@Semiclassical now we're being circular :P

@LeakyNun I don't care for the problem at all, let alone thinking about overkilling

overkills are fun

as a joke

or as an alternative proof

I think only IMO people and competitive math students think about these types of issues 24/7

They are good computational tricks, but... that's it

Upon finishing the standard proof of something, your professor says "Okay guys check this out, it's gonna be hilarious"

yes, I find computional tricks boring as well

relevant:

smbc-comics.com/?id=1777

@Daminark and then use exact sequences and abstract nonsense

exact sequences are strange when you see them first, but if you have some experience with them, they are a quite natural tool

in the sense that knowing how to do these computational tricks is fun, but by and large it's really not relevant to actual research problems

@TedShifrin I'm with you. Were it ym course Id do it differently. I had a bad habit as a tutor though of trying to re-teach the material the way I would do it and I've found there isn't enough time and the students don't like it.

It's not a very healthy practice IMO

@BalarkaSen IMO :P

as a way of exposing people to math they haven't seen before, math competitions are fine

@BalarkaSen well it does help to train the brain

@LeakyNun Well I don't fucken want to train the brain!

it trains the brain in a specific way. doesn't mean it's in a good way

@BalarkaSen doesn't mean it isn't healthy for other people

I just want to learn new things, and get exposed to newer ideas

live and let live

@Leaky It really isn't. It brings forth an eternal rot of

the brain

Ask any practicing professional mathematician

@Semiclassical That's true. Although I will say I have a whole paper where essentially the punch line is 'I figured out that this expression cna be re-written as a convolution in the Mellin sense, which makes this thing trivial to calculate.' I'd call that a computational trick.

@TedShifrin :P

ehhh

i'd call that a computational insight

Mathematics is not just practicing ad hoc competition problems

@Leaky sometimes the abstract business is illuminating

The opposite of "ad hoc calculations" is also not ncatlab abstractness

that is also harmful for the brain

more so

@BalarkaSen the caveat i'd attach to that is that there's no easy way to demarcate what level of abstractness is unhealthy

That's probably true

at least not while you're in the trenches trying to figure out the problem

It's really easy to get caught up in a philosophy of doing things a certain way with no willingness to engage in other modes of thought. That's all fine and great if it's a question of, you process things in a way that more or less compatible with various ideas

I like situations where the right amount of theory (which does not have to be a lot!) can replace a computional proof. i.e. there's a completely elementary proof that for a commutative ring $R$, a polynomial in $R[x]$ is a unit iff the constant term is a unit in R and the other coefficients are nilpotent

but it's really annoying, there's a also a conceptual proof that requires one of the first lemmas you prove in commutative algebra

I think a good balance has been striked up by eg the Canada/USA mathcamp program

Being eclectic is ideal but some people are better off with a certain focus. But when people try to jump from their own tastes and thought processes to making normative claims on how others should approach their stuff, it's really not good

^

From what I heard of it, that is the perfect mathematical exposure a mathematically motivated kid should get

I find it strange that Atiyah-Macdonald suggest the computational proof in their hint to that problem

you run the risk of circularity, though

i.e. how does one prove the lemma in the first place

(I stress that it's a risk, not a certainty.)

I know the proof of that lemma, it does not require anything about units in a polynomial ring

kk

the example I was running into lately was trying to find a simple argument for the normalization of the Fourier transform. people i asked just kept saying "do the Fourier inversion theorem"

yeah, you have to be wary about circularity, that's true, but mostly only if you're doing really overkill stuff

but that basically just says "the normalization is this". that on its own is not really illuminating.

