@RandomVariable $\log \Gamma(z)$ sure has poles at the negative integers. Why should that mean it's a multivalued function?

@BalarkaSen do you see a very brilliant way of proving that?

@Chris'ssis Haven't tried it yet.

Been busy doing other stuff.

@RandomVariable have you seen this one? $$\sum_{k=1}^{\infty} (-1)^{k+1}(H_{\large (k+1)^2}-H_{\large k^2})$$

@r9m maybe you find a brilliant way for the series above ... :D

@Chris'ssis Did you find it ?

@Hippalectryon Sure.

@BalarkaSen $\log \Gamma (z)$ does not have poles at the negative integers. It has branch points there.

@Chris'ssis How fast does it converge ?

Gah. Yes, it does.

@Hippalectryon I'll show you the details when I put all on paper (if I didn't do mistakes).

@Chris'ssis Thanks, i'd be interested. The curve in wolfram looked like a very slow convergence

@Chris'ssis the $\displaystyle \sum\limits_{n=1}^{\infty} \psi'(n)^3$ is giving me a headache ... =P its almost done .. till now only elementary highschool stuff .. I'm left with another euler sum (last bit of the puzzle) .. after that I think 'll be done =)

Not sure if it's $\ln$ or lower

@r9m That can be shot down pretty easily.

@Chris'ssis oh God :(( .. I couldn't manage it easily :(

@BalarkaSen The paper I was reading contends it has no branch points anywhere.

@r9m It can be done with generating functions.

@r9m Well, you also need to use some clever manipulations. The point is that all flows naturally.

(smoothly I mean)

@RandomVariable WAT. Clearly if $z$ approaches $-1$ from the left then $\Gamma(z)$ is positive so imaginary part of log-gamma is $0$ while if it approaches from right side $\Gamma(z)$ is negative so the imaginary part is $i\pi$.

@BalarkaSen ya sure ... It can be done in a few lines if you lemme use residue thm .. the point being finding an elementary solution :))

using only series manipulations ..

@r9m Who cares.

@hipp, @Balarka \o

@BalarkaSen I do :P .. just or the sake of it =P

anyway bbl .. dinner time :)

I am not sure why generating functions are considered nonelementary here.

HI @DanielFischer

$\psi$-type generating functions can be derived elementarily.

@Studentmath ?

@Hipp fixed your laptop yet?

@Studentmath Not really :/

As long as it don't blow up, everything is fine.

It will eventually though

:)

I am reading a proof of Cauchy-Shwartz inequality (in order to prove that the standard definition of distance over $R^n$ is a metric), and in the proof they talk about discriminant of a polynom being non-positive. I tried looking up what is this 'discriminant' but couldn't find anything. Anyone can link me something relevant?

@BalarkaSen Really? How so?

mathworld.wolfram.com/PolynomialDiscriminant.html ? @Studentmath (not sure)

I think it is that @Hipp! thanks

Should I give 200 points to "you can't really do it."

@Studentmath Btw i just searched 'discriminant polynom' on google ʕ•ᴥ•ʔ

@Alizter For example, you can elementarily calculate $\sum_{n = 0}^\infty \psi(n) x^n$

@BalarkaSen: School sucks, I agree.

@Nick >:c

hi

could anyone take a look at math.stackexchange.com/questions/915033/… and tell me if it is unclear?

It isn't getting much love :(

@Nick School is great ᕙ(⇀‸↼‶)ᕗ

@Alizter math.stackexchange.com/questions/475491/…

Did anyone mention how stupid and unfair and boring and unproductive school is?

Because I can write an essay.

@ParthKohli unreproductive? :)

@ParthKohli School is productive when it's a good school

@ParthKohli what are you studying?

why do you think no one is answering math.stackexchange.com/questions/915033/… ?

@Hippalectryon: Dear ElectricHippo, you do not know the horrors I have to face in school. The Father who is the principal of the school is also the administrator. And he administrates everyday, with a long sharp wooden stick.

are we talking university or high school here?

Although he has a bias towards me, it pains me to hear to the cries of (like 17-18 year old ) young adults who are treated like babies.

@Nick He could have used a knife :P Jokes apart, it all depends of the school

It's much waste of a time.

I think you should change your attitude to school. First imagine you weren't allowed to go to school, then imagine someone let you and think of all the things you could get from it

@felix: You get to leave a university before the last bell.

To me, of course.

and then grab those things

@Hippalectryon Unfortunately, you have to get adjusted to the curriculum for the lowest common denominator.

I see mystical greetings is catching on. I dig it.

@ParthKohli I guess things vary a lot in different countries anyway

@felix Highschool... doesn't really teach you anything.

