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7:00 AM
@KajHansen for all $m$.
 
Oh cool
 
take the point at infinity. ba dum tss.
 
Actually, I think I got it @MikeMiller
 
nice
 
Do you see the basic idea?
 
7:03 AM
now @Kaj. if you have your elliptic curve y^2 = x^3 + ax + b over rationals, i.e., a, b \in Q, then action by Gal(\bar Q/Q) will leave y^2 = x^3 + ax + b fixed. however, the torsion points would be switched.
 
haha, no
well, maybe the basic idea is one of the various dead ends I ran into
 
You're supposed to use $a_1,\dots,a_n$ as your basis.
 
if it is, then I saw the basic idea at one point
 
Well, for $J/J^2$.
You can extend to $J^k/J^{k+1}$ in the same way
 
I suppose you want all $k$-products of the basis elements as the new basis?
mod permutation, of course
 
7:05 AM
yeah for the latter
 
this was what I figured was the right basis but didn't write down a proof
maybe I just wasn't persistent enough
 
@KajHansen do you agree with me about the last statement made?
 
@MikeMiller it took me a while to see that it wasn't wrong
 
I'll have to think about it at some point @BalarkaSen
 
well torsion points are algebraic. Gal(\bar Q/Q) acts on them...
 
7:07 AM
not difficult to show it works for $J/J^2$
 
that the action leaves order fixed follows from the fact that they are automorphisms
 
why do we care about if something is $+\infty$ or $-\infty$?
 
seems tedious for higher powers
 
Oh ok got it.
 
@MikeMiller Really isn't any more work
 
7:07 AM
I'm doing French at the same time we're talking. Not really in math mode.
 
Gal(\bar Q/Q) acts on the torsion points (the group Z/m \times Z/m) by automorphismsm. so there is a homomorph Gal(\bar Q/Q) \to Aut(Z/m \times Z/m)
 
I'll try writing it down in a bit
I'm finishing up a couple other problems
 
Aut(Z/m \times Z/m) is just GL_2(F_m), so you have a representation Gal(\bar Q/Q) \to GL_2(F_m)
 
want one I found fun that doesn't really require any machinery?
 
7:09 AM
Getting lazy with your dollar signs I see. That last line sense though.
 
then I'll go back and see if I unstuck myself on what I was originally doing :)
 
that's the general idea @Kaj. that's why the representation theory of teh abs galois group is interesting.
@KajHansen glad to be very unmotivating :P
 
haha
 
given an ideal $I$ of a ring $A$, we can define the associated graded ring $\text{gr}_IA$ by giving it the group structure $\bigoplus I^n/I^{n+1}$ and product $(x+I^{n+1})(y+I^{m+1}) = xy+I^{n+m+1}$ (and extending, ofc)
Matsumura has a theorem that $\text{dim }\text{gr}_IA \leq \text{dim } A$
the exercise was to find a situation where the inequality is strict
it won't work for any geometric picture you try to draw, so you have to make your rings pretty messy
what're you working on lately? i think i've asked before
 
@KajHansen someone said teach someone the stuff you don't understand. i'm just using you as a bait. :P
 
7:14 AM
@BalarkaSen Doesn't that strongly suggest you will be teaching him something wrong?
 
I'm not good bait when I'm preoccupied with finals :/
 
@KajHansen Yeah, this chat should be reserved for non-math topics, lol.
 
@Committing sure. i am dumping the wrong stuff in him, and understanding the correct stuff myself. :P
for example i was not sure why Gal(\bar Q/Q) acts on the torsion points by auts. i even patched that up by naive arguments above. but now i'm gonna go do the calculations :P
waits for a smack from @Kaj
 
Smacking should be banned from this chat.
2
 
@Will how about bashing then?
 
7:19 AM
Thanks a lot @BalarkaSen. Now you're going to distract me once my finals are over because you could be just babbling for all I know.
 
@BalarkaSen Not funny, lol.
@KajHansen How was your Putnam?
 
@BalarkaSen That's really bad to be honest
 
Now I'll have to delay my post-finals partying and debauchery until I can figure out this math
:P
 
i know @Committingtoachallenge. but good for self-learning :P
 
@MikeMiller I'm not focusing effort on any one project over a long period of time. Now, I'm playing around with something related to Hilbert rings.
 
7:21 AM
@Committingtoachallenge, not really. I didn't assign any truth value to his stuff anyways given the lack of proof.
But it does give me stuff to consider.
 
