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19:00
@Alyosha I think you wanna see something ... (are you around?)
@DanielFischer I just computed the series above (however I needed a walk before doing that).
Hi
Can anyone help me why K Gamma[-K, x] = -Gamma[1 - K, x] ? where K is positive integer and gearter than zero
@Chris'ssis Cool. Using only stuff ordinary mortals know, or some weird esoteric identities?
@Chris'ssis Thank you. Yes, though I doubt I'll be able to do it.
@DanielFischer By doing a clever trick :-). Well, we only need to wisely add a parameter and differentiate with respect to it. The rest is an easy job.
:15465023
@Chris'ssis What series?
19:06
Neato.
@PedroTamaroff $$\sum_{n=1}^{\infty}\left(\frac{n}{n^2+1}- \log\left(1+\frac{n}{n^2+1}\right)\right)=\log\left(\sinh(\pi) \operatorname{sech}\left(\frac{\sqrt{3}}{2}\pi\right)\right)-\psi(1+i)$$
Sometimes I just look at that and go "Meh." I mean it's cool but seems very irrelevant.
4
Irrelevant to what?
19:08
It strikes me as a particular calculation with little significance. As opposed to say $$\sum n^{-s}=\prod (1-p^{-s})^{-1}$$
Significant to what?
@skullpatrol The theory.
@PedroTamaroff I'm attending a divine art, this is what I like to believe, and I don't know what you mean by "irrelevant".
2
@Chris'ssis Oh, I thought you meant a problem. Yes, the solution is very nice, it's yours? Also, how would you go about formally proving that the Riemann sum is equal to the integral (I know it's obvious without proof)?
19:11
@Alyosha Sure, both the problem and the solution are created by me.
(Unrelated question): must an essential singularity have arbitrarily high negative powers of $z$ in its Laurent series?
@Alyosha What do you think?
For the singularity to be non-removable, I'd say yes, but there could be a strange counterexample.
@Chris'ssis he might mean that these problems you create and solve "are not natural". theory-sticking guys think like that (i fall into that category, sometimes)
Sticking blindly to a theory is not what math teaches.
19:17
@Alyosha The singularity is removable if and only if the principal part of the Laurent series is $0$. It is a pole if and only if the principal part is nonzero and contains only finitely many nonzero terms. It is essential if and only if the principal part contains infinitely many nonzero terms.
@skullpatrol That's not the point.
@skullpatrol well, i like to think that natural questions deserve better than that.
The point is some things strike us as "made up"; and other arise naturally.
@PedroTamaroff yes, i agree with you.
If someone is interested in a discussion, do not flag. You can highlight a discussion you liked!
Else we'll die crushed by stars.
Do not star the above!
19:18
lol
great, someone starred the below.
Duh.
"Please do not spam the star queue"...
But what did you expect, Pedro?
@MattN. Hello there.
19:18
Hey there.
@DanielFischer Yes, I shouldn't have said non-removable, that wasn't right.
@MikeMiller Tempted to star.
Don't succumb to temptation, @DanielFischer
@MikeMiller I'm on page 50 of Great.
Hi @MattN. how are you my friend?
19:19
@MattN. Isn't it... great?
'Tis.
I have the books. A pricy but worthwhile investment.
: )
Just like Kreyszig.
How is your flu?
I think they should rename the skullpatrol account to ELIZA.
19:22
Please stay away from horses for awhile :-)
"I could never resist an integral." - G. H. Hardy
3
@Chris'ssis actually, that does not mean all integrals represent something deep.
integrals are passtime funs.
@BalarkaSen This is just your interpretation.
@Chris'ssis and many others.
19:24
Only Hardy can tell you what he meant.
@BalarkaSen What does "not all" exactly mean? How many exactly then?
@skullpatrol He can't.
@Chris'ssis the ones arising from the theory.
@DanielFischer why?
@skullpatrol He's dead. Dead as a Dodo.
6
19:25
@DanielFischer I can't resist starring that.
@DanielFischer True dat :-)
My point is we all study math for our own reasons.
That^ is the point @PedroTamaroff
7 mins ago, by Matt N.
I think they should rename the skullpatrol account to ELIZA.
