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

Oct 9, 2015

See explanation...

#### Explanation:

Here's one way of thinking about it.

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

(i.e. the Taylor series for ${e}^{z}$, but pretend we don't know that yet).

Then we find:

$e \left(0\right) = 1$

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 ${e}^{0} = 1$, ${e}^{1} = e$ and $\frac{d}{\mathrm{dz}} {e}^{z} = {e}^{z}$

Essentially this $e \left(z\right)$ is doing a very good impression of ${e}^{z}$

More formally:

Let $f \left(z\right) = {e}^{z}$

Then using $\frac{d}{\mathrm{dz}} {e}^{z} = {e}^{z}$ and ${e}^{0} = 1$, the Taylor expansion at $0$ is:

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!)