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Expand $[2e^{rx^2}-e^{mx^2-\sigma x} -e^{mx^2+\sigma x}]/[{e^{mx^2+\sigma x}-e^{mx^2-\sigma x}}]$ around $x=0$ and simplify (second order expansion)

I'm a bit stumped here. I get something like $2+2rx^2+o(x^2)-1-mx^2+\sigma x - o(x^2) - mx^2 - \sigma x - o(x^2)$ for the numerator -- am I on the right track?

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    it should be a second order expansion.2012-05-29

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HINT: Try simplifying the expression first. The last two terms of the numerator can be simplified as follows: $-\mathrm{e}^{m x^2 - \sigma x} - \mathrm{e}^{m x^2 + \sigma x} = -\mathrm{e}^{m x^2} \left( \mathrm{e}^{\sigma x} + \mathrm{e}^{-\sigma x} \right) = - 2 \mathrm{e}^{m x^2} \cosh(\sigma x)$ Similarly the denominator will becomes $2 \mathrm{e}^{m x^2} \sinh(\sigma x)$. Now: $ \frac{\mathrm{e}^{(r-m) x^2}-\cosh(\sigma x)}{\sinh(\sigma x)} $

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Expanding about $x = 0$, one has \begin{align} \frac{2e^{r x^{2}} - e^{mx^{2} - \sigma x} - e^{m x^{2} + \sigma x} }{e^{m x^{2} + \sigma x} - e^{m x^{2} - \sigma x}} = \left( \frac{r - m}{s} - \frac{s}{2} \right) x + O(x^{3}). \end{align} Let $f(x) = \sum_{i} a_{i} x^{i}$ denote the Taylor series of the above function. Clear the denominator by bringing it to the right side (as a multiplier of $f(x)$). Expand all exponentials about $x = 0$. Now match coefficients to your desired degree of accuracy.