Infinite product representations of theta functions<\/b><\/p>\n
\nLet $f(z) = \\prod_{n=1}^{\\infty} (1-q^{2n-1}e^{2iz})(1-q^{2n-1}e^{-2iz})$. Each of these two products converge absolutely and uniformly in any bounded domain of values of $z$. Hence $f(z)$ is analytic throughout the finite part of the $z$ plane. The zeros of $f(z)$ are simple zeros at the points $z=\\frac{2n+1}{2}\\pi \\tau i + m\\pi i$ where $m,n\\in \\mathbb{Z}$. So, the zeros of $f(z)$ coincide with the zeros of $\\vartheta_4(z)$. Therefore, the function $\\frac{\\vartheta_4(z)}{f(z)}$ has no poles or zeros in the finite part of the plane. Also, it is easy to see that $f(z+\\pi) = f(z)$ and\n$$\n\\begin{aligned}\nf(z+\\pi \\tau) &= \\prod_{n=1}^\\infty (1-q^{2n+1}e^{2iz})(1-q^{2n-3}e^{-2iz}) \\\\\n&= f(z) \\frac{1-q^{-1}e^{-2iz}}{1-qe^{2iz}} \\\\\n&= -q^{-1}e^{-2iz} f(z)\n\\end{aligned}\n$$\nThis is, $\\frac{\\vartheta_4(z)}{f(z)}$ is a doubly periodic function with periods $\\pi,\\pi\\tau$ and has no poles and zeros. By section 20.12 of [2], it is simply a constant, say $G$. We have\n$$\n\\vartheta_4(z) = G \\prod_{n=1}^{\\infty} (1-q^{2n-1}e^{2iz})(1-q^{2n-1}e^{-2iz})\\quad (1)\n$$\nIncrementing $z$ by the half periods $\\frac{\\pi}{2}, \\frac{\\pi \\tau}{2}$ and $\\frac{\\pi +\\pi\\tau}{2}$, we get the product representations for the other theta functions:\n$$\n\\begin{aligned}\n\\vartheta_3(z) &= G \\prod_{n=1}^{\\infty} (1+q^{2n-1}e^{2iz})(1+q^{2n-1}e^{-2iz})\\quad (2) \\\\\n\\vartheta_1(z) &= 2G q^{\\frac{1}{4}}\\sin(z) \\prod_{n=1}^{\\infty} (1-q^{2n}e^{2iz})(1-q^{2n}e^{-2iz}) \\quad (3) \\\\\n\\vartheta_2(z) &= 2G q^{\\frac{1}{4}}\\cos(z) \\prod_{n=1}^{\\infty} (1+q^{2n}e^{2iz})(1+q^{2n}e^{-2iz}) \\quad (4)\n\\end{aligned}\n$$\nNow, all that left is to find the constant $G$. From identity (3), it easy to see that\n$$\n\\vartheta_1′(0) = 2G q^{\\frac{1}{4}}\\prod_{n=1}^{\\infty} (1-q^{2n})^2\n$$\nNow, using the identity $\\vartheta_1′(0)=\\vartheta_2(0)\\vartheta_3(0)\\vartheta_4(0)$, we get:\n$$\n\\begin{aligned}\n2G q^{\\frac{1}{4}}\\prod_{n=1}^{\\infty} (1-q^{2n})^2 &= \\left\\{2G q^{\\frac{1}{4}}\\prod_{n=1}^{\\infty}(1+q^{2n})^2 \\right\\} \\times \\left\\{G\\prod_{n=1}^{\\infty}(1+q^{2n-1})^2 \\right\\} \\\\ &\\quad \\times \\left\\{G \\prod_{n=1}^{\\infty}(1-q^{2n-1})^2 \\right\\} \\\\\n\\implies G &= \\prod_{n=1}^\\infty (1-q^{2n})\n\\end{aligned}\n$$\nNote that the rearrangements are justified since all products converge absolutely. Finally, we have\n$$\n\\begin{aligned}\n\\vartheta_1(z) &= 2 q^{\\frac{1}{4}}\\sin(z) \\prod_{n=1}^{\\infty} (1-q^{2n})(1-2q^{2n} \\cos(2z)+ q^{4n}) \\quad (5) \\\\\n\\vartheta_2(z) &= 2 q^{\\frac{1}{4}}\\cos(z) \\prod_{n=1}^{\\infty} (1-q^{2n})(1+2q^{2n} \\cos(2z) + q^{4n}) \\quad (6) \\\\\n\\vartheta_3(z) &= \\prod_{n=1}^{\\infty} (1-q^{2n})(1+2q^{2n-1} \\cos(2z) + q^{4n-2}) \\quad (7) \\\\\n\\vartheta_4(z) &= \\prod_{n=1}^{\\infty} (1-q^{2n})(1-2q^{2n-1} \\cos(2z) + q^{4n-2}) \\quad (8)\n\\end{aligned}\n$$\n<\/p>\n\n\n\n
