Mathematics - 2017-07-06 (page 1 of 10)

Simply Beautiful Art

00:00

Wondrous patterns, it just so happens that the rules to the Goodstein sequence are pretty much equivalents to Hardy hierarchy :-)

From the Hardy hierarchy, we can build the fast growing hierarchy.

And from there you can reference any notation scheme, such as Conway's chained arrows (which only go up to about $f_{\omega^3}$ as far as I know) or BEAF (Bower's exploding array notation)

Sir Cumference

I'm letting my sister use my account to ask a calc question.

Simply Beautiful Art

lol, okay

Sir Cumference

So yeah

Akiva Weinberger

Is this sister better or worse than you at math

(Just kidding, don't answer that)

Sir Cumference

Worse, I'm above calc :P

Akiva Weinberger

00:10

Ah, OK :P

You can't answer her question, then?

John Locke

I have a question. In fact, I did ask it about an hour ago on the main site (math.stackexchange.com/questions/2347794/…) but am still struggling with figuring it out. Can anyone help me?

Sir Cumference

@JohnLocke I'm a fan of your work, Mr. Locke

John Locke

Haha, funny enough I chose the name because I enjoy political science and theory and struggle profoundly with mathematics

Simply Beautiful Art

Oh yes, and my question still hangs. If $x=a/b$ and $\operatorname{gcd}(a,b)=1$, then we define $f(x)=a\cdot b$. Does the sum converge?
$$\sum_{x\in\mathbb Q^+}\frac1{[f(x)]^2}$$

@JohnLocke I do not believe you struggling as much as some others :)

Sir Cumference

Hi quick question

Can someone show me how to find lim(x to 0) sin(x)/x?

Akiva Weinberger

00:20

That's a famous one

Sir Cumference

yeah, my prof was somehow using squeeze theorem

Akiva Weinberger

I'd guess you've already learned at least one proof of it in class

I see

Sir Cumference

Completely lost me

Akiva Weinberger

There's a standard proof using the unit circle and areas. Does that sound familiar?

Sir Cumference

Yeah

Simply Beautiful Art

00:21

Famous indeed... Here's the math.se link to that question:

283

Q: How to prove that $\lim\limits_{x\to0}\frac{\sin x}x=1$?

How can one prove the statement $$\lim\limits_{x\to 0}\frac{\sin x}x=1$$ without using the Taylor series of $\sin$, $\cos$ and $\tan$? Best would be a geometrical solution. This is homework. In my math class, we are about to prove that $\sin$ is continuous. We found out, that proving the above s...

Akiva Weinberger

Sir Cumference

so wait, does squeeze theorem deal with graphed equations or areas? I mean, if i took the integral of a function and got its area, could I use it in squeeze theorem?

Akiva Weinberger

Here's part of a unit circle. The angle of that sector (at $A$) is $x$.

Ted Shifrin

I saw your earlier page, DogAteMy. growl

Simply Beautiful Art

@SirCumference Well the first question is how are you defining sin(x). If you are defining it geometrically, then this is the way we should go.

And we're not dealing with area. the picture just happens to have color. Focus on the lines on the right side, there is a vertical line, a slanted line, and an arc.

Akiva Weinberger

00:23

@SirCumference The squeeze theorem applies whenever you know that the thing you're takin the limit of is in between two other things that go to the same limit

John Locke

@Akiv

Sir Cumference

Akiva even if we were talking about volumes, etc.?

Ted Shifrin

That is how Archimedes did everything

squeezing from inside and out

John Locke

**sorry I sent something before I meant to. @AkivaWeinberger can you help me understand further? Your hint is helping me a little more but not quite there yet. When you are done explaining to Sir Cumference, can you help me here too?

Akiva Weinberger

I might actually need to get off chat and start dinner

Ted Shifrin

00:26

you've become indispensable, DogAteMy — no escaping now

Akiva Weinberger

Hopefully Ted and Simply can help Cumference

@TedShifrin Oh, no!

Simply Beautiful Art

@TedShifrin xD Escaping? From what?

Akiva Weinberger

(and that link above)

Daminark

Hai waves

Sir Cumference

So wait Simply, I remember areas being involved

Akiva Weinberger

00:26

And, uh, John, it ends up being bijective with something easy to count

Ted Shifrin

You can do that limit in different ways, @SirCumference. Most calculus books base the argument on area.

