Mathematics - 2014-03-20 (page 4 of 4)

6:00 PM

I have to think.

You can't expect me to be good at these things, I have no practice.

I can give you a hint: show that if $s\in (1,+\infty)$ there is a compact neighborhood of $s$ bounded away from $1$, and use that you have a sum of continuous functions and the sum is bounded above by a convergent series.

Balarka Sen

But I will try.

Mats Granvik

{0., ComplexInfinity, ComplexInfinity, ComplexInfinity, ComplexInfinity, 0.}
For zeta zeros 1 to 6

robjohn

@MatsGranvik I doubt that you could use $\infty$ and $-1$ interchangeably.

Mats Granvik

@robjohn ok

robjohn

6:03 PM

@MatsGranvik I do not understand what that means. Are they residues?

Mats Granvik

This is the code:

Clear[s, n]
Monitor[Table[
N[Limit[Zeta[s]*(Exp[Total[1/Divisors[2]^(s - 1)]/Zeta[s]] - 1),
s -> ZetaZero[n]]], {n, 1, 6}], n]

I set divisors to 2, just to see if there is any pattern.

Balarka Sen

6:17 PM

@Pedro I have an idea.

Pedro Tamaroff

@BalarkaSen OK.

Balarka Sen

I will post it after I am done eating.

Just hang on for that while.

skullpatrol

1 hour ago, by Pedro Tamaroff

hi @Sawarnik

Sawarnik

hi

skullpatrol

are you mad at me?

Sawarnik

6:27 PM

why? o.O

r9m

@Sawarnik not talking to me .. whats up ?

skullpatrol

dunno, you stopped talking to me

Balarka Sen

@Pedro Hang on, just there, eating Mike at super fast speed.

Sawarnik

@r9m Nah...I was talking to others in the other room.

r9m

@Sawarnik okway :)

Sawarnik

6:30 PM

@skullpatrol More or less, I stopped coming to the main chat here, nothing else. :)

skullpatrol

ic ic

:-)

robjohn

@MatsGranvik so you are trying to evaluate the limit of $\displaystyle\zeta(s)\left(\exp\left(\frac{1+2^{1-s}}{\zeta(s)}\right)-1\right)‌$ as $s$ tends to a root of $\zeta(s)$.

It seems to depend on the direction that $s$ approaches the root

Mats Granvik

yes, that is the second limit. The starting point is this:
Clear[s, n]
Monitor[Table[
Limit[Zeta[s]*(Total[MoebiusMu[Divisors[n]]/Divisors[n]^(s - 1)]),
s -> 1], {n, 1, 23}], n]

Which I modified just by inspecting the numbers into this:

Clear[s, n]
Monitor[Table[
Limit[Zeta[s]*(Exp[Total[1/Divisors[n]^(s - 1)]/Zeta[s]] - 1),
s -> 1], {n, 1, 23}], n]

Balarka Sen

@Pedro I am most certainly sure that this is flawed, but let's see :

A plausible thing to do here is to consider uniform convergence, as that clearly implies continuity.

Mats Granvik

@robjohn I see now, Mathematica chooses somehow the direction of the limit.

Balarka Sen

6:39 PM

So we consider $f_n(x) = \sum_{k \leq n} \frac{\log k}{k^s}$

robjohn

@MatsGranvik The limit seems to depend on how $s$ approaches the root. I don't think the limit exists in all directions.

Balarka Sen

Following, @PedroTamaroff, or gone?

Pedro Tamaroff

@BalarkaSen Here I am.

Balarka Sen

OK, OK.

$f(x) = \sum_{k \leq n} \log k/k^x$ clearly converges in the given interval.

Ryker

Hi guys, quick question. Is the direct sum unique? That is, is it true that $H = A \bigoplus B = A \bigoplus C \implies A = C$?

It seems to me it is, but I just wanted to check.

I'm sure it's true in finite cases, but would infinite dimension mess things up?

Balarka Sen

6:43 PM

Now, @Pedro, I remember you did something with the M-test here for uniform convergence, correct?

Now I believe I don't need it. Am I wrong?

Pedro Tamaroff

@BalarkaSen You do need it! =D

Balarka Sen

@PedroTamaroff

Pedro Tamaroff

@Ryker Direct sum of what?

Balarka Sen

@PedroTamaroff In my derivation, I didn't use it, I think.

Pedro Tamaroff

@BalarkaSen You need to show uniform convergence somehow.

