# taylor series at b=1

• Oct 20th 2008, 12:21 PM
khuezy
taylor series at b=1
hi,
how would i find the taylor series for xe^x at b=1
i know the known taylor series for e^x but what do I do if there's an x in front and b=1

thanks
• Oct 20th 2008, 12:44 PM
Spec
Just write it as $x(e^x) = x(1 + x + ...)$

EDIT: I used the Maclaurin series, but you get the idea.
• Oct 20th 2008, 12:48 PM
khuezy
how
how would this look in the sigma form? I check the answer and it had some 'e' there. how about b=1
• Oct 20th 2008, 12:55 PM
Spec
If $f(x)=xe^x$, and $g(x)=e^x$:

$g(1+t)=f(1) + f'(1)t + \frac{f''(1)t^2}{2!} + O(t^3) = e^1 + {e^1}t + {e^1}t^2/2 + O(t^3)$

Since $x = 1+t$, we get $e(1+t)(1+ t + t^2/2 + O(t^3))$

Or in Sigma form:

$\sum_{n=0}^{\infty}e(t+1) \frac{t^{n}}{(n)!}$
• Oct 20th 2008, 01:09 PM
khuezy
?
where did u get 2+t?
• Oct 20th 2008, 01:12 PM
Spec
Quote:

Originally Posted by khuezy
where did u get 2+t?

Sorry, I read your original post as $b=2$ for some reason.
• Oct 20th 2008, 01:19 PM
khuezy
$

\sum_{n=0}^{\infty}e(t+1) \frac{t^{n}}{(n)!}
$

this isn't in the taylor series formation right?
• Oct 20th 2008, 01:26 PM
Spec
Quote:

Originally Posted by khuezy
$

\sum_{n=0}^{\infty}e(t+1) \frac{t^{n}}{(n)!}
$

this isn't in the taylor series formation right?

It's the same as $

\sum_{n=0}^{\infty}xe \frac{(x-1)^{n}}{(n)!}
$
since $x=1+t \implies t=x-1$
• Oct 20th 2008, 01:35 PM
khuezy
thanks
if b=2 then
$

\sum_{n=0}^{\infty}xe^{2} \frac{(x-2)^{n}}{(n)!}
$

?
• Oct 20th 2008, 01:38 PM
Spec
Quote:

Originally Posted by khuezy
if b=2 then
$

\sum_{n=0}^{\infty}xe^{2} \frac{(x-2)^{n}}{(n)!}
$

?

Correct.
• Oct 20th 2008, 01:44 PM
khuezy
$

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

that's the answer key to the problem. How come they look so different?