# Some Ramanujan-type circular summation formulas

## Abstract

In this paper, we give two Ramanujan-type circular summation formulas by applying the way of elliptic functions and the properties of theta functions. As applications, we obtain the corresponding imaginary transformation formulas for Ramanujan-type circular summations and some theta function identities.

## Introduction, preparation, and motivation

The classical Jacobi four theta functions $$\vartheta _{i} (z|\tau )$$, $$i=1,2,3,4$$, with the notation of Tannery and Molk, are defined as follows.

### Definition 1.1

(see, e.g., [3, 12, 25])

For $$q=e^{\pi {i}\tau }$$, $$\operatorname{Im} (\tau )>0$$, $$z \in \mathbb{C}$$.

\begin{aligned}& \vartheta _{1} (z|\tau ) =-iq^{\frac{1}{4}}\sum _{n=-\infty }^{\infty } (-1 )^{n}q^{n ( n+1 )}e^{ (2n+1 )iz}, \end{aligned}
(1.1)
\begin{aligned}& \vartheta _{2} (z|\tau ) =q^{\frac{1}{4}}\sum _{n=-\infty }^{ \infty } q^{n (n+1 )}e^{ (2n+1 )iz}, \end{aligned}
(1.2)
\begin{aligned}& \vartheta _{3} (z|\tau ) =\sum _{n=-\infty }^{\infty }q^{n^{2}}e^{2niz}, \end{aligned}
(1.3)
\begin{aligned}& \vartheta _{4} (z|\tau ) =\sum _{n=-\infty }^{\infty } (-1 )^{n}q^{n^{2}}e^{2niz}. \end{aligned}
(1.4)

From the Jacobi theta functions (1.1)–(1.4), via the direct calculation, we have the following properties, respectively.

### Proposition 1.2

\begin{aligned} & \vartheta _{1} (z+\pi |\tau )=-\vartheta _{1} (z| \tau ),\qquad \vartheta _{1} (z+\pi \tau |\tau )=-q^{-1}e^{-2iz} \vartheta _{1} (z|\tau ), \end{aligned}
(1.5)
\begin{aligned} & \vartheta _{2} (z+\pi |\tau )=-\vartheta _{2} (z| \tau ),\qquad \vartheta _{2} (z+\pi \tau |\tau )=q^{-1}e^{-2iz} \vartheta _{2} (z|\tau ), \end{aligned}
(1.6)
\begin{aligned} & \vartheta _{3} (z+\pi |\tau )=\vartheta _{3} (z| \tau ),\qquad \vartheta _{3} (z+\pi \tau |\tau )=q^{-1}e^{-2iz} \vartheta _{3} (z|\tau ), \end{aligned}
(1.7)
\begin{aligned} & \vartheta _{4} (z+\pi |\tau )=\vartheta _{4} (z| \tau ),\qquad \vartheta _{4} (z+\pi \tau |\tau )=-q^{-1}e^{-2iz} \vartheta _{4} (z|\tau ). \end{aligned}
(1.8)

### Proposition 1.3

\begin{aligned} & \vartheta _{1} \biggl(z+{\frac{\pi }{2}}\Big| \tau \biggr)=\vartheta _{2} (z|\tau ),\qquad \vartheta _{1} \biggl(z+\frac{\pi \tau }{2}\Big|\tau \biggr)=iq^{-\frac{1}{4}}e^{-iz} \vartheta _{4} (z| \tau ), \end{aligned}
(1.9)
\begin{aligned} & \vartheta _{2} \biggl(z+\frac{\pi }{2}\Big| \tau \biggr)=-\vartheta _{1}(z| \tau ),\qquad \vartheta _{2} \biggl(z+\frac{\pi \tau }{2}\Big|\tau \biggr)=q^{-\frac{1 }{4}}e^{-iz} \vartheta _{3} (z|\tau ), \end{aligned}
(1.10)
\begin{aligned} & \vartheta _{3} \biggl(z+\frac{\pi }{2}\Big| \tau \biggr)=\vartheta _{4}(z| \tau ),\qquad \vartheta _{3} \biggl(z+\frac{\pi \tau }{2}\Big|\tau \biggr)=q^{-\frac{1 }{4}}e^{-iz} \vartheta _{2} (z|\tau ), \end{aligned}
(1.11)
\begin{aligned} & \vartheta _{4} \biggl(z+\frac{\pi }{2}\Big| \tau \biggr)=\vartheta _{3}(z| \tau ),\qquad \vartheta _{4} \biggl(z+\frac{\pi \tau }{2}\Big|\tau \biggr)=iq^{-\frac{1 }{4}}e^{-iz} \vartheta _{1} (z|\tau ). \end{aligned}
(1.12)

From (1.5)–(1.8), by applying induction, we easily obtain the following.

### Lemma 1.4

For n is a nonnegative integer, we have

\begin{aligned} & \vartheta _{1} (z+n \pi |\tau )=(-1)^{n} \vartheta _{1} (z|\tau ),\qquad \vartheta _{1} (z+n \pi \tau | \tau )=(-1)^{n}q^{-n^{2}}e^{-2niz} \vartheta _{1} (z|\tau ), \end{aligned}
(1.13)
\begin{aligned} & \vartheta _{2} (z+n \pi |\tau )=(-1)^{n} \vartheta _{2} (z|\tau ),\qquad \vartheta _{2} (z+n\pi \tau |\tau )=q^{-n^{2}}e^{-2niz} \vartheta _{2} (z|\tau ), \end{aligned}
(1.14)
\begin{aligned} & \vartheta _{3} (z+n \pi |\tau )=\vartheta _{3} (z| \tau ),\qquad \vartheta _{3} (z+n\pi \tau | \tau )=q^{-n^{2}}e^{-2niz}\vartheta _{3} (z|\tau ), \end{aligned}
(1.15)
\begin{aligned} & \vartheta _{4} (z+n \pi |\tau )=\vartheta _{4} (z| \tau ),\qquad \vartheta _{4} (z+n\pi \tau | \tau )=(-1)^{n}q^{-n^{2}}e^{-2niz}\vartheta _{4} (z|\tau ). \end{aligned}
(1.16)

From (1.9)–(1.16), we have the following lemmas.

### Lemma 1.5

For n is any positive integer, we have

\begin{aligned} & \vartheta _{1} \biggl(z+\frac{n \pi }{2}\Big| \tau \biggr)= \textstyle\begin{cases} i^{n} \vartheta _{1} (z|\tau ), & n \textit{ is even}, \\ i^{n-1} \vartheta _{2} (z|\tau ), & n \textit{ is odd}, \end{cases}\displaystyle \end{aligned}
(1.17)
\begin{aligned} & \vartheta _{2} \biggl(z+\frac{n \pi }{2}\Big| \tau \biggr)= \textstyle\begin{cases} i^{n} \vartheta _{2} (z|\tau ), & n \textit{ is even}, \\ -i^{n-1} \vartheta _{1} (z|\tau ), & n \textit{ is odd}, \end{cases}\displaystyle \end{aligned}
(1.18)
\begin{aligned} & \vartheta _{3} \biggl(z+\frac{n \pi }{2}\Big| \tau \biggr)= \textstyle\begin{cases} \vartheta _{3} (z|\tau ), & n \textit{ is even}, \\ \vartheta _{4} (z|\tau ), & n \textit{ is odd}, \end{cases}\displaystyle \end{aligned}
(1.19)
\begin{aligned} & \vartheta _{4} \biggl(z+\frac{n \pi }{2}\Big| \tau \biggr)= \textstyle\begin{cases} \vartheta _{4} (z|\tau ), & n \textit{ is even}, \\ \vartheta _{3} (z|\tau ), & n \textit{ is odd}. \end{cases}\displaystyle \end{aligned}
(1.20)

### Lemma 1.6

For n is any positive integer, we have

\begin{aligned} & \vartheta _{1} \biggl(z+\frac{n \pi \tau }{2}\Big| \tau \biggr)= \textstyle\begin{cases} i^{n} q^{-\frac{n^{2}}{4}}e^{-niz}\vartheta _{1} (z|\tau ), & n \textit{ is even}, \\ i^{n} q^{-\frac{n^{2}}{4}}e^{-niz}\vartheta _{4} (z|\tau ), & n \textit{ is odd}, \end{cases}\displaystyle \end{aligned}
(1.21)
\begin{aligned} & \vartheta _{2} \biggl(z+\frac{n \pi \tau }{2}\Big| \tau \biggr)= \textstyle\begin{cases} q^{-\frac{n^{2}}{4}}e^{-niz}\vartheta _{2} (z|\tau ), & n \textit{ is even}, \\ q^{-\frac{n^{2}}{4}}e^{-niz}\vartheta _{3} (z|\tau ), & n \textit{ is odd}, \end{cases}\displaystyle \end{aligned}
(1.22)
\begin{aligned} & \vartheta _{3} \biggl(z+\frac{n \pi \tau }{2}\Big| \tau \biggr)= \textstyle\begin{cases} q^{-\frac{n^{2}}{4}}e^{-niz}\vartheta _{3} (z|\tau ), & n \textit{ is even}, \\ q^{-\frac{n^{2}}{4}}e^{-niz}\vartheta _{2} (z|\tau ), & n \textit{ is odd}, \end{cases}\displaystyle \end{aligned}
(1.23)
\begin{aligned} & \vartheta _{4} \biggl(z+\frac{n \pi \tau }{2}\Big| \tau \biggr)= \textstyle\begin{cases} i^{n} q^{-\frac{n^{2}}{4}}e^{-niz}\vartheta _{4} (z|\tau ), & n \textit{ is even}, \\ i^{n} q^{-\frac{n^{2}}{4}}e^{-niz}\vartheta _{1} (z|\tau ), & n \textit{ is odd}. \end{cases}\displaystyle \end{aligned}
(1.24)

On page 54 in Ramanujan’s lost notebook (see [21, p. 54, Entry 9.1.1], [2, p. 337]), Ramanujan recorded the following claim (without proof), which is now well known as Ramanujan’s circular summation. The appellation circular summation was initiated by Son (see [2, p. 338]).

