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