Ryker

6:45 PM

Well, anything.

Balarka Sen

I used summation by parts.

Ryker

Two sets.

Balarka Sen

(yeah, that's how a number theorist think)

@Ryker So that's direct product, no?

Pedro Tamaroff

For example, if $\varepsilon>0,s>1+\varepsilon>1$ and $s\in[1+\varepsilon,s+\varepsilon]$, $$\sum_{n\geqslant 1}\frac{\log n}{n^{s}}\leqslant \sum_{n\geqslant 1}\frac{\log n}{n^{1+\varepsilon}}<+\infty$$

Balarka Sen

Oof, too much analysis. I simply used summation by parts.

Ryker

6:47 PM

No, it's the direct sum.

Balarka Sen

Nothing less nothing more.

Pedro Tamaroff

@BalarkaSen Too much analysis? Stop complaining! It is a simple inequality. If $s>1+\varepsilon$, $n^s>n^{1+\varepsilon}$.

Then you can use the Weiertrass $M$-test.

Balarka Sen

I won't. Won't won't won't won't won't.

See my summation by parts derivation.

Pedro Tamaroff

And you get the sum converges uniformly on $1+\varepsilon\leqslant s\leqslant s+\varepsilon$, hence continuous at $s$.

@BalarkaSen OK.

@BalarkaSen So?

Balarka Sen

Writing, wait up.

$$\sum_{m \leq k \leq n} \frac{\log k}{k^x} = \frac{1}{(n+1)^{x+1}}\sum_{m \leq k \leq n} \log k - \frac{1}{m^x}\sum_{m \leq k \leq n} \log k - \sum_{m \leq k \leq n} \left [ \frac1{k^x} - \frac1{(k+1)^x}\right ] \sum_{m \leq k \leq i} \log i $$

My internet connection is not incorporating.

Pedro Tamaroff

6:59 PM

@BalarkaSen OK, and what will you do with that?

Balarka Sen

Now we take absolute values on both sides.

Pedro Tamaroff

Note the log gets turned into $\log(n!/m!)$

I don't find that particularily useful.

Balarka Sen

Oh, damn. I wrote up wrong there.

I meant

$$\sum_{m \leq k \leq n} \frac{\log k}{k^x k^y} = \frac{1}{(n+1)^{x+1}}\sum_{m \leq k \leq n} \frac1{k^y} \log k - \frac{1}{m^x}\sum_{m \leq k \leq n} frac1{k^y} \log k - \sum_{m \leq k \leq n} \left [ \frac1{k^x} - \frac1{(k+1)^x}\right ] \sum_{m \leq k \leq i} \frac1{k^y} \log i $$

It's just that the LaTeX is playing trickseys on me.

And so is the internet connection.

Pedro Tamaroff

Too complicated.

Balarka Sen

Now you do that absolute value thingy.

Pedro Tamaroff

7:03 PM

$$\sum_{n\geqslant 1}\frac{\log n}{n^{s}}\leqslant \sum_{n\geqslant 1}\frac{\log n}{n^{1+\varepsilon}}<+\infty$$

Done. =)

Balarka Sen

@PedroTamaroff Stop complaining, sire!

=)

Pedro Tamaroff

@BalarkaSen I don't see how your method will yield anything useful! It just too messy. Let's agree on this one.

Balarka Sen

$$\left | \sum_{m \leq k \leq n} \frac{\log k}{k^x} \right | \leq \frac{1}{(n+1)^{x+1}} \left | \sum_{m \leq k \leq n} \log k \right | + \frac{1}{m^x} \left | \sum_{m \leq k \leq n} \log k \right | + \left | \sum_{m \leq k \leq n} \left [ \frac1{k^x} - \frac1{(k+1)^x}\right ] \sum_{m \leq i \leq k} \log i \right |$$

@PedroTamaroff Will you just see without complaining? Point out if I am wrong.

Pedro Tamaroff

@BalarkaSen OK, what happens now?

Balarka Sen

@PedroTamaroff Now we choose an $\epsilon$ that bounds the both of the $$\sum_{m \leq k \leq n} \frac1{k^y} \log k$$ terms.

Pedro Tamaroff

7:08 PM

No, you don't choose $\varepsilon$.

You choose $m,n$.

Balarka Sen

Yes, we find $\epsilon$

That is satisfied by certain large $m, n$

Correct?

Pedro Tamaroff

@BalarkaSen No, you're given $\varepsilon$!

