Using numerical methods we construct evaluation formulas for the Jacobi $\theta_4$ functions. Our results are conjectures, but are verified numerically.
1503.01141.tex
\documentclass{article}\begin{document}
\centerline{\bf\large On the Complete Evaluation of Jacobi Theta Functions}
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\textbf{\centerline{N.D.Bagis}}
\textbf{\centerline{Stenimahou 5 Edessa}}
\textbf{\centerline{Pellas 58200, Greece}}
\textbf{\centerline{email-nikosbagis@hotmail.gr}}
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\]
\begin{abstract}
Using numerical methods we construct evaluation formulas for the Jacobi $\theta_4$ functions. Our results are conjectures, but are verified numerically.
\end{abstract}

\section{Introduction}

Let $K(x)$ be the complete elliptic integral of the first kind
\begin{equation}
K(x)=\frac{\pi}{2}{}_2F_1\left(\frac{1}{2},\frac{1}{2};1;x^2\right)
\end{equation}
and $k_r$, the elliptic singular modulus, solution of the equation
\begin{equation}
\frac{K\left(\sqrt{1-k_r^2}\right)}{K(k_r)}=\sqrt{r}
\end{equation} 
When $r$ is positive rational the function $k_r$ take algebraic values.\\
The 4th-Jacobian theta function is given by  
\begin{equation}
\vartheta_4(z,q)=\sum^{\infty}_{n=-\infty}(-1)^nq^{n^2}e^{2niz}
\end{equation}
where $|q|<1$ and $z\in \textbf{C}$. Avoiding this definition we will use the equivalent notation
\begin{equation}
\theta(a,b;q):=\sum^{\infty}_{n=-\infty}(-1)^nq^{an^2+bn}
\end{equation}
and we shall restrict to the case of $a$ positive rational and $b$ general rational.\\ 
In [Bag,Glas] we have shown that if $|q|<1$ and 
\begin{equation}
A(a,p;q):=q^{\frac{p}{12}-\frac{a}{2}+\frac{a^2}{2p}}\prod^{\infty}_{n=1}\left(1-q^{np+a}\right)\left(1-q^{np+p-a}\right)
\end{equation}
with $p>0$, then
\begin{equation}
A(a,p;q)=\frac{q^{\frac{p}{12}-\frac{a}{2}+\frac{a^2}{2p}}}{\eta(q^p)}\theta\left(\frac{p}{2},\frac{p}{2}-a;q\right)
\end{equation}
where $\eta$ is the Dedekind-Ramanujan eta function
\begin{equation}
\eta(q):=\prod^{\infty}_{n=1}(1-q^n)\textrm{, }|q|<1 
\end{equation} 
The above equality (6) is simple consequence of the Jacobi triple product formula (see [2]).   

\section{Conjecture}

Having define what we need we state our\\
\\
\textbf{Conjecture.}\\
Let $q=e^{-\pi\sqrt{r}}$ with $r>0$, then for $a,p$ rationals and $p>0$ there always exist algebraic function $Q(x)$, $0<x<1$, such that 
\begin{equation}
A(a,p,q)=Q(k_r)\textrm{, }\forall r>0
\end{equation}
\\
\textbf{Corollary.}\\
If $q=e^{-\pi\sqrt{r}}$ and $r,|a|,p\in\bf Q^{*}_{+}\rm$, then 
\begin{equation}
A(a,p;q)=Algebraic\textrm{  }Number
\end{equation}