@KajHansen I've been drinking and mathing; there's one problem I only solved post-beer, so I can firmly say that it's thanks to beer that I could solve it.
@KarlKronenfeld Motivated by Pedro?
 
@KajHansen That's good xD. I try not to distract people before their finals personally.
 
I totally distracted @Kaj from doing french.
 
@MikeMiller Not seeing any solutions right now, I'll probably come up with something later.
@MikeMiller Yeah. I also wanted to study Hilbert rings before reading his notes, so this was a decent opportunity to do so.
 
There's no accounting for taste...
 
7:24 AM
OK, I gotta go. Good luck for the finals and just don't let the torsion points bite Kaj.
:P
 
@BalarkaSen Goodbye Balarka-san[intentionally sic]
 
It's 24 days to the new year.
 
 
1 hour later…
Anonymous
8:46 AM
@Committingtoachallenge saw the post?
 
@Ashwin You should make messages like that in my chat room so we don't clutter :)
 
Guys, I will be deleting my account soon. For those of you who know my email address, we will keep in touch via email. If I change my email, I will let you know via email again. I wish all of you happiness in this life and all future lives, until the end of this cycle of universe expansion and contraction.
3
 
@WillHunting That is sad to hear. I would be interested in reading a blog if you decided to start one for your studies.
 
Anonymous
No contraction @WillHunting
 
Anonymous
I don't believe in the big crunch :D
 
8:55 AM
@Ashwin The unification of gravity? I do
 
Anonymous
I think he meant the Universe will end in a big crunch
 
@Ashwin Yeah I haven't read too much about it, I assumed the big crunch was gravity compressing to a point
 
Anonymous
That's not what physicists think @Committingtoachallenge
 
Anonymous
The last phase of the Universe probably is the Black Hole era-where everything that there is are Black Holes
 
@Ashwin I am not a physics student, so I will definitely have some reading to do before I make more comments :)
 
Anonymous
8:59 AM
I was wrong.The last phase would be composed of dark matter
 
Anonymous
and photons?
 
Anonymous
@Committingtoachallenge By the time our star-the Sun gets exploded,we should move to some other habitable place.@WillHunting is gonna search one for us :D
 
@pleasedeleteme Goodbye dear 'Please delete me'
 
He'll come back.
 
9:18 AM
@MikeMiller Our magical numbers have a deep connection. 4! = 24
 
9:44 AM
What's up with these spam questions? Another one just showed up.
 
10:02 AM
Is there a better title for my question?
 
@MikeMiller I lied about it being as easy in J^k/J^{k+1} as it is in J/J^2. However, after filling in all of the details I still haven't used the hypothesis that A is Noetherian and local.
 
Hi, guys. Does anyone here know what this means: $x\in\mathbb{C}\setminus\{1\}$?
 
$x$ is a member of the set of complex numbers with the element $1$ removed.
 
10:22 AM
@BalarkaSen Probably within a week
@KajHansen Do you track your workouts online?
 
10:44 AM
@KajHansen Thanks
 
@robjohn Is there any solution for this
 
@Integrator It seems that you've picked up some friends. I have some, too. If they serially downvote, their votes will be reversed. However, if they don't go away within 24 hours, let me or any other mod know and we will have a community manager look into things.
 
11:17 AM
@robjohn Okay!
@robjohn Actually a down-vote just take away 2 rep, no big deal. But that -2 in achievement tab seem to frustrate me! :(
@robjohn btw thanks!
 
@Integrator Yeah, downvotes are not that bad in a numerical sense, but the psychological effect is far worse, especially on a good answer. Most of those answers are highly upvoted and I think most are accepted.
 
@robjohn They've taken my one nice answer badge! :(
 
11:37 AM
Can someone give me reference for construction of Green function for second order non-homogeneous linear differential equation? Thanks
 
12:26 PM
@UserX you're account is suspended!!!, Why?
 
Hey guys - I have a question on something that is quite trivial, but I seem to be missing out on a conceptual issue. I am not quite sure how to calculate the value of a non-zero sum, Bimatrix game. I have looked throughout my text (and the web) for examples of actually 'working it out', however I have found none - most of them just go along the lines of "Here is the minmax theorem, here is our bimatrix game, and wa-la, here is the value".
Any help with this, would be very much appreciated. :)
 
Greetings
 
Hello!
 
