# Elementary Approach to Modular Equations: Hypergeometric Series 1

For quite some time I have been studying Ramanujan's Modular Equations and Approximations to $\pi$ and in this series of posts I will try to present my understanding of the modular equations. Ramanujan's work on modular equations was brought to limelight by Borwein brothers in their famous book Pi and the AGM and later on by Bruce C. Berndt through Ramanujan Notebooks. Much of what I present here would also be based on the material presented in these books. However my approach here is going to be elementary and requires at best a working knowledge of calculus. Apart from this reader is expected to have some background on elliptic functions and theta functions as presented in my previous series of posts (here and here).

Modular equations form the most interesting parts of the theory about the modulus $k$ (hence the name modular) which is used in the definitions of elliptic functions and integrals. The elliptic integral $K$ with modulus $k$ is expressible in the form of what is called a Hypergeometric Series: \begin{align} K &= \int_{0}^{\pi / 2}\frac{d\theta}{\sqrt{1 - k^{2}\sin^{2}\theta}}\notag\\ &= \frac{\pi}{2}\left(1 + \left(\frac{1}{2}\right)^{2}k^{2} + \left(\frac{1\cdot 3}{2\cdot 4}\right)^{2}k^{4} + \cdots\right)\notag\\ &= \frac{\pi}{2}\sum_{n = 0}^{\infty}\frac{1^{2}\cdot 3^{2}\cdots (2n - 1)^{2}}{2^{2n}(n!)^{2}}k^{2n}\notag\end{align} The series on the right is expressed in the form of hypergeometric function of Gauss $_{2}F_{1}(a, b; c; z)$ defined by $${}_{2}F_{1}(a, b; c; z) = \sum_{n = 0}^{\infty}\frac{(a)_{n}(b)_{n}}{(c)_{n}}\frac{z^{n}}{n!}$$ where $$(a)_{n} = a(a + 1)(a + 2)\cdots (a + n - 1)$$ Clearly here $c$ can not be equal to zero or a negative integer.

The subscripts $2$ and $1$ in the notation are used to refer to the fact that there are two factors $(a)_{n}, (b)_{n}$ in the numerator and one factor $(c)_{n}$ in the denominator of each term in the series apart from the multiplier $z^{n}/n!$. There are hypergeometric series of type $_{p}F_{q}$ with $p$ factors in numerator and $q$ factors in denominator of each term and we shall have occasion to deal with such generalized hypergeometric series later. For the time being lets work with the ${}_{2}F_{1}$ type series and then we make a simplification in the notation by dropping the subscripts and just using $F(a, b; c; z)$ to denote such a series.  Also note that the semicolons used in the notation help to separate the factors in numerator from those in denominator and also from variable $z$.

In general the series $F(a, b; c; z)$ is absolutely convergent for $|z| < 1$ and under suitable conditions it is also convergent when $|z| = 1$. Using this definition of the hypergeometric series we can express $K$ as $$K = \frac{\pi}{2}F\left(\frac{1}{2}, \frac{1}{2}; 1; k^{2}\right)$$ In this post we will study some basic properties and transformation of the hypergeometric srries. First we begin with the differential equation which is satisfied by such eries.

### Differential Equation for the Hypergeometric Function

To obtain the differential equation satisfied by the hypergeometric function it is convenient to introduce an operator which we denote by $\Theta$ as follows: $$\Theta g(z) = z\frac{d}{dz}g(z)$$ The convenience offered by this operator is that $\Theta z^{n} = nz^{n}$ so that it provides easy manipulation of series containing terms of the form $z^{n}/n!$. Therefore if we apply this operator on the hypergeometric function $F(a, b; c; z)$ we get $$\Theta F(a, b; c; z) = \sum_{n = 0}^{\infty}\frac{(a)_{n}(b)_{n}}{(c)_{n}}\frac{nz^{n}}{n!}$$ and therefore we have \begin{align} \Theta(\Theta + c - 1)F &= \sum_{n = 0}^{\infty}\frac{n(n + c - 1)(a)_{n}(b)_{n}}{(c)_{n}}\frac{z^{n}}{n!}\notag\\ &= \sum_{n = 1}^{\infty}\frac{(a)_{n}(b)_{n}}{(c)_{n - 1}}\frac{z^{n}}{(n - 1)!}\notag\\ &= \sum_{n = 0}^{\infty}\frac{(a)_{n + 1}(b)_{n + 1}}{(c)_{n}}\frac{z^{n + 1}}{n!}\notag\\ &= z\sum_{n = 0}^{\infty}\frac{(a + n)(b + n)(a)_{n}(b)_{n}}{(c)_{n}}\frac{z^{n}}{n!}\notag\\ &= z(\Theta + a)(\Theta + b)F\notag\end{align} Now we have \begin{align} \Theta(\Theta + c - 1)F &= \Theta(zF' + (c - 1)F)\notag\\ &= z(zF' + (c - 1)F)'\notag\\ &= z(F' + zF'' + (c - 1)F')\notag\\ &= z(zF'' + cF')\notag\end{align} and \begin{align}z(\Theta + a)(\Theta + b)F &= z(\Theta + a)(zF' + bF)\notag\\ &= z(z(zF' + bF)' + a(zF' + bF))\notag\\ &= z(z(F' + zF'' + bF') + azF' + abF)\notag\\ &= z(z^{2}F'' + (a + b + 1)zF' + abF)\notag\end{align} Hence we finally arrive at the following differential equation $$z(1 - z)F'' + (c - (a + b + 1)z)F' - abF = 0$$ or in full notation $$z(1 - z)\frac{d^{2}F}{dz^{2}} + \{c - (a + b + 1)z\}\frac{dF}{dz} - abF = 0$$ This differential equation turns out to be very useful in establishing certain transformation identities related to the hypergeometric function. Essentially the idea is that if two functions satisfy this above equation and are analytic in a certain region and agree on certain point in this region then they are equal identically in that region. We don't justify this procedure in detail, but it can be found in books dedicated to solution of linear differential equations.