One can do it with lengths, too.

Simply Beautiful Art

Hm, I don't think area is involved in this limit @SirCumference ... lemme read the proof a bit closer lol

Akiva Weinberger

Oh, also, your sibling says not to flip out on him/her

Simply Beautiful Art

Yeah, no area required.

Sir Cumference

I would like to understand the area argument

Akiva What?

Ted Shifrin

00:28

@SirCumference: Call the point between A and B (below C) D.

Sir Cumference

@TedShifrin Yeah

Ted Shifrin

No, E. Sorry. Write down formulas for the area of $\triangle AEC$, sector $ABC$, and $\triangle ABD$, and use the fact that they're related by inequality.

Sir Cumference

Wait what?

Ted Shifrin

I changed the name to E. I realized D was already used.

E is the point below C on AB.

Sir Cumference

Oka

Ugh *okay

Ted Shifrin

00:31

Howdy, Demonark. So how did the first bootcamp meeting go?

Simply Beautiful Art

I think one can do the limit without any area. You just need to consider lengths and recall that the circumference of a circle is $2\pi$.

Sir Cumference

@AkivaWeinberger for what?

Ted Shifrin

I said you could do it with lengths or with areas. Most calculus books use area, because length is a much more subtle notion. And you need to know things like: The line segment joining two points is the shortest path between them.

Try proving that. :)

Daminark

It was interesting. For the first day, one of last year's people gave an introductory lecture, started off on limit switching, then the basics of what it means to be a holomorphic function and why it's that much more powerful than just being $\mathbb{R}^2\to\mathbb{R}^2$ differentiable

Sir Cumference

I'm still confused on how area is related to the limit

gtg be back in 20 mins

Daminark

00:33

Did the complex exponential and log, along with a branch

Ted Shifrin

It's important to understand the distinction, Demonark. And real differentiable + conformal derivative isn't sufficient.

Simply Beautiful Art

Nice @Daminark

Daminark

Oh really?

Ted Shifrin

Not just at a point, nope, Demonark. If you assume $C^1$, then sure.

Daminark

I mean we said that the real and imaginary parts were $C^1$

Ted Shifrin

00:35

(I think. I'd have to rethink it all.)

Daminark

Sniped, but yeah

Ted Shifrin

Ah.

Simply Beautiful Art

lol

Daminark

Hmm, I wonder what's a counterexample

Ted Shifrin

There's always a definition debate, too. Can you be holomorphic (complex differentiable) just at a point, or must it be on an open set?

Daminark

00:36

We defined it as being alright at a point

Ted Shifrin

So $|z|^2$ is complex differentiable at the origin, but nowhere else.

Eric Silva

In my head its weird to just be at a point

Simply Beautiful Art

Hm, my definition is that a function $f$ is holomorphic at $z_0$ if $\lim_{z\to z_0}\frac{f(z)-f(z_0)}{z-z_0}$ exists.

Daminark

That's how we did it. I guess it'd correspond to like, a power series with radius of convergence 0, which is bizarre to say the least

Or at least it registers to me somewhat bizzarely, but I guess that can work?

Ted Shifrin

No, no, holomorphic and analytic are not equivalent unless you're on an open set.

Simply Beautiful Art

00:39

^

Ted Shifrin

It's important to understand the distinction.

Daminark

Oh, I see

Ted Shifrin

And hi @EricS

Eric Silva

I think the reason I think at a point is weird is because I want holomorohic to mean analytic

Ted Shifrin

oh damn, still pings both EricS's.

Eric Silva

00:39

Hi Ted

Ted Shifrin

No, they're conceptually totally different, ERic.

Eric Silva

Lol

Simply Beautiful Art

My 'other' form of being holomorphic is if Cauchy's equations are satisfied at $z_0$, then $f$ is holomorphic at $z_0$.

Ted Shifrin

That requires some care (like assuming $C^1$), @Simply.

Simply Beautiful Art

yeah

Eric Silva

00:40

Ik that they mean different things

Ted Shifrin

oh, no, sorry. C-R equations do not.

Daminark

I think I remember someone mentioning some time back in the chat that if the function is continuous and satisfies C-R, that should do it

Simply Beautiful Art

No? Hm... well, too busy to think too much

Ted Shifrin

Eric, I always want to keep the definitions distinct. It's sort of like the definition of invertble and the definition of nonsingular, even though a theorem says they're equivalent.

