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This solves this post and is also related to this MO post by involving $\tfrac12,\tfrac13,\tfrac14,\tfrac16$.

$p=2$

The singly-periodic trigonometric functions and the doubly-periodic Jacobi elliptic functions parameterize the circle so,

$$\sin^2 z+\cos^2 z=1\tag1$$ $$\mathrm{sn}^2(u)+\mathrm{cn}^2(u)=1\tag2$$

If we use these as arguments for hypergeometric ratios, then for $0<z<1$,

$$\frac{_2F_1\Big(\tfrac12,\tfrac12;\tfrac32; \cos^2 z \Big)}{ _2F_1\Big(\tfrac12,\tfrac12;\tfrac32; \sin^2 z \Big)}=\Big(\frac{\pi_\color{blue}2}{2z}-1\Big) \frac{\sin z}{\cos z}\tag3$$

and elliptic integral of the first kind $u = F(\phi,k)$ for $\phi>0$ and any $k$,

$$\frac{_2F_1\Big(\tfrac12,\tfrac12;\tfrac32; \mathrm{cn}^2(u) \Big)}{ _2F_1\Big(\tfrac12,\tfrac12;\tfrac32; \mathrm{sn}^2(u) \Big)}=\Big(\frac{\pi_\color{blue}2}{2\phi}-1\Big) \frac{\mathrm{sn}(u) }{\mathrm{cn}(u) }\tag4$$

where $\pi_2 = B\big(\tfrac12,\tfrac12\big)=\pi$. It turns out this can be generalized.

$p=3$

The doubly-periodic Dixonian elliptic functions $\mathrm{sm}(z), \mathrm{cm}(z)$ parameterize the Fermat cubic so

$$\mathrm{sm}^{3}(z)+\mathrm{cm}^{3}(z)=1\tag5$$

In terms of the Weierstrass elliptic functions,

$$\mathrm{sm}(z)=\frac{-6\wp\left(z;0,\frac1{27}\right)}{3\wp^\prime\left(z;0,\frac1{27}\right)-1},\quad \mathrm{cm}(z)=\frac{3\wp^\prime\left(z;0,\frac1{27}\right)+1}{3\wp^\prime \left(z;0,\frac1{27}\right)-1}$$

After some experimentation with Mathematica, we find, for $0<z<1$, the analogous ratio,

$$\frac{_2F_1\Big(\tfrac13,\tfrac23;\tfrac43;\mathrm{cm}^3(z) \Big)}{_2F_1\Big(\tfrac13,\tfrac23;\tfrac43;\mathrm{sm}^3(z) \Big)}=\Big(\frac{\pi_\color{blue}3}{3z}-1\Big) \frac{\mathrm{sm(z)}}{\mathrm{cm(z)}}\tag6$$

where $\pi_3 = B\big(\tfrac13,\tfrac13\big)=\frac{\sqrt3}{2\pi}\Gamma^3\big(\tfrac13\big)$.

Question for $p=4,6$

What suitable functions $\alpha(z),\beta(z)$ with, $$\alpha^k(z)+\beta^k(z)=1$$ will solve the hypergeometric ratios below in a similar manner?:

$$\frac{_2F_1\Big(\tfrac14,\tfrac34;\tfrac54;\beta^k(z)\Big)} {_2F_1\Big(\tfrac14,\tfrac34;\tfrac54;\alpha^k(z)\Big)}=\Big(\frac{\pi_{\color{blue}4}}{4z}-1\Big) \frac{\alpha(z)}{\beta(z)}\tag7$$ or,

$$\frac{_2F_1\Big(\tfrac16,\tfrac56;\tfrac76;\beta^k(z)\Big)} {_2F_1\Big(\tfrac16,\tfrac56;\tfrac76;\alpha^k(z)\Big)}=\Big(\frac{\pi_{\color{blue}6}}{6z}-1\Big) \frac{\alpha(z)}{\beta(z)}\tag8$$

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  • $\begingroup$ @Nemo: That occurred to me as well. Just like the one-variable trigonometrics are generalized by the two-variable Jacobi elliptics, the one-variable Dixonians might be generalized by a two-variable something. (Those strictly speaking the Jacobi elliptics are just a function of the variable $u$.) $\endgroup$
    – Tito Piezas III
    Commented Jan 11, 2017 at 9:48
  • $\begingroup$ Dixon actually considered more general versions of $\operatorname{sm}$ and $\operatorname{cm}$ in his paper, associated with the cubic $x^3+y^3-3\alpha xy=1$, and thus take $\alpha$ as a second parameter. (The usual ones of course correspond to $\alpha=0$.) $\endgroup$
    – J. M. isn't a mathematician
    Commented Jan 12, 2017 at 16:05

