Elliptic Functions: Infinite Products

Preliminary Results

Let us consider the ascending Landen sequence of moduli $$\cdots < k_{-n} < k_{-(n - 1)} < \cdots < k_{-2} < k_{-1} < k_{0} = k < k_{1} < k_{2} < \cdots < k_{n} < \cdots$$ where $$k_{n + 1} = \frac{2\sqrt{k_{n}}}{1 + k_{n}},\,\, k_{n} = \frac{1 - k_{n + 1}'}{1 + k_{n + 1}'}$$ Then it can be checked easily that the sequence of complementary moduli in reverse order $$\cdots < k_{n}' < k_{n - 1}' < \cdots < k_{2}' < k_{1}' < k_{0}' = k' < k_{-1}' < k_{-2}' < \cdots < k_{-n}' < \cdots$$ also forms an ascending Landen sequence.

Also let us put $K_{n} = K(k_{n}), K_{n}' = K(k_{n}')$. We then have $K_{n + 1} = (1 + k_{n})K_{n}$ for all $n$. Hence we can write \begin{align} K_{0} &= (1 + k_{-1})K_{-1} = (1 + k_{-1})(1 + k_{-2})K_{-2}\notag\\ &= (1 + k_{-1})(1 + k_{-2}) \cdots (1 + k_{-n})K_{-n}\notag\\ \Rightarrow \frac{K_{0}}{K_{-n}} &= (1 + k_{-1})(1 + k_{-2}) \cdots (1 + k_{-n})\notag \end{align} and letting $n \to \infty$ and noting that $k_{-n} \to 0$ so that $K_{-n} \to \pi / 2$ we get $$\frac{2K}{\pi} = (1 + k_{-1})(1 + k_{-2})(1 + k_{-3}) \cdots$$ and more generally if $\lambda_{0} > \lambda_{1} > \lambda_{2} > \cdots$ forms a decreasing Landen sequence then $$\frac{2\Lambda}{\pi} = \frac{2K(\lambda_{0})}{\pi} = (1 + \lambda_{1})(1 + \lambda_{2})(1 + \lambda_{3}) \cdots$$ In particular we have $$\frac{2K'}{\pi} = (1 + k_{1}')(1 + k_{2}')(1 + k_{3}')\cdots$$ Again we note that \begin{align} (1 + k_{1}')K_{1} &= 2K\notag\\ (1 + k_{2}')K_{2} &= 2K_{1}\notag\\ (1 + k_{3}')K_{3} &= 2K_{2}\notag\\ (1 + k_{4}')K_{4} &= 2K_{3}\notag\\ (1 + k_{n}')K_{n} &= 2K_{n - 1}\notag \end{align} Multiplying theabove equations we get $$(1 + k_{1}')(1 + k_{2}')(1 + k_{3}') \cdots (1 + k_{n}')K_{n} = 2^{n}K$$ or $$\frac{K_{n}}{2^{n}K} = \frac{1}{(1 + k_{1}')(1 + k_{2}')(1 + k_{3}') \cdots (1 + k_{n}')}$$ Taking limits as $n \to \infty$ we get $$\lim_{n \to \infty}\frac{K_{n}}{2^{n}K} = \frac{\pi}{2K'}$$

Infinite Product Expansion of Elliptic Functions

