I would like to pose the following conjecture.Given

$$\phi(q) =\cfrac{1}{1-q+\cfrac{q(1-q)^2}{1-q^3+\cfrac{q^3(1-q^2)^2}{1-q^5+\cfrac{q^5(1-q^3)^2}{1-q^7+\ddots}}}}$$



Then prove/disprove that the two continued fractions are equivalent



1 Answer 1


So with formal manipulation:

$$\psi(1/q)= \frac{-q^{-1}}{1-q^{-1}+ \frac{q^{-1}(1-q^{-1})^2}{1-q^{-3}+\ldots}}$$

Times top and bottom by $-q$ to give:

$$\psi(1/q)= \frac{1}{1-q+ \frac{-qq^{-3}(1-q)^2}{1-q^{-3}+\frac{q^{-1}(1-q^{-2})^2}{1-q^{-5}+\ldots}}}$$

Then mutiply top and bottom of the next fraction by $-q^3$ to give:

$$\psi(1/q)= \frac{1}{1-q+ \frac{q(1-q)^2}{1-q^{3}+\frac{-q^3q^{-5}(1-q^{2})^2}{1-q^{-5}+\ldots}}}$$

Then times $-q^5$ gives

$$\psi(1/q)= \frac{1}{1-q+ \frac{q(1-q)^2}{1-q^{3}+\frac{q^3(1-q^{2})^2}{1-q^{5}+\frac{-q^5q^{-7}(1-q^3)^2}{1-q^{-7}+\ldots}}}}$$

So letting the first denominator be line $1$, then in line $n$ we assume we carry a factor of $-q^{2n-1}$ which corrects the first term. In the second term, ie the top of the new denominator which is line $n+1$, we pull out a factor of $q^{-n}$ which times the $1\over q$ becomes $q^{-2n-1}$. Then carry this factor with the minus sign, through to the next line.

I imagine however this is actually why you assumed it was true in the first place and are looking more for a proof where you use series and convergence properties of generalised continued fractions. But from the formal algebraic standpoint above it seems true.


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