Mathematics

robjohn

10:13 AM

Good morning.

$$
\frac12\zeta(2)^2-\frac12\zeta(4)
$$

No paper or pencil :-)

@Chris'ssis what? that's ridiculous! I did it while staring at it on the screen. (if my answer is correct :-)

Plugging in values, it is $\dfrac{\pi^4}{120}$

I just checked with Mathematica, and it agrees with me :-) Sum[HurwitzZeta[2,n+1]/n^2, {n, 1, Infinity}]

Note that
$$
\begin{align}
\cot^{-1}(n)-\frac1n
&=\tan^{-1}\left(\frac1n\right)-\frac1n\\
&=\int_0^{1/n}\frac{\mathrm{d}x}{1+x^2}-\frac1n\\
&=-\int_0^{1/n}\frac{x^2\,\mathrm{d}x}{1+x^2}\\
&=-\int_0^1\frac{x^2\,\mathrm{d}x}{n(n^2+x^2)}\tag{1}
\end{align}
$$
and partial fractions gives
$$
\frac{x^2}{n(n^2+x^2)}=\mathrm{Re}\left(\frac1n-\frac1{n-ix}\right)\tag{2}
$$
Combining $(1)$ and $(2)$ yields
$$
\begin{align}
\sum_{n=1}^{\infty}(\cot^{-1}(n)-1/n)
&=-\mathrm{Re}\left(\int_0^1\sum_{n=1}^\infty\left(\frac1n-\frac1{n-ix}\right)\,\mathrm{d}x\right)\\

Astrum

10:32 AM

@robjohn do you have any advice for working without solutions? The texts I'm using for math right now (Hoffman and Kunze, Spivak) have no solution manuals

robjohn

@Astrum if you are unsure of your answer ask Wolfram Alpha if the problem can be answered that way, or ask here.

Astrum

@robjohn all of the problems are proofs, wolfram isn't any help with that

robjohn

@Astrum then that leaves the second option. Try your hardest first. That way the help here will mean more to you.

Astrum

@robjohn it seems to be helpful in a way, because I don't have a choice but to solve it myself (or ask for help). Math comes much harder to me than physics =/

you guys have been a lot of help so far, really been a great thing for me

robjohn

@Astrum that's what MSE is for.

Ethan

10:35 AM

heres a similar way

wait wait

let me try at it

Astrum

the chat at physics SE is less populated, and their rules on HW help are a lot stricter

Ethan

$$\sum_{a=1}^\infty\sum_{b=1}^\infty \frac{1}{a^2(a+b)^2}=\sum_{a=1}^\infty\sum_{b=1}^\infty \frac{1}{b^2(a+b)^2}=s_1$$

By changing the order of the summations

$$2s_1=\sum_{a=1}^\infty\sum_{b=1}^\infty \frac{1}{a^2(a+b)^2}+\frac{1}{b^2(a+b)^2}$$

Chris's sis

@robjohn without pen and paper? That's good ...

Ethan

ahh screw it

its 2 am here

Astrum

@Ethan it's 5:40 here, I'm going to sleep soon

robjohn

10:39 AM

@Chris'ssis It seems really pretty easy...

The sum of $\frac1{n^2}\zeta(2)$ is easy $\zeta(2)^2$

The other part is $\frac12(\zeta(2)^2+\zeta(4))$

Ethan

oh no I got it

$$\sum_{(a,b)\in \mathbb{N^2}}\frac{1}{a^2(a+b)^2}=\sum_{(a,b)\in \mathbb{N^2}}\frac{1}{b^2(a+b)^2}=A$$

$$2A=\sum_{(a,b)\in \mathbb{N^2}}\frac{1}{a^2(a+b)^2}+\frac{1}{b^2(a+b)^2}$$

$$2A=\sum_{(a,b)\in \mathbb{N^2}}\frac{1}{a^2b^2}-\frac{1}{ab(a+b)^2}$$

Chris's sis

@robjohn I also have 3, 4 ways, but nothing without pen and paper.

Ethan

$$2A=\zeta(2)^2-\sum_{n=2}^\infty\frac{1}{n^2}\sum_{k=1}^{n-1}\frac{1}{k(n-k)}$$

alright I don't know what i'm doing anymore

I took adevan to go to sleep but its not working

robjohn

@Chris'ssis I pictured it on the integer grid

Ethan

lol

I did something similar a while back

Define $\beta_s$:
$$\beta_s(x)=\sum_{n=1}^\infty\frac{1}{n^{s}+x^s}$$

$$\sum_{m=1}^\infty\frac{\beta_s(mx)}{m^s}+\sum_{m=1}^\infty \frac{\beta_s(mx^{-1})}{m^s}=\zeta(s)^2$$

$$\beta_2(x)=\frac{\pi}{x}\frac{1}{e^{2\pi x}-1}+\frac{\pi}{2x}-\frac{1}{2x^2}$$

------------------------------------------------------------------------------------------------------------$$\frac{\pi^3}{180}(x^{-2}+x^2+5)-\frac{1}{2}\zeta(3)(x^{-1}+x)=$$
$$\frac{1}{x}\sum_{m=1}^\infty\frac{1}{m^3(e^{2\pi mx}-1)}+\frac{1}{x^{-1}}\sum_{m=1}^\infty\frac{1}{m^3(e^{2\pi mx^{-1}}-1)}$$

