The first three derivatives for \(f(x) = \frac{1}{1-x}\) are \[f'(x) = \frac{1}{(1-x)^2} \\ f''(x) = \frac{2}{(1-x)^3} \\ f'''(x) = \frac{6}{(1-x)^4}\]
This can be generalized as \[f^{(n)}(x) = \frac{n!}{(1-x)^{n+1}}\]
Substituting this into the general equation, \[\frac{1}{1-x} \approx \sum_{n=0}^{\infty} \frac{n!}{(1-a)^{n+1}} \times \frac{(x-a)^n}{n!} = \frac{(x-a)^n}{(1-a)^n}\]
At \(a=0\), this becomes \[\frac{1}{1-x} \approx \sum_{n=0}^{\infty} x^n\]
Using the ratio series for infinite series: \[\frac{A_{n+1}}{A_n} = \frac{x^{n+1}}{x^n} = x \\ \lim_{n \to \infty} x = x\]
So the series converges for \(|x| < 1\).
The derivative of \(f(x) = e^x\) is \(e^x\), so any derivative \(f^{(n)}(x) = e^x\).
Substituting this into the general equation,
\[e^x \approx \sum_{n=0}^{\infty} e^x \times \frac{(x-a)^n}{n!} = \frac{e^x (x-a)^n}{n! (1-a)^n}\]
At \(a=0\), this becomes \[\frac{1}{1-x} \approx \sum_{n=0}^{\infty} \frac{x^n}{n!}\]
Using the ratio series for infinite series: \[\frac{A_{n+1}}{A_n} = \frac{x^{n+1}}{(n+1)!} \frac{n!}{x^n} = \frac{x}{n+1} \\ \lim_{n \to \infty} \frac{x}{n+1} = 0\]
So the series converges for all values of x.
The first four derivatives for \(f(x) = \ln(x)\) are \[f^{(1)}(x) = \frac{1}{1+x} \\ f^{(2)}(x) = \frac{-1}{(1+x)^2} \\ f^{(3)}(x) = \frac{2}{(1+x)^3} \\ f^{(4)}(x) = \frac{-6}{(1+x)^4}\]
This can be generalized as \[f^{(n)}(x) = (-1)^{n-1} \frac{(n-1)!}{(1+x)^{n}}\]
Substituting this into the general equation, \[\ln(1+x) \approx \sum_{n=0}^{\infty} (-1)^{n-1} \frac{(n-1)!}{(1+a)^{n}} \times \frac{(x-a)^n}{n!} = (-1)^{n-1} \frac{(x-a)^n}{n (1+a)^n}\]
At \(a=0\), this becomes \[\ln(1+x) \approx \sum_{n=0}^{\infty} (-1)^{n-1} \frac{x^n}{n}\]
Using the ratio series for infinite series: \[\frac{A_{n+1}}{A_n} = - \frac{x^{n+1}}{n} \frac{n}{x^n} = - \frac{-xn}{n+1} \\ \lim_{n \to \infty} \frac{-xn}{n+1} = -x\]
So the series converges for \(|x| < 1\).