1. ## taylor series

Find the taylor series centered at c.

f(x)= 1/(1-x) c=5

Find the taylor series centered at c.

f(x)= 1/(1-x) c=5

First write the Taylor series for $f(x)=\frac{1}{1-x}$.

$f(x)=f(c)+f'(c)(x-c)+\frac{f''(c)}{2!}(x-c)^2+..........$

$f'(c)=\frac{1}{(1-c)^2}~,~f''(c)=\frac{2}{(1-c)^3}........$

Therefore, $f(x)=\frac{1}{1-c}+\frac{1}{(1-c)^2}(x-c)+\frac{2}{2!(1-c)^3}(x-c)^2+...........$

Since c=5,

$f(x)=\frac{1}{1-5}+\frac{1}{(1-5)^2}(x-5)+\frac{2}{2!(1-5)^3}(x-5)^2+\frac{2\times{3}}{3!(1-5)^4}(x-5)^3+\frac{2\times{3}\times{4}}{4!(1-5)^5}(x-5)^4...........$

$f(x)=\frac{1}{1-5}+\frac{1}{(1-5)^2}(x-5)+\frac{1}{(1-5)^3}(x-5)^2+\frac{1}{(1-5)^4}(x-5)^3+\frac{1}{(1-5)^5}(x-5)^4...........$

$f(x)=-\frac{1}{4}+\frac{1}{4^2}(x-5)-\frac{1}{4^3}(x-5)^2+\frac{1}{4^4}(x-5)^3-\frac{1}{4^5}(x-5)^4...........$

Hope this helps you.

Find the taylor series centered at c.

f(x)= 1/(1-x) c=5
An alternative method is to use a geometric series expansion like this:

$\frac{1}{1-x}=\frac{1}{1-5-(x-5)}=-\frac{1}{4}\cdot \frac{1}{1+\frac{x-5}{4}}=-\frac{1}{4}\cdot\sum\limits_{n=0}^\infty \left(-\frac{1}{4}\right)^n(x-5)^n=\sum\limits_{n=0}^\infty\left(-\frac{1}{4}\right)^{n+1}(x-5)^n$

4. Originally Posted by Sudharaka

First write the Taylor series for $f(x)=\frac{1}{1-x}$.

$f(x)=f(c)+f'(c)(x-c)+\frac{f''(c)}{2!}(x-c)^2+..........$

$f'(c)=\frac{1}{(1-c)^2}~,~f''(c)=\frac{2}{(1-c)^3}........$
The signs of the terms must alternate in the taylor series, which might seem a little surprising since
$\frac{1}{1-x}=1+x+x^2+\cdots$
for $|x|<1$, but such is life...

5. Originally Posted by Failure
Almost, but not quite: $f''(c)=-\frac{2}{(1-c)^3}$
The signs of the terms must alternate in the taylor series, which might seem a little surprising since
$\frac{1}{1-x}=1+x+x^2+\cdots$
for $|x|<1$, but such is life...
Dear Failure,

Thanks. Corrected the earlier post.