Would it help to consider the more general. Actually, it probably wouldn't. I know hardly anything about hyperbolic functions anyway. $$ \prod_{n \geq 1} \dfrac{\sqrt[a]{\tanh a}}{\sqrt[b]{\tanh b }}$$

Well, there is some typo above.

It's $$\prod_{n=1}^{\infty}\frac{\sqrt[\Large 2^{2n}]{\tanh(2^{\Large2n}})}{\sqrt[\Large 2^{2n-1}]{\displaystyle \tanh (2^{\Large2n-1}})}$$

It looks like a marvellous piece.

Oh, yea that's what I thought you meant! I didn't pay attention to the argument of the hyperbolic tangent function in the numerator.

Yea, it looks like a beast!

@Khallil Glad you like my creations. By the way, it has a very nice closed form. :-)

@Chris'ssis Hey Chris, do you solve all of your problems like the one you sent me yesterday? I mean, do you always write it down neatly on your computer like that? I learned some good new tricks from that image u sent me

^_^

@rehband Well, I try to do my best. When I fail, I do it again and again until I manage to get what I want. :-)

@Chris'ssis I think I need to start doing that haha. Awesome!

@rehband :-)

When I fail, I watch some anime and procrastinate as long as I can!

@Chris'ssis There was a step which I did not understand. May I ask you about it?

@Khallil Haha

@rehband OK

@Chris'ssis: May I see the image rehband referred to?

Well, if you refer to the error term, then that part was a bit too optimistic. I did all in a hurry.

@Chris'ssis Near the bottom had the term $$O\Big(\lim_{n\to\infty}\sum_{k=1}^{n}\frac{1}{(1+k/n)^2}\Big)$$ Then in the next equality, you pulled out $$1/n^2$$ is that right?

@rehband Yeah.

Why is the sum now $$\sum_{k=1}^{n} \frac{1}{(1+k/n)^2}$$?

@rehband $$O\Big(\lim_{n\to\infty}1/n^2\sum_{k=1}^{n}\frac{1}{(1+k/n)^2}\Big)$$

@rehband Well, I used the fact that $$\lim_{n\to\infty}1/n \sum_{k=1}^{n}\frac{1}{(1+k/n)^2}$$ is a Riemann sum.

Oh yeah, nevermind, I see what you did :)

And this Riemann sum is the same as which integral?

$$\int_0^1 \frac{1}{(1+x)^2} \ dx=\frac{1}{2}$$

$\int_{0}^{1} \frac{1}{(1+x)^2}$ ?

yes

Alright, got it, thank you!

@rehband The point is that there I had $1/n$ that tends to $0$ when $n$ is large that is multiplied then by a finite value that comes from the Riemann sum. There was no problem to split the limit into 2 limits as I did.

Right, so whatever is in the $O(...)$ vanishes in the limit

and on the left there is another riemann sum

Awesome

Yes.

@Chris'ssis Great solution

@rehband Thank you! :-)

@rehband Did you see this one I previously posted? i.stack.imgur.com/HcXlg.jpg

@Chris'ssis No, not yet. Where do you post these?

@rehband It was initially posted here (the question I mean) math.stackexchange.com/questions/389991/…

When I found the solution there was only one answer involving Taylor series. I knew there must be an elementary way.

I'm trying to calculate $$\lim_{n\to\infty} n\prod_{m=1}^{n}\Big(1-\frac{1}{m}+\frac{5}{4m^2}\Big)$$ at the moment :)

@Chris'ssis Oh nice, I'll look at your solution in a few minutes

@rehband This one seems known to me. I think I computed it in the past and even had some kind of generalization.

Let me see if I have a good memory ...

@Chris'ssis Wow nice!

<^.^<

@rehband $\displaystyle \frac{\cosh(\pi)}{\pi}$

@Chris'ssis Alright :)

@Chris'ssis What methods did you use to solve it?

@N3buchadnezzar

@BalarkaSen ?

@rehband I used some nice tricks. I need to find my solution.

@N3buchadnezzar Should I go about proving analyticity of zeta by noting that $\zeta(s)$ with it's usual p-series definition is uniformly convergent on $\Re[s] > 1$ and thus one can interchange the integral and the summation for arbitrary closed contour $C$ $$\oint_C \zeta(s) ds = \oint_{C} \sum_{n = 1}^\infty \frac1{n^s} ds = \sum_{n \geq 1} \oint \frac1{n^s} ds$$ which is $0$ as $1/n^s$ is analytic for any positive integer $n$ and thus by Morera $\zeta(s)$ is analytic?

on the region $\Re[s] > 1$

@rehband By the way, it also appears in Ovidiu Furdui's book. Well, it initially appeared in AMM if I'm not wrong.

