\jyear

2021

[1,2]\fnmHemant \surMasal

[1]\orgdivDepartment of First Year Engineering, \orgnamePune Institute of Computer Technology, \orgaddress\streetDhankawadi, \cityPune, \postcode411 043, \stateMaharashtra, \countryIndia

2]\orgdivDepartment of Mathematics, \orgnameSavitribai Phule Pune University, \orgaddress\streetGaneshkhind, \cityPune, \postcode411 007, \stateMaharashtra, \countryIndia

Ramanujan Theta Function Identities and Quadratic Numbers

[email protected] \fnmHemant \surBhate [email protected] \fnmSubhash \surKendre [email protected] * [

Abstract

Eigenvectors of the discrete Fourier transform can be expressed using Ramanujan theta functions. New theta function identities, Ramanujan theta function identities, and generating functions for the quadratic numbers are a consequence.

keywords:

Discrete Fourier transform, eigenvectors, Ramanujan theta function identities, Modular equations

pacs:

[

MSC Classification]11F03, 11F27, 15A18, 33F05

1 Introduction

Let $A$ denote the matrix of the discrete Fourier transform. The $(j,k)^{th}$ entry of the DFT matrix $A$ of size $n$ is $A_{jk}=\frac{1}{\sqrt{n}}e^{\frac{2\pi ijk}{n}},~{}~{}j,k\in\mathbb{Z}_{n}$ . It is clear that $A^{4}=I$ , and hence the eigenvalues are $1,-1,i,-i$ with non negative multiplicities $1,-1,i,-i$ are $[(n+4)/4],[(n+2)/4],[(n+1)/4],[(n-1)/4]$ respectively matveev2001intertwining ; mehta1987eigenvalues , where $[x]$ denotes greatest integer not greater than $x.$

Recall following Theorem from Matveev matveev2001intertwining :

Theorem 1.

Let $\sum_{m\in\mathbb{Z}}g_{m}$ be any absolutely convergent series. Then the vector whose $j^{th}$ component is

v_{j}(k)=\sum_{m\in\mathbb{Z}}(g_{mn+j}+(-1)^{k}g_{mn-j})+\frac{1}{\sqrt{n}}\sum_{m\in\mathbb{Z}}\left[(-i)^{k}g_{m}+(i)^{k}g_{-m}\right]e^{\frac{2\pi imj}{n}}

is an eigenvector of DFT of order $n$ corresponding to eigenvalue $i^{k}$ .

2 The Ramanujan theta function

Definition 1.

The Ramanujan theta function is defined by berndt2012ramanujan ; Bruce ; Robert ,

f(a,b)=\sum_{m\in\mathbb{Z}}a^{\frac{m(m+1)}{2}}b^{\frac{m(m-1)}{2}},~{}~{}~{}~{}~{}~{}\mid ab\mid<1.

(1)

The Ramanujan theta function satisfy berndt2012ramanujan

f(a,b)=f(b,a),f(1,a)=2f(a,a^{3}),f(-1,a)=0.

Also, for fixed integer $n,$

f(a,b)=a^{\frac{n(n+1)}{2}}b^{\frac{n(n-1)}{2}}f(a(ab)^{n},b(ab)^{-n}).

Recall that the notation for the infinite product is,

(\alpha:\beta)_{\infty}=\prod_{k=0}^{\infty}(1-\alpha\beta^{k}),~{}~{}\text{and}~{}(\alpha,\gamma,\cdots,\zeta:\beta)_{\infty}=(\alpha:\beta)_{\infty}(\gamma:\beta)_{\infty}\cdots(\zeta:\beta)_{\infty}

The well known Jacobi triple product identity berndt2012ramanujan ; murty2015problems gives us following

\begin{split}f(a,b)=(-a:ab)_{\infty}(-b:ab)_{\infty}(ab:ab)_{\infty}=(-a,-b,-ab:ab)_{\infty}.\end{split}

(2)

We now introduce the Ramanujan theta function with characteristics $(\alpha,\beta,c)$ as

f_{(\alpha,\beta,c)}(a,b,x)=\sum_{m\in\mathbb{Z}}a^{\frac{(m+\alpha)(m+\alpha+1/c)}{2}}b^{\frac{(m+\alpha)(m+\alpha-1/c)}{2}}e^{2\pi i(m+\alpha)(x+\beta)}.

