# Question 2eeda

##### 1 Answer
Jan 24, 2017

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

#### Explanation:

We have that:

$\frac{d}{\mathrm{dx}} {e}^{x} = {e}^{x}$

and therefore for higher orders:

${d}^{n} / \left({\mathrm{dx}}^{n}\right) {e}^{x} = {e}^{x}$

and for $x = 1$

${\left[{d}^{n} / \left({\mathrm{dx}}^{n}\right) {e}^{x}\right]}_{x = 1} = e$ for $n = 0 , 1 , 2 , \ldots$

The expression for the Taylor series is then:

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

We can note that in general we have that for every point ${x}_{0}$:

${e}^{x} = {e}^{x - {x}_{0} + {x}_{0}} = {e}^{x - {x}_{0}} {e}^{{x}_{0}}$

and so if we start from the MacLaurin series:

e^x = sum_(n=0)^oo x^n/(n!)

the Taylor series around a point ${x}_{0}$ always have the form:

e^x = sum_(n=0)^oo x^n/(n!) = e^(x_0) sum_(n=0)^oo (x-x_0)^n/(n!)#