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.

  • 4
    $\begingroup$ Most mathematicians use the term "Taylor series" even in the special case $a=0$. In that special case, the series could also be called the MacLaurin series, but usually poor MacLaurin gets forgotten. $\endgroup$ – André Nicolas Jan 24 '12 at 4:36
  • $\begingroup$ By definition, Maclaurin series of $f$ is Taylor series of $f$ at $a=0$: en.wikipedia.org/wiki/Taylor_series#Definition $\endgroup$ – Paul Jan 24 '12 at 4:36
  • $\begingroup$ But what would those expansions look like? $\endgroup$ – user906153 Jan 24 '12 at 4:51
  • $\begingroup$ The full Taylor series about $a=0$ of the polynomial $P(x)$ is just $P(x)$, with constant term first, then the "$x$" term, and so on. The point of the exercise, though, is for you to discover this, so you are expected to compute the derivatives at $0$. $\endgroup$ – André Nicolas Jan 24 '12 at 6:47

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 $f(0)+\displaystyle\frac{f'(0)}{1!}x+\frac{f''(0)}{2!}x^2=-3+2x-2$


  • $\begingroup$ Thank you so much! But, one more thing. How would I use this estimation for Δx=0.5? Would I be inserting .5 for the x term in each series? $\endgroup$ – user906153 Jan 24 '12 at 5:30
  • $\begingroup$ Yes, exactly. Normally we write $\Delta x=x-a$. For Maclaurin series, we have $a=0$, so $\Delta x=x$. $\endgroup$ – Paul Jan 24 '12 at 5:32

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