One can learn far, far better with far, far less help of the dratted school.

@BalarkaSen 9th and 10th don't.

@felix: ... I get nothing from school. I have to sit there from 9 AM to 4PM everyday except sunday. We barely have 3 full descent periods. The teachers don't come and when they do, they don't even teach as good as it is written in the textbook!

11th and 12th do.

@ParthKohli Like calculus snort

@BalarkaSen Calculus isn't the only interesting thing that you get to learn in +2 (a term many people use for the last two years).

@ParthKohli What else do you get to learn? I can't imagine me benefiting from 11th and 12th.

And the physics and chemistry - yes!

@BalarkaSen: if I one day go into politics and stand up for child rights, will you support me in saying a word about how homeschooling is a valid mode of education?

@ParthKohli Those aren't subjects, right?

@Nick Sure.

@ParthKohli: I'm an organism chemically decomposing in Orgainic Reaction Mechanisms.

@BalarkaSen Uh, physics and chemistry are subjects.

@Nick Ratta

@ParthKohli Bleh. Not to me.

@BalarkaSen Excuse me - I believe you back up the usability of your theories using chemists and physicists.

@ParthKohli Last time I looked, math doesn't need chemistry and physics.

Look better :P

The way calculus is taught is horrible as well. It takes the teacher 40 minutes to do 3 problems on differentiation. Those problems can't even be called problems. It's a waste of time. And one would be whacked for not sincerely paying "attention"

I have no theory, @Parth

Let alone theories.

@BalarkaSen: But Physics and Chem need Math.

@Nick So what.

@BalarkaSen math needs chemistry and physics in the sense that the three stand hand-in-hand.

@ParthKohli WAT

You're kidding me.

@BalarkaSen: So, if everyone on earth neglected that fact, Einstein could never have been born.

@BalarkaSen No, xkcd is NOT a reference ლ(ಠ益ಠლ)

@Hippalectryon I know, I know.

@RandomVariable Yes. The results may (in part) still be right, even if the argument is nonsense, though.

I was making a pun.

@Hippalectryon: Maybe not in the 21st century atleast.

You need some refinement in your definition of a pun.

@Hippalectryon: Where do you get those faces from?

@Nick 2183 : Oh no I have meme class at 3pm then trolling class at 4

Hi @DanielFischer what do you think of my comment, does it make sense?

@Nick unicodeemoticons.com and other websites, I even have a git with 150+

@Parth Anyways, math do not need phys and chem.

@BalarkaSen One needs phys and chem to need math.

@BalarkaSen: ... Um, theoretically you're right. Financially, you're wrong.

@Nick Check my pseudo description for more :P

@ParthKohli OK, back yourself up.

@BalarkaSen Is more pure good?

@DanielFischer At least some of the results are correct because the branch points of $\ln \Gamma \left( \frac{z+ia}{2 \pi i}\right)$ for $a>0$ are in the lower half-plane.

@Hippalectryon: (づ｡◕‿‿◕｡)づ Thanks!

@JohnDoe Which comment?

@Nick No problem (ᵔᴥᵔ)

what's going on with these strange emoticons .. how do you generate them ? :o

Yay! (｡◕‿‿◕｡)

@r9m: Copypasta ^

@r9m textfac.es

@r9m Here (ﾉ◕ヮ◕)ﾉ*:･ﾟ✧ emoticons for you

@JohnDoe You need a little bit more work, since you only have one $m\geqslant n$ (guaranteed) with $\lvert x_n - y_m\rvert \leqslant 1/n$. But the principle is right.

That site isn't working for me anymore :\

@ParthKohli: ﴾͡๏̯͡๏﴿ O'RLY?

@Hippalectryon Flipping table thingy is fun.

Post it on the chat

@DanielFischer Okay but that should be enough since lim n implies lim m.

* short story *

What's wonderful is that on my screen, it fits exactly into the default desc screen on the user page

@Hippalectryon There's a reddit account dedicated to those.

@ParthKohli :o

reddit.com/user/PleaseRespectTables

:O

I still prefer the axeGiraff though

          ("\ '' /").___..--' ' "`-._
           `ಠ_ ಠ  )   `-.  (       ).`-.__.`)
           (_Y_.)'    ._   )  `._ `. ``-..-'
         _..`--'_. .-_/  /--'_.'  .'
        (i l).-' '    ((i). '   ((!.-'

@DanielFischer Can I ask you something.

@JohnDoe You can try.

The Bolzano Weiestrass Theorem states that Every bounded infinite set in R^n has an accumulation point. Why does it not then follow that every bounded sequence in R^n converges? Is this because an accumulation point is not necessarily a limit?

