What is the Taylor Series generated by f(x) = x - x^3, centered around a = -2?

Jun 14, 2015

The answer is $6 - 11 \left(x + 2\right) + 6 {\left(x + 2\right)}^{2} - {\left(x + 2\right)}^{3}$

Explanation:

There are two methods to get the answer.

1) Use the expression f(-2)+f'(-2)(x+2)+(f''(-2))/(2!)(x+2)^2+(f'''(-2))/(3!)(x+2)^3+\cdots

Since $f \left(x\right) = x - {x}^{3}$, we get $f \left(- 2\right) = - 2 + 8 = 6$, $f ' \left(x\right) = 1 - 3 {x}^{2}$ so that $f ' \left(- 2\right) = 1 - 12 = - 11$, $f ' ' \left(x\right) = - 6 x$ so that $f ' ' \left(- 2\right) = 12$, and $f ' ' ' \left(x\right) = - 6$ so that $f ' ' ' \left(- 2\right) = - 6$. All higher-order derivatives are zero since $f$ is a cubic.

The answer is therefore $6 - 11 \left(x + 2\right) + 6 {\left(x + 2\right)}^{2} - {\left(x + 2\right)}^{3}$.

2) Let $u = x + 2$ so that $x = u - 2$ and expand $f \left(x\right) = f \left(u - 2\right)$ before replacing $u$ with $x + 2$ at the end.

Here are the details, which follow from the binomial theorem (Pascal's triangle ).

$f \left(u - 2\right) = \left(u - 2\right) - {\left(u - 2\right)}^{3}$

$= u - 2 - \left({u}^{3} - 6 {u}^{2} + 12 u - 8\right) = 6 - 11 u + 6 {u}^{2} - {u}^{3}$.

The answer is therefore, as we saw before,

$6 - 11 \left(x + 2\right) + 6 {\left(x + 2\right)}^{2} - {\left(x + 2\right)}^{3}$.