\nThe Jacobi theta functions are defined for all complex variables of $z$ and $q$ such that $|q| < 1$, as follows:\n$$\n\\begin{aligned}\n\\vartheta_1 (z,q) &= -i \\sum_{n=-\\infty}^{\\infty} (-1)^n q^{(n+1\/2)^2} e^{i(2n+1)z} \\\\\n\\vartheta_2 (z,q) &= \\sum_{n=-\\infty}^{\\infty} q^{(n+1\/2)^2} e^{i(2n+1)z} \\\\\n\\vartheta_3 (z,q) &= \\sum_{n=-\\infty}^{\\infty} q^{n^2} e^{2inz} \\\\ \n\\vartheta_4 (z,q) &= \\sum_{n=-\\infty}^{\\infty} (-1)^n q^{n^2} e^{2inz} \\\\ \n\\end{aligned}\n$$\nThe parameter $q$ is called the nome. Let $\\tau$ be a complex number whose imaginary part is positive and write $q=e^{i\\pi \\tau}$ so that $|q| < 1$. Sometimes, $q$ will not be specified, so that $\\vartheta_r (z)$ is written for $\\vartheta_r(z,q)$ where $r=1,2,3,4$. It is easy to see that\n$$\n\\begin{aligned}\n\\vartheta_1(z+\\pi) &= -i e^{i \\pi}\\sum_{n=-\\infty}^{\\infty}(-1)^n q^{\\left(n+\\frac{1}{2}\\right)^2}e^{i(2n+1)z} = -\\vartheta_1(z) \\\\\n\\vartheta_1(z+\\pi \\tau) &= i e^{-i\\pi \\tau -2iz} \\sum_{n=-\\infty}^{\\infty} (-1)^{n+1} e^{i\\pi \\tau \\left( n+\\frac{3}{2}\\right)^2 + (2n+3) iz} = – (q e^{2iz})^{-1} \\vartheta_1(z)\n\\end{aligned}\n$$\nSimilarly, the periodicity laws for the other theta functions can be derived as well. The multipliers of the theta functions associated with the periods $\\pi$ and $\\pi \\tau$ are summarized in the following table:\n$$\n\\begin{array}{|c|c|c|c|c|}\n\\hline\n\\omega & \\vartheta_1(z+\\omega) & \\vartheta_2(z+\\omega) & \\vartheta_3(z+\\omega) & \\vartheta_4(z+\\omega) \\\\ \\hline \\hline\n\\pi & -1 & -1 & 1 & 1 \\\\ \\hline\n\\pi \\tau & -\\lambda & \\lambda & \\lambda & -\\lambda \\\\ \\hline\n\\end{array}\n$$\nwhere $\\lambda = (q e^{2iz})^{-1}$. Furthermore, incrementation of $z$ by the half periods $\\frac{\\pi}{2}, \\frac{\\pi \\tau}{2}, \\frac{\\pi +\\pi \\tau}{2}$ yields the following identities:\n$$\n\\begin{array}{|c|c|c|c|c|}\n\\hline\n\\omega & \\vartheta_1(z+\\omega) & \\vartheta_2(z+\\omega) & \\vartheta_3(z+\\omega) & \\vartheta_4(z+\\omega) \\\\ \\hline \\hline\n\\frac{\\pi}{2} & \\vartheta_2(z) & -\\vartheta_1(z) & \\vartheta_4(z) & \\vartheta_3(z) \\\\ \\hline\n\\frac{\\pi \\tau}{2} & i \\mu \\vartheta_4(z) & \\mu \\vartheta_3(z) & \\mu \\vartheta_2(z) & i\\mu \\vartheta_1(z) \\\\ \\hline\n\\frac{\\pi + \\pi \\tau}{2} & \\mu \\vartheta_3(z) & -i\\mu \\vartheta_4(z) & i\\mu \\vartheta_1(z) & \\mu \\vartheta_2(z) \\\\ \\hline\n\\end{array}\n$$\nwhere $\\mu = (q^{\\frac{1}{4}}e^{iz})^{-1}$.\n<\/p>\n\n\n\n
