\(f(x) = \sum_{n=0}^{\infty}\frac{f^{(n)}(a)}{n!}(x-a)^n\)
\(= f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^2+\frac{f'''(a)}{3!}(x-a)^3+...\)
\(f(x) = \frac{1}{(1-x)}\)
\(f'(x) = \frac{1}{(1-x)^2}\) \(f''(x) = \frac{2}{(1-x)^3}\) \(f'''(x) = \frac{6}{(1-x)^4}\)
can be generalized as,
\(f^{(n)}(x) = \frac{n!}{(1-x)^{n+1}}\)
substitute to the general equation,
\(\frac{1}{1-x} \approx \sum_{n=0}^{\infty}\frac{n!}{(1-a)^{n+1}}\times \frac{(x-a)^n}{n!} = \frac{(x-a)^n}{(1-a)^n}\)
when \(a=0\) this becomes,
\(\frac{1}{1-x} \approx \sum_{n=0}^{\infty}x^n\)
\(f(x) = e^x\)
The derivative of \(f(x) = e^x\) is \(e^x\) so any derivative of \(f^{(n)}(x)=e^x\)
At \(a=0\) \(\frac{1}{1-x} \approx \sum_{n=0}^{\infty}\frac{x^n}{n!}\)
\(f(x) = ln(1+x)\)
The first four derivatives of \(f(x) = ln(x)\),
\(f^{(1)}(x) = \frac{1}{1+x}\)
\(f^{(2)}(x) = \frac{-1}{(1+x)^2}\)
\(f^{(3)}(x) = \frac{2}{(1+x)^3}\)
\(f^{(4)}(x) = \frac{-6}{(1+x)^4}\)
can be generalized as,
\(f^{(n)}(x) = (-1)^{n-1}\frac{(n-1)!}{(1+x)^n}\)
substitute to the general equation,
\(ln(1+x) \approx \sum_{n=0}^{\infty} (-1)^{n-1} \frac{(n-1)!}{(1+a)^n}\times\frac{(x-a)^a}{n!}=(-1)^{n-1}\frac{(x-a)^n}{n(1+a)^n}\)
when \(a =0\) this becomes,
\(ln(1+x) \approx \sum_{n=0}^{\infty}(-1)^{n-1}\frac{x^n} {n}\)