We shall now use the standard Landen's ascending transformation \begin{align} \text{sn}(u, k) &= \dfrac{(1 + k_{1}')\,\text{sn}\left(\dfrac{(1 + k)u}{2}, k_{1}\right)\,\text{cn}\left(\dfrac{(1 + k)u}{2}, k_{1}\right)}{\text{dn}\left(\dfrac{(1 + k)u}{2}, k_{1}\right)}\notag\\ &= \dfrac{(1 + k_{1}')\,\text{sn}\left(\dfrac{K_{1}u}{2K}, k_{1}\right)\,\text{cn}\left(\dfrac{K_{1}u}{2K}, k_{1}\right)}{\text{dn}\left(\dfrac{K_{1}u}{2K}, k_{1}\right)}\notag\\ &= (1 + k_{1}')\,\text{sn}\left(\dfrac{K_{1}u}{2K}, k_{1}\right)\,\text{sn}\left(\dfrac{K_{1}u}{2K} + K_{1}, k_{1}\right)\notag\\ &= (1 + k_{1}')\,\text{sn}\left(\dfrac{K_{1}u}{2K}, k_{1}\right)\,\text{sn}\left(\dfrac{K_{1}}{2K}(u + 2K), k_{1}\right)\notag \end{align} Using the same formula again we get \begin{align} \text{sn}\left(\dfrac{K_{1}u}{2K}, k_{1}\right) &= (1 + k_{2}')\,\text{sn}\left(\dfrac{K_{2}}{2K_{1}}\dfrac{K_{1}u}{2K}, k_{2}\right)\,\text{sn}\left(\dfrac{K_{2}}{2K_{1}}\left(\dfrac{K_{1}u}{2K} + 2K_{1}\right), k_{2}\right)\notag\\ &= (1 + k_{2}')\,\text{sn}\left(\dfrac{K_{2}u}{2^{2}K}, k_{2}\right)\,\text{sn}\left(\dfrac{K_{2}}{2^{2}K}(u + 4K), k_{2}\right)\notag \end{align} and by the same token $$\text{sn}\left(\dfrac{K_{1}}{2K}(u + 2K), k_{1}\right)= (1 + k_{2}')\,\text{sn}\left(\dfrac{K_{2}}{2^{2}K}(u + 2K), k_{2}\right)\,\text{sn}\left(\dfrac{K_{2}}{2^{2}K}(u + 6K), k_{2}\right)$$ Hence we have \begin{align}\text{sn}(u, k) &= (1 + k_{1}')(1 + k_{2}')^{2}\,\text{sn}\left(\dfrac{K_{2}u}{2^{2}K}, k_{2}\right)\,\text{sn}\left(\dfrac{K_{2}}{2^{2}K}(u + 2K), k_{2}\right)\notag\\ &\,\,\,\,\,\,\,\,\text{sn}\left(\dfrac{K_{2}}{2^{2}K}(u + 4K), k_{2}\right)\,\text{sn}\left(\dfrac{K_{2}}{2^{2}K}(u + 6K), k_{2}\right)\notag \end{align} We can write the last term as $$\text{sn}\left(\dfrac{K_{2}}{2^{2}K}(u + 6K), k_{2}\right) = \text{sn}\left(2K_{2} - \dfrac{K_{2}}{2^{2}K}(u + 6K), k_{2}\right) = \text{sn}\left(\dfrac{K_{2}}{2^{2}K}(2K - u), k_{2}\right)$$ and thus we can write \begin{align}\text{sn}(u, k) &= (1 + k_{1}')(1 + k_{2}')^{2}\,\text{sn}\left(\dfrac{K_{2}u}{2^{2}K}, k_{2}\right)\,\text{sn}\left(\dfrac{K_{2}}{2^{2}K}(2K + u), k_{2}\right)\notag\\ &\,\,\,\,\,\,\,\,\text{sn}\left(\dfrac{K_{2}}{2^{2}K}(2K - u), k_{2}\right)\,\text{sn}\left(\dfrac{K_{2}}{2^{2}K}(4K + u), k_{2}\right)\notag \end{align} This can be continued further in the same way to yield \begin{align} \text{sn}(u, k) &= (1 + k_{1}')(1 + k_{2}')^{2}(1 + k_{3}')^{4}\,\text{sn}\left(\dfrac{K_{3}u}{2^{3}K}, k_{3}\right)\notag\\ &\,\,\,\,\,\,\,\,\text{sn}\left(\dfrac{K_{3}}{2^{3}K}(2K + u), k_{3}\right)\,\text{sn}\left(\dfrac{K_{3}}{2^{3}K}(2K - u), k_{3}\right)\notag\\ &\,\,\,\,\,\,\,\,\text{sn}\left(\dfrac{K_{3}}{2^{3}K}(4K + u), k_{3}\right)\,\text{sn}\left(\dfrac{K_{3}}{2^{3}K}(4K - u), k_{3}\right)\notag\\ &\,\,\,\,\,\,\,\,\text{sn}\left(\dfrac{K_{3}}{2^{3}K}(6K + u), k_{3}\right)\,\text{sn}\left(\dfrac{K_{3}}{2^{3}K}(6K - u), k_{3}\right)\notag\\ &\,\,\,\,\,\,\,\,\text{sn}\left(\dfrac{K_{3}}{2^{3}K}(8K + u), k_{3}\right)\notag \end{align} Finally if we continue till the modulus $k_{n}$ we get \begin{align} \text{sn}(u, k) &= (1 + k_{1}')(1 + k_{2}')^{2}(1 + k_{3}')^{4}\cdots (1 + k_{n}')^{2^{n - 1}}\,\text{sn}\left(\dfrac{K_{n}u}{2^{n}K}, k_{n}\right)\notag\\ &\,\,\,\,\,\,\,\,\text{sn}\left(\dfrac{K_{n}}{2^{n}K}(2K + u), k_{n}\right)\,\text{sn}\left(\dfrac{K_{n}}{2^{n}K}(2K - u), k_{n}\right)\notag\\ &\,\,\,\,\,\,\,\,\text{sn}\left(\dfrac{K_{n}}{2^{n}K}(4K + u), k_{n}\right)\,\text{sn}\left(\dfrac{K_{n}}{2^{n}K}(4K - u), k_{n}\right)\notag\\ &\,\,\,\,\,\,\,\,\text{sn}\left(\dfrac{K_{n}}{2^{n}K}(6K + u), k_{n}\right)\,\text{sn}\left(\dfrac{K_{n}}{2^{n}K}(6K - u), k_{n}\right)\notag\\ &\,\,\,\,\,\,\,\,\cdots \cdots\notag\\ &\,\,\,\,\,\,\,\,\text{sn}\left(\dfrac{K_{n}}{2^{n}K}((2^{n} - 2)K + u), k_{n}\right)\,\text{sn}\left(\dfrac{K_{n}}{2^{n}K}((2^{n} - 2)K - u), k_{n}\right)\notag\\ &\,\,\,\,\,\,\,\,\text{sn}\left(\dfrac{K_{n}}{2^{n}K}(2^{n}K + u), k_{n}\right)\notag \end{align} Now we can take limits as $n \to \infty$ (assuming that we have convergence for the entities we are concerned with here, this assumption being justified at a later stage) and note that $k_{n} \to 1$ and the ratio $K_{n} / (2^{n}K) \to \pi / (2K')$. Thus we arrive at $$\text{sn}(u, k) = A'\,\text{sn}\left(\frac{\pi u}{2K'}, 1\right) \prod_{n = 1}^{\infty}\,\text{sn}\left(\frac{\pi}{2K'}(2nK + u), 1\right)\,\text{sn}\left(\frac{\pi}{2K'}(2nK - u), 1\right)$$ where $$A' = \lim_{n \to \infty} (1 + k_{1}')(1 + k_{2}')^{2}(1 + k_{3}')^{4}\cdots (1 + k_{n}')^{2^{n - 1}}$$ Since $$\text{sn}(v, 