\nMany theta function identities can be derived by multiplying their series and rearranging the terms. This is ok to do since both the series are absolutely convergent. For e.g., we have\n$$\n\\begin{aligned}\n\\vartheta_3(x,q)\\vartheta_3(y,q) &= \\left(\\sum_{n=-\\infty}^{\\infty} q^{n^2} e^{2inx} \\right)\\left(\\sum_{m=-\\infty}^{\\infty} q^{m^2} e^{2imy} \\right) \\\\\n&= \\sum_{n=-\\infty}^\\infty \\sum_{m=-\\infty}^\\infty q^{n^2+m^2}e^{2i(nx+my)}\n\\end{aligned}\n$$\nNow, we change the summation indices from $(m,n)$ to $(r,s)$ using the following transformation:\n$$\n\\begin{aligned}\nr &= m+n \\\\\ns &= m-n\n\\end{aligned}\n$$\nIf $(m,n)$ are both even then $(r,s)$ will both be even and if $(m,n)$ have opposite parity then $(r,s)$ will both be odd. Therefore, we can rearrange the series as follows:\n$$\n\\begin{aligned}\n\\vartheta_3(x,q)\\vartheta_3(y,q) &= \\sum_{r=-\\infty}^{\\infty} \\sum_{s=-\\infty}^{\\infty} q^{2(r^2 +s^2)}e^{2i(r(x+y)+s(x-y))} \\\\ &\\quad + \\sum_{r=-\\infty}^{\\infty} \\sum_{s=-\\infty}^{\\infty}q^{2\\left[\\left(r+\\frac{1}{2} \\right)^2 + \\left(s+\\frac{1}{2} \\right)^2\\right]} e^{i(2r+1)(x+y) + i(2s+1)(x-y)} \\\\\n&= \\vartheta_3(x+y,q^2)\\vartheta_3(x-y,q^2) + \\vartheta_2(x+y,q^2)\\vartheta_2(x-y,q^2) \\quad (1)\n\\end{aligned}\n$$\nSome similar identities that can be derived using this method are:\n$$\n\\begin{aligned}\n\\vartheta_1(x,q) \\vartheta_1(y,q) &= \\vartheta_3(x+y,q^2)\\vartheta_2(x-y,q^2) – \\vartheta_2(x+y,q^2)\\vartheta_3(x-y,q^2) \\quad (2) \\\\\n\\vartheta_2(x,q) \\vartheta_2(y,q) &= \\vartheta_2(x+y,q^2)\\vartheta_3(x-y,q^2) + \\vartheta_3(x+y,q^2)\\vartheta_2(x-y,q^2) \\quad (3) \\\\\n\\vartheta_4(x,q) \\vartheta_4(y,q) &= \\vartheta_3(x+y,q^2)\\vartheta_3(x-y,q^2) – \\vartheta_2(x+y,q^2)\\vartheta_2(x-y,q^2) \\quad (4) \\\\\n\\vartheta_1(x,q) \\vartheta_2(y,q) &= \\vartheta_1(x+y,q^2)\\vartheta_4(x-y,q^2) + \\vartheta_4(x+y,q^2)\\vartheta_1(x-y,q^2) \\quad (5) \\\\\n\\vartheta_3(x,q) \\vartheta_4(y,q) &= \\vartheta_4(x+y,q^2)\\vartheta_4(x-y,q^2) – \\vartheta_1(x+y,q^2)\\vartheta_1(x-y,q^2) \\quad (6)\n\\end{aligned}\n$$\nSquaring and subtracting identities (4) and (2) gives us:\n$$\n\\begin{aligned}\n&\\; \\vartheta_4^2(x,q) \\vartheta_4^2(y,q) – \\vartheta_1^2(x,q) \\vartheta_1^2(y,q)\\\\ \n&= \\left[\\vartheta_3^2(x+y,q^2)-\\vartheta_2^2(x+y,q^2) \\right]\\times \\left[\\vartheta_3^2(x-y,q^2)-\\vartheta_2^2(x-y,q^2) \\right] \\quad (7)\n\\end{aligned}\n$$\nPutting $y=0$ in the above equation gives the result:\n$$\n\\vartheta_3^2(x,q^2)-\\vartheta_2^2(x,q^2) = \\vartheta_4(x,q)\\vartheta_4(0,q)\n$$\nNow, using the above result in equation (7) gives us:\n$$\n\\vartheta_4(x+y,q)\\vartheta_4(x-y,q)\\vartheta_4^2(0,q) = \\vartheta_4^2(x,q) \\vartheta_4^2(y,q) – \\vartheta_1^2(x,q) \\vartheta_1^2(y,q) \\quad (8)\n$$\nNote that all theta functions in this equation have the same nome, $q$. More identities of this type can be derived by