Simply Beautiful Art

Can a function be analytic at one point?

Secret

00:42

How can one possibly have something that is nonsingular, but not invertible?

I need to revise what singular means, does it involve singularities as part of its definition

Daminark

@Secret I think the point was more to keep the fact that they're equivalent as the theorem, and to emphasize that a priori they're not the same

Ted Shifrin

@Secret: I was referring to square matrices. Sorry.

Eric Silva

I think that's fair. But I think im not trying to say that i want to think of them as being the same thing, i think it's more that i want my definition of holomorphic to imply harmonicity -> analytic

Ted Shifrin

Well, I'm now officially listed as an instructor :P

3

Simply Beautiful Art

@TedShifrin Doesn't Cauchy's equations come directly from the limit definition of the derivative without assumptions?

Eric Silva

00:45

i think if i wanted "holomorphic at a point" i would use complex differentiable

Simply Beautiful Art

@TedShifrin :P You were always an instructor here

Daminark

@Ted You mean for aops?

Ted Shifrin

Not exactly, Simply.

Yes, Demonark. They put my name up.

Daminark

Nice

Ted Shifrin

Yeah, @Simply. I retracted my negation.

Secret

00:46

Daminark: That true, though I am thinking something more wild. I am thinking about what happens if we nuke an important step that leads to that theorem to establish, then what it will mean if those two become different. (because the theorem is nuked)

hmm, rank is not really well defined for infinite matrix...

Simply Beautiful Art

x.x you can't be responding so vaguely to multiple questions like that @TedShifrin

Ted Shifrin

@SimplyBeautifulArt THAT.

Simply Beautiful Art

Which response though x.x

Daminark

womp womp

Ted Shifrin

ignores Simply

Eric Silva

00:48

@Daminark did you guys decide who's lecturing yet

Simply Beautiful Art

lol, okay

Eric Silva

also did you get told your dynamics hw

Daminark

Yeah, the two Chriss are lecturing on Friday

And we weren't given homework, just the lecture

Eric Silva

ok well im p sure we're just gonna tell you guys to do everything in the assigned sections

cause there arent that many problems

Ted Shifrin

Be a mean teacher, Eric. Oh, write more problems, then. :D

Daminark

00:49

Makes sense

Eric Silva

the thing is i dont really know dynamics so i dont really have much that's good to give them

Secret

Hmm...
Singular = not have full rank
Invertible = $A^{-1}$ exists
now what is the name of that theorem that ensure they are equivalent...

Eric Silva

we have a guy who knows a looot of dynamics who will be in charge of being mean to them

except when we get to the diff geo i plan on being very mean... :)

Ted Shifrin

It's not got a name, @Secret. But it's a theorem nevertheless.

Daminark

Yeah it's a bit of a chain. Probably the best place to nuke would be rank-nullity, which is a thing when you pass to infinite dimensions, but then that gets messy with the whole, writing transformations as a matrix

Eric Silva

00:52

what do you mean nuke

Daminark

Presumably, if we were to just make the theorem false

Secret

@EricSilva Brief intro: One thing that I am famous on this chat is exploring mathematical structures by relaxing conditions to make some proofs fail

Eric Silva

oh i mean it's false in infinite dim

Secret

@Daminark Ah, then I can kinda expect a countably infinite matrix that is nonsingular but not invertible, I guess

Daminark

I mean, your problem is that coordinatization and matrices are not something you can naively deal with in infinite dimensions

Eric Silva

00:55

the most ik off the top of my head you could recover is for operators of the form $Id - K$ where $K$ is a compact operator from a hilbert space to itself

rank nullity is basically true for those things

Daminark

Matrices which aren't full rank correspond to transformations which aren't surjective, and it's easy to find surjective operators which aren't invertible in infinite dimensions

Left shift on $\ell^2(\mathbb{N})$

Secret

yup, that's the classical example

Daminark

That's something you should actually be able to write as a matrix well enough

Eric Silva

@Daminark i would say things that aren't full rank are things that aren't injective but ok

Secret

If I recall, the shift operator under the standard basis is a lower subdiagonal of 1s. In iinfinite dimensions, however, you have the problem that you cannot use a hamel basis

Daminark

00:59

Ah, yeah that'd be an issue

@Eric I've seen some sources define rank of a matrix as dimension of the image, the other as being the number of linearly independent columns, and then each said that the other was the "theorem"

Eric Silva

in finite dim it's the same, but i think that in infinite dim it usually makes way more sense to talk about kernels than ranges

so i prefer to think about it through the kernel

Typhon

@TedShifrin you still here?

