Originally Posted by

**davesface** Obtain the Taylor series $\displaystyle e^z=e \sum_{n=0}^{\infty}\frac{(z-1)^n}{n!} (\left |z-1 \right |< \infty)$ for the function $\displaystyle f(z)=e^z$ by using $\displaystyle f^{(n)}(1) (n=0,1,2,...)$.

So it would seem to me that $\displaystyle f^{(n)}(1)=e, \forall n$, but clearly that can't be the case. Clearly that's the correct Taylor series, since it's just e times the Taylor series for $\displaystyle e^{z-1}$, but I'm not seeing at all how to get from point A to point B here.