### Theorem 1.7

(Ramanujan’s circular summation)

For each positive integer n and $$\vert ab \vert < 1$$,

$$\sum_{-n/2< r \leq n/2} \Biggl(\sum _{ \substack{k=-\infty \\ k \equiv r (\operatorname{mod} n)}}^{\infty } a^{{k(k+1)}/{(2n)}}b^{{k(k-1)}/{(2n)}} \Biggr)^{n}=f(a,b)F_{n}(ab),$$
(1.25)

where

$$F_{n}(q):=1+2nq^{(n-1)/2}+\cdots , \quad n \geq 3.$$

Ramanujan’s theta function $$f(a,b)$$ is defined by

$$f(a,b)=\sum_{n=-\infty }^{\infty }a^{{n(n+1)}/{2}}b^{{n(n-1)}/{2}},\quad \vert ab \vert < 1.$$

Chan, Liu, and Ng  proved that Theorem 1.7 is equivalent to the following form.

### Theorem 1.8

(Ramanujan’s circular summation)

For each positive integer n,

$$\sum_{k=0}^{n-1}q^{k^{2}}e^{2k iz} \vartheta ^{n}_{3}(z+k\pi \tau |n \tau )=\vartheta _{3}(z|\tau )F_{n}(\tau ),$$
(1.26)

where for $$n \geq 3$$,

$$F_{n}(\tau )=1+2nq^{n-1}+\cdots .$$

Chan, Liu, and Ng  also showed that Theorem 1.8 is an equivalent of the theorem below by applying the Jacobi imaginary transformation formulas [25, p. 475]. They also proved that Theorem 1.9 is equivalent to Theorem 1.7.

### Theorem 1.9

(Ramanujan’s circular summation)

For any positive integer n, there exists a quantity $$G_{n}(\tau )$$ such that

$$\sum_{k=0}^{n-1} \vartheta ^{n}_{3} \biggl(z+\frac{k\pi }{n}\Big|\tau \biggr)=G_{n}(\tau )\vartheta _{3}(nz|n\tau ),$$
(1.27)

where

$$G_{n}(\tau )=\sqrt{n} (-i \tau )^{(1-n)/2}F_{n} \biggl(- \frac{1}{n \tau } \biggr).$$
(1.28)

Ramanujan’s circular summation is an interesting subject in his notebook. On the subject of Ramanujan’s circular summation and related theta function identities and their various extensions, a remarkably large number of investigations have appeared in the literature (see, for example, Andrews, Berndt, Rangachari, Ono, Ahlgren, Chua, Murayama, Son, Chan, Liu, Ng, Chan, Shen, Cai, Zhu, and Xu et al. [1, 2, 411, 13, 14, 16, 1820, 2224, 26, 27, 29, 30]).

Recently, Liu and Luo  obtained the alternating circular summation formulas of theta function $$\vartheta _{3} (z|\tau )$$. Luo  further generalized the results of Chan and Liu on Ramanujan’s circular summation formula for theta functions $$\vartheta _{3} (z|\tau )$$ and deduced some alternating summation formulas of theta functions $$\vartheta _{1} (z|\tau )$$ and $$\vartheta _{2} (z|\tau )$$. Zhou and Luo  studied a variation for Ramanujan’s circular summation of theta function $$\vartheta _{4} (z|\tau )$$, which here we call the Ramanujan-type circular summation.

Motivated by [10, 11], and [15, 17, 28], by applying the theory of elliptic functions, we further investigate other two Ramanujan-type circular summations for theta functions $$\vartheta _{1} (z|\tau )$$ and $$\vartheta _{2} (z|\tau )$$, which are two variations of Ramanujan’s circular summations (noting that is not alternating).

The paper is organized as follows: In the first section we display the definitions and properties of four theta functions. In the second section we show and prove two Ramanujan-type circular summation formulas for theta functions $$\vartheta _{1}(z|\tau )$$ and $$\vartheta _{2}(z|\tau )$$ based on the theory and method of elliptic functions and properties of theta functions. In the third section we derive the corresponding imaginary transformation formulas of circular summation formulas by using the imaginary transformation formulas of $$\vartheta _{1}(z|\tau )$$ and $$\vartheta _{2}(z|\tau )$$. In the fourth section we give some further results and remarks.

## Two Ramanujan-type circular summation formulas

In the present section, by applying the method for elliptic functions, we obtain two Ramanujan-type circular summation formulas of theta functions $$\vartheta _{1}(z|\tau )$$ and $$\vartheta _{2}(z|\tau )$$, respectively, which are two variations for Ramanujan’s circular summation formulas. We now state our main results as follows.

### Theorem 2.1

Let n be even, m be any positive integer, and p be any integer. Also let $$y_{1}, y_{2}, \ldots , y_{n}$$ be any complex numbers. Then:

• When $$y_{1}+y_{2}+\cdots +y_{n}=\frac{p \pi }{m}$$, we have

\begin{aligned} \sum_{k=0}^{mn-1}\prod _{j=1}^{n}\vartheta _{1} \biggl(z+y_{j}+ \frac{k \pi }{mn}\Big|\tau \biggr)=R_{1,1}^{(1)} (y_{1}, y_{2}, \ldots , y_{n};m,n,p;\tau ) \vartheta _{3} \bigl(mnz|m^{2}n \tau \bigr), \end{aligned}
(2.1)

where

$$R_{1,1}^{(1)} (y_{1}, y_{2}, \ldots , y_{n};m,n,p;\tau )=mnq^{- \frac{n}{4}} e^{\frac{p \pi {i}}{m}}\sum_{ \substack{r_{1}, \ldots , r_{n}=-\infty \\ r_{1}+\cdots +r_{n}=\frac{n}{2}}}^{\infty }q^{r^{2}_{1}+\cdots +r^{2}_{n}}e^{-2i (r_{1}y_{1}+\cdots +r_{n}y_{n} )}.$$
(2.2)
• When $$y_{1}+y_{2}+\cdots +y_{n}=\frac{(2p+1)\pi }{2m}$$, we have

$$\sum_{k=0}^{mn-1}\prod _{j=1}^{n}\vartheta _{1} \biggl(z+y_{j}+ \frac{k \pi }{mn} \Big|\tau \biggr)=R_{1,1}^{(2)} (y_{1}, y_{2}, \ldots , y_{n};m,n,p;\tau ) \vartheta _{4} \bigl(mnz|m^{2}n \tau \bigr),$$
(2.3)

where

\begin{aligned}& R_{1,1}^{(2)} (y_{1}, y_{2}, \ldots , y_{n};m,n,p;\tau ) \\& \quad =mnq^{- \frac{n}{4}} e^{\frac{(2p+1)\pi {i}}{2m}}\sum_{ \substack{r_{1}, \ldots , r_{n}=-\infty \\ r_{1}+\cdots +r_{n}=\frac{n}{2}}}^{\infty }q^{r^{2}_{1}+\cdots +r^{2}_{n}}e^{-2i (r_{1}y_{1}+\cdots +r_{n}y_{n} )}. \end{aligned}
(2.4)

### Proof

Let $$f(z)$$ be the left-hand side of (2.1) with $$z \longmapsto \frac{z}{mn}$$, $$\tau \longmapsto \frac{\tau }{m^{2}n}$$. We have

$$f (z )= \sum_{k=0}^{mn-1} \prod_{j=1}^{n}\vartheta _{1} \biggl(\frac{z}{mn}+y_{j}+ \frac{k\pi }{mn} \Big|\frac{\tau }{m^{2}n} \biggr).$$
(2.5)

By (1.5), we easily obtain

\begin{aligned} f(z+\pi )=& \sum_{k=0}^{mn-1}\prod _{j=1}^{n}\vartheta _{1} \biggl( \frac{z+\pi }{mn}+y_{j}+\frac{k\pi }{mn} \Big| \frac{\tau }{m^{2}n} \biggr) \\ =& \sum_{k=1}^{mn-1} \prod_{j=1}^{n}\vartheta _{1} \biggl( \frac{z}{mn}+y_{j}+ \frac{k \pi }{mn} \Big|\frac{\tau }{m^{2}n} \biggr)+(-1)^{n}\prod _{j=1}^{n}\vartheta _{1} \biggl(\frac{z}{m n}+y_{j} \Big|\frac{\tau }{m^{2}n} \biggr). \end{aligned}
(2.6)

Comparing (2.5) and (2.6), when n is only even, we have

$$f(z)=f(z+\pi ).$$
(2.7)

