientist mindset more than an artist. I try to understand how integrals works in a deeper level

Waiting

@Zee It is something common to hear in this place.

Zee

@Waiting talk is cheap

Secret

@Waiting Identities involving both integrals and series are common in the domin of integro differential equations, so it is nice to be able to expand on the knowledge base on that

Zee

4:43 PM

@Secret what does the "unknown" link?

Waiting

@Zee also your opinion is cheap for me, you can say anything. ;)

Secret

@Zee The unknown represents the direct pathway to go from that integral to its antiderivative closed form. I highly doublt people can do that by inspection

which is why when solving integrals and series, we mostly go round about

and that is only possible because of those functional identities

what I ilustrared up there, however is one of the many pathways to the closed form of this integral, and for that I have a conjecture

Zee

@Waiting I can say anything yet I said what I said, don't discount people

@Secret what's a functional identity?

Akiva Weinberger

Something like $\Gamma(n+1)=n\cdot\Gamma(n)$

Secret

A functional identity is any relation that relates one or more functions. Examples include simple things like $(a+b)^2=a^2+2ab+b^2$, to more complicated things like differential equations

Akiva Weinberger

4:48 PM

Identities using functions at different values

Semiclassical

$\cos^2x+\sin^2x=1$ being another obvious one.

Akiva Weinberger

I was just about to bring up the sum formulae for sine and cosine

Secret

they are what we believed to be the source of the symmetry of integrands and summands. That is, we conjectured that they generate the symmetries that make close forms exists for them

Semiclassical

typically one means relations between special functions, though.

trig function relations being a case where familiarity breeds contempt.

Akiva Weinberger

It's the sort of thing that helps you convert one sequence of symbols into another

Secret

4:49 PM

The space of all functional identities is a collection of every of those things, which Leaky Lun had determined to be countably infinitely large via a finite alphabet arguement

Zee

What's the difference between a functional identity and a formula

Or equation

Semiclassical

well, an equation typically only has to be valid for some specific solutions (or may not be valid at all)

Akiva Weinberger

Usually you specifically want it to say somethin nontrivial about a special function (or a few special functions)

Secret

but functional identities have to be valid for all points of the functions involved

Semiclassical

I tend to associate formulas as well with writing one complicated thing in terms of some combination of simpler things.

That doesn't have to be the case for functional identities. But at some level it does seem rather semantic.

Secret

4:52 PM

e.g. $(a+b)^2=a^2+2ab+b^2$ is valid for all $a,b$, as long the ring commutes

Zee

I don't like jargon so am just gonna refer to them as formulas

Semiclassical

sorta like the difference between 'formula' and 'identity' in trig.

which is to say, not much of one.

Hippalectryon

@Waiting o/ sorry I was away

Secret

well, they are technically formulae under the language of {+,-,*,/,f,\circ,etc.} so you are correct

Waiting

@Secret there will appear soon such a book, containing outstanding functional identities like the ones posed by Ramanujan (I heard it, I have my sources).

Secret

4:54 PM

cool, cannot wait to see the region of this space you explored and how it connects to the grand scheme of things

Zee

no source identified, did not happen

@Secret what are you trying to achieve with this?

Waiting

@Hippalectryon How is it going? :-)

@Secret Hope you'll be lucky enough and get a copy.

Hippalectryon

@Waiting fine :D I've got a problem about integrals that's bothering me, it's probably fairly simple for you but somehow I haven't got a conclusive answer :( mind if I share it ?

Waiting

@Hippalectryon share it

Secret

@Zee Well, the initial motivation is I want to be stop being mysterified by how people come up with nontrivial substitution that solves integrals by as if magic happens. I want t understand why I am confused and I really hate being confused, thus my response to the unknown is to solve it, so I can be at peace

Hippalectryon

4:57 PM

I'm trying to show that $\int_1^\infty\sqrt{\frac1{(1-t^2)^2}-\frac{(n+1)^2t^{2n}}{(1-t^{2n+2})^2}}\sim \log(n)/2$

Secret

The attitude of abhor any unexplained phenomenon is the main driving force of most of my researches and dedictation. As I always said to my friends. My purpose of life is to unlock the mysteries of nature

Zee

@Secret and how are you planning to go about this?

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$Mathematics$ Mathematics