Chris's sis

10:49 AM

@Ethan maybe you like this one. Compute
$$\sum_{k=1}^{\infty} \frac{\zeta(2n)}{n(2n+1)2^{2n}}$$

Ethan

hmm

I am really screwing up my sleep schedule

Astrum

ok guys, I'm goin to sleep, see you all later

Ethan

I should be studying for the SAT

Astrum

@robjohn bye, and thanks for the help earlier

robjohn

$$\begin{align}
A
&=\sum_{a=1}^\infty\sum_{b=1}^\infty\frac1{a^2(a+b)^2}\\
&=\sum_{a=1}^\infty\sum_{b=a+1}^\infty\frac1{a^2b^2}\\
&=\sum_{b=2}^\infty\sum_{a=1}^{b-1}\frac1{a^2b^2}\\
&=\sum_{b=1}^\infty\sum_{a=1}^{b-1}\frac1{a^2b^2}
\end{align}$$
So twice $A$ plus $\zeta(4)$ (the $b=a$ terms) is $\zeta(2)^2$

@Astrum you're welcome

Ethan

10:53 AM

I had alot of interest in these sorts of double sums a while back

can't find my stuff

why are you still up robjohn?

robjohn

@Ethan What do you mean?

Ethan

its like 3 am in the morning

robjohn

@Chris'ssis did you see the answer $-\gamma-\arg(i!)$ ?

Chris's sis

@robjohn where?

robjohn

@Chris'ssis here

Chris's sis

10:56 AM

@robjohn Ah, beautiful!!! That's a really nice closed form!

robjohn

@Chris'ssis Nicer than the one with the Gammas and logs

Chris's sis

@robjohn Yeah, far nicer.

Ethan

I think I remember seeing a question just like that on the main

robjohn

N[-EulerGamma - Arg[I!], 20] gives $-0.27557534443399966272$

Vrouvrou

hello

Ethan

11:03 AM

good night everyone

Vrouvrou

can someone help me on relative homology

???

robjohn

@Chris'ssis: I thought this answer and this answer were kind of neat.

@Ethan Good night!

@Vrouvrou Sorry, above my pay grade...

Chris's sis

@robjohn Indeed. I like them.

robjohn

@Chris'ssis This morning has been a good one so far :-) good problems

Chris's sis

@robjohn You seem to be in a very good shape. :-)

Vrouvrou

11:11 AM

@anon can you help ?

{math.stackexchange.com/questions/586679/…}

robjohn

@Vrouvrou I said above that that is not my area. Sorry.

Vrouvrou

i understand but i say to to @anon

robjohn

@Vrouvrou Oh, geez. I'm sorry. I saw Chris'sis ping to me just above yours and mistook it :-)

Vrouvrou

no problem !^^

Mats Granvik

I wish I had found this triangle earlier: oeis.org/A042977 It is the generalization of the LambertW power series.

Dated at Sat Dec 11 03:00:00 EST 1999.

robjohn

11:26 AM

@MatsGranvik does this triangle bear on what you are doing?

Mats Granvik

@robjohn yes it is exactly what I have been doing. Only that it is expressed through differentiation, and not power series reversion.

robjohn

@MatsGranvik Ah. Reversion is often easier. You can get some nice recursive formulae that way.

Mats Granvik

@robjohn My trouble now seems to be that I expand exp(-x) around Log(z), but tautologically expanding exp(-x) around LambertW(z) seems to work too.

In fact it is identity. Expanding exp(-x) around LambertW(z) gives LambertW(z). How weird. Or true. But Wouter Meeussen differentiates and gets expansion around Log(z).

Through series reversion.

Vrouvrou

@Rasmus

Mats Granvik

The reason must be that Log(z) and ProductLog(z) are close to each other, as I think I have seen in a comment by Charles Greathouse.

