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12:00 AM
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)
 
I'm letting my sister use my account to ask a calc question.
 
lol, okay
 
So yeah
 
Is this sister better or worse than you at math
(Just kidding, don't answer that)
 
Worse, I'm above calc :P
 
12:10 AM
Ah, OK :P
You can't answer her question, then?
 
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?
 
@JohnLocke I'm a fan of your work, Mr. Locke
 
Haha, funny enough I chose the name because I enjoy political science and theory and struggle profoundly with mathematics
 
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 :)
 
Hi quick question
Can someone show me how to find lim(x to 0) sin(x)/x?
 
12:20 AM
That's a famous one
 
yeah, my prof was somehow using squeeze theorem
 
I'd guess you've already learned at least one proof of it in class
I see
 
Completely lost me
 
There's a standard proof using the unit circle and areas. Does that sound familiar?
 
12:21 AM
Famous indeed... Here's the math.se link to that question:
283
Q: How to prove that $\lim\limits_{x\to0}\frac{\sin x}x=1$?

FUZxxlHow 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...

 
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?
 
Here's part of a unit circle. The angle of that sector (at $A$) is $x$.
 
I saw your earlier page, DogAteMy. growl
 
@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.
 
12:23 AM
@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
 
@Akiv
 
Akiva even if we were talking about volumes, etc.?
 
That is how Archimedes did everything
squeezing from inside and out
 
**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?
 
I might actually need to get off chat and start dinner
 
12:26 AM
you've become indispensable, DogAteMy — no escaping now
 
Hopefully Ted and Simply can help Cumference
@TedShifrin Oh, no!
 
@TedShifrin xD Escaping? From what?
 
(and that link above)
 
Hai waves
 
So wait Simply, I remember areas being involved
 
12:26 AM
And, uh, John, it ends up being bijective with something easy to count
 
You can do that limit in different ways, @SirCumference. Most calculus books base the argument on area.
One can do it with lengths, too.
 
Hm, I don't think area is involved in this limit @SirCumference ... lemme read the proof a bit closer lol
 
Oh, also, your sibling says not to flip out on him/her
 
Yeah, no area required.
 
I would like to understand the area argument
Akiva What?
 
12:28 AM
@SirCumference: Call the point between A and B (below C) D.
 
@TedShifrin Yeah
 
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.
 
Wait what?
 
I changed the name to E. I realized D was already used.
E is the point below C on AB.
 
Oka
Ugh *okay
 
12:31 AM
Howdy, Demonark. So how did the first bootcamp meeting go?
 
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 $2\pi$.
 
@AkivaWeinberger for what?
 
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. :)
 
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
 
I'm still confused on how area is related to the limit
gtg be back in 20 mins
 
12:33 AM
Did the complex exponential and log, along with a branch
 
It's important to understand the distinction, Demonark. And real differentiable + conformal derivative isn't sufficient.
 
Nice @Daminark
 
Oh really?
 
Not just at a point, nope, Demonark. If you assume $C^1$, then sure.
 
I mean we said that the real and imaginary parts were $C^1$
 
12:35 AM
(I think. I'd have to rethink it all.)
 
Sniped, but yeah
 
Ah.
 
Hmm, I wonder what's a counterexample
 
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?
 
12:36 AM
We defined it as being alright at a point
 
So $|z|^2$ is complex differentiable at the origin, but nowhere else.
 
In my head its weird to just be at a point
 
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.
 
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?
 
No, no, holomorphic and analytic are not equivalent unless you're on an open set.
 
12:39 AM
^
 
It's important to understand the distinction.
 
Oh, I see
 
And hi @EricS
 
I think the reason I think at a point is weird is because I want holomorohic to mean analytic
 
oh damn, still pings both EricS's.
 
12:39 AM
Hi Ted
 
No, they're conceptually totally different, ERic.
 
Lol
 
My 'other' form of being holomorphic is if Cauchy's equations are satisfied at $z_0$, then $f$ is holomorphic at $z_0$.
 
That requires some care (like assuming $C^1$), @Simply.
 
