Find $h$ such that $h(x^{-1} - x ) = f(x)$

Question

I am given that $f(-x) = f(x^{-1})$ for some $f \in \mathbb{Z}[x, x^{-1}]$.

I am told it is possible to find $h \in \mathbb{Z}[z]$ such that $h(x^{-1} - x) = f(x)$.

Any pointers would be much appreciated, thanks

Answer 1

$$f(x)=f(-x^{-1})$$ is easier to work with. This means $$f(x)=a_0+\sum_{k=1}^\infty{a_kx^k}+\sum_{k=1}^\infty{a_k(-x)^{-k}}=a_0+\sum_{k=1}^\infty{a_k(x^k+(-1)^kx^{-k})}$$ Let $n$ be the maximum positive exponent in this sum. Clearly $f(x)-a_n(x^n+(-1)^nx^{-n})$ satisfies the same identity, so it is sufficient by induction to prove that $$x^n+(-1)^nx^{-n}=h(x^{-1}-x)$$ for some $h$. This is clear if $0\le n\le 1$. Otherwise note that $$(x^{n-1}+(-1)^{n-1}x^{-n+1})(x^{-1}-x)=(x^{n-2}+(-1)^{n-2}x^{-n+2})-(x^n+(-1)^nx^{-n})$$ Thus $$x^n+(-1)^nx^{-n}=(x^{n-2}+(-1)^{n-2}x^{-n+2})-(x^{n-1}+(-1)^{n-1}x^{-n+1})(x^{-1}-x)$$ By the induction hypothesis the right hand side is a polynomial in $x^{-1}-x$, hence the result follows by induction.