Find the taylor series centered at c.
f(x)= 1/(1-x) c=5
Dear Link88,
First write the Taylor series for $\displaystyle f(x)=\frac{1}{1-x}$.
$\displaystyle f(x)=f(c)+f'(c)(x-c)+\frac{f''(c)}{2!}(x-c)^2+..........$
$\displaystyle f'(c)=\frac{1}{(1-c)^2}~,~f''(c)=\frac{2}{(1-c)^3}........$
Therefore, $\displaystyle f(x)=\frac{1}{1-c}+\frac{1}{(1-c)^2}(x-c)+\frac{2}{2!(1-c)^3}(x-c)^2+...........$
Since c=5,
$\displaystyle f(x)=\frac{1}{1-5}+\frac{1}{(1-5)^2}(x-5)+\frac{2}{2!(1-5)^3}(x-5)^2+\frac{2\times{3}}{3!(1-5)^4}(x-5)^3+\frac{2\times{3}\times{4}}{4!(1-5)^5}(x-5)^4...........$
$\displaystyle f(x)=\frac{1}{1-5}+\frac{1}{(1-5)^2}(x-5)+\frac{1}{(1-5)^3}(x-5)^2+\frac{1}{(1-5)^4}(x-5)^3+\frac{1}{(1-5)^5}(x-5)^4...........$
$\displaystyle f(x)=-\frac{1}{4}+\frac{1}{4^2}(x-5)-\frac{1}{4^3}(x-5)^2+\frac{1}{4^4}(x-5)^3-\frac{1}{4^5}(x-5)^4...........$
Hope this helps you.
An alternative method is to use a geometric series expansion like this:
$\displaystyle \frac{1}{1-x}=\frac{1}{1-5-(x-5)}=-\frac{1}{4}\cdot \frac{1}{1+\frac{x-5}{4}}=-\frac{1}{4}\cdot\sum\limits_{n=0}^\infty \left(-\frac{1}{4}\right)^n(x-5)^n=\sum\limits_{n=0}^\infty\left(-\frac{1}{4}\right)^{n+1}(x-5)^n$