We illustrate the above technique by establishing the quadratic transformation of Gauss. To begin with let's put $c = 2b$ so that $F(a, b; 2b; z)$ is a solution of $$z(1 - z)F'' + (2b - (a + b + 1)z)F' - abF = 0$$ We now use the quadratic transformation $z = 4x/(1 + x)^{2}$ and note that $$\frac{dF}{dz} = \dfrac{\dfrac{dF}{dx}}{\dfrac{dz}{dx}} = \frac{(1 + x)^{4}}{(1 + x)^{2}\cdot 4 - 4x \cdot 2(1 + x)}\frac{dF}{dx} = \frac{(1 + x)^{3}}{4(1 - x)}\frac{dF}{dx}$$ \begin{align}\frac{d^{2}F}{dz^{2}} &= \dfrac{\dfrac{d}{dx}\left(\dfrac{(1 + x)^{3}}{4(1 - x)}\dfrac{dF}{dx}\right)}{\dfrac{dz}{dx}}\notag\\ &= \frac{(1 + x)^{3}}{4(1 - x)}\left(\frac{(1 + x)^{3}}{4(1 - x)}\frac{d^{2}F}{dx^{2}} + \frac{1}{4}\frac{(1 - x)3(1 + x)^{2} + (1 + x)^{3}}{(1 - x)^{2}}\frac{dF}{dx}\right)\notag\\ &= \frac{(1 + x)^{3}}{4(1 - x)}\left(\frac{(1 + x)^{3}}{4(1 - x)}\frac{d^{2}F}{dx^{2}} + \frac{(2 - x)(1 + x)^{2}}{2(1 - x)^{2}}\frac{dF}{dx}\right)\notag\\ &= \frac{(1 + x)^{5}}{16(1 - x)^{3}}\left((1 - x^{2})\frac{d^{2}F}{dx^{2}} + (4 - 2x)\frac{dF}{dx}\right)\notag\end{align} and $$z(1 - z) = \frac{4x(1 - x)^{2}}{(1 + x)^{4}}$$ Putting pieces together we have the equation $$x(1 - x)(1 + x)^{2}\frac{d^{2}F}{dx^{2}} + 2(1 + x)(b - 2ax + bx^{2} - x^{2})\frac{dF}{dx} - 4ab(1 - x)F = 0$$ which is having the solution $F(a, b; 2b; 4x/(1 + x)^{2})$. If we put $F = (1 + x)^{2a}G$ then after some manipulation we get $$x(1 - x^{2})\frac{d^{2}G}{dx^{2}} + 2\{b - (2a - b + 1)x^{2}\}\frac{dG}{dx} - 2ax(1 + 2a - 2b)G = 0$$ The above equation does not change when we replace $x$ by $-x$ and therefore it turns out that $G$ is an even function of $x$ and therefore a function of $x^{2}$. Putting $t = x^{2}$ and simplifying like before we get $$t(1 - t)\frac{d^{2}G}{dt^{2}} + \left(b + \frac{1}{2} - \left(2a - b + \frac{3}{2}\right)t\right)\frac{dG}{dt} - a\left(a - b + \frac{1}{2}\right)G = 0$$ Comparing with the standard differential equation satisfied by $F(a, b; c; z)$ we find that $G = F(a, a - b + 1/2; b + 1/2; t) = F(a, a - b + 1/2; b + 1/2; x^{2})$ is a solution. By the earlier substitutions used we can also see that $G = (1 + x)^{-2a}F(a, b; 2b; 4x/(1 + x)^{2})$ is also a solution. Both these solutions are analytic in neighborhood of $x = 0$ and are also equal to $1$ at $x = 0$. It turns out that we have the following quadratic transformation of Gauss: $$F\left(a, b; 2b; \frac{4x}{(1 + x)^{2}}\right) = (1 + x)^{2a}F\left(a, a - b + \frac{1}{2}; b + \frac{1}{2}; x^{2}\right)$$ The conditions for validity of the above relation turn out to be the ones which make the corresponding series convergent, i.e. $2b$ should not be zero or negative integer and we must have $|x| < 1, |4x/(1 + x)^{2}| < 1$.

The reader who is familiar with the theory of elliptic integrals presented in my posts will not fail to notice that this is a generalization of the Landen's Transformation which is obtained by putting $a = b = 1/2$ and replacing $x$ by $k$. This leads to another proof of the formula $$K\left(\frac{2\sqrt{k}}{1 + k}\right) = (1 + k)K(k)$$ We will use the same technique to derive two more transformations of the hypergeometric functions, but let's keep that for the next post.