Question

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

Answer 1

`f(x) = e^2( 1 + (x-2) + (x-2)^2/2 + (x-2)^3/6 +...)`

Explanation:

Did you mean find the Taylor series of `f(x)` 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(x) = e^2( 1 + (x-2) + (x-2)^2/2 + (x-2)^3/6 +...)`

as the derivative of `e^x` is simply `e^x`.