« first day (1491 days earlier)   

12:13 AM
@Nick what did this mean? Yeah, I know I've asked twice ...
 
r9m
12:42 AM
@RandomVariable do you have some free time
 
@r9m Yes.
 
r9m
@RandomVariable I wanted to ask how you came up with the function here ?
 
@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.
 
r9m
@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 .. ) .. :-)
 
12:59 AM
@r9m Sometimes the contour integration approach is tedious and not very fun.
 
r9m
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 ? :-)
 
1:14 AM
@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.
 
r9m
@RandomVariable oh ! link please :D
 
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?
 
@TheSubstitute I don't understand what you mean.
 
leo
@TheSubstitute do you mean, to create subgroups?
 
1:24 AM
yes sorry I meant subgroups
 
@r9m It's probably not unique. But we can pretend that it is. :)
 
I still don't understand what you mean. Are you asking about the converse, i.e., that if $d \mid |G|$, then there is a subgroup of order $d$?
 
@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.
 
How do you propose it would?
 
leo
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
 
1:29 AM
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?
 
leo
ah, now I understand
yes kind of converse
 
leo
in general the answer is no
 
do you have a counter example in mind?
 
1:39 AM
@r9m As for creasson, he never responds to questions. And on another forum he basically called me stupid.
 
@Leo thanks
 
r9m
@RandomVariable sorry I went afk .. thanks !! :D I read that solution in the past and (+1)'ed it :-) Incredible ! :D
 
leo
@TheSubstitute counterexample
 
r9m
@RandomVariable :( hmm .. dosen't seem to be a nice guy then :-(
 
@r9m What was more incredible was Tunk-Fey's evaluation of that log integral.
 
r9m
1:47 AM
@RandomVariable which one ?
 
@r9m The really long one. You haven't seen it?
 
r9m
@RandomVariable maybe .. but a link might help :-)
 
r9m
holy moly !! its so long ... I have to read that one !! :) I missed that .. :O
 
It already has 16 more upvotes than anything I've ever posted.
@r9m I just realized that I haven't posted much of anything for quite some time.
 
r9m
2:02 AM
@RandomVariable :) !! then break the silence and post something !! :)
 
@r9m There is something I want to post, but it's based on something I read in a paper recently. I might post it as a community wiki.
 
@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...
 
r9m
@RandomVariable why as community wiki ?
 
@r9m Because it's not something I thought of on my own.
 
r9m
@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 ..
 

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