Find a Maclaurin series for f(x)=(6)/((1-x)^4)
looks simple, but I know I am not getting the right answer
Suppose that You have...
$\displaystyle f(x)=\frac{1}{1-x}$ (1)
... and You derive it three times. You obtain...
$\displaystyle \frac{d^{3}}{dx^{3}} f(x)= \frac{6}{(1-x)^4}$ (2)
Now if You know the McLaurin expansion of $\displaystyle f(x)$ what is the procedure to obtain the McLaurin expansion of $\displaystyle \frac{d^{3}}{dx^{3}} f(x)$ ?...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$
Because for $\displaystyle |x|<1$ is...
$\displaystyle \frac{1}{1-x} = 1+x+x^{2}+ \dots + x^{n} + \dots = \sum_{n=0}^{\infty} x^{n}$ (1)
... deriving 'term by term' we have in sequence...
$\displaystyle \frac{d}{dx} \frac{1}{1-x} = \frac{1}{(1-x)^{2}} = 1+2x+3x^{2}+\dots+(n+1)\cdot x^{n} + \dots= \sum_{n=0}^{\infty} (n+1)\cdot x^{n}$ (2)
$\displaystyle \frac{d^{2}}{dx^{2}} \frac{1}{1-x} = \frac{2}{(1-x)^{3}} = 2+6x+12x^{2}+\dots+(n+1)\cdot(n+2)\cdot x^{n} + \dots= \sum_{n=0}^{\infty} (n+1)\cdot(n+2)\cdot x^{n}$ (3)
$\displaystyle \frac{d^{3}}{dx^{3}} \frac{1}{1-x} = \frac{6}{(1-x)^{4}} = 6+24x+120 x^{2}+\dots+(n+1)\cdot(n+2)\cdot (n+3)\cdot x^{n} + \dots=$
$\displaystyle = \sum_{n=0}^{\infty} (n+1)\cdot(n+2)\cdot (n+3)\cdot x^{n}$ (4)
If necessary of course we can proceed further ... so that we obtain in general...
$\displaystyle \frac{1}{(1-x)^{k}} = \frac{1}{(k-1)!}\cdot \sum_{n=0}^{\infty} (n+1)\cdot (n+2)\dots (n+k-1)\cdot x^{n}$ (5)
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$