(3)

For $\alpha=0,\beta=0,x=0,c=1$ we will get Ramanujan theta function.

3 The Eigenvectors of DFT

The eigenvalue and eigenvectors decomposition of DFT is studied by McClellan and Parks mcclellan1972eigenvalue . Mehta mehta1987eigenvalues studied eigenvctors of DFT using Hermite functions whereas Matveev matveev2001intertwining proved that Jacobi theta functions also gives eigenvectors of the DFT. Following Theorem gives the eigenvectors of the DFT in the form of Ramanujan theta function.

Theorem 2.

For $n$ fixed, $a,b\in\mathbb{C}$ with $\mid ab\mid<1$ and for any $x\in\mathbb{C}$ , the $j^{th},$ component with $j\in\{0,1,2,\cdots,n-1\},$ of the eigenvector corresponding to eigenvalue $i^{k}$ of the DFT is

\begin{split}v_{j}(k)=&f_{(\frac{j}{n},0,n)}(a,b,x)+(-1)^{k}f_{(\frac{-j}{n},0,n)}(a,b,x)\\ +&\frac{1}{\sqrt{n}}\left[(-i)^{k}f_{(0,\frac{j}{n},1)}(a^{1/n^{2}},b^{1/n^{2}},x/n)+(-i)^{3k}f_{(0,\frac{-j}{n},1)}(a^{1/n^{2}},b^{1/n^{2}},x/n)\right].\end{split}

Remark 1.

This follows by taking

g_{m}=a^{\frac{m(m+1)}{2n^{2}}}b^{\frac{m(m-1)}{2n^{2}}}e^{2\pi i\frac{m}{n}x}

(4)

and applying Theorem 1.

4 Functional Identities

In this section we established some functional relations between Ramanujan theta functions.

Lemma 3.

The following functional equations holds for the DFT of size $2$

1.

$2f_{(0,0,2)}(a,b,x)+f_{(\frac{1}{2},0,2)}(a,b,x)+f_{(\frac{-1}{2},0,2)}(a,b,x)=2f_{(0,0,1)}(a^{1/4},b^{1/4},x/2)$
2.

$2f_{(0,0,2)}(a,b,x)-[f_{(\frac{1}{2},0,2)}(a,b,x)+f_{(\frac{-1}{2},0,2)}(a,b,x)]=f_{(0,\frac{1}{2},1)}(a^{1/4},b^{1/4},x/2)+f_{(0,\frac{-1}{2},1)}(a^{1/4},b^{1/4},x/2).$

Proof: The DFT $A$ of size $2$ has two eigenvalues $1,-1$ . Both the eigenvalues are non degenerate. Let $v_{1},v_{2}$ be the eigenvectors corresponding to the eigenvalues $1,-1$ respectively. By using Theorem 2, we have

v_{1}=\begin{pmatrix}v_{0}(0)\\ v_{1}(0)\end{pmatrix}~{}~{}\text{and}~{}~{}v_{2}=\begin{pmatrix}v_{0}(2)\\ v_{1}(2)\end{pmatrix},

where,

\begin{split}v_{0}(0)=&2f_{(0,0,2)}(a,b,x)+\sqrt{2}f_{(0,0,1)}(a^{1/4},b^{1/4},x/2)\\ v_{1}(0)=&f_{(\frac{1}{2},0,2)}(a,b,x)+f_{(\frac{-1}{2},0,2)}(a,b,x)\\ &+\frac{1}{\sqrt{2}}\left[f_{(0,\frac{1}{2},1)}(a^{1/4},b^{1/4},x/2)+f_{(0,\frac{-1}{2},1)}(a^{1/4},b^{1/4},x/2)\right],\\ v_{0}(2)=&2f_{(0,0,2)}(a,b,x)-\sqrt{2}f_{(0,0,1)}(a^{1/4},b^{1/4},x/2),\\ v_{1}(2)=&f_{(\frac{1}{2},0,2)}(a,b,x)+f_{(\frac{-1}{2},0,2)}(a,b,x)\\ &-\frac{1}{\sqrt{2}}\left[f_{(0,\frac{1}{2},1)}(a^{1/4},b^{1/4},x/2)+f_{(0,\frac{-1}{2},1)}(a^{1/4},b^{1/4},x/2)\right].\\ \end{split}

So,

\displaystyle v_{1}+v_{2}=\begin{pmatrix}4f_{(0,0,2)}(a,b,x)\\ 2\left[f_{(\frac{1}{2},0,2)}(a,b,x)+f_{(\frac{-1}{2},0,2)}(a,b,x)\right]\end{pmatrix}.

