# q-analogizing Glasser’s hypergeometric solution to the quintic

(assume that you’ve already performed the Tschirnhausen transformation) $F_1(t) = \,_4F_3\left(\frac{-1}{20}, \frac{3}{20}, \frac{7}{20}, \frac{11}{20} ;\frac{1}{4}, \frac{1}{2}, \frac{3}{4}; \frac{3125t^4}{256}\right)$ $F_2(t) = \,_4F_3\left(\frac{1}{5}, \frac{2}{5}, \frac{3}{5}, \frac{4}{5} ;\frac{1}{2}, \frac{3}{4}, \frac{5}{4}; \frac{3125t^4}{256}\right) \\[6pt]$ $F_3(t) = \,_4F_3\left(\frac{9}{20}, \frac{13}{20}, \frac{17}{20}, \frac{21}{20} ;\frac{3}{4}, \frac{5}{4}, \frac{3}{2}; \frac{3125t^4}{256}\right)$ $F_4(t) = \,_4F_3\left(\frac{7}{10}, \frac{9}{10}, \frac{11}{10}, \frac{13}{10} ;\frac{5}{4}, \frac{3}{2}, \frac{7}{4}; \frac{3125t^4}{256}\right)$

(the signs are wrong on the wikipedia page. they have been corrected below) $x_1 = {} tF_2(t)$ $x_2 = {} F_1(t) - \frac{1}{4}tF_2(t) - \frac{5}{32}t^2F_3(t) - \frac{5}{32}t^3F_4(t)$ $x_3 = -F_1(t) - \frac{1}{4}tF_2(t) + \frac{5}{32}t^2F_3(t) - \frac{5}{32}t^3F_4(t)$ $x_4 = {} +{\mathrm{i}}F_1(t) - \frac{1}{4}tF_2(t) + \frac{5}{32}{\mathrm{i}}t^2F_3(t) + \frac{5}{32}t^3F_4(t)$ $x_5 = -{\mathrm{i}}F_1(t) - \frac{1}{4}tF_2(t) + \frac{5}{32}{\mathrm{i}}t^2F_3(t) + \frac{5}{32}t^3F_4(t)$

I was reading Gasper’s Basic Hypergeometric Series, and since I’ve been playing around (amongst other things) with the hypergeometric solution to the quintic, and in the first few pages Gasper mentions ${}_a\phi_{b}$ $q$-hypergeometric functions, the question leapt into my mind: What would happen if we take the $q$-analogue of the Glasser’s method hypergeometric solution, in other words: $\Phi_1(t,q) = \,_4\phi_3\left(q^{\frac{-1}{20}}, q^{\frac{3}{20}}, q^{\frac{7}{20}}, q^{\frac{11}{20}} ;q^{\frac{1}{4}}, q^{\frac{1}{2}}, q^{\frac{3}{4}}; q ;\frac{3125t^4}{256}\right)$ $\Phi_2(t,q) = \,_4\phi_3\left(q^{\frac{1}{5}}, q^{\frac{2}{5}}, q^{\frac{3}{5}}, q^{\frac{4}{5}} ;q^{\frac{1}{2}}, q^{\frac{3}{4}}, q^{\frac{5}{4}}; q ; \frac{3125t^4}{256}\right) \\[6pt]$ $\Phi_3(t,q) = \,_4\phi_3\left(q^{\frac{9}{20}}, q^{\frac{13}{20}}, q^{\frac{17}{20}}, q^{\frac{21}{20}}; q^{\frac{3}{4}}, q^{\frac{5}{4}}, q^{\frac{3}{2}};q; \frac{3125t^4}{256}\right)$ $\Phi_4(t,q) = \,_4\phi_3\left(q^{\frac{7}{10}}, q^{\frac{9}{10}}, q^{\frac{11}{10}}, q^{\frac{13}{10}} ;q^{\frac{5}{4}}, q^{\frac{3}{2}}, q^{\frac{7}{4}};q \frac{3125t^4}{256}\right)$ $x_1(q) = {} t\Phi_2(t,q)$ $x_2(q) = {} \Phi_1(t,q) - \frac{1}{4}t\Phi_2(t,q) - \frac{5}{32}t^2\Phi_3(t,q) - \frac{5}{32}t^3\Phi_4(t,q)$ $x_3(q) = -\Phi_1(t,q) - \frac{1}{4}t\Phi_2(t,q) + \frac{5}{32}t^2\Phi_3(t,q) - \frac{5}{32}t^3\Phi_4(t,q)$ $x_4(q) = {} +{\mathrm{i}}\Phi_1(t,q) - \frac{1}{4}t\Phi_2(t,q) + \frac{5}{32}{\mathrm{i}}t^2\Phi_3(t,q) + \frac{5}{32}t^3\Phi_4(t,q)$ $x_5(q) = -{\mathrm{i}}\Phi_1(t,q) - \frac{1}{4}t\Phi_2(t,q) + \frac{5}{32}{\mathrm{i}}t^2\Phi_3(t,q) + \frac{5}{32}t^3\Phi_4(t,q)$

The idea here would be to choose a fixed $t$, one whose roots were all within the unit disk, and then to make five phase portraits of their $q$-analogues

Remark: I am tempted to look at cases where $q=g^{20n}, n\in\mathbb{N}$ because all of those fractional powers will vanish

Additionally, suppose we were to get $\prod_{n=1}^{5} (x_{n}(q)-t)$ into Bring form and then solve that with the non- $q$-analogized version.(matrix identity to be worked out soon)