Ramanujan developed his theory of modular equations using the theory of theta functions independently of Jacobi. A complete understanding of his approach is unfortunately not possible till now because he did not publish something like Fundamenta Nova containing detailed explanations of his approach. What we have today is his Notebooks edited by Bruce C. Berndt and his Collected Papers. His Notebooks are just statements of various mathematical formulas without any proof. A large part of these notebooks is concerned with modular equations and modern authors have not been able to discern his methods fully. Hence I will not be able to present a true picture of his approach. Rather I will try to present whatever I understand from his Collected Papers and his Notebooks and only focus on the elementary aspects.
Ramanujan's Theta Functions
Ramanujan started his theory of modular equations from the relation $ L'/L = nK'/K$ and worked with the parameter $ q = e^{-\pi K'/K}$. The modulus $ k$ can be expressed in form of theta functions of parameter $ q$. Thus we have $ k = k(q) = \theta_{2}^{2}(q)/\theta_{3}^{2}(q)$. From the equation $ L'/L = nK'/K$ we immediately see that $ l = \theta_{2}^{2}(q^{n})/\theta_{3}^{2}(q^{n})$. Thus an algebraic relation between $ k$ and $ l$ is actually a relation between theta functions of parameter $ q$ and $ q^{n}$. Ramanujan was such
an expert in formal manipulations of series, products and continued fractions that he readily obtained many relations between theta functions of $ q$ and $ q^{n}$ and then transformed them into the traditional relation between $ k, l$. Thus his modular equations are quite diverse in nature and sometimes they also contain more than 2 moduli.
Following Ramanujan we define the Ramanujan Theta Function $ f(a, b)$ by $$f(a, b) = \sum_{n = -\infty}^{\infty}a^{n(n + 1)/2}b^{n(n - 1)/2} = 1 + \sum_{n = 1}^{\infty}(ab)^{n(n - 1)/2}(a^{n} + b^{n})$$ for all complex numbers $ a, b$ with $ |ab| < 1$. This function has the following elementary properties
For the second property we have \begin{align} f(-1, a) &= f(a, -1) = \sum_{n = -\infty}^{\infty}a^{n(n + 1)/2}(-1)^{n(n - 1)/2}\notag\\ &= \sum_{n = -\infty}^{n = 0}a^{n(n + 1)/2}(-1)^{n(n - 1)/2} + \sum_{n = 1}^{\infty}a^{n(n + 1)/2}(-1)^{n(n - 1)/2}\notag\\ &= \sum_{n = 0}^{\infty}(-1)^{n(n + 1)/2}a^{n(n - 1)/2} + \sum_{n = 1}^{\infty}(-1)^{n(n - 1)/2}a^{n(n + 1)/2}\notag\\ &= 1 - 1 + \sum_{n = 2}^{\infty}(-1)^{n(n + 1)/2}a^{n(n - 1)/2} + \sum_{n = 1}^{\infty}(-1)^{n(n - 1)/2}a^{n(n + 1)/2}\notag\\ &= \sum_{n = 1}^{\infty}(-1)^{(n + 1)(n + 2)/2}a^{n(n + 1)/2} + \sum_{n = 1}^{\infty}(-1)^{n(n - 1)/2}a^{n(n + 1)/2}\notag\end{align} Now $$ \frac{(n + 1)(n + 2)}{2} - \frac{n(n - 1)}{2} = \frac{n^{2} + 3n + 2 - n^{2} + n}{2} = 2n + 1$$ so that $ (-1)^{(n + 1)(n + 2)/2}$ and $ (-1)^{n(n - 1)/2}$ are of opposite signs and hence the sum cancels to zero.
