Taylor Expansion

**Zocken** · July 27th 2009, 11:13 PM

Write the taylor expansion of $x/(1-x)^2$

**Random Variable** · July 27th 2009, 11:44 PM

I'm assuming you mean the Taylor series about x=0.

We know from the sum of a geometric series that $1+x+x^{2}+x^{3}+ ... = \frac {1}{1-x}$

then $\frac{d}{dx} (1+x+x^{2}+x^{3}+ ...) = 1+2x+3x^{2}+4x^{3} +... = \frac{d}{dx} \frac{1}{1-x} = \frac{1}{(1-x)^{2}}$

so $\frac{x}{(1-x)^{2}} = x (1+2x+3x^{2}+4x^{3}+...) = (x+2x^{2}+3x^{3}+4x^{4} + ... ) = \sum_{n=1}^{\infty}nx^{n}$

**chisigma** · July 27th 2009, 11:56 PM

The binomial series...

$(a+x)^{n} = a^{n} + \binom{n}{1} a^{n-1}x + \binom{n}{2} a^{n-2}x^{2} + \binom{n}{3} a^{n-3} x^{3} + \dots$ (1)

... in the case $a=1$ , $n=-2$ and changing $x$ into $-x$ becomes...

$(1-x)^{-2} = 1 + 2x + 3 x^{2} + 4 x^{3} + \dots$ (2)

If you multiply each term of (2) by $x$ you obtain what you have requested. That isn't the only way to arrive to the result...

Kind regards

$\chi$ $\sigma$

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