@Chris'ssis I've never seen that one or anything similar.

@JohnDoe Yes. In general, a bounded sequence in $\mathbb{R}^n$ has many accumulation points.

@Chris'ssis I just found for free the pdf of the book, after having ordered it -__-

@DanielFischer Okay kewl, in general also are the following correct: a limit of a sequence is both an accumulation point and a limit point....accumulation point implies limit point but not conversely in general topological space?

@Hippalectryon Did you? Do you like the problems in it?

@Chris'ssis i'm still reading the intro :)

@Hippalectryon :-))))

@Chris'ssis I noticed it's pretty recent, how did you hear of it ?

@Hippalectryon From a friend of mine.

@Chris'ssis Grr that's not rigorous 'Landau’s notations f(x) = o(g(x)), as x → x0, if f(x)/g(x) → 0, as x → x0

@Chris'ssis And also $0\notin \mathbb{N}$ uh ... is that the usual way to define $\mathbb{N}$ in Romania/America ? (for me $0\in\mathbb{N}$)

@JohnDoe Accumulation point and limit point of sequences and accumulation/limit points of subsets of topological spaces are different things. If $y$ is an accumulation point of the set $\{ x_n : n\in\mathbb{N}\}$ (the underlying set of the sequence $(x_n)$), then it is also an accumulation point of the sequence, but a point can be an accumulation point of the sequence without being an accumulation point of the underlying set. Consider $(-1)^n$ for an easy example.

@Hippalectryon Not really. In our country $0\in\mathbb{N}$ too.

Uh weird

@Hippalectryon Even if I'm from Romania I consider the version without $0$, it seems this form is used more (at least in this site).

@DanielFischer How do their definitions differ?

@Hippalectryon Someone here suggested in the past that I might be Ovidiu Furdui, but this would be a terrible mistake.

@Chris'ssis Also, in $A$, is it $A=\underset{n\rightarrow\infty}{\lim}\dfrac{H(n)}{n^{n^2/2+n/2+1/12\cdot e^{-n^2/4}}}$ or $A=\underset{n\rightarrow\infty}{\lim}\dfrac{H(n)}{n^{n^2/2+n/2+1/12}\cdot e^{-n^2/4}}$ ?

@Chris'ssis xD

@Chris'ssis You're not that old are you ;)

@Hippalectryon Where do you have that from? Which page?

xvii

@Hippalectryon I can tell you for sure that guy is pretty gifted at this stuff.

In that page it seems that the exp is not in the power

But on the internet it seemed that it was in the power

@Hippalectryon let me check it

@Hippalectryon I don't find anything like that. Could you give me the number of the page one more time? I have a hard copy here.

@Chris'ssis prntscr.com/4injw2

@Hippalectryon That version in the book is correct.

@Chris'ssis Thanks

@Hippalectryon This is a basic fact, well-known.

@Chris'ssis You're back! :) What have you been working on lately?

@rehband I only created some stuff, but these days since my dog passed away, I haven't had such a pleasure for math as before ...

I'm still very creative ... (that's good)

@Chris'ssis Sorry, I didn't know your dog passed :(

@rehband Yeah ... :-(

I have a question regarding this problem: math.stackexchange.com/questions/905672/…

Can someone help me understand an element of the answer

@Hippalectryon I can also tell you for sure that many problems from that book would put down dozens of professors. Especially when you ask them to finish all in the spirit of the art.

@Chris'ssis i'll give that to my fellow classmates :)

@Hippalectryon I've never seen before such a book. By the way, is there any book better than that? One specialized on limits, series and integrals? I wanna buy it immediately.

@Chris'ssis You should know better than me :)

Anyone ?

@Hippalectryon I read some books , but I like more to do research and discover things on my own.

Oh wait

Nvm

@Hippalectryon Do you know what is the meaning of $B$?

@Hippalectryon Do you want me to teach you something nice, btw?

I just know its expression, but i had read $B_1$ instead of $B_0$

@Chris'ssis Sure :D

@Hippalectryon you can impress many classmates if you know this formula mathworld.wolfram.com/DobinskisFormula.html. You only have to know some of the Bell numbers that you multiply by $e$, and some series are immediately done. :D

Dobiński's Formula

There was such a question some years ago in one of the Harvard-MIT tournaments that I solved like that, without pen and paper.

@Chris'ssis I think I prefer reading books to doing research, so maybe I should not go to graduate school.

Well it can reduce $\sum\frac{k^n}{k!}$ to a Bell function, but then I need to know how to simplify $B_n$ @Chris'ssis

@Chris'ssis: good day! It's a holiday here. I have to go help my wife get some new walking shoes this morning.

