Define the coefficient $B = (b_0, \cdots b_n)$ of elementarty polynomial product of $A=(a_0 \cdots, a_n)$. $B = \sum_{i=0}^N b_ix^i = \prod_{i=0}^N(1+a_ix) $. Use $f(A) = B$ to get it's coefficient.
Given a length $n$ sequence $A$, i need to compute $g(A) := \frac{1}{n+1}\langle f(A), 1/f(1_n)\rangle$. Here $1_n$ denote length $n$, 1 sequence. $1/f(1_n)$ is the element-wise reciprocal. and $\langle, \rangle$ is the inner product.
My question is: is it possible to factor $g(A)$? E.g. compute some quantity using the first half of $A$ and the second half of $A$ independently and then join them together.
The reason why I want to do this is when $n$ is big, I can easily run into a precision problem. My code would overflow :(.
(Too long for a comment.)
The coefficients $b_i$ are the elementary symmetric polynomials $\,b_i=e_i\left(a_1, a_2, \dots,a_n\right)\,$, which can be calculated more efficiently than direct brute force, see for example the answers and links under Algorithm(s) for computing an elementary symmetric polynomial.
Let $\displaystyle\,f(1_n) = \sum_{i=0}^n c_ix^i = (1+x)^n\,$ then $\displaystyle\,c_i = \binom{n}{i} = \frac{(n)_i}{i!}\,$ where $\,(n)_i\,$ is the falling factorial, which can be calculated iteratively as $\displaystyle\,c_{i+1}=\frac{n-i}{i+1}\,c_i\,$.
Then $\displaystyle\,(n+1)\,g(A) = \langle f(A), 1/f(1_n)\rangle = \sum_{i=0}^n \frac{b_i}{c_i}\,$ where $\,b_i, \,c_i\,$ can be calculated as above.