On Parameterized Theta Function Identities

1. Main results

In this section, we apply logarithmic differerentiation to compute the residue of elliptic functions at high order poles. Then we obtain an theta identity with five parameters.

Theorem 2.1. 

Suppose $f\left(z\right)$ is an entire function satisfying the functional equations

$$\begin{array}{c}\hfill f(z+\pi )={(-1)}^{6}f\left(z\right)\mathrm{and}f(z+\pi \tau )={\left(-q^{-1}{e}^{-2iz}\right)}^{6}f\left(z\right),\end{array}$$

(2.1)

then, we have

$$\begin{array}{c}\hfill \frac{f^{\prime \prime }\left(0\right)}{2{\theta }_1^{\prime }{(0\mid \tau )}^2}-\frac{f\left(0\right)\left(1-24{\sum }_{n=1}^{\infty }\frac{q^{2n}}{1-q^{2n}}·n\right)}{{\theta }_1^{\prime }{(0\mid \tau )}^2}+\frac{q^{\frac{3}{2}}f\left(\frac{\pi \tau }{2}\right)}{{\theta }_4^2(0\mid \tau )}+\frac{q^{\frac{3}{2}}f\left(\frac{\pi +\pi \tau }{2}\right)}{{\theta }_3^2(0\mid \tau )}=\frac{f\left(\frac{\pi }{2}\right)}{{\theta }_2^2(0\mid \tau )}.\end{array}$$

(2.2)

Proof. 

We consider the following function

$$\begin{array}{c}\hfill g\left(z\right)=\frac{f\left(z\right)}{{\theta }_1^2(z\mid \tau ){\theta }_1(2z\mid \tau )},\end{array}$$

From (6) and (14), we find that $g\left(z\right)$ satisfies

$$\begin{array}{c}\hfill g(z+\pi )=g\left(z\right)\mathrm{and}g(z+\pi \tau )=g\left(z\right).\end{array}$$

Thus, $g\left(z\right)$ is an elliptic function with the periods $\pi$ and $\pi \tau$. The poles of $g\left(z\right)$ are 0, $\frac{\pi }{2}$,$\frac{\pi \tau }{2}$,$\frac{\pi +\pi \tau }{2}$, all of which are simple poles.

Based on Theorem 0.3, we have

$$\begin{array}{c}\hfill Res(g;0)+Res\left(g;\frac{\pi }{2}\right)+Res\left(g;\frac{\pi \tau }{2}\right)+Res\left(g;\frac{\pi +\pi \tau }{2}\right)=0.\end{array}$$

(2.3)

Now let us computer the above residues, respectively.

Suppose $G\left(z\right)=z^{3}g\left(z\right)\mathrm{and}\varphi \left(z\right)=\frac{G^{\prime }\left(z\right)}{G\left(z\right)}$. Notice that

$$\begin{array}{cc}\hfill \underset{z\to 0}{lim}{\theta }_1(z\mid \tau )& =\underset{z\to 0}{lim}z{\theta }_1^{\prime }(z\mid \tau ),\hfill \\ \hfill \underset{z\to 0}{lim}{\theta }_1(2z\mid \tau )& =\underset{z\to 0}{lim}2z{\theta }_1^{\prime }(2z\mid \tau ).\hfill \end{array}$$

By using L’Hoptital’rule, we derive

$$\begin{array}{cc}\hfill G\left(0\right)& =\underset{z\to 0}{lim}\frac{z^{3}f\left(z\right)}{{\theta }_1^2(z\mid \tau ){\theta }_1(2z\mid \tau )}\hfill \\ \hfill & =\underset{z\to 0}{lim}\frac{z^{3}f\left(z\right)}{2z^3{\theta }_1^{\prime }{(z\mid \tau )}^2{\theta }_1^{\prime }(2z\mid \tau )}\hfill \\ \hfill & =\frac{f\left(0\right)}{2{\theta }_1^{\prime }{(0\mid \tau )}^3}.\hfill \end{array}$$

(2.4)

From the expression of $G\left(z\right)$, we have

$$lnG\left(z\right)=ln{z}^{3}g\left(z\right)=3lnz+lng\left(z\right)$$

and

$$\frac{G^{\prime }\left(z\right)}{G\left(z\right)}=\frac{3}{z}+\frac{g^{\prime }\left(z\right)}{g\left(z\right)}.$$

