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In this answer, it says that one may take the logarithmic derivative w.r.t. $z$ on both sides of the equation $$(q^4;q^4)_\infty\left\{z(-z^4q^3;q^4)_\infty(-z^{-4}q;q^4)_\infty-z^{-1}(-z^4q;q^4)_\infty(-z^{-4}q^3;q^4)_\infty\right\}\\=(z-z^{-1})(q;q)_\infty(z^2q;q)_\infty(z^{-2}q;q)_\infty,\tag{1}$$ to get $$1+4\sum_{n\ge1}(-1)^n\frac{q^{2n-1}}{1-q^{2n-1}}=\frac{(q;q)_\infty^3}{(q^4;q^4)_\infty(-q^3;q^4)_\infty(-q;q^4)_\infty},\tag{2}$$ after having set $z=1$.

I am having some trouble with doing this.

I decided to simplify things a bit so I came up with $$z^2(-z^4q^3;q^4)_\infty(-z^{-4}q;q^4)_\infty-(-z^4q;q^4)_\infty(-z^{-4}q^3;q^4)_\infty=\frac{(q;q)_\infty}{(q^4;q^4)_\infty}(z^2-1)(z^2q;q)_\infty(z^{-2}q;q)_\infty.\tag{3}$$ The logarithmic derivative of the R.H.S. of $(3)$ is easy, and it evaluates to $$\frac{\partial}{\partial z}\log\left\{\frac{(q;q)_\infty}{(q^4;q^4)_\infty}(z^2-1)(z^2q;q)_\infty(z^{-2}q;q)_\infty\right\}=\frac{2z}{z^2-1}+\frac{2z}{(z^2q;q)_\infty}A(1,z^2;q)-\frac{2z^{-3}}{(z^{-2}q;q)_\infty}A(1,z^{-2};q),$$ where $$A(a,b;q)=\sum_{n\ge1}\frac{(aq)^n}{1-bq^n}.$$ But, simplifying the logarithm of the left hand side of $(3)$ is proving to be a difficult task. Is there a better way to go about this? Thanks.

clathratus
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1 Answers1

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Let \begin{align*} f(z)&=z\prod_{n = 1}^\infty (1 - q^{4n})(1 + q^{4n - 1} z^4)(1 + q^{4n - 3} z^{-4})\\ g(z)&=z^{-1}\prod_{n = 1}^\infty (1 - q^{4n})(1 + q^{4n - 3} z^4)(1 + q^{4n - 1} z^{-4})\\ h(z)&= (z - z^{-1})\prod_{n = 1}^\infty (1 - q^n)(1 - q^n z^2)(1 - q^n z^{-2}). \end{align*} Then \begin{align*} \log f(z)&=\log z+\sum_{n = 1}^\infty\left[\log(1 - q^{4n})+\log(1 + q^{4n - 1} z^4)+\log(1 + q^{4n - 3} z^{-4})\right]\\ \log g(z)&=-\log z+\sum_{n = 1}^\infty\left[\log(1 - q^{4n})+\log(1 + q^{4n - 3} z^4)+\log(1 + q^{4n - 1} z^{-4})\right]\\ \log h(z)&=\log(z - z^{-1})+\sum_{n = 1}^\infty\left[\log(1 - q^n)+\log(1 - q^n z^2)+\log(1 - q^n z^{-2})\right] \end{align*} so \begin{align*} \frac{f'(z)}{f(z)}&=\frac1z+\sum_{n = 1}^\infty\left[\frac{4q^{n-1}z^3}{1 + q^{4n - 1} z^4}-\frac{4q^{4n-3}z^{-5}}{1 + q^{4n - 3} z^{-4}}\right]\\ \frac{g'(z)}{g(z)}&=-\frac1z+\sum_{n = 1}^\infty\left[\frac{4q^{4n-3}z^3}{1 + q^{4n - 3} z^4}-\frac{4q^{4n-1}z^{-5}}{1 + q^{4n - 1} z^{-4}}\right]\\ \frac{h'(z)}{h(z)}&=\frac{1+z^{-2}}{z - z^{-1}}+\sum_{n = 1}^\infty\left[\frac{-2q^nz}{1 - q^n z^2}+\frac{2q^nz^{-3}}{1 - q^n z^{-2}}\right]. \end{align*} This gives \begin{align*} f'(1)&=f(1)\left(1+\sum_{n = 1}^\infty\left[\frac{4q^{n-1}}{1 + q^{4n - 1}}-\frac{4q^{4n-3}}{1 + q^{4n - 3}}\right]\right)\\&=\left(\prod_{n = 1}^\infty (1 - q^{4n})(1 + q^{4n - 1})(1 + q^{4n - 3})\right)\left(1 + 4\sum_{n = 1}^\infty (-1)^n \frac{q^{2n - 1}}{1 + q^{2n - 1}}\right)\\ g'(1)&=g(1)\left(-1+\sum_{n = 1}^\infty\left[\frac{4q^{4n-3}}{1 + q^{4n - 3}}-\frac{4q^{4n-1}}{1 + q^{4n - 1}}\right]\right)\\&=\left(\prod_{n = 1}^\infty (1 - q^{4n})(1 + q^{4n - 1})(1 + q^{4n - 3})\right)\left(-1-4\sum_{n = 1}^\infty (-1)^n \frac{q^{2n - 1}}{1 + q^{2n - 1}}\right)\\ h'(1)&=\left(\lim_{z\to1}\frac{h(z)}{z-z^{-1}}\right)\times2=2\prod_{n = 1}^\infty (1 - q^n)^3 \end{align*} so $f'(1)-g'(1)=h'(1)$ is equivalent to $$1 + 4\sum_{n = 1}^\infty (-1)^n \frac{q^{2n - 1}}{1 + q^{2n - 1}}=\frac{\prod\limits_{n = 1}^\infty (1 - q^n)^3}{\prod\limits_{n = 1}^\infty (1 - q^{4n})(1 + q^{4n - 1})(1 + q^{4n - 3})}.$$

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