(5)

and

\displaystyle v_{1}-v_{2}=\begin{pmatrix}2\sqrt{2}f_{(0,0,1)}(a^{1/4},b^{1/4},x/2)\\ \sqrt{2}\left[f_{(0,\frac{1}{2},1)}(a^{1/4},b^{1/4},x/2)+f_{(0,\frac{-1}{2},1)}(a^{1/4},b^{1/4},x/2)\right]\end{pmatrix}.

(6)

It is clear that

A(v_{1}+v_{2})=v_{1}-v_{2}.

(7)

Equations (5),(6) and (7) give the required results.

The immediate consequence of the lemma above is the next corollary.

Corollary 4.1.

The following equations hold.

1.

$4f_{(0,0,2)}(a,b,x)=2f_{(0,0,1)}(a^{1/4},b^{1/4},x/2)+f_{(0,\frac{1}{2},1)}(a^{1/4},b^{1/4},x/2)+f_{(0,\frac{-1}{2},1)}(a^{1/4},b^{1/4},x/2),$
2.

$2\left[f_{(\frac{1}{2},0,2)}(a,b,x)+f_{(\frac{-1}{2},0,2)}(a,b,x)\right]=2f_{(0,0,1)}(a^{1/4},b^{1/4},x/2)-[f_{(0,\frac{1}{2},1)}(a^{1/4},b^{1/4},x/2)+f_{(0,\frac{-1}{2},1)}(a^{1/4},b^{1/4},x/2)].$

The quadratic identity: This identity can be considers as product identity for Ramanujan theta functions berndt2012ramanujan .

\begin{split}&4f^{2}_{(0,0,2)}(a,b,x)-\left[f_{(\frac{1}{2},0,2)}(a,b,x)+f_{(\frac{-1}{2},0,2)}(a,b,x)\right]^{2}\\ &=2f_{(0,0,1)}(a^{1/4},b^{1/4},x/2)\left[f_{(0,\frac{1}{2},1)}(a^{1/4},b^{1/4},x/2)+f_{(0,\frac{-1}{2},1)}(a^{1/4},b^{1/4},x/2)\right].\end{split}

We now derive the identity corresponding to the DFT of size $3.$

Lemma 4.

\begin{split}&\frac{2\sqrt{3}}{1+\sqrt{3}}\sum_{m=0}^{\infty}a^{\frac{3m(3m+1)}{2}}b^{\frac{3m(3m-1)}{2}}e^{2\pi imx}\\ -\sqrt{3}&\left[\sum_{m=0}^{\infty}a^{\frac{(3m+1)(3m+2)}{2}}b^{\frac{3m(3m+1)}{2}}e^{2\pi i(m+1/3)x}+\sum_{m=0}^{\infty}a^{\frac{3m(3m-1)}{2}}b^{\frac{(3m-1)(3m-2)}{2}}e^{2\pi i(m-1/3)x}\right]\\ =&\left(ab,-ae^{\frac{2\pi i(x+1)}{3}},-be^{\frac{-2\pi i(x+1)}{3}}:ab\right)_{\infty}+\left(ab,-ae^{\frac{2\pi i(x-1)}{3}},-be^{\frac{-2\pi i(x-1)}{3}}:ab\right)_{\infty}\\ &-\frac{-2}{1+\sqrt{3}}\left(ab,-ae^{2\pi ix},-be^{-2\pi ix}:ab\right)_{\infty}\end{split}

(8)

