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