To handle $ f(1, a)$ we have \begin{align} f(1, a) &= \sum_{n = -\infty}^{\infty}a^{n(n + 1)/2}\notag\\ &= 1 + 1 + \sum_{n = -\infty}^{-2}a^{n(n + 1)/2} + \sum_{n = 1}^{\infty}a^{n(n + 1)/2}\notag\\ &= 2 + \sum_{n = 1}^{\infty}a^{n(n + 1)/2} + \sum_{n = 2}^{\infty}a^{n(n - 1)/2}\notag\\ &= 2 + \sum_{n = 1}^{\infty}a^{n(n + 1)/2} + \sum_{n = 1}^{\infty}a^{n(n + 1)/2}\notag\\ &= 2\left(1 + \sum_{n = 1}^{\infty}a^{n(n + 1)/2}\right)\notag\end{align} Splitting the sum on right into even and odd indices we get \begin{align} f(1, a) &= 2\left(1 + \sum_{n = 1}^{\infty}a^{n(2n + 1)} + \sum_{n = 1}^{\infty}a^{n(2n - 1)}\right)\notag\\ &= 2\left(1 + \sum_{n = 1}^{\infty}a^{n(n - 1)/2}(a^{3})^{n(n + 1)/2} + \sum_{n = 1}^{\infty}a^{n(n + 1)/2}(a^{3})^{n(n - 1)/2}\right) = 2f(a, a^{3})\notag\end{align} The last equation can be handled as follows: \begin{align} f(a, b) &= \sum_{k = -\infty}^{\infty}a^{k(k + 1)/2}b^{k(k - 1)/2}\notag\\ &= \sum_{k = -\infty}^{\infty}a^{(k + n)(k + n + 1)/2}b^{(k + n)(k + n - 1)/2}\notag\\ &= a^{n(n + 1)/2}b^{n(n - 1)/2}\sum_{k = -\infty}^{\infty}a^{k(k + 2n + 1)/2}b^{k(k + 2n - 1)/2}\notag\\ &= a^{n(n + 1)/2}b^{n(n - 1)/2}\sum_{k = -\infty}^{\infty}\{a(ab)^{n}\}^{k(k + 1)/2}\{b(ab)^{-n}\}^{k(k - 1)/2}\notag\\ &= a^{n(n + 1)/2}b^{n(n - 1)/2}f(a(ab)^{n}, b(ab)^{-n})\notag\end{align} If we put $ a = qe^{2iz}, b = qe^{-2iz}$ we obtain $ f(a, b) = \sum_{n = -\infty}^{\infty}q^{n^{2}}e^{2inz} = \theta_{3}(z, q)$ which gives the link between Ramanujan's theta function and the classical Jacobi's theta function.
The Jacobi's Triple Product identity is given by $$\theta_{3}(z, q) = \sum_{n = -\infty}^{\infty}q^{n^{2}}e^{2inz} = \prod_{n = 1}^{\infty}(1 - q^{2n})(1 + q^{2n - 1}e^{2iz})(1 + q^{2n - 1}e^{-2iz})$$ and this transforms to the following when we put $ a = qe^{2iz}, b = qe^{-2iz}, q^{2} = ab$ \begin{align}f(a, b) &= \prod_{n = 1}^{\infty}(1 - (ab)^{n})(1 + q^{2n - 2}a)(1 + q^{2n - 2}b)\notag\\ &= \prod_{n = 1}^{\infty}(1 - (ab)^{n})(1 + a(ab)^{n - 1})(1 + b(ab)^{n - 1})\notag\\ &= (-a;ab)_{\infty}(-b;ab)_{\infty}(ab;ab)_{\infty}\notag\end{align} where we have the notation \begin{align}(a; q)_{n} &= \prod_{k = 1}^{n}(1 - aq^{k - 1})\notag\\ (a; q)_{\infty} &= \prod_{k = 1}^{\infty}(1 - aq^{k - 1})\notag\end{align} Next Ramanujan defines his other theta functions in terms of $ f(a, b)$:
Also it should be noted that the function $ \psi(q)$ is the equivalent of classical $ \theta_{2}(q)$ and we have $ \theta_{2}(q) = 2q^{1/4}\psi(q^{2})$. The functions $ \chi(q)$ and $ \chi(-q)$ will be used later in the definitions Ramanujan's invariants.