12:40 AM
Ik that they mean different things
 
oh, no, sorry. C-R equations do not.
 
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
 
No? Hm... well, too busy to think too much
 
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.
 
Can a function be analytic at one point?
 
12:42 AM
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
 
@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
 
@Secret: I was referring to square matrices. Sorry.
 
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
 
Well, I'm now officially listed as an instructor :P
3
 
@TedShifrin Doesn't Cauchy's equations come directly from the limit definition of the derivative without assumptions?
 
12:45 AM
i think if i wanted "holomorphic at a point" i would use complex differentiable
 
@TedShifrin :P You were always an instructor here
 
@Ted You mean for aops?
 
Not exactly, Simply.
Yes, Demonark. They put my name up.
 
Nice
 
Yeah, @Simply. I retracted my negation.
 
12:46 AM
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...
 
x.x you can't be responding so vaguely to multiple questions like that @TedShifrin
 
@SimplyBeautifulArt THAT.
 
Which response though x.x
 
womp womp
 
ignores Simply
 
12:48 AM
@Daminark did you guys decide who's lecturing yet
 
lol, okay
 
also did you get told your dynamics hw
 
Yeah, the two Chriss are lecturing on Friday
And we weren't given homework, just the lecture
 
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
 
Be a mean teacher, Eric. Oh, write more problems, then. :D
 
12:49 AM
Makes sense
 
the thing is i dont really know dynamics so i dont really have much that's good to give them
 
Hmm...
Singular = not have full rank
Invertible = $A^{-1}$ exists
now what is the name of that theorem that ensure they are equivalent...
 
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... :)
 
It's not got a name, @Secret. But it's a theorem nevertheless.
 
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
 
12:52 AM
what do you mean nuke
 
Presumably, if we were to just make the theorem false
 
@EricSilva Brief intro: One thing that I am famous on this chat is exploring mathematical structures by relaxing conditions to make some proofs fail
 
oh i mean it's false in infinite dim
 
@Daminark Ah, then I can kinda expect a countably infinite matrix that is nonsingular but not invertible, I guess
 
I mean, your problem is that coordinatization and matrices are not something you can naively deal with in infinite dimensions
 
12:55 AM
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
 
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 $\ell^2(\mathbb{N})$
 
yup, that's the classical example
 
That's something you should actually be able to write as a matrix well enough
 
@Daminark i would say things that aren't full rank are things that aren't injective but ok
 
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
 
12:59 AM
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"
 
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
 
@TedShifrin you still here?
 
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
 
ranges usually just dont behave as well
 
I missed him
darn
 
1:03 AM
[Unrelated pondering]
 
@Daminark plus i reaaaaally dislike the idea that a linear embedding of one space into another would not be full rank
 
@Secret do you understand the geometry I was describing before? You can see how some rigid might be one way, right?
 
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
 
@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
 
@Secret are you still talking about rank-nullity
 
1:08 AM
@EricSilva nah, something more general, any maths
 
ah then this statement is vague enough that i have no idea what it could mean
 
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
 
It saddens me that we're not Lebesgue-ing the stuff
 
why
 
Fubini/LDCT would be very convenient for the early stuff on limit interchanging
 
1:13 AM
i mean it makes it like
not hard
 
@Daminark lol
 
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:
 
Let the children learn the fundamentals and their challenges and let them truly relish in their victories rather than handing them over so easily.
 
@Secret fair enough. It might be a geometry worth bringing up next semester.
 
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
 
1:17 AM
@Eric We'll get to other hard things later I imagine or something
 
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
 
Fun fact: If $a\cdot$ distributes over $b+c$, then $a\cdot$ is a homomorphism over +
(distributive law in a nutshell)
 
hi chat
 
hi chat
 
@EricSilva Whereas hearing about it is meant to put the fear of Titchmarsh in you
 
1:21 AM
lol
 
Isn't that what the word "distribute" means? @Secret
 
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.
 
@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
 
@Semi the man can manipulate trigonometric series like a monster
 
neat.
 