Some verifications (proofs) of the above conjecture have given in [8] for the case of theta functions of the form
\begin{equation}
\sum^{\infty}_{n=-\infty}q^{n^2+mn}\textrm{, } m\in\textbf{Z} 
\end{equation}
Moreover it has been shown that:\\
\textbf{i)} if $m=2s$ (even), then
\begin{equation}
\sum^{\infty}_{n=-\infty}q^{n^2+2sn}=q^{-s^2}\sqrt{\frac{2K(k_r)}{\pi}}
\end{equation}
\textbf{ii)} if $m=2s+1$ (odd), then
\begin{equation}
\sum^{\infty}_{n=-\infty}q^{n^2+(2s+1)n}=2^{5/6}q^{-(2s+1)^2/4}\frac{(k_{11}k_{12}k_{21})^{1/6}}{k_{22}^{1/3}}\sqrt{\frac{K(k_{11})}{\pi}}
\end{equation}
where
$k_{11}=k_r$, $k_{12}=\sqrt{1-k_{11}^2}$, $k_{21}=\frac{2-k_{11}^2-2k_{12}}{k_{11}^2}$, $k_{22}=\sqrt{1-k_{21}^2}$ and in view of the evaluation formula (see [8]):
\begin{equation}
\eta(q)^8=\frac{2^{8/3}}{\pi^4}q^{-1/3}k_r^{2/3}(k'_r)^{8/3}K(k_r)^4
\end{equation}
Examples of the above conjecture can also be found if we consider the function
\begin{equation}
r=k_i(x):=\left(\frac{K\left(\sqrt{1-x^2}\right)}{K(x)}\right)^2
\end{equation} 
which, is the inverse function of the singular modulus $k_r$. Our method consist of inserting the value $r=k_i\left(\frac{m}{n}\right)$, where $0<m<n$, $m,n$ integers into the form $A(a,p;q)$ and get numerically, using the routine $RootApproximant$ of the program Mathematica, a minimal polynomial which is esentialy the value of an algebraic number. This lead us to conclude that beneath (any theta function) exists minimal polynomials with coefficients rational polynomials of $k_r$ (in all cases if Conjecture holds) and for fixed $a,p$, unique algebraic function $Q_{\{a,p\}}(x)$, such that these polynomials have solution the functions $Q_{\{a,p\}}(k_r)$.\\A very easy example to see someone is with $a=1$ and $p=4$. In this case all the values of $A\left(1,4,e^{-\pi\sqrt{k_i(\frac{m}{n})}}\right)^{24}$ are rationals. With a simple algorithm one can see that
\begin{equation}
A\left(1,4,e^{-\pi\sqrt{k_i\left(r\right)}}\right)^{24}=\frac{16(1-r^2)}{r^2}
\end{equation}
Hence 
\begin{equation}
A\left(1,4,e^{-\pi\sqrt{r}}\right)=\sqrt[24]{\frac{16(1-k_r^2)}{k_r^2}}
\end{equation} 
and by relation (6)\\
\\
\textbf{Theorem 1.}\\
If $q=e^{-\pi\sqrt{r}}$, $r>0$
\begin{equation}
\sum^{\infty}_{n=-\infty}(-1)^nq^{2n^2+n}=q^{1/24}\eta(q^4)\sqrt[12]{\frac{4(1-k_r^2)}{k_r}}\textrm{, }\forall r>0   
\end{equation}      

The continuation follows from the validity of (6) and (8) in rationals which are dense in reals.\\
Another example is with $p=2$ and $a=1/2$, where we have
\begin{equation}
A\left(1/2,2,e^{-\pi\sqrt{r}}\right)=\sqrt[24]{\frac{4(1-k_r)^4}{k_r(1+k_r)^2}}
\end{equation} 
and is the same theta function as (16) by changing $q\rightarrow q^{1/2}$. 
For avoiding these cases it is useful to know that
\begin{equation}
\sum^{\infty}_{n=-\infty}(-1)^nq^{an^2+bn}=\sum^{\infty}_{n=-\infty}q^{an^2-bn}
\end{equation}
and also that if $s>0$
\begin{equation}
\sum^{\infty}_{n=-\infty}(-1)^nq^{an^2+bn}\textrm{, }\sum^{\infty}_{n=-\infty}(-1)^nq^{asn^2+bsn}
\end{equation}
are equivalent.\\
\\
\textbf{Theorem 2.}\\
For $q=e^{-\pi\sqrt{r}}$, $r>0$, we have
\begin{equation}
\sum^{\infty}_{n=-\infty}(-1)^nq^{2n^2+3n/2}=q^{-11/96}\eta(q^4)\sqrt[48]{\frac{4(1-k_r)^4(2+k_r-2\sqrt{1+k_r})^{12}}{k_r^{13}(1+k_r)^2}}
\end{equation}

