On $x^3+y^3=z^3$, the Dixonian elliptic functions, and the Borwein cubic theta functions

The Borwein cubic theta functions $a(q),b(q),c(q)$ and the doubly-periodic Dixonian elliptic functions $\mathrm{sm}(z), \mathrm{cm}(z)$ parameterize the Fermat cubic $x^3+y^3=1$, so

$$\frac{b^3(q)}{a^3(q)}+\frac{c^3(q)}{a^3(q)} =1\tag1$$

$$\mathrm{sm}^{3}(z)+\mathrm{cm}^{3}(z)=1\tag2$$

This post asks if there is any relation between the Borwein and Dixonian versions. It is known that the former is associated with $_2F_1\big(\tfrac13,\tfrac23;\color{blue}1;u\big)$ while the latter is with $_2F_1\big(\tfrac13,\tfrac23;\color{blue}{\tfrac43};v\big)$.

For argument $q=e^{2\pi i \sqrt{-n}}$, we have the nice relationship,

$$\large \frac{_2F_1\Big(\tfrac13,\tfrac23;1;\tfrac{b^3(q)}{a^3(q)} \Big)}{_2F_1\Big(\tfrac13,\tfrac23;1;\tfrac{c^3(q)}{a^3(q)} \Big)}=\sqrt{3n}$$

Q: By analogy, does the ratio, $$\frac{_2F_1\Big(\tfrac13,\tfrac23;\tfrac43;\mathrm{sm}^3(z) \Big)}{_2F_1\Big(\tfrac13,\tfrac23;\tfrac43;\mathrm{cm}^3(z) \Big)}=\,??\tag3$$ evaluate to anything "meaningful" for appropriate choice of argument $z$?

P.S. To test this conjecture, for the circle $x^2+y^2=1$, we have the Dedekind eta quotients $\alpha(q), \beta(q)$ parameterization (given here) and the singly-periodic trigonometric functions,

$$\alpha^8(q)+\beta^8(q)=1\tag4$$ $$\sin^2 z+\cos^2 z=1\tag5$$

The analogous ratios, $$\frac{_2F_1\Big(\tfrac12,\tfrac12;1; \beta^8(q) \Big)}{ _2F_1\Big(\tfrac12,\tfrac12;1; \alpha^8(q)\Big)}=\sqrt{4n}$$ and choice of $0<z<1$,

$$\frac{_2F_1\Big(\tfrac12,\tfrac12;\tfrac32; \cos^2 z \Big)}{ _2F_1\Big(\tfrac12,\tfrac12;\tfrac32; \sin^2 z \Big)}=\Big(\frac{\pi}{2z}-1\Big) \tan z$$ so $(3)$ might indeed evaluate to some closed-form, perhaps involving the fundamental constant of Dixonian functions $\large \pi_3=\frac{\sqrt3}{2\pi}\Gamma^3\big(\tfrac13\big)$.

• As soon as I saw the title of this question, I thought that you were probably the author. I admire the work you do here. – marty cohen Jan 9 '17 at 17:19
• @martycohen: Thanks, Marty. :) – Tito Piezas III Jan 9 '17 at 17:35
• After a good night's rest, I figured out the answer. See this MO post for a possible generalization – Tito Piezas III Jan 10 '17 at 5:36

Recall that

$$\int_0^z\frac{\mathrm du}{\left(1-u^3\right)^{2/3}}=\operatorname{sm}^{(-1)}(z)=z\;{}_2 F_1\left({{\frac13,\frac23}\atop{\frac43}}\middle|z^3\right)$$

and

$$\int_z^1\frac{\mathrm du}{\left(1-u^3\right)^{2/3}}=\operatorname{cm}^{(-1)}(z)$$

along with

$$\frac{\pi_3}{3}=\int_0^1\frac{\mathrm du}{\left(1-u^3\right)^{2/3}}=\frac13 B\left(\frac13,\frac13\right)$$

so within a limited domain, your expression ought to simplify to

$$\frac{{}_2 F_1\left({{\frac13,\frac23}\atop{\frac43}}\middle|\operatorname{sm}^3(z)\right)}{{}_2 F_1\left({{\frac13,\frac23}\atop{\frac43}}\middle|\operatorname{cm}^3(z)\right)}=\frac{\frac{z}{\operatorname{sm}(z)}}{\frac{\frac{\pi_3}{3}-z}{\operatorname{cm}(z)}}=\frac{z}{\frac{\pi_3}{3}-z}\frac{\operatorname{cm}(z)}{\operatorname{sm}(z)}$$

Note the phrase "limited domain". This is expected, in the same way that $\arcsin\sin z=z$ and $\arccos\cos z=z$ are also only valid within a limited domain. In particular, the relation only works within the intersection of the domains of validity of $\operatorname{sm}^{(-1)}(\operatorname{sm}(z))$ and $\operatorname{cm}^{(-1)}(\operatorname{cm}(z))$ (both of which are, perhaps not surprisingly, hexagons), which is the rhombus with corners at $0,\frac{\pi_3}{3}\exp\left(\frac{\pi i}{3}\right),\frac{\pi_3}{3},\frac{\pi_3}{3}\exp\left(-\frac{\pi i}{3}\right)$.

• @JM: Yes, I believe it is $0<z<1$. This morning I posted such relations in MO for $\sin, \mathrm{sn}, \mathrm{sm}$ and asked if it could be generalized from $a=1/2$ and $a=2/3$, to $a=1/4$ and $a=1/6$. – Tito Piezas III Jan 10 '17 at 16:37
• @JM: Perhaps you can answer that MO post as well for $a=1/4$? :) – Tito Piezas III Jan 10 '17 at 16:52
• The cases $a=1/4,1/6$ have been posted here mathoverflow.net/a/259267/82588 – Nemo Jan 10 '17 at 22:20
• @Tito, $0<z<1$ is but a small cut of the rhombus I mentioned in my answer. :) – J. M. is a poor mathematician Jan 12 '17 at 15:42
• @J.M.: And someone also did work on $x^5+y^5=1$. – Tito Piezas III Jan 12 '17 at 15:49