Question

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

Answer 1

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(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 `e^0 = 1`, `e^1 = e` and `d/(dz) e^z = e^z`

Essentially this `e(z)` is doing a very good impression of `e^z`

More formally:

Let `f(z) = e^z`

Then using `d/(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!)`