Taylor Series, Complex

**eskimo343** · November 6th 2009, 12:03 AM

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

I do not see how to do this. I was able to prove that by writing $e^z=e^{z-1}e$ but I don't see how to do it that way. Thanks in advance for help.

**tonio** · November 6th 2009, 02:17 AM

Originally Posted by eskimo343

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

I do not see how to do this. I was able to prove that by writing $e^z=e^{z-1}e$ but I don't see how to do it that way. Thanks in advance for help.

Well, since $f^{(n)}(z)=f(z)\,\,\forall\,z\in \mathbb{C}$ , we get that $f^{(n)}(1)=e\,\,\forall n\in \mathbb{N}$ , so...
Tonio

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