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Find the Maclaurin series for $f(x)=(x^2+4)e^{2x}$ and use it to calculate the 1000th derivative of $f(x)$ at $x=0$. Is it possible to just find the Maclaurin series for $e^{2x}$ and then multiply it by $(x^2+4)$? I've tried to take multiple derivatives and find a pattern in order to express it as a sum, but I can't find the pattern for part of the derivative.

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up vote 2 down vote accepted

$$(x^2+4)e^{2x}=(x^2+4)\sum_{n\ge 0}\frac{(2x)^n}{n!} = \sum_{n\ge 2}x^n\cdot \frac{2^{n-2}}{(n-2)!}+\sum_{n\ge 0}x^n\cdot\frac{2^n\cdot 4}{n!}=\\ = \sum_{n\ge 0}x^n\cdot \left(\frac{n(n-1)2^{n-2}+2^{n+2}}{n!}\right)\,.$$

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Thanks Berci, that is super helpful. – Chance Mar 15 '13 at 2:29
Why can't you keep the limit of the sum 0 and then just combine the two sums that way. – Chance Mar 15 '13 at 2:42
Because that would give you a sum of the form $$ \sum_n a_n x^n + b_n x^{n+2} $$ and not a MacLaurin series. – Hurkyl Mar 15 '13 at 3:05

Hint: $$e^{2x}=\sum_{n=0}^{\infty}\frac{(2x)^n}{n!}$$ Multiply it by $x^2+4$.

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Ok thank you that answers my question – Chance Mar 15 '13 at 2:28

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