Expand function into a Maclaurin's series The function is given with:
$f(x)=\dfrac{x^{2012}}{(1-x^3)^2}$
I have no idea what to do here and honestly, I don't really understand what a Maclaurin's series is. I know the definition but I don't understand the concept enough to be able to solve problems like this.
Any help would be appreciated.
 A: You are undoubtedly familiar with the geometric series expansion 
$$\frac{1}{1-t}=1+t+t^2+t^3+t^4+\cdots,$$
valid for $|t|\lt 1$. Differentiate with repect to $t$. We get
$$\frac{1}{(1-t)^2}=1+2t+3t^2+4t^3+\cdots.$$
Put $t=x^3$. We get the expansion of $\dfrac{1}{(1-x^3)^2}$.  
Then multiply term by term by $x^{2012}$.
Remark: In principle, we could also find the series expansion by differentiating our function $f(x)$ repeatedly, and using the fact that the Maclaurin series is 
$$\sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!}x^n,$$
where $f^{(n)}(x)$ is the $n$-th derivative of $f(x)$. That would in this case be quite a bit harder than the approach we took above.
Once we know the Maclaurin series for a few familiar and important functions, most of the time we use manipulations of known series instead of going back to fundamentals.
A: First try to find the Taylor expansion for $\frac{1}{1-x^3}$ (use substitution and the Taylor expansion for $\frac{1}{1-x}$). Now differentiate term-by-term to get an expression  for $\frac{3x^2}{(1-x^3)^2}$. Finally, multiply through by some power of $x$ (2010 I think) and divide by 3.  
A: We know that the Taylor series, at some point $a$, is given by
$$f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x - a)^2 + \frac{f'''(a)}{3!}(x - a)^3 + \cdots,$$
then, the MacLaurin Series is simply the Taylor series at $a = 0$, i.e
$$f(0) + f'(0)(x-0) + \frac{f''(0)}{2!}(x - 0)^2 + \frac{f'''(0)}{3!}(x - 0)^3 + \cdots$$
$$= f(0) + f'(0)(x) + \frac{f''(0)}{2!}(x)^2 + \frac{f'''(0)}{3!}(x)^3 + \cdots.$$
So now, to answer your question, you need to simply "fill in the gaps" to the above series and see if you can put it in some kind of summation form.