incrementing $x$ and\/or $y$ by the half periods $\\frac{\\pi}{2},\\frac{\\pi \\tau}{2},\\frac{\\pi+\\pi\\tau}{2}$:\n$$\n\\begin{aligned}\n\\vartheta_1(x+y)\\vartheta_1(x-y)\\vartheta_4^2(0) &= \\vartheta_3^2(x) \\vartheta_2^2(y) – \\vartheta_2^2(x) \\vartheta_3^2(y) \\\\ &= \\vartheta_1^2(x) \\vartheta_4^2(y) – \\vartheta_4^2(x) \\vartheta_1^2(y)\\quad (9) \\\\\n\\vartheta_2(x+y)\\vartheta_2(x-y)\\vartheta_4^2(0) &= \\vartheta_4^2(x) \\vartheta_2^2(y) – \\vartheta_1^2(x) \\vartheta_3^2(y) \\\\ &= \\vartheta_2^2(x) \\vartheta_4^2(y) – \\vartheta_3^2(x) \\vartheta_1^2(y)\\quad (10) \\\\\n\\vartheta_3(x+y)\\vartheta_3(x-y)\\vartheta_4^2(0) &= \\vartheta_4^2(x) \\vartheta_3^2(y) – \\vartheta_1^2(x) \\vartheta_2^2(y) \\\\ &= \\vartheta_3^2(x) \\vartheta_4^2(y) – \\vartheta_2^2(x) \\vartheta_1^2(y)\\quad (11) \\\\\n\\vartheta_4(x+y)\\vartheta_4(x-y)\\vartheta_4^2(0) &= \\vartheta_3^2(x) \\vartheta_3^2(y) – \\vartheta_2^2(x) \\vartheta_2^2(y) \\\\ &= \\vartheta_4^2(x) \\vartheta_4^2(y) – \\vartheta_1^2(x) \\vartheta_1^2(y)\\quad (12) \\\\\n\\end{aligned}\n$$\n\nOne can keep on deriving similar identities by squaring and adding equations (1) and (2) and squaring and subtracting equations (3) and (2). For a complete list of such identities, refer to [1 sec 1.4]. \n\n\n<\/p>\n\n\n\n
\nAnother way to derive theta function identities is to utilize properties of doubly periodic functions. More details on this method are presented in [2]. Consider the function:\n$$\\frac{a \\vartheta_1^2(z) + b \\vartheta_4^2(z)}{\\vartheta_2^2(z)}$$\nThis is a doubly periodic function with periods $\\pi$ and $\\pi \\tau$. Furthermore, we can choose the constants $a$ and $b$ such that there are no poles in each cell. By Liouville’s Theorem, such a function is a constant. By appropriately scaling the constants $a$ and $b$, we can make the function equal to 1. Therefore, there exists a relationship of the form:\n$$\na \\vartheta_1^2(z) + b \\vartheta_4^2(z) = \\vartheta_2^2(z)\n$$\nTo find $a$ and $b$, we can put $z = 0, \\frac{\\pi\\tau}{2}$:\n$$\n\\begin{aligned}\nb \\vartheta_4^2(0) &= \\vartheta_2^2(0) \\; \\implies b = \\frac{\\vartheta_2^2(0)}{\\vartheta_4^2(0)} \\\\\na (i \\mu\\vartheta_4(0))^2 &= (\\mu \\vartheta_3(0))^2 \\; \\implies a = -\\frac{\\vartheta_3^2(0)}{\\vartheta_4^2(0)}\n\\end{aligned}\n$$\nThus, we have obtained the identity:\n$$ \\vartheta_2^2(0) \\vartheta_4^2(z) -\\vartheta_3^2(0) \\vartheta_1^2(z) = \\vartheta_4^2(0) \\vartheta_2^2(z) \\quad (13) $$\nA similar technique yields the identity:\n$$ \\vartheta_3^2(0) \\vartheta_4^2(z) – \\vartheta_2^2(0) \\vartheta_1^2(z) = \\vartheta_4^2(0) \\vartheta_3^2(z) \\quad (14) $$\nIncrementing $z$ by $\\frac{\\pi}{2}$ in (11) and (12) gives two