@Chris'ssis I got this problem from Furdui's book :)

@BalarkaSen dunno

@rehband What level is Furdui's book?

@N3buchadnezzar and it took you 3 minutes to say that?

oye.

@Khallil The problems are hard for me, I fiddle around with them for several hours, but I learn a lot from them :)

how would you go about proving the analyticity of zeta?

@N3buchadnezzar

@Khallil finish your set theory first.

I just browsed through Lang's Fundamentals of Differential Geometry. It seems terribly difficult.

@rehband Isn't there a solution?

Lang also wrote about differential geometry?!?

@BalarkaSen Yes, he uses Banach manifolds which may be infinite-dimensional.

@Chris'ssis Yeah, I haven't looked at the solution yet hehe. There's also a hint for this one, I'm trying to use that. Do you still know if your solution was different from the one in the book?

@rehband let me look at the solution in the book. Which problem?

@Chris'ssis 1.11

@BalarkaSen Fine!

just finished writing another draft about RH. this is a tedious job!

@BalarkaSen When you win a million dollars, you can give me some of it.

i am just writing a survey, Jasper

@rehband I have a huge number of problems and solutions ... (just wait a bit)

@BalarkaSen Is that Riemann Hypotheisis?

i am not mental enough to try it

@rehband yeah

@Chris'ssis Take your time!

3 very long posts and yet i haven't introduced all the tools needed to describe the mere statement of the conjecture.

@rehband Found it.

@Chris'ssis Sweet

@rehband I'm very talkative in my proofs, sometimes I write novels there. Let me cut some. :-)

@Chris'ssis I really like the talkative style! It makes things nice and clear. The downside is that there is no work left to be done :D

@Chris'ssis better than saying "this is obvious".

The software Mathematica is much better than Mathematics Stackexchange, it does not criticize you for asking questions.

@BalarkaSen lol :-)

@rehband take it

@Chris'ssis Got it, thanks a ton!

@rehband Welcome!

@Chris'ssis That's really a cool way to compute it.

@BalarkaSen Thank you! :-)

@Anthony

@Chris'ssis Back to the old limit problem: If you had used the other approximation with $\log^2(k)$ in the numerator, how would you have dealt with the term $$O\Big( \lim_{n\to\infty} \sum_{k=1}^{n} \frac{\log^2(n+k)}{(n+k)^2} \Big)$$

Can we make a Riemann sum of this?

@rehband It can be done nicely tough. How about thinking of it for a while? Maybe you find a nice way? :-)

Just give it a try.

@Chris'ssis Okay

@rehband HINT: factorize $n$ both in numerator and denominator.

(use the square in the numerator, split the sum, compute each some, done)

@Chris'ssis How do you want to factorize $n$ in the numerator?

@rehband $$\log^2(n(1+k/n))=(\log(n)+\log(1+k/n))^2$$

@Chris'ssis Oh ok, I see

@rehband The rest of the job is a piece of cake. All flows naturally.

@Chris'ssis Thanks :)

@rehband I mean when you go to school and someone asks you this one, you can tell them the solution without pen and paper.

They will be impressed! :-)

@Chris'ssis I hope that happens! hehe

Hmm ...

AFK

Today someone asked me about my dreams, and I think the first 3 that came to mind were: 1). publishing a book (that I firstly wanna dedicate to my mother) 2). leaving my country and never returning back 3). becoming like Ramanujan one day

you mean dying early?

Of course, there are more dreams. We are obliged to have dreams.

@robjohn Can you evaluate me $$\int_0^1 \frac{\{x\}}{x^{s+1}} \mathrm{d}x$$?

I think it's $1/(s - 1)$

God how I suck at integrals

@BalarkaSen on $[0,1]$, $\{x\}=x$

Yes.

I know.

Oh, wait, ah.

Thanks.

@robjohn I got a lot of nice results. :-)

@Chris'ssis do we get to see them?

@robjohn Here is one, a cute one $$\prod_{n=1}^{\infty}\frac{\sqrt[\Large 2^{2n}]{\tanh(2^{\Large2n}})}{\sqrt[\Large 2^{2n-1}]{\displaystyle \tanh (2^{\Large2n-1}})}=1+e^{-4}$$

Hello @TedShifrin

hi @Balarka ... how're you?

fine. as i pinged you twice the day before and the day before the day before, and received no reply, i thought you were ignoring me

@robjohn $$\prod_{n=1}^{\infty} \sqrt[\Large 2^{2n}]{\tanh(2^{\Large2n}})=\sqrt{1-e^{-8}}$$

@TedShifrin How about you? You are rarely seen in chat nowadays

@Chris'ssis Ah, those look familiar. I think I've done something like those.

