# How do you find the Taylor's formula for f(x)=e^x for x-2?

Sep 2, 2017

$f \left(x\right) = {e}^{2} \left(1 + \left(x - 2\right) + {\left(x - 2\right)}^{2} / 2 + {\left(x - 2\right)}^{3} / 6 + \ldots\right)$

#### Explanation:

Did you mean find the Taylor series of $f \left(x\right)$ about $x = 2$ ?

The Taylor expansion of a function about a point a is defined as

f(x) = f(a) + f'(a) (x-a) + (f''(a))/(2!) (x-a)^2 + (f'''(a))/(3!) (x-a)^3 + ... = sum_(n=0)^(oo) (f^((n))(a))/(n!) (x - a)^n

Taking $a = 2$ we have that

$f \left(x\right) = {e}^{2} \left(1 + \left(x - 2\right) + {\left(x - 2\right)}^{2} / 2 + {\left(x - 2\right)}^{3} / 6 + \ldots\right)$

as the derivative of ${e}^{x}$ is simply ${e}^{x}$.