1:24 AM
i thought i hated it when i read it but a year down the line i appreciate the chops it gave me
 
It might not be obvious, but distributive law actually constraint the algebraic structure a lot
(as well make it nicer to handle)
 
It's a property of real numbers.
 
Lmao, I guess I'm just kinda fond of nuking potatoes from orbit sometimes, something something efficiency something something
 
for example, ordinals violate the right distributive law
 
Woohoo for ordinals lol
(the one thing I managed to notice in the chat while doing my homework lol)
 
1:27 AM
@Daminark there's a place and a time for it
 
@Secret Just the transfinite ordinals. Finite ordinals follow the 'normal' laws of arithmetic
 
shrugs
 
Potatoes? @Daminark
 
@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
 
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
 
1:30 AM
[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)
 
Yeah, I guess I haven't really been in that context much before
 
I've definitely been somewhat iffy on machinery sometimes in subjects like graph theory
 
what do you mean by machinery
do you mean techniques or theory or like pulleys and wheels
 
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
 
1:31 AM
idk what that means
 
I think I may, though the example I know isn't quite 'math'
 
0
Q: What kind of geometry can I classify as "3D modeling" and why do I get surfaces with *branching lines*. (Geometry paradox)

TyphonMost 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...

 
Namely, Sudoku.
 
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
 
@Semiclassical I think I found 2d wormhole geometry. Any comments?
 
1:32 AM
What I always start with there is just trying to proceed by inspection to make as much progress as I can.
 
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
 
or confirmations?
 
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"
 
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.
 
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.
 
1:34 AM
@Daminark you would be surprised how far you can get sometimes in analysis by just finding the right thing to integrate by parts
 
In the set of real numbers @Secret
 
So my attitude has traditionally been to try to do things using high-powered theorems fitting together
 
So I proceed by inspection+intuition as much as I can, but I'm willing to proceed more systematically if it becomes necessary.
 
@Justwinbaby I have got around that, but unless you have nonassociativity, your structures are still quite trivial
 
@Typhon Huh.
 
1:35 AM
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
 
@Daminark conceptual problems can be straight computations man
 
Lol actually one of the things in Titchmarsh was something where I actually was trying to parts it when it was easier to substitute
 
I have a hard time spotting integration by parts.
 
I mean I think my idea for partsing it out would've worked nicely
It was like, $\int_0^1 n^2x(1-x)^n dx$
 
The only places where I usually remember it is when I'm doing something like "oh, I can turn $(\partial_x \psi)^2$ to $-\psi\partial_{xx}\psi$"
 
1:36 AM
I usually try to substitute before taking parts unless i already know from inspection what to do
 
@Daminark $n\to\infty$?
 
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 $(1-x)^n$
 
Just win baby:
 
Just that Chris found the easy substitution $u = 1-x$ before I hashed out much on that front
 
iusually end up guessing right but i usually do it by accident
 
1:38 AM
@Simply Nah, just leave $n$ there, it's a closed form expression
 
The one I'm looking at now is interesting
 
@Daminark just looking at i have no idea why you would try parts tbf
 
One problem I had at one point which took me a while to understand
 
1:39 AM
Say, does the notation:
$$\sum_{x\in\mathbb Q^+}f(x)$$
make sense?
 
yeah, I'd say so
 
not like the ibp is actually bad
 
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 $\frac{x^n}{n!}$
 
but the substitution is like very immediate
why would you not differentiate the x
 
I'm pretty sure I've seen $\sum_{n\in \mathbb{N}}f(n)=\sum_{n=0}^\infty f(n)$ for instance
 
1:40 AM
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
 
My initial response to that integral would just to use the Beta function -> Gamma function -> recursive formula -> closed form
 
That method was just the first one to come to mind for some reason
 
(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)
 
@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 $\mathbb N$.
 
Sure, but the principle is the same: You evaluate the function at each element of the set, and then sum those up
 
1:41 AM
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 $\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}$
 
not sure if the sum will be divergent though, it depends on $f$
 
Well, sure.
 