\section{The Algorithm}

In this section we give the algorithm for finding the expression $Q_{\{a,p\}}\left(k_r^2\right)$. Our method is based on interpolation. We  find the minimal polynomial
\begin{equation}
P\left(A(a,p;q),m(q)\right)=0,
\end{equation} 
with $m(q)=k_r^2$ and then solve with respect to $A(a,p;m(q))$, if $P(x,y)=0$ is solvable.\\
\\  
\textbf{The algorithm} (In Mathematica Program) 
$$
Clear[A]
$$
$$
eta[q]:=QPochhammer[q,q]
$$
$$
A[a,p,q]:=q^{p/12-a/2+a^2/(2 p)}eta[q^p]^{-1} Sum[(-1)^n q^{n^2 p/2+(p-2a) n/2},\{n,-100,100\}]
$$
$$
Clear[q,x,y,u,v]
$$
$$
m[q]:=InverseEllipticNomeQ[q]
$$
$$
x=Series[A[a,p,q]^{12},\{q,0,M\}];
$$
$$
y=Series[m[q],\{q,0,M\}];
$$
$$
t=Table[Coefficient[Sum[c[i,j]x^iy^j,q^{n}]==0,\{n,1,s^2\}];
$$
$$
rr=Table[c[i,j],\{i,0,s\},\{j,0,s\}];
$$
$$
rr1=Table[u^i v^j,\{i,0,s\},\{j,0,s\}];
$$
$$
mm=Normal[Extract[CoefficientArrays[t//Flatten,rr//Flatten],2]];
$$
$$
m0=Normal[mm];
$$
$$
r1=Take[NullSpace[m0],1] . Flatten[rr]
$$
$$
Take[NullSpace[m0],1] . Flatten[rr1]//Factor
$$


\section{Theoretical Results and Directions}


In this section we will try to characterize these functions $Q_{\{a,p\}}(x)$. For this, assume that $P_{n}$ is the $n$-th modular equation of $A(a,p;q)$, then 
\begin{equation}
A(a,p;q^n)=P_{n}\left(A(a,p;q)\right)
\end{equation}
Also assume that our conjecture (8) hold, then 
$$
Q_{\{a,p\}}\left(k_{n^2r}\right)=P_{n}\left(Q_{\{a,p\}}(k_r)\right)
$$
By using (14), we get
$$
Q_{\{a,p\}}\left(k_{n^2k_i(x)}\right)=P_n\left(Q_{\{a,p\}}(x)\right)
$$
Setting 
\begin{equation}
S_n(x):=k_{n^2k_i(x)}
\end{equation}
we have the next\\
\\ 
\textbf{Theorem 3}\\
If the $n$-th modular equation of $A(a,p;q)$ is that of (23),  then 
\begin{equation}
k_{n^2k_i(x)}=S_n(x)=Q_{\{a,p\}}{}^{(-1)}\left(P_{n}\left(Q_{\{a,p\}}(x)\right)\right),\textrm{ }n=2,3,4,...
\end{equation}
\\

If one manages to solve equation (25) with respect to $Q_{\{a,p\}}(x)$ for given $a,p$, then  
\begin{equation}
\sum^{\infty}_{n=-\infty}(-1)^nq^{pn^2/2+(p-2a)n/2}=q^{-\frac{p}{12}+\frac{a}{2}-\frac{a^2}{2p}}\eta(q^p)Q_{\{a,p\}}(k_r), \forall r>0
\end{equation}
and $Q_{\{a,p\}}(x)$ will be  a root of a minimal polynomial of degree $\nu=\nu(a,p,x)$. Note that in case of rational $x\in(0,1)$ and $a,p$ rational with $0<a,p$, then the degree $\nu$ is independent of $x$ and the minimal polynomial of $Q_{\{a,p\}}(x)$ will have integer coefficients.\\
\\
\textbf{Example.}\\
The 2nd degree modular equation of $A(1,4;q)$ is
\begin{equation}
16 u^8+u^{16}v^8-v^{16}=0
\end{equation} 
If we solve with respect to $v$ we get $v=P_2(u)$, where $v=A(1,4;q^2)$ and $u=A(1,4;q)$. Moreover
\begin{equation}
P_2(w)=\frac{\left(w^{16}+w^4\sqrt{64+w^{24}}\right)^{1/8}}{2^{1/8}}
\end{equation}
It is $n=2$ and hold (see [9])
\begin{equation}
k_{4r}=\frac{1-\sqrt{1-k_r^2}}{1+\sqrt{1-k_r^2}}
\end{equation}
Hence
\begin{equation}
S_2(x)=k_{4k_i(x)}=\frac{1-\sqrt{1-x^2}}{1+\sqrt{1-x^2}}
\end{equation}
Finally we get from the relation (25) of Theorem 4:
\begin{equation}
Q_{\{1,4\}}^{(-1)}\left(\frac{\sqrt[8]{Q_{\{1,4\}}(x)^{16}+Q_{\{1,4\}}(x)^4\sqrt{Q_{\{1,4\}}(x)^{24}+64} }}{\sqrt[8]{2}}\right)=\frac{1-\sqrt{1-x^2}}{1+\sqrt{1-x^2}}
\end{equation}
which has indeed a solution 
$$
Q_{\{1,4\}}(x)=\sqrt[12]{\frac{4(1-x^2)}{x}}.
$$  
\\
\textbf{Note.}\\
We note that function $m(q)=k_r^2$ is implemented in program Mathematica. However a useful expansion is
\begin{equation}
k_r=\sqrt{m(q)}=4q^{1/2}\exp\left(-4\sum^{\infty}_{n=1}q^n\sum_{d|n}\frac{(-1)^{d+n/d}}{d}\right),
\end{equation}        
where $q=e^{-\pi\sqrt{r}}$, $r>0$.

