Elementary Approach to Modular Equations: Ramanujan's Theory 2

Ramanujan's Theory of Elliptic Functions

Ramanujan used the letter $ x$ in place of $ k^{2}$ and studied the function $ {}_{2}F_{1}(1/2, 1/2; 1; x)$ in great detail and developed his theory of elliptic integrals and functions.

Following Ramanujan let's define \begin{align} x &= k^{2}\notag\\ z &=\,{}_{2}F_{1}\left(\frac{1}{2},\frac{1}{2}; 1; x\right)\notag\\ y &= \pi\cdot\dfrac{{}_{2}F_{1}\left(\dfrac{1}{2},\dfrac{1}{2}; 1; 1 - x\right)}{{}_{2}F_{1}\left(\dfrac{1}{2},\dfrac{1}{2}; 1; x\right)}\notag\end{align} For Ramanujan elliptic function theory consisted of finding relations among $ x, y, z$. As can be easily seen from the above definitions $ z$ corresponds to $ K$ and $ y$ corresponds to $ \pi K'/K$ so that the nome $ q$ is given by $ q = e^{-y}$.

Ramanujan in fact studied the function $ F(x) = e^{-y}$ and obtained the fundamental formulas connecting $ x, y, z$. In doing so Ramanujan went far ahead of his predecessors and developed the theory of theta functions to alternative bases. Using various identities in the theory of hypergeometric functions he studied the properties of $ u_{x} =\,_{2}F_{1}(1 - n, n; 1; x)$ for non-integral $ n$ and the associated function (which is nome) $$\exp\left(-\frac{\pi}{\sin(\pi n)}\frac{u_{1 - x}}{u_{x}}\right)$$ For $ n = 1/2$ this reduces to the classical theory. Also the factor $ \pi/\sin(\pi n)$ is actually $ \Gamma(n)\Gamma(1 - n)$. This is the source of the constant $ \pi$ in the formula for nome. Apart from the classical case $ n = 1/2$ Ramanujan developed the theories for $ n = 1/3, 1/4, 1/6$.

We will restrict ourselves however to the classical case when $ n = 1/2$. Let us then define along the lines of Ramanujan the function $ F(x)$ by $$F(x) = e^{-y} = \exp\left(-\pi\,\dfrac{{}_{2}F_{1}\left(\dfrac{1}{2},\dfrac{1}{2};1; 1 - x\right)}{{}_{2}F_{1}\left(\dfrac{1}{2},\dfrac{1}{2};1; x\right)}\right)$$

Behavior of $ F(x)$ Near $ x = 0$

We first establish the important result that $ F(x)/x$ tends to positive constant value as $ x \to 0+$. In other words we wish to prove that $ F(x) = x(A + o(1))$ where $ A$ is some positive constant. This is equivalent to proving $$-\pi\,\dfrac{{}_{2}F_{1}\left(\dfrac{1}{2},\dfrac{1}{2};1; 1 - x\right)}{{}_{2}F_{1}\left(\dfrac{1}{2},\dfrac{1}{2};1; x\right)} = \log x + \log (A + o(1)) = \log A + \log x + o(1)$$ Since the denominator on the left tends to $ 1$ as $ x \to 0+$ it is sufficient to prove that $$-\pi\,{}_{2}F_{1}\left(\dfrac{1}{2},\dfrac{1}{2};1; 1 - x\right) = \log x + C + o(1)$$ where $ C = \log A$ is some constant. Replacing $ x$ by $ (1 - x)$ we see that we need to prove $$\pi\,{}_{2}F_{1}\left(\dfrac{1}{2},\dfrac{1}{2};1; x\right) = -\log (1 - x) - C + o(1)$$ when $ x \to 1-$. In the language of elliptic integrals this means that $$2K(\sqrt{x}) = -\log(1 - x) - C + o(1)$$ Replacing $ x$ by $ 1 - k^{2}$ we see that $$2K(k') = -2\log k - C + o(1) $$ where $ k \to 0+$ or in other words $$K(k') = K' = -\log k - (C/2) + o(1)$$ We will now establish that $$ K' = \log \left(\frac{4}{k}\right) + o(1)$$ so that $ C = -\log 16$ and $ A = e^{C} = 1/16$. We proceed by starting with the definition of $ K'$: \begin{align} K' &= K(k') = \int_{0}^{\pi / 2} \frac{d\theta}{\sqrt{1 - (1 - k^{2})\sin^{2}\theta}}\notag\\ &= \int_{0}^{\pi / 2} \frac{k'\sin\theta\,d\theta}{\sqrt{1 - (1 - k^{2})\sin^{2}\theta}} + \int_{0}^{\pi / 2} \frac{1 - k'\sin\theta\,d\theta}{\sqrt{1 - (1 - k^{2})\sin^{2}\theta}}\notag\\ &= \int_{0}^{\pi / 2} \frac{k'\sin\theta\,d\theta}{\sqrt{k^{2} + k'^{2}\cos^{2}\theta}} + \int_{0}^{\pi / 2} \frac{1 - k'\sin\theta\,d\theta}{\sqrt{1 - k'^{2}\sin^{2}\theta}}\notag\\ &= \int_{0}^{k'}\frac{dt}{\sqrt{t^{2} + k^{2}}} + \int_{0}^{\pi / 2}\sqrt{\frac{1 - k'\sin\theta}{1 + k'\sin\theta}}\,d\theta\notag\\ &= \log\left(\frac{1 + k'}{k}\right) + \int_{0}^{\pi / 2}\sqrt{\frac{1 - k'\sin\theta}{1 + k'\sin\theta}}\,d\theta\notag\end{align} If we denote the integral on the right as $ a(k)$ we can see that by uniformity arguments $ a(k) \to a(0)$ as $ k \to 0+$. And value of $ a(0)$ is easily calculated as: $$a(0) = \int_{0}^{\pi / 2}\sqrt{\frac{1 - \sin\theta}{1 + \sin\theta}}\,d\theta = \int_{0}^{\pi / 2}\frac{\cos\theta}{1 + \sin\theta}\,d\theta = \log 2$$ Therefore we have the following relation \begin{align}K' &= \log\left(\frac{2(1 + k')}{k}\right) + o(1)\notag\\ &= \log\left(\frac{4}{k}\right) + \log\left(\frac{1 + k'}{2}\right) + o(1)\notag\\ &= \log\left(\frac{4}{k}\right) + o(1) \notag\end{align} It now follows that $$\lim_{x \to 0+}\frac{F(x)}{x} = \frac{1}{16}$$ or in a grander form $$\lim_{x \to 0+}\frac{1}{x}\,\exp\left(-\pi\,\dfrac{{}_{2}F_{1}\left(\dfrac{1}{2},\dfrac{1}{2};1; 1 - x\right)}{{}_{2}F_{1}\left(\dfrac{1}{2},\dfrac{1}{2};1; x\right)}\right) = \frac{1}{16}$$ This is the key result which is used in Ramanujan's elliptic function theory. In the next post we use this result and the transformation fomulas of hypergeometric functions to develop Ramanujan's theory.

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