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Given the j-function $j(\tau)$ and elliptic lambda function $\lambda(\tau)$. Define,

$$g = -1+2\frac{\,_2F_1\big(\tfrac{1}{2},\tfrac{1}{2},1,\,\lambda(2\tau)\big) }{\,_2F_1\big(\tfrac{1}{2},\tfrac{1}{2},1,\,\lambda(\tau)\big) }$$

then,

$$j(\tau) = \frac{4^4(g^4-g^2+1)^3 }{(g^4-g^2)^2}\tag{1}$$

Question: Any other formula that uses $\,_2F_1\big(a,b;c;z\big)$ for other $a,b,c$?

$\color{blue}{Edit}$: (In response to ccorn's answer.)

For any non-zero constant $N(N-1728)\neq0$, one can solve the following two equations,

$$N = \frac{(x^2+10x+5)^3}{x}\tag{2}$$

$$N = \frac{-(r^{20} - 228r^{15} + 494r^{10} + 228r^5 + 1)^3}{r^5(r^{10} + 11r^5 - 1)^5}\tag{3}$$

for unknowns $x,r$ as,

$$x = \frac{-125r^5}{r^{10}+11r^5-1}$$

where,

$$r = \frac{N\,^{-11/60}\,_2F_1\left(\tfrac{31}{60},\tfrac{11}{60},\tfrac{6}{5},\tfrac{1728}{N}\right)}{N\,^{1/60}\,_2F_1\left(\tfrac{19}{60},\tfrac{-1}{60},\tfrac{4}{5},\tfrac{1728}{N}\right)} = \frac{(N-1728)^{-11/60}\,_2F_1\left(\tfrac{41}{60},\tfrac{11}{60},\tfrac{6}{5},\tfrac{1728}{1728-N}\right)}{(N-1728)^{1/60}\,_2F_1\left(\tfrac{29}{60},\tfrac{-1}{60},\tfrac{4}{5},\tfrac{1728}{1728-N}\right)}$$

The eqns (2) and (3) are the Jacobi sextic and icosahedral equation, respectively, both of which do not have a solvable Galois group. (Eq. (1) has a solvable group.)

The utility of the question is then it finds an equation, with one free parameter $N$ (the special case $N = j(\tau)$ being only a subset), that can be solved in terms of $_2F_1(a,b;c;z)$ where $z$ is a function of $N$. (This also implies the general quintic, via the Jacobi sextic, is solvable in terms of $r$.)

So it would be nice to find more polynomial examples like (1), but has a higher degree.

  • All but one of them involve $a,b = 1/2, 1/2$. My question is about *other* $a,b$. – Tito Piezas III Jan 27 '14 at 17:57
  • Yes, my question was explicit for other $a,b,c$, (not the argument $z$). I believe one can also find for $a,b = 1/3,2/3$ and $a,b =1/4,3/4$ (which probably implies another $z$ as well). – Tito Piezas III Jan 27 '14 at 18:06

2 Answers2

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The following is an too-long-for-a-comment remark of mine to the original question. It seems outdated now.

Sorry, I do not get the point of this question.

In your example, $\lambda$ is a modular function for some subgroup of the full modular group, and therefore $j$ and $\lambda$ fulfill some bivariate polynomial equation. In fact, $j$ can be expressed as a rational function of $\lambda$. Plumbing hypergeometrics in such a simple relation looks like an obfuscated math contest to me. In particular, the given example obfuscates that $$g=k'=\sqrt{1-\lambda}$$

So probably my two cents here will not be able to assist your quest. Of course, I could provide another obfuscated math example: $$j=\gamma_3^2\left(\frac{{}_2F_1\left(\frac{1}{12},\frac{5}{12};1;\frac{12^3}{\gamma_2^3}\right)}{{}_2F_1\left(\frac{1}{12},\frac{7}{12};1;-\frac{12^3}{\gamma_3^2}\right)}\right)^{12}$$ where $\gamma_2$, $\gamma_3$ are Weber functions. If $\tau+m$ is in the full modular group's fundamental domain for some $m\in\mathbb{Z}$, the numerator in the fraction is $\sqrt[4]{E_4}$, and the denominator is $\sqrt[6]{E_6}$, where $E_4=\gamma_2\eta^8$ and $E_6=\gamma_3\eta^{12}$ are Eisenstein series and $\eta$ is the Dedekind eta function. This thing just obfuscates $$j=\gamma_2^3=\gamma_3^2+12^3$$ Therefore I do not see the point of such jugglings.