Chris's sis

11:43 AM

@robjohn I think also this one works with some integer grid $$\sum_{n=1}^{\infty}\frac{1}{n}\left(1-\frac{1}{2}+\frac{1}{3}-...+\frac{(-1)^{‌n+1}}{n}-\log(2)\right)$$

@robjohn I finally managed to compute one question without pen and paper (the first one this day)

Mats Granvik

Note to self. Expanding exp(-x) around LambertW(z) gives exact answer while expanding around Log(z) is only approximate.

user2369284

hello

robjohn

@Chris'ssis Is it $\frac12\log(2)^2-\frac12\zeta(2)$?

Chris's sis

@robjohn No.

robjohn

@Chris'ssis Ah, there is only one sign alternation. Sorry. I was trying just looking at it.. I need to look harder :-)

Chris's sis

11:52 AM

@robjohn hehe, I like this kind of approach! :D

(brb - I have lunch)

robjohn

@Chris'ssis Is it $\frac12\log(2)^2-\zeta(2)$?

I think that is it

Chris's sis

@robjohn No.

robjohn

really? Hmmm

Chris's sis

@robjohn It's a very precious question. I love that.

robjohn

12:23 PM

Ah, $\frac12\log(2)^2$

Chris's sis

@robjohn No.

Pablo Rotondo

Hello

robjohn

@Chris'ssis Are you sure? I actually used pencil and paper and $\sum\limits_n(-1)^{n-1}\frac{H_n}{n}$...

Chris's sis

@PabloRotondo Hi!

robjohn

I will check it over and write it up in LaTeX

was the sign off?

Chris's sis

12:27 PM

@robjohn Indeed.

Pablo Rotondo

It would be convenient to write the difference of the sum and the log as an integral I think.

robjohn

Okay

@PabloRotondo I do that in computing $\gamma$

Pablo Rotondo

@Chris'ssis Is this your problem, or from your students?

Chris's sis

@PabloRotondo No, this one comes from a professor that challenges me once in a while.

Pablo Rotondo

The terms are $\frac{(-1)^{n+1}}{n}\int_0^1 \frac{x^{n+1}}{1+x}dx$, right? or something like that

Chris's sis

12:30 PM

(brb - lunch time)

Pablo Rotondo

@Chris'ssis Britain?

@Chris'ssis Okay, I think I have computed it correctly, is it $$ \frac{\log^2(2)}{2}$$ ?

Mimi

Anyone know the official definition of a pointed polyhedron?

robjohn

$$
\begin{align}
\sum_{n=1}^\infty\sum_{k=n+1}^\infty\frac1n\frac1k(-1)^k
&=\sum_{k=2}^\infty\sum_{n=1}^{k-1}\frac1n\frac1k(-1)^k\\
&=\sum_{k=1}^\infty\sum_{n=1}^{k-1}\frac1n\frac1k(-1)^k\\
&=\tfrac12\zeta(2)+\sum_{k=1}^\infty\sum_{n=1}^k\frac1n\frac1k(-1)^k\\
&=\tfrac12\zeta(2)+\sum_{k=1}^\infty(-1)^k\frac{H_k}k\\
&=\tfrac12\zeta(2)+\tfrac12\log(2)^2-\tfrac12\zeta(2)\\
&=\tfrac12\log(2)^2
\end{align}
$$

Pablo Rotondo

@robjohn Nice.

robjohn

@PabloRotondo I had previously computed the sum with the harmonic numbers

Pablo Rotondo

12:39 PM

@robjohn I see

I did it by noticing $$ \left(1-\frac{1}{2}+\frac{1}{3}-...+\frac{(-1)^{‌n+1}}{n}-\log(2)\right) = (-1)^{n}\int_0^1 \frac{x^{n}}{1+x}dx $$

(integrating the tail of the geometric series)

robjohn

12:55 PM

@PabloRotondo Yes, I use that in equation $(13)$ of this answer to accelerate convergence.

badass

1:25 PM

@robjohn Hey, how's it going?

Alizter

@robjohn What about this one? $$\sum^\infty_{n=0}\frac{(n-1)^{1-n}\;\Gamma(n+1, n-1)}{n!}$$

I got it "down" to $$\sum^\infty_{n=0}\left((n-1)e\right)^{1-n}\sum^n_{k=0}\frac{(n-1)^k}{k!}$$

Nick

@alizter: hi there Al. Um, since when did the gamma function take in two inputs?

Alizter

@Nick It's the incomplete gamma function

Nick

1:41 PM

oh, you mathematicians with your new-fangled functions. How adorable.

Alizter

Incomplete gamma function

In mathematics, the upper and the lower incomplete gamma functions are respectively as follows: : \Gamma(s,x) = \int_x^{\infty} t^{s-1}\,e^{-t}\,{\rm d}t ,\,\! \qquad \gamma(s,x) = \int_0^x t^{s-1}\,e^{-t}\,{\rm d}t .\,\! Properties In both cases s is a complex parameter, such that the real part of s is positive. By integration by parts we find the recurrence relations :\Gamma(s,x)= (s-1)\Gamma(s-1,x) + x^{s-1} e^{-x} and conversely : \gamma(s,x) =(s-1)\gamma(s-1,x) - x^{s-1} e^{-x} Since the ordinary gamma function is defined as : \Gamma(s) = \int_0^{\infty} t^{s-1}\,e^{-t}\,{\...

