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Let $f$ be an entire function without zeroes. I have to show that there exists an entire function $g$ such that $f = g^n$ for some $n \in \mathbb{N}$. I tried proving this by constructing explicitly such a $g$:

If $f = g^n$, we have $f g^{-n} \equiv 1 =: h(z)$. For this to hold, we must have $h'(z) \equiv 0$. We compute $$h'(z) = f' g^{-n} + f \cdot (-n) g^{-n-1} g' = f' g^{-n} - n f g^{-n-1} g' = 0$$ which is equivalent to $\frac{g'}{g} = \frac{f'}{n f}$. Now I don't really know how to solve this differential equation for $g$, but I think $g(z) := f(z/n)$ should be a solution to it, as $g'(z) = f'(z/n)/n$. But is this really a solution? Because on the left hand side, the argument is $z$ whereas on the right hand side it is $z/n$. This doesn't seem correct to me.

In case my solution is incorrect, how would I actually solve this differential equation? Or is there an even easier approach to prove this claim?

Thanks in advance for any help.

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$n=1$ and $g=f$ is a trivial example. So for the problem to be interesting at all, you'd better require $n>1$. –  Harald Hanche-Olsen Nov 7 '12 at 15:32
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don't you mean "forall $n$ there is an $f$ such that ..." ? because right now I can only think of "well let's pick n=1 and f=g, we're done !" –  mercio Nov 7 '12 at 15:33
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What is really wanted here, is the $\sqrt[n]f$. In other words, $\sqrt[n]f=f^{1/n} = \exp(\displaystyle\frac1n\log f)$, and you need to ensure a regular branch for $\log f$, using that $f$ has no zero.. –  Berci Nov 7 '12 at 15:33
    
Shouldn't the problem be: let $f$ entire without zeros, and $n\geqslant 2$ an integer. Show that we c an find $g$ entire such that $f=g^2$? –  Davide Giraudo Nov 7 '12 at 15:34
    
I am sorry, what I is actually wanted is what Berci wrote. I don't know why I wrote it the other way, I actually understood the problem correctly. –  studeth Nov 8 '12 at 14:00

1 Answer 1

up vote 1 down vote accepted

Hint: Look for a branch of $\log f$, that is a function $h$ so that $\exp(h(z))=f(z)$. Derive a differential equation for $h$.

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