I am implementing a floating point function to compute the q-Pochhammer Euler function (1, 2) $$\phi(q)=(q;q)_{\infty}=\prod_{k=1}^{\infty}(1-q^k)$$ for real $-1 \le q \le 1$. The basic algorithm uses the Euler identity (Pentagonal number theorem) $$\phi(q)=\sum_{n=-\infty}^{\infty}(-1)^n q^{(3n^2-n)/2}$$
For $|q| \approx 1$ the summation suffers from rounding errors and cancellation (which cannot be avoided because $\phi \rightarrow 0$.)
For $q \rightarrow 1$ there is the asymptotic expression (2, formula 8)
$$\phi(q)=\sqrt{\frac{2\pi}{t}} \exp\left( -\frac{\pi^2}{6t} + \frac{t}{24}\right)$$ with $t=-\ln q.$
This expression is remarkably efficient with a relative error $\le 10^{-18}$ for $q>0.5$ (I guess it comes from the Jacobi Theta function representation of $\phi$ in [2], formulas 6,7.)
Question: Is there a similar asymptotic expression for $\phi(q)$ for $q\rightarrow -1?$
Unfortunately I could not find such a result, and Maple or Wolfram Alpha refuse to help me.
Update: With the result of @Professor Vector I get (using algebraic manipulations) the asymptotic expression
$$\phi(-q)=\frac{\phi(q^2)^3}{\phi(q)\phi(q^4)} \sim \sqrt{\frac{\pi}{t}} \exp\left( -\frac{\pi^2}{24t} + \frac{t}{24}\right)$$
From the product, you can easily derive $$\phi(-q)=\frac{\phi(q^2)^3}{\phi(q)\,\phi(q^4)},$$ due to the elementary identity $(1+x)(1-x)=1-x^2$, so you can use the asymptotics for $q\rightarrow1$ to get those for $q\rightarrow-1$.
EDIT: No problem, I'll add some details. It's clear that with $q\rightarrow -q$, only the terms with odd powers of $q$ will change. Let's define $$\phi_{odd}(q)=\prod^\infty_{k=1}(1-q^{2k-1}).$$ Then, it's obvious that $$\phi_{odd}(q)=\phi(q)/\phi(q^2),$$ and from the identity I mentioned, $$\phi_{odd}(q)\,\phi_{odd}(-q)=\phi_{odd}(q^2).$$ Thus, we have $$\phi(-q)=\phi(q^2)\,\phi_{odd}(-q)=\phi(q^2)\,\phi_{odd}(q^2)/\phi_{odd}(q),$$ and substituting the above expression for $\phi_{odd}$ will give the result.
Don't you worry, I found that a bit tricky myself, and since I couldn't find a reference, either, I did make some numerical checks myself, just to be on the safe side. :)