additional identities:\n$$\n\\begin{aligned}\n\\vartheta_4^2(0) \\vartheta_1^2(z) &= \\vartheta_2^2(0) \\vartheta_3^2(z) – \\vartheta_3^2(0) \\vartheta_2^2(z) \\quad (15) \\\\\n\\vartheta_4^2(0) \\vartheta_4^2(z) &= \\vartheta_3^2(0) \\vartheta_3^2(z) – \\vartheta_2^2(0) \\vartheta_2^2(z) \\quad (16)\n\\end{aligned}\n$$\nWith these relations one can express any theta function in terms of any other pair of theta functions. Putting $z=0$ in (16) gives the famous identity:\n$$ \\vartheta_4^4(0)+\\vartheta_2^4(0) = \\vartheta_3^4(0) $$\n<\/p>\n\n\n\n
\nThis particular proof is from [1]. Differentiating equation (5) with respect to $x$ and substituting $x=y=0$, we get:\n$$\n\\vartheta_1′(0,q) \\vartheta_2(0,q) = 2 \\vartheta_1′(0,q^2)\\vartheta_4(0,q^2) \\quad (17)\n$$\nSubstituting $x=y=0$ in equations (3) and (6) gives:\n$$\n\\begin{aligned}\n\\vartheta_2^2(0,q) &= 2\\vartheta_2(0,q^2) \\vartheta_3(0,q^2) \\quad (18) \\\\\n\\vartheta_3(0,q)\\vartheta_4(0,q) &= \\vartheta_4^2(0,q^2) \\quad (19)\n\\end{aligned}\n$$\nNow, dividing (17) by both (18) and (19) gives\n$$\n\\frac{\\vartheta_1′(0,q)}{\\vartheta_2(0,q)\\vartheta_3(0,q)\\vartheta_4(0,q)} = \\frac{\\vartheta_1′(0,q^2)}{\\vartheta_2(0,q^2)\\vartheta_3(0,q^2)\\vartheta_4(0,q^2)} \n$$\nThe repeated application of this result gives:\n$$\n\\frac{\\vartheta_1′(0,q)}{\\vartheta_2(0,q)\\vartheta_3(0,q)\\vartheta_4(0,q)} = \\frac{\\vartheta_1′(0,q^{2^n})}{\\vartheta_2(0,q^{2^n})\\vartheta_3(0,q^{2^n})\\vartheta_4(0,q^{2^n})} \n$$\nfor all positive integers $n$. Letting $n\\to\\infty$ in the above equation, we find that:\n$$\n\\frac{\\vartheta_1′(0,q)}{\\vartheta_2(0,q)\\vartheta_3(0,q)\\vartheta_4(0,q)} = \\lim_{q\\to 0}\\frac{\\vartheta_1′(0,q)}{\\vartheta_2(0,q)\\vartheta_3(0,q)\\vartheta_4(0,q)} \\quad (20)\n$$\nFrom the definitions of the theta functions, it is evident that\n$$\n\\begin{aligned}\n\\vartheta_1′(0,q) &= 2q^{\\frac{1}{4}} + \\mathcal{O}(q^{\\frac{9}{4}}) \\\\\n\\vartheta_2(0,q) &= 2q^{\\frac{1}{4}} + \\mathcal{O}(q^{\\frac{9}{4}}) \\\\\n\\vartheta_3(0,q) &= 1 + \\mathcal{O}(q) \\\\\n\\vartheta_4(0,q) &= 1 + \\mathcal{O}(q) \\\\\n\\end{aligned}\n$$\nTherefore, the limit in equation (20) equals 1 and\n$$\\vartheta_1′(0)=\\vartheta_2(0)\\vartheta_3(0)\\vartheta_4(0)$$\nis established as desired.\n<\/p>\n\n\n\n
References<\/b><\/p>\n <\/p>\n","protected":false},"excerpt":{"rendered":" The Jacobi theta functions are defined for all complex variables of $z$ and $q$ such that $|q| < 1$, as follows: $$ \\begin{aligned} \\vartheta_1 (z,q) &= -i \\sum_{n=-\\infty}^{\\infty} (-1)^n q^{(n+1\/2)^2} e^{i(2n+1)z} \\\\ \\vartheta_2 (z,q) &= \\sum_{n=-\\infty}^{\\infty} q^{(n+1\/2)^2} e^{i(2n+1)z} \\\\ \\vartheta_3 … Continue reading \n