On the other hand, we have

$$\begin{array}{cc}\hfill lng\left(z\right)& =ln\frac{f\left(z\right)}{{\theta }_1^2(z\mid \tau ){\theta }_1(2z\mid \tau )}\hfill \\ & =lnf\left(z\right)-2ln{\theta }_1(z\mid \tau )-ln{\theta }_1(2z\mid \tau ),\hfill \end{array}$$

and

$$\frac{g^{\prime }\left(z\right)}{g\left(z\right)}=\frac{f^{\prime }\left(z\right)}{f\left(z\right)}-2\frac{{\theta }_1^{\prime }(z\mid \tau )}{{\theta }_1(z\mid \tau )}-2\frac{{\theta }_1^{\prime }(2z\mid \tau )}{{\theta }_1(2z\mid \tau )}.$$

Thus,

$$\frac{G^{\prime }\left(z\right)}{G\left(z\right)}=\frac{3}{z}+\frac{f^{\prime }\left(z\right)}{f\left(z\right)}-2\frac{{\theta }_1^{\prime }(z\mid \tau )}{{\theta }_1(z\mid \tau )}-2\frac{{\theta }_1^{\prime }(2z\mid \tau )}{{\theta }_1(2z\mid \tau )}.$$

In view of (1), we arrive at

$$\begin{array}{cc}\hfill \varphi \left(z\right)=& \frac{G^{\prime }\left(z\right)}{G\left(z\right)}=\frac{3}{z}+\frac{f^{\prime }\left(z\right)}{f\left(z\right)}-2\left(\frac{1}{z}-\frac{z}{3}-O\left(z^3\right)+4\sum _{n=1}^{\infty }\frac{q^{2n}}{1-q^{2n}}sin2nz\right)\hfill \\ \hfill & -2\left(\frac{1}{2z}-\frac{2z}{3}-O\left(z^3\right)+4\sum _{n=1}^{\infty }\frac{q^{2n}}{1-q^{2n}}sin4nz\right)\hfill \\ \hfill =& 2z+\frac{f^{\prime }\left(z\right)}{f\left(z\right)}-8\sum _{n=1}^{\infty }\frac{q^{2n}}{1-q^{2n}}(sin2nz+sin4nz)+O\left(z^3\right).\hfill \end{array}$$

(2.5)

Furthermore, we have

$$\begin{array}{c}\hfill {\varphi }^{\prime }\left(z\right)={\left(\frac{G^{\prime }\left(z\right)}{G\left(z\right)}\right)}^{\prime }=2+{\left(\frac{f^{\prime }\left(z\right)}{f\left(z\right)}\right)}^{\prime }-8\sum _{n=1}^{\infty }\frac{q^{2n}}{1-q^{2n}}(2ncos2nz+\mid 4ncos4nz)+O\left(z^2\right).\end{array}$$

(2.6)

Based on Theorem 0.2, (17), (18), and (19), we have

$$\begin{array}{cc}\hfill Res(g;0)& =\frac{1}{2}G\left(0\right)\left({\varphi }^2\left(0\right)+{\varphi }^{\prime }\left(0\right)\right)\hfill \\ \hfill & =\frac{1}{2}·\frac{f\left(0\right)}{2{\theta }_1^{\prime }{(0\mid \tau )}^3}·\left({\left(\frac{f^{\prime }\left(0\right)}{f\left(0\right)}\right)}^2+{\left(\frac{f^{\prime }\left(0\right)}{f\left(0\right)}\right)}^{\prime }+2-48\sum _{n=1}^{\infty }\frac{q^{2n}}{1-q^{2n}}·n\right)\hfill \\ \hfill & =\frac{f^{\prime \prime }\left(0\right)}{4{\theta }_1^{\prime }{(0\mid \tau )}^3}-\frac{f\left(0\right)\left(1-24{\sum }_{n=1}^{\infty }\frac{q^{2n}}{1-q^{2n}}·n\right)}{2{\theta }_1^{\prime }{(0\mid \tau )}^3}.\hfill \end{array}$$