12:45 PM
$$\int_0^1 \operatorname{li}^4(x) \ dx=?$$
 
What is $\li^4$?
 
To the power of 4?
 
@regret Yes.
@regret Do you know what is the value for the cubic version?
 
No clue!
 
12:49 PM
@regret You might like to know?
 
I would!
Oh no!
 
@regret It's not that hard.
 
@Chris'ssis what is the next series, i know partial fractions now
 
@beginner $$\sum_{n=1}^{\infty}\frac{9n^2+12n+5}{(3n+2)!} $$
 
@Chris'ssis woah that is bigger. thank you!
 
12:54 PM
@beginner Welcome :-)
 
@Chris'ssis About that sum you posted yesterday, I realised afterwards that it can be solved fairly quickly using only the generating function of $\dfrac{H_n}{n^2}$.
 
@M.N.C.E. Yeah, that is needed. You refer to the squared version, right? Not the cubic one.
 
@Chris'ssis Yes. But I think the cubic one can be solved in the same way.
 
@M.N.C.E. there is needed the generating function of $\displaystyle \frac{H_n}{n^3}$.
 
@Chris'ssis It's
\begin{align}
\sum^\infty_{n=1}\frac{H_n}{n^3}z^n
=&2{\rm Li}_4(z)+{\rm Li}_4\left(\tfrac{z}{z-1}\right)-{\rm Li}_4(1-z)-{\rm Li}_3(z)\ln(1-z)-\frac{1}{2}{\rm Li}_2^2\left(\tfrac{z}{z-1}\right)\\
&+\frac{1}{2}{\rm Li}_2(z)\ln^2(1-z)+\frac{1}{2}{\rm Li}_2^2(z)+\frac{1}{6}\ln^4(1-z)-\frac{1}{6}\ln{z}\ln^3(1-z)\\
&+\frac{\pi^2}{12}\ln^2(1-z)+\zeta(3)\ln(1-z)+\frac{\pi^4}{90}
\end{align}
 
1:06 PM
@M.N.C.E. Yeah. I have it in a form less simplified.
 
Is this the right direction, or should I go back to the start?
$$\sum_{n=1}^{\infty}\frac{9n^2+12n+5}{(3n+2)!} = \sum_{n=1}^{\infty} \frac{1}{(3n)!} + \sum_{n=1}^{\infty} \frac{1}{(3n+2)(3n+1)(3n-1)!} + \sum_{n=1}^{\infty} \frac{3}{(3n+2)!}$$
 
@Chris'ssis I'm just curious. Are you able to compute the generating function of $\dfrac{H_n}{n^4}$?
My attempts failed so far.
 
@M.N.C.E. I didn't try that yet, but I'll give it some tries soon.
@M.N.C.E. Happily there are other ways to establish their closed forms for $x=-1$ and $x=1/2$
@M.N.C.E. like in one of the problems I proposed to myself. Let me check my proofs.
 
@Chris'ssis What is closed-form of $$\int_0^1\int_0^y\, \frac{y^{a-1}}{\ln x\sqrt{-\ln y}}\,dx\,dy$$
 
@Venus $\frac{e^3}{\pi\zeta(3)}$ something like that probably :)
 
1:15 PM
@Venus Just a bit to show something to M.N.C.E.
 
@HatMan the hatman doesnt exist
 
r9m
@Chris'ssis I have seen that series somewhere ! :o
 
@M.N.C.E. take all pictures and after that I delete them.
@r9m Where? It's an elementary series, maybe it's everywhere.
 
r9m
@Chris'ssis everywhere .. ! wait lemme check sth :-)
 
@Chris'ssis Thanks for sharing.
 
1:20 PM
@robjohn could you delete my pictures above? They passed 2 minutes on chat.
@M.N.C.E. Welcome.
@M.N.C.E. That integral allows us to establish a connection to $\displaystyle (-1)^{n+1} \frac{H_n}{n^3}$ and $\displaystyle \frac{H_n}{2^n n^3}$ and compute them both without using the generating function you posted above.
@robjohn thank you
 
Heya @Alizter
 
@Chris'ssis Yup. I recall doing something similar for $\sum^\infty_{n=1}\frac{(-1)^nH_n}{n^4}$ and $\sum^\infty_{n=1}\frac{H_n}{2^nn^4}$.
9
A: Find the closed form of $\sum_{n=1}^{\infty} \frac{H_{ n}}{2^nn^4}$

M.N.C.E.Here is a solution that does not rely (too much) on softwares. I will be using the known values of the sums $\small{\displaystyle \sum^\infty_{n=1}\frac{H_n}{n2^n},\ \sum^\infty_{n=1}\frac{H_n}{n^22^n},\ \sum^\infty_{n=1}\frac{H_n}{n^32^n}}$. Let $$\mathcal{S}=\sum^\infty_{n=1}\frac{H_n}{n^42^n}...

