nth derivative taylor series

**polymerase** · April 21st 2008, 08:28 AM

Let $f^{(n)}(a)$ denote the n-th derivative of $f$ at $a$ . if $f(x)=xe^{2x}$ , then $f^{(64)}(0)=$

I can get the answer using the old fashion way, but when I lookeed at the solution using power series, I don't understand one of the steps. heres the soultion my prof has:

$e^x=\sum_{n=0}^\infty \frac{x^n}{n!}$
$f(x)=xe^{2x}=\sum_{n=0}^\infty \frac{2^nx^{n+1}}{n!}=\sum_{m=1}^\infty \frac{2^{m-1}x^{m}}{(m-1)!}= \boxed{\sum_{m=1}^\infty \frac{f^{(m)}(0)}{m!}x^m}$

hence, $f^{(m)}(0)=\frac{2^{m-1}m!}{(m-1)!}$

I understand every step except the step to the result in the box. How are they equal?

**Moo** · April 21st 2008, 08:30 AM

Hello,

The thing in the box is the result of the Taylor series : Taylor series - Wikipedia, the free encyclopedia
It is said that the sum begins at m=1, but here it doesn't matter as f(0)=0

**polymerase** · April 21st 2008, 08:42 AM

Originally Posted by Moo

Hello,

The thing in the box is the result of the Taylor series : Taylor series - Wikipedia, the free encyclopedia
It is said that the sum begins at m=1, but here it doesn't matter as f(0)=0

nvm i figured it out...
What I was really trying to get at was how does $\sum_{m=1}^\infty \frac{f^{(m)}(0)}{m!}x^m=\sum_{n=0}^\infty \frac{2^nx^{n+1}}{n!}$

**Moo** · April 21st 2008, 08:44 AM

nvm ?

The left hand part comes from the Taylor series
The right hand part comes from the power series :-)

$e^x=\sum_{n \geq 0} \frac{x^n}{n!}$

Hence $e^{2x}=\sum_{n \geq 0} \frac{(2x)^n}{n!}=\sum_{n \geq 0} \frac{2^n \ x^n}{n!}$

-> $xe^{2x}=\sum_{n \geq 0} \frac{2^n \ x^{n+1}}{n!}$

**polymerase** · April 21st 2008, 09:20 AM

Originally Posted by Moo

nvm ?

never mind

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