Daminark

I guess I've heard of module rank so the latter might be better, though the former is what I thought of first

I guess that's fair, yeah

Eric Silva

ranges usually just dont behave as well

Typhon

I missed him

darn

Secret

01:03

[Unrelated pondering]

Eric Silva

@Daminark plus i reaaaaally dislike the idea that a linear embedding of one space into another would not be full rank

Typhon

@Secret do you understand the geometry I was describing before? You can see how some rigid might be one way, right?

Secret

I wonder if it is possible to find two definitions which are shown to be equivalent via a theorem, but the theorem is "unnukable" because any attempt in doing so you end up throwing away so much that the resulting structure is not useful

Typhon

@Secret imagine if one side of that protrusion for the "T" from before simply didn't exist.

The one side did

but the other didnt

the resulting geometric paradox is.. terrifying

Scary enough: that can actually be a valid 3D model

Eric Silva

@Secret are you still talking about rank-nullity

Secret

01:08

@EricSilva nah, something more general, any maths

Eric Silva

ah then this statement is vague enough that i have no idea what it could mean

Secret

It's one of those "meta blow out of proportions" questions of mine that I think I will only be able to find the answer much later

Daminark

It saddens me that we're not Lebesgue-ing the stuff

Eric Silva

why

Daminark

Fubini/LDCT would be very convenient for the early stuff on limit interchanging

Eric Silva

01:13

i mean it makes it like

not hard

Simply Beautiful Art

@Daminark lol

Secret

I am honestly not very sure whether one can determine which triangle the geodesic can end up at unless the geodesic is consists of a bunch of line segments and has coordinates at each junction (then you might be able to use the criteria of whether a segment of the geodesic (determined by a pair of coordinates) passes through a triangle's side

If that still end up twisting your quesiton, then I think my mind is tying up into knots that I think I don't know if I have enough background to answer your question

In pictures:

Simply Beautiful Art

Let the children learn the fundamentals and their challenges and let them truly relish in their victories rather than handing them over so easily.

Typhon

@Secret fair enough. It might be a geometry worth bringing up next semester.

Secret

In other news, I just sent my first of my two troubleshootnig reports to the software company to let them sort it out

Now preparing the 2nd one

Daminark

01:17

@Eric We'll get to other hard things later I imagine or something

Eric Silva

the whole thing is supposed to be hard

it's why schlag uses the book

Titchmarsh is meant to put the fear of god in you

Secret

Fun fact: If $a\cdot$ distributes over $b+c$, then $a\cdot$ is a homomorphism over +

(distributive law in a nutshell)

Semiclassical

hi chat

Just win baby

hi chat

Semiclassical

@EricSilva Whereas hearing about it is meant to put the fear of Titchmarsh in you

Eric Silva

01:21

lol

Just win baby

Isn't that what the word "distribute" means? @Secret

Semiclassical

I've never actually seen Titchmarsh, I'll admit.

I saw references to it while reading some spectral theory papers, but never have had reason to consult it myself.

Secret

@Justwinbaby well yes, but in a more abstract level, obeying the distributive law means all elements of the ring under multiplication is a homomorphism over +

and homomorphisms are very useful in exploring algebraic structures and proving things in abstract algebra

Eric Silva

@Semi the man can manipulate trigonometric series like a monster

Semiclassical

neat.

Eric Silva

01:24

i thought i hated it when i read it but a year down the line i appreciate the chops it gave me

Secret

It might not be obvious, but distributive law actually constraint the algebraic structure a lot

(as well make it nicer to handle)

Just win baby

It's a property of real numbers.

