# $\displaystyle \prod_{n=1}^{\infty} (1-x^n)^{24}$ coefficients

$P(x) = \displaystyle \prod_{n=1}^{\infty} (1-x^n)^{24}$

"Unwrapping" the product, you get a polynomial with coefficients $a_n \in \mathbb{R}$ :

$P(x) = a_0 + a_1 x + a_2 x^2 + ...$

Prove whether or not there exists a coefficient $a_n = 0$ in this polynomial. This is a problem I've been attempting to solve for over a week now but have made no significant progress on after turning the product (without the 24th power) into an infinite series. Could anyone offer some insight, please?

Note by Arsenii Zharkov
6 months ago

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So far I think my most significant progress has been getting it into the form $\displaystyle \prod_{n=1}^{\infty} (1-x^n)^{24} = \sum_{n=-\infty}^{\infty} (-1)^n x^{\frac {3n^2-n}{2} }$

- 6 months ago

If the statement you said that$P(x)=a_0+a_1x+a_2 x^2+a_3 x^3+....$ then , $P(0)=a_0,(\frac{d}{dx}P(0)) =a_1 ,(\frac{d^2}{dx^2}\frac{P(0)}{2!}) =a_2,...$and so on. So if any $a_n$ to be zero the derivative of the product defined at zero must be zero.This is my insight and I hope you will get some idea , personal I feel it won't have a zero coefficient.

- 5 months, 4 weeks ago

Here due to my bad knowledge of latex the derivative of the product has to be calculated first then put x=0.It's a pretty good question has many applications.

- 5 months, 4 weeks ago

Plus the basic idea is using the Taylor series expantion,to denote the coefficient.

- 5 months, 4 weeks ago

Unfortunately I haven't managed to find a good way to take the derivative of the function, but I did figure out the first 5 coefficients using combinatorics. For x, there's 24 "blocks" to choose from, For x^2, either 24 x^2's to choose from or 24 x's to choose 2 of from, etc., but I've been struggling with coming up with any form of general formula that actually confirms the absence of an $a_n=0$

- 5 months, 4 weeks ago

- 6 months ago

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- 5 months, 4 weeks ago