1) = \tanh v = \frac{e^{v} - e^{-v}}{e^{v} + e^{-v}}$$ we can write \begin{align} &\text{sn}\left(\frac{\pi}{2K'}(2nK + u), 1\right)\,\text{sn}\left(\frac{\pi}{2K'}(2nK - u), 1\right)\notag\\ &\,\,\,\,\,\,\,\,= \tanh\left(\frac{n\pi K}{K'} + \frac{\pi u}{2K'}\right)\tanh\left(\frac{n\pi K}{K'} - \frac{\pi u}{2K'}\right)\notag \end{align} Let us put $$q' = \exp\left(-\frac{\pi K}{K'}\right),\,\,\, x = \frac{\pi u}{2K'}$$ and then the above expression becomes \begin{align} &\left(\frac{q'^{-n}e^{x} - q'^{n}e^{-x}}{q'^{-n}e^{x} - q'^{n}e^{-x}}\right)\left(\frac{q'^{-n}e^{-x} - q'^{n}e^{x}}{q'^{-n}e^{-x} + q'^{n}e^{x}}\right)\notag\\ &\,\,\,\,\,\,\,\,= \frac{q'^{-2n} + q'^{2n} - (e^{2x} + e^{-2x})}{q'^{-2n} + q'^{2n} + (e^{2x} + e^{-2x})}\notag\\ &\,\,\,\,\,\,\,\,= \frac{q'^{-2n} + q'^{2n} - 2\cos(2ix)}{q'^{-2n} + q'^{2n} + 2\cos(2ix)}\notag\\ &\,\,\,\,\,\,\,\,= \frac{1 - 2q'^{2n}\cos(2ix) + q'^{4n}}{1 + 2q'^{2n}\cos(2ix) + q'^{4n}}\notag\\ &\,\,\,\,\,\,\,\,= \dfrac{1 - 2q'^{2n}\cos\left(\dfrac{\pi iu}{K'}\right) + q'^{4n}}{1 + 2q'^{2n}\cos\left(\dfrac{\pi iu}{K'}\right) + q'^{4n}}\notag \end{align} Hence we finally have $$\text{sn}(u, k) = A'\tanh\left(\dfrac{\pi u}{2K'}\right)\prod_{n = 1}^{\infty}\dfrac{1 - 2q'^{2n}\cos\left(\dfrac{\pi iu}{K'}\right) + q'^{4n}}{1 + 2q'^{2n}\cos\left(\dfrac{\pi iu}{K'}\right) + q'^{4n}}$$ Noting that $\tanh x = -i \tan(ix)$ and $\text{sn}(u, k) = -i \,\text{sc}(iu, k')$ we get $$\dfrac{\text{sn}(iu, k')}{\text{cn}(iu, k')} = A'\tan\left(\dfrac{\pi iu}{2K'}\right)\prod_{n = 1}^{\infty}\dfrac{1 - 2q'^{2n}\cos\left(\dfrac{\pi iu}{K'}\right) + q'^{4n}}{1 + 2q'^{2n}\cos\left(\dfrac{\pi iu}{K'}\right) + q'^{4n}}$$ Replacing $iu$ by $u$ and $k'$ by $k$ we get $$\dfrac{\text{sn}(u, k)}{\text{cn}(u, k)} = A\tan\left(\dfrac{\pi u}{2K}\right)\prod_{n = 1}^{\infty}\dfrac{1 - 2q^{2n}\cos\left(\dfrac{\pi u}{K}\right) + q^{4n}}{1 + 2q^{2n}\cos\left(\dfrac{\pi u}{K}\right) + q^{4n}}$$ where $A$ is some constant and $q = \exp(-\pi K' / K)$.

The above equation marks the birth of the parameter $q$ and all things q-otic (not chaotic!!) like the q-series, q-products. We shall have occasion to deal with lot of such q-otic stuff in later posts.

