Show fnfinite products are equal

• Nov 17th 2010, 03:35 AM
TheEmptySet
Show fnfinite products are equal
I have been asked to show that for $|z|<1$ that

$\displaystyle \left[ \prod_{j=0}^{\infty}(1-z^{2j+1})\right]^{-1}=\prod_{j=0}^{\infty}(1+z^{j+1})$

I know since $\sum_{j=0}^{\infty}|z^{j+1}|$ and $\sum_{j=0}^{\infty}|z^{2j+1}|$
Converge uniformly on compact subsets of the unit disk that both of the infinite products converge to a holomorphic function of the open disk.

So here is my idea let

$f= \prod_{j=0}^{\infty}(1-z^{2j+1})$ and

$g=\prod_{j=0}^{\infty}(1+z^{j+1})$

I would be sufficient to show that $fg=1$

My idea was to take the deriative and show that it is equal to $0$

let $h=fg \implies \ln(h)=\ln(fg)$

This gives

$\displaystyle \frac{h'(z)}{h(z)}=\frac{d}{dz} \left( \sum_{j=0}^{\infty}\ln(1+z^{j+1})+\ln(1-z^{2j+1})\right)=\sum_{j=0}^{\infty}\frac{(j+1)z^j }{1+z^{j+1}}-\frac{(2j+1)z^{2j}}{1-z^{2j+1}}$

I cannot sum this series or show that it is equal to 0. Also if you see a different approach please let me know. (Clapping)

Thanks
• Nov 17th 2010, 12:06 PM
Opalg
Quote:

Originally Posted by TheEmptySet
I have been asked to show that for $|z|<1$ that

$\displaystyle \left[ \prod_{j=0}^{\infty}(1-z^{2j+1})\right]^{-1}=\prod_{j=0}^{\infty}(1+z^{j+1})$

I know since $\sum_{j=0}^{\infty}|z^{j+1}|$ and $\sum_{j=0}^{\infty}|z^{2j+1}|$
Converge uniformly on compact subsets of the unit disk that both of the infinite products converge to a holomorphic function of the open disk.

So here is my idea let

$f= \prod_{j=0}^{\infty}(1-z^{2j+1})$ and

$g=\prod_{j=0}^{\infty}(1+z^{j+1})$

It would be sufficient to show that $fg=1$.

With that notation $f(z)= \prod_{j=0}^{\infty}(1-z^{2j+1})$ and $g(z)=\prod_{j=0}^{\infty}(1+z^{j+1})$, both infinite products converge uniformly to analytic (therefore continuous) functions on the open unit disk. Also, any rearrangement of the product converges to the same limit. It follows that

\begin{aligned}f(z)g(z) &= \prod_{j=0}^{\infty}(1-z^{2j+1})\prod_{j=0}^{\infty}(1+z^{2j+1})\prod_{j= 0}^{\infty}(1+z^{2j+2}) \\ &= \prod_{j=0}^{\infty}(1-z^{4j+2})\prod_{j=0}^{\infty}(1+z^{2j+2}) = f(z^2)g(z^2).\end{aligned}

By induction, $f(z)g(z) = f(z^{2^n})g(z^{2^n})$ for all n. Then by continuity, as $n\to\infty$ we get $f(z)g(z) = f(0)g(0) = 1$.