$\{x\}$ is the fractional part of $x$, right?

yes

So in general, it's $\{x\} = x - \lfloor x \rfloor$?

yes

Cool, I picked that one up from here a while back.

Often I just appear for a few seconds to see if there's anything interesting going on ... So I'm not ignoring you, just not stopping. Classes have started, so I'm insanely busy. Plus trying to learn probability ahead of teaching it :D

Yea, over $[ -1, 1 ]$, $\{x\} = x$, right?

No, @Khallil.

@TedShifrin Oof, that must be tiresome.

I have a problem exactly like this, youtube.com/watch?v=yF2oCqt2-jU except distance to sattellite is 13000 instead of 12000. Anybody feel like running the calculation?

@Khallil Nope

Is it just over $[0,1]$?

No @Joe.

Yes.

$\{-1\} = 0 \neq -1$

In particular, @Khallil, $0\le\{x\}<1$.

@TedShifrin: sorry, I cannot get it to come out right. I've tried multiple variations.

Hold on. Then what about $\{1\}$?

@robjohn Do you have mathematica at hand?

Oh, thanks @TedShifrin!

@Khallil Oops. it should be $[0, 1)$

oh oh ... mathjax is having serious issues today.

Got it, @BalarkaSen.

Back to set theory.

(i.e. Back to Naruto.)

@robjohn $$\prod_{n=1}^{\infty} \sqrt[\Large 2^{2n+1}]{\tanh(2^{\Large2n+1}})=\tanh(2)$$

@robjohn This is a day that I also dedicate to a special product I did in the past, the one we attended in the past.

I can produce lots of marvellous products.

@Chris'ssis I just read your computation of $$\prod_{n=1}^{\infty} \sqrt[2^n]{\tanh(2^n)}$$. Truly awe some.

@rehband Thank you. I produced many fantastic results. See some above.

@Chris'ssis Wow. Impressive

These products are simply pure art.

I need to learn more elementary knowledge of infinite products. I had no idea that $$\cosh(z) = \prod_{k=1}^{\infty}\Big( 1+ \frac{4z^2}{(2k-1)^2\pi^2} \Big)$$

It looks so easy reading through your proofs lol

:D

@BalarkaSen sorry, I am proctoring an exam at the moment, so I am not going to be very responsive. What did you need Mathematica for?

Nevermind my complaints here earlier today. The feedback is actually the best thing about the site.

@rehband This is more easily shown with trig functions. A similar identity exists for cosine, and this one is gotten via $\cosh(x)=\cos(ix)$

True. There are lots (tons) of papers about these products.

What is the good supplementary book for a linear algebra course?

@Chris'ssis

@rehband Where is computation method link?

@MrWho I don't think I'm the right person to give you the best advice.

@Chris'ssis You can be one of the rights :D

I think @robjohn @DanielFischer @TedShifrin might give you a better advice here.

(and many others I didn't mention here since I don't know them well)

@Chris'ssis What are you working on now? :D

@MrWho I'm creating some special products. :-)

♫ moderatin' on the bus ♫

@Chris'ssis Have you written any book?

@MrWho Not yet ...

@Chris'ssis That's seriously sad, write a book for kids like me :D

@MrWho lol :-) It's not a bad idea to write a book for kids.

Them let me show you the way we represent and solve the infinite series ...

Wait ...

@Chris'ssis One problem I face, reading SE users solutions, which makes it hard for me to learn their techniques, it seems they have chewed both Complex and Real Analysis several times, they use as much as special functions like drinking a cup of water.

Ok

@MrWho see above: you have the question, and you have the solution.

yo @Alex

@Chris'ssis Fractal geometry?

@Chris'ssis Sierpinski.

@BalarkaSen Do you see what infinite series I can compute with that image?

@robjohn Nevermind. I was checking something from a book.

@Chris'ssis Yes.

The area.

@BalarkaSen Great. :-)

It's a pretty standard stuff.

Where can I find stuff about sierpinski triangle except wikipedia?

try mathworld.

i dunno of any standard book on fractal geometry

@MrWho why the hatred towards wikipedia though? just curious.

@MrWho I did it in Python some months ago :)

Wiki is the best resource. I learned the algo from there and coded it

@BalarkaSen Wikipedia is good, but not good enough, you know!

I hate wiki too, but I have reasons for that.