 
@M.N.C.E. You need no residue there since $$\sum^\infty_{n=1}\frac{(-1)^nH_n}{n^4}$$ comes from a class of series that can be computed by series manipulations only (when the power in denominator is even).
I have somewhere that paper ...
@M.N.C.E. that is $$S(k)=\sum^\infty_{n=1}\frac{(-1)^nH_n}{n^{2k}}$$
@M.N.C.E. the nice nice part is that we can always establish a relation between these series at points $x=-1$ and $x=1/2$.
 
r9m
@Chris'ssis via euler transforms ? :-)
 
@beginner $$-2\sqrt{\frac{\pi}{a}}\ln\left(\,\sqrt{a}+\sqrt{a+1}\,\right)$$
 
1:32 PM
@r9m via integral manipulations at least. @robjohn did something like that using euler transform.
 
r9m
@Chris'ssis yes ! @robjohn's method ! :) but I failed to get a general relation like for the one you posted above !
 
@r9m Well, I had in mind to try to find such a generalization, but it seems I didn't do it so far. The truth is that I have many interesting research subjects, hard to choose which one to work on.
 
r9m
@robjohn sensei is there a way to get the general relation using euler transforms (via the binomial identity you used for one of the answers .. $\displaystyle \sum\limits_{n=1}^{\infty} \dfrac{H_n}{2^nn^2}$) :)
@Chris'ssis oh ! okay ! :D please lemme know if you write a paper on it :D I am dying to find a generalization ever since I saw the case for $k=1$ :D
 
I don't think that is possible for powers higher than $4$ since that will involve sums like $\displaystyle\sum^\infty_{n=1}\frac{H_n}{n^{2q+1}2^n}$...
 
@r9m I'm still waiting for the first paper to be released ... it seems there are great odds to happen this soon.
 
r9m
1:38 PM
@Chris'ssis :D Cool !!
 
@r9m That paper is terribly awesome ... hope all will be fine ...
 
r9m
@Chris'ssis :D
 
@r9m :D
 
Hello everyone I'm new here. I posted a question regarding triangles, can you please check it out? Thanks :)
 
r9m
@user35828 link please :)
 
1:41 PM
Yes one second, sorry
0
Q: Triangles area question

AyanThis question came in RMO, an olympiad in India. I solved it but with the assumption that the lines are parallel, though we are not given this info in the question. Let ABC be a triangle. AD be a perpendicular from A to BC. Let there be 3 other lines intersecting AD at K,L,M such that AK = KL = ...

 
Hello everyone,when I post, then appearthis :We are currently offline for maintenance
We are currently offline for maintenance

Routine maintenance usually takes less than an hour. If this turns into an extended outage, we will tweet updates from @StackStatus or post details on the status blog.
 
It was there a few minutes ago, but it's fine for me now..try again?
 
r9m
@Chris'ssis this on is interesting :-) ... is there a nice closed form ?
 
@r9m Anything is possible ... :-)
 
r9m
@Chris'ssis wo wo wo ! what does that mean ?! :D there is ? :D
 
1:45 PM
@r9m Let me quote Oloa: "Maybe someday we will be able to say something deeper..." :-)
 
r9m
@Chris'ssis hee :) okay !
 
@Chris'ssis You're mean. at least you can give upvotes to those who answer your question :D
 
@Venus lol, I'm not really mean. Some upvotes can be given sometime later. I need to convince myself that there is no closed form ... :-)
 
@Chris'ssis Oloa's answer seems correct to me
 
1:50 PM
@Venus Yeah, it's correct for a certain level of knowledge.
 
@Chris'ssis I give him upvote
I dont get, why this question get high votes
5
Q: How to Evaluate $\int\frac1{x \ln x+ 7 \ln x} \,\mathrm dx$

AfrojackI have tried many methods but do not know how to integrate this: $$ \int \frac{1}{x\ln x + 7\ln x} dx $$ with respect to x.

I doubt it can be done without using machines
 

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