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Suppose that power series function $a_0 + a_1x + a_2x^2 + a_3x^3 + ...$ is constantly zero on a bounded non-empty open interval $I$, which may or may not contain $0$.

Prove that $a_j = 0$ for every $j$, so that the power series is constantly zero on its whole domain (which is of course centered at $0$).

Here is my thought: let us call power series function given in the problem $f$. Let $I = (a, b)$ (where $a < b$, of course), and let $c= (a+b)/2=$ the center of $I$. I don't know how to go from here. Any help?

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  • $\begingroup$ I've noticed that you have asked 6 questions in two days. I wanted to make sure that you are aware of the quotas 50 questions/30 days and 6 questions/24 hours, so that you can plan posting your questions accordingly. (If you try to post more questions, StackExchange software will not allow you to do so.) For more details see meta. $\endgroup$ – Martin Sleziak Mar 30 '16 at 16:41
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Complex analytic solution:

$f$ is a power series, so it is analytic. Since $f$ is $0$ in a uncountable set, it agrees with $0$ everywhere in its domain.

Real analytic solution:

There exists a unique set of coefficients $(b_j)_{j=0}^\infty$ such that $$ f(x) = \sum_{j=0}^\infty b_j (x - c)^j = \sum_{j=0}^\infty \frac{f^{(j)}(c)}{j!} (x - c)^j $$ Since the derivatives of $f$ at $c$ are all $0$, all $b_j = 0$. $f$ is a power series, so $(a_j)_{j=0}^\infty$ is a unique set of coefficients so that $$ \sum_{j=0}^\infty a_j x^j = \sum_{j=0}^\infty b_j (x-c)^j $$ and it turns out that $a_j = 0$.

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  • $\begingroup$ You have to justify why you can use the identity principle, however. $\endgroup$ – Pedro Tamaroff Mar 30 '16 at 0:43
  • $\begingroup$ @PedroTamaroff Thanks. It is fixed. $\endgroup$ – Henricus V. Mar 30 '16 at 0:44
  • $\begingroup$ I havent done any complex analysis. Can you explain the first step a little more? $\endgroup$ – user326676 Mar 30 '16 at 0:45
  • $\begingroup$ @THT16224 The first way uses the identity theorem. It is justified since every power series is analytic (a.k.a. holomorphic) $\endgroup$ – Henricus V. Mar 30 '16 at 0:46
  • $\begingroup$ I meant the first step in the real analytic way $\endgroup$ – user326676 Mar 30 '16 at 0:52

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