The Jacobi theta function is given by \begin{align} \theta_1(\tau|z)&=\theta_1(q,y)=-iq^{\frac{1}{8}}y^{\frac{1}{2}}\prod_{k=1}^{\infty}(1-q^k)(1-yq^k)(1-y^{-1}q^{k-1}) \\ &= -i\sum_{n\in \mathbb{Z}}(-1)^n e^{2\pi i z(n+\frac{1}{2})} e^{\pi i \tau(n+\frac{1}{2})^2} \end{align} Where $q=e^{2\pi i \tau}$, and $y=e^{2\pi i z}$

I want to show the first line is equivalent to the second line, i just plug $y$ and $q$, with given parameters, but i could not obtain the desired results.

• You essentially need the proof for the Jacobi Triple Product namely $$\sum_{n=-\infty}^{\infty}x^{n}q^{n^{2}}=\prod_{n=1}^{\infty}(1-q^{2n})(1+xq^{2n-1})(1+x^{-1}q^{2n-1})$$ Check Jacobi's original proof presented at paramanands.blogspot.com/2011/02/… You can get the above product formula if you replace $q$ by $q^{2}$ and put $y=-x/q$ in your formula. – Paramanand Singh Sep 27 '14 at 3:05
• @Paramanand Singh, Thank you, i tried to proof the Jacobi Triple Product. – phy_math Sep 27 '14 at 3:20
• I expanded my comment into bit of an answer. – Paramanand Singh Sep 27 '14 at 3:30

You essentially need the proof for the Jacobi Triple Product namely $$\sum_{n=-\infty}^{\infty}x^{n}q^{n^{2}}=\prod_{n=1}^{\infty}(1-q^{2n})(1+xq^{2n-1})(1+x^{-1}q^{2n-1})$$ Check Jacobi's original proof presented at http://paramanands.blogspot.com/2011/02/elliptic-functions-theta-functions-contd.html

If you cancel the factor $-iq^{1/8}y^{1/2}$ from first and second line of your formula we are left with \begin{aligned}\prod_{k=1}^{\infty}(1-q^{k})(1-yq^{k})(1-y^{-1}q^{k-1})&=\sum_{n\in\mathbb{Z}}(-1)^{n}e^{2\pi inz}e^{\pi i\tau(n^{2}+n)}\\ &=\sum_{n\in\mathbb{Z}}(-1)^{n}y^{n}q^{(n^{2}+n)/2}\\\end{aligned} You can get the Jacobi triple product formula if you replace $q$ by $q^{2}$ and put $y=-x/q$ in your formula.