Finally \begin{align} \phi(-q)\psi^{2}(q) &= \frac{(q;q^{2})_{\infty}(q^{2};q^{2})_{\infty}}{(-q;q^{2})_{\infty}(-q^{2};q^{2})_{\infty}}\frac{(q^{2};q^{2})_{\infty}^{2}}{(q;q^{2})_{\infty}^{2}}\notag\\ &= \frac{(q^{2};q^{2})_{\infty}^{3}}{(-q;q)_{\infty}(q;q^{2})_{\infty}} = (q^{2};q^{2})_{\infty}^{3} = f^{3}(-q^{2})\notag\end{align} What we see in the proofs of the above results is that they are really elementary and depend upon formal manipulation of series and products. The only non-trivial identity we have used till now is the Jacobi's Triple product which forms the basis of all the above product expansions and the relations between these Ramanujan theta functions.
Next we establish some identities connecting theta functions of $ q, q^{2}$ and $ q^{4}$. These can be seen as equivalents of the classical theta function identities established here.
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Ramanujan's Theta Functions
Ramanujan started his theory of modular equations from the relation $ L'/L = nK'/K$ and worked with the parameter $ q = e^{-\pi K'/K}$. The modulus $ k$ can be expressed in form of theta functions of parameter $ q$. Thus we have $ k = k(q) = \theta_{2}^{2}(q)/\theta_{3}^{2}(q)$. From the equation $ L'/L = nK'/K$ we immediately see that $ l = \theta_{2}^{2}(q^{n})/\theta_{3}^{2}(q^{n})$. Thus an algebraic relation between $ k$ and $ l$ is actually a relation between theta functions of parameter $ q$ and $ q^{n}$. Ramanujan was such
an expert in formal manipulations of series, products and continued fractions that he readily obtained many relations between theta functions of $ q$ and $ q^{n}$ and then transformed them into the traditional relation between $ k, l$. Thus his modular equations are quite diverse in nature and sometimes they also contain more than 2 moduli.Following Ramanujan we define the Ramanujan Theta Function $ f(a, b)$ by $$f(a, b) = \sum_{n = -\infty}^{\infty}a^{n(n + 1)/2}b^{n(n - 1)/2} = 1 + \sum_{n = 1}^{\infty}(ab)^{n(n - 1)/2}(a^{n} + b^{n})$$ for all complex numbers $ a, b$ with $ |ab| < 1$. This function has the following elementary properties
- $ f(a, b) = f(b, a)$
- $ f(-1, a) = 0$
- $ f(1, a) = 2f(a, a^{3})$
- $ f(a, b) = a^{n(n + 1)/2}b^{n(n - 1)/2}f(a(ab)^{n}, b(ab)^{-n})$ where $ n$ is an integer
For the second property we have \begin{align} f(-1, a) &= f(a, -1) = \sum_{n = -\infty}^{\infty}a^{n(n + 1)/2}(-1)^{n(n - 1)/2}\notag\\ &= \sum_{n = -\infty}^{n = 0}a^{n(n + 1)/2}(-1)^{n(n - 1)/2} + \sum_{n = 1}^{\infty}a^{n(n + 1)/2}(-1)^{n(n - 1)/2}\notag\\ &= \sum_{n = 0}^{\infty}(-1)^{n(n + 1)/2}a^{n(n - 1)/2} + \sum_{n = 1}^{\infty}(-1)^{n(n - 1)/2}a^{n(n + 1)/2}\notag\\ &= 1 - 1 + \sum_{n = 2}^{\infty}(-1)^{n(n + 1)/2}a^{n(n - 1)/2} + \sum_{n = 1}^{\infty}(-1)^{n(n - 1)/2}a^{n(n + 1)/2}\notag\\ &= \sum_{n = 1}^{\infty}(-1)^{(n + 1)(n + 2)/2}a^{n(n + 1)/2} + \sum_{n = 1}^{\infty}(-1)^{n(n - 1)/2}a^{n(n + 1)/2}\notag\end{align} Now $$ \frac{(n + 1)(n + 2)}{2} - \frac{n(n - 1)}{2} = \frac{n^{2} + 3n + 2 - n^{2} + n}{2} = 2n + 1$$ so that $ (-1)^{(n + 1)(n + 2)/2}$ and $ (-1)^{n(n - 1)/2}$ are of opposite signs and hence the sum cancels to zero.
