Question

How do you find the third degree Taylor polynomial for `f(x)= ln x`, centered at a=2?

Answer 1

`ln(2)+1/2(x-2)-1/8(x-2)^2+1/24(x-2)^3`.

Explanation:

The general form of a Taylor expansion centered at `a` of an analytical function `f` is `f(x)=sum_{n=0}^oof^((n))(a)/(n!)(x-a)^n`. Here `f^((n))` is the nth derivative of `f`.

The third degree Taylor polynomial is a polynomial consisting of the first four (`n` ranging from `0` to `3`) terms of the full Taylor expansion.

Therefore this polynomial is `f(a)+f'(a)(x-a)+(f''(a))/2(x-a)^2+(f'''(a))/6(x-a)^3`.

`f(x)=ln(x)`, therefore `f'(x)=1/x`, `f''(x)=-1/x^2`, `f'''(x)=2/x^3`. So the third degree Taylor polynomial is:
`ln(a)+1/a(x-a)-1/(2a^2)(x-a)^2+1/(3a^3)(x-a)^3`.

Now we have `a=2`, so we have the polynomial:
`ln(2)+1/2(x-2)-1/8(x-2)^2+1/24(x-2)^3`.