\section{Table}

Here we give a table of evaluations, which does not include Theorems 1 and 2, for certain lower values $a,p$ of theta functions.\\
\\
\textbf{1.}
\begin{equation}
\sum^{\infty}_{n=-\infty}(-1)^nq^{3/2n^2+n/2}=q^{1/12}\eta(q^3)Q(k_r^2)
\end{equation}
The polynomial which relates $u=A(1,3,q)^{12}$ with $v=m(q)=k_r^2$ is
$$
u^4 v^5-4 u^4 v^4+6 u^4 v^3-4 u^4 v^2+u^4 v-16 u^3 v^6+84 u^3 v^5
-12480 u^3 v^4-
$$
$$
-40712 u^3 v^3-12480 u^3 v^2+84 u^3 v-16 u^3+196830 u^2 v^5-787320 u^2 v^4+1180980 u^2 v^3-
$$
$$
-787320 u^2 v^2+196830 u^2 v+19131876 u v^5-76527504 u v^4+114791256 u v^3-
$$
$$
-76527504 u v^2+19131876 u v+387420489 v^5-1549681956 v^4+2324522934 v^3-
$$
\begin{equation}
-1549681956 v^2+387420489 v=0
\end{equation}
\\
\textbf{2.}
\begin{equation}
\sum^{\infty}_{n=-\infty}(-1)^nq^{3n^2-5n}=q^{-11/6}\eta(q^6)Q(k_r)
\end{equation} 
The polynomial equation which relates $u=A(8,6;q)^6$ with $v=m(q)^{1/2}=k_r$ is 
$$
u^8 v^4-u^8 v^2+16 u^6 v^6-24 u^6 v^4-24 u^6 v^2+16 u^6-486 u^4 v^4+
$$
\begin{equation}
+486 u^4 v^2-19683 v^4+19683 v^2=0
\end{equation} 
\\
\textbf{3.}
\begin{equation}
\sum^{\infty}_{n=-\infty}(-1)^nq^{3n^2+4n}=q^{-13/12}\eta(q^6)Q(k_r)
\end{equation} 
The polynomial equation which relates $u=A(-1,6,q)^6$, with $v=m(q)^{1/2}=k_r$ is
\begin{equation}
u^4 v^3-u^4 v+16 u^3 v^2-18 u^2 v^3+18 u^2 v+4 u v^4-8 u v^2+4 u+v^3-v=0
\end{equation}
\\
\textbf{4.}
\begin{equation}
\sum^{\infty}_{n=-\infty}(-1)^nq^{4n^2+6n}=q^{-23/12}\eta(q^8)Q(k_r)
\end{equation}
The polynomial equation which relates $u=A(-2,8;q)^{12}$, with $v=m(q^2)^2=k_{4r}^4$ is
\begin{equation}
-u^4v-64 u^2 v+256 v^2-512 v+256=0
\end{equation}
\\
\textbf{5.}
\begin{equation}
\sum^{\infty}_{n=-\infty}(-1)^nq^{5/2n^2+3/2n}=q^{-1/60}\eta(q^5)Q(\eta_5(q^4)^5)  
\end{equation}
where 
\begin{equation}
\eta_5(q)=\frac{1}{2}\left(-1-h_5(q)+\sqrt{5+2h_5(q)+h_5(q)^2}\right)
\end{equation}
and 
\begin{equation}
h_5(q)=\frac{\eta(q^{1/5})}{q^{1/5}\eta(q^5)}
\end{equation} 
The polynomial equation which relates $u=A(1,5,q^2)^{15}$ with $v=\eta_5(q^4)^5$ is
$$
u^4+v^{11}+55 v^{10}+1205 v^9+13090 v^8+69585 v^7+134761 v^6-69585 v^5+
$$
\begin{equation}
+13090 v^4-1205 v^3+55 v^2-v=0
\end{equation}
Note that
$$
h_5(q^5)=\frac{\eta(q)}{q\eta(q^{25})}=\frac{1}{\sqrt{M_5(q)M_5(q^5)}}\left(\frac{m(q)}{m(q^{25})}\right)^{1/24}\left(\frac{m'(q)}{m'(q^{25})}\right)^{1/6}
$$
is function of $k_r=\sqrt{m(q)}$, since (see [14])
\begin{equation}
\left(5M_5(q)-1\right)^5\left(1-M_5(q)\right)=256\cdot m(q)m'(q)M_5(q)
\end{equation}
and $m'(q)=1-m(q)$.\\
\\
For tables of singular modulus one can see [10],[11],[14].
\[
\]
\centerline{\textbf{References}}
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\\