To handle $ f(1, a)$ we have \begin{align} f(1, a) &= \sum_{n = -\infty}^{\infty}a^{n(n + 1)/2}\notag\\ &= 1 + 1 + \sum_{n = -\infty}^{-2}a^{n(n + 1)/2} + \sum_{n = 1}^{\infty}a^{n(n + 1)/2}\notag\\ &= 2 + \sum_{n = 1}^{\infty}a^{n(n + 1)/2} + \sum_{n = 2}^{\infty}a^{n(n - 1)/2}\notag\\ &= 2 + \sum_{n = 1}^{\infty}a^{n(n + 1)/2} + \sum_{n = 1}^{\infty}a^{n(n + 1)/2}\notag\\ &= 2\left(1 + \sum_{n = 1}^{\infty}a^{n(n + 1)/2}\right)\notag\end{align} Splitting the sum on right into even and odd indices we get \begin{align} f(1, a) &= 2\left(1 + \sum_{n = 1}^{\infty}a^{n(2n + 1)} + \sum_{n = 1}^{\infty}a^{n(2n - 1)}\right)\notag\\ &= 2\left(1 + \sum_{n = 1}^{\infty}a^{n(n - 1)/2}(a^{3})^{n(n + 1)/2} + \sum_{n = 1}^{\infty}a^{n(n + 1)/2}(a^{3})^{n(n - 1)/2}\right) = 2f(a, a^{3})\notag\end{align} The last equation can be handled as follows: \begin{align} f(a, b) &= \sum_{k = -\infty}^{\infty}a^{k(k + 1)/2}b^{k(k - 1)/2}\notag\\ &= \sum_{k = -\infty}^{\infty}a^{(k + n)(k + n + 1)/2}b^{(k + n)(k + n - 1)/2}\notag\\ &= a^{n(n + 1)/2}b^{n(n - 1)/2}\sum_{k = -\infty}^{\infty}a^{k(k + 2n + 1)/2}b^{k(k + 2n - 1)/2}\notag\\ &= a^{n(n + 1)/2}b^{n(n - 1)/2}\sum_{k = -\infty}^{\infty}\{a(ab)^{n}\}^{k(k + 1)/2}\{b(ab)^{-n}\}^{k(k - 1)/2}\notag\\ &= a^{n(n + 1)/2}b^{n(n - 1)/2}f(a(ab)^{n}, b(ab)^{-n})\notag\end{align} If we put $ a = qe^{2iz}, b = qe^{-2iz}$ we obtain $ f(a, b) = \sum_{n = -\infty}^{\infty}q^{n^{2}}e^{2inz} = \theta_{3}(z, q)$ which gives the link between Ramanujan's theta function and the classical Jacobi's theta function.
The Jacobi's Triple Product identity is given by $$\theta_{3}(z, q) = \sum_{n = -\infty}^{\infty}q^{n^{2}}e^{2inz} = \prod_{n = 1}^{\infty}(1 - q^{2n})(1 + q^{2n - 1}e^{2iz})(1 + q^{2n - 1}e^{-2iz})$$ and this transforms to the following when we put $ a = qe^{2iz}, b = qe^{-2iz}, q^{2} = ab$ \begin{align}f(a, b) &= \prod_{n = 1}^{\infty}(1 - (ab)^{n})(1 + q^{2n - 2}a)(1 + q^{2n - 2}b)\notag\\ &= \prod_{n = 1}^{\infty}(1 - (ab)^{n})(1 + a(ab)^{n - 1})(1 + b(ab)^{n - 1})\notag\\ &= (-a;ab)_{\infty}(-b;ab)_{\infty}(ab;ab)_{\infty}\notag\end{align} where we have the notation \begin{align}(a; q)_{n} &= \prod_{k = 1}^{n}(1 - aq^{k - 1})\notag\\ (a; q)_{\infty} &= \prod_{k = 1}^{\infty}(1 - aq^{k - 1})\notag\end{align} Next Ramanujan defines his other theta functions in terms of $ f(a, b)$:
- $ \displaystyle \phi(q) = f(q, q) = \sum_{n = -\infty}^{\infty}q^{n^{2}} = \theta_{3}(q) = \frac{(-q;q^{2})_{\infty}(q^{2};q^{2})_{\infty}}{(q;q^{2})_{\infty}(-q^{2};q^{2})_{\infty}}$
- $ \displaystyle \psi(q) = f(q, q^{3}) = \sum_{n = 0}^{\infty}q^{n(n + 1)/2} = \frac{(q^{2};q^{2})_{\infty}}{(q;q^{2})_{\infty}}$
- $ \displaystyle f(-q) = f(-q, -q^{2}) = \sum_{n = -\infty}^{\infty}(-1)^{n}q^{n(3n + 1)/2} = (q;q)_{\infty}$
- $ \displaystyle \chi(q) = (-q;q^{2})_{\infty} = \prod_{n = 1}^{\infty}(1 + q^{2n - 1})$
Also it should be noted that the function $ \psi(q)$ is the equivalent of classical $ \theta_{2}(q)$ and we have $ \theta_{2}(q) = 2q^{1/4}\psi(q^{2})$. The functions $ \chi(q)$ and $ \chi(-q)$ will be used later in the definitions Ramanujan's invariants.
