In this question f and g are the functions given by

$\displaystyle f(x)=\frac{6}{6-x}$

$\displaystyle g(x)=\frac{6}{1-x}$

(a) By writing

$\displaystyle \frac{6}{6-x}=\frac{1}{1-\frac{1}{6}x}$

and using substitution in one of the standard Taylor series given in the course, find the Taylor series about 0 for f. Give explicitly all terms up to the term in $\displaystyle x^3$. Determine a range of validity for this Taylor series.

(b) Use your answer to part (a), and the fact that 1-x=6-(x+5), to find the Taylor series about -5 for g. Give explicitly the same numbers of terms as in part (a). Determine a range of validity for this Taylor series.

(c) Check the first four terms in the Taylor series that you found in part (b) by finding the first, second and third derivatives of g, and using these to find the cubic Taylor polynomial about -5 for g.

For (a) I have:

The Taylor series about 0 for $\displaystyle \frac{1}{1-x}$ is:

$\displaystyle \frac{1}{1-x}=1+x+x^2+x^3+...$ for -1<x<1

Using this series and replacing x by $\displaystyle \frac{1}{6}x$ I get:

$\displaystyle \frac{6}{6-x}=\frac{1}{1-\frac{1}{6}x}=1+\frac({1}{6}x)+\frac({1}{6}x)^2+\f rac({1}{6}x)^3+...$

$\displaystyle =1+\frac{1}{6}x+\frac{1}{36}x^2+\frac{1}{216}x^3+. ..$

Since the Taylor series about 0 for $\displaystyle \frac{1}{1-x}$ is valid for -1<x<1, the above series is valid for $\displaystyle -1<\frac{1}{6}x<1$, i.e. it is valid for

-6<x<6.

For (b)

I know that $\displaystyle g(x)=\frac{6}{1-x}$ and 1-x=6-(x+5), so

$\displaystyle g(x)=\frac{6}{1-x}=\frac{6}{6-(x+5)}=\frac{1}{1-\frac{(x+5)}{6}}$, but I am not sure where to go from here.

Any help would be gratefully received