@r9m Sometimes it's a bit of trial and error. But generally if the sum involves $H_{n}$ you want to use $\psi(-z) + \gamma $. And if the sum involves $H_{n}^{(2)}$, you want to use $\psi_{1}(-z)$. I just combined the two.
@RandomVariable okay thanks :-) .. I must say the evaluation of the series is really incredible :D
also I guess I have an idea with computing $\displaystyle \sum_{n=1}^{\infty} \frac{H_{n}^{(2)}}{n^{4}}$ without using contour integration (just manipulating the series .. ) .. :-)
I was guessing writing $\displaystyle \sum_{n=1}^{\infty} \frac{H_{n}^{(4)}}{n^{2}} = \zeta(6) + \displaystyle \sum_{n=1}^{\infty} \frac{\psi^{(1)}(n)}{n^{4}}$, might help .. the later series could be connected to $\displaystyle \sum_{k\geq 1}\frac{H^{(2)}_k H_k }{k^2}$ and $\displaystyle \sum_{k\geq 1}\frac{H_k^{(3)}}{k^2}$ and computed ..
@RandomVariable well its much better imo to have an identity that connecting the series with other known/easier to evaluate series ... the function you chose brilliantly reduced the effort by $1/100$ :-) .. which is why I was amazed ..
contour integration approach looks more promising than just trying to manipulate the series (the later somehow makes mee feel like I am calling shots in the dark) ..
reason why I'm blabbering : I'm poor with contour integrals .. @RandomVariable can you give me advice on where I should start reading from ? :-)
@r9m I have one on MSE where I use contour integration to evaluate a sum involving $H_{4n}$. I'm more proud of that one because I think it might be unique.
Hi off topic. Can Lagrange's theorem (the order of a subgroup divides the order of the group) be used to create subsets of a given group if we are given some element of the group?
@MikeMiller I'm wondering if the proof of Lagrange's theorem gives us a way to find a particular subgroup $H$ if we are given some element $a$ in the original group.
given an element of a group you can always use it to create a subgroup, namely, the subgroup generated by that element. But it doesn't involves Lagrange Theorem at all
If I put the relation $~$ on a given subgroup $H$ where $a~b$ iff $ab^{-1} \in H$ then the maps from $H$ to the equivalence classes of this relation gives bijections to each of the equivalences classes with the group. On the other hand, can I start off with some relation such that each class has the same number of elements and then work backward to find a group?
@MJD Sounds like the site needs a catalog of binomial sums. All the questions matching $... \sum ... \binom ... $ pattern in the first formula in the body of the post...
@RandomVariable okay :-) .. but I guess its okay to post it normally if you put the source as reference along with your insight :) .. I have seen many people do that ..
hmm ... the last time I posted something useful was 9th Aug .. :{
@r9m This fall I was going to take a graduate course in Galois theory as a graduate student-at-large. But they canceled it right before classes started.
@r9m After I graduated I took an undergraduate class in abstract algebra and an undergraduate class in theory of equations. But I don't feel like I learned much. The classes moved very slowly.
Hey, does anyone know a quicker way to get a Lefschetz decomposition of some de Rham form than by doing a wedge product with powers of the symplectic form to get each part of the decomposition one piece at a time?
Let $x_n$ be a real number sequence. There are infinitely many subsequences of $x_n$ that converge to any of its subsequential limit ( including lim superior of $x_n$ ). Is that a correct statement ?
I learnt we have different definitions for the total variation for functions of the form $f:\mathbb{R}^2\to\mathbb{R}$ which are in some way analogous to the total variation of functions of one variable.
For $y = f(x)$, where $f$ is smooth, the total variation is equal to the arc length of the c...
@RandomVariable Sure, first we want to do good things, but there is also an inner need for doing great things, at least to try to do them. It's in human nature to desire more and more ... Nowadays we take benefit of this technology because some believed in great things and produced them.
It's the miracle they produced such that I can read you, and you can read me right now.
@RandomVariable If referring to the series, well, a great thing, because I considered it like that, was the fact that I managed along with another mathematician to create a high school level solution to the Au-Yeung series. This is absolutely amazing to me, kids in high school would be able to solve that series elementarily.
I hope our solutions will fill the pages of the textbooks in the future with more and more elementary solutions to problems where at first sight there is no such hope.
@Chris'ssis Do you have any tips for periodic integrals?
I have a really disgusting solution for one where I managed to integrate the first half of the first cycle, then I just did some Janky Shit flipping and mirroring to make it "loop"
@robjohn I already mentioned this to Daniel Fischer, but I was reading a paper in which the author incredibly states that $\log \Gamma (z)$ has no branch points because $\Gamma (z)$ has no zeros. The paper is about evaluating integrals attributed to Carl Malmsten using contour integration. In most cases the results are still correct because the branch points of the functions he's integrating are in the lower half-plane and he's integrating in the upper half-plane.
And I wouldn't like to be misunderstood, I don't wanna be great for the sake of being great, but if I'm great it also means that I understood amazing things, I created amazing things, and I had fun, enjoyed a lot the beauty of mathematics as long as I did mathematics. After all, I need no appreciation from the others. The appreciation I receive from my dogs is enough for me.
@RandomVariable So, I wonder how he explains that $-\log(\Gamma(z))=\log\left(\frac1{\Gamma(z)}\right)$ has branch points at all the non-positive integers since $\frac1{\Gamma(z)}$ has zeros at all the non-positive integers.
@JasperLoy Intro to Real Analysis, Number Systems and the Foundations of Analysis, Urban Diversity, Comparative Politics, Poetry, Intro to Anthropology, and Acting for non-majors
@Khallil Was that for me? If that was for me, I've been fine... Job interviews this afternoon, now getting back into the maths. Got a couple of weird questions to run by here soon
in some physics setting there is a solution to a problem that can be calculated numerically because no integral can be found, and therefore apparently you cannot revert the function
just asking if this is true, sorry for poor wording
i personally don't think it be like this but they say it do, I'd like an opinion before I waste my time
@user35945 Let $f$ be strictly positive and continuous. Then $F(a) = \int_0^a f(x)dx$ is a monotone continuous function with $F(0)=0$. Then $F$ has a continuous inverse
actually, even if $f$ is not continuous this is true. But the positivity is essential.
@BalarkaSen When I heard you only used logarithms, my heart almost stopped for a few seconds ... I was about to faint ... for pleasure, of course :-))))
