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(a,b;2b;4x(1+x)2)=(1+x)2aF(a,ab+12;b+12;x2) where F is the hypergeometric function 2F1. Putting a=b=1/2 we get 2F1(12,12;1;4x(1+x)2)=(1+x)2F1(12,12;1;x2) or 2F1(12,12;1;1(1x1+x)2)=(1+x)2F1(12,12;1;x2)
If we set 1x1+x=ϕ2(q)ϕ2(q) where ϕ(q)=qn2 is one of the theta functions defined by Ramanujan (see this post) so that x=ϕ2(q)ϕ2(q)ϕ2(q)+ϕ2(q)=4qψ2(q4)ϕ2(q2)1+x=2ϕ2(q)ϕ2(q)+ϕ2(q)=ϕ2(q)ϕ2(q2) and x2=16q2ψ4(q4)ϕ4(q2)=ϕ4(q2)ϕ4(q2)ϕ4(q2)=1ϕ4(q2)ϕ4(q2) Therefore we finally obtain: 2F1(12,12;1;1ϕ4(q)ϕ4(q))=ϕ2(q)ϕ2(q2)2F1(12,12;1;1ϕ4(q2)ϕ4(q2)) 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 q2. Also the first factor on the right consists of parameter q in numerator and q2 in denominator so that the recursion can be utilized quite nicely to lead us to the following result: 2F1(12,12;1;1ϕ4(q)ϕ4(q))=ϕ2(q)ϕ2(q2n)2F1(12,12;1;1ϕ4(q2n)ϕ4(q2n)) When n,q2n0 and therefore ϕ(q2n)1,ϕ(q2n)1 so that we obtain ϕ2(q)=2F1(12,12;1;1ϕ4(q)ϕ4(q)) We can use the Gauss transformation formula again by setting x=ϕ2(q)ϕ2(q) and noting that 4x(1+x)2=4ϕ2(q)ϕ2(q)ϕ4(q)(ϕ2(q)+ϕ2(q))2=ϕ2(q)ϕ2(q)ϕ4(q2)=ϕ4(q2)ϕ4(q2) and 1+x=2ϕ2(q2)ϕ2(q) Thus we arrive at 2F1(12,12;1;ϕ4(q2)ϕ4(q2))=2ϕ2(q2)ϕ2(q)2F1(12,12;1;ϕ4(q)ϕ4(q)) or 2F1(12,12;1;ϕ4(q)ϕ4(q))=ϕ2(q)2ϕ2(q2)2F1(12,12;1;ϕ4(q2)ϕ4(q2)) By applying recursion we get 2F1(12,12;1;ϕ4(q)ϕ4(q))=ϕ2(q)2nϕ2(q2n)2F1(12,12;1;ϕ4(q2n)ϕ4(q2n)) Dividing (1) by (3) and replacing 2n by m we get 2F1(12,12;1;1ϕ4(q)ϕ4(q))2F1(12,12;1;ϕ4(q)ϕ4(q))=m2F1(12,12;1;1ϕ4(qm)ϕ4(qm))2F1(12,12;1;ϕ4(qm)ϕ4(qm)) Mutiplying by π and taking exponentials on both sides we get F(ϕ4(q)ϕ4(q))={F(ϕ4(qm)ϕ4(qm))}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 {F(1ϕ4(q)ϕ4(q))}m=F(1ϕ4(qm)ϕ4(qm))

The Inversion Fomula

Using the above result we can prove the fundamental inversion formula F(1ϕ4(q)ϕ4(q))=q Clearly if we set xm=ϕ4(qm)ϕ4(qm) then xm1 as m and hence 1xm0. Clearly then we have F(1xm)1xm16 and hence F(1ϕ4(q)ϕ4(q))=mF(1xm)=lim 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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