@Semiclassical the motivation I was given is that you want it to be unitary for $L^2$

sure, but you're still faced with proving (for instance) that $f(x)=\int_{-\infty}^\infty\int_{-\infty}^\infty e^{2\pi i k(x'-x)}f(x')\,dx\,dk$

and not C times that, for instance

I don't really see motivation as a circular problem per se. Some stuff just makes sense a posteriori

I mean, you should expect a motivation for concepts and definitions eventually, but not always immediately

oh, blah, should've been $dx'$

The other thing I've got for you, @Leaky, is that while it's definitely good to invest in your strengths, if you shun other things too much, even for yourself, you might have trouble. I've gotten the vibe that you are definitely into getting down on the scene and finding some clever manipulation that solves something without too much machinery. That stuff can be good fun, and is nice when applicable, but you don't want to find yourself unable to at least start trying something else if that fails

Sorry it took me a while, I thought I sent the message but didn't

I just think Leaky cares too much about completely ad hoc exercise problems.

@Daminark was that level of seriousness necessary :o

@BalarkaSen wat @_@

in particular, it's not exactly obvious that $$\int_{-\infty}^\infty \int_{-\infty}^\infty e^{2\pi ik x}\,dxdk=1$$

@LeakyNun Is seriousness ever unnecessary?

psh, greybeard :P

@ACuriousMind yes, when you are being light-hearted about a certain topic

@Semiclassical I'm probably younger than you :P

yeah, that was me trying to reference someting else

@LeakyNun Comically responding to our messages is probably not going to help getting the point across.

but i don't think i did it right...

ah, Greyface

(discordianism is very silly)

@Semiclassical SPeaking of that paper one of the insights happens to be that if you have the difference equation

oops

A bunch of good games are coming out lately

going back to what i was saying earlier, the easiest thing I know to prove is that the Fourier integral identity holds when $f(x)$ is Gaussian

since then the integrals involved are nice and convergent.

and I'm pretty sure I have to be happy with that, since this is really a story about distributions in general

But apparently all of them are as hard as Dark Souls

$g(\nu +1) = -\nu \cot{\frac{\pi \nu}{2}}\left(1- \frac{8 \sin{\frac{\pi \nu}{6}}}{\sqrt{3}\nu \cos{\frac{\pi \nu}{2}}} \right) g(\nu)$

oh lawd

This is a nasty recurrence relation

yes, yes it is

@BalarkaSen DS isn't hard. :P

But somehow $g(\frac{1}{2} + \nu) g(\frac{1}{2}- \nu) = \frac{\pi^3}{\cosh{2 \pi s_0}-\cosh{2 \pi \nu}}$ can be derived after doing some very hard work to find the full solution

huh, that's a hell of a reflection formula

and the question for me is, is there are way to see that this recurrence relation implies this nice result without knowing the complete solution

$s_0$ is some real numer thats a root to a particular transcendental equation

@ACuriousMind Game journalists seem to think it is

It just has no difficulty slider, no way for people who think it's too hard to tone it down. Older games - like from the 90s - often were way more unforgiving and no one thought second thoughts about that because it was just the way things were.

They are saying every game is like Dark Souls now

@ACuriousMind well, there was also the extent to which that reflected the old arcade style of gaming. in that context it was a Good Thing if the player died a bunch

(why else did Super Mario Bros have a lives system)

I think Dark Souls is even a bit different from older punishing games. Becasue they largely used 1-hit kills and lives to limit the number of times a player can retry particular sections. Dark SOuls tends not to do that (although there are some 'runs' in the game from bonfires to bosses that are long).

@Semiclassical Yeah, but people wouldn't have said these games were "too hard" - it just was accepted as part of what games were.

true

That's right. I'd just say Dark Souls is hard in a somewhat orthogonal way

the way I've heard it put is that Dark Souls is, above all things, a very deep game

both in terms of gameplay and in terms of lore/atmosphere

DS just features a throwback to an old paradigm - if you die, you gotta start over. And it's even merciful by having checkpoint (bonfires) compared to some platformers of the old brigade.

Ghosts and Goblins lol

The old games often did have continues though

(Not always though)

It's not better nor worse - it just features a paradigm of gaming that wasn't very well known anymore. And it executes it superbly

Hi ; I have a question on what counts on being rude. I posted my wrong answer to a problem to Math Stack Exchange and somebody posted where I went wrong. I have fixed several errors in my problem but it still comes out wrong. Is it rude to just give up?