@MattN. if you wanna go to a private room I'm ready pal.
wating...
still waiting
@skullpatrol True but irrelevant. Hardy has written some texts. He does not need to live. Open "A mathematician's apology", and he talks to you.
@skullpatrol Are you some kind of troubleseeker?
@PedroTamaroff no, trouble seeks me.
@PedroTamaroff No, it's just an ELIZA bot.
19:35
@MattN. Do you think so?
Maybe.
I can buy that.
@PedroTamaroff Yes. That's what it seems to me, anyway.
@MattN. Did you ever watch Blade Runner?
@skullpatrol What do you remember of your mother?
@MattN. still waiting...
Haha.
19:37
@skullpatrol I just killed a cat. How does that me you feel?
But I mean a real ELIZA not some futuristic thing called android.
I'm serious.
Here is another sweet version I just created $$\sum_{n=1}^{\infty} \log\left(1+\frac{1}{n+n^2+n^3}\right)$$
@PedroTamaroff Have you every had a conversation with ELIZA? It feels exactly the same.
@MattN. Nope. Link me.
@MattN. I'm lovin this.
@PedroTamaroff Try this for example.
19:40

 0

4 mins ago, by skullpatrol
@MattN. still waiting...
@PedroTamaroff Not my kind of thing. Too jerky, not smooth enough.
2
Q: Dealing with offensive users on chat

Matt N.If someone uses swear words in chat then they might get flagged by other chat users and subsequently (and automatically) banned. If someone personally offends a user without using any swear words to the point where the attacked users decides to leave chat for good, nothing happens. I think th...

13 mins ago, by skullpatrol
7 mins ago, by Matt N.
I think they should rename the skullpatrol account to ELIZA.
Those are the facts @PedroTamaroff
@skullpatrol You're always bothering people pal.
You contribute to nothing.
3
You're noise.
8
You've pushed Asaf and Jonas away.
So I will side with Matt here.
@PedroTamaroff you know where the ignore button is pal.
@skullpatrol I think I may be joining you :( People don't like us in general.
19:47
@skullpatrol Still, that's not the point.
You're being obnoxious to people. You can't have everyone ignore you.
The easier thing is you to ignore people, meaning leave the chat.
Wait, I come up with a marvellous version ...
$$\sum_{n=1}^{\infty} \log\left(1+\frac{1}{n^2+n^4+n^6}\right)$$
I think I just developed a new series family ...
@Sawarnik who cares what people like or don't like?
That is their problem.
Not actually.
And I'm out of this.
"I don't like being stabbed in the face."
"Well that's your problem." *stab stab stab...*
You're not making sense. =)
Your opinion means nothing to me pal.
19:52
stabs Pedro in the face
@skullpatrol U MAD?
@MattN. I love the Anna and Maria arc. "HAVING SO MUCH FUN!! >:("
I'm still waiting @MattN.
Let's take it outside of this room.
Gezzz, I'm so creative today ... (I created more than 50 questions)
19:58
@Chris'ssis What is your average?
In say a month.
@hichris123 Hello there. I take it you came in for the flags?
... yup.
@skullpatrol hmmm, it depends. I'd say at least 600-700 questions a month.
@Chris'ssis :-O
@PedroTamaroff who flagged what?
20:00
@MikeMiller Sorry, was afk. What's the Anna and Maria arc?
@skullpatrol I often said here that I have (many) thousands of questions. It wasn't a joke.
i think it's not actually a good sign that we are having so much flags these days.
@hichris123 Well, some locals (at least four due to starring) that @skullpatrol is well... noise. (S)he usually finds ways to bother people. (S)he has drawn two good users away. Maybe you can talk some sense into him/her.
@Chris'ssis I always thought you were joking.
reads transcript
20:01
With a baseball bat.
@hichris123 Well, it is not new.
@skullpatrol No. This month I think I'll create more than 1000 questions, somewhere around 1000-1200 questions.
When he makes math questions it looks like he's trolling. Like $0^0$ questions and whatnot.
@BalarkaSen This room is renowned for that.