Proof: The DFT of size $n=3$ has eigenvalues $1,-1,i$ and all non-degenerate. The eigenvectors corresponding to eigenvalue $1$ are,

v_{1}=\begin{bmatrix}1+\sqrt{3}\\ 1\\ 1\end{bmatrix},

Let,

\begin{split}v_{0}(0)=&2f_{(0,0,3)}(a,b,x)+\frac{2}{\sqrt{3}}f_{(0,0,1))}(a^{1/9},b^{1/9},x/3),\\ v_{1}(0)=&f_{(\frac{1}{3},0,3)}(a,b,x)+f_{(\frac{-1}{3},0,3)}(a,b,x)\\ &+\frac{1}{\sqrt{3}}\left[f_{(0,\frac{1}{3},1)}(a^{1/9},b^{1/9},x/3)+f_{(0,\frac{-1}{3},1)}(a^{1/9},b^{1/9},x/3)\right],\\ v_{2}(0)=&f_{(\frac{2}{3},0,3)}(a,b,x)+f_{(\frac{-2}{3},0,3)}(a,b,x)\\ &+\frac{1}{\sqrt{3}}\left[f_{(0,\frac{2}{3},1)}(a^{1/9},b^{1/9},x/3)+f_{(0,\frac{-2}{3},1)}(a^{1/9},b^{1/9},x/3)\right]\end{split}

v_{2}=\begin{bmatrix}v_{0}(0)\\ v_{1}(0)\\ v_{2}(0)\end{bmatrix}.

It is clear that any $2\times 2$ minor of $[v_{1},v_{2}]$ vanishes. So,

\det\begin{bmatrix}1+\sqrt{3}&v_{0}(0)\\ 1&v_{1}(0)\end{bmatrix}=0

\begin{split}&2f_{(0,0,3)}(a,b,x)+\frac{2}{\sqrt{3}}f_{(0,0,1))}(a^{1/9},b^{1/9},x/3)\\ =&(1+\sqrt{3})\left[f_{(\frac{1}{3},0,3)}(a,b,x)+f_{(\frac{-1}{3},0,3)}(a,b,x)\right]\\ &+\frac{1+\sqrt{3}}{\sqrt{3}}\left[f_{(0,\frac{1}{3},1)}(a^{1/9},b^{1/9},x/3)+f_{(0,\frac{-1}{3},1)}(a^{1/9},b^{1/9},x/3)\right]\end{split}

(9)

i.e.

\begin{split}&\frac{2\sqrt{3}}{1+\sqrt{3}}f_{(0,0,3)}(a,b,x)-\sqrt{3}\left[f_{(\frac{1}{3},0,3)}(a,b,x)+f_{(\frac{-1}{3},0,3)}(a,b,x)\right]\\ =&f_{(0,\frac{1}{3},1)}(a^{1/9},b^{1/9},x/3)+f_{(0,\frac{-1}{3},1)}(a^{1/9},b^{1/9},x/3)-\frac{2}{1+\sqrt{3}}f_{(0,0,1))}(a^{1/9},b^{1/9},x/3)\end{split}

(10)

This leads to,

\begin{split}&\frac{2\sqrt{3}}{1+\sqrt{3}}\sum_{m=0}^{\infty}a^{\frac{m(m+1/3)}{2}}b^{\frac{m(m-1/3)}{2}}e^{2\pi imx}\\ &-\sqrt{3}\sum_{m=0}^{\infty}a^{\frac{(m+1/3)(m+2/3)}{2}}b^{\frac{m(m+1/3)}{2}}e^{2\pi i(m+1/3)x}\\ &-\sqrt{3}\sum_{m=0}^{\infty}a^{\frac{m(m-1/3)}{2}}b^{\frac{(m-1/3)(m-2/3)}{2}}e^{2\pi i(m-1/3)x}\\ =&\sum_{m=0}^{\infty}a^{\frac{m(m+1)}{18}}b^{\frac{m(m-1)}{18}}e^{2\pi im(x+1)/3}+\sum_{m=0}^{\infty}a^{\frac{m(m+1)}{18}}b^{\frac{m(m-1)}{18}}e^{2\pi im(x-1)/3}\\ &-\frac{2}{1+\sqrt{3}}\sum_{m=0}^{\infty}a^{\frac{m(m+1)}{18}}b^{\frac{m(m-1)}{18}}e^{2\pi imx}\end{split}

(11)

