How can you prove that

$\displaystyle \prod_{n=1}^{\infty}(1-x^{n})^{\mu(n)/n}=e^{-x}$

for $\displaystyle |x|<1$.

Here, $\displaystyle \mu(n)$ is the Mobius function.

Thanks in advance.

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- Jul 18th 2011, 12:08 PMCairoHelp with proof.
How can you prove that

$\displaystyle \prod_{n=1}^{\infty}(1-x^{n})^{\mu(n)/n}=e^{-x}$

for $\displaystyle |x|<1$.

Here, $\displaystyle \mu(n)$ is the Mobius function.

Thanks in advance. - Jul 18th 2011, 07:33 PMDrexel28Re: Help with proof.
What have you tried? Begin by taking the logarithm of both sides to get that, if $\displaystyle F(x)$ is our function, $\displaystyle \displaystyle \log(F(x))=\sum_{n=1}^{\infty}\frac{\mu(n)}{n}\log \left(1-x^n\right)=-\sum_{n=1}^{\infty}\sum_{m=1}^{\infty}\frac{\mu(n) }{n}\frac{x^{mn}}{m}$. But show that the coefficient of $\displaystyle x^n$ in this is $\displaystyle \displaystyle \frac{-1}{n}\sum_{d\mid n}\mu(d)$ which I'm sure you know is $\displaystyle -1$ if $\displaystyle n=1$ and $\displaystyle 0$ otherwise. Thus, $\displaystyle \log(F(x))=-x$.

- Jul 18th 2011, 09:49 PMCairoRe: Help with proof.
Thanks for this, Drexel.

I was going down a very different road and trying to solve a first order differential equation by considering generating functions.

Your approach looks much more easier to prove.

(Rofl)