Another "simple looking" Maclaurin Series

**Grimey** · November 15th 2009, 06:40 PM

Find a Maclaurin series for f(x)=(6)/((1-x)^4)

looks simple, but I know I am not getting the right answer

**chisigma** · November 15th 2009, 06:57 PM

Suppose that You have...

$f(x)=\frac{1}{1-x}$ (1)

... and You derive it three times. You obtain...

$\frac{d^{3}}{dx^{3}} f(x)= \frac{6}{(1-x)^4}$ (2)

Now if You know the McLaurin expansion of $f(x)$ what is the procedure to obtain the McLaurin expansion of $\frac{d^{3}}{dx^{3}} f(x)$ ?...

Kind regards

$\chi$ $\sigma$

**Grimey** · November 15th 2009, 07:49 PM

Still confused, on expansion.

**chisigma** · November 16th 2009, 12:01 AM

Because for $|x|<1$ is...

$\frac{1}{1-x} = 1+x+x^{2}+ \dots + x^{n} + \dots = \sum_{n=0}^{\infty} x^{n}$ (1)

... deriving 'term by term' we have in sequence...

$\frac{d}{dx} \frac{1}{1-x} = \frac{1}{(1-x)^{2}} = 1+2x+3x^{2}+\dots+(n+1)\cdot x^{n} + \dots= \sum_{n=0}^{\infty} (n+1)\cdot x^{n}$ (2)

$\frac{d^{2}}{dx^{2}} \frac{1}{1-x} = \frac{2}{(1-x)^{3}} = 2+6x+12x^{2}+\dots+(n+1)\cdot(n+2)\cdot x^{n} + \dots= \sum_{n=0}^{\infty} (n+1)\cdot(n+2)\cdot x^{n}$ (3)

$\frac{d^{3}}{dx^{3}} \frac{1}{1-x} = \frac{6}{(1-x)^{4}} = 6+24x+120 x^{2}+\dots+(n+1)\cdot(n+2)\cdot (n+3)\cdot x^{n} + \dots=$

$= \sum_{n=0}^{\infty} (n+1)\cdot(n+2)\cdot (n+3)\cdot x^{n}$ (4)

If necessary of course we can proceed further

... so that we obtain in general...

$\frac{1}{(1-x)^{k}} = \frac{1}{(k-1)!}\cdot \sum_{n=0}^{\infty} (n+1)\cdot (n+2)\dots (n+k-1)\cdot x^{n}$ (5)

Kind regards

$\chi$ $\sigma$

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