Elementary Approach to Modular Equations: Ramanujan's Theory 3

Connection between Theta Functions and Hypergeometric Functions

Let's recall the Gauss Transformation formula from an earlier post: $$F\left(a, b; 2b; \frac{4x}{(1 + x)^{2}}\right) = (1 + x)^{2a}F\left(a, a - b + \frac{1}{2}; b + \frac{1}{2}; x^{2}\right)$$ where $F$ is the hypergeometric function ${}_{2}F_{1}$. Putting $a = b = 1/2$ we get $${}_{2}F_{1}\left(\frac{1}{2}, \frac{1}{2}; 1; \frac{4x}{(1 + x)^{2}}\right) = (1 + x)\,{}_{2}F_{1}\left(\frac{1}{2}, \frac{1}{2}; 1; x^{2}\right)$$ or $${}_{2}F_{1}\left(\frac{1}{2}, \frac{1}{2}; 1; 1 - \left(\frac{1 - x}{1 + x}\right)^{2}\right) = (1 + x)\,{}_{2}F_{1}\left(\frac{1}{2}, \frac{1}{2}; 1; x^{2}\right)$$
If we set $$\frac{1 - x}{1 + x} = \frac{\phi^{2}(-q)}{\phi^{2}(q)}$$ where $\phi(q) = \sum_{-\infty}^{\infty}q^{n^{2}}$ is one of the theta functions defined by Ramanujan (see this post) so that $$\begin{align}x &= \frac{\phi^{2}(q) - \phi^{2}(-q)}{\phi^{2}(q) + \phi^{2}(-q)} = \frac{4q\psi^{2}(q^{4})}{\phi^{2}(q^{2})}\notag\\ \Rightarrow 1 + x &= \frac{2\phi^{2}(q)}{\phi^{2}(q) + \phi^{2}(-q)} = \frac{\phi^{2}(q)}{\phi^{2}(q^{2})}\notag\end{align}$$ and $$x^{2} = \frac{16q^{2}\psi^{4}(q^{4})}{\phi^{4}(q^{2})} = \frac{\phi^{4}(q^{2}) - \phi^{4}(-q^{2})}{\phi^{4}(q^{2})} = 1 - \frac{\phi^{4}(-q^{2})}{\phi^{4}(q^{2})}$$ Therefore we finally obtain: $${}_{2}F_{1}\left(\frac{1}{2}, \frac{1}{2}; 1; 1 - \frac{\phi^{4}(-q)}{\phi^{4}(q)}\right) = \frac{\phi^{2}(q)}{\phi^{2}(q^{2})}\,{}_{2}F_{1}\left(\frac{1}{2}, \frac{1}{2}; 1; 1 - \frac{\phi^{4}(-q^{2})}{\phi^{4}(q^{2})}\right)$$ The beauty of this formula is that it can be applied recursively. The hypergeometric function on the right is the same as that on the left except that the parameter $q$ is replaced by $q^{2}$. Also the first factor on the right consists of parameter $q$ in numerator and $q^{2}$ in denominator so that the recursion can be utilized quite nicely to lead us to the following result: $${}_{2}F_{1}\left(\frac{1}{2}, \frac{1}{2}; 1; 1 - \frac{\phi^{4}(-q)}{\phi^{4}(q)}\right) = \frac{\phi^{2}(q)}{\phi^{2}(q^{2^{n}})}\,{}_{2}F_{1}\left(\frac{1}{2}, \frac{1}{2}; 1; 1 - \frac{\phi^{4}(-q^{2^{n}})}{\phi^{4}(q^{2^{n}})}\right)\tag{1}$$ When $n \to \infty, q^{2^{n}} \to 0$ and therefore $\phi(q^{2^{n}}) \to 1,\,\phi(-q^{2^{n}}) \to 1$ so that we obtain $$\phi^{2}(q) =\,{}_{2}F_{1}\left(\frac{1}{2}, \frac{1}{2}; 1; 1 - \frac{\phi^{4}(-q)}{\phi^{4}(q)}\right)\tag{2}$$ We can use the Gauss transformation formula again by setting $$x = \frac{\phi^{2}(-q)}{\phi^{2}(q)}$$ and noting that $$\frac{4x}{(1 + x)^{2}} = 4\,\frac{\phi^{2}(-q)}{\phi^{2}(q)}\frac{\phi^{4}(q)}{(\phi^{2}(q) + \phi^{2}(-q))^{2}} = \frac{\phi^{2}(q)\phi^{2}(-q)}{\phi^{4}(q^{2})} = \frac{\phi^{4}(-q^{2})}{\phi^{4}(q^{2})}$$ and $$1 + x = \frac{2\phi^{2}(q^{2})}{\phi^{2}(q)}$$ Thus we arrive at $${}_{2}F_{1}\left(\frac{1}{2}, \frac{1}{2}; 1; \frac{\phi^{4}(-q^{2})}{\phi^{4}(q^{2})}\right) = \frac{2\phi^{2}(q^{2})}{\phi^{2}(q)}\,{}_{2}F_{1}\left(\frac{1}{2}, \frac{1}{2}; 1; \frac{\phi^{4}(-q)}{\phi^{4}(q)}\right)$$ or $${}_{2}F_{1}\left(\frac{1}{2}, \frac{1}{2}; 1; \frac{\phi^{4}(-q)}{\phi^{4}(q)}\right) = \frac{\phi^{2}(q)}{2\phi^{2}(q^{2})}\,{}_{2}F_{1}\left(\frac{1}{2}, \frac{1}{2}; 1; \frac{\phi^{4}(-q^{2})}{\phi^{4}(q^{2})}\right)$$ By applying recursion we get $${}_{2}F_{1}\left(\frac{1}{2}, \frac{1}{2}; 1; \frac{\phi^{4}(-q)}{\phi^{4}(q)}\right) = \frac{\phi^{2}(q)}{2^{n}\phi^{2}(q^{2^{n}})}\,{}_{2}F_{1}\left(\frac{1}{2}, \frac{1}{2}; 1; \frac{\phi^{4}(-q^{2^{n}})}{\phi^{4}(q^{2^{n}})}\right)\tag{3}$$ Dividing $(1)$ by $(3)$ and replacing $2^{n}$ by $m$ we get $$\dfrac{{}_{2}F_{1}\left(\dfrac{1}{2}, \dfrac{1}{2}; 1; 1 - \dfrac{\phi^{4}(-q)}{\phi^{4}(q)}\right)}{{}_{2}F_{1}\left(\dfrac{1}{2}, \dfrac{1}{2}; 1; \dfrac{\phi^{4}(-q)}{\phi^{4}(q)}\right)} = m\,\dfrac{{}_{2}F_{1}\left(\dfrac{1}{2}, \dfrac{1}{2}; 1; 1 - \dfrac{\phi^{4}(-q^{m})}{\phi^{4}(q^{m})}\right)}{{}_{2}F_{1}\left(\dfrac{1}{2}, \dfrac{1}{2}; 1; \dfrac{\phi^{4}(-q^{m})}{\phi^{4}(q^{m})}\right)}\tag{4}$$ Mutiplying by $-\pi$ and taking exponentials on both sides we get $$F\left(\frac{\phi^{4}(-q)}{\phi^{4}(q)}\right) = \left\{F\left(\frac{\phi^{4}(-q^{m})}{\phi^{4}(q^{m})}\right)\right\}^{m}$$ where $m$ is a positive integral power of $2$ and $F(x)$ represents the Ramanujan's function defined in previous post.

