Let's suppose we have a polytope P with $dim(P)=d$ and the h vector $ h(P,x)=\sum\limits_{i = 0}^{n} h_ix^{d-i}$ i have to prove that if $h_{k}=h_{d-k}$(simplicial polytope) then $xh(P,x)=h(P,1/x)$
but i think it's wrong for example if $dim(P)=2$ then we have $h(P,x)=h_0x^2+h_1x+h_2$ and because $(h_k=h_{d-k})$ we have $h(P,x)=h_0x^2+h_1x+h_0$ thus $xh(P,x)=h_0x^3+h_1x^2+h_0x$ and ofcourse is not equal to $h(P,1/x)=h_0x^{-2}+h_1x^{-1}+h_0$ So the reasonable is to prove that $h(P,x)=x^{d}h(P,1/x)$ What are your opinion about that have you ever seen $xh(P,x)=h(P,1/x)$ ? Sorry for my english
You are correct, the $h$-vector for a simple polytope $P$ of $dim(P)=d$ satisfies $h(P,x)=x^dh(P,1/x)$. In particular, the $h$-vector is complementarily symmetric, with $h_i = h_{d-i}$.