By (1.13) and noting that n is even, we obtain

\begin{aligned} f (z+\pi \tau )&= \sum_{k=0}^{mn-1} \prod_{j=1}^{n} \vartheta _{1} \biggl(\frac{z+\pi \tau }{mn}+y_{j}+ \frac{k\pi }{mn} \Big|\frac{\tau }{m^{2}n} \biggr) \\ &= \sum_{k=0}^{mn-1}\prod _{j=1}^{n}\vartheta _{1} \biggl( \frac{z}{mn}+y_{j}+\frac{k\pi }{mn}+m \pi \frac{\tau }{m^{2}n}\Big| \frac{\tau }{m^{2}n} \biggr) \\ &=\sum_{k=0}^{mn-1}\prod _{j=1}^{n}(-1)^{m}q^{-\frac{1}{n}}e^{-2im( \frac{z}{mn}+y_{j}+\frac{k\pi }{mn})} \vartheta _{1} \biggl( \frac{z}{mn}+y_{j}+ \frac{k\pi }{mn} \Big|\frac{\tau }{m^{2}n} \biggr) \\ &=(-1)^{mn}q^{-1}e^{-2iz}e^{-2k\pi {i}}e^{-2im(y_{1}+y_{2}+\cdots +y_{n})}f(z) \\ &= q^{-1}e^{-2iz}e^{-2im(y_{1}+y_{2}+\cdots +y_{n})}f(z). \end{aligned}
(2.8)
• When $$y_{1}+y_{2}+\cdots +y_{n}=\frac{p \pi }{m}$$ in (2.8), we have

$$f(z+\pi \tau )=q^{-1}e^{-2iz}f(z).$$
(2.9)

We construct the function $$\frac{f(z)}{\vartheta _{3}(z|\tau )}$$. By (1.7), (2.7), and (2.9), we find that the function $$\frac{f(z)}{\vartheta _{3}(z|\tau )}$$ is an elliptic function with double periods π and πτ, and has only a simple pole at $$z=\frac{\pi }{2}+\frac{\pi \tau }{2}$$ in the period parallelogram. Hence the function $$\frac{f(z)}{\vartheta _{3}(z|\tau )}$$ is a constant, say $$C_{1,1}^{(1)}(y_{1}, y_{2}, \ldots , y_{n};\tau )$$, we have

$$f(z)=C_{1,1}^{(1)}(y_{1}, y_{2}, \ldots , y_{n};\tau )\vartheta _{3}(z| \tau ),$$

or, alternatively,

$$\sum_{k=0}^{mn-1}\prod _{j=1}^{n}\vartheta _{1} \biggl(\frac{z}{mn}+y_{j}+ \frac{k\pi }{mn} \Big| \frac{\tau }{m^{2}n} \biggr)=C_{1,1}^{(1)} (y_{1}, y_{2}, \ldots , y_{n};\tau ) \vartheta _{3} (z| \tau ).$$
(2.10)

Letting

$$z \longmapsto mnz \quad \text{and}\quad \tau \longmapsto {m^{2}n \tau }$$

in (2.10), and then setting

$$R_{1,1}^{(1)} (y_{1}, y_{2}, \ldots , y_{n};m,n,p;\tau )=C_{1,1}^{(1)} \bigl(y_{1}, y_{2}, \ldots ,y_{n};{m^{2}n \tau } \bigr),$$

we arrive at (2.1).

Setting

$$z \longmapsto z+y_{j}+\frac{k \pi }{mn}$$

in (1.1), by some simple calculation and noting that n is even, we obtain

\begin{aligned}[b] \prod_{j=1}^{n} \vartheta _{1} \biggl(z+y_{j}+ \frac{k \pi }{mn} \Big| \tau \biggr)={}& i^{n}q^{\frac{n}{4}}e^{i(y_{1}+y_{2}+ \cdots +y_{n})}\sum _{r_{1},\ldots ,r_{n}=-\infty }^{\infty }(-1) ^{r_{1}+ \cdots +r_{n}}q^{r_{1}^{2}+\cdots +r_{n}^{2}+r_{1}+\cdots +r_{n}} \\ &{} \times e^{(2r_{1}+\cdots +2r_{n}+n)iz} e^{2i (r_{1}y_{1}+\cdots +r_{n}y_{n})}e^{ \frac{k\pi {i}}{mn}(2r_{1}+\cdots +2r_{n}+n)}. \end{aligned}
(2.11)

Setting

$$z \longmapsto mnz\quad \text{and}\quad \tau \longmapsto {m^{2}n \tau }$$

in (1.3), we get

$$\vartheta _{3} \bigl(mnz|m^{2}n \tau \bigr)=\sum_{r=-\infty }^{ \infty }q^{m^{2}nr^{2}}e^{2mnriz}.$$
(2.12)

Substituting (2.11) and (2.12) into (2.1), we have

\begin{aligned} &R_{1,1}^{(1)} (y_{1}, y_{2}, \ldots , y_{n} ;m,n,p;\tau ) \sum _{r=-\infty }^{\infty }q^{m^{2}nr^{2}}e^{2mnriz} \\ &\quad = i^{n}q^{\frac{n}{4}}e^{i(y_{1}+y_{2}+\cdots +y_{n})}\sum _{k=0}^{mn-1} \sum_{r_{1},\ldots ,r_{n}=-\in fty}^{\infty }(-1)^{r_{1}+\cdots +r_{n}}q^{r_{1}^{2}+ \cdots +r_{n}^{2}+r_{1}+\cdots +r_{n}} \\ &\qquad {} \times e^{(2r_{1}+\cdots +2r_{n}+n)iz} e^{2i (r_{1}y_{1}+\cdots +r_{n}y_{n})}e^{ \frac{k\pi {i}}{mn}(2r_{1}+\cdots +2r_{n}+n)}. \end{aligned}
(2.13)

Equating the constants of both sides of (2.13) and noting the condition $$y_{1}+y_{2}+\cdots +y_{n}=\frac{p \pi }{m}$$, we get (2.2).

• When $$y_{1}+y_{2}+\cdots +y_{n}=\frac{(2p+1) \pi }{2m}$$ in (2.8), we have

$$f(z+\pi \tau )=-q^{-1}e^{-2iz}f(z).$$
(2.14)

We construct the function $$\frac{f(z)}{\vartheta _{4}(z|\tau )}$$. By (1.8), (2.7), and (2.14), we find that the function $$\frac{f(z)}{\vartheta _{4}(z|\tau )}$$ is an elliptic function with double periods π and πτ, and has only a simple pole at $$z=\frac{\pi \tau }{2}$$ in the period parallelogram. Hence the function $$\frac{f(z)}{\vartheta _{4}(z|\tau )}$$ is a constant, say $$C_{1,1}^{(2)}(y_{1}, y_{2}, \ldots , y_{n};\tau )$$, we have

$$f(z)=C_{1,1}^{(2)}(y_{1}, y_{2}, \ldots , y_{n};\tau )\vartheta _{4}(z| \tau ),$$

or, equivalently,

$$\sum_{k=0}^{mn-1}\prod _{j=1}^{n}\vartheta _{1} \biggl(\frac{z}{mn}+y_{j}+ \frac{k\pi }{mn}\Big| \frac{\tau }{m^{2}n} \biggr)=C_{11}^{(1)} (y_{1}, y_{2}, \ldots , y_{n};\tau ) \vartheta _{4} (z| \tau ).$$
(2.15)

Letting

$$z \longmapsto mnz \quad \text{and}\quad \tau \longmapsto {m^{2}n \tau }$$

in (2.15), and then setting

$$R_{1,1}^{(2)} (y_{1}, y_{2}, \ldots , y_{n};m,n,p;\tau )=C_{11}^{(2)} \bigl(y_{1}, y_{2}, \ldots ,y_{n};{m^{2}n \tau } \bigr),$$

we arrive at (2.3).

A similar proof as that of (2.2). By using (1.1) and (1.4) in (2.3), and noting that n is even and the condition $$y_{1}+y_{2}+\cdots +y_{n}=\frac{(2p+1)\pi }{2m}$$, we can obtain (2.4). This proof is complete. □

### Theorem 2.2

Let n be even, m be any positive integer, and p be any integer. Also let $$y_{1}, y_{2}, \ldots , y_{n}$$ be any complex numbers. Then

• When $$y_{1}+y_{2}+\cdots +y_{n}=\frac{p \pi }{m}$$, we have

$$\sum_{k=0}^{mn-1}\prod _{j=1}^{n}\vartheta _{2} \biggl(z+y_{j}+ \frac{k \pi }{mn} \Big|\tau \biggr)=R_{2,2}^{(1)} (y_{1}, y_{2}, \ldots , y_{n};m,n,p;\tau ) \vartheta _{3} \bigl(mnz|m^{2}n \tau \bigr),$$
(2.16)

where

$$R_{2,2}^{(1)} (y_{1}, y_{2}, \ldots , y_{n};m,n,p;\tau )=mnq^{- \frac{n}{4}} e^{\frac{p \pi {i}}{m}}\sum_{ \substack{r_{1}, \ldots , r_{n}=-\infty \\ r_{1}+\cdots +r_{n}=\frac{n}{2}}}^{\infty }q^{r^{2}_{1}+\cdots +r^{2}_{n}}e^{-2i (r_{1}y_{1}+\cdots +r_{n}y_{n} )}.$$
(2.17)
• When $$y_{1}+y_{2}+\cdots +y_{n}=\frac{(2p+1)\pi }{2m}$$, we have

$$\sum_{k=0}^{mn-1}\prod _{j=1}^{n}\vartheta _{2} \biggl(z+y_{j}+ \frac{k \pi }{mn} \Big|\tau \biggr)=R_{2,2}^{(2)} (y_{1}, y_{2}, \ldots , y_{n};m,n,p;\tau ) \vartheta _{4} \bigl(mnz|m^{2}n \tau \bigr),$$
(2.18)

where

\begin{aligned}& R_{2,2}^{(2)} (y_{1}, y_{2}, \ldots , y_{n};m,n,p;\tau ) \\& \quad =mnq^{- \frac{n}{4}} e^{\frac{(2p+1)\pi {i}}{2m}}\sum_{ \substack{r_{1}, \ldots , r_{n}=-\infty \\ r_{1}+\cdots +r_{n}=\frac{n}{2}}}^{\infty }q^{r^{2}_{1}+\cdots +r^{2}_{n}}e^{-2i (r_{1}y_{1}+\cdots +r_{n}y_{n} )}. \end{aligned}
(2.19)

### Proof

For n is even, hence $$\frac{mn}{2}$$ is a positive integer.

