1. ## Exponential Taylor Series

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.

2. 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.

That is a weird way to put things... : $\displaystyle f(z)=e^z\Longrightarrow f^{(n)}(z)=e^z\Longrightarrow f^{(n)}(0)=1\,\,\,\forall\,n\in\mathbb{N}$ , so $\displaystyle e^z=1+\frac{z}{1!}+\frac{z^2}{2!}+\frac{z^3}{3!}+\ ldots =\sum^\infty_{k=0}\frac{z^k}{k!}$ $\displaystyle \Longrightarrow e^z=e\cdot e^{z-1}=e\sum^\infty_{k=0}\frac{(z-1)^k}{k!}$ ...weird! Why would anyone want to represent $\displaystyle e^z\,\,\,as\,\,\,e\cdot e^{z-1}$ without given any further reason? Beats me.

Tonio