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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 $f(z)=zg(z^2)\forall z\in \Delta_{\delta}$

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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

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what is $k(x)$? –  Une Femme Douce Jun 2 '13 at 13:46
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@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.

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