The right hand side of the equation (11) can be expressed as follows,

\begin{split}f(a^{(1/9)}e^{2\pi ix},b^{(1/9)}e^{-2\pi ix})=&\sum_{m=0}^{\infty}(a^{(1/9)}e^{2\pi ix})^{m(m+1)/2}(b^{(1/9)}e^{-2\pi ix})^{m(m-1)/2}\\ &=\sum_{m=0}^{\infty}a^{\frac{m(m+1)}{18}}b^{\frac{m(m-1)}{18}}e^{2\pi imx}.\end{split}

The equation (11) becomes,

\begin{split}&\frac{2\sqrt{3}}{1+\sqrt{3}}\sum_{m=0}^{\infty}a^{\frac{m(m+1/3)}{2}}b^{\frac{m(m-1/3)}{2}}e^{2\pi imx}\\ &-\sqrt{3}\sum_{m=0}^{\infty}a^{\frac{(m+1/3)(m+2/3)}{2}}b^{\frac{m(m+1/3)}{2}}e^{2\pi i(m+1/3)x}\\ &-\sqrt{3}\sum_{m=0}^{\infty}a^{\frac{m(m-1/3)}{2}}b^{\frac{(m-1/3)(m-2/3)}{2}}e^{2\pi i(m-1/3)x}\\ =&f\left(a^{(1/9)}e^{2\pi i(x+1)/3},b^{(1/9)}e^{-2\pi i(x+1)/3}\right)+f\left(a^{(1/9)}e^{2\pi i(x-1)/3},b^{(1/9)}e^{-2\pi i(x-1)/3}\right)\\ &-\frac{2}{1+\sqrt{3}}f\left(a^{(1/9)}e^{2\pi ix},b^{(1/9)}e^{-2\pi ix}\right).\end{split}

(12)

The Jacobi theta function identities can be derived from these identities of Ramanujan theta function.

Lemma 5.

For any function $f$ from the upper half plane to itself following identities hold.

1.

$2\theta\left(\frac{f(\tau)-\tau}{4}+x,\tau\right)-2\theta\left(\frac{f(\tau)-\tau}{8}+\frac{x+1}{2},\frac{\tau}{4}\right)\\ =e^{\pi i(\frac{f(\tau)}{4}+x)}\theta\left(\frac{f(\tau)+\tau}{4}+x,\tau\right)+e^{\pi i(\frac{2\tau-f(\tau)}{4}-x)}\theta\left(\frac{f(\tau)-3\tau}{4}+x,\tau\right),$
2.

$\frac{2\sqrt{3}}{1+\sqrt{3}}\theta\left(\frac{f(\tau)-\tau}{6}+x,\tau\right)\\ -\sqrt{3}\left[e^{\frac{\pi i}{3}\left(\frac{f(\tau)}{3}+2x\right)}\theta\left(\frac{f(\tau)+\tau}{6}+x,\tau\right)+e^{\frac{\pi i}{3}\left(\frac{2\tau-f(\tau)}{3}-2x\right)}\theta\left(\frac{f(\tau)-3\tau}{6}+x,\tau\right)\right]\\ =\theta\left(\frac{f(\tau)-\tau}{18}+\frac{x+1}{3},\frac{\tau}{9}\right)+\theta\left(\frac{f(\tau)-\tau}{18}+\frac{x-1}{3},\frac{\tau}{9}\right)-\frac{2}{1+\sqrt{3}}\theta\left(\frac{f(\tau)-\tau}{18}+\frac{x}{3},\frac{\tau}{9}\right).$

Proof:

Let $a=e^{\pi if(\tau)},b=e^{\pi ig(\tau)}$ where $f,g$ are the functions from upper half of complex plane to itself.

\begin{split}f_{(0,0,2)}(a,b,x)=&\sum_{m\in\mathbb{Z}}e^{\pi if(\tau)\left(\frac{m^{2}}{2}+\frac{m}{4}\right)}e^{\pi ig(\tau)\left(\frac{m^{2}}{2}-\frac{m}{4}\right)}e^{2\pi imx}\\ =&\sum_{m\in\mathbb{Z}}e^{\pi i\left(\frac{(f+g)(\tau)}{2}\right)m^{2}}e^{2\pi im\left(\frac{(f-g)(\tau)}{8}+x\right)}\\ =&\theta\left(\frac{(f-g)(\tau)}{8}+x,\frac{(f+g)(\tau)}{2}\right),\end{split}