Derivatives of Ratios of Theta Functions<\/b><\/p>\n
\nConsider the function $\\phi(z) = \\frac{\\vartheta_1′(z)\\vartheta_4(z) – \\vartheta_1(z)\\vartheta_4′(z)}{\\vartheta_2(z)\\vartheta_3(z)}$. Now, $\\phi(z)$ is doubly periodic with periods $\\pi$ and $\\frac{\\pi\\tau}{2}$. Relative to these periods, the only possible poles of $\\phi(z)$ are at points congruent to $\\pi\\over 2$. It is easy to verify that $\\frac{\\pi}{2}$ is a removable singularity. This means that $\\phi(z)$ is a constant. Letting $z\\to 0$, we see that $\\phi(z)=\\vartheta_4^2(0)$. It is therefore established that \n$$\\frac{d}{dz}\\left(\\frac{\\vartheta_1(z)}{\\vartheta_4(z)} \\right)=\\vartheta_4^2(0) \\frac{\\vartheta_2(z)\\vartheta_3(z)}{\\vartheta_4^2(z)} \\quad (9)$$\nTwo other identities of this kind can be derived using the same technique:\n$$\n\\begin{aligned}\n\\frac{d}{dz}\\left(\\frac{\\vartheta_2(z)}{\\vartheta_4(z)} \\right) &=-\\vartheta_3^2(0) \\frac{\\vartheta_1(z)\\vartheta_3(z)}{\\vartheta_4^2(z)} \\quad (10) \\\\\n\\frac{d}{dz}\\left(\\frac{\\vartheta_3(z)}{\\vartheta_4(z)} \\right) &=-\\vartheta_2^2(0) \\frac{\\vartheta_1(z)\\vartheta_2(z)}{\\vartheta_4^2(z)} \\quad (11)\n\\end{aligned}$$\nBy writing $\\xi = \\frac{\\vartheta_1(z)}{\\vartheta_4(z)}$ and making use of identities (13) and (14) from part 1<\/a>, we obtain:\n$$\\left(\\frac{d\\xi}{dz}\\right)^2 = (\\vartheta_2^2(0) – \\xi^2 \\vartheta_3^2(0))(\\vartheta_3^2(0)-\\xi^2 \\vartheta_2^2(0))$$\nUsing the change of variables $y = \\frac{\\vartheta_3(0)}{\\vartheta_2(0)}\\xi$ and $u = \\vartheta_3^2(0) z$, the above differential equation can be written as:\n$$\n\\left(\\frac{dy}{du}\\right)^2 = (1-y^2)(1-k^2 y^2)\n$$\nwhere $k = \\frac{\\vartheta_2^2(0)}{\\vartheta_3^2(0)}$. This differential equation has the particular solution:\n$$y = \\frac{\\vartheta_3(0)}{\\vartheta_2(0)} \\frac{\\vartheta_1(u \\vartheta_3^{-2}(0))}{\\vartheta_4(u \\vartheta_3^{-2}(0))}$$\nWe can write $y$ as a function of $u$ and $k$ as $y = \\text{sn}(u,k)$ or simply $\\text{sn}(u)$. Clearly, $\\text{sn}(u,k)$ has periods $2\\pi \\vartheta^2_3(0)$ and $\\pi \\tau \\vartheta_3^2(0)$. It has two simple poles congruent to $\\frac{1}{2}\\pi\\tau \\vartheta_3^2(0)$ and $\\pi \\vartheta_3^2(0) + \\frac{1}{2}\\pi\\tau \\vartheta_3^2(0)$. Also, it is easy to see that\n$$\\int_0^1 \\frac{dy}{\\sqrt{(1-y^2)(1-k^2 y^2)}} = \\text{sn}^{-1}(1) = \\frac{\\pi}{2}\\vartheta_3^2(0)$$\nThe integral on the left is denoted by complete elliptic integral of the first kind and is denoted by $K(k)$.\n<\/p>\n\n\n\n References<\/b><\/p>\n Infinite product representations of theta functions Let $f(z) = \\prod_{n=1}^{\\infty} (1-q^{2n-1}e^{2iz})(1-q^{2n-1}e^{-2iz})$. Each of these two products converge absolutely and uniformly in any bounded domain of values of $z$. Hence $f(z)$ is analytic throughout the finite part of the $z$ plane. … Continue reading \n