Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
4  
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. –  André Nicolas Jan 24 '12 at 4:36
    
By definition, Maclaurin series of $f$ is Taylor series of $f$ at $a=0$: en.wikipedia.org/wiki/Taylor_series#Definition –  Paul Jan 24 '12 at 4:36
    
But what would those expansions look like? –  user906153 Jan 24 '12 at 4:51
    
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 $0$. –  André Nicolas Jan 24 '12 at 6:47

1 Answer 1

up vote 3 down vote accepted

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

...

share|improve this answer
    
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? –  user906153 Jan 24 '12 at 5:30
    
Yes, exactly. Normally we write $\Delta x=x-a$. For Maclaurin series, we have $a=0$, so $\Delta x=x$. –  Paul Jan 24 '12 at 5:32

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.