You have to choose $m,n$ given $\varepsilon$.

Balarka Sen

@PedroTamaroff let me put it this way : We find some large $m ,n $ s.t. a small $\epsilon$ bounds the sum.

Right or wrong?

Pedro Tamaroff

@BalarkaSen The point is not that $\varepsilon$ bounds the sum, rather, the sum can be made smaller than $\varepsilon$ choosing $n,n$ large for any $\varepsilon$. Look it from right to left (i.e. the importance is in the sum) not from left to right (i.e. the importance is on somethign that bounds it)

Balarka Sen

I surely know this much calculus, don't you think, to put it straight?

@PedroTamaroff In any case, $m, n$ is the ones that we need to choose.

Pedro Tamaroff

7:12 PM

@BalarkaSen Exactly.

Balarka Sen

(I might be bad at these representating things time to time, bear with me)

Now, where were we?

Aha.

Pedro Tamaroff

I have to go.

Sorry.

Balarka Sen

Fine.

Pedro Tamaroff

The point is the sum is continuous on $(1,+\infty)$. Yay.

Balarka Sen

But you can see where this will lead, no?

@PedroTamaroff Of course, so is any Dirichlet series with certain condiitons imposed (in which case, Perron! Fun, fun!)

But my point is that we don't need M-test.

No weirdo stuff.

Hello, @N3buchadnezzar

Matt N.

7:25 PM

Lol. Just read this thread on Meta.

We are amused.

@DanielFischer Hello.

Daniel Fischer

Hi @MattN. Which thread on meta?

Matt N.

@DanielFischer Oh, I already closed the tab. Someone seriously posted a thread complaining that some questions were too long. But funnier than that there was a (now deleted) answer suggesting adding a new close reason "too much effort".

Even typing it still gives me the smiles : D

Daniel Fischer

Ah, that one. I remember ;)

Matt N.

But, funnier than the already funny thread, the poster of the deleted question (too bad I didn't get a chance to downvote) is not a new user and whatever name he puts on his account his unpleasant personality shines through brightly.

And now this guy posted a comment telling someone else that they were unpleasant.

Not in these words but that's the gist.

So now I am quite amused.

Balarka Sen

@DanielFischer Got an ENT for you. Interested?

Matt N.

7:31 PM

And I just couldn't help posting a comment. I felt that Bill was wronged there.

I mean, Bill is quirky. At best. But this other guy is properly unpleasant.

Daniel Fischer

@BalarkaSen Can't tell without seeing it.

Balarka Sen

Count the number of solutions $(x, y)$ of $$\frac1x + \frac1y = \frac1N$$ for some fixed $N$.

@DanielFischer Combinatorial problem mostly.

Daniel Fischer

A classic, @BalarkaSen. $$\frac{1}{N+d} + \frac{1}{N + \frac{N^2}{d}} = \frac{1}{N}.$$

Matt N.

@DanielFischer You don't have to answer if you don't like but I was wondering if you ever did a PhD.

Daniel Fischer

I didn't, @MattN.

Matt N.

7:37 PM

Oh wow.

Scary. I was pretty sure the answer would be yes.

@DanielFischer Now I can't ask you about your PhD :)) Too bad.

Oh well. I'll find another victim. : )

Daniel Fischer

@MattN. You could ask Ted, I'm pretty sure he has one.

Balarka Sen

Darn my internet.

@DanielFischer So the solution is.

$\tau(N^2)$

Matt N.

@DanielFischer Thanks, I will -- next time I see him here.

Balarka Sen

Well done.

I did a bit differently though.

Daniel Fischer

@BalarkaSen Or $(\tau(N^2)+1)/2$, if you don't distinguish $(x,y)$ and $(y,x)$.

Balarka Sen

7:41 PM

Yes, modulo the symmetric equivalence $\sim$, that.

But I didn't want that.

I did it by counting $$\sum_{a|N} \sum_{b | N/a} 1$$

(At least, one of the many solutions)

Another classic is $(N - x)(N - y) = N^2$

But I don't like trick theory.

My counting trick is the most analytic non-tricky way I can think of.

Vrouvrou

7:58 PM

hello, i'm sorry to do this but can you help me :math.stackexchange.com/questions/718143/…

?

thank you

Chris's sis

8:09 PM

I'm skimming the table of integrals series and products by Gradshteyn and Ryzhik for half an hour and nothing interesting. I wonder if it's something wrong with me ...

skullpatrol

light reading :D

Chris's sis

@skullpatrol :-)

skullpatrol

@Chris'ssis There is definitely nothing wrong with you, it sounds like you need something more challenging.

