# Proof of Jacobi triple product using Ramanujan's notation

Inspired by this proof in MathWorld, I rewrote the proof in terms of the Ramanujan theta function.

Define the function $$M(c)=\prod_{n = 1}^{\infty}(1 +a^{n}b^{n-1}c)\left(1+\frac{a^{n-1}b^{n}}{c}\right)\tag1$$

then $$M(abc)=\prod_{n = 1}^{\infty}(1 +a^{n+1}b^{n}c)\left(1+\frac{a^{n-2}b^{n-1}}{c}\right)\tag2$$

$$M(abc)=(1+a^2bc)\left(1+\frac{1}{ac}\right)(1+a^{3}b^{2}c)\left(1+\frac{b}{c}\right)(1+a^{4}b^{3}c)\left(1+\frac{ab^{2}}{c}\right)\cdots$$

$$M(c)=(1+ac)\left(1+\frac{b}{c}\right)(1+a^{2}bc)\left(1+\frac{ab^{2}}{c}\right)(1+a^{3}b^{2}c)\left(1+\frac{a^{2}b^{3}}{c}\right)\cdots$$

Taking $$\frac{M(abc)}{M(c)}=\left(1+\frac{1}{ac}\right)\left(\frac{1}{1+ac}\right)=\frac{1}{ac}$$

yields the following relation $$M(c)=acM(abc).$$

Now define $$N(c)=M(c)\prod_{n = 1}^{\infty}(1 -(ab)^{n}).$$

Then $$N(abc)=M(abc)\prod_{n = 1}^{\infty}(1 -(ab)^{n})$$

which becomes $$N(c)=acN(abc).$$

Now expand $N(c)$ in a Laurent series $$N(c)=\sum_{n=-\infty}^{\infty}u_{n}c^{n}.$$

Using the fundamental relation, we have $$\sum_{n=-\infty}^{\infty}u_{n}c^{n}=ac\sum_{n=-\infty}^{\infty}u_{n}(abc)^{n}$$

$$=\sum_{n=-\infty}^{\infty}u_{n}a^{n+1}b^{n}c^{n+1}$$

$$=\sum_{n=-\infty}^{\infty}u_{n-1}a^{n}b^{n-1}c^{n}.$$

Which leads to the recurrence relation $$u_{n}=u_{n-1}a^{n}b^{n-1}.$$

$$u_{1}=u_{0}a,$$ $$u_{2}=u_{1}a^{2}b=u_{0}a^{3}b,$$ $$u_{3}=u_{2}a^{3}b^{2}=u_{0}a^{6}b^{3},$$ $$u_{4}=u_{3}a^{4}b^{3}=u_{0}a^{10}b^{6}.$$

Which in general form is $$u_{n}=u_{0}a^{n(n+1)/2}b^{n(n-1)/2}.$$

Now substituting back into the original Laurent series, we obtain $$N(c)=u_{0}\sum_{n=-\infty}^{\infty}a^{n(n+1)/2}b^{n(n-1)/2}c^{n}.$$

It can easily be shown that $u_{0}=1$, so that we have the Jacobi triple product in terms of the Ramanujan theta function $$N(c)=\sum_{n=-\infty}^{\infty}a^{n(n+1)/2}b^{n(n-1)/2}c^{n}.$$

Q:Is this generalisation of the proof correct?

• Looks OK to me as far as the calculation goes. However if we have $A = ac, B = b/c$ then we automatically get $$A^{n(n + 1)/2}B^{n(n - 1)/2} = a^{n(n + 1)/2}b^{n(n - 1)/2}c^{n}$$ and the generalization obtained is equivalent to the same old Jacobi Triple product. So at best its a new proof similar to the one given in MathWorld. – Paramanand Singh Apr 3 '16 at 4:44
• How do you know that $N(c)$ has a Laurent series? How do you know that $M(c)$ infinite product converges? Also it would help if you gave a complete statement of exactly what you are proving. – Somos Jul 8 '17 at 20:18
• @Somos :As you can see I adapted the method of proof exactly from the linked mathworld site above and used a different notation instead.In case there are missing important details,I hope someone will fill them in – Nicco Jul 9 '17 at 8:36

Unfortunately, that proof is flawed. The crux is the step from $M(c)$ to $N(c)$ in your proof. It comes from nowhere. Why did you multiply by that infinite product? The same problem arises in the Mathworld proof going from $F(z)$ to $G(z)$. Without being able to justify $a_0=1$ the proofs are not complete. Of course, the result is still true, but the justification is missing.
• Perhaps. Still, the problem remains to justify the Laurent series $a_0=1$. – Somos Jul 19 '17 at 16:57