We can again go back to the equation $$\text{sn}(u, k) = A'\,\text{sn}\left(\frac{\pi u}{2K'}, 1\right) \prod_{n = 1}^{\infty}\,\text{sn}\left(\frac{\pi}{2K'}(2nK + u), 1\right)\,\text{sn}\left(\frac{\pi}{2K'}(2nK - u), 1\right)$$ Replacing $u$ by $u + K$ we get \begin{align} \text{sn}(u + K, k) &= A'\,\text{sn}\left(\frac{\pi (u + K)}{2K'}, 1\right)\notag\\ &\,\,\,\,\,\,\,\, \prod_{n = 1}^{\infty}\,\text{sn}\left(\frac{\pi}{2K'}((2n + 1)K + u), 1\right)\,\text{sn}\left(\frac{\pi}{2K'}((2n - 1)K - u), 1\right)\notag\\ \Rightarrow \dfrac{\text{cn}(u, k)}{\text{dn}(u, k)} &= A'\prod_{n = 1}^{\infty}\,\text{sn}\left(\frac{\pi}{2K'}((2n - 1)K + u), 1\right)\,\text{sn}\left(\frac{\pi}{2K'}((2n - 1)K - u), 1\right)\notag \end{align} and like the previous case we have \begin{align} \dfrac{\text{cn}(u, k)}{\text{dn}(u, k)} &= A'\prod_{n = 1}^{\infty}\dfrac{1 - 2q'^{2n - 1}\cos\left(\dfrac{\pi iu}{K'}\right) + q'^{4n - 2}}{1 + 2q'^{2n - 1}\cos\left(\dfrac{\pi iu}{K'}\right) + q'^{4n - 2}}\notag\\ \Rightarrow \dfrac{1}{\text{dn}(iu, k')} &= A'\prod_{n = 1}^{\infty}\dfrac{1 - 2q'^{2n - 1}\cos\left(\dfrac{\pi iu}{K'}\right) + q'^{4n - 2}}{1 + 2q'^{2n - 1}\cos\left(\dfrac{\pi iu}{K'}\right) + q'^{4n - 2}}\notag\\ \Rightarrow \text{dn}(iu, k') &= \dfrac{1}{A'}\prod_{n = 1}^{\infty}\dfrac{1 + 2q'^{2n - 1}\cos\left(\dfrac{\pi iu}{K'}\right) + q'^{4n - 2}}{1 - 2q'^{2n - 1}\cos\left(\dfrac{\pi iu}{K'}\right) + q'^{4n - 2}}\notag \end{align} Replacing $iu$ by $u$ and $k'$ by $k$ we get $$\text{dn}(u, k) = \dfrac{1}{A}\prod_{n = 1}^{\infty}\dfrac{1 + 2q^{2n - 1}\cos\left(\dfrac{\pi u}{K}\right) + q^{4n - 2}}{1 - 2q^{2n - 1}\cos\left(\dfrac{\pi u}{K}\right) + q^{4n - 2}}$$ To evaluate the constant $A$ we put $u = 0$ in the above equation and get $$A = \prod_{n = 1}^{\infty}\dfrac{1 + 2q^{2n - 1} + q^{4n - 2}}{1 - 2q^{2n - 1} + q^{4n - 2}} = \prod_{n = 1}^{\infty}\left(\dfrac{1 + q^{2n - 1}}{1 - q^{2n - 1}}\right)^{2}$$ Again putting $u = K$ we get $$k' = \dfrac{1}{A}\prod_{n = 1}^{\infty}\dfrac{1 - 2q^{2n - 1} + q^{4n - 2}}{1 + 2q^{2n - 1} + q^{4n - 2}} = \frac{1}{A^{2}}$$ Thus we get $$\frac{1}{A} = \sqrt{k'},\,\,\,\,\,\, k' = \prod_{n = 1}^{\infty}\left(\frac{1 - q^{2n - 1}}{1 + q^{2n - 1}}\right)^{4}$$ It now follows that \begin{align} \text{dn}(u, k) &= \sqrt{k'}\prod_{n = 1}^{\infty}\dfrac{1 + 2q^{2n - 1}\cos\left(\dfrac{\pi u}{K}\right) + q^{4n - 2}}{1 - 2q^{2n - 1}\cos\left(\dfrac{\pi u}{K}\right) + q^{4n - 2}}\notag\\ \dfrac{\text{sn}(u, k)}{\text{cn}(u, k)} &= \frac{1}{\sqrt{k'}}\tan\left(\dfrac{\pi u}{2K}\right)\prod_{n = 1}^{\infty}\dfrac{1 - 2q^{2n}\cos\left(\dfrac{\pi