(13)

\begin{split}f_{(\frac{1}{2},0,2)}(a,b,x)=&\sum_{m\in\mathbb{Z}}e^{\pi if(\tau)\left(\frac{m^{2}}{2}+\frac{3m}{4}+\frac{1}{4}\right)}e^{\pi ig(\tau)\left(\frac{m^{2}}{2}+\frac{m}{4}\right)}e^{2\pi i(m+1/2)x}\\ =&e^{\pi i(\frac{f(\tau)}{4}+x)}\sum_{m\in\mathbb{Z}}e^{\pi i\left(\frac{(f+g)(\tau)}{2}\right)m^{2}}e^{2\pi im\left(\frac{(3f+g)(\tau)}{8}+x\right)}\\ =&e^{\pi i(\frac{f(\tau)}{4}+x)}\theta\left(\frac{(3f+g)(\tau)}{8}+x,\frac{(f+g)(\tau)}{2}\right),\end{split}

(14)

\begin{split}f_{(\frac{-1}{2},0,2)}(a,b,x)=&e^{\pi i(\frac{g(\tau)}{4}-x)}\sum_{m\in\mathbb{Z}}e^{\pi i\left(\frac{(f+g)(\tau)}{2}\right)m^{2}}e^{2\pi im\left(\frac{-(f+3g)(\tau)}{8}+x\right)}\\ =&e^{\pi i(\frac{g(\tau)}{4}-x)}\theta\left(\frac{-(f+3g)(\tau)}{8}+x,\frac{(f+g)(\tau)}{2}\right),\end{split}

(15)

\begin{split}f_{(0,1/2,1)}(a^{1/4},b^{1/4},x/2)=&\sum_{m\in\mathbb{Z}}e^{\pi i\left(\frac{(f+g)(\tau)}{8}\right)m^{2}}e^{2\pi im\left(\frac{(f-g)(\tau)}{16}+\frac{x+1}{2}\right)}\\ =&\theta\left(\frac{(f-g)(\tau)}{16}+\frac{x+1}{2},\frac{(f+g)(\tau)}{8}\right)\end{split}

(16)

Using equations (13), (14), (15) and (16) in the second identity of lemma 3 we have,

\begin{split}&2\theta\left(\frac{(f-g)(\tau)}{8}+x,\frac{(f+g)(\tau)}{2}\right)-2\theta\left(\frac{(f-g)(\tau)}{16}+\frac{x+1}{2},\frac{(f+g)(\tau)}{8}\right)\\ =&e^{\pi i(\frac{f(\tau)}{4}+x)}\theta\left(\frac{(3f+g)(\tau)}{8}+x,\frac{(f+g)(\tau)}{2}\right)\\ &+e^{\pi i(\frac{g(\tau)}{4}-x)}\theta\left(\frac{-(f+3g)(\tau)}{8}+x,\frac{(f+g)(\tau)}{2}\right)\end{split}

(17)

In particular, we will choose $f(\tau),g(\tau)$ such that $f(\tau)+g(\tau)=2\tau$ .

\begin{split}&2\theta\left(\frac{f(\tau)-\tau}{4}+x,\tau\right)-2\theta\left(\frac{f(\tau)-\tau}{8}+\frac{x+1}{2},\frac{\tau}{4}\right)\\ =&e^{\pi i(\frac{f(\tau)}{4}+x)}\theta\left(\frac{f(\tau)+\tau}{4}+x,\tau\right)+e^{\pi i(\frac{2\tau-f(\tau)}{4}-x)}\theta\left(\frac{f(\tau)-3\tau}{4}+x,\tau\right).\end{split}

(18)

2.

Similarly, the identity can be obtained from lemma 4.

The identities discussed in lemma gives new insights to the Landen transformations.

5 Modular equations: Generating function for quadratic numbers

Following lemma gives the generating function for the quadratic numbers.

Lemma 6.

