Jacobi Triple Product

The Jacobi triple product is the beautiful identity

$$\prod_{n=1}^\infty (1-x^{2n})(1+x^{2n-1}z^2)\left({1+{x^{2n-1}\over z^2}}\right)= \sum_{m=-\infty }^\infty x^{m^2}z^{2m}.$$ (1)

In terms of the Q-Function, (1) is written

$$Q_1Q_2Q_3=1,$$ (2)

which is one of the two Jacobi Identities. For the special case of $z = 1$, (1) becomes

$$\varphi(x) \equiv G(1)=\prod_{n=1}^\infty (1+x^{2n-1})^2(1-x^{2n})$$

$$= \sum_{m=-\infty}^\infty x^{m^2} = 1+2\sum_{m=1}^\infty x^{m^2},$$ (3)

where $\varphi(x)$ is the one-variable Ramanujan Theta Function.

To prove the identity, define the function

$$F(z) \equiv \prod_{n=1}^\infty (1+x^{2n-1}z^2)\left({1 + {x^{2n-1}\over z^2}}\right)$$

$$= (1+xz^2)\left({1 + {x\over z^2}}\right)(1+x^3z^2)\left({1 + {x^3\over z^2}}\right)(1+x^5z^2)\left({1 + {x^5\over z^2}}\right)\cdots.$$ (4)

Then

$$F(xz) = (1+x^3z^2)\left({1 + {1\over xz^2}}\right)(1+x^5z^2)... ...z^2}}\right)(1+x^7z^2)\left({1 + {x^3\over z^2}}\right)\cdots.$$ (5)

Taking (5) $\div$ (4),

$${F(xz)\over F(z)} = \left({1 + {1\over xz^2}}\right)\left({1\over 1+xz^2}\right)$$

$$= {xz^2+1\over xz^2} {1\over 1+xz^2} = {1\over xz^2},$$ (6)

which yields the fundamental relation

$$xz^2F(xz) = F(z).$$ (7)

Now define

$$G(z) \equiv F(z)\prod_{n=1}^\infty (1-x^{2n})$$ (8)

$$G(xz) = F(xz) \prod_{n=1}^\infty (1-x^{2n}).$$ (9)

Using (7), (9) becomes

$$G(xz) = {F(z)\over xz^2} \prod_{n=1}^\infty (1-x^{2n}) = {G(z)\over xz^2},$$ (10)

so

$$G(z) = xz^2G(xz).$$ (11)

Expand $G$ in a Laurent Series. Since $G$ is an Even Function, the Laurent Series contains only even terms.

$$G(z) =\sum_{m=-\infty}^\infty a_mz^{2m}.$$ (12)

Equation (11) then requires that

$$\sum_{m=-\infty}^\infty a_mz^{2m} = xz^2\sum_{m=-\infty }^\infty a_m(xz)^{2m}$$

$$= \sum_{m=-\infty}^\infty a_mx^{2m+1}z^{2m+2}.$$ (13)

This can be re-indexed with $m'\equiv m-1$ on the left side of (13)

$$\sum_{m=-\infty}^\infty a_mz^{2m} = \sum_{m=-\infty}^\infty a_mx^{2m-1}z^{2m},$$ (14)

which provides a Recurrence Relation

$$a_m = a_{m-1}x^{2m-1},$$ (15)

so

$$a_1 = a_0x$$ (16)

$$a_2 = a_1x^3 = a_0x^{3+1} = a_0x^4 =a_0x^{2^2}$$ (17)

$$a_3 = a_2x^5 = a_0x^{5+4} = a_0x^9 = a_0x^{3^2}.$$ (18)

The exponent grows greater by $(2m-1)$ for each increase in $m$ of 1. It is given by

$$\sum_{n=1}^m (2m-1) = 2 {m(m+1)\over 2}-m = m^2.$$ (19)

Therefore,

$$a_m = a_0x^{m^2}.$$ (20)

This means that

$$G(z) = a_0\sum_{m=-\infty }^\infty x^{m^2}z^{2m}.$$ (21)

The Coefficient $a_0$ must be determined by going back to (4) and (8) and letting $z = 1$. Then

$$F(1) = \prod_{n=1}^\infty (1+x^{2n-1})(1+x^{2n-1})$$

$$= \prod_{n=1}^\infty (1+x^{2n-1})^2$$ (22)

$$G(1) = F(1) \prod_{n=1}^\infty (1-x^{2n})$$

$$= \prod_{n=1}^\infty (1+x^{2n-1})^2\prod_{n=1}^\infty (1-x^{2n})$$

$$= \prod_{n=1}^\infty (1+x^{2n-1})^2(1-x^{2n}),$$ (23)

since multiplication is Associative. It is clear from this expression that the $a_0$ term must be 1, because all other terms will contain higher Powers of $x$. Therefore,

$$a_0 = 1,$$ (24)

so we have the Jacobi triple product,

$$G(z) = \prod_{n=1}^\infty (1-x^{2n})(1+x^{2n-1}z^2)\left({1 + { x^{2n-1}\over z^2}}\right)$$

$$= \sum_{m=-\infty }^\infty x^{m^2}z^{2m}.$$ (25)

See also Euler Identity, Jacobi Identities, Q-Function, Quintuple Product Identity, Ramanujan Psi Sum, Ramanujan Theta Functions, Schröter's Formula, Theta Function

References

Andrews, G. E. $q$-Series: Their Development and Application in Analysis, Number Theory, Combinatorics, Physics, and Computer Algebra. Providence, RI: Amer. Math. Soc., pp. 63-64, 1986.

Borwein, J. M. and Borwein, P. B. ``Jacobi's Triple Product and Some Number Theoretic Applications.'' Ch. 3 in Pi & the AGM: A Study in Analytic Number Theory and Computational Complexity. New York: Wiley, pp. 62-101, 1987.

Jacobi, C. G. J. Fundamentia Nova Theoriae Functionum Ellipticarum. Regiomonti, Sumtibus fratrum Borntraeger, p. 90, 1829.

Whittaker, E. T. and Watson, G. N. A Course in Modern Analysis, 4th ed. Cambridge, England: Cambridge University Press, p. 470, 1990.