Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

If $f$ is analytic in a nbd $\Delta_{\delta}$ of $0$ and $f(z)=-f(-z)\forall z\in\Delta_{\delta} $ Then there exist an analytic function $g\in \Delta_{\delta}$ such that $f(z)=zg(z^2)\forall z\in \Delta_{\delta}$

share|cite|improve this question
up vote 1 down vote accepted

Expand $f(x)$ at its taylor series you get $\sum_{i=0}^{\infty} a_i x^i=-\sum_{i=0}^{\infty} a_i (-x)^i=\sum_{i=0}^{\infty} a_i (-1)^{i+1}x^i \Rightarrow \sum_{i=0}^{\infty} a_{2i} x^{2i}=0$ the last equation from analytic continuity and the fact the $k(x)=0$ is analytics means that $a_{2i}=0$ for $i\in \mathbb{N}$ hence $f(x)=\sum_{i=0}^{\infty} a_{2i+1} x^{2i+1}= x\sum_{i=0}^{\infty} a_{2i+1} x^{2i}=xg(x^2)$ and $g(x)$ is analytics because it is a taylor series

share|cite|improve this answer
what is $k(x)$? – Un Chien Andalou Jun 2 '13 at 13:46
@TaxiDriver $k(x)= \sum _{i=0}^{\infty}a_{2i}x^{2i}$, we have that $k(x)=0$ and since the zero function is $0= \sum _{i=0}^{\infty}b_i x^i,\mathrm{\,with\,} b_i=0$ from analytic continuation the coefficients of $k(x)$ equal to zero's function which are zero. – clark Jun 2 '13 at 16:29

Hint: Consider the expansion series of $f$ at zero.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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