After multiplying out and simplifying polynomial $ (x - 1)(x^2 - 1)(x^3 - 1)\cdots(x^{2007} - 1),$ getting rid of all terms whose powers are greater than $ 2007,$ we acquire a new polynomial $ f(x).$ Find its degree and the coefficient of the term having the highest power. Find the degree of $ f(x) = (1 - x)(1 - x^{2})...(1 - x^{2007})$ $ (mod$ $ x^{2008}).$
Problem
Source: Chinese TST 2007 4th quiz P2
Tags: algebra, polynomial, Euler, number theory unsolved, number theory
me@home
27.03.2007 12:34
For instance $(1-x)(1-x^{2})=1-x-x^{2}+x^{3}\equiv 1-x+x^{2}\mod{x^{3}}$ has degree $2$...
maxal
27.03.2007 16:49
The question is equivalent to finding such maximum $d<2008$ that A010815(d) is non-zero (follow the link for further references and formulas).
Rust
27.03.2007 19:21
Result is \[f(x)=1+\sum_{k=1}^{36}(-1)^{k}(x^{k(3k-1)/2}+x^{k(3k+1)/2}). \] Therefore $deg(f(x))=1962$.
shyong
28.03.2007 09:10
Rust wrote: Result is \[f(x)=1+\sum_{k=1}^{36}(-1)^{k}(x^{k(3k-1)/2}+x^{k(3k+1)/2}). \] Therefore $deg(f(x))=1962$. how did you get that expansion ? Thanks
Rust
28.03.2007 09:36
It is known, that \[\prod_{k=1}^{\infty }(1-x^{k})=1+\sum_{k=1}^{\infty}(-1)^{k}(x^{k(3k-1)/2}+x^{k(3k+1)/2}).\]
shencaili
28.03.2007 11:06
In fact,it's also bacause the professor was born in 1962.
ZetaX
29.03.2007 10:52
The known fact is called Eulers Pentagonal number theorem.
malinger
01.04.2007 19:38
I don't understand, why this theorem has to state the sum to infinity, not finite.
N.T.TUAN
01.04.2007 20:21
malinger wrote: I don't understand, why this theorem has to state the sum to infinity, not finite. I searched and I have http://en.wikipedia.org/wiki/Pentagonal_number_theorem That is all.
malinger
02.04.2007 14:03
N.T.TUAN wrote: I searched and I have http://en.wikipedia.org/wiki/Pentagonal_number_theorem That is all. That's all right, but just wondering, does anyone know the reason why to infinity?
ZetaX
02.04.2007 14:06
I don't see the sense of your question... Proofing it gives us this, so what's to discuss¿
Albanian Eagle
08.05.2007 21:57
One could form the problem with $f(x)=\prod (1-x^{i})^{3}$ instead. This time the solution needs another nice identity.