Question

Given `f(x)= x^2 + 3x + 8` how do you determine the first three Taylor polynomials at x=0?

Answer 1

`f(x)= x^2 + 3x + 8`

Explanation:

A Taylor Series about `x=0` is known as a Maclaurin Series. Any series expansion is unique to the function.

They both use a polynomials to estimate the function. The more terms you add, the closer the polynomial becomes to the function

A polynomial is therefore its own series

Hence the first three terms (and the only terms) of the series for `f(x)` are the terms we already have;

ie

`f(x)= x^2 + 3x + 8`

Proof

If you are not convinced by the above explanation we can see by direct proof using the definition of the Maclaurin series (The TS about `x=0`) as follows:

By definition:

`f(x) = f(0) + f'(0)x + (f''(0))/(2!)x^2 + (f^((3))(0))/(3!)x^3 + ... + (f^((n))(0))/(n!)x^n + ...`

We have:

`\ \ \ \ \ f(x) = x^2 + 3x + 8`
`:. f(0) = 8`

Differentiate wrt to get the first derivative:

`\ \ \ \ \ f'(x) = 2x+3`
`:. f'(x) = 3`

Differentiate again wrt to get the second derivative:

`\ \ \ \ \ f''(x) = 2`
`:. f''(0) = 2`

Differentiate again wrt to get the third derivative:

`\ \ \ \ \ f^((3))(x) = 0`

All further derivatives are zero

So we can now form the Maclaurin series:

`f(x) = 8 + 3x + 2x^2/2 + 0 x^3 + 0x^4 +...`
`\ \ \ \ \ \ \ = 8 + 3x + x^2`

Which is exactly the same polynomial that we started with.