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Using binomial thaorem (http://en.wikipedia.org/wiki/Binomial_theorem) find the general formula for the coefficients of the expantion: $ \left(\sum_{i=0}^{\infty}\frac{t^{2i}}{n^i6^ii!}\left(1-\frac{t^2}{6}+\frac{t^4}{120}\right)\right)^n $

Thank you for your help.

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    Seems like a tedious exercise, but why would you get stuck? What have you tried?2012-06-19

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The inner sum is $ \sum_{i=0}^\infty\frac{x^{2i}}{n^i6^ii!}=e^{\frac{x^2}{6n}}\tag{1} $ When raised to the $n^{\text{th}}$ power, $(1)$ is $e^{x^2/6}$. So it remains to compute $ e^{x^2/6}\left(1-\frac{t^2}{6}+\frac{t^4}{120}\right)^n =e^{x^2/6}\sum_{j=0}^n\binom{n}{j}\left(-\frac{t^2}{6}+\frac{t^4}{120}\right)^j\tag{2} $ If the desire is to $(2)$ up to $O(t^{2k})$, one only need sum the first $k$ terms; that is, $ e^{x^2/6}\left(1-\frac{t^2}{6}+\frac{t^4}{120}\right)^n=e^{x^2/6}\sum_{j=0}^{k-1}\binom{n}{j}\left(-\frac{t^2}{6}+\frac{t^4}{120}\right)^j+e^{x^2/6}O(t^{2k})\tag{3} $

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    Thank you for your eplanations.2012-06-23