Question

How do you determine the third and fourth Taylor polynomials of `x^3 + 9x - 1` at x = -1?

Answer 1

`T_3(x) = -11 + 12(x+1) -3(x+1)^2`

`T_4(x) = -11 + 12(x+1) -3(x+1)^2 + (x+1)^3`

Explanation:

Firstly note that as we have a polynomial of degree `3` then its TS will be exact for `O(x^3)`.

Let use define the function, `f(x)`, by:

`f(x) = x^3+9x-1`

Let us find all the derivatives:

`f^((1))(x) = 3x^2+9`
`f^((2))(x) = 6x`
`f^((3))(x) = 6`
`f^((4))(x) = 0`, along with all higher derivatives:

Now let us find the values of the above at `a=-1`

`f^((0))(-1) = 11`
`f^((1))(-1) = 12`
`f^((2))(-1) = -6`
`f^((3))(-1) = 6`
`f^((4))(-1) = 0`, along with all higher derivatives:

The the TS about `x=a` is given by:

`f(x) = f(a) + f^((1))(x)(x-a) + (f^((2))(x))/(2!)(x-a)^2 + (f^((3))(x))/(3!)(x-a)^3 + (f^((4))(x))/(4!)(x-a)^4 + ...`

So, if we want to write the truncated TS, we can just truncate the series as required. Thus the `3^(rd)` order TS about `a=-1` is:

`T_3(x) = (-11) + (12)(x+1) + (-6)/(2!)(x+1)^2`
`" " = -11 + 12(x+1) -6/(2)(x+1)^2`
`" " = -11 + 12(x+1) -3(x+1)^2`

And similarly, truncating at the next term we have:

`T_4(x) = T_3(x) + (6)/(3!)(x+1)^3`
`" " = T_3(x) + 6/(6)(x+1)^3`
`" " = -11 + 12(x+1) -3(x+1)^2 + (x+1)^3`

And as initially, as the starting function is a polynomial of degree `3` then this truncated polynomial `T_4(x)` is in fact exact as all higher derivatives (and therefore terms) are zero.