If one were going to withdraw, it would be better to do so in the middle rather than after the fact, so they can go down to the suitable next candidate.
@TedShifrin It's not that. If I truly did want to not be a mod, I wouldn't have nominated. It's a thing I'm willing to do. I'm just not hot on it. But somebody has to do it.
@DanielFischer If we consider that the elements have to be different, what would we have to check? I haven't understood why this for loop doesn't work for this case...
@DanielFischer Now, when you say you are not hot on it, it sounds like you have a hidden problem somewhere. So I think it's best that you calm down and think it over again.
@Alizter Now we have no idea what you are talking about, if you are referring to a day old problem.
BTW, @Mike, in case I haven't made it abundantly clear: You should read anything appropriate that you find Blaine Lawson has written. He is a superb expositor and mathematician. He was almost my Ph.D. adviser. So both his CBMS notes and the Spin Geometry book should be on your list.
@evinda Consider A = [1,2]. A[1] > 1*A[0] leads to incrementing i, so next you check A[1] == 1*A[1], which would tell you "yes", but we want different elements.
@TedShifrin This is what I am trying to solve using that generating function: Let $P(x)$ be the partition generating function, and let $R(x)=\sum_{n=0}^{\infty} r_nx^n$, where $r_n$ is the number of partitions of n containing no $1s$ or $2s$.
Then $\frac{R(x)}{P(x)}$ is a polynomial. What polynomial is it? (Enter your answer in expanded form.)
@Alizter The weird thing for me is that when I was studying for my A level exams, I just read through all the lecture notes. I did not do a single problem for math, further math, physics, or chem, lol. But I guess not everyone is like that.
Interesting, @Mathy, but you're never going to get it by trying to figure out numerator and denominator in closed form. My hint would be: Try to figure out for what $Q(x)$ you have $P(x)Q(x)=R(x)$.
@pedro The problem is this: "Let $P(x)$ be the partition generating function, and let $R(x)=\sum_{n=0}^{\infty} r_nx^n$, where $r_n$ is the number of partitions of n containing no $1s$ or $2s$.
Then $\frac{R(x)}{P(x)}$ is a polynomial. What polynomial is it? (Enter your answer in expanded form.)". I simplified it down to $\frac{1+x^3+x^4+x^5+2x^6+2x^7+3x^8+....}{1+x+2x^2+3x^3+5x^4+7x^5+11x^6+...}$, but I can't seem to get a polynomial out of it. I am trying to turn the infinite power series into power series representations (like $\frac{1}{1-x}$), and then simplify from there..but now I am…
@DanielFischer The loop is exited when j=m or i=m or A[j]==y*A[i]. So, if we go out of this while loop we will know if there are two elements at the part with the positive integers such that their quotient is equal to $y$, right?
@PedroTamaroff The idea is this: when you publish things to the journal you give all the details, of course, this might take some space since you wanna explain things in details. However, using the core idea of the proof I can prove it in one line, without explaining the obvious things.
I was going to give you a mumbling structure group answer to your question about why people care about Spin, @Mike, but I'm sure you already know that.
So I think your $w_2$ question is answered very simply from playing around with the appropriate exact sequence for cohomology with values in groups ... But I haven't sat down to think about it.
@BalarkaSen Why is that silly? Just think about it: It can be added to many textbooks and it can be easily understood by many students in high school and uni. It only uses elementary steps, all is clear, concise, fast.
@DanielFischer We have the same case.. We could consider the absolute value of these numbers since both elements will be negative and that what interests us is their quotient. We just have to begin from the end of the first part of the array and continue till we reach the first position.. Or am I wrong?
My favorite book on the analysis of Riemann Surfaces is Forster. There are, of course, all sorts of more algebro-geometric treatments. Springer is a classic, too.
@JasperLoy I have your e-mail, I'll show it to you privately after I receive the some feedback from some professors (and after I publish it in any case - unless someone already found it).
@evinda That's one way to do it. There are other ways of course (which could also have been used for the positive part), but it's best to use the same principle for both parts. Now you just need to be careful that you have your inequalities in the right sense.
@BalarkaSen It might sound disrespectul to some of you if I compared my proof to the ones of Euler, but I wanna say only one thing: it's totally amazing. I'll show it to you at a certain point and let you conclude alone how good it is.
For example, $2$ has $2$ partitions, because we have the $x^2$ in $(1-x)^{-1}$ which is $1+1$, and we have the $x^2$ in $(1-x^2)^{-1}$ which is just $2$.
So to remove $1$ and $2$s -- I was mistaken! -- we just need to remove those two first factors.
More generally, if $A=\{k_1,\ldots,k_j,\ldots\}$ is a subset of $\Bbb N$, partitions containing only elements of $A$ are obtained from $\prod (1-x^{k_j})^{-1}$.
@DanielFischer And do we have to check the part of the array with the negative integers in each case or only if we haven't found two elements from the part with the positive integers of which the quotient is equal to y?
Yeah, I wanna make the statement though, with any risk: my proof is simpler, easier, faster than the ones of Euler (I think). I said it! (closing the eyes here)
@evinda This is like debugging imaginary code. You already have the right ideas where things might go wrong and how to fix them, so I'd say now is the time for you to go ahead and actually program, test and debug the thing :-)
