Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $f:\mathbb C \rightarrow \mathbb C$ be holomorphic and $f(z)=f(-z)$ for all $z\in \mathbb C$. Show that there exists a holomorphic function $g$ such that $g(z^2)=f(z)$.

If I take $g(z):=(f(z)+f(-z))/2$ then I can prove that $g(z)=\sum_0^\infty a_{2n}z^n$. Of course $g$ thus defined is holomorphic. But how does this show that $g(^2)=f(z)$ as well?. I mean the coefficients of the two are different when you compare term by term. Can you guys help?. Or should this $g$ be defined differently? Thanks for your help.

share|improve this question

1 Answer 1

up vote 4 down vote accepted

If $f$ is even, the Maclaurin series of $f$ contains only even powers. (Fill in the details of this!)

Hence, $$ f(z) = a_0 + a_2z^2 + a_4z^4 + a_6z^6 +\cdots = a_0 + a_2(z^2) + a_4(z^2)^2 + a_6(z^2)^3 + \cdots$$

Set $$g(z) = a_0 + a_2z + a_4z^2 + a_6z^3 + \cdots$$

share|improve this answer
    
Thanks I got it –  Jack Dawkins Jan 5 '13 at 9:03

Your Answer

 
discard

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.