Heine's transformation

Remarks

Proof idea: write $\frac{(b; q)_n}{(c; q)_n} = \frac{(b; q)_\infty}{(c; q)_\infty} \cdot \frac{(c q^n; q)_\infty}{(b q^n; q)_\infty}$. Expand $\frac{(c q^n; q)_\infty}{(b q^n; q)_\infty}$ by the q-binomial theorem as a series in $b q^n$, exchange the two sums, and sum over $n$ with the $q$-binomial theorem again. Example: for $c = b$ the left side is ${}_1\phi_0(a; -; q, z) = \frac{(az; q)_\infty}{(z; q)_\infty}$, and on the right ${}_2\phi_1(1, z; az; q, b) = 1$ since $(1; q)_n = 0$ for $n \ge 1$; both sides agree. Iterating the transformation gives two more forms, and the three are often called Heine's first, second and third transformations. The third form is ${}_2\phi_1(a, b; c; q, z) = \frac{(abz/c; q)_\infty}{(z; q)_\infty}\, {}_2\phi_1(c/a, c/b; c; q, abz/c)$, valid when both series converge.

Going deeper

For a student

It is the first and most used transformation of the ${}_2\phi_1$ series, and its third form is the $q$-analogue of Euler's transformation of the hypergeometric series ${}_2F_1$. Setting $z = c/(ab)$ and summing the right side with the $q$-binomial theorem gives the $q$-Gauss sum ${}_2\phi_1(a, b; c; q, c/(ab)) = \frac{(c/a, c/b; q)_\infty}{(c, c/(ab); q)_\infty}$ for $\abs{c/(ab)} < 1$. Its terminating case leads to the q-Vandermonde identity.