The theorem that relates modular forms with functions like ${}_2F_1$ is given, for example, in

  1. Don Zagier: Elliptic modular forms and their applications. In: Kristian Ranestad (ed.): The 1-2-3 of modular forms. Springer 2008, DOI: 10.1007/978-3-540-74119-0.

It implies that a weight-$1$ modular form (in the wider sense, that is, with respect to subgroups of the full modular group, or with a multiplier system) fulfills an ordinary second-order linear differential equation if the independent variable of that equation is chosen to be a suitable modular function. Often, the differential equation turns out to be the hypergeometric one, and with a proper handling of initial conditions, the modular form can be expressed locally in terms of ${}_2F_1$, whose main argument is then necessarily some modular function.

The key point here is that the main argument of ${}_2F_1$, let us name it $t$, must itself be a modular function and thus have weight zero. But then $j$ and $t$ are algebraically related over $\mathbb{C}$. Therefore, I expect every expression of $j$ in terms of ratios of ${}_2F_1$ to be just obfuscations of a simpler algebraic relation. In some cases, it may serve as an illustration of hypergeometric transformations though.

ccorn
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  • The original motivation in it weak form was to determine if $j$ can be expressed in terms of $,_2F_1(a,b;c;z)$ for some function $z$. It can be but, as pointed out by Balarka, any argument $z$ seems to have a polynomial relationship with $j$. In its strong form then, one can specify the additional constraint that $j$ and $z$ should not be algebraically related over $\mathbb{C}$. I believe that there are also for $,_2F_1(1/3,2/3;1;t)$ and $,_2F_1(1/4,3/4;1;t)$ but, as you mentioned, these probably will involve $t$ algebraically similar (but not identical) to $\lambda$. – Tito Piezas III Jan 29 '14 at 01:33
  • If there turns out to be anything like that, I would find that very interesting myself. It might imply that modular forms can fulfill linear ODEs over variables that are not modular functons. However, the usual way that makes ${}_2F_1$ appear requires modular functions and thereby seems to enforce an algebraic relation, so I'm skeptic. – ccorn Jan 29 '14 at 01:41
  • ccorn, I've added an edit. I hope you can see why I find polynomial relations involving $j$ can be quite fruitful. :) – Tito Piezas III Jan 29 '14 at 03:21
  • Thanks. To pre-empt comments/answers similar to mine, I let it stand. In case I find something helpful, I'll edit of course. – ccorn Jan 29 '14 at 06:35
  • @BalarkaSen: Back to the topic: The theorem I mentioned has an if, not an iff, so I do not have a proof that no ${}_2F_1$ expressions for $j$ exist that are not obfuscations of algebraic relations. I just do not expect them to exist, so I have gone out on a limb and stated that. By the way, an answer to this question might bring us one step further. – ccorn Jan 30 '14 at 21:34
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Note the following relation

$$ _2F_1\left( 1/2,1/2,1, x \right)= \frac{2}{\pi} K(\sqrt{x}). $$

where $K(x)$ is the Complete elliptic integral of the first kind and $_2F_1$ is the hypergeometric function.

  • Yes. However, I think there should also be formulas using ratios of $,_2F_1\big(\tfrac{1}{3},\tfrac{2}{3},1,,z\big)$ or $,_2F_1\big(\tfrac{1}{4},\tfrac{3}{4},1,,z\big)$ though I'll need some time to find them. – Tito Piezas III Jan 27 '14 at 06:39