Question

Question #2eeda

Answer 1

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

Explanation:

We have that:

`d/(dx) e^x = e^x`

and therefore for higher orders:

`d^n/(dx^n) e^x = e^x`

and for `x=1`

`[d^n/(dx^n) e^x]_(x=1) = e` for `n=0,1,2,...`

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!)`