Here is the link: math.stackexchange.com/questions/2477631/…

I feel the problem is over my head

Is there a good way of seeing why $\displaystyle \sum_{n=0}^\infty \frac{(-1)^n}{x-n} \binom x n = \frac \pi {\sin \pi x}$?

@Bob No, it's not rude to admit you can't answer a question.

But if you really think you're not answering the question, you should probably delete your answer.

I posted the question

@bob You know bets what you can do. Don't leave incorrect calculations about without a note that they are incorrect, but if you've given it your bst shot dont feel obligated to clean things up.

if I delete the question, the person who answered it does not get the points right?

@LeakyNun that looks like the gamma reflection identity

@Semiclassical right

@Bob Wait, so...you're asking whether it's rude of someone trying to answer it to try to answer it?

@bob I think so yes. That would be a little bad if you deleted someone else's karma

If I just leave my wrong answer there, that is okay, right?

what I’d start with is trading the binomial coefficient for factorials

Here is the link: math.stackexchange.com/questions/2477631/…

You can edit out the wrong calculations if you like and leave the question itself. Or at least put a note saying that the calculations are wrong.

Or even just x(x-1)...(x-n+1)/n!

You gave the link already, but I'm still confused about what your question is. Are you asking whether it's rude of someone to try to answer it but be wrong? Are you asking whether it's rude to show an attempt at solving a question that is wrong?

it should be clear that the calculations are wrong, the post says so

if I do nothing else, is that rude?

I think the group is telling me it is not

I don't think it's rude to show your attempt at solving a question, even if it's wrong. Asking to "show my error" would be off-topic at physics.SE, but I don't know math.SE policies well enough to say whether there's something wrong with it unrelated to rudeness.

I think I am just going to leave it for now.

thanks for your comments and have a nice day

@leaky one line of attack

Both sides have simple poles at integer x

...actually, does the LHS have simple poles at negative integers?

The RHS certainly does

hmm

Oh. Yes, it does: the series diverges when x is a negative integer

Ya each term doesn't but the whole thing does Id guess

right

Though to make this precise we’d want to examine the behavior near, say, x=-1

$\phi(0) = |(\Bbb Z/0\Bbb Z)^*| = |\Bbb Z^*| = 2$

lol

$\displaystyle \frac{1}{x-n} = \int_0^1 t^{x-n-1} \, dt$

Oh, nice

Yeah, then you can formally exchange the sum and integrand, resum the series, and hope the resulting integral is doable

@Semiclassical near $x=-1$ the binomial factor just gives $+1$ or $-1$ depending on $n$ , cancelling the $(-1)^n$ out front and making it essentially the harmonic series

Yeah

On the other hand, it should be true that (x+1)*LHS -> finite as x->-1

After switching the order, the sum is the binomial series - then the integral is the beta function!

There you go

One should justify the exchange of order...by which I mean, someone else should

@LeakyNun I don't think you want to extend arithmetic functions to $0$. Under multiplication, $\Bbb N$ behaves more nicely without $0$.

@MatheiBoulomenos allegedly it appeared in a recent conversation in the mailing list of pari/gp

and we decided to reply to the message at the same time

ProofWiki community promotion ad uploaded! math.meta.stackexchange.com/a/27255/43288

@barto nice

When I’m thinking of n as an index or just addition, I want n=0

When I’m doing multiplicative stuff, I don’t

Yes, but arithmetic functions like $\varphi$ want to capture arithmetic, i.e. multiplicative properties of the argument

Right

So power series contain exponents starting from n=0, but not Dirichlet series

Can we make the class of all sets a group?

AoC can make every set a group

@LeakyNun symmetric difference

@MatheiBoulomenos nice

@Leaky and if you define multiplication as intersection, you even get a ring strucure

@MatheiBoulomenos :o

but a ring without unit, sadly

@MatheiBoulomenos there is a unity

alright, it isn't a set

@Leaky it's really even a vector space over $\Bbb F_2$

@MatheiBoulomenos ...

there's a funny proof that $|\mathcal P (X)| = 2^{|X|}$ for finite sets $X$ along those lines.

