I am always having a hard time when I am dealing with expressions of the form :
$$B(x) = \prod_{i = 1}^N \sum_{k = 1}^{L_i} a_{i, k}(x)$$ where $(L_n)$ is a sequence of naturals numbers such that : $L_i \geq 1$ and $(a_{i,j})$ is a sequence of functions.
Most of the time in order to get a better understanding we need to somehow do some tricks in order to invert $\sum$ and $\prod$ or in order to simply get rid of the $\prod$.
Yet I am always having a hard time getting $B$ into this form. So I am wondering is there a general formula or a strategy to get $B$ into this form ?
For example when dealing with power series we have the Cauchy-product which alows us to put an expression of the form :
$$\prod_{i = 1}^2 \sum_{k = 0}^n a_nx^n $$
into the form :
$$\sum_{k = 0}^{2n} x^k \sum_{i = 0}^k a_{k -i}a_i$$
Thank you !
We have $$\prod_{i = 1}^N \sum_{k = 1}^{L_i} a_{i, k}=\sum_{\ell}\prod_{i=1}^Na_{i,\ell(i)}$$ where the sum on the right runs over the functions $$\ell:\{1,\ldots,N\}\to\Bbb N$$ such that $\ell(i)\leq L_i$ for every $i$.
More generally, we have for sets $I$ and $K_i$ for $i\in I$: $$\prod_{i\in I}\sum_{k\in K_i}a_{i,k}=\sum_\ell\prod_{i\in I}a_{i,\ell(i)}$$ where the sum on the right runs over $$\ell\in\prod_{i\in I}K_i$$ that's $$\ell:I\to\bigcup_{i\in I}K_i$$ such that $\ell(i)\in K_i$ for every $i\in I$.