@DanielFischer There is a theorem that a powerseries $\sum a_kx^k$ is a rational function iff the $a_k$ satisfy a linear relation $a_{n+k}=b_{k-1}a_{n+k-1}+\cdots b_0 a_n$
@Behaviour I guess some would do it for the need of power over the others, for some kind of fame, but my needs couldn't ever be such needs. If I go out now and ask someone to help me with something, after the job that someone will expect some payment from me. It's interesting how people behave in general.
@PedroTamaroff That's meta voting, it reflects the opinions of meta population. Election results are determined by the main site population, most of which is barely aware of meta or of any issues on the site at all.
Asking questions and answering questions is enough to pay back to the community in my opinion. Well, it's fair to help the others since you are also helped more or less. Yeah, there is a lot of stuff one can learn from. If there is room to improve that amount stuff, it's OK.
You two will end up marrying. I am back to my Algebra
Hrm, I am trying to show $D_{2n}$ is isomorphic to $D_n \times Z_2$. So, they are both normal subgroups of $D_{2n}$, and I know their orders fit - so according to something I proved before, all that's left to do is to show their intersection includes only $1$, or that $D_{2n}=D_nZ_2$. But I don't think either is true.
Ugh... I'm stuck on the integral that Venus posted earlier... $$\int_0^{\infty}\frac{x}{\sqrt{e^x-1}}\mathbb{d}x =4\int_0^{\frac{\pi}{2}}\log(\sec\theta)\mathbb{d}\theta$$
@teadawg1337 there is a nice expansion of $\log(\cos(\theta))$ that might give you that integral: $$\int_0^{\pi/2}\log(\sec(\theta))\,\mathrm{d}\theta=\frac\pi2\log(2)$$ as you already know :-)
This is more common in other fields, where the communicating author is the one to which communication about the article should go; often this is an older faculty member who's not moving any time soon (so that the contact information would stay accurate). This is somewhat of a holdover from when it was harder to find someone's contact information than it is now.
Ugh... This question can be easily explained using polylogarithms, specifically $\displaystyle \operatorname{Li}_{-1}(z)$, but I feel that this is beyond the poster's expertise...
@MathyPerson Hint Suppose you have a number $n>6$, and take a composition $k_1+\cdots+k_j=n$ with largest part at most $6$ (this is what you're counting!). Then $k_j=k$ is a number in $1,\ldots,6$, so you obtain a composition of $n-k$ with largest part at most $6$. You need to show this establishes a bijection when $k$ ranges through $k=1,\ldots,6$.
Alternatively, consider the series $1+\alpha+\alpha^2+\cdots$ with $\alpha=x+x^2+\cdots+x^6$.
The coefficient of $x^n$ is the number of ways we can write $x$ as a sum of $1,2,3,4,5,6$. Why?
This means the polynomial is $1-x-x^2-x^3-x^4-x^5-x^6$.
Looks nasty, but try to understand the big picture, don't lose sleep on the nifty details. They are important, of course, but everything in its own time.
ok, but you said it would diverge if one part of that equation converged and another part diverged. Why would the entire equation diverge if there is a part that converges?
^^^^ People, Look at this user's attitude :o I am soooo >.< right now
@robjohn @Integrator ^^^^^ shouldn't someone teach him ???
@robjohn , I was just trying to help him to get an answer because he was getting downvotes :o and I was very polite :o I don't understand this people -.-
@Integrator but he's been selfish :o I'm not saying moderators should do something...just someone needs to teach him about this site, he has so many misconceptions about how the site works. And I tried to explain to him , he doesn't understand me, so someone needs to explain him.
@Integrator it's no longer about someone else editing it for him , it's that he can do it but he doesn't, and talks about how it's simple to him so he doesn't care about others...
Hello @robjohn !! Do you mean that we should use the following theorem: "If $A$ is measurable then $\forall \epsilon>0 \exists F \subset A$ closed with $m(A \setminus F)<\epsilon$." ??
@user2179021 Well, with publications there is no concern about having a math background, but for getting a uni job one needs to have some math background I think. :-)
@user2179021 I already have some results that look like those by Ramanujan, seriously speaking (I don't care too much what the others think of them, I consider them totally amazing)
@user2179021 Sincere people really appreciate my work, but that doesn't mean that I only want to be appreciated, since the criticism is a heathy part of anyone's evolution. Actually, the positive criticism is mandatory! The one that allows you to become amazing on your way!