@MatheiBoulomenos nice

define a $\Bbb F_2$ vector space structure on $\mathcal P (X)$ with symmetric difference as addition and the obvious thing to do for scalar multiplication. Observe that the singletons form a basis. Now use linear algebra

I'll admit that it's really an obfuscation of a combinatorial argument in some way, but I find it kind of neat

@MatheiBoulomenos Or that a finite topological space in which every open set is closed, has $2^n$ open sets for some $n$

@barto why? (wouldn't that be circular?)

@LeakyNun One shows that open sets form a vector space over F2

lol

One can probably do it too by tinkering around with partitions and inelegant combinatorics

There was some graph-theoretical application of proving that the order of some set is a power of two by proving it is a vector space over $\Bbb F_2$ as well, but I don't remember the statement exactly.

@BalarkaSen For 4-manifolds even intersection form is enough. $w_2(x)= Q(x,x)$ is due to Wu, which tells you w_2=0 iff Q(x,x)=0 mod 2. Milnor showed that a spin structure is equivalent to a trivialization of the stabilized tangent bundle, and in dimension 4 the only invariant for this should be w_2.

@MatheiBoulomenos I think I assumed G is finite. though you should be able to extend it to G f.p.

But it's already interesting to me when the universal cover is compact and trying to construct these finite group actions.

There's no diffeomorphism between a connected submanifold and a disconnected submanifold of $R^n$, right? If so, is there some really obvious invariant you can compute thats different between the 2?

The number of connected components?

@PVAL-inactive Ah interesting

@PVAL-inactive Yeah, I was replying to you talking about removing the finiteness assumption

Yeah I mean f.p.

when I talk about fundamental groups.

The finiteness assumption is mainly because I wan

't compact universal covers.

@BalarkaSen Ya but.... how do you like count that? Like if someone handed me a manifold and I didnt know if it was connected or not, is there some operation I could do to it to compute such a number?

or at least understanding that case seems easier.

@Kevin $H_0$ :P

How are you "handed" a manifold?

If you are given a triangulation you can compute H_0.

What is $H_0$?

@KevinDriscoll I second PVAL's question.

It depends on how are you presented the manifold.

If you're given it as a zero set of a bunch of real polynomials, good luck counting the connected components

So I'm new to this, so I don't know how people usually come about manifolds 'in the wild' but I was thinking about the presentation as being quite abstract. Like something related to the solutions to some PDE or something.

What if your manifold is the set of lines which contain multiple solutions to the riemann-zeta equation?

Actually that might not be a manifold.

but suppose it is...

Hahah, right, how about the zero set of the Riemann zeta function on a certain line on the critical strip?

Which is not Re[s] = 1/2

OH! Sorry I meant smooth manifold by the way, not just any manifold

I apologize Im so used to diff top class where 'smooth' is just understood to be inserted every place it makes sense to

The thing I mentioned is a manifold

It's a 0-manifold

ya that was clear

But you don't know if it's the empty manifold or not; that's Riemann Hypothesis (or weaker than that)

if it's empty, it's a $n$-manifold for every $n$

I should rather say the zero set of Riemann zeta on critical strip minus the Re[s] = 1/2 axis

Hilarious example, @PVAL, by the way

Let me ask a related question

Can you come up with a similar example where there's a manifold with Riemannian metric, but there's no obvious way to compute its euler characteristic?

@Kevin if you're interested in manifolds, I'd really suggest learning about homology.

If you have a Riemannian metric I think you can triangulate the manifold by hand

Which should enable you to compute $\chi$. I am not 100% sure if the triangulation procedure is "useful" in the sense that the triangulation is computable

@MatheiBoulomenos If Kevin's in a smooth manifolds course, he'll eventually learn de Rham cohomology

I was taking intro diff geo which was largely smooth manifolds (for the latter half with Riemmanian metrics) and we didn't do (co)homology at all

@MatheiBoulomenos I think perhaps we will be doing some derham cohomology near the end of the semester but Im not sure

I hear a lot of buzz with Tyler the Creator

is he actually good

or what

Im not a fan really

@BalarkaSen how much do you know about $\beta \Bbb N$ or are you not into point-set stuff?

ah, stone-cech

that monster

I must say not much

THe motivation for my question though was just that we've just started to develop some invariants that can distinguish manifolds. Like we used mod2 intersection theory to show $S^n \times S^1$ is not simply connected. And so I can kind of think about associating $0$ to all the simply connected spaces, because the mod2 intersection of any paths there is $0$.