(2.7)

In terms of the definition of residue, we have the following results.

$$\begin{array}{cc}\hfill Res\left(g;\frac{\pi }{2}\right)& =\underset{z\to \frac{\pi }{2}}{lim}\frac{\left(z-\frac{\pi }{2}\right)f\left(z\right)}{{\theta }_1^2(z\mid \tau ){\theta }_1(2z\mid \tau )}\hfill \\ \hfill & =\underset{z\to 0}{lim}\frac{zf\left(z+\frac{\pi }{2}\right)}{{\theta }_1^2\left(z+\frac{\pi }{2}\mid \tau \right){\theta }_1(2z+\pi \mid \tau )}\hfill \\ \hfill & =-\frac{f\left(\frac{\pi }{2}\right)}{2{\theta }_1^{\prime }(0\mid \tau ){\theta }_2^2(0\mid \tau )},\hfill \end{array}$$

(2.8)

$$\begin{array}{c}\hfill Res\left(g;\frac{\pi \tau }{2}\right)=\frac{q^{\frac{3}{2}}f\left(\frac{\pi \tau }{2}\right)}{2{\theta }_1^{\prime }(0\mid \tau ){\theta }_4^2(0\mid \tau )},\end{array}$$

(2.9)

and

$$\begin{array}{c}\hfill Res\left(g;\frac{\pi +\pi \tau }{2}\right)=\frac{q^{\frac{3}{2}}f\left(\frac{\pi +\pi \tau }{2}\right)}{2{\theta }_1^{\prime }(0\mid \tau ){\theta }_3^2(0\mid \tau )}.\end{array}$$

(2.10)

Substituting the (20)–(23) into (16), we obtain (15). We complete the proof.□

Theorem 1.2. 

For any complex numbers $a,b,c,d,e$, let $f\left(z\right)={\theta }_1(z+a\mid \tau ){\theta }_1(z+b\mid \tau ){\theta }_1(z+c\mid \tau ){\theta }_1(z+d\mid \tau ){\theta }_1(z+$$e\mid \tau ){\theta }_1(z-a-b-c-d-e\mid \tau )$, we have

$$\begin{array}{cc}\hfill & \frac{f^{\prime \prime }\left(0\right)}{2{\theta }_1^{\prime }{(0\mid \tau )}^2}\hfill \\ \hfill & +\frac{{\theta }_1(a\mid \tau ){\theta }_1(b\mid \tau ){\theta }_1(c\mid \tau ){\theta }_1(d\mid \tau ){\theta }_1(e\mid \tau ){\theta }_1(a+b+c+d+e\mid \tau )\left(1-24{\sum }_{n=1}^{\infty }\frac{q^{2n}}{1-q^{2n}}·n\right)}{{\theta }_1^{\prime }{(0\mid \tau )}^2}\hfill \\ \hfill & +\frac{{\theta }_3(a\mid \tau ){\theta }_3(b\mid \tau ){\theta }_3(c\mid \tau ){\theta }_3(d\mid \tau ){\theta }_3(e\mid \tau ){\theta }_3(a+b+c+d+e\mid \tau )}{{\theta }_3^2(0\mid \tau )}\hfill \\ \hfill & =\frac{{\theta }_4(a\mid \tau ){\theta }_4(b\mid \tau ){\theta }_4(c\mid \tau ){\theta }_4(d\mid \tau ){\theta }_4(e\mid \tau ){\theta }_4(a+b+c+d+e\mid \tau )}{{\theta }_4^2(0\mid \tau )}\hfill \\ \hfill & +\frac{{\theta }_2(a\mid \tau ){\theta }_2(b\mid \tau ){\theta }_2(c\mid \tau ){\theta }_2(d\mid \tau ){\theta }_2(e\mid \tau ){\theta }_2(a+b+c+d+e\mid \tau )}{{\theta }_2^2(0\mid \tau )}.\hfill \end{array}$$

(2.11)

Proof. 