Daminark

Lmao, I guess I'm just kinda fond of nuking potatoes from orbit sometimes, something something efficiency something something

Secret

for example, ordinals violate the right distributive law

Simply Beautiful Art

Woohoo for ordinals lol

(the one thing I managed to notice in the chat while doing my homework lol)

Eric Silva

01:27

@Daminark there's a place and a time for it

Simply Beautiful Art

@Secret Just the transfinite ordinals. Finite ordinals follow the 'normal' laws of arithmetic

Daminark

shrugs

Just win baby

Potatoes? @Daminark

Secret

@SimplyBeautifulArt That's true, though when I talk abotu ordinals, I tend to not talk about the finites and focus on the more interesting transfinites

Eric Silva

i try not to use fancier things than the author expects of me @Daminark, especially when im reading a book to gain techniques as opposed to raw info

Secret

01:30

[Warning, currently non mainstream maths] Distributive law is also one of the two reasons why it is so hard to have division by zero (following showed one of the workings I had from last year)

Daminark

Yeah, I guess I haven't really been in that context much before

Secret

Daminark

I've definitely been somewhat iffy on machinery sometimes in subjects like graph theory

Eric Silva

what do you mean by machinery

do you mean techniques or theory or like pulleys and wheels

Daminark

Like, in graph theory, if a problem was more "Ah-hah!", I'd be less inclined to try to throw, say, linear algebra at it

Eric Silva

01:31

idk what that means

Semiclassical

I think I may, though the example I know isn't quite 'math'

Typhon

0

Q: What kind of geometry can I classify as "3D modeling" and why do I get surfaces with *branching lines*. (Geometry paradox)

Most people assume that 3D models are just a collection of triangles. This is false. They are a collection of one sided triangles. In other words, one side exists but the other does not. This is known as backface culling. I can thereby construct the following geometric puzzle regarding rigid moti...

geometry riemannian-geometry 3d surfaces paradoxes

Semiclassical

Namely, Sudoku.

Daminark

There are definitely some problems where doing something by linear algebra allows you to just pull off some matrix thing and then the result reveals itself, but that there's a combinatorial insight in doing it directly

Typhon

@Semiclassical I think I found 2d wormhole geometry. Any comments?

Semiclassical

01:32

What I always start with there is just trying to proceed by inspection to make as much progress as I can.

Secret

in simple terms, no matter what you structure is, a one sided additive identity multiplied to anything will itself be a one sided additive identity if the distributive law holds

Typhon

or confirmations?

Daminark

In calculus and analysis so far I've definitely found that most of the problems were either conceptual ones, or "I want to make sure you don't forget what an integral looks like"

Just win baby

Indeed @Secret the distributive property is used in the proof of the multiplicative property of zero. Which is why you can not divide by zero.

Semiclassical

But if I find I can't make progress, I eventually do start annotating each cell in order to keep track of which numbers are possible and to find patterns.

Eric Silva

01:34

@Daminark you would be surprised how far you can get sometimes in analysis by just finding the right thing to integrate by parts

Just win baby

In the set of real numbers @Secret

Daminark

So my attitude has traditionally been to try to do things using high-powered theorems fitting together

Semiclassical

So I proceed by inspection+intuition as much as I can, but I'm willing to proceed more systematically if it becomes necessary.

Secret

@Justwinbaby I have got around that, but unless you have nonassociativity, your structures are still quite trivial

Semiclassical

@Typhon Huh.

Julius

01:35

Hi. I'm dealing with a graph of some entities, with $K$ vertices in total. I'm looking for examples of such entities that it would make sense for $K\to\infty$ but also that there would be some restrictions; e.g., that $K$ would always have to be even or odd.. any ideas? As it's a graph, something like a number of cells would be better than a number of divisors

Eric Silva

@Daminark conceptual problems can be straight computations man

Daminark

Lol actually one of the things in Titchmarsh was something where I actually was trying to parts it when it was easier to substitute

Semiclassical

I have a hard time spotting integration by parts.

Daminark

I mean I think my idea for partsing it out would've worked nicely

It was like, $\int_0^1 n^2x(1-x)^n dx$

Semiclassical

The only places where I usually remember it is when I'm doing something like "oh, I can turn $(\partial_x \psi)^2$ to $-\psi\partial_{xx}\psi$"

Eric Silva

01:36

I usually try to substitute before taking parts unless i already know from inspection what to do

Simply Beautiful Art

@Daminark $n\to\infty$?