@Daminark fun one
i remember how to do it so i wont say anything
 
But then $\int_\mathbb{R} f(x)\,dx$ doesn't need to be convergent for us to understand what the notation intends
 
yup
 
1:43 AM
@Daminark lol, that's how Euler proved it
 
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 $\log(1-x)$
 
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...
 
It'll give you a double sum which should be absolutely convergent?
 
@Daminark The intention is right, but you're making things too complicated.
 
@Semi a lot of titchmarsh problems in complex is stuff euler did
it's p great
 
1:44 AM
@SimplyBeautifulArt If the sum converges, then the ordering does not matter?
 
huh, nice.
 
@Daminark Extract the $x^2$ coefficient from the product:
$$\prod_{n=1}^k\left(1-\frac{x^2}{n^2\pi^2}\right)$$
 
@SimplyBeautifulArt tsh, giving the game away :P
 
@Secret That's absolute convergence
D: Oops, well, it still leaves a bit of work.
Anyways, g'night folks.
 
@Daminark if you ever have time read Euler Master of us all
it's great
 
1:45 AM
For what it's worth, this is on the section about double sums
 
I saw Euler's proof in "Journey through Genius"
 
$$\int_0^1 n^2x(1-x)^n dx$$ This integral has one obvious symmetry: polynomials are nilpotent under differentiation, thus the IBP will terminate
 
Great little book.
 
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)
 
@Secret Isn't doing $y=1-x$ simpler?
 
1:47 AM
@Secret you think like I do
 
Though I guess maybe the point is just that it can be done
 
I fully acknowledge that it seems like the "morally correct" way to think about it is through the substitution
 
ah yes, that too (thinking too much about IBP, I tend to just try to IBP everything)
 
But for some reason this IBP just stood out to me
 
._. No one thinks Gamma functions and beta functions?
 
1:48 AM
eh, no harm there.
 
The $y=1-x$ provides the shortest path in computing that integral, for reasons I still yet to understood
 
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.
 
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
 
it;s just cause it lets you avoid expanding a binomial
 
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.
 
1:50 AM
That one I will just IBP it
 
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
 
that one is waaay harder
iirc that can be written in terms of $\Gamma$
 
Yeah.
 
Now what would you do if we replaced the integral with $\int_0^1\sqrt x (1- x)^n~\mathrm dx$?
 
that looks like gamma function thingy
 
1:53 AM
My inclination would be to try to get rid of the square root by doing $x=y^2$ first.
 
That results in something that reminds me of hypergeometric functions
 
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.
 
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
 
@Secret What I'd add to that is that they have representations as contour integrals in the complex plane
 
1:56 AM
@Secret it's just a power series where the coefficients are a ratio of falling factorials/pochhammers
 
their names, however always give me a feeling I will get something spherical from them, but I never found any
 
Lol, same
 
Never thought I'd say this but maybe doing some of these integrals is actually fun
 
I mean, the following is what I pictured when I first heard of hypergeometric functions:
 
But Demonark: You know only how to integrate the 0 function.
 
1:57 AM
At least as long as it's not like, integrating by parts 20 times where the expression refuses to cancel
 
Picture a towering sphere many million times of your height right in front of you and you gaze upwards
 
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.
 
Hi @TedShifrin
 
@Ted This is my side dealing in integrals, don't tell Peter May
 
Which means they can be recast as Riemann-Hilbert problems.
And from there a whole hell of a lot of possibilities open up.
 
1:57 AM
Well, Peter and I are not on a first-name basis, Demonark, so you're safe.
Hi @Justwinbaby.
 
whew
 
Painleve transcendents, oh me oh my
 
@Daminark the key is to integrate indefinitely until you derive Taylor's theorem. Exercise for the reader.
 
Hello again
 
Maybe that's a bit much for me
 
1:59 AM
(Riemann-Hilbert problems are a vast country which I have only barely visited.)
 
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
 
Oh yes, and *ahem* goodnight
 
So I stick with my conjecture from my earlier discussion with Eric Silva, but I don't know how to prove it
 
But yeah Peter May does actually believe in not being totally ignorant of the computations, as it goes
 

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