Setting

$$z \longmapsto z+\frac{\pi }{2}$$

in equation (2.1) of Theorem 2.1 and applying properties (1.9) and (1.19), we arrive at formula (2.16) of Theorem 2.2.

Setting

$$z \longmapsto z+\frac{\pi }{2}$$

in equation (2.3) of Theorem 2.1 and applying properties (1.9) and (1.20), we arrive at formula (2.18) of Theorem 2.2.

Clearly, we consider that both $$R_{1,1}^{(1)} (y_{1}, y_{2}, \ldots , y_{n};m,n,p;\tau )$$ and $$R_{1,1}^{(2)} (y_{1}, y_{2}, \ldots , y_{n};m,n,p;\tau )$$ are independent of z, therefore we have

$$R_{2,2}^{(1)} (y_{1}, y_{2}, \ldots , y_{n};m,n,p;\tau )=R_{1,1}^{(1)} (y_{1}, y_{2}, \ldots , y_{n};m,n,p;\tau )$$

and

$$R_{2,2}^{(2)} (y_{1}, y_{2}, \ldots , y_{n};m,n,p;\tau )=R_{1,1}^{(2)} (y_{1}, y_{2}, \ldots , y_{n};m,n,p;\tau ).$$

The proof is complete. □

## The imaginary transformation formulas for Ramanujan-type circular summations

In the present section, we first derive the corresponding imaginary transformation formulas of Theorem 2.1 by applying the imaginary transformations for theta functions. Some theta function identities are also shown.

### Theorem 3.1

Let n be even, m be any positive integer, and p be any integer. Also let $$y_{1}, y_{2}, \ldots , y_{n}$$ be any complex numbers. Then

• When $$y_{1}+y_{2}+\cdots +y_{n}=mnp$$, we have

\begin{aligned}& \sum_{k=0}^{mn-1}q^{k^{2}+2kp}e^{2(k+p)iz} \prod_{j=1}^{n}\vartheta _{1} \bigl(mz+y_{j} \pi \tau +mk \pi \tau |m^{2}n \tau \bigr) \\& \quad =F_{1,1}^{(1)} (y_{1}, y_{2}, \ldots , y_{n};m,n,p;\tau ) \vartheta _{3} (z|\tau ), \end{aligned}
(3.1)

where

\begin{aligned}& F_{1,1}^{(1)} (y_{1}, y_{2}, \ldots , y_{n};m,n,p;\tau ) \\& \quad =\frac{(\sqrt{-i})^{1-3n}\sqrt{\tau ^{1-n}}}{(m \sqrt{n})^{n}} q^{- \frac{y_{1}^{2}+\cdots +y_{n}^{2}}{m^{2}n}} R_{1,1}^{(1)} \biggl( \frac{y_{1}\pi }{m^{2}n}, \frac{y_{2}\pi }{m^{2}n}, \ldots , \frac{y_{n}\pi }{m^{2}n};m,n,p;-\frac{1}{m^{2}n\tau } \biggr), \end{aligned}
(3.2)
\begin{aligned}& F_{1,1}^{(1)} (y_{1}, y_{2}, \ldots , y_{n};m,n,p;\tau ) \\& \quad =i^{n} \sum_{k=0}^{mn-1}q^{- (k+\frac{mn}{2} )^{2}} \\& \qquad {}\times \sum_{ \substack{r_{1}, \ldots , r_{n}=-\infty \\ 2m(r_{1}+\cdots +r_{n})=mn+2(k+p)}}^{\infty }(-1)^{r_{1}+\cdots +r_{n}}q^{m^{2}n(r_{1}^{2}+ \cdots +r_{n}^{2})-2(r_{1}y_{1}+\cdots +r_{n}y_{n})}. \end{aligned}
(3.3)
• When $$y_{1}+y_{2}+\cdots +y_{n}=\frac{(2p+1)mn}{2}$$, we have

\begin{aligned}& \sum_{k=0}^{mn-1}q^{k^{2}+k(2p+1)}e^{(2k+2p+1)iz} \prod_{j=1}^{n} \vartheta _{1} \bigl(mz+y_{j} \pi \tau +mk \pi \tau |m^{2}n \tau \bigr) \\& \quad =F_{1,1}^{(2)} (y_{1}, y_{2}, \ldots , y_{n};m,n,p; \tau )\vartheta _{2} (z|\tau ), \end{aligned}
(3.4)

where

\begin{aligned}& F_{1,1}^{(2)} (y_{1}, y_{2}, \ldots , y_{n};m,n,p;\tau ) \\& \quad =\frac{(\sqrt{-i})^{1-3n}\sqrt{\tau ^{1-n}}}{(m \sqrt{n})^{n}} q^{- \frac{y_{1}^{2}+\cdots +y_{n}^{2}}{m^{2}n}} R_{1,1}^{(2)} \biggl( \frac{y_{1}\pi }{m^{2}n}, \frac{y_{2}\pi }{m^{2}n}, \ldots , \frac{y_{n}\pi }{m^{2}n};m,n,p;-\frac{1}{m^{2}n\tau } \biggr), \end{aligned}
(3.5)
\begin{aligned}& F_{1,1}^{(2)} (y_{1}, y_{2}, \ldots , y_{n};m,n,p;\tau ) \\& \quad =i^{n} \sum_{k=0}^{mn-1}q^{- (k+\frac{mn}{2}-\frac{1}{2} )^{2}} \\& \qquad {}\times \sum_{ \substack{r_{1}, \ldots , r_{n}=-\infty \\ 2m(r_{1}+\cdots +r_{n})=mn+2(k+p)}}^{\infty }(-1)^{r_{1}+\cdots +r_{n}}q^{m^{2}n(r_{1}^{2}+ \cdots +r_{n }^{2})-2(r_{1}y_{1}+\cdots +r_{n}y_{n})}. \end{aligned}
(3.6)

### Proof

In (2.1) making the transformations $$\tau \longmapsto -\frac{1}{m^{2}n \tau }$$, and then $$z \longmapsto \frac{z}{mn \tau }$$ and $$y_{j} \longmapsto \frac{y_{j} \pi }{m^{2}n}$$ for $$j=1,2,\ldots , n$$, then equation (2.1) becomes

\begin{aligned}& \sum_{k=0}^{mn-1}\prod _{j=1}^{n}\vartheta _{1} \biggl( \frac{mz+y_{j} \pi \tau +mk \pi \tau }{m^{2}n \tau } \Big|{-}\frac{1}{m^{2}n \tau } \biggr) \\& \quad =R_{1,1}^{(1)} \biggl(\frac{y_{1} \pi }{m^{2}n}, \ldots , \frac{y_{n} \pi }{m^{2}n};m,n,p;- \frac{1}{m^{2}n \tau } \biggr)\vartheta _{3} \biggl( \frac{z}{\tau } \Big|{-}\frac{1}{\tau } \biggr). \end{aligned}
(3.7)

Applying the imaginary transformations formulas (see, e.g., [3, 12, 25])

$$\vartheta _{1} \biggl(\frac{z}{\tau } \Big|{-} \frac{1}{\tau } \biggr)=-i \sqrt{-i \tau } e^{\frac{iz^{2}}{\pi \tau }}\vartheta _{1} (z|\tau ) \quad \text{and}\quad \vartheta _{3} \biggl( \frac{z}{\tau } \Big|{-}\frac{1}{\tau } \biggr)=\sqrt{-i \tau } e^{ \frac{iz^{2}}{\pi \tau }}\vartheta _{3} (z|\tau )$$

to the above equation (3.7), via the suitable substitutions of the variables z and τ and noting that $$y_{1}+y_{2}+\cdots +y_{n}=mnp$$ and simplifying, we thus obtain (3.1) and (3.2). Applying the series expressions of $$\vartheta _{1} (z|\tau )$$ and $$\vartheta _{3} (z|\tau )$$ in (3.1), via direct calculation, we obtain (3.3).

In the same manner, using the imaginary transformations formulas

$$\vartheta _{1} \biggl(\frac{z}{\tau } \Big|{-} \frac{1}{\tau } \biggr)=-i \sqrt{-i \tau } e^{\frac{iz^{2}}{\pi \tau }}\vartheta _{1} (z|\tau )\quad \text{and}\quad \vartheta _{4} \biggl( \frac{z}{\tau }\Big|{-}\frac{1}{\tau } \biggr)=\sqrt{-i \tau } e^{ \frac{iz^{2}}{\pi \tau }}\vartheta _{2} (z|\tau ),$$

to (2.3) and noting that $$y_{1}+y_{2}+\cdots +y_{n}=\frac{(2p+1)mn}{2}$$, we can prove formulas (3.4), (3.5), and (3.6), respectively. Therefore we complete the proof of Theorem 3.1. □

Similarly, by applying the imaginary transformations formulas

$$\vartheta _{2} \biggl(\frac{z}{\tau } \Big|{-}\frac{1}{\tau } \biggr)=\sqrt{-i \tau } e^{\frac{iz^{2}}{\pi \tau }}\vartheta _{4} (z|\tau ) \quad \text{and}\quad \vartheta _{3} \biggl( \frac{z}{\tau }\Big|{-} \frac{1}{\tau } \biggr)=\sqrt{-i \tau } e^{ \frac{iz^{2}}{\pi \tau }}\vartheta _{3} (z|\tau ),$$

we can obtain the following imaginary transformation formulas of Theorem 2.2.

