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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 $f(x)$:

$f'(x)=-1+(1-x)^{-2}$

$f''(x)=0+(-2)(1-x)^{-3}(-1)=(-1)^22!(1-x)^{-3}$

$...................................................$

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

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

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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
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2 Answers 2

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 $x^2$?

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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 $a$).

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