Elliptic functions from $F(\frac{1}{3}, \frac{2}{3} ; \frac{1}{2} ; \bullet)$

TL;DR

This paper discusses the development of elliptic functions derived from a specific hypergeometric function, providing new proofs and insights into Li-Chien Shen's work.

Contribution

It offers new proofs and commentary on the construction of elliptic functions from the hypergeometric function $_2F_1(1/3, 2/3; 1/2; ullet)$ by Li-Chien Shen.

Findings

New proofs of elliptic functions from hypergeometric functions

Enhanced understanding of Shen's elliptic function family

Clarification of the mathematical structure involved

Abstract

Li-Chien Shen developed a family of elliptic functions from the hypergeometric function $_2F_1(\frac{1}{3}, \frac{2}{3} ; \frac{1}{2} ; \bullet)$. We comment on this development, offering some new proofs.

Equations

$$u={\int }_0^{\sin \varphi }F(\frac{1}{3},\frac{2}{3};\frac{1}{2};k^2{t}^2)\frac{\mathrm{d}t}{\sqrt{1-t^2}}$$

$$\frac{\mathrm{d}u}{\mathrm{d}\varphi }=F(\frac{1}{3},\frac{2}{3};\frac{1}{2};k^2{\sin }^2\varphi ).$$

$$s(u)=\sin \varphi (u)$$

$$c(u)=\cos \varphi (u)$$

$$d(u)=\varphi {}^{\prime }(u)=1/F(\frac{1}{3},\frac{2}{3};\frac{1}{2};k^2{s}^2(u)).$$

$$s^2+c^2=1.$$

$$F(\frac{1}{3},\frac{2}{3};\frac{1}{2};{\sin }^{2}z)=\frac{\cos \frac{1}{3}z}{\cos z}$$

$$4{\cos }^3\frac{1}{3}z-3\cos \frac{1}{3}z=\cos z$$

$$d^3+3d^2=4(1-k^2{s}^2)$$

$$4k^2{s}^2=(1-d)(2+d{)}^2.$$

$$s{}^{\prime }=cd$$

$$c{}^{\prime }=-sd$$

$$d{}^{\prime }=-\frac{8}{3}{k}^2\frac{sc}{2+d}.$$

$$(d{}^{\prime }{)}^2=\frac{4}{9}(1-d)(d^3+3d^2+4k^2-4).$$

$$g_2=\frac{4}{27}(9-8k^2)$$

$$g_3=\frac{8}{27^2}(8k^4-36k^2+27).$$

$$(d{}^{\prime }{)}^2=\frac{4}{9}(1-d)(d^3+3d^2+4k^2-4)$$

$$(r\,^{\prime})^{2}=\frac{4k^{2}}{9}\Big{(}4r^{3}-\frac{9}{k^{2}}r^{2}+\frac{6}{k^{2}}r-\frac{1}{k^{2}}\Big{)}.$$

$$(q{}^{\prime }{)}^2=4q^3-4q^2+\frac{2^5}{3^3}{k}^{2}q-\frac{2^6}{3^6}{k}^4$$

$$(p{}^{\prime }{)}^2=4p^3-g_{2}p-g_3$$

$$g_2^3-27g_3^2=\frac{16^3}{27^3}{k}^6(1-k^2).$$

$$d=1-\frac{4}{9}{k}^2(\wp +\frac{1}{3}{)}^{-1}.$$

$$\wp(2a)+2\,\wp(a)=\frac{1}{4}\Big{\{}\frac{\wp^{\prime\prime}(a)}{\wp^{\prime}(a)}\Big{\}}^{2}$$

$${\wp }^{\prime \prime }(a)=6\wp (a{)}^2-\frac{1}{2}{g}_2$$

$${\wp }^{\prime }(a{)}^2=4\wp (a{)}^2-g_2\wp (a)-g_3.$$

$$2a=\frac{4}{3}\mathrm{i}{K}^{\prime }\equiv -\frac{2}{3}\mathrm{i}{K}^{\prime }=-a$$

$$\wp (2a)=\wp (-a)=\wp (a)$$

$$3b=\frac{1}{4}\frac{(6b^2-\frac{1}{2}{g}_2{)}^2}{4b^3-g_{2}b-g_3}.$$

$$f(z)=12z(4z^3-g_{2}z-g_3)-(6z^2-\frac{1}{2}{g}_2{)}^2.$$

$$\frac{27}{4}f(\frac{w}{3})=w^4-\frac{2}{3}(9-8k^2)w^2-\frac{8}{27}(8k^4-36k^2+27)w-\frac{1}{27}(9-8k^2{)}^2$$

Full text

Shen [1] has presented an interesting construction of elliptic functions based on the hypergeometric function $F(\frac{1}{3},\frac{2}{3};\frac{1}{2};\bullet)$. We open our commentary with a brief review of his construction, making some minor notational changes (most of which amount to the dropping of suffixes).

