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S. Carnahan
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$S(x) = -\frac12 \theta(1/2, 2x) - 1$ I don't know what you mean by "polynomial series" since your function doesn't seem to have much to do with polynomials (perhaps you could elaborate?).

$S(x) = -\frac12 \theta(\frac12, \frac{\log x}{\pi i }) - 1$, where $\theta$ is Jacobi's theta function. I'm I don't know of any nice algebraic methods to take inverses of modular forms (even if you restrict to real $x$, i.e., $\tau$ purely imaginary), so I'm not sure if this is the sort of answer you want.

$S(x) = -\frac12 \theta(1/2, 2x) - 1$, where $\theta$ is Jacobi's theta function. I'm not sure if this is the sort of answer you want.

I don't know what you mean by "polynomial series" since your function doesn't seem to have much to do with polynomials (perhaps you could elaborate?).

$S(x) = -\frac12 \theta(\frac12, \frac{\log x}{\pi i }) - 1$, where $\theta$ is Jacobi's theta function. I don't know of any nice algebraic methods to take inverses of modular forms (even if you restrict to real $x$, i.e., $\tau$ purely imaginary), so I'm not sure if this is the sort of answer you want.

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S. Carnahan
  • 45.7k
  • 6
  • 114
  • 220

$S(x) = -\frac12 \theta(1/2, 2x) - 1$, where $\theta$ is Jacobi's theta function. I'm not sure if this is the sort of answer you want.