@Chris'ssis You need write a textbook :D
20:04
@skullpatrol Yeah, but it's hard for someone without a math background :-(. Anyway, I'm sure if I went to math uni, I'd be able to finish the first 3 years in one year. (if they let me do that)
@Chris'ssis That's quite the claim!
@PedroTamaroff I don't understand you either. Why is it not on your ignore list if it bothers you?
@MattN. I wanted to take action now.
If things fails, I will use it.
@PedroTamaroff I'm kinda lost. What do you want me to do about "it"? And what even is "it"? (examples plz?)
@PedroTamaroff Well, I'm used to the very hard work.
20:06
@hichris123 The ignore feature.
@hichris123 Dunno, advice him to change his behaviour.
@PedroTamaroff I have no idea why you haven't already. Years ago.
@PedroTamaroff What behavior though? I'm not a Math.SE regular... so I'm kinda lost on the context.
@hichris123 Dis.-
See badp.
@PedroTamaroff I think badp's answer is entirely unhelpful.
20:10
@MattN. Well, it's just an opinion.
Like anything else he's ever done while he was here in the chat room.
@hichris123 Well, maybe he should take a break. How many flags so far?
sigh I'm going to be going through the transcript for a bit. Gimme a few minutes (I have other stuff to do, too).
@PedroTamaroff Quite a few. :P
We like numbers. What do you got?
over 9,000
20:12
I'm... not sure if I can make that public, sorry.
@hichris123 No worries. I can't quite understand why anyone would be even asking such a question.
But hey, anyone may waste their time if they wish.
I guess.
Okay guys, enough "fun" for today, be seeing you.
later pal
@MattN the arc where the two women decide to live fulfilling lives
You just missed him, @Mike
y'all need to chill
20:23
@Sawarnik I just realised what your avatar reminds me of, "The Ring"
Ooooookay, I am out of here for the day.. peace.
20:39
Can anyone recommend a book that introduces (not necessarily exclusively) theta (and similar) functions?
@robjohn when you have some time, take a look of this since it's mind-blowing and beyond $$\sum_{n=1}^{\infty} \log\left(1+\frac{1}{n+n^2+n^3}\right)$$
@robjohn I created it today.
@Chris do you create an interesting integral daily
@ಠ_ಠ Thank you! :-)
@Chris'ssis What would you choose as your first book to write?
@skullpatrol Integrals, series and limits (again and again). (with at least 1000 very nice questions)
20:49
@Chris'ssis You would be good at it imo :-)
@skullpatrol Thanks! :-)
@Chris'ssis have you thought about taking some writing courses?
@Alyosha Do you mean some undergrad textbook on modular forms?
=P
Even people with disgustingly huge reps sometimes give unhelpful and wrong responses. Sigh.
pedro in 3, 2, 1
20:50
Hello, Professor @TedShifrin
@skullpatrol Not yet.
hi @skull, @Mike, @Balarka, @chris'ssis, mr eyeglasses
No, the current thing on modular functions I've got doesn't cover much the theta functions.
@TedShifrin Hello! :-)
Hello Professor @TedShifrin
20:51
@Alyosha Ah. Try Tata lectures on Theta (though it's hardly undergrad).
Um, scratch that.
Why? It looks OK.
The only book I have on modforms is Diamond & Shurman, so I don't think I can recommend anything useful, @Alyosha
And hello mr @Pedro, if you're lurking.
@Alyosha OK? Then OK. =D. I have never read all of it, just Umemura's article.
Just back from the first two rounds of the NCAA tennis ... the main tournament starts later in the week.
20:53
nice
@TedShifrin I just finished my combinatorics presentation.
You probably don't know what I'm talking about ... That's the American collegiate sports system ... so the national tennis tournament for college players (many of whom are ex-pats ...)
@TedShifrin how long have you played?
To whom did you present, @Pedro?
Have you seen the content of the poster you told me about, professor? @TedShifrin
20:54
no, @Balarka. It's been the weekend.
@TedShifrin I mean, no. I mean I just finished writing it down.
@PedroTamaroff grombinatorics, you mean.
@BalarkaSen No, no.
Ah, cool @Pedro ... @skull, since college ... but I'm just a poor hack.
@TedShifrin It's ok. Whenever you have time.
