\[f(x) = \frac{1}{(1-a)0!}(x-a)^{0} + \frac{1