Daminark

And my intuitive response was to say that if you sorta integrated by parts $n$ times, you'd pull up a $\frac{1}{n!}$ from $x$ but then multiply by $n!$ thanks to differentiating $(1-x)^n$

Secret

Just win baby:

Daminark

Just that Chris found the easy substitution $u = 1-x$ before I hashed out much on that front

Eric Silva

iusually end up guessing right but i usually do it by accident

Daminark

01:38

@Simply Nah, just leave $n$ there, it's a closed form expression

Simply Beautiful Art

Oh, okay

Secret

Daminark

The one I'm looking at now is interesting

Eric Silva

@Daminark just looking at i have no idea why you would try parts tbf

Semiclassical

One problem I had at one point which took me a while to understand

Simply Beautiful Art

01:39

Say, does the notation:
$$\sum_{x\in\mathbb Q^+}f(x)$$
make sense?

Semiclassical

yeah, I'd say so

Eric Silva

not like the ibp is actually bad

Daminark

I mean, if you look at it, you'll take the primitive of $x$, and then that, and so forth, $n$ times, so that'll lead eventually to $\frac{x^n}{n!}$

Eric Silva

but the substitution is like very immediate

why would you not differentiate the x

Semiclassical

I'm pretty sure I've seen $\sum_{n\in \mathbb{N}}f(n)=\sum_{n=0}^\infty f(n)$ for instance

Daminark

01:40

Then on the other hand, when you keep differentiating $(1-x)^n$, you'll keep multiplying until you get $n!(1-x)$, so that stuff cancels

Simply Beautiful Art

My initial response to that integral would just to use the Beta function -> Gamma function -> recursive formula -> closed form

Daminark

That method was just the first one to come to mind for some reason

Semiclassical

(and if anyone wants to argue about 0 being in N...please don't, it's a tiresome argument and I don't really care)

Simply Beautiful Art

@Semiclassical Yes, but it helps that $n\in\mathbb N$ translates into $n=0,1,2,3,\dots$

There is no simple ordering to $\mathbb Q^+$ when compared to $\mathbb N$.

Semiclassical

Sure, but the principle is the same: You evaluate the function at each element of the set, and then sum those up

Daminark

01:41

But yeah the problem I have in mind right now is this

It says that assuming $\frac{\sin(x)}{x} = \prod_{n=1}^{\infty} (1-\frac{x^2}{n^2\pi^2})$, show that $\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}$

Secret

not sure if the sum will be divergent though, it depends on $f$

Semiclassical

Well, sure.

Eric Silva

@Daminark fun one

i remember how to do it so i wont say anything

Semiclassical

But then $\int_\mathbb{R} f(x)\,dx$ doesn't need to be convergent for us to understand what the notation intends

Secret

yup

Semiclassical

01:43

@Daminark lol, that's how Euler proved it

Daminark

Right now my inclination is to take a log of the product, because that becomes a sum of logs, and then take the power series of $\log(1-x)$

Simply Beautiful Art

But who is to say all orderings provide the same sum? I suppose we can't invoke the rearrangement theorem, as this series is either absolutely convergent or divergent...

Daminark

It'll give you a double sum which should be absolutely convergent?

Semiclassical

@Daminark The intention is right, but you're making things too complicated.

Eric Silva

@Semi a lot of titchmarsh problems in complex is stuff euler did

it's p great

Secret

01:44

@SimplyBeautifulArt If the sum converges, then the ordering does not matter?

Semiclassical

huh, nice.

Simply Beautiful Art

@Daminark Extract the $x^2$ coefficient from the product:
$$\prod_{n=1}^k\left(1-\frac{x^2}{n^2\pi^2}\right)$$

Semiclassical

@SimplyBeautifulArt tsh, giving the game away :P

Simply Beautiful Art

@Secret That's absolute convergence

D: Oops, well, it still leaves a bit of work.

Anyways, g'night folks.

Eric Silva

@Daminark if you ever have time read Euler Master of us all

it's great

Daminark

01:45

For what it's worth, this is on the section about double sums

Semiclassical

I saw Euler's proof in "Journey through Genius"

Secret

$$\int_0^1 n^2x(1-x)^n dx$$ This integral has one obvious symmetry: polynomials are nilpotent under differentiation, thus the IBP will terminate

Semiclassical

Great little book.

Daminark

And this is $\log(\sin(x)) - \log(x) = \sum_{m=1}^{\infty} \frac{1}{m} \sum_{n=1}^{\infty} (\frac{x^2}{n^2\pi^2})^m$ (after doing the sum swap)

Semiclassical

@Secret Isn't doing $y=1-x$ simpler?