u}{K}\right) + q^{4n}}{1 + 2q^{2n}\cos\left(\dfrac{\pi u}{K}\right) + q^{4n}}\notag\\ \Rightarrow \dfrac{\text{sn}(u, k)}{\text{cn}(u, k)\,\tan\left(\dfrac{\pi u}{2K}\right)} &= \frac{1}{\sqrt{k'}}\prod_{n = 1}^{\infty}\dfrac{1 - 2q^{2n}\cos\left(\dfrac{\pi u}{K}\right) + q^{4n}}{1 + 2q^{2n}\cos\left(\dfrac{\pi u}{K}\right) + q^{4n}}\notag \end{align} Taking limits as $u \to 0$ we get \begin{align} \dfrac{2K}{\pi} &= \frac{1}{\sqrt{k'}}\prod_{n = 1}^{\infty}\dfrac{1 - 2q^{2n} + q^{4n}}{1 + 2q^{2n} + q^{4n}} = \frac{1}{\sqrt{k'}}\prod_{n = 1}^{\infty}\left(\frac{1 - q^{2n}}{1 + q^{2n}}\right)^{2}\notag\\ \Rightarrow \dfrac{2\sqrt{k'}K}{\pi} &= \prod_{n = 1}^{\infty}\left(\frac{1 - q^{2n}}{1 + q^{2n}}\right)^{2}\notag \end{align} Multiplying by $\sqrt{k'}$ (and its infinite product equivalent) we get $$\frac{2k'K}{\pi} = \prod_{n = 1}^{\infty}\left(\frac{1 - q^{n}}{1 + q^{n}}\right)^{2}$$ Again \begin{align} \frac{2K}{\pi} &= \frac{1}{\sqrt{k'}}\prod_{n = 1}^{\infty}\left(\frac{1 - q^{2n}}{1 + q^{2n}}\right)^{2} = A\prod_{n = 1}^{\infty}\left(\frac{1 - q^{2n}}{1 + q^{2n}}\right)^{2}\notag\\ &= \prod_{n = 1}^{\infty}\left(\dfrac{1 + q^{2n - 1}}{1 - q^{2n - 1}}\right)^{2}\left(\frac{1 - q^{2n}}{1 + q^{2n}}\right)^{2}\notag\\ \Rightarrow \frac{2K}{\pi} &= \prod_{n = 1}^{\infty}\{(1 - q^{2n})(1 + q^{2n - 1})^{2}\}^{2}\notag\end{align} To summarize: \begin{align} \dfrac{\text{sn}(u, k)}{\text{cn}(u, k)} &= \frac{1}{\sqrt{k'}}\tan\left(\dfrac{\pi u}{2K}\right)\prod_{n = 1}^{\infty}\dfrac{1 - 2q^{2n}\cos\left(\dfrac{\pi u}{K}\right) + q^{4n}}{1 + 2q^{2n}\cos\left(\dfrac{\pi u}{K}\right) + q^{4n}}\notag\\ \text{dn}(u, k) &= \sqrt{k'}\prod_{n = 1}^{\infty}\dfrac{1 + 2q^{2n - 1}\cos\left(\dfrac{\pi u}{K}\right) + q^{4n - 2}}{1 - 2q^{2n - 1}\cos\left(\dfrac{\pi u}{K}\right) + q^{4n - 2}}\notag\\ \frac{2K}{\pi} &= \prod_{n = 1}^{\infty}\{(1 - q^{2n})(1 + q^{2n - 1})^{2}\}^{2}\notag\\ \frac{2k'K}{\pi} &= \prod_{n = 1}^{\infty}\left(\frac{1 - q^{n}}{1 + q^{n}}\right)^{2}\notag\\ \dfrac{2\sqrt{k'}K}{\pi} &= \prod_{n = 1}^{\infty}\left(\frac{1 - q^{2n}}{1 + q^{2n}}\right)^{2}\notag\\ k' &= \prod_{n = 1}^{\infty}\left(\frac{1 - q^{2n - 1}}{1 + q^{2n - 1}}\right)^{4}\notag \end{align} As we can see above the elliptic functions have been expressed as ratio of two infinite products. Each of the infinite products in numerator and denominator belongs to a class of functions known as the Theta Functions which will form the main topic of the next post. In that post we will also justify the results which we have obtained here (by assuming the desired convergence of the infinite products involved).