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

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Comments

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

Arsenii Zharkov - 10 months, 3 weeks 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.

Aruna Yumlembam - 10 months, 3 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.

Aruna Yumlembam - 10 months, 3 weeks ago

@Aruna Yumlembam – Plus the basic idea is using the Taylor series expantion,to denote the coefficient.

Aruna Yumlembam - 10 months, 3 weeks ago

@Aruna Yumlembam – 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 $a_n=0$

Arsenii Zharkov - 10 months, 3 weeks ago

@Zakir Husain