It may seem trivial, but I just cannot find a proper $b_n \space (n \in N)$ sequence as a function of $a_n \space (n \in N)$ that satisfies the following equation:
$ \left(\displaystyle\sum_{n=1}^{\infty}a_n\sin(n \pi x)\right)^3 = \displaystyle\sum_{n=1}^{\infty}b_n\sin(n \pi x) $
I know I need to expand the cube and use trigonometric identities to get the following sum:
$ \left(\displaystyle\sum_{n=1}^{\infty}a_n\sin(n \pi x)\right)^3 = \displaystyle\sum_{i=1}^{\infty} \sum_{j=1}^{\infty} \sum_{k=1}^{\infty} a_i a_j a_k \sin(i \pi x)\sin(j \pi x)\sin(k \pi x) = \\ =\displaystyle\sum_{i=1}^{\infty} \sum_{j=1}^{\infty} \sum_{k=1}^{\infty} \frac{1}{4} a_i a_j a_k \Big(\sin\big((i+j-k)\pi x\big) + \sin\big((i-j+k)\pi x\big) - \\ - \sin\big((i-j-k)\pi x\big) - \sin\big((i+j+k)\pi x\big)\Big) $
But after that I always get lost in reindexing this expression (especially in obtaining the summation boundaries and multinomial coefficients). Is there an easy process for solving this problem? If so, can it be generalized for other multiple composite sums?