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

May 22, 2017

${T}_{3} \left(x\right) = - 11 + 12 \left(x + 1\right) - 3 {\left(x + 1\right)}^{2}$

${T}_{4} \left(x\right) = - 11 + 12 \left(x + 1\right) - 3 {\left(x + 1\right)}^{2} + {\left(x + 1\right)}^{3}$

Explanation:

Firstly note that as we have a polynomial of degree $3$ then its TS will be exact for $O \left({x}^{3}\right)$.

Let use define the function, $f \left(x\right)$, by:

$f \left(x\right) = {x}^{3} + 9 x - 1$

Let us find all the derivatives:

${f}^{\left(1\right)} \left(x\right) = 3 {x}^{2} + 9$
${f}^{\left(2\right)} \left(x\right) = 6 x$
${f}^{\left(3\right)} \left(x\right) = 6$
${f}^{\left(4\right)} \left(x\right) = 0$, along with all higher derivatives:

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

${f}^{\left(0\right)} \left(- 1\right) = 11$
${f}^{\left(1\right)} \left(- 1\right) = 12$
${f}^{\left(2\right)} \left(- 1\right) = - 6$
${f}^{\left(3\right)} \left(- 1\right) = 6$
${f}^{\left(4\right)} \left(- 1\right) = 0$, 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 ${3}^{r d}$ order TS about $a = - 1$ is:

 T_3(x) = (-11) + (12)(x+1) + (-6)/(2!)(x+1)^2
$\text{ } = - 11 + 12 \left(x + 1\right) - \frac{6}{2} {\left(x + 1\right)}^{2}$
$\text{ } = - 11 + 12 \left(x + 1\right) - 3 {\left(x + 1\right)}^{2}$

And similarly, truncating at the next term we have:

 T_4(x) = T_3(x) + (6)/(3!)(x+1)^3
$\text{ } = {T}_{3} \left(x\right) + \frac{6}{6} {\left(x + 1\right)}^{3}$
$\text{ } = - 11 + 12 \left(x + 1\right) - 3 {\left(x + 1\right)}^{2} + {\left(x + 1\right)}^{3}$

And as initially, as the starting function is a polynomial of degree $3$ then this truncated polynomial ${T}_{4} \left(x\right)$ is in fact exact as all higher derivatives (and therefore terms) are zero.