20:55
You should google Wolfram and quintic and see what comes up, @Balarka. They were selling this poster or giving it away years ago.
@PedroTamaroff OK, so it's been some real combinatorics, eh?
@BalarkaSen Ah, come on man don't be a douche.
Is their no special program for gifted students where you live @BalarkaSen?
@TedShifrin Well, I suppose it will only give Mathworld's article on quintics, which is highly uninformative.
20:56
@PedroTamaroff Don't call him a "douche"
Hmm, I dunno.
@skullpatrol It's ok. I don't mind if it's Peter.
=D
Only I am not allowed to call @Balarka names ...
@skullpatrol No, though I think it should be for the ones who are interested, not just gifted (I belong to the former team)
20:57
Name calling is not good.
Wow, that analysis problem got yet another wrong solution posted and removed. It's a killer, and I gather @DanielF still hasn't a solution.
Right, @Ted. It's not particularly easy.
I am aware. I had thought about it a few hours before I gave in and asked you. :) But I've never before seen a problem on SE generate so many mistaken solutions!!
@TedShifrin There's an integral question that got more. Hang on a minute, searching.
21:06
LOL, @DanielF ... I didn't mean for you to take my statement as a challenge :P
Mathematicians tend to do that a lot.
You speak from experience, @skull?
Just using what I have observed professor @TedShifrin :-)
So how many years, exactly, have you observed?
And how many real mathematicians?
Is this a test sir?
21:11
Sure.
You can fail me right now then.
@DanielFischer Don't we have some theorems about uniform convergence and $f_n(x_n)\to f(x)$, $x_n\to x$?
@PedroTamaroff We sure do. Why do you ask?
@DanielF: All I can say is WOW. I don't remember if I happened upon that post ...
Yes, we do.
21:14
@DanielFischer Well, let $$f_N(x)=\sum_{n\leqslant N}\frac{x^n}{(1+x^n)^n}$$
I think Ted suggested indirectly to look at $f_N(N^{-1/N})$
Ted pointed out something like $(N-1)^{-1/N}$ as the maximum point of $f_N$.
@DanielFischer Wow I have never seen a -16 voted answer :-O
@skullpatrol Go to meta.se, you can find -100 there ;)
I just deleted two answers there.
LULZ.
@DanielFischer That is just plain nasty imo
21:17
@skullpatrol Well, that was probably to Jeff, so... :P
I need to start downvoting more generously :P
I might as well turn into a total curmudgeon here.
@hichris123 Shouldn't Evan Carroll also have a few in that region? Or were they all deleted?
@DanielFischer Probably all deleted.
like really people climb down from your ivory towers...-193
The only ones of his that are not deleted are the non-trolling ones, @DanielFischer.
21:19
You talkin' to me, @skull?
@TedShifrin no
@hichris123 Which aren't many, Evan being Evan.
-193
A: Can we have the ability to retract a close vote before it closes?

Jeff Atwooddeclining -- you can always cast a reopen vote if the post gets closed. Also note that all close votes automatically expire after two days. (and for that matter reopen votes, or any other vote that attempts to reach a threshold -- otherwise, over an absurdly long period of time, say 10 years, e...

^ That's what he's talking about.
@DanielFischer yup, pretty much.
Ah, thanks @hichris
@DanielFischer Let $g_n$ be the individual terms in the sum. Maybe one should look at $$\sum_{n\leqslant k\leqslant 2n}g_k(n^{-1/n})$$ to get bad behaviour.
That's my hunch.
21:21
@TedShifrin I would never say that to such a gifted educator :-)
You are rare.
I only say I'm a gifted educator; my students might disagree :D
@TedShifrin Youtube does not lie.
@DanielFischer Note that $g_n(x)\geqslant g_{n+1}(x)$.
@PedroTamaroff Yes, looking at something like that may be instructive. Can't go thinking about that problem while being online, though, too much distraction.
@DanielFischer Do you think convergence is uniform or not? I am not sure.
21:24
@Pedro: It is alleged (and I do believe) that the series does converge uniformly.
I confess that early on I cheated and drew some graphs with Mathematica. Usually that's an immediate clue to the answer.
mr eyeglasses: Times have changed. Even the purest of mathematicians now use technology. To discover theorems, verify conjectures that they then try to prove, etc.
