0
$\begingroup$

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

$\endgroup$
  • 1
    $\begingroup$ How about using geometric series to convert the summation into rational function? The problem will be changed into easier problem for polynomial. $\endgroup$ – Seewoo Lee Jul 1 '16 at 14:18
0
$\begingroup$

I just proved it. Guess it would be better to write my own answer than to delete the post. Any other proof is still welcome.

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.