### Theorem 3.2

Let n be even, m be any positive integer, and p be any integer. Also let $$y_{1}, y_{2}, \ldots , y_{n}$$ be any complex numbers. Then

• When $$y_{1}+y_{2}+\cdots +y_{n}=mnp$$, we have

\begin{aligned}& \sum_{k=0}^{mn-1}q^{k^{2}+2kp}e^{2(k+p)iz} \prod_{j=1}^{n}\vartheta _{4} \bigl(mz+y_{j} \pi \tau +mk \pi \tau |m^{2}n \tau \bigr) \\& \quad =F_{2,2}^{(1)} (y_{1}, y_{2}, \ldots , y_{n};m,n,p;\tau ) \vartheta _{3} (z|\tau ), \end{aligned}
(3.8)

where

\begin{aligned}& F_{2,2}^{(1)} (y_{1}, y_{2}, \ldots , y_{n};m,n,p;\tau ) \\& \quad =\frac{(-i\tau )^{\frac{1-n}{2}}}{(m^{2} n)^{\frac{n}{2}}} q^{- \frac{y_{1}^{2}+\cdots +y_{n}^{2}}{m^{2}n}}R_{2,2}^{(1)} \biggl( \frac{y_{1}\pi }{m^{2}n}, \frac{y_{2}\pi }{m^{2}n}, \ldots , \frac{y_{n}\pi }{m^{2}n};m,n,p;-\frac{1}{m^{2}n\tau } \biggr), \end{aligned}
(3.9)
\begin{aligned}& F_{2,2}^{(1)} (y_{1}, y_{2}, \ldots , y_{n};m,n,p;\tau ) \\& \quad =\sum_{k=0}^{mn-1}\sum _{ \substack{r_{1}, \ldots , r_{n}=-\infty \\m(r_{1}+\cdots +r_{n})=k+p}}^{ \infty }(-1)^{r_{1}+\cdots +r_{n}}q^{m^{2}n(r_{1}^{2}+\cdots +r_{n}^{2})-2(r_{1}y_{1}+ \cdots +r_{n}y_{n})-k^{2}}. \end{aligned}
(3.10)
• When $$y_{1}+y_{2}+\cdots +y_{n}=\frac{(2p+1)mn}{2}$$, we have

\begin{aligned}& \sum_{k=0}^{mn-1}q^{k^{2}+k(2p+1)}e^{(2k+2p+1)iz} \prod_{j=1}^{n} \vartheta _{4} \bigl(mz+y_{j} \pi \tau +mk \pi \tau |m^{2}n \tau \bigr) \\& \quad =F_{2,2}^{(2)} (y_{1}, y_{2}, \ldots , y_{n};m,n,p;\tau ) \vartheta _{2} (z|\tau ), \end{aligned}
(3.11)

where

\begin{aligned}& F_{2,2}^{(2)} (y_{1}, y_{2}, \ldots , y_{n};m,n,p;\tau ) \\& \quad =\frac{(-i\tau )^{\frac{1-n}{2}}}{(m^{2} n)^{\frac{n}{2}}} q^{- \frac{y_{1}^{2}+\cdots +y_{n}^{2}}{m^{2}n}} R_{2,2}^{(2)} \biggl( \frac{y_{1}\pi }{m^{2}n}, \frac{y_{2}\pi }{m^{2}n}, \ldots , \frac{y_{n}\pi }{m^{2}n};m,n,p;-\frac{1}{m^{2}n\tau } \biggr), \end{aligned}
(3.12)
\begin{aligned}& F_{2,2}^{(2)} (y_{1}, y_{2}, \ldots , y_{n};m,n,p;\tau ) \\& \quad =\sum_{k=0}^{mn-1} \sum _{ \substack{r_{1}, \ldots , r_{n}=-\infty \\m(r_{1}+\cdots +r_{n})=k+p}}^{ \infty }(-1)^{r_{1}+\cdots +r_{n}}q^{m^{2}n(r_{1}^{2}+\cdots +r_{n}^{2})- 2(r_{1}y_{1}+\cdots +r_{n}y_{n})-(k-\frac{1}{2})^{2}}. \end{aligned}
(3.13)

Taking $$p=0$$ in Theorem 3.1.

### Corollary 3.3

Let n be even, m be any positive integer. Also let $$y_{1}, y_{2}, \ldots , y_{n}$$ be any complex numbers. Then

• When $$y_{1}+y_{2}+\cdots +y_{n}=0$$, we have

\begin{aligned}& \sum_{k=0}^{mn-1}q^{k^{2}}e^{2kiz} \prod_{j=1}^{n}\vartheta _{1} \bigl(mz+y_{j} \pi \tau +mk \pi \tau |m^{2}n \tau \bigr) \\& \quad =F_{1,1}^{(1)} (y_{1}, y_{2}, \ldots , y_{n};m,n;\tau ) \vartheta _{3} (z|\tau ), \end{aligned}
(3.14)

where

\begin{aligned}& F_{1,1}^{(1)} (y_{1}, y_{2}, \ldots , y_{n};m,n;\tau ) \\& \quad =\frac{(\sqrt{-i})^{1-3n}\sqrt{\tau ^{1-n}}}{(m \sqrt{n})^{n}} q^{- \frac{y_{1}^{2}+\cdots +y_{n}^{2}}{m^{2}n}} R_{1,1}^{(1)} \biggl( \frac{y_{1}\pi }{m^{2}n}, \frac{y_{2}\pi }{m^{2}n}, \ldots , \frac{y_{n}\pi }{m^{2}n};m,n;-\frac{1}{m^{2}n\tau } \biggr), \end{aligned}
(3.15)
\begin{aligned}& F_{1,1}^{(1)} (y_{1}, y_{2}, \ldots , y_{n};m,n;\tau ) \\& \quad =i^{n} \sum_{k=0}^{mn-1}q^{- (k+\frac{mn}{2} )^{2}} \sum_{ \substack{r_{1}, \ldots , r_{n}=-\infty \\2m(r_{1}+\cdots +r_{n})=mn+2k}}^{ \infty }(-1)^{r_{1}+\cdots +r_{n}}q^{m^{2}n(r_{1}^{2}+\cdots +r_{n}^{2})-2(r_{1}y_{1}+ \cdots +r_{n}y_{n})}. \end{aligned}
(3.16)
• When $$y_{1}+y_{2}+\cdots +y_{n}=\frac{mn}{2}$$, we have

\begin{aligned}& \sum_{k=0}^{mn-1}q^{k^{2}+k}e^{(2k+1)iz} \prod_{j=1}^{n}\vartheta _{1} \bigl(mz+y_{j} \pi \tau +mk \pi \tau |m^{2}n \tau \bigr) \\& \quad =F_{1,1}^{(2)} (y_{1}, y_{2}, \ldots , y_{n};m,n;\tau ) \vartheta _{2} (z|\tau ), \end{aligned}
(3.17)

where

\begin{aligned}& F_{1,1}^{(2)} (y_{1}, y_{2}, \ldots , y_{n};m,n;\tau ) \\& \quad =\frac{(\sqrt{-i})^{1-3n}\sqrt{\tau ^{1-n}}}{(m \sqrt{n})^{n}} q^{- \frac{y_{1}^{2}+\cdots +y_{n}^{2}}{m^{2}n}} R_{1,1}^{(2)} \biggl( \frac{y_{1}\pi }{m^{2}n}, \frac{y_{2}\pi }{m^{2}n}, \ldots , \frac{y_{n}\pi }{m^{2}n};m,n;-\frac{1}{m^{2}n\tau } \biggr), \end{aligned}
(3.18)
\begin{aligned}& F_{1,1}^{(2)} (y_{1}, y_{2}, \ldots , y_{n};m,n;\tau ) \\& \quad =i^{n} \sum_{k=0}^{mn-1}q^{- (k+\frac{mn}{2}-\frac{1}{2} )^{2}} \\& \qquad {}\times \sum_{ \substack{r_{1}, \ldots , r_{n}=-\infty \\ 2m(r_{1}+\cdots +r_{n})=mn+2k}}^{\infty }(-1)^{r_{1}+\cdots +r_{n}}q^{m^{2}n(r_{1}^{2}+ \cdots +r_{n}^{2 })-2(r_{1}y_{1}+\cdots +r_{n}y_{n})}. \end{aligned}
(3.19)

### Corollary 3.4

Let n be even and m be any positive integer. Then

\begin{aligned} & \sum_{k=0}^{mn-1}q^{k^{2}}e^{2kiz} \vartheta _{1}^{n} \bigl(mz+mk \pi \tau |m^{2}n \tau \bigr)=F_{1,1}^{(1)} (m,n; \tau )\vartheta _{3} (z|\tau ), \end{aligned}
(3.20)
\begin{aligned} & \sum_{k=0}^{mn-1}q^{k^{2}+k}e^{(2k+1)iz} \prod_{j=1}^{n}\vartheta _{1} \biggl(mz+\frac{m \pi \tau }{2}+mk \pi \tau |m^{2}n \tau \biggr)=F_{1,1}^{(2)} (m,n;\tau )\vartheta _{2} (z| \tau ), \end{aligned}
(3.21)

where

\begin{aligned} & F_{1,1}^{(1)} (m,n;\tau )= \frac{(\sqrt{-i})^{1-3n}\sqrt{\tau ^{1-n}}}{(m \sqrt{n})^{n}} R_{1,1}^{(1)} \biggl(m,n;- \frac{1}{m^{2}n\tau } \biggr), \end{aligned}
(3.22)
\begin{aligned} & F_{1,1}^{(1)} (m,n;\tau )= i^{n} \sum_{k=0}^{mn-1}q^{- (k+\frac{mn}{2} )^{2}} \sum_{ \substack{r_{1}, \ldots , r_{n}=-\infty \\ 2m(r_{1}+\cdots +r_{n})=mn+2k}}^{\infty }(-1)^{r_{1}+\cdots +r_{n}}q^{m^{2}n(r_{1}^{2}+ \cdots +r_{n}^{2 })}, \end{aligned}
(3.23)
\begin{aligned} & F_{1,1}^{(2)} (m,n;\tau )= \frac{(\sqrt{-i})^{1-3n}\sqrt{\tau ^{1-n}}}{(m \sqrt{n})^{n}} q^{- \frac{1}{4}} R_{1,1}^{(2)} \biggl(m,n;-\frac{1}{m^{2}n\tau } \biggr), \end{aligned}
(3.24)
\begin{aligned} & F_{1,1}^{(2)} (m,n;\tau )=i^{n} \sum_{k=0}^{mn-1}q^{- (k+\frac{mn}{2}+\frac{1}{2} )^{2}} \sum_{ \substack{r_{1}, \ldots , r_{n}=-\infty \\ 2m(r_{1}+\cdots +r_{n})=mn+2k}}^{\infty }(-1)^{r_{1}+\cdots +r_{n}}q^{m^{2}n(r_{1}^{2}+ \cdots +r_{n}^{2 })}. \end{aligned}
(3.25)