@robjohn Hi. No problem. I posted some very delicious questions, don't miss them. :-)

@robjohn What holidays?

:17448330$$ \sum\frac{k^n}{k!}=B_n \cdot e$$

$$\sum\frac{k^5}{k!}=?$$

@Chris'ssis can you tell me the closed form of $\displaystyle\sum_{k=1}^{\infty }\left(\Psi^{(1)}(k)\right)^3$ :-) .. I need to verify sth ;)

@r9m Didn't I tell it to you?

@Chris'ssis no :( .. I'm not sure if all my calcs are right :|

@Chris'ssis $B_5\cdot e$.... what then ?

@Hippalectryon $$52 e$$

I have to learn that by heart right

1,2,5,15,52,203,877

@Hippalectryon ;)

Is there a cool way to learn that ?

877 seems ugly

@Hippalectryon It's not that bad to know the first few Bell numbers.

@Chris'ssis Well, thanks :)

I'll write them on a sticker

Stick it on my door

@r9m OK, but you got the right answer or not yet? That one I want to add to my book.

I'm off now :) see you

(´・ω・)っ由

@Chris'ssis I don't know the answer yet .. all I have now is $\displaystyle\sum_{k=1}^{\infty }\left(\Psi^{(1)}(k)\right)^3 = 11\zeta(2)\zeta(3) - \frac{37}{2}\zeta(5) - \zeta^2(2) + 2(E)$ .. E being an euler sum that I wanna reverse engineer the proof if I had the solution :)

@r9m don't you wanna think more of it?

@Chris'ssis I do .. :| but you know how lazy I am .. :P

not making excuses .. but I guess it might help to know the end result ..

hmmghh !! :( okay I'll try use the residues then .. sigh :{

I hate it when papers prove pretty much every single proposition and statement they have, then they state some major proposition and leave it proofless

"We allow ourselves some slack here"

@r9m Will you put that result on your blog before publishing my book? :D As I said I plan to add that one and 2 more from the same family in my book.

@Chris'ssis nah :( .. not just like that :P .. but If I'm able to complete the proof elementarily .. then maybe =P

Hi @DanielFischer

@JasperLoy Labor Day

@robjohn That is first of May here, lol.

@Chris'ssis posted on main or chat?

@robjohn on chat

@r9m did you read?

@JasperLoy the first day of May is May Day here, but it is not a holiday, at least, it wasn't for us.

@Chris'ssis ya :)

@r9m Done. ;)

@Chris'ssis got it ... :D

@r9m Crazy awesome or what? :-)

hahaha, you need to check to be sure. :-)

@Chris'ssis at this point I usually hug ppl .. but I guess an internet hug will do =D

@r9m :-))))))

@r9m Now, a crazy challenge $$\displaystyle\sum_{k=1}^{\infty }\left(\Psi^{(1)}(k)\right)^4$$

@Chris'ssis nah, can't accept that ... that will be the end of me =P

@r9m I'm not challenging you, but I only said that is a crazy challenge - - you'll also find it in my book.

o/w my grave will look like .. x_X R.I.P. r9m --- Here lies in unrest the soul of r9m crushed by sheer candy weight :P

@Chris'ssis: does this cover the limits you were talking about yesterday?

@Chris'ssis ya .. I'm totally waiting for The Book :D

@r9m :D

@Chris'ssis any idea when it will come out?

My books will take another decade at least to come out, lol.

@robjohn Well, probably somewhere during the next year. I have much work to do. The book must be completely crazy awesome.

@robjohn it seems it works that way.

It is very confusing that skullpatrol has Daniel Fischer's avatar.

@JasperLoy only if they are not the same person :-D

I am wondering whether to get Gelbaum's Modern Real and Complex Analysis. I have browsed through the book. I should read some reviews.

It seems that no graduate school uses this book, lol.

@JohnDoe Been eating. For accumulation points of sequences, it's the set of indices that matters; $x$ is an accumulation point of the sequence $(x_n)$, if for every neighbourhood $U$ of $x$ there are infinitely many $n$ such that $x_n\in U$. The point $x_n$ may be the same point for all these $n$. A point $x$ is an accumulation point of a set $S$, if for every neighbourhood $U$ of $x$ the intersection $U\cap S$ contains infinitely many points.

hi @DanielF, @robjohn, @chris'ssis, @Jasper

@TedShifrin Hi

@TedShifrin Hi. Has school started?

Hi @TedS.

@TedShifrin having a good holiday?

@chris'ssis: I sent you sincere condolences a few days ago.

@Jasper: Into the third week.

Yes, thanks, @robjohn.

