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

Mar 1, 2017

$f \left(x\right) = {x}^{2} + 3 x + 8$

#### 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 $f \left(x\right)$ are the terms we already have;

ie

$f \left(x\right) = {x}^{2} + 3 x + 8$

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:

$\setminus \setminus \setminus \setminus \setminus f \left(x\right) = {x}^{2} + 3 x + 8$
$\therefore f \left(0\right) = 8$

Differentiate wrt to get the first derivative:

$\setminus \setminus \setminus \setminus \setminus f ' \left(x\right) = 2 x + 3$
$\therefore f ' \left(x\right) = 3$

Differentiate again wrt to get the second derivative:

$\setminus \setminus \setminus \setminus \setminus f ' ' \left(x\right) = 2$
$\therefore f ' ' \left(0\right) = 2$

Differentiate again wrt to get the third derivative:

$\setminus \setminus \setminus \setminus \setminus {f}^{\left(3\right)} \left(x\right) = 0$

All further derivatives are zero

So we can now form the Maclaurin series:

$f \left(x\right) = 8 + 3 x + 2 {x}^{2} / 2 + 0 {x}^{3} + 0 {x}^{4} + \ldots$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = 8 + 3 x + {x}^{2}$

Which is exactly the same polynomial that we started with.