Elliptic Functions: Theta Functions Contd.

The values of theta functions at the point $ z = 0$ are themselves very important and surprisingly have deep connections with number theory. For example consider $$\theta_{3}(q) = \sum_{n = -\infty}^{\infty} q^{n^{2}}$$ Then clearly $$\theta_{3}^{2}(q) = \sum_{i, j = -\infty}^{\infty} q^{i^{2} + j^{2}} = \sum_{n = 0}^{\infty} r(n)q^{n}$$ where $ r(n) = r_{2}(n)$ represents the number of ways in which integer $ n$ can be expressed as the sum of two squares (here we count order as well as sign separately).

Similarly we can see that $$\theta_{3}^{j}(q) = \sum_{n = 0}^{\infty}r_{j}(n)q^{n}$$ where $ r_{j}(n)$ is the number of ways in which $ n$ can be expressed as the sum of $ j$ squares. Thus the theta functions can be used as generating functions for many important number theoretic functions. In general there are various kinds of $ q$-series and products (which are related with theta functions) which can be used in studying number theory.

Relation Between Values of Theta Functions at $ z = 0$

Hence it makes sense to study these special values $ \theta_{j}(0, q) = \theta_{j}(q) = \theta_{j}$ of the theta functions. Unfortunately we have $ \theta_{1}(0, q) = 0$ and hence it does not give us anything worth noticing. However we can try to salvage the situation here by studying the derivative $ \theta_{1}'(0, q) = \theta_{1}'$ and we then have a very deep relation between these values: $$\theta_{1}' = \theta_{2}\theta_{3}\theta_{4}$$ We will establish this result here and thereby obtain the value of the constant $ G$ referred to in the last post. If we look at the series and product expansions of the theta functions it follows that \begin{align} \theta_{1}' &= 2q^{1/4}\sum_{n = 0}^{\infty}(-1)^{n}(2n + 1)q^{n(n + 1)} = 2q^{1/4}G\prod_{n = 1}^{\infty}(1 - q^{2n})^{2}\notag\\ \theta_{2} &= 2q^{1/4}\sum_{n = 0}^{\infty}q^{n(n + 1)}=2q^{1/4}G\prod_{n = 1}^{\infty}(1 + q^{2n})^{2}\notag\\ \theta_{3} &= 1 + 2\sum_{n = 1}^{\infty}q^{n^{2}} = G\prod_{n = 1}^{\infty}(1 + q^{2n - 1})^{2}\notag\\ \theta_{4} &= 1 + 2\sum_{n = 1}^{\infty}(-1)^{n}q^{n^{2}} = G\prod_{n = 1}^{\infty}(1 - q^{2n - 1})^{2}\notag \end{align} Another thing we can notice from the formula for $ \text{sn}(u, k)$ $$\text{sn}(u, k) = \frac{\theta_{3}}{\theta_{2}}\frac{\theta_{1}(z, q)}{\theta_{4}(z, q)}$$ is that if we divide both sides by $ u$ and take limits as $ u \to 0$ then we get \begin{align} 1 &= \frac{\theta_{3}}{\theta_{2}}\frac{\pi}{2K}\frac{\theta_{1}'}{\theta_{4}}\notag\\ \Rightarrow \frac{2K}{\pi} &= \frac{\theta_{1}'}{\theta_{2}\theta_{3}\theta_{4}}\theta_{3}^{2}\notag \end{align} and hence we shall have (once we prove the identity connecting the values of theta functions at $ z = 0$) $$\frac{2K}{\pi} = \theta_{3}^{2}$$ Now we can rewrite the product expansions for $ \theta_{1}'$ as \begin{align} \theta_{1}' &= 2q^{1/4}G\prod_{n = 1}^{\infty}(1 - q^{2n})^{2} = 2q^{1/4}G\prod_{n = 1}^{\infty}(1 + q^{n})^{2}(1 - q^{n})^{2}\notag\\ &= 2q^{1/4}G\prod_{n = 1}^{\infty}(1 + q^{2n})^{2}\prod_{n = 1}^{\infty}(1 + q^{2n - 1})^{2}\prod_{n = 1}^{\infty}(1 - q^{2n - 1})^{2}\prod_{n = 1}^{\infty}(1 - q^{2n})^{2}\notag \end{align} and $$\theta_{2}\theta_{3}\theta_{4} = 2q^{1/4}G^{3}\prod_{n = 1}^{\infty}(1 + q^{2n})^{2}\prod_{n = 1}^{\infty}(1 + q^{2n - 1})^{2}\prod_{n = 1}^{\infty}(1 - q^{2n - 1})^{2}$$ so that we get $$\alpha(q) := \frac{\theta_{1}^{\prime}}{\theta_{2}\theta_{3}\theta_{4}} = \frac{1}{G^{2}}\prod_{n = 1}^{\infty}(1 - q^{2n})^{2}$$ Since $ G \to 1$ as $ q \to 0$ it follows that $ \alpha(q) \to 1$ as $ q \to 0$. Hence we can write $$G = A(q)\prod_{n = 1}^{\infty}(1 - q^{2n})$$ where $ A(q) \to 1$ as $ q \to 0$. Our job will be done if we show that $ A(q) = 1$ for all values of $ q$. This requires a reasonable amount of investigation and we follow the path taken by Jacobi. First of all we can see that \begin{align} \theta_{2}(2z, q^{4}) &= 2\sum_{n = 0}^{\infty}q^{(2n + 1)^{2}}\cos 2(2n + 1)z\notag\\ \theta_{3}(2z, q^{4}) &= 1 + 2\sum_{n = 1}^{\infty}q^{(2n)^{2}} \cos 2(2n)z\notag \end{align} and therefore on adding these we get $$\theta_{3}(z, q) = \theta_{2}(2z, q^{4}) + \theta_{3}(2z, q^{4})$$ and on putting $ z = \pi / 4$ we get $$\theta_{3}\left(\frac{\pi}{4}, q\right) = \theta_{3}\left(\frac{\pi}{2}, q^{4}\right)$$ Since $$\theta_{3}(z, q) = A(q)\prod_{n = 1}^{\infty}(1 - q^{2n})\prod_{n = 1}^{\infty}(1 + 2q^{2n - 1}\cos 2z + q^{4n - 2})$$ therefore \begin{align} \theta_{3}\left(\frac{\pi}{4}, q\right) &= A(q)\prod_{n = 1}^{\infty}(1 - q^{2n})\prod_{n = 1}^{\infty}(1 + q^{4n - 2})\notag\\ &= A(q)\prod_{n = 1}^{\infty}(1 - q^{4n})\prod_{n = 1}^{\infty}(1 - q^{4n - 2})\prod_{n = 1}^{\infty}(1 + q^{4n - 2})\notag\\ &= A(q)\prod_{n = 1}^{\infty}(1 - q^{8n})\prod_{n = 1}^{\infty}(1 - q^{8n - 4})\prod_{n = 1}^{\infty}(1 - q^{8n - 4})\notag\\ &= A(q)\prod_{n = 1}^{\infty}(1 - q^{8n})\prod_{n = 1}^{\infty}(1 - q^{8n - 4})^{2}\notag \end{align} and \begin{align} \theta_{3}\left(\frac{\pi}{2}, q^{4}\right) &= A(q^{4})\prod_{n = 1}^{\infty}(1 - q^{8n})\prod_{n = 1}^{\infty}(1 - 2q^{8n - 4} + q^{16n - 8})\notag\\ &= A(q^{4})\prod_{n = 1}^{\infty}(1 - q^{8n})\prod_{n = 1}^{\infty}(1 - q^{8n - 4})^{2}\notag \end{align} Comparing the above expressions for $ \theta_{3}(\pi / 4, q)$ and $ \theta_{3}(\pi / 2, q^{4})$ we get $ A(q) = A(q^{4})$ and therefore $ A(q) = A(q^{4^{n}})$. Taking limits when $ n \to \infty$ we get $$A(q) = \lim_{n \to \infty}A(q^{4^{n}}) = \lim_{ q \to 0}A(q) = 1$$ Thus we achieve the following results: $$G = \prod_{n = 1}^{\infty}(1 - q^{2n}),\,\,\, \theta_{1}^{\prime} = \theta_{2}\theta_{3}\theta_{4},\,\,\, \frac{2K}{\pi} = \theta_{3}^{2}$$ We can now have the product expansions of the theta functions as follows \begin{align} \theta_{1}(z, q) &= 2q^{1/4}\sin z\prod_{n = 1}^{\infty}(1 - q^{2n})(1 - 2q^{2n}\cos 2z + q^{4n})\notag\\ \theta_{2}(z, q) &= 2q^{1/4}\cos z\prod_{n = 1}^{\infty}(1 - q^{2n})(1 + 2q^{2n}\cos 2z + q^{4n})\notag\\ \theta_{3}(z, q) &= \prod_{n = 1}^{\infty}(1 - q^{2n})(1 + 2q^{2n - 1}\cos 2z + q^{4n - 2})\notag\\ \theta_{4}(z, q) &= \prod_{n = 1}^{\infty}(1 - q^{2n})(1 - 2q^{2n - 1}\cos 2z + q^{4n - 2})\notag \end{align}