0x5f3759df

Is there a noun associated with usages of capital pi notation, the way sigma notation has 'summation'? I'm tempted to call it a 'productation' but google only finds a few references to this term ever having been used...

Chris's sis

@skullpatrol There is a thing I admire for some time, this one: math.stackexchange.com/questions/464769/…

@skullpatrol I hope to be able to prove that sometime by real analysis.

skullpatrol

8:19 PM

@Chris'ssis Thanks :-)

@0x5f3759df how about "factoration"?

0x5f3759df

I thought about that, but every reference to that I can find is just a misspelling of 'factorization'. Productation actually has a few uses in literature but it's nowhere near common. I was hoping there was a non-obvious standard I wasn't aware of.

skullpatrol

producation sounds interesting :-)

Jessy Cat

Why can't people write properly anymore?

Seriously, if what you're writing wouldn't make sense if you walked up to somebody on the street and said it to them, it's probably wrong.

Just saying.

Okay, rant done.

This is Math StackExchange after all, and not the grammar police.

Yeah...

skullpatrol

8:34 PM

too fast

Karl Kronenfeld

Hey.. I was playing a game while reading that. Unfortunately my opponent was still more error-prone than me.

Vrouvrou

@robjohn can you help me ?

robjohn

@Vrouvrou what's up?

Vrouvrou

it's about series !

{math.stackexchange.com/questions/718143/…}

AndrewG

9:26 PM

What's the relationship between modules and non-abelian groups? (since abelian groups are just Z-modules, etc.)

Karl Kronenfeld

@AndrewG A way to express $M$ as an (abelian) $R$-module, is to give a ring homomorphism from $R$ to the endomorphism ring of $M$. Since the set of endomorphisms of a non-abelian group does not necessarily have commutative "addition", it would be incorrect to say it is a module over a ring.

(I think one of the distributive properties fails as well, but this is a result of the addition being noncommutative)

robjohn

@Vrouvrou Is there some reason it is not as simple as my answer?

Karl Kronenfeld

9:45 PM

@AndrewG That said, one can feel free to define "modules" over a weaker type of ring by using the definition I gave and defining homomorphisms the same way we do for rings. The problem is that the homomorphic image of an actual ring will be an actual ring, so I do not see any real gain here.

Ted Shifrin

Howdy @robjohn @Karl @AndrewG

Karl Kronenfeld

@TedShifrin Hello

AndrewG

10:02 PM

@Ted hi :) @Karl thanks

Ted Shifrin

Haven't seen you in a while, @Andrew!

AndrewG

Yeah, I've been knee-deep in classes. How have you been?

Ted Shifrin

Doing ok, thanks ... Torturing students as always :)

Mike

Hi @Andrew, been a while

AndrewG

:) glad to hear it. I've been mathing on the side, but gen eds are eating up most of my semester. Have to read another 100 pages of lit tonight.

heyo Mike!

Mike

10:10 PM

Yuck gen eds. I feel ya.

Ted Shifrin

I still believe in liberal arts education, sorry, guys.

Should spread it out liberally, however.

AndrewG

I'm actually enjoying it -- got Joseph Conrad's Heart of Darkness and Kafka's Metamorphosis tonight -- it's just really time-consuming.

Mike

@Ted I believe in it. It falls apart when I'm forced to take my third damn religion class.

Ted Shifrin

Math is way time-consuming for most ...

We've been around this bush already, @Mike :D

Mike

@Ted I have a trivial topology question for you.

Ted Shifrin

10:14 PM

Trivial topology is boring.

AndrewG

Well, i'm off for dinner. Good seeing you both again :)

Ted Shifrin

See ya @Andrew

Mike

I can't prove locally path connected and connected implies path connected.

The intuitive picture is obvious but I can't make it a proof.

user58869

what does proof mean?

Ted Shifrin

Oh, standard connectedness game. Look at the set of points pathable from $x$.

Mike

10:17 PM

Oh. Done.

Ted Shifrin

Yup. :)

Karl Kronenfeld

:)

user58869

also, how can i learn to understand the mathematical notation used so much by people on the mathematics stack exchange site and also no wikipedia?

