Question

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

Answer 1

The answer is `6-11(x+2)+6(x+2)^2-(x+2)^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(x)=x-x^3`, we get `f(-2)=-2+8=6`, `f'(x)=1-3x^2` so that `f'(-2)=1-12=-11`, `f''(x)=-6x` so that `f''(-2)=12`, and `f'''(x)=-6` so that `f'''(-2)=-6`. All higher-order derivatives are zero since `f` is a cubic.

The answer is therefore `6-11(x+2)+6(x+2)^2-(x+2)^3`.

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

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

`f(u-2)=(u-2)-(u-2)^3`

`=u-2-(u^3-6u^2+12u-8)=6-11u+6u^2-u^3`.

The answer is therefore, as we saw before,

`6-11(x+2)+6(x+2)^2-(x+2)^3`.