1 Answer 1

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Define the generalized trigonometic functions (discussion of these functions is given in this answer) $$ z=\int_0^{\sin_{pr}z}\frac{dt}{\sqrt[p]{1-t^r}},\qquad \cos_{pr}z=\sqrt[r]{1-(\sin_{pr}z)^r},\qquad \frac{1}{p}=1-\frac{1}{r}. $$ It was shown by Edmunds,Gurka, and Lang (Properties of generalized trigonometric functions, J. Approx. Theory 164 (2012), no. 1, 47-56.) that when $r=4$ $$ \sin_{\frac{4}{3},4}z=\sqrt{\frac{1-\text{cn}^4\left(z,\frac{1}{\sqrt{2}}\right)}{1+\text{cn}^4\left(z,\frac{1}{\sqrt{2}}\right)}}\tag{*} $$ Using formulas for the incomplete beta function http://dlmf.nist.gov/8.17 one has $$ B\Big(z;\tfrac14,\tfrac14\Big)=4\int_0^{\sqrt[4]{z}}\frac{dt}{(1-t^4)^{3/4}}=4z^{1/4}{}_2F_1(1/4,3/4,5/4;z), $$ $$ B\Big(x;\tfrac14,\tfrac14\Big)=\pi_4-B\Big(1-x;\tfrac14,\tfrac14\Big). $$ From these formulas we immediately see that $$ 4z=B\Big(\sin^4_{\frac{4}{3}}z;\tfrac14,\tfrac14\Big)=4\sin_{\frac{4}{3},4}z\cdot{}_2F_1\Big(1/4,3/4,5/4;\sin^4_{\frac{4}{3},4}z\Big), $$ $$ \pi_4-4z=B\Big(\cos^4_{\frac{4}{3},4}z;\tfrac14,\tfrac14\Big)=4\cos_{\frac{4}{3},4}z\cdot{}_2F_1\Big(1/4,3/4,5/4;\cos^4_{\frac{4}{3},4}z\Big). $$ This gives the required parametrization $$ \frac{_2F_1\Big(\tfrac14,\tfrac34;\tfrac54;\cos^4_{\frac{4}{3},4}z\Big)} {_2F_1\Big(\tfrac14,\tfrac34;\tfrac54;\sin^4_{\frac{4}{3},4}z\Big)}=\Big(\frac{\pi_{\color{blue}4}}{4z}-1\Big) \frac{\sin_{\frac{4}{3},4}z}{\cos_{\frac{4}{3},4}z}\tag{7a}, $$ with $~\sin^4_{\frac{4}{3},4}z+\cos^4_{\frac{4}{3},4}z=1$ and (*), or equivalently in terms of Weierstrass elliptic functions, $$\sin^2_{\frac43,4}z=\frac{4\wp (z;1,0)}{4\wp^2(z;1,0)+1},\quad\cos^2_{\frac43,4}z=\frac{4\wp^2 (z;1,0)-1}{4\wp^2(z;1,0)+1}\tag{i}$$

with (*) implying, $$ \sin_{\frac{4}{3},4}z=\sqrt{\frac{4\wp (z;1,0)}{4\wp^2(z;1,0)+1}}=\sqrt{\frac{1-\text{cn}^4\left(z,\frac{1}{\sqrt{2}}\right)}{1+\text{cn}^4\left(z,\frac{1}{\sqrt{2}}\right)}}\tag{ii} $$ with the $\mathrm{cn}(u)$ in Mathematica syntax as JacobiCN[z,1/2] since it uses the parameter $m=k^2$.

The case $r=6$ is considered similarly and the equation analogous to (*) is given by $$ \sin_{\frac{6}{5},6}z=\left(\frac{1}{2}-\frac{1}{2}\sqrt{1-\left(1+\sqrt{3}\frac{1+\text{cn}\left(3^{1/4} 2^{4/3} z,k\right)}{1-\text{cn}\left(3^{1/4} 2^{4/3} z,k\right)}\right)^{-3}}\right)^{1/6},\tag{**} $$ where the modulus of elliptic function is $k=\sin\left(\frac{\pi }{12}\right)$. Also equivalent form of (**) is $\Big(1-2\sin^6_{\frac56,6}z\Big)^2=1-\frac{4}{\wp^3(z;0,16)}$. Then

$$ \frac{_2F_1\Big(\tfrac16,\tfrac56;\tfrac76;1-\sin^6_{\frac{6}{5},6}z\Big)} {_2F_1\Big(\tfrac16,\tfrac56;\tfrac76;\sin^6_{\frac{6}{5},6}z\Big)}=\Big(\frac{\pi_{\color{blue}6}}{6z}-1\Big) \frac{\sin_{\frac{6}{5},6}z}{\cos_{\frac{6}{5},6}z}\tag{7b}. $$

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  • $\begingroup$ For completeness, since we already have $r=3$, kindly edit your answer and include the Weierstrass formulas for $r=4$ and $r=6$. Thanks. $\endgroup$
    – Tito Piezas III
    Commented Jan 11, 2017 at 7:25
  • $\begingroup$ I note that those Weierstrass functions for $r=4$ are of the "lemnisactic" type. That's why it's relatively easy to interconvert between them and the Jacobi ones. $\endgroup$
    – J. M. isn't a mathematician
    Commented Jan 12, 2017 at 16:06
  • $\begingroup$ Also, note that $$\frac{1+\operatorname{cn}(u,k)}{1-\operatorname{cn}(u,k)}=\frac{\operatorname{cn}^2\left(\frac{u}{2},k\right)}{\operatorname{sn}^2\left(\frac{u}{2},k\right)\operatorname{dn}^2\left(\frac{u}{2},k\right)}$$ $\endgroup$
    – J. M. isn't a mathematician
    Commented Jan 12, 2017 at 16:09
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    $\begingroup$ @Nemo, sounds right; in any event, one could use $\wp^\prime(z;1,0)^2=4\wp(z;1,0)^3-\wp(z;1,0)$ to simplify things further. $\endgroup$
    – J. M. isn't a mathematician
    Commented Jan 15, 2017 at 15:27
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    $\begingroup$ @Nemo: Now that has been clarified, so it turns out the Weierstrass form for $r=4$ just simply uses a Pythagorean triple, $$\mathrm{sin}_4^2\,z = \frac{2\lambda}{\lambda^2+1},\quad \mathrm{cos}_4^2\,z = \frac{\lambda^2-1}{\lambda^2+1}$$ with $\lambda=2\wp(z;1,0)$. $\endgroup$
    – Tito Piezas III
    Commented Jan 16, 2017 at 16:18

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