# How to prove that $h''(x)$ has at most one zero on $(0,1)$.

$h(x)=1-\sum_{i=1}^{k-1}x^i+a_kx^k+\sum_{i=k+1}^\infty x^i$, where $|a_k|\le1$, is the power series of an analytic function. Prove that $h''(x)$ has at most one zero on $(0,1)$.

• How about using geometric series to convert the summation into rational function? The problem will be changed into easier problem for polynomial. – Seewoo Lee Jul 1 '16 at 14:18

So $h''(x)$ is of the form $-\sum_{i=1}^l\alpha_ix^i+\sum_{i=l+1}^\infty\beta_i x^i$, where $\alpha_i,\beta_i>0$, denote it by $H(x)$. All the discussion below is on $(0,1)$.
Notice whenever $H(x)\ge0$, we have $H'(x)\ge0$. In fact, if $\sum_{i=l+1}^\infty\beta_ix^i\ge\sum_{i=1}^l\alpha_ix^i$, $$\sum_{i=l+1}^\infty i\beta_ix^i\ge(l+1)\sum_{i=l+1}^\infty\beta_ix^i\ge(l+1)\sum_{i=1}^l\alpha_ix^i\ge\sum_{i=1}^l i\alpha_ix^i.$$ Thus, whenever $H(x)$ reaches $0$, it will increase and cannot reach $0$ again.