Mike

@Karl I have a question for you

Karl Kronenfeld

@Mike sure

user58869

10:19 PM

every time i try to read something on wikipedia, i come up to an image that i dont understand and i am lost at that point

Ted Shifrin

Hmm, is there an on-line notation glossary? I put this sort of thing as an appendix in my books.

Mike

@Karl I was gonna make a 5#2 joke but meh

Karl Kronenfeld

meh

Mike

@Alex what math do you know? The usual answer is 'read a book', but without more information I can't recommend one.

user58869

0

Q: how can i learn about mathematical notation

i am baffled by mathematical notation. i think that if i could understand it then lots of things that i read would make sense. how does one learn this? i went to school, i attended maths lessons, but perhaps i didn't listen at the right time or maybe i was at a dentists appointment when they d...

Mike

10:23 PM

You're going to need to answer my question, either there or here.

user58869

@Mike which book? i imagine that once upon a time you did not know how to understand the mathematical notation, but now you do, so i need to get from where you were to where you are now in that respect

Mike

@Alex My point is I can't say. What math do you know? That's our starting point.

user58869

@Mike it's hard to answer that, i can tell you what i dont know, i dont know the notation

user58869

i read it all the time, but it means nothing

user58869

how can i understand it

user58869

10:25 PM

the symbols

user58869

wait i'll find an example

skullpatrol

@Alex Have you used the glossary or index to look up notation?

Karl Kronenfeld

...

user58869

this

Chris's sis

@robjohn Have you seen this one before? $$\int_0^1\int_0^1 \log\left(\log\left(\frac{1}{x y}\right)\right) \log\left(\frac{1}{xy}\right)^{s} \ dx \ dy$$ I just created it (that doesn't mean I'm the first one that did that).

user58869

10:26 PM

$user image$

user58869

what is that?

Mike

@Ted I did universal coefficients for homology today. Tor is scary.

Karl Kronenfeld

@Alex It could be anything.

user58869

@skullpatrol what glossary?

skullpatrol

@Alex In textbooks.

user58869

10:27 PM

@KarlKronenfeld that's exactly what goes through my mind when i look at it

Karl Kronenfeld

@Alex You have to take into account its context.

user58869

@skullpatrol where do i begin

skullpatrol

At the beginning :-)

user58869

ok... i'm sorry everyone, i'm an idiot

skullpatrol

of the book

user58869

10:28 PM

bye

skullpatrol

later

:(

Mike

I assume he's a kid, he'll learn

skullpatrol

the problem is too common

Mike

the issue is you can't say where to start if you don't know where he is

user4140

@Mike mabye he just doesn't have experience reading math books.

Mike

10:32 PM

that has nothing to do with what I said

he has experience with some education system (I hope)

skullpatrol

But to get gain from experience you need experience...that's what you wanted to know: how much experience do you have ie What math do you know?

user4140

@Mike Having experience with an educational system and having experience self studying from a math textbook are not the same thing.

skullpatrol

true dat^

robjohn

@Chris'ssis never seen it before. Does it have a nice closed form?

Mike

@user4140 which is, again, not what I said.

Chris's sis

10:43 PM

@robjohn It should be $\Gamma(s+2) \psi(s+2)$ if I didn't miss anything there ... although I have a strange feeling and want to see again my calculations.

robjohn

@Chris'ssis have you checked these numerically?

user4140

Oh, I thought your last comment was aimed at me

robjohn

@Chris'ssis that is a good test

Chris's sis

@robjohn I was inclined to believe this is true for all $s>-2$, but around the point $-1.4$ I begin to have some issues there.

robjohn

@user4140 in a busy chat like this one, it is best to use the arrows that link to the comment you are replying to.

user4140

10:45 PM

@robjohn thanks for the tip

Chris's sis

@robjohn Numerically, for $s\ge-1.3$ (approximately) things are OK.

robjohn

@Chris'ssis I imagine it is probably okay for $\mathrm{Re}(s)\gt-2$, but numerically, it may be unstable near there

Chris's sis

@robjohn Yeah. At any rate tomorrow morning I'll check again all my work to see if I missed any important point.

skullpatrol

Hey @JasperLoy what's going on?

Jasper Loy

@skullpatrol I just woke up from a nap.