Theta Function Identities
Theformulas presented above provide the series as well as product expansions of these functions and it turns out that their product expansions are quite helpful in deriving an amazing number of results. Ramanujan uses this idea to the fullest extent as is shown by the following results:- $ \displaystyle \frac{f(q)}{f(-q)} = \frac{\psi(q)}{\psi(-q)} = \frac{\chi(q)}{\chi(-q)} = \sqrt{\frac{\phi(q)}{\phi(-q)}}$
- $ \displaystyle f^{3}(-q) = \phi^{2}(-q)\psi(q) = \sum_{n = 0}^{\infty}(-1)^{n}(2n + 1)q^{n(n + 1)/2}$
- $ \displaystyle \chi(q) = \frac{f(q)}{f(-q^{2})} = \sqrt[3]{\frac{\phi(q)}{\psi(-q)}} = \frac{\phi(q)}{f(q)} = \frac{f(-q^{2})}{\psi(-q)}$
- $ \displaystyle f^{3}(-q^{2}) = \phi(-q)\psi^{2}(q),\,\,\, \chi(q)\chi(-q) = \chi(-q^{2})$
Finally \begin{align} \phi(-q)\psi^{2}(q) &= \frac{(q;q^{2})_{\infty}(q^{2};q^{2})_{\infty}}{(-q;q^{2})_{\infty}(-q^{2};q^{2})_{\infty}}\frac{(q^{2};q^{2})_{\infty}^{2}}{(q;q^{2})_{\infty}^{2}}\notag\\ &= \frac{(q^{2};q^{2})_{\infty}^{3}}{(-q;q)_{\infty}(q;q^{2})_{\infty}} = (q^{2};q^{2})_{\infty}^{3} = f^{3}(-q^{2})\notag\end{align} What we see in the proofs of the above results is that they are really elementary and depend upon formal manipulation of series and products. The only non-trivial identity we have used till now is the Jacobi's Triple product which forms the basis of all the above product expansions and the relations between these Ramanujan theta functions.
Next we establish some identities connecting theta functions of $ q, q^{2}$ and $ q^{4}$. These can be seen as equivalents of the classical theta function identities established here.
- $ \displaystyle \phi(q) + \phi(-q) = 2\phi(q^{4})$
- $ \displaystyle \phi(q) - \phi(-q) = 4q\psi(q^{8})$
- $ \displaystyle \phi(q)\phi(-q) = \phi^{2}(-q^{2})$
- $ \displaystyle \psi(q)\psi(-q) = \psi(q^{2})\phi(-q^{2})$
- $ \displaystyle \phi(q)\psi(q^{2}) = \psi^{2}(q)$
- $ \displaystyle \phi^{2}(q) - \phi^{2}(-q) = 8q\psi^{2}(q^{4})$
- $ \displaystyle \phi^{2}(q) + \phi^{2}(-q) = 2\phi^{2}(q^{2})$
- $ \displaystyle \phi^{4}(q) - \phi^{4}(-q) = 16q\psi^{4}(q^{2})$
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