@BalarkaSen I completely agree on it.

@BalarkaSen hi

i am more or less showing this to every algebraist i know of in this room. must. stop. showing. off.

Is there an easier way to do this?

@BalarkaSen: Can you tell me if the result is rational or irrational?

One of my products above is not correctly stated.

@BalarkaSen Hang on, is the left exact sequence not just always true for any chain of subgroups N, H, G? Since if the map from Aut_N(G) to Aut_N(H) is just restriction, the kernel is always going to be Aut_H(G) by definition, giving the short exact sequence?

@BalarkaSen Tell me the reason to learn group theory?

What are the applications of it to physical world?

@MrWho: I think our questions made him jump back into the real world. Nice Going.

@Nick I don't know what the hell I'm supposed to do with papers messing around me! :|

@Nick It should diverge?!

@MrWho: The papers around you started to diverge?

@Nick LoL, no , integral should diverge

Harmonic series?

$1+1/2+1/3+1/4+1/5+...$

yes Surely.

@Nick Why probably?

So, there's no answer?

@Nick :D

@Nick Just break it into several pieces - sum rule.

since the integrand has a floor function, it behaves like harmonic series.We know that Harmonic series diverge.@robjohn Verify my statement?

Would if I could

Yes it does diverge.

But integrating it is a hassle.

@Chris'ssis You never gave me a clue about knowing whether an improper integral diverges or not?how can we find out at the first place?no clue for it?

@Chris'ssis: When you said I should go learn some more. You didn't give me any direction.

@Nick No, not seriously, it's clear and nice integral.

Just another way to represent the series.

I love it.

@MrWho Does it converge or no? If yes, why? If no, why? $$\int_0^{\pi/2} \frac{x}{\sin(x)} \ dx$$

What kind a test one would imagine here?

@Chris'ssis It should converge.

@Chris'ssis I have no idea of a particular test, but boundary :-|

@Chris'ssis: Bud, any resources you can link me to?

@Nick What sort of resources are you looking for?

@MrWho: Integration from bottom to top

@Nick No particular resource which wraps up everything.You have to dig in tons of problems and learn from mistakes.

@Chris'ssis Am I wrong?

@Khallil: I didn't see

@MrWho Solving problems is one of the best way to learn things.

It was basically the same advice that @MrWho gave. (Splitting the integral up into the sum of rectangles.)

@Nick read and learn all the stuff you come across ...

:D Good enough!

@Chris'ssis: And if I have problems I can come to you, right?

@Chris'ssis Not really for the advanced parts, learning from fragments is good but when it gets to serious topics you will be lost desperately.

@Nick Trust him, he shuts the shit(integrals) down. :D

@MrWho Of course, you need to know the basics, but if you wanna become good and attend the hardest problems then you need to learn a lot, to study, to do research.

You create you own problems, you challenge yourself, you push your limits every day.

@Chris'ssis The problem is, I spend two days to solve a particular integral and after two days I still doubt about the whole thing.

@Chris'ssis I have lots of flaws in Series Convergence Tests.I just cannot get along with it.

@MrWho I wouldn't recommend you to spend a lot of time on an integral when you're not experienced enough.

:D

@Chris'ssis Then what should I do to warm up?

@MrWho when you don't know, ask for help.

ok

@MrWho I mean it's also a matter of time management. You need to learn fast, you cannot afford to spend so much time on an integral now.

There is much stuff around, so try to be fast. No worry for the things you don't fully understand. You'll came back to them after a while.

disagrees

@Chris'ssis I tried several times to pass the first few boring chapters of Apostol, But I couldn't, I need to learn Analysis, but it's kind of bad.

@Chris'ssis Okay.

@MrWho Study the answers posted on MSE, there are lots. Study them deeply and learn.

agrees with @Karl

@Chris'ssis Is there any category to focus on?

@MrWho If you like the limits, series and integrals, there are lots of such questions and answers around.

@Chris'ssis I like them, but I want to get to 3 dimensions problems.Specially the ones which are applied in electromagnetic theory.I want to have mathematical peace when beating electromagnetic theory.

@Chris'ssis Something like elliptic integrals and their behaviors, I want to learn those stuff , I need them.

@MrWho Maybe before getting there you might like to know how to compute an integral like the one I posted above.

@Chris'ssis I'm trying atm.

@KarlKronenfeld You should expose your way, maybe I also learn a better way. I usually do as I described and had some succes (preparing to write a book now).

I'm sure there is always room for improvements.

