# Thread: taylor series at b=1

1. ## 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

2. Just write it as $\displaystyle x(e^x) = x(1 + x + ...)$

EDIT: I used the Maclaurin series, but you get the idea.

3. ## how

how would this look in the sigma form? I check the answer and it had some 'e' there. how about b=1

4. If $\displaystyle f(x)=xe^x$, and $\displaystyle g(x)=e^x$:

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

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

Or in Sigma form:

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

5. ## ?

where did u get 2+t?

6. Originally Posted by khuezy
where did u get 2+t?
Sorry, I read your original post as $\displaystyle b=2$ for some reason.

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

this isn't in the taylor series formation right?

8. Originally Posted by khuezy
$\displaystyle \sum_{n=0}^{\infty}e(t+1) \frac{t^{n}}{(n)!}$

this isn't in the taylor series formation right?
It's the same as $\displaystyle \sum_{n=0}^{\infty}xe \frac{(x-1)^{n}}{(n)!}$ since $\displaystyle x=1+t \implies t=x-1$

9. ## thanks

if b=2 then
$\displaystyle \sum_{n=0}^{\infty}xe^{2} \frac{(x-2)^{n}}{(n)!}$
?

10. Originally Posted by khuezy
if b=2 then
$\displaystyle \sum_{n=0}^{\infty}xe^{2} \frac{(x-2)^{n}}{(n)!}$
?
Correct.

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

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