
Originally Posted by
macca101 Hello !!
Give the Taylor series about 0 for f and determine a range of validity for the series
$\displaystyle
f(x)=\frac{1}{(1+\frac{1}{6}x)^2}
$
Using the standard taylor series
$\displaystyle
f(x)=\frac{1}{(1-x)}= 1+ x+x^2+x^3+.... for -1<x<1
$
Substituting $\displaystyle
x = -\frac{1}{6}x
$
And taking 2 in the binomial series gives
$\displaystyle
1+(2)*(-\frac{1}{6}x)+\frac{(2)(3)}{2!}(-\frac{1}{6}x)^2+\frac{(2)(3)(4)}{3!}(-\frac{1}{6}x)^3
$
Gives
$\displaystyle
1+-\frac{1}{3}x+\frac{1}{12}x^2-\frac{1}{54}x^3
$
as
$\displaystyle
f(x)=\frac{1}{(1-x)}= 1+ x+x^2+x^3+.... for -1<x<1
$
Then
$\displaystyle
\frac{1}{(1+\frac{1}{6}x)^2}
$
is valid for
$\displaystyle
-1<-\frac{1}{6}x<1
$
Thanks Macca