@Chris'ssis I think I have to use differentiation under the integral sign, some sort of laplace transform

not sure, though

@Nick For $n\in\mathbb{N}$, do you see what $$\int_n^{n+1}\frac1{\lfloor x\rfloor}\,\mathrm{d}x$$ is? What is $\lfloor x\rfloor$ on $[n,n+1]$?

success

@Chris'ssis ?

Can someone help me try to determine whether one of the answers to my question is correct? The question is here: math.stackexchange.com/questions/900921 and it's Fly by Night's answer. Algebraically it looks fine, but I'm an amateur so I could easily be wrong

agrees with @Khallil

agrees with all who agree

that would make you agreeable :)

@robjohn I corrected my word above. I initially wrote "succes".

(I was out for some jogging)

@Chris'ssis You're connected to internet while jogging?

She said "was" :)

@skullpatrol You mean HE said "was" ?

Perhaps

@skullpatrol Surely, he. She can't be ... !!

@skullpatrol What was your major?

I'm not a major, I'm just a soldier :D

Army?

OR body guard? :D

lol

I guard the weak.

@skullpatrol So you're Robinhood :D

Sorta

@MrWho Yeah.

So you are connected to the internet while jogging? @Chris?

@skullpatrol I mean my internet connection is active, but I'm not in front of the monitor, of course.

oh

(I don't close my connection when I go out)

Do you have mobile access?

@skullpatrol No, I don't have at the moment.

icic

To make it slightly simpler, @Nick. $$ \begin{aligned} \int_{1}^{\infty} \dfrac{1}{\lfloor x \rfloor} \text{ d}x \ \ & = \lim_{n \to \infty} \left( \int_{1}^{n} \dfrac{1}{\lfloor x \rfloor} \text{ d}x \right) \\ & = \lim_{n \to \infty} \left( \int_{1}^{2} \dfrac{1}{\lfloor x \rfloor} \text{ d}x + \int_{2}^{3} \dfrac{1}{\lfloor x \rfloor} \text{ d}x + \dots + \int_{n-1}^{n} \dfrac{1}{\lfloor x \rfloor} \text{ d}x \right) \end{aligned} $$ Now follow @robjohn's hint, I think.

@Nick open problem.

@TomOldfield if H is not characteristic to G then it is not always true that every elt of Aut(G) can be restricted to H. In fact, Aut(H) can be even larger than Aut(G).

@MrWho loads application to chemistry, mostly coming from symmetry. loads application to physics, mostly coming from abstract topological groups. but all of them are too technical to explain or understand in full.

@BalarkaSen It's kind of load of notation rather than beautiful thought.Kind of organized system.

that's just the treatment of textbooks.

there is a great intuitive way to understand groups.

they come out naturally from set theory. there is nothing something as "abstract groups".

@BalarkaSen I don't like set theory, it makes mathematics organized but boring.

you haven't studied combinatorics much then?

@BalarkaSen I have, but depends on what you mean by "Much" ?

well, set theory is the basis of combinatorics.

There is no doubt that discrete math is one of the most applicable branches in math but I don't like those stuff, specially probability, although I have studied decently about it.

I like intuition-tended precision :-)

i don't like probability either, but i think i soon might, seeing that there is a vast connection with number theory.

btw, i haven't read decently about probability. i only know the basics.

@BalarkaSen Discrete math is not for human mind, it's kind of theology :-)

the only branch i have rigorously studied are nt and algebra

@MrWho i dunno about discrete math but set theory and all the algebra coming our from it are much intuitive

start by looking at the set-theoretic definition of groups, you might enjoy it.

Algebra is a system, the one which dominates , it's accepted, intuition is not needed, it comes from the safest part of brain :D

i mean modern algebra of course when i mean algebra

group, ring, field stuff

I will check it out.Right now, I'm shooting techniques to unlock the integral chris gave me.

I haven't even read a page about modern algebra, but I know, algebra is one of the safest places in math to walk in.

right, that's kinda true.

if it's one thing i hate totally about math, it;s the existence of integrals and summation though

@BalarkaSen Integral idea is beautiful, solving them is :-( !

haha yeah

hello @AntonioVargas!

@MrWho How much number theory have you studied so far?

Hallo Hallo!

So I'm trying to figure out a way to work in both Apostol and Spivak at the same time

Any idea how to go around this?

Basically, I want to start from 0 and do the 2 books in 2 months

Point to note: I did read parts of the books and did some problems in them before(as a supplement to Stewart(bad book))

Salut @Sab ... Do just one, not both.

Too much in either one to do just one in 2 months.

Salut @TedShifrin :)

Do you think it's doable in 2 months though?