Cheating is allowed in math, it's the Physicists who can't cheat in the labs :-)
Okay, accusation rescinded
@PedroTamaroff I tend to think that yes. But I wouldn't bet any body parts.
LOL, mr eyeglasses, you can accuse me all you want :P
LOL @DanielF ... I think I would.
Unless I am doing poop, I get
$$\sum\limits_{n \leqslant k \leqslant 2n} {{g_k}} \left[ {{{\left( {2n} \right)}^{ - 1/2n}}} \right] \geqslant \frac{1}{2}\frac{1}{{{{\left( {1 + \frac{1}{{2n}}} \right)}^{2n}}}}$$
I used that $g_k\geqslant g_{2n}$.
Where $g_k=x^k/(1+x^k)^k$.
21:28
@PedroTamaroff I don't think that's true.
Well, that would seem to contradict the Cauchy criterion ... so I don't believe it. :P
@DanielFischer Yes, if $k>j$, $g_j>g_k$.
@TedShifrin what kind of mathematician would you consider the "purest"?
Peter opens Mathematica 9.
How ze hell do I input an infinite sum?
LOL ... If you put the infinite sum in there with functions, you'll wait a long time.
I plotted partial sums up to 1000, and then going to 10000 and beyond made it cranky.
21:33
Can you give me the code?
Try this, @Pedro:
g[k_] := x^k/(1 + x^k)^k;
Manipulate[Plot[Sum[g[k], {k, 0, N}], {x, 0, 1}], {N, 10, 1000}]
And how do I make it plot it?
That's in my code, silly. Manipulate gives you a slider so you can vary the N and watch.
Ah good.
@685-252: I don't play those games :)
21:36
meta-philosophical questions :D
@TedShifrin I think my calculations are correct.
Well, because Mathematica is confirming it.
But hmmmm ...
Btw, thanks @hichris123 for the time-out :-)
21:43
@TedShifrin $$\sum\limits_{n \leqslant k \leqslant 2n} {{g_k}\left( x \right)} \geqslant \sum\limits_{n \leqslant k \leqslant 2n} {{g_{2n}}\left( x \right)} = n\frac{{{x^{2n}}}}{{{{\left( {1 + {x^{2n}}} \right)}^{2n}}}}$$
@skullpatrol Wasn't me. It was an auto-ban because 10k+ users/mods thought one of your messages was inappropriate for chat.
Now put $x_n=(2n)^{-1/2n}$.
$$\sum\limits_{n \leqslant k \leqslant 2n} {{g_k}\left( {{x_n}} \right)} \geqslant \frac{1}{2}\frac{1}{{{{\left( {1 + \frac{1}{{2n}}} \right)}^{2n}}}}$$
@hichris123 Orly? a 10k user can ban?
21:46
@skullpatrol Flag.
Then people vote if the flag is valid or not.
^ that
I think I was flagged in here by a person of very little rep :P
6 flags from 10k users, or 1 flag from a mod
@TedShifrin they don't like "top of the world ma" by jimmy cagney
21:48
cagney?
yes
@TedShifrin you've seen it?
classic stuff
on YouTube
So, @Pedro, this means I lose my body parts?
The OP was told to prove it converges uniformly. Very interesting.
@TedShifrin Well, no. I spare you, of course. Unless that is wrong. Doesn't look like it, really.
21:50
BTW, to be pedantic (as you are wont to be, @Pedro), you need $\sum_{n<k\le 2n}$.
I'm afraid I'll post it and the post will turn my solution into something wrong.
The post is cursed!
Yes, I want to think about this very carefully.
Shhh...let him think.
So, @Pedro, to be specific, you're claiming that $\lim_{x\to 1^-} f(x) > f(1)$.
@TedShifrin Certainly! Look at the plots! =D
21:54
Well, I didn't perceive the discontinuity last night.
It was dark.
I guess the nontrivial part, perhaps, is to show $g_n\geqslant g_{n+1}$.
Oh, I thought you'd done so.
Well, it is true.
That's for sure.
SHOOT.
=D
Who shot?

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