Fix $0<k<1$ and write

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so that

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In a (connected) neighbourhood of the origin, the relation $\phi\mapsto u$ inverts to $u\mapsto\phi$ fixing [math]. Define functions $s,c,d$ by

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and

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Plainly, $s$ and $c$ satisfy the Pythagorean relation

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Shen uses the hypergeometric identity

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and the trigonometric triplication formula

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to show that $s$ and $d$ satisfy the relation

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or equivalently

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By differentiation,

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and

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By means of the $(s,d)$ and $(s,c)$ relations, it follows that $d$ satisfies the differential equation

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and

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suggests the substitution $r=(1-d)^{-1}$: this has the effect of removing the explicit linear factor, thus

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Next, the rescaling $q=\frac{4k^{2}}{9}r$ leads to

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and the shift $p=q-\frac{1}{3}$ removes the quadratic term on the right side, yielding

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with $g_{2}$ and $g_{3}$ as stated in the theorem. Finally, the initial condition $d(0)=1$ gives $p$ a pole at [math]; thus $p$ is the Weierstrass function $\wp$ and so $d$ is as claimed. ∎

We remark that $\wp$ has discriminant

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Now, recall that $0<k<1$. It follows that the Weierstrass function $\wp$ has real invariants and positive discriminant. Consequently, the period lattice of $\wp$ is rectangular; let $2K$ and $2{\rm i}K^{\prime}$ be fundamental periods, with $K>0$ and $K^{\prime}>0$. We may take the period parallelogram to have vertices $0,2K,2K+2{\rm i}K^{\prime},2{\rm i}K^{\prime}$ (in counter-clockwise order); alternatively, we may take it to have vertices $\pm K\pm{\rm i}K^{\prime}$ (with all four choices of sign). The values of $\wp$ around the rectangle $0\to K\to K+{\rm i}K^{\prime}\to{\rm i}K^{\prime}\to 0$ strictly decrease from $+\infty$ to $-\infty$; moreover, the extreme ‘midpoint values’ satisfy $\wp(K)>0>\wp({\rm i}K^{\prime})$. In particular, $\wp$ is strictly negative along the purely imaginary interval $(0,{\rm i}K^{\prime})$.

As the Weierstrass function $\wp$ is elliptic of order two, with $2K$ and $2{\rm i}K^{\prime}$ as fundamental periods, the same is true of the function

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One of the most substantial efforts undertaken in [1] is the task of locating the poles of $d$. As a preliminary step, in [1] Lemma 3.1 it is shown (using conformal mapping theory) that $d$ has a pole in the interval $(0,{\rm i}K^{\prime})$; as $d$ is even, it also has a pole in $(-{\rm i}K^{\prime},0)$. The precise location of the poles of $d$ is announced in [1] Lemma 3.2; the proof of this Lemma is prepared in [1] Section 4 and takes up essentially the whole of [1] Section 5. The approach taken in [1] rests heavily on the theory of theta functions and does more than just locate the poles of $d$. Our approach to locating the poles of $d$ will be more direct: we work with $\wp$ alone, without the need for theta functions. The following is our version of [1] Lemma 3.2.

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where

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and

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Here,

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where the middle congruence is modulo the period $2{\rm i}K^{\prime}$ of $\wp$; consequently,

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because $\wp$ is even. The left side of the duplication formula thus reduces to $3\,\wp(a)$ and we deduce that $b=\wp(a)$ satisfies

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Otherwise said, $\wp(a)$ is a zero of the quartic $f$ defined by

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For convenience we work with

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which factorizes as

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Here, the cubic factor has discriminant

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and so has just one real zero, which is clearly positive. Thus, $f$ has four zeros: a conjugate pair of non-real zeros, a positive zero and $-1/3$. As the values of $\wp$ along $(0,{\rm i}K^{\prime})$ are strictly negative, it follows that $\wp(a)=-1/3$ as claimed.

∎

As an even function, $d$ also has a pole at $-\frac{2}{3}{\rm i}K^{\prime}$. Both of the poles $\pm\frac{2}{3}{\rm i}K^{\prime}$ lie in the period parallelogram with vertices $\pm K\pm{\rm i}K^{\prime}$ and $d$ has order two, so each pole is simple and the accounting of poles (modulo periods) is complete; of course, this may be verified otherwise.

We close our commentary with a couple of remarks.

In [1] it is mentioned that the squares $s^{2}$ and $c^{2}$ are elliptic: as $d$ is elliptic, these facts follow in turn from the $(s,d)$ relation $4k^{2}s^{2}=(1-d)(2+d)^{2}$ and the $(s,c)$ relation $c^{2}=1-s^{2}$. As $d$ has simple poles, it follows that the poles of $s^{2}$ and $c^{2}$ are triple; thus, $s$ and $c$ themselves are not elliptic, as is also mentioned in [1]. Beyond this, the product $s\,c$ is elliptic with triple poles: indeed,

[TABLE]

REFERENCES

[1] Li-Chien Shen, On the theory of elliptic functions based on ${}_{2}F_{1}(\frac{1}{3},\frac{2}{3};\frac{1}{2};z)$, Transactions of the American Mathematical Society 357 (2004) 2043-2058.