The same equation $(4)$ on taking reciprocals leads us to $$\left\{F\left(1 - \frac{\phi^{4}(-q)}{\phi^{4}(q)}\right)\right\}^{m} = F\left(1 - \frac{\phi^{4}(-q^{m})}{\phi^{4}(q^{m})}\right)\tag{5}$$

The Inversion Fomula

Using the above result we can prove the fundamental inversion formula $$F\left(1 - \frac{\phi^{4}(-q)}{\phi^{4}(q)}\right) = q\tag{6}$$ Clearly if we set $$x_{m} = \frac{\phi^{4}(-q^{m})}{\phi^{4}(q^{m})}$$ then $x_{m} \to 1$ as $m \to \infty$ and hence $1 - x_{m} \to 0$. Clearly then we have $$F(1 - x_{m}) \sim \frac{1 - x_{m}}{16}$$ and hence $$\begin{align}F\left(1 - \frac{\phi^{4}(-q)}{\phi^{4}(q)}\right) &= \sqrt[m]{F(1 - x_{m})}\notag\\ &= \lim_{m \to \infty}\sqrt[m]{F(1 - x_{m})}\notag\\ &= \lim_{m \to \infty}\sqrt[m]{\frac{1 - x_{m}}{16}}\notag\\ &= \lim_{m \to \infty}\sqrt[m]{1 - x_{m}} = A\notag\end{align}$$ Then we have $$\begin{align} \log A &= \lim_{m \to \infty}\frac{1}{m}\log(1 - x_{m})\notag\\ &= \lim_{m \to \infty}\frac{1}{m}\log\left(1 - \frac{\phi^{4}(-q^{m})}{\phi^{4}(q^{m})}\right)\notag\\ &= \lim_{m \to \infty}\frac{1}{m}\log\left(\frac{\phi^{4}(q^{m}) - \phi^{4}(-q^{m})}{\phi^{4}(q^{m})}\right)\notag\\ &= \lim_{m \to \infty}\frac{1}{m}\log\left(\frac{16q^{m}\psi^{4}(q^{2m})}{\phi^{4}(q^{m})}\right)\notag\\ &= \lim_{m \to \infty}\log q + \frac{\log 16}{m} + \log\left(\frac{\psi^{4}(q^{2m})}{\phi^{4}(q^{m})}\right)\notag\\ &= \log q \notag\end{align}$$ And thus we have $A = q$ and the inversion formula is established. In other words if $$x = 1 - \frac{\phi^{4}(-q)}{\phi^{4}(q)}$$ then $$q = F(x)$$ Using $(6)$ in $(2)$ we get $$\phi^{2}\left\{F\left(1 - \frac{\phi^{4}(-q)}{\phi^{4}(q)}\right)\right\} =\,{}_{2}F_{1}\left(\frac{1}{2}, \frac{1}{2}; 1; 1 - \frac{\phi^{4}(-q)}{\phi^{4}(q)}\right)$$ or $$\phi^{2}(F(x)) =\,{}_{2}F_{1}\left(\frac{1}{2}, \frac{1}{2}; 1; x\right)$$ where $$x = 1 - \frac{\phi^{4}(-q)}{\phi^{4}(q)}$$ Using the Ramanujan's notation $x, y, z$ we now have $$\phi(F(x)) = \phi(e^{-y}) = \sqrt{z}$$

Transformation Formula for $\phi(q)$

If we have a look at the definitions of $x, y$ and $F(x)$ then we can see that $$\log F(x) = -y$$ and $$\log F(x) \cdot \log F(1 - x) = \pi^{2}$$ so that $$\log F(1 - x) = -\pi^{2} / y$$ or $$F(1 - x) = e^{-\pi^{2} / y}$$ Let $\alpha,\beta$ be positive numbers such that $\alpha\beta = \pi$ and let $y = \alpha^{2}$ so that $\pi^{2}/y = \beta^{2}$. Then we have $$\frac{\phi^{2}(F(1 - x))}{\phi^{2}(F(x))} = \dfrac{{}_{2}F_{1}\left(\dfrac{1}{2},\dfrac{1}{2}; 1; 1 - x\right)}{{}_{2}F_{1}\left(\dfrac{1}{2},\dfrac{1}{2}; 1; x\right)} = \frac{y}{\pi} = \frac{\alpha^{2}}{\alpha\beta} = \frac{\alpha}{\beta}$$ or $$\frac{\phi^{2}(e^{-\beta^{2}})}{\phi^{2}(e^{-\alpha^{2}})} = \frac{\alpha}{\beta}$$ or $$\sqrt{\alpha}\,\phi(e^{-\alpha^{2}}) = \sqrt{\beta}\,\phi(e^{-\beta^{2}})$$ In the classical notation this is $$\sqrt{s}\,\theta_{3}(e^{-\pi s}) = \theta_{3}(e^{-\pi / s})$$ (see this post) and therefore the above constitutes a proof of the transformation formula for theta functions without the use of Poisson Summation formula.

In the next post we will have a look at various identities relating the theta functions of $q$ and $q^{n}$ and these will be finally transcribed in the form of modular equations. Most of the identities will be derived using Lambert series for the theta functions.

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