### Proof

Putting $$y_{1}=y_{2}=\cdots =y_{n}$$ in Corollary 3.3, we obtain Corollary 3.4. □

### Corollary 3.5

For n is even, we have

\begin{aligned} & \sum_{k=0}^{n-1}q^{k^{2}}e^{2kiz} \vartheta _{1}^{n} (z+k \pi \tau |n \tau )=F_{1,1}^{(1)} (n;\tau )\vartheta _{3} (z| \tau ), \end{aligned}
(3.26)
\begin{aligned} & \sum_{k=0}^{n-1}q^{k^{2}+k}e^{(2k+1)iz} \prod_{j=1}^{n}\vartheta _{1} \biggl(z+\frac{ \pi \tau }{2}+k \pi \tau \Big|n \tau \biggr)=F_{1,1}^{(2)} (n;\tau )\vartheta _{2} (z|\tau ), \end{aligned}
(3.27)

where

\begin{aligned} & F_{1,1}^{(1)} (n;\tau )= \frac{(\sqrt{-i})^{1-3n}\sqrt{\tau ^{1-n}}}{(\sqrt{n})^{n}} R_{1,1}^{(1)} \biggl(n;- \frac{1}{n\tau } \biggr), \end{aligned}
(3.28)
\begin{aligned} & F_{1,1}^{(1)} (n;\tau )= \sum _{k=0}^{n-1}(-1)^{k}q^{- (k+\frac{n}{2} )^{2}} \sum_{ \substack{r_{1}, \ldots , r_{n}=-\infty \\ 2(r_{1}+\cdots +r_{n})=n+2k}}^{\infty }q^{n(r_{1}^{2}+\cdots +r_{n}^{2})}, \end{aligned}
(3.29)
\begin{aligned} & F_{1,1}^{(2)} (n;\tau )= \frac{(\sqrt{-i})^{1-3n}\sqrt{\tau ^{1-n}}}{( \sqrt{n})^{n}} q^{- \frac{1}{4}} R_{1,1}^{(2)} \biggl(n;-\frac{1}{n\tau } \biggr), \end{aligned}
(3.30)
\begin{aligned} & F_{1,1}^{(2)} (n;\tau )=\sum _{k=0}^{n-1}(-1)^{k}q^{- (k+\frac{n}{2}+\frac{1}{2} )^{2}} \sum_{ \substack{r_{1}, \ldots , r_{n}=-\infty \\ 2(r_{1}+\cdots +r_{n})=n+2k}}^{\infty }q^{n(r_{1}^{2}+\cdots +r_{n}^{2})}. \end{aligned}
(3.31)

### Proof

Taking $$m=1$$ in Corollary 3.4, we obtain Corollary 3.5. □

### Corollary 3.6

Setting $$m=1$$, $$n=2$$ in (3.1), $$y_{1}=y_{2}=p$$, we have

\begin{aligned}& e^{2piz} \vartheta _{1}^{2}(z+p \pi \tau |2\tau )+q^{2p+1}e^{2(p+1)iz} \vartheta _{1}^{2}(z+p \pi \tau +\pi \tau |2\tau ) \\& \quad =(-1)^{p} \bigl[\vartheta _{2}(2p \pi \tau |4\tau )-q^{2p+1}\vartheta _{2} \bigl((2p+1) \pi \tau |4\tau \bigr)\vartheta _{3}(z|\tau ) \bigr]. \end{aligned}
(3.32)

### Corollary 3.7

Setting $$m=1$$, $$n=2$$ in (3.4), $$y_{1}=y_{2}=\frac{2p+1}{2}$$, we have

\begin{aligned}& e^{(2p+1)iz} \vartheta _{1}^{2} \biggl(z+p \pi \tau + \frac{\pi \tau }{2}\Big|2\tau \biggr)+q^{2p+2}e^{(2p+3)iz} \vartheta _{1}^{2} \biggl(z+p \pi \tau + \frac{3\pi \tau }{2}\Big|2\tau \biggr) \\& \quad =(-1)^{p}q^{-\frac{1}{4}} \bigl[q^{-p}\vartheta _{2}(2p \pi \tau |4\tau )-q^{p+1} \vartheta _{2} \bigl((2p+1) \pi \tau |4\tau \bigr)\vartheta _{2}(z|\tau ) \bigr]. \end{aligned}
(3.33)

Taking $$p=0$$ in (3.32) and (3.33), respectively, we deduce the following theta function identities:

\begin{aligned} & \vartheta _{1}^{2}(z|2\tau )+qe^{2iz} \vartheta _{1}^{2}(z+\pi \tau |2 \tau )= \vartheta _{2}(0|4\tau )-q \vartheta _{2}(\pi \tau |4 \tau )\vartheta _{3}(z|\tau ), \end{aligned}
(3.34)
\begin{aligned} & e^{iz} \vartheta _{1}^{2} \biggl(z+\frac{\pi \tau }{2}\Big|2\tau \biggr)+q^{2}e^{3iz} \vartheta _{1}^{2} \biggl(z+\frac{3\pi \tau }{2}\Big|2 \tau \biggr) \\ &\quad = q^{- \frac{1}{4}} \bigl[\vartheta _{2}(0|4\tau )-q \vartheta _{2}(\pi \tau |4 \tau )\vartheta _{2}(z|\tau ) \bigr]. \end{aligned}
(3.35)

Taking $$p=1$$ in (3.32) and (3.33), respectively, we deduce the following theta function identities:

\begin{aligned} & \vartheta _{4}^{2}(z|2\tau )- \vartheta _{1}^{2}(z|2\tau )= q \vartheta _{2}(2 \pi \tau |4\tau )-q^{4}\vartheta _{2}(3 \pi \tau |4 \tau )\vartheta _{3}(z|\tau ), \end{aligned}
(3.36)
\begin{aligned} & \vartheta _{4}^{2} \biggl(z+ \frac{\pi \tau }{2}\Big|2\tau \biggr)- \vartheta _{1}^{2} \biggl(z+\frac{\pi \tau }{2}\Big|2\tau \biggr) \\ &\quad =q^{ \frac{3}{4}}e^{-iz} \bigl[\vartheta _{2}(2 \pi \tau |4\tau )-q^{3} \vartheta _{2}(3 \pi \tau |4\tau )\vartheta _{2}(z|\tau ) \bigr]. \end{aligned}
(3.37)

Taking $$p=-1$$ in (3.32) and (3.33), respectively, we deduce the following theta function identities:

\begin{aligned} & e^{-2iz} \vartheta _{1}^{2}(z- \pi \tau |2\tau )+q^{-1} \vartheta _{1}^{2}(z|2 \tau )= \vartheta _{2}(-2 \pi \tau |4\tau )+q^{-1} \vartheta _{2}(- \pi \tau |4\tau )\vartheta _{3}(z|\tau ), \end{aligned}
(3.38)
\begin{aligned} & e^{-iz} \vartheta _{1}^{2} \biggl(z-\frac{\pi \tau }{2}\Big|2\tau \biggr)+e^{iz} \vartheta _{1}^{2} \biggl(z+\frac{\pi \tau }{2}\Big|2\tau \biggr) \\ &\quad = -q^{- \frac{1}{4}} \bigl[q \vartheta _{2}(-2 \pi \tau |4\tau )- \vartheta _{2}(- \pi \tau |4\tau )\vartheta _{2}(z|\tau ) \bigr]. \end{aligned}
(3.39)

### Remark 3.8

Corollary 3.5 is an analogue of Ramanujan’s circular summation formula Theorem 1.8.

### Remark 3.9

From Theorem 3.1 and Theorem 3.2 we may obtain more theta function identities.

## Further results and remarks

In the present section, we give some special cases of Theorem 2.1 and derive some theta function identities.

Setting $$p=0$$ in Theorem 2.1, we have the following.

### Corollary 4.1

Suppose that n is even, m is any positive integer; $$y_{1}, y_{2}, \ldots , y_{n}$$ are any complex numbers.