[2]: J.V. Armitage and W.F. Eberlein. 'Elliptic Functions'. 2006 ed., New York: Cambridge University, (2006) 
\\

[3]: Nikos Bagis. 'On a general sextic equation solved by the Rogers-Ramanujan continued fraction'. arXiv:1111.6023v2, (2012)
\\

[4]: Nikos Bagis. 'Evaluation of Fifth Degree Elliptic Singular Moduli'. arXiv:1202.6246v3, (2015)
\\

[5]: N.D. Bagis and M.L. Glasser. 'Ramanujan type 1/$\pi$ Approximation Formulas'. Journal of Number Theory, Vol(133), 2013 pp. 3453-3469. 
\\

[6]: Nikos Bagis. 'Parametric Evaluations of the Rogers Ramanujan Continued Fraction'. International Journal of Mathematics and Mathematical Sciences. Vol(2011). doi:101155/2011/940839.
\\

[7]: Nikos Bagis and M.L. Glasser. 'Conjectures on the evaluation of alternative modular bases and formulas approximating 1/$\pi$'. Journal of Number Theory, Vol(132), 2012 pp. 2353-2370. 
\\

[8]: Nikos Bagis and M.L. Glasser. 'Jacobian Elliptic Functions, Continued Fractions and Ramanujan Quantities'. arXiv:1001.2660v1, (2010)
\\ 

[9]: B.C. Berndt. 'Ramanujan's Notedbooks Part III'. 1991 ed., New York: Springer-Verlag. (1991) 
\\

[10]: B.C. Berndt. 'Ramanujan's Notedbooks Part V'. 1998 ed., New York: Springer-Verlag. (1998) 
\\

[11]: J.M. Borwein and P.B. Borwein. 'Pi and the AGM'. 1987 ed., New York: John Wiley and Sons Inc., (1987) 
\\

[12]: D. Broadhurst. 'Solutions by Radicals at Singular Values $k_N$ from New Class Invariants for $N\equiv3mod8$'. arXiv:0807.2976 (math-phy), (2008) 
\\

[13]: ET. Whittaker and G.N. Watson. 'A course on Modern Analysis'. 4th ed., Cambridge University Press. (1927)
\\

[14]: J.M. Borwein, M.L. Glasser, R.C. McPhedran, J.G. Wan, I.J. Zucker. 'Lattice Sums Then and Now'. Cambridge University Press. New York, (2013).
\\

[15]: N.D. Bagis. 'On Generalized Integrals and Ramanujan-Jacobi Special Functions'. arXiv:1309.7247v2

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