I did my grilling last night and had friends over, @robjohn.

@TedShifrin Thank you for that. I think I missed that message.

@TedShifrin That's always a relaxing thing.

I thought you might have, @chris'ssis ... so I repeated myself ...

@TedShifrin She was like a human being to me ...

@DanielFischer How are those different when considering a sequence as a set of points.

I understand, @chris'ssis ... I have lost all of my cats and was very, very upset.

I have not lost any cats. I have lost my mind, however.

@TedShifrin That's a hard moment.

@Jasper: No point belaboring the obvious :)

@JohnDoe Consider $x_n = (-1)^n$. The underlying set of the sequence is finite. Hence it has no accumulation points. The sequence however has two accumulation points.

Confusing use of language there, @DanielF.

Yes, confusing indeed.

@DanielFischer I see so it is the fact that the points need to be distinct for the set definition that makes the definitions...well...distinct?

@TedShifrin Well, not the only one. And I didn't invent it, so I take no responsibility.

I don't actually ever use the term "accumulation point."

@TedShifrin What do you use?

limit point, convergent subsequence :P

I use limit point of a set and subsequential limit of a sequence.

high-fives @Jasper

@TedShifrin But, not all accumulation points of a sequence need to be the limit of a subsequence.

Oh?

Then what in the world does it mean?

Of course if you only deal with first countable spaces ...

Now that is a confusing def.

Daniel vs Ted fight fight fight

I told you: I don't use "accumulation point." I agree that in a non-first-countable sequence one can have a point of the closure that is the limit of no sequence.

I don't think @DanielF and I have fought. Can't say the same for some other members of this august chatroom.

@Daniel Do you distinguish between limit points for sets and sequences as well in the same way?

@Ted Oh but if you did who would win?

Depends on what the fight concerns, I imagine :P

Between Rudin, Royden and Folland, which is your favourite for real analysis? Just a survey.

@Ted You see its smart comments like that that get you in trouble with Daniel.

Folland was written after my days of studying were over. You should also include Stein-Stakarchi.

@DanielF: Am I in trouble with you?

I guess so ...

@JohnDoe A limit point of a set $S$ is a point $x$ such that for every neighbourhood $U$ of $x$ the set $S\cap (U\setminus\{x\})$ is non-empty. For sequences, I have seen two uses of the term limit point; one, if the sequence is convergent, every limit of the sequence [the limit need not be unique in non-Hausdorff spaces] is a limit point of the sequence, and no other point is one; two, a limit point of a sequence is a point $x$ such that some subsequence converges to $x$.

@TedShifrin Do you intend to enter the homework-done-no-effort-required camp? I didn't think so, so I guess you're not in trouble with me.

Well, I did give a hint to someone with a Riemann-Roch Riemann surfaces question with no effort.

I hardly answer anything any more :(

Hint is no problem.

I have a feeling that the site will continue to deteriorate to the point where most of us will quit coming ...

Has anyone here seen Gelbaum's Modern Real and Complex Analysis? I would like an opinion of the book. There are none online, lol.

Nope, @Jasper. Don't know it. When was it written?

@DanielFischer So Accumlation point of a set is clearly a limit point of a set...and the first definition of limit point of a sequence you gave is an accumulation point of a sequence?

@TedShifrin 1995.

I think these mixed real/complex analysis books are even more unpopular than the undergraduate rings-before-groups algebra books (such as mine).

It even covers Riemann surfaces and several complex variables, slightly.

@JohnDoe In both uses of limit point for sequences, a limit point of the sequence is an accumulation point of the sequence.

It's just that essentially no graduate curriculum has a 2-to-3-semester integrated course in that material.

Sorry, I keep asking about books because I am still deciding on the perfect reading list for when I begin to study math again.

@Jasper: These discussions are sort of silly. You have a very specific taste in style, which disagrees with mine. So my opinions are de facto not useful to you ... even if I had them.

Hmm, OK.

Ah, I wondered if he was the Gelbaum from Counterexamples in Analysis. He is.

@DanielFischer Could I ask how you would change the proof to make it complete, the one that I showed you earlier: Say sequences ${x_n}$ and ${y_n}$ are Cauchy such that for every n there exists $m \geq n$ such that $|x_n-y_m| \leq \frac{1}{n}$ , can we show that $\lim\limits_{n} x_n = \lim\limits_{n} y_n$ by noting that they both converge.

Note that $n \rightarrow \infty$ implies $m \rightarrow \infty$ and we have $y_m - \frac{1}{n} <= x_n <= y_m + \frac{1}{n}$, taking the limits as $n \rightarrow \infty$ gives the result by squeeze theorem.