• When $$y_{1}+y_{2}+\cdots +y_{n}=0$$, we have

$$\sum_{k=0}^{mn-1}\prod _{j=1}^{n}\vartheta _{1} \biggl(z+y_{j}+ \frac{k \pi }{mn} \Big|\tau \biggr)=R_{1,1}^{(1)} (y_{1}, y_{2}, \ldots , y_{n};m,n;\tau ) \vartheta _{3} \bigl(mnz|m^{2}n \tau \bigr),$$
(4.1)

where

$$R_{1,1}^{(1)} (y_{1}, y_{2}, \ldots , y_{n};m,n;\tau )=mnq^{- \frac{n}{4}} \sum_{ \substack{r_{1}, \ldots , r_{n}=-\infty \\ r_{1}+\cdots +r_{n}=\frac{n}{2}}}^{\infty }q^{r^{2}_{1}+\cdots +r^{2}_{n}}e^{-2i (r_{1}y_{1}+\cdots +r_{n}y_{n} )}.$$
(4.2)
• When $$y_{1}+y_{2}+\cdots +y_{n}=\frac{\pi }{2m}$$, we have

$$\sum_{k=0}^{mn-1}\prod _{j=1}^{n}\vartheta _{1} \biggl(z+y_{j}+ \frac{k \pi }{mn} \Big|\tau \biggr)=R_{1,1}^{(2)} (y_{1}, y_{2}, \ldots , y_{n};m,n;\tau ) \vartheta _{4} \bigl(mnz|m^{2}n \tau \bigr),$$
(4.3)

where

$$R_{1,1}^{(2)} (y_{1}, y_{2}, \ldots , y_{n};m,n;\tau )=mnq^{- \frac{n}{4}} e^{\frac{\pi {i}}{2m}}\sum_{ \substack{r_{1}, \ldots , r_{n}=-\infty \\ r_{1}+\cdots +r_{n}=\frac{n}{2}}}^{\infty }q^{r^{2}_{1}+\cdots +r^{2}_{n}}e^{-2i (r_{1}y_{1}+\cdots +r_{n}y_{n} )}.$$
(4.4)

### Corollary 4.2

Suppose that n is even, m is any positive integer, then

\begin{aligned} & \sum_{k=0}^{mn-1} \vartheta _{1}^{n} \biggl(z+\frac{k \pi }{mn} \Big|\tau \biggr)=R_{1,1}^{(1)} (m,n;\tau ) \vartheta _{3} \bigl(mnz|m^{2}n \tau \bigr), \end{aligned}
(4.5)
\begin{aligned} & \sum_{k=0}^{mn-1} \vartheta _{1}^{n} \biggl(z+\frac{\pi }{2mn}+ \frac{k \pi }{mn} \Big|\tau \biggr)=R_{1,1}^{(2)} (m,n; \tau )\vartheta _{4} \bigl(mnz|m^{2}n \tau \bigr), \end{aligned}
(4.6)

where

$$R_{1,1}^{(1)} (m,n;\tau )=R_{1,1}^{(2)} (m,n;\tau )=mnq^{-\frac{n}{4}} \sum _{ \substack{r_{1}, \ldots , r_{n}=-\infty \\ r_{1}+\cdots +r_{n}=\frac{n}{2}}}^{\infty }q^{r^{2}_{1}+\cdots +r^{2}_{n}}.$$
(4.7)

### Proof

Taking $$y_{1}=y_{2}=\cdots =y_{n}$$ in Corollary 4.1, we get Corollary 4.2. □

### Corollary 4.3

For m is any positive integer, we have

\begin{aligned} & \sum_{k=0}^{2m-1} \vartheta _{1}^{2} \biggl(z+\frac{k \pi }{2m} \Big|\tau \biggr)=2m \vartheta _{2} (0|2 \tau ) \vartheta _{3} \bigl(2mz|2m^{2} \tau \bigr), \end{aligned}
(4.8)
\begin{aligned} & \sum_{k=0}^{2m-1} \vartheta _{1}^{2} \biggl(z+\frac{\pi }{4m}+ \frac{k \pi }{2m} \Big|\tau \biggr)=2m \vartheta _{2} (0|2 \tau ) \vartheta _{4} \bigl(2mz|2m^{2} \tau \bigr). \end{aligned}
(4.9)

### Proof

Putting $$n=2$$ in Theorem 4.2 and noting that $$R_{1,1}^{(1)} (m,n;\tau )=R_{1,1}^{(2)} (m,n;\tau )=2m \vartheta _{2} (0| 2 \tau )$$, we get Corollary 4.3. □

### Remark 4.4

Corollary 4.3 is an analogue of Boon’s result [6, p. 3440, Eq. (10)].

### Corollary 4.5

For m is any positive integer, we have

\begin{aligned} & \sum_{k=0}^{4m-1} \vartheta _{1}^{4} \biggl(z+\frac{k \pi }{4m} \Big| \tau \biggr)=R(m;\tau )\vartheta _{3} \bigl(4mz|4m^{2} \tau \bigr), \end{aligned}
(4.10)
\begin{aligned} & \sum_{k=0}^{4m-1} \vartheta _{1}^{4} \biggl(z+\frac{\pi }{8m}+ \frac{k \pi }{4m} \Big|\tau \biggr)=R(m;\tau )\vartheta _{4} \bigl(4mz|4m^{2} \tau \bigr), \end{aligned}
(4.11)

where

$$R(m;\tau )=4mq^{-1} \sum _{ \substack{r_{1}, \ldots , r_{4}=-\infty \\ r_{1}+\cdots +r_{4}=2}}^{\infty }q^{r^{2}_{1}+\cdots +r^{2}_{4}}.$$
(4.12)

### Corollary 4.6

For n is even, we have

\begin{aligned} & \sum_{k=0}^{n-1} \vartheta _{1}^{n} \biggl(z+\frac{k \pi }{n} \Big| \tau \biggr)=R(\tau )\vartheta _{3} (nz|n \tau ), \end{aligned}
(4.13)
\begin{aligned} & \sum_{k=0}^{n-1} \vartheta _{1}^{n} \biggl(z+\frac{\pi }{2n}+ \frac{k \pi }{n} \Big|\tau \biggr)=R(\tau )\vartheta _{4} (nz|n \tau ), \end{aligned}
(4.14)

where

$$R(\tau )=nq^{-\frac{n}{4}}\sum _{ \substack{r_{1}, \ldots , r_{n}=-\infty \\ r_{1}+\cdots +r_{n}=\frac{n}{2}}}^{\infty }q^{r^{2}_{1}+\cdots +r^{2}_{n}}.$$
(4.15)

### Proof

Taking $$m=1$$ in Corollary 4.2, we get Corollary 4.6. □

Setting $$m=1$$ in Theorem 2.1, we have the following.

### Corollary 4.7

Suppose that n is even, p is any integer; $$y_{1}, y_{2}, \ldots , y_{n}$$ are any complex numbers.

• When $$y_{1}+y_{2}+\cdots +y_{n}=p \pi$$, we have

$$\sum_{k=0}^{n-1}\prod _{j=1}^{n}\vartheta _{1} \biggl(z+y_{j}+ \frac{k \pi }{n} \Big|\tau \biggr)=R_{1,1}^{(1)} (y_{1}, y_{2}, \ldots , y_{n};n,p;\tau ) \vartheta _{3} (nz|n \tau ),$$
(4.16)

where

$$R_{1,1}^{(1)} (y_{1}, y_{2}, \ldots , y_{n};n,p;\tau )=(-1)^{p}nq^{- \frac{n}{4}} \sum_{ \substack{r_{1}, \ldots , r_{n}=-\infty \\ r_{1}+\cdots +r_{n}=\frac{n}{2}}}^{\infty }q^{r^{2}_{1}+\cdots +r^{2}_{n}}e^{-2i (r_{1}y_{1}+\cdots +r_{n}y_{n} )}.$$
(4.17)
• When $$y_{1}+y_{2}+\cdots +y_{n}=\frac{(2p+1)\pi }{2}$$, we have

$$\sum_{k=0}^{n-1}\prod _{j=1}^{n}\vartheta _{1} \biggl(z+y_{j}+ \frac{k \pi }{n} \Big|\tau \biggr)=R_{1,1}^{(2)} (y_{1}, y_{2}, \ldots , y_{n};n,p;\tau ) \vartheta _{4} (nz|n \tau ),$$
(4.18)

where

$$R_{1,1}^{(2)} (y_{1}, y_{2}, \ldots , y_{n};n,p;\tau )=(-1)^{p}niq^{- \frac{n}{4}} \sum_{ \substack{r_{1}, \ldots , r_{n}=-\infty \\ r_{1}+\cdots +r_{n}=\frac{n}{2}}}^{\infty }q^{r^{2}_{1}+\cdots +r^{2}_{n}}e^{-2i (r_{1}y_{1}+\cdots +r_{n}y_{n} )}.$$
(4.19)

Setting $$p=0$$ in Corollary 4.7, we have the following.

### Corollary 4.8

Suppose that n is even; $$y_{1}, y_{2}, \ldots , y_{n}$$ are any complex numbers.

• When $$y_{1}+y_{2}+\cdots +y_{n}=0$$, we have

$$\sum_{k=0}^{n-1}\prod _{j=1}^{n}\vartheta _{1} \biggl(z+y_{j}+ \frac{k \pi }{n} \Big|\tau \biggr)=R_{1,1}^{(1)} (y_{1}, y_{2}, \ldots , y_{n};n;\tau ) \vartheta _{3} (nz|n \tau ),$$
(4.20)

where

$$R_{1,1}^{(1)} (y_{1}, y_{2}, \ldots , y_{n};n;\tau )=nq^{- \frac{n}{4}} \sum _{ \substack{r_{1}, \ldots , r_{n}=-\infty \\ r_{1}+\cdots +r_{n}=\frac{n}{2}}}^{\infty }q^{r^{2}_{1}+\cdots +r^{2}_{n}}e^{-2i (r_{1}y_{1}+\cdots +r_{n}y_{n} )}.$$
(4.21)
• When $$y_{1}+y_{2}+\cdots +y_{n}=\frac{\pi }{2}$$, we have

$$\sum_{k=0}^{n-1}\prod _{j=1}^{n}\vartheta _{1} \biggl(z+y_{j}+ \frac{k \pi }{n} \Big|\tau \biggr)=R_{1,1}^{(2)} (y_{1}, y_{2}, \ldots , y_{n};n;\tau ) \vartheta _{4} (nz|n \tau ),$$
(4.22)

where

$$R_{1,1}^{(2)} (y_{1}, y_{2}, \ldots , y_{n};n;\tau )=niq^{- \frac{n}{4}} \sum _{ \substack{r_{1}, \ldots , r_{n}=-\infty \\ r_{1}+\cdots +r_{n}=\frac{n}{2}}}^{\infty }q^{r^{2}_{1}+\cdots +r^{2}_{n}}e^{-2i (r_{1}y_{1}+\cdots +r_{n}y_{n} )}.$$
(4.23)

### Corollary 4.9

Suppose that n is even, m is any positive integer, a, b are any nonnegative integers and $$a+b=n$$, p is any integer; x, y are any complex numbers.