\sum_{m=-\infty}^{\infty}q^{k_{1}m^{2}+k_{2}m}=\frac{\left(q^{\frac{k_{1}+2k_{2}}{4}},q^{\frac{k_{1}-2k_{2}}{4}},q^{\frac{k_{1}}{2}}:q^{\frac{k_{1}}{2}}\right)_{\infty}+\left(-q^{\frac{k_{1}+2k_{2}}{4}},-q^{\frac{k_{1}-2k_{2}}{4}},q^{\frac{k_{1}}{2}}:q^{\frac{k_{1}}{2}}\right)_{\infty}}{2}

Proof: Evaluate first identity of the corollary 4.1 at $x=-\tau,a=q^{k_{1}+2k_{2}+4},b=q^{k_{1}-2k_{2}-4}.$ This gives,

4f_{(0,0,2)}(a,b,x)=4f_{(0,0,2)}(q^{k_{1}+2k_{2}+4},q^{k_{1}-2k_{2}-4},-\tau)=4\sum_{m=-\infty}^{\infty}q^{k_{1}m^{2}+k_{2}m}.

and

\begin{split}&2f_{(0,0,1)}(a^{1/4},b^{1/4},x/2)+[f_{(0,\frac{1}{2},1)}(a^{1/4},b^{1/4},x/2)+f_{(0,\frac{-1}{2},1)}(a^{1/4},b^{1/4},x/2)]\\ =&2\sum_{m=-\infty}^{\infty}a^{\frac{m(m+1)}{8}}b^{\frac{m(m-1)}{8}}e^{\pi imx}+2\sum_{m=-\infty}^{\infty}a^{\frac{m(m+1)}{8}}b^{\frac{m(m-1)}{8}}e^{\pi imx}(-1)^{m},\\ =&2\left[f\left(q^{\frac{k_{1}+2k_{2}}{4}},q^{\frac{k_{1}-2k_{2}}{4}}\right)+f\left(-q^{\frac{k_{1}+2k_{2}}{4}},-q^{\frac{k_{1}-2k_{2}}{4}}\right)\right].\end{split}

Using equation (2) we obtain the quadratic numbers identity:

\sum_{m=-\infty}^{\infty}q^{k_{1}m^{2}+k_{2}m}=\frac{\left(q^{\frac{k_{1}+2k_{2}}{4}},q^{\frac{k_{1}-2k_{2}}{4}},q^{\frac{k_{1}}{2}}:q^{\frac{k_{1}}{2}}\right)_{\infty}+\left(-q^{\frac{k_{1}+2k_{2}}{4}},-q^{\frac{k_{1}-2k_{2}}{4}},q^{\frac{k_{1}}{2}}:q^{\frac{k_{1}}{2}}\right)_{\infty}}{2}

(19)

Remark 2.

Different choices of $x$ and corresponding choices of $a$ and $b$ give the alternate expression for the generating functions for quadratic numbers.

The generating functions for the polygonal numbers are given in the following corollary.

Corollary 5.1.

The generating function for triangular numbers

\begin{split}&2\sum_{m=-\infty}^{\infty}q^{\frac{m(m+1)}{2}}\\ =&\prod_{k=0}^{\infty}(1-q^{\frac{k+1}{4}})\left[\prod_{k=0}^{\infty}(1-q^{\frac{2k+3}{8}})(1-q^{\frac{2k-1}{8}})+\prod_{k=0}^{\infty}(1+q^{\frac{2k+3}{8}})(1+q^{\frac{2k-1}{8}}))\right]\end{split}

The generating function for square numbers

2+4\sum_{m=1}^{\infty}q^{m^{2}}=\prod_{k=0}^{\infty}(1-q^{\frac{k+1}{2}})\left[\prod_{k=0}^{\infty}(1-q^{\frac{2k+1}{4}})^{2}+\prod_{k=0}^{\infty}(1+q^{\frac{2k+1}{4}})^{2}\right]

The generating function for Pentagonal numbers

\begin{split}&2\sum_{m=-\infty}^{\infty}q^{\frac{m(3m-1)}{2}}\\ =&\prod_{k=0}^{\infty}(1-q^{\frac{3(k+1)}{4}})\left[\prod_{k=0}^{\infty}(1-q^{\frac{6k+1}{8}})(1-q^{\frac{6k+5}{8}})+\prod_{k=0}^{\infty}(1+q^{\frac{6k+1}{8}})(1+q^{\frac{6k+5}{8}})\right]\end{split}

4.

