Question

If `f(x) = ln(1+2x)`, where a = 2 and n = 3 how do you approximate f by a Taylor polynomial with degree n at the number a?

Answer 1

`= ln 5 + 2/5 (x-2) - 2/25 (x-2)^2 + 8/375 (x-2)^3`

Explanation:

The value of `f(a)` is straightforward ... it's `ln (1 + 2a)`.

So maybe you want to approximate the behaviour of the function around `x = a`?

If that is the case, you can apply the Taylor Series formulation to third degree:

`f(x-a) = f(a) + ((x-a) f'(a))/(1!) + ((x-a)^2 f''(a))/(2!) + ((x-a)^3 f'''(a))/(3!)`

And so we need these:

`f = ln (1 + 2x)`

`f' = 2/(1 + 2x)`

`f'' = - 4/(1 + 2x)^2`

`f''' = 16/(1 + 2x)^3`

We can then say that:

`f(x-2) = ln 5 + ((x-2) (2/5))/(1!) + ((x-2)^2 (- 4/25))/(2!) + ((x-2)^3 (16/125))/(3!)`

`= ln 5 + 2/5 (x-2) - 2/25 (x-2)^2 + 8/375 (x-2)^3`