First four Taylor Series Expansions

Question

I'm supposed to write the first four taylor series expansions of $f(x=0)$ using: one term, two terms, three terms, four terms

This is the function: $$f(x) = x^3 - 2x^2 + 2x - 3$$

Should I be using MacLaurin series since I'm supposed to be using $f(x=0)$?

Then, I'm supposed to give an estimation for the 4 Taylor Series expansion for $Δx=0.5$

In this case, should I be working with Taylor Series, not MacLaurin this time? And the $Δx=0.5$ would used in the $(x-c)$ parts.

Answer 1

In general, the Maclaurin series of $f$ is: (which is the Taylor series of $f$ at $a=0$) $$f(x)=\sum_{n=0}^\infty\frac{f^{(n)}(0)}{n!}x^n.$$ Therefore, the first four terms of the Maclaurin series of $f$ are: $$\tag{1}\sum_{n=0}^3\frac{f^{(n)}(0)}{n!}x^n=f(0)+\frac{f'(0)}{1!}x+\frac{f''(0)}{2!}x^2+\frac{f'''(0)}{3!}x^3.$$

In your case, $f(x)=x^3 - 2x^2 + 2x - 3$. Then you can calculate $f(0)$, $f'(0)$, $f''(0)$, and $f'''(0)$ and put them back into $(1)$ to get the answer. For example, $$f(0)=0^3 - 2\cdot0^2 + 2\cdot 0- 3=-3, f'(0)=(3x^2-4x+2)\big|_{x=0}=3\cdot 0^2-4\cdot 0+2=2.$$ I will let you do $f''(0)$ and $f'''(0)$.

The Maclaurian series of $f$ (Taylor series of $f$ at $0$) with one term is $f(0)=-3$.

The Maclaurian series of $f$ with two terms is $f(0)+\displaystyle\frac{f'(0)}{1!}x=-3+2x$.

The Maclaurian series of $f$ with three terms is $f(0)+\displaystyle\frac{f'(0)}{1!}x+\frac{f''(0)}{2!}x^2=-3+2x-2$

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