# Show fnfinite products are equal

• Nov 17th 2010, 02:35 AM
TheEmptySet
Show fnfinite products are equal
I have been asked to show that for $\displaystyle |z|<1$ that

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

I know since $\displaystyle \sum_{j=0}^{\infty}|z^{j+1}|$ and $\displaystyle \sum_{j=0}^{\infty}|z^{2j+1}|$
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

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

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

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

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

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

This gives

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

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, 11:06 AM
Opalg
Quote:

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

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

I know since $\displaystyle \sum_{j=0}^{\infty}|z^{j+1}|$ and $\displaystyle \sum_{j=0}^{\infty}|z^{2j+1}|$
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

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

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

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

With that notation $\displaystyle f(z)= \prod_{j=0}^{\infty}(1-z^{2j+1})$ and $\displaystyle g(z)=\prod_{j=0}^{\infty}(1+z^{j+1})$, 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

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