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.
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
@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
@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.
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.
@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
@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.
@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.
@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.
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. :)
@Chris'ssis About that sum you posted yesterday, I realised afterwards that it can be solved fairly quickly using only the generating function of $\dfrac{H_n}{n^2}$.
Is this the right direction, or should I go back to the start? $$\sum_{n=1}^{\infty}\frac{9n^2+12n+5}{(3n+2)!} = \sum_{n=1}^{\infty} \frac{1}{(3n)!} + \sum_{n=1}^{\infty} \frac{1}{(3n+2)(3n+1)(3n-1)!} + \sum_{n=1}^{\infty} \frac{3}{(3n+2)!}$$
@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.
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 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.
@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
This 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 = ...
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