Solving a formal power series equation

Solution 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 $f(x,y)$.