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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?

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  • $\begingroup$ 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. $\endgroup$ – Paramanand Singh Apr 3 '16 at 4:44
  • $\begingroup$ 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. $\endgroup$ – Somos Jul 8 '17 at 20:18
  • $\begingroup$ @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 $\endgroup$ – Nicco Jul 9 '17 at 8:36
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The proof you have is essentially the same as the MathWorld proof with a few reparametrizing of terms. It stands or falls with the one you are modeling it on. It is similar in the way that Ramanujan's theta function differs from Jacobi's theta function and I commend you for that.

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.

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  • $\begingroup$ Perhaps the author of the original proof from mathworld may give his/her side. $\endgroup$ – Nicco Jul 19 '17 at 15:14
  • $\begingroup$ Perhaps. Still, the problem remains to justify the Laurent series $a_0=1$. $\endgroup$ – Somos Jul 19 '17 at 16:57

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