Chris's sis

2:33 PM

@robjohn did you manage to attend that Fibonacci series?

Alizter

@Chris'ssis HI

Chris's sis

@Alizter Hello

@robjohn btw, if you compute it, no need to post the answer (solution). I'm only interested to see that value you get.

Alizter

$$\sum_{k=0}^n \frac{e^{1-n}(n-1)^{1+k-n}}{k!}$$

robjohn

@Chris'ssis I haven't worked on it yet.

Chris's sis

@robjohn ok

@robjohn I'm writing up the proof immediately.

robjohn

2:50 PM

@Chris'ssis Found it... I will work on it.

@Chris'ssis Got it... That's cool. The partial sum is $\tan^{-1}(F_{2n})$

Chris's sis

@robjohn hehe, glad you like it.

robjohn

Ergo, the sum is $\frac\pi2$

Chris's sis

@robjohn :D

@robjohn I hope your son knows what kind of nice things his father can do!

:-)

I've always wanted to have someone like you around ...

out for a while

Alyosha

@Alizter, have you encountered using operators to simplify series?

There's a small chance it may work with your problem.

Alizter

@Alyosha Do go on

Alyosha

3:03 PM

I have no idea why, but you can treat operators like numbers if you are summing them so that they form $e^{\text{Operator} ...}$ (see page 3 here 129.81.170.14/~vhm/papers_html/rmt-final.pdf for an example).

robjohn

Note that
$$
\begin{align}
\frac{F_{2n-2}+\frac1{F_{2n-1}}}{1-\frac{F_{2n-2}}{F_{2n-1}}}
&=\frac{F_{2n-2}F_{2n-1}+1}{F_{2n-1}-F_{2n-2}}\\
&=\frac{F_{2n}F_{2n-3}}{F_{2n-3}}\\
&=F_{2n}
\end{align}
$$
Therefore,
$$
\sum_{k=1}^n\tan^{-1}\left(\frac1{F_{2n-1}}\right)=\tan^{-1}(F_{2n})
$$

Alizter

@Alyosha It looks interesting but so far it only looks like it works for special cases.

Alyosha

In the example, the operator $E$ is defined such that $E(f(n))=f(n+1)$, and $E^n(f(0))=f(n)$. $f(0)$ is moved out of the integral, so I assume it doesn't matter whether you operate before or after the integral.

There's a chance that your series could be converted into that special case, although with the emergence of the integral it seems unlikely.

It depends whether one can find a nice enough operator.

Alizter

@Alyosha This is beyond me right now :P

Alyosha

3:21 PM

This hardly helps, but $\sum_{n=0}^\infty \frac{e^{-n}}{n!}=e^{e^{-1}}$.

Alizter

@Alyosha So the two ways I can turn the function into an integral is by looking at the gamma version in which i steered away from in the first place. The problem is I either get an incomplete gamma function integral or a reciprocal of the gamma function. Both of which I have no idea what to do with.

Alyosha

You could first split the integral into $\int_0^\infty e^{-x}x^ndx- \int_{0}^{n-1} e^{-x}x^ndx=n!-\int_{0}^{n-1} e^{-x}x^n$ for integral n.

i love maths

sos440 , user : 9340 , Byrom Schmuland user : 940

Alizter

@Alyosha Lets work on this $$\sum^\infty_{n=0
}\frac{(n-1)^{1-n}\;\Gamma(n+1, n-1)}{n!}$$

Alyosha

3:37 PM

$\sum_{n=0}^\infty\frac{(n-1)^{1-n}}{n!}\left(\int_{0}^\infty e^{-x}x^{n}dx-\int_0^{n-1}e^{-x}x^ndx\right)
=\sum_{n=0}^\infty(n-1)^{1-n}-\sum_{n=0}^\infty\frac{(n-1)^{1-n}}{n!}\int_0^{n-1}e^{-x}x^ndx$

As the first integral is simply $n!$.

The rightmost integral is probably doable by IBP, it will form a sum, which may not be nice.

Vrouvrou

4:34 PM

help please math.stackexchange.com/questions/586972/relative-homology

user2369284

4:51 PM

Can anyone help me with my coordinate geometry problem ?

Will anyone be kind enough to help me ?

robjohn

@user2369284 It's best if you just ask. Then, if someone comes along who can answer, they will.

No one knows ahead of time if they can help.