Based on (6)-() and the following

$$\begin{array}{cc}\hfill & {\theta }_1(-z\mid \tau )=-{\theta }_1(z\mid \tau ),\hfill \\ \hfill & {\theta }_2(-z\mid \tau )={\theta }_2(z\mid \tau ),\hfill \\ \hfill & {\theta }_3(-z\mid \tau )={\theta }_3(z\mid \tau )\hfill \\ \hfill & {\theta }_4(-z\mid \tau )={\theta }_4(z\mid \tau ),\hfill \\ \hfill & {\theta }_1^{\prime }(-z\mid \tau )=-{\theta }_1^{\prime }(z\mid \tau ),\hfill \\ \hfill & {\theta }_1^{\prime \prime }(-z\mid \tau )={\theta }_1^{\prime \prime }(z\mid \tau ),\hfill \end{array}$$

we can derive

$$\begin{array}{cc}\hfill f(z+\pi )& ={(-1)}^{6}f\left(z\right)=f\left(z\right)\hfill \\ \hfill f(z+\pi \tau )& ={\left(-q^{-1}{e}^{-2iz}\right)}^{6}f\left(z\right)\hfill \end{array}$$

so, $f\left(z\right)$ satisfies (14).

On the other hand, we have

$$\begin{array}{cc}\hfill & f\left(\frac{\pi }{2}\right)={\theta }_2(a\mid \tau ){\theta }_2(b\mid \tau ){\theta }_2(c\mid \tau ){\theta }_2(d\mid \tau ){\theta }_2(e\mid \tau ){\theta }_2(a+b+c+d+e\mid \tau ),\hfill \\ \hfill & f\left(\frac{\pi \tau }{2}\right)=-q^{-\frac{3}{2}}{\theta }_4(a\mid \tau ){\theta }_4(b\mid \tau ){\theta }_4(c\mid \tau ){\theta }_4(d\mid \tau ){\theta }_4(e\mid \tau ){\theta }_4(a+b+c+d+e\mid \tau ),\hfill \\ \hfill & f\left(\frac{\pi +\pi \tau }{2}\right)=q^{-\frac{3}{2}}{\theta }_3(a\mid \tau ){\theta }_3(b\mid \tau ){\theta }_3(c\mid \tau ){\theta }_3(d\mid \tau ){\theta }_3(e\mid \tau ){\theta }_3(a+b+c+d+e\mid \tau ).\hfill \end{array}$$

Substituting the above three identities into (15), we derive (24). We compete the proof.□

Let $a=b=c=d=e=0$ in (24), we obtain the famous Jacobi identity [13].

$$\begin{array}{c}\hfill {\theta }_2^4\left(0\right|\tau )+{\theta }_4^4\left(0\right|\tau )={\theta }_3^4\left(0\right|\tau ).\end{array}$$

Let $a=0,b=c=d=\frac{\pi }{3},e=0$ in (24), we have the following theta identity.

$${\theta }_2^3\left(\frac{\pi }{3}\mid \tau \right){\theta }_2(0\mid \tau )+{\theta }_3^3\left(\frac{\pi }{3}\mid \tau \right){\theta }_3(0\mid \tau )={\theta }_4^3\left(\frac{\pi }{3}\mid \tau \right){\theta }_4(0\mid \tau ).$$

Based on (2)-(), we have

$${\theta }_2(0\mid 3\tau )\sqrt{\frac{{\theta }_2(0\mid 3\tau )}{{\theta }_2(0\mid \tau )}}+{\theta }_3(0\mid 3\tau )\sqrt{\frac{{\theta }_3(0\mid 3\tau )}{{\theta }_3(0\mid \tau )}}={\theta }_4(0\mid 3\tau )\sqrt{\frac{{\theta }_4(0\mid 3\tau )}{{\theta }_4(0\mid \tau )}}.$$

Let $\tau \to -\frac{1}{3\tau }$, we get

$${\theta }_3(0\mid \tau )\sqrt{\frac{{\theta }_3(0\mid \tau )}{{\theta }_3(0\mid 3\tau )}}+{\theta }_4(0\mid \tau )\sqrt{\frac{{\theta }_4(0\mid \tau )}{{\theta }_4(0\mid 3\tau )}}={\theta }_2(0\mid \tau )\sqrt{\frac{{\theta }_2(0\mid \tau )}{{\theta }_2(0\mid 3\tau )}}.$$

Notice that the above two identities are relative with modular equations which are found by Ramanujan [9].