The generating function for Hexagonal numbers

$2\sum_{m=-\infty}^{\infty}q^{m(2m-1)}=\prod_{k=0}^{\infty}(1-q^{2k+2})(1+q^{k})$

The generating function for Heptagonal numbers

\begin{split}&2\sum_{m=-\infty}^{\infty}q^{\frac{m(5m-3)}{2}}\\ =&\prod_{k=0}^{\infty}(1-q^{\frac{5(k+1)}{4}})\left[\prod_{k=0}^{\infty}(1-q^{\frac{10k-1}{8}})(1-q^{\frac{10k+11}{8}})+\prod_{k=0}^{\infty}(1+q^{\frac{10k-1}{8}})(1+q^{\frac{10k+11}{8}})\right]\end{split}

The generating function for Octagonal numbers

\begin{split}&2\sum_{m=-\infty}^{\infty}q^{3m^{2}-2m}\\ =&\prod_{k=0}^{\infty}(1-q^{\frac{3(k+1)}{2}})\left[\prod_{k=0}^{\infty}(1-q^{\frac{6k-1}{4}})(1-q^{\frac{6k+7}{4}})+\prod_{k=0}^{\infty}(1+q^{\frac{6k-1}{4}})(1+q^{\frac{6k+7}{4}})\right]\end{split}

The $m^{th}$ $r$ -gonal number is defined by $\frac{(r-2)m^{2}+(4-r)m}{2}$ . The generating function is

\begin{split}&2\sum_{m=-\infty}^{\infty}q^{\frac{(r-2)m^{2}+(4-r)m}{2}}\\ =&\left(q^{\frac{r-2}{4}},q^{\frac{6-r}{8}},q^{\frac{3r-10}{8}}:q^{\frac{r-2}{4}}\right)_{\infty}+\left(q^{\frac{r-2}{4}},-q^{\frac{6-r}{8}},-q^{\frac{3r-10}{8}}:q^{\frac{r-2}{4}}\right)_{\infty}\end{split}

Remark 3.

A fifth power of the third identity of the corollary 5.1 is

\left(\sum_{m=-\infty}^{\infty}q^{\frac{m(3m-1)}{2}}\right)^{5}=\sum_{n=0}^{\infty}p(n)q^{n}

where $p(n)$ is the number of ways by which a non negative integer can be expressed as a sum of five pentagonal numbers, and

\sum_{n=0}^{\infty}p(n)q^{n}=1+5q+15q^{2}+30q^{3}+45q^{4}+56q^{5}+65q^{6}+85q^{7}+115q^{8}+150q^{9}+\cdots.

2.

Lemma 6 also gives a generating function for the numbers of the form $k_{1}m^{2}+k_{2}m+k_{3}$ .

6 Conclusion

A new technique to obtain the functional relations between Ramanujan theta functions has been developed. This idea can be extended to higher dimensional DFT’s to obtain some significant identities between Jacobi theta functions.

References

(1) Mehta, M., Eigenvalues and eigenvectors of the finite Fourier transform, J. Math. Phys., 28(4) (1987),pp.781-785 .
(2) Bruce C. Berndt, Ramanujan’s notebooks: Part III, Springer Science & Business Media, 2012
(3) Bruce C. Berndt, Örs Rebák.:Explicit values for Ramanujan’s theta function $\phi(q)$ , Hardy-Ramanujan Journal, 44, 2021.
(4) Matveev, V.B., Intertwining relations between the Fourier transform and discrete Fourier transform, the related functional identities and beyond, Inverse Problems,17(4) (2001), pp.633.
(5) McClellan, J. and Parks, T., Eigenvalue and eigenvector decomposition of the discrete Fourier transform, IEEE Transactions on Audio and Electroacoustics, 20(1) (1972), pp.66-74.
(6) Murty, MR., Dewar, M. and Graves, H., Problems in the theory of modular forms, Springer, 2015.
(7) Robert C., On Ramanujan’s definition of mock theta function, Proceedings of the National Academy of Sciences, 10(19), 7592–7594, 2013