# How is the taylor series for #e^z# derived?

##### 1 Answer

Oct 9, 2015

See explanation...

#### Explanation:

Here's one way of thinking about it.

Let

(i.e. the Taylor series for

Then we find:

#e(0) = 1#

#e(1) = sum_(n=0)^oo 1/(n!) = e#

#d/(dz) e(z) = sum_(n=0)^oo n*(z^(n-1))/(n!) =sum_(n=1)^oo (z^(n-1)/((n-1)!))#

#=sum_(n=0)^oo z^n/(n!) = e(z)#

Compare with

Essentially this

More formally:

Let

Then using

#sum_(n=0)^oo (f^((n))(0) z^n/(n!))=sum_(n=0)^oo e^0 z^n/(n!)=sum_(n=0)^oo z^n/(n!)#