Chris's sis

10:52 PM

@robjohn For $s=-1.5$, Mathematica says NIntegrate::inumri: "The integrand Log[Log[1/(x y)]]/Log[1/(x y)]^1.5 has evaluated to Overflow, Indeterminate, or Infinity for all sampling points in the region with boundaries {{0.,3.97545*10^-31},{0,1}}"

mga

Hello nice people. May I ask a quick question about self-learning?

skullpatrol

askaway

:-)

Chris's sis

@robjohn This evening I conjectured an interesting fact about this integral family, namely: if we consider the multiple integral version with $n, \space n\ge2$ variables, then we get something like $\displaystyle \frac{1}{n-1} \Gamma(s+n) \psi(s+n)$. I'll have to check that one more time to be sure it's like that. (tomorrow)

Karl Kronenfeld

@mga ~~Of course not.~~

mga

Cheers. So I have an engineering-style math education. I just want to know how monumentous a task it would be to go through a good Real Analysis course, say Rudin. Is it doable, or am I being over ambitious?

Jasper Loy

11:00 PM

@mga Which Rudin?

mga

Baby Rudin, as I understand it's called :)

skullpatrol

To be perfectly honest, you're in way over your head, just kidding :-)

Jasper Loy

@mga If you have some talent in math, it's OK. Otherwise, it's tough working on your own.

mga

I always got very high grades in math courses in engineering, but whether that will actually translate into doing proper math I guess remains to be seen.

skullpatrol

How much learning have you done on your own?

mga

11:03 PM

But I am correct that analysis is an absolute must for any one wanting to learn mathematics properly, right?

Jasper Loy

@mga Well, nobody can answer your question here, because we don't know your math skills.

@mga I would say it is fundamental.

mga

@JasperLoy Sure I appreciate that you can't give me a tailored answer. :-)

Chris's sis

@robjohn For the general case with $n$ variables I think it's $$\displaystyle \frac{1}{(n-1)!} \Gamma(s+n) \psi(s+n), \space n\ge1$$

mga

I'm decent, but I don't think I'm a prodigy. I have a PhD (in engineering) though, so I'm used to having to learn stuff on my own.

skullpatrol

Electrical?

mga

11:05 PM

Yes.

Jasper Loy

@mga Hmm, why don't you just pick up the book and start and see how you go?

mga

@JasperLoy Sure. Just wanted a quick opinion of others first.

skullpatrol

I agree with Jasper :-)

Jasper Loy

@mga Anyway, what made you choose Rudin? Just want you to know there are many equally good alternatives.

mga

Everyone seems to rant on and on about Rudin and sing its praises, it seems to be "the" textbook.

Do you have any other favourites?

(Another factor is that I could find several sets of solutions.)

Jasper Loy

11:07 PM

Well, you can look at Browder's Mathematical Analysis, Pugh's Real Mathematical Analysis or Protter's A First Course in Real Analysis as well.

mga

To be honest when doing self learning going through books is hard; choosing the books is harder.

So cheers for the suggestions.

skullpatrol

It takes a lot of determination to stick with one book.

So choose carefully.

mga

I'll try Rudin first - if I find I can't handle it, I'll consider alternatives. Thanks for your help both.

skullpatrol

2 mins ago, by mga

(Another factor is that I could find several sets of solutions.)

this ^ is important

when you're on your own

mga

Indeed - probably one of the most important factors in fact.

skullpatrol

11:11 PM

:)

mga

Though math.se always helps.

Anyway, enough wasting your time. Thanks again.

skullpatrol

good luck!

Jasper Loy

That will be 100 dollars.

YoungGrasshopper

I was wondering if anyone could help me with my problem. math.stackexchange.com/q/720284/128346

skullpatrol

each

+ tip :D

mga

11:15 PM

@JasperLoy If you close your eyes and believe really really hard it should appear on your desk.

Jasper Loy

@mga There can be miracles when you believe.

mga

;)

Good night!

skullpatrol

later

user4140

@JasperLoy Isn't that anons phrase?

Jasper Loy

@user4140 It is my phrase, sorry.

user4140

11:17 PM

really?

Sorry in that case

Jasper Loy

It is well known in this chat that it is my phrase.

skullpatrol

no it's Mariah Carey's phrase

Jasper Loy

Yes, and then it became mine too, lol.

skullpatrol

you just want her, admit it

Jasper Loy

True dat.

skullpatrol

11:19 PM

:D

Alyosha

How does one evaluate finite sums using residues?

Mike

I can't advise it

Alyosha

I've heard that it's doable, though I still don't see how it makes anything easier.

Transcript for

$Mathematics$ Mathematics