And I was wondering if theres something similar for other invariants which we have really talked baout as invariants yet, like connectedness

as a set, it is actually somewhat naturally in bijection to some prime spectrum. I was just wondering if the topology is the same, but I suspect not

@KevinDriscoll The thing is if you're given a disconnected manifold in a non-contrived way it's pretty easy to recognize it. Eg $S^n \times S^0$

The number of connected components are usually computable by hand most of the time once you see the manifold

@BalarkaSen Ya indeed. Im just not sure if there is a difference in kind between invariants where theres some algorithmic way to computer them versus ones where there is no

Somewhat related, after one-point compactifying, the Jordan-Curve theorem is basically a statement about the connected components of an open submanifold of $S^2$ and that's not trivial at all

Bernoulli -> binomial -> poisson -> exponential -> gamma

distributions are interesting

@Mathei is right, for example.

one-point compactifying means send everybody ont he boundary to a point, right?

Is there a way to prove that $O(n)$ has two connected components without relying on the determinant?

@KevinDriscoll it means add one point for which the only open set containing it is the whole space

so that the space becomes compact

@Daminark what is O? orthogonal matrix?

@daminark probably not a better way.

Yup

@Daminark Well, think about O(n) as Isom(S^(n-1))

@LeakyNun that's not right, the complement of each compact set will be an open neighborhood of the point at infinity

And yeah I definitely imagine that it's not gonna be cleaner, I'm just wondering if that's in principle a thing

tfw you interpret matrices as groups and then as topology

Then you end up proving orientation-reversing and preserving isometries are not homotopic

That's a degree argument

@LeakyNun Okay, thanks. I will have to think about the proof that the result is compact

You can maybe do something with the fibratiton involving the sphere and O(n) and O(n+1)

That proves O(n) is path disconnected

and then just use the fact that O(1) is disconnected

and use induction.

@KevinDriscoll apparently Mathei says I'm wrong

That seems to work

but it's stupid.

I was wondering because the thing you said seems ot ruin Haudorff-ness

becuase there not disjoint open sets containing the point you added and any other point, given what you said

@KevinDriscoll You need the space you are one pt cptifying to be locally compact Hausdorff I think

S^{n-1} \to O(n) \to O(n-1)

and use the the homotopy exact sequence.

btw my book writes "if you can always draw a curve in D from P to Q for an arbitrary pair of points P and Q, we say that D is connected"

what should I do with it lol

and use induction.

@BalarkaSen @KevinDriscoll you need Hausdorff of the base space for the topology to be defined, you need local compactness in addition of the base space for the resulting space to be Hausdorff

kind of funny

Right

@LeakyNun D is connected iff there's a smooth math $F: [a,b] \to D$ st $F(a) = p$ and $F(b) = q$???

thats how I would try to formalize that sentence, dunno fi it scorrect

They're calling path connectedness as connectedness

I mean, of course my book isn't right

@BalarkaSen do you approve?

it doesn't sound like a rigorous book, tbh

depending on the book

it's an informal treatment on covering space theory and differential galois theory

not entirely sure I approve but go with it

which probably doesn't care about topologist's sine curve

Whom is this written by

the legendary late Michio Kuga

you want path-connectedness for the covering space business anyway

I see

@MatheiBoulomenos Is every smooth manifold locally compact?

in 1967(!)

I have not heard of him

No one cares about what Arnol'd would call "pathological topology"/

@KevinDriscoll yes

becuase theyre locally euclidian

Even if you remove the smoothness part

indeed