Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

First off all, I am sorry if my english is not perfect. I need help again for this exercise:

Find Maclaurin series expansion for $f(x)=\frac{x^2}{1-x}$.

That's what I did:

$f(x)=\frac{x^2}{1-x}=\frac{x^2+1-1}{1-x}=\frac{(x+1)(x-1)}{1-x}+\frac{1}{1-x}= -x-1+\frac{1}{1-x}$.

I know that $\sum\limits_{n=0}^{\infty}x^n=\frac{1}{1-x}$ for $|x|<1 $. But What to do with $-x-1$?

$(-x-1)'=-1$ and $ (-x-1)''=0$.

If I chose to find the nth derivative for $f(x)$:




$f^{(n)}(x)=(-1)^{2n}(1-x)^{-(n+1)}n! \Rightarrow f^{(n)}(0)=n! \Rightarrow f(x)=\sum\limits_{n=0}^{\infty} x^n$

It's this correct? I am a little confused and I hope someone could help me again..

share|cite|improve this question
I am a little disappointed: Very recently, you asked a question of a similar nature. I answered, stressing the importance of recycling known expansions, in particular the expansion of $1/(1-z)$. The answer stressed the usefulness of avoiding derivative calculations if possible. This answer was accepted, and yet a day later you are attacking a very similar, indeed easier, problem by repeated differentiation. – André Nicolas Aug 17 '11 at 14:28
If you didn't understand @André's (wonderful and applicable) answer to your previous question, you should not have accepted so quickly. Accept answers only when you have digested them fully and are fully satisfied. – J. M. Aug 18 '11 at 17:32
up vote 5 down vote accepted

You say that you know $\sum\limits_{n=0}^{\infty}x^n=\frac{1}{1-x}$. What happens if you multiply both sides by $x^2$?

share|cite|improve this answer

Think about it for a minute - $-x-1$ is already an infinite series $\sum_{n=1}^\infty a_nx^n$; in this case, $a_0 = -1$ and $a_1 = -1$ and $a_n=0$ for $n>1$.

(The fact that the series is McLaurin lies in that we are seeking a series of the form $\sum_{n=1}^\infty a_nx^n$; for other Taylor expansions we want a series of the form $\sum_{n=1}^\infty a_n(x-a)^n$ for some $a$).

share|cite|improve this answer

Your Answer


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.