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I want to find a function $f(x,y)$ which can satisfy the following equation,

$$\prod _{n=1} ^{\infty} \frac{1+x^n}{(1-x^{n/2}y^{n/2})(1-x^{n/2}y^{-n/2})} = \exp \left[ \sum _{n=1} ^\infty \frac{f(x^n,y^n)}{n(1-x^{2n})}\right]$$

  • I would like to know how this is solved.

In a certain paper where I ran into this, it is claimed that the function is,

$$f(x,y) = \sqrt{x}(y + 1/y) + x(1+y^2 + 1/y) + x^{3/2}(y^3+1/y^3) + x^2(y^4+1/y^4) + \sum _{n=5}^\infty x^{n/2}(y^n + 1/y^n - y^{n-4} - 1/y^{n-4})$$

The paper doesn't state any proof or explanation for how this was obtained but perturbatively the above can be checked to be correct!


Now I tried to do something obvious but it didn't work!

\begin{eqnarray} \prod _ {n=1} ^{\infty} \frac{ (1+x^n) }{1+x^n -x^{\frac{n}{2}} \left(y^{\frac{n}{2}} + y^{-\frac{n}{2}}\right) } = \exp \left[ \sum _ {n=1} ^{\infty} \frac{ I_{ST}(x^n,y^n) } {n (1-x^{2n}) } \right] \\ \Rightarrow \sum_{n=1}^{\infty} \left\{ \ln (1+x^n) - \ln(1-(\sqrt{xy})^n) - \ln\left(1- \left(\sqrt{\frac{x}{y}}\right)^n\right) \right\} = \sum_{n=1}^\infty \frac{I_{ST}(x^n,y^n)} {n(1-x^{2n})} \end{eqnarray}

Now we expand the logarithms and we have,

\begin{eqnarray} \sum _ {n=1} ^ {\infty} \left \{ \sum _{a=1}^{\infty} (-1)^{a+1} \frac{x^{na}}{a} + \sum_{b=1} ^{\infty} \frac{ (\sqrt{xy})^{nb} } {b} + \sum _{c=1}^{\infty} \frac{ \left(\sqrt{\frac{x}{y}}\right)^{nc} }{c} \right \} = \sum _{n=1} ^\infty \frac{I_{ST}(x^n,y^n)} {n(1-x^{2n})} \\ \Rightarrow \sum _{a=1} ^{\infty} \frac{1}{a} \left\{ \sum _{n=1} ^{\infty} \left( (-1)^{a+1}x^{na} + (xy)^{\frac{na}{2}} + \left(\frac{x}{y}\right)^{\frac{na}{2}} \right) \right\} = \sum _{n=1} ^\infty \frac{I_{ST}(x^n,y^n)} {n(1-x^{2n})} \end{eqnarray}

By matching the patterns on both sides one sees that one way this equality can hold is if, $$ \begin{eqnarray} I_{ST}(x,y) = (1-x^2) \sum _{n=1} ^{\infty} \left\{ x^n + (xy)^{\frac{n}{2}} + (\frac{x}{y})^{\frac{n}{2}} \right\} \\ \Rightarrow I_{ST} (x,y) = (1-x^2) \left(-1 + \frac{1}{1-x} -1 + \frac{1}{1-\sqrt{xy}} - 1 + \frac{1}{1-\sqrt{\frac{x}{y}} } \right) \end {eqnarray} $$ But this solution doesn't satisfy the original equation!

TMM
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Student
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1 Answers1

1

What a goofy looking function. Why do they need this anyway? Taking the log of both sides, you should get:

$$ \log(1+x^n) - \log(1 - x^{n/2}y^{n/2}) - \log(1 - x^{n/2}y^{-n/2}) $$

Then

$$ \log ( 1 + x^n) = 1 - x^n + \frac{1}{2}x^{2n} - \frac{1}{3}x^{3n} + \dots $$

and also

$$ \log (1 - x^{n/2}y^{n/2}) = 1 + (xy)^{n/2} + \frac{1}{2} (xy)^n + \dots $$

and

$$ \log (1 - x^{n/2}y^{-n/2}) = 1 + (x/y)^{n/2} + \frac{1}{2} (x/y)^n + \dots $$

I suppose if you add from $n=1 \to \infty$ you will get the correct $f(x,y)$.

cactus314
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