Jacobi's Triple Product Identity

Moreover if we look at the series and product expansions of $ \theta_{3}(z, q)$ and put $ w = e^{2iz}$ we get the famous Jacobi's Triple Product identity:
$$\prod_{n = 1}^{\infty}(1 - q^{2n})(1 + q^{2n - 1}w)(1 + q^{2n - 1}w^{-1}) = \sum_{n = -\infty}^{\infty}q^{n^{2}}w^{n},\,\,\, w \neq 0$$ This is such an important identity that it is worthwhile giving another proof. There are many proofs in literature for this celebrated identity but the one which I like most is by Jacobi given in Fundamenta Nova. There Jacobi just multiplies the factors on the left and obtains the series on the right. This is sheer genius. That way Jacobi sees the identity as a direct multiplication like $ 2 \cdot 3 = 6$.

Jacobi begins with the series expansion $$F(z) = \prod_{n = 1}^{\infty} (1 + q^{2n - 1}z) = 1 + A_{1}z + A_{2}z^{2} + \cdots$$ where the coefficients $ A_{i}$ can be found by noting that $ F(z) = (1 + qz)F(q^{2}z)$ or $$1 + A_{1}z + A_{2}z^{2} + \cdots = (1 + qz)(1 + A_{1}q^{2}z + A_{2}q^{4}z^{2} + \cdots)$$ Comparing coefficients of $z^{n}$ on both sides we get \begin{align} A_{n} &= A_{n}q^{2n} + A_{n - 1}q^{2n - 1}\notag\\ \Rightarrow A_{n} &= \frac{q^{2n - 1}}{1 - q^{2n}}A_{n - 1}\notag\\ &= \dfrac{q^{n^{2}}}{(1 - q^{2})(1 - q^{4})\cdots(1 - q^{2n})}\text{ (via recursion)}\notag \end{align} Thus we have finally $$F(z) = \prod_{n = 1}^{\infty}(1 + q^{2n - 1}z) = 1 + \frac{q}{1 - q^{2}}z + \frac{q^{4}}{(1 - q^{2})(1 - q^{4})}z^{2} + \cdots$$ Next Jacobi uses the following expansion \begin{align}G(z, q) &= \frac{1}{(1 - qz)(1 - q^{2}z)(1 - q^{3}z)\cdots}\notag\\ &= 1 + \frac{B_{1}z}{1 - qz} + \frac{B_{2}z^{2}}{(1 - qz)(1 - q^{2}z)} + \cdots\notag\end{align} and finds the coefficients $ B_{i}$ through the relation $ G(z, q) = G(qz, q) / (1 - qz)$ or $$1 + \frac{B_{1}z}{1 - qz} + \frac{B_{2}z^{2}}{(1 - qz)(1 - q^{2}z)} + \cdots = \frac{1}{1 - qz} + \frac{B_{1}qz}{(1 - qz)(1 - q^{2}z)} + \cdots$$ Now we need a little bit of manipulation to get things done. We multiply $ n^{\text{th}}$ term on the right by $ (1 - q^{n}z)$ and then and also add the term multiplied by $ q^{n}z$. Thus for example we have \begin{align} \frac{1}{1 - qz} &= \frac{1 - qz}{1 - qz} + \frac{qz}{1 - qz} = 1 + \frac{qz}{1 - qz}\notag\\ \frac{B_{1}qz}{(1 - qz)(1 - q^{2}z)} &= \frac{B_{1}qz(1 - q^{2}z)}{(1 - qz)(1 - q^{2}z)} + \frac{B_{1}q^{3}z^{2}}{(1 - qz)(1 - q^{2}z)}\notag\\ &= \frac{B_{1}qz}{1 - qz} + \frac{B_{1}q^{3}z^{2}}{(1 - qz)(1 - q^{2}z)}\notag\end{align} It now follows that $$1 + \frac{B_{1}z}{1 - qz} + \frac{B_{2}z^{2}}{(1 - qz)(1 - q^{2}z)} + \cdots = 1 + \frac{(q + B_{1}q)z}{1 - qz} + \frac{(B_{1}q^{3} + B_{2}q^{2})z^{2}}{(1 - qz)(1 - q^{2}z)} + \cdots$$ and therefore by comparing coefficients we get $ B_{n} = B_{n - 1}q^{2n - 1} + B_{n}q^{n}$ so that $$B_{n} = \frac{q^{2n - 1}}{1 - q^{n}}B_{n - 1} = \frac{q^{n^{2}}}{(1 - q)(1 - q^{2})\cdots(1 - q^{n})}\text{ (using recursion)}$$ Therefore it follows that \begin{align}G(z, q) &= \frac{1}{(1 - qz)(1 - q^{2}z)(1 - q^{3}z)\cdots}\notag\\ &= 1 + \frac{q}{1 - q}\frac{z}{1 - qz} + \frac{q^{4}}{(1 - q)(1 - q^{2})}\frac{z^{2}}{(1 - qz)(1 - q^{2}z)} + \cdots\notag\end{align} Now like Jacobi we consider the product \begin{align} F(z)F\left(\frac{1}{z}\right) = &\left(1 + \frac{q}{1 - q^{2}}z + \frac{q^{4}}{(1 - q^{2})(1 - q^{4})}z^{2} + \cdots \right)\notag\\ &\left(1 + \frac{q}{1 - q^{2}}\frac{1}{z} + \frac{q^{4}}{(1 - q^{2})(1 - q^{4})}\frac{1}{z^{2}} + \cdots \right)\notag\end{align} Since replacing $ z$ by $ 1/z$ does not change the above product it is clear that the coefficient of $ z^{n}$ in the product on the right side will be same as the coefficient of $ 1/z^{n}$. Let us denote this coefficient by $ C_{n}$ then we can write $$F(z)F\left(\frac{1}{z}\right) = C_{0} + C_{1}\left(z + \frac{1}{z}\right) + C_{2}\left(z^{2} + \frac{1}{z^{2}}\right) + \cdots$$ Clearly the term $ z^{n}$ comes a result of multiplying a term containing $ z^{n + r}$ and a term containing $ 1/z^{r}$ where $ r$ is any non-negative integer. Therefore we have \begin{align}C_{n} &= \sum_{r = 0}^{\infty}\frac{q^{(n + r)^{2}}}{(1 - q^{2})(1 - q^{4})\cdots(1 - q^{2n + 2r})}\frac{q^{r^{2}}}{(1 - q^{2})(1 - q^{4})\cdots(1 - q^{2r})}\notag\\ &= \frac{q^{n^{2}}}{(1 - q^{2})(1 - q^{4})\cdots(1 - q^{2n})}\notag\\ &\,\,\,\,\,\left(1 + \sum_{r = 1}^{\infty}\frac{q^{2r^{2}}}{(1 - q^{2})(1 - q^{4})\cdots(1 - q^{2r})}\frac{q^{2nr}}{(1 - q^{2n + 2})(1 - q^{2n + 4})\cdots(1 - q^{2n + 2r})}\right)\notag\end{align} The expression in the large brackets on the right is clearly seen to be $ G(q^{2n}, q^{2})$ and hence is equal to $$\frac{1}{(1 - q^{2n + 2})(1 - q^{2n + 4})(1 - q^{2n + 6})\cdots}$$ It now follows that $$C_{n} = \frac{q^{n^{2}}}{(1 - q^{2})(1 - q^{4})(1 - q^{6})\cdots}$$ This clearly leads us to $$F(z)F(1/z)\prod_{n = 1}^{\infty}(1 - q^{2n}) = 1 + \sum_{n = 1}^{\infty}q^{n^{2}}(z^{n} + z^{-n})$$ or $$\prod_{n = 1}^{\infty}(1 - q^{2n})(1 + q^{2n - 1}z)(1 + q^{2n - 1}z^{-1}) = \sum_{n = -\infty}^{\infty}q^{n^{2}}z^{n}$$ and the Jacobi's proof is completed.

The above product leads to a beautiful formula of Euler (better known as Euler's Pentagonal Theorem) when we replace $ q$ by $ q^{3/2}$ and $ z$ by $ -q^{1/2}$: $$\prod_{n = 1}^{\infty}(1 - q^{3n})(1 - q^{3n - 1})(1 - q^{3n - 2}) = \sum_{n = -\infty}^{\infty}(-1)^{n}q^{(3n^{2} + n)/2}$$ i.e. $$\prod_{n = 1}^{\infty}(1 - q^{n}) = \sum_{n = -\infty}^{\infty}(-1)^{n}q^{(3n^{2} + n)/2}$$ For another proof (based on completely different ideas) of Euler's Pentagonal theorem see this post. Since the elliptic functions are intimately related to the theta functions, the identities involving the elliptic functions can be transformed into corresponding identities involving theta functions. Such a transition is sometimes just a matter of simple substitution and sometimes requires the use of Liouville's theorem. We will have a look at some such theta function identities in the next post.

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