What is the taylor series of ##xe^x##?

aaaah now here’s a fun question. I kinda wanna thank you for asking it!
Here’s my answer, if you just want this in a rush:
If ##f(x) = xe^x##, then you can actually write an exact sum:
##f(x)=sum_[n=0]^infty (x-a)^n (a+n)/[n!]e^a##
Proof
step (1) ‘##[d^n]/[dx^n] xe^x = (x+n)e^x##’
We prove this step by induction.
Firstly, it is trivially true for n=0:
##[d^0]/[dx^0] f(x) = f(x) = xe^x = (x+0)e^x##
Secondly, the induction step:
Suppose ##[d^nf]/[dx^n] = (x+n)e^x##, then
##[d^[n+1]f]/[dx^[n+1]]=d/dx [d^nf]/[dx^n]=d/dx (x+n)e^x##
##=d/dx (xe^x +n e^x) = n e^x + d/dx xe^x = n e^x + xe^x + e^x ##
##= (x+(n+1))e^x##
Where on the last step of the second line we applied the product derivative rule to ##xe^x##
Thus by induction, (1) is proven
step (2): apply to taylor expansion
The general taylor expansion is
##f(x) = sum_[n=0]^[infty] [f^[(n)] (a)] [(x-a)^n]/[n!]##
substituting in (1) we get
##f(x) = sum_[n=0]^[infty] [(a+n)e^a] [(x-a)^n]/ [n!]##
which trivially rearranges to the form I gave ##square##
As a specific instance, take the maclaurin series (a=0):
##f(x) = sum_[n=0]^[infty] [(0+n)e^0] [(x-0)^n]/[n!] =sum_[n=0]^[infty]n/[n!] x^n##
##xe^x=sum_[n=1]^[infty][x^n]/[(n-1)!]##
Nicely, this is just ##x## times the maclaurin expansion for ##e^x##, confirming that this is a good expansion for ##xe^x##. (The n=0 term was removed because it was equal to zero, and one cannot divide by zero to neaten the factorial).

Posted in Uncategorized

Leave a Reply

Your email address will not be published. Required fields are marked *

You may use these HTML tags and attributes:

<a href="" title=""> <abbr title=""> <acronym title=""> <b> <blockquote cite=""> <cite> <code> <del datetime=""> <em> <i> <q cite=""> <s> <strike> <strong>