• When $$ax+by=\frac{p \pi }{m}$$, we have

\begin{aligned}& \sum_{k=0}^{mn-1} \vartheta _{1}^{a} \biggl(z+x+\frac{k \pi }{mn} \Big| \tau \biggr) \vartheta _{1}^{b} \biggl(z+y+ \frac{k \pi }{mn} \Big|\tau \biggr) \\& \quad =R_{1,1}^{(1)} (x,y;m,n,p; \tau )\vartheta _{3} \bigl(mnz|m^{2}n \tau \bigr), \end{aligned}
(4.24)

where

\begin{aligned}& R_{1,1}^{(1)} (x,y;m,n,p;\tau ) \\& \quad =mnq^{-\frac{n}{4}} e^{ \frac{p \pi {i}}{m}-nyi}\sum_{ \substack{r_{1}, \ldots , r_{a}, s_{1}, \ldots , s_{b}=-\infty \\ r_{1}+\cdots +r_{a}+s_{1}+\cdots +s_{b}=\frac{n}{2}}}^{\infty }q^{r_{1}^{2}+ \cdots +r_{a}^{2}+s_{1}^{ 2}+\cdots +s_{b}^{2}}e^{-2i (r_{1}+\cdots +r_{a})(x-y) }. \end{aligned}
(4.25)
• When $$ax+by=\frac{(2p+1)\pi }{2m}$$, we have

\begin{aligned}& \sum_{k=0}^{mn-1} \vartheta _{1}^{a} \biggl(z+x+\frac{k \pi }{mn} \Big| \tau \biggr) \vartheta _{1}^{b} \biggl(z+y+ \frac{k \pi }{mn} \Big|\tau \biggr) \\& \quad =R_{1,1}^{(2)} (x,y;m,n,p; \tau )\vartheta _{4} \bigl(mnz|m^{2}n \tau \bigr), \end{aligned}
(4.26)

where

\begin{aligned}& R_{1,1}^{(2)} (x,y;m,n,p;\tau ) \\& \quad =mnq^{-\frac{n}{4}} e^{ \frac{(2p+1)\pi {i}}{2m}-nyi}\sum_{ \substack{r_{1}, \ldots , r_{a}, s_{1}, \ldots , s_{b}=-\infty \\ r_{1}+\cdots +r_{a}+s_{1}+\cdots +s_{b}=\frac{n}{2}}}^{\infty }q^{r_{1}^{2}+ \cdots +r_{a}^{2}+s_{1}^{ 2}+\cdots +s_{b}^{2}}e^{-2i (r_{1}+\cdots +r_{a})(x-y)}. \end{aligned}
(4.27)

### Proof

Setting $$y_{1}=y_{2}=\cdots =y_{a}=x$$ and $$y_{a+1}=y_{a+2}=\cdots =y_{n}=y$$ with $$a+b=n$$ in Theorem 2.1, we obtain Corollary 4.9 □

### Corollary 4.10

Suppose that m is any positive integer.

\begin{aligned} & \sum_{k=0}^{2m-1} \vartheta _{1} \biggl(z+x+\frac{k \pi }{2m}\Big| \tau \biggr) \vartheta _{1} \biggl(z-x+\frac{k \pi }{2m} \Big|\tau \biggr)=2m \vartheta _{2} (2x|2 \tau )\vartheta _{3} \bigl(2mz|2m^{2} \tau \bigr), \end{aligned}
(4.28)
\begin{aligned} & \sum_{k=0}^{2m-1} \vartheta _{1} \biggl(z+x+\frac{k \pi }{2m} \Big| \tau \biggr) \vartheta _{1} \biggl(z-x+\frac{(k+1) \pi }{2m} \Big|\tau \biggr) \\ &\quad =2m \vartheta _{2} \biggl(2x- \frac{\pi }{2m}\Big|2 \tau \biggr) \vartheta _{4} \bigl(2mz|2m^{2} \tau \bigr). \end{aligned}
(4.29)

### Proof

Taking $$p=0$$, $$a=b=1$$ in Corollary 4.9, we have

\begin{aligned} & R_{1,1}^{(1)} (x;m;\tau )=2mq^{-\frac{1}{2}} e^{2xi} \sum_{\substack{r_{1}, s_{1}=-\infty \\ r_{1}+s_{1}=1}}^{\infty }q^{r_{1}^{2}+s_{1}^{2}}e^{-4i r_{1}x}=2m \vartheta _{2} (2x|2 \tau ), \\ & R_{1,1}^{(2)} (x,y;m;\tau )=2mq^{-\frac{1}{2}} e^{2ix- \frac{\pi {i}}{2m}}\sum_{ \substack{r_{1}, s_{1}=-\infty \\ r_{1}+s_{1}=1}}^{\infty }q^{r_{1}^{2}+s_{1}^{2}}e^{-2i r_{1}(2x- \frac{\pi }{2m})}=2m \vartheta _{2} \biggl(2x-\frac{\pi }{2m}\Big|2 \tau \biggr) \end{aligned}

and noting that $$x+y=0$$ and $$x+y=\frac{\pi }{2m}$$, we obtain Corollary 4.10. □

Taking $$m=1$$ in Corollary 4.10, we have

\begin{aligned} & \vartheta _{1} (z+x |\tau ) \vartheta _{1} (z-x | \tau ) +\vartheta _{2} (z+x |\tau ) \vartheta _{2} (z-x |\tau ) =2 \vartheta _{2} (2x|2 \tau )\vartheta _{3} (2z|2 \tau ), \end{aligned}
(4.30)
\begin{aligned} & \vartheta _{1} (z+x |\tau ) \vartheta _{2} (z-x | \tau ) - \vartheta _{2} (z+x |\tau ) \vartheta _{1} (z-x |\tau ) =2 \vartheta _{1} (2x|2 \tau ) \vartheta _{4} (2z|2 \tau ). \end{aligned}
(4.31)

Taking $$m=2$$ in Corollary 4.10, we have

\begin{aligned} & \vartheta _{1} (z+x |\tau ) \vartheta _{1} (z-x | \tau ) +\vartheta _{2} (z+x |\tau ) \vartheta _{2} (z-x |\tau ) \\ &\qquad {}+\vartheta _{1} \biggl(z+x+\frac{\pi }{4} \Big|\tau \biggr) \vartheta _{1} \biggl(z-x+\frac{\pi }{4} \Big|\tau \biggr) +\vartheta _{1} \biggl(z+x-\frac{\pi }{4} \Big|\tau \biggr) \vartheta _{1} \biggl(z-x-\frac{\pi }{4} \Big|\tau \biggr) \\ &\quad =4 \vartheta _{2} (2x|2 \tau )\vartheta _{3} (4z|8 \tau ), \end{aligned}
(4.32)
\begin{aligned} & \vartheta _{1} (z+x |\tau ) \vartheta _{1} \biggl(z-x+\frac{\pi }{4} \Big|\tau \biggr) +\vartheta _{1} \biggl(z+x+\frac{\pi }{4} \Big|\tau \biggr) \vartheta _{2} (z-x | \tau ) \\ &\qquad {} -\vartheta _{2} (z+x |\tau ) \vartheta _{1} \biggl(z-x-\frac{ \pi }{4} \Big|\tau \biggr) +\vartheta _{1} \biggl(z+x-\frac{\pi }{4} \Big|\tau \biggr) \vartheta _{1} (z-x |\tau ) \\ &\quad =4 \vartheta _{2} \biggl(2x-\frac{\pi }{4}\Big|2 \tau \biggr) \vartheta _{4} (4z|8 \tau ). \end{aligned}
(4.33)

### Remark 4.11

We can obtain the corresponding results of $$\vartheta _{2} (z|\tau )$$ from Theorem 2.2, we here omit them.

### Remark 4.12

No doubt more theta function identities may be formulated from Theorem 2.2 and other theorems and corollaries.

## Conclusion

Ramanujan’s circular summation is an interesting subject in his notebook. In this paper, we obtain two Ramanujan-type circular summation formulas. We also give the corresponding imaginary transformation formulas for Ramanujan-type circular summations and some identities of the classical Jacobi theta functions $$\vartheta _{1} (z|\tau )$$ and $$\vartheta _{2} (z|\tau )$$.

Not applicable.

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## Acknowledgements

The authors greatly appreciate two referees for their valuable comments and suggestions.

## Funding

The present investigation was supported by the Natural Science Foundation General Project of Chongqing, China under grant cstc2019jcyj-msxmX0143 and the Science and Technology Research Program of Chongqing Municipal Education Commission, China under grant No. KJZD-K201901701.

## Author information

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### Contributions

There was an equal amount of contributions from all authors. All authors read and approved the final manuscript.

### Corresponding author

Correspondence to Qiu-Ming Luo.

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