Daminark

01:47

@Secret you think like I do

Semiclassical

Though I guess maybe the point is just that it can be done

Daminark

I fully acknowledge that it seems like the "morally correct" way to think about it is through the substitution

Secret

ah yes, that too (thinking too much about IBP, I tend to just try to IBP everything)

Daminark

But for some reason this IBP just stood out to me

Simply Beautiful Art

._. No one thinks Gamma functions and beta functions?

Semiclassical

01:48

eh, no harm there.

Secret

The $y=1-x$ provides the shortest path in computing that integral, for reasons I still yet to understood

Semiclassical

Eh, nothing terribly mysterious. It retains the symmetry of the integral but swaps the exponents.

And small powers of 1-y are easier to deal with than large powers of 1-x.

Secret

I guess it might be because one can exploit the following functional identity $(1-y)y^n=y^n-y^{n+1}$ which is absent for the unsubstituted case

Eric Silva

it;s just cause it lets you avoid expanding a binomial

Semiclassical

The case of $\int_0^1 x^n(1-x)^n\,dx$ is harder, by contrast.

precisely because the above symmetry leaves the problem just as hard as it was originally.

Secret

01:50

That one I will just IBP it

Semiclassical

I tend to do the beta function for that one

Not because I think it's the right thing to do but because it works and I'm lazy

Eric Silva

that one is waaay harder

iirc that can be written in terms of $\Gamma$

Semiclassical

Yeah.

Simply Beautiful Art

Now what would you do if we replaced the integral with $\int_0^1\sqrt x (1- x)^n~\mathrm dx$?

Secret

that looks like gamma function thingy

Semiclassical

01:53

My inclination would be to try to get rid of the square root by doing $x=y^2$ first.

Simply Beautiful Art

That results in something that reminds me of hypergeometric functions

Semiclassical

well, it's even more like hypergeometric functions if you do x=1-y.

then the integral is $\int_0^1 y^n (1-y)^{1/2}\,dy$

So expand that in a power series and integrate term-by-term.

Secret

I still know nothing about hypergeometric functions other than 1) it's a power series, 2) it is a solution of some ODE, 3) it has a lot of parameters which special cases will reduce to some special functions

Simply Beautiful Art

Yeah

Semiclassical

@Secret What I'd add to that is that they have representations as contour integrals in the complex plane

Simply Beautiful Art

01:56

@Secret it's just a power series where the coefficients are a ratio of falling factorials/pochhammers

Secret

their names, however always give me a feeling I will get something spherical from them, but I never found any

Simply Beautiful Art

Lol, same

Daminark

Never thought I'd say this but maybe doing some of these integrals is actually fun

Secret

I mean, the following is what I pictured when I first heard of hypergeometric functions:

Ted Shifrin

But Demonark: You know only how to integrate the 0 function.

Daminark

01:57

At least as long as it's not like, integrating by parts 20 times where the expression refuses to cancel

Secret

Picture a towering sphere many million times of your height right in front of you and you gaze upwards

Semiclassical

Part of the reason I like to mention the contour integral representations is because they can be readily interpreted as jump conditions for sectionally analytic functions.

Just win baby

Hi @TedShifrin

Daminark

@Ted This is my side dealing in integrals, don't tell Peter May

Semiclassical

Which means they can be recast as Riemann-Hilbert problems.

And from there a whole hell of a lot of possibilities open up.

Ted Shifrin

01:57

Well, Peter and I are not on a first-name basis, Demonark, so you're safe.

Hi @Justwinbaby.

Daminark

whew

Semiclassical

Painleve transcendents, oh me oh my

Simply Beautiful Art

@Daminark the key is to integrate indefinitely until you derive Taylor's theorem. Exercise for the reader.

Akiva Weinberger

Hello again

Daminark

Maybe that's a bit much for me

Semiclassical

01:59

(Riemann-Hilbert problems are a vast country which I have only barely visited.)

Daminark

I mean the problem I was trying to integrate $n$ times on at least had a cancellation thing, you pulled out minuses twice, and then the spare term died

Simply Beautiful Art

Oh yes, and *ahem* goodnight

Just win baby

cya

Akiva Weinberger

So I stick with my conjecture from my earlier discussion with Eric Silva, but I don't know how to prove it

Daminark

But yeah Peter May does actually believe in not being totally ignorant of the computations, as it goes

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$Mathematics$ Mathematics