Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

2 Answers 2

up vote 1 down vote accepted

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

share|improve this answer
    
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 $x^2+4$.

share|improve this answer
    
Ok thank you that answers my question –  Chance Mar 15 '13 at 2:28

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.