Elementary Approach to Modular Equations: Ramanujan's Theory 3

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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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