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I am implementing a floating 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

$$\phi(-q)=\frac{\phi(q^2)^3}{\phi(q)\phi(q^4)} \sim \sqrt{\frac{\pi}{t}} \exp\left( -\frac{\pi^2}{24t} + \frac{t}{24}\right)$$

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

$$\phi(-q)=\frac{\phi(q^2)^3}{\phi(q)\phi(q^4)} \sim \sqrt{\frac{\pi}{t}} \exp\left( -\frac{\pi^2}{24t} + \frac{t}{24}\right)$$

I am implementing a floating 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.

I am implementing a floating 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

$$\phi(-q)=\frac{\phi(q^2)^3}{\phi(q)\phi(q^4)} \sim \sqrt{\frac{\pi}{t}} \exp\left( -\frac{\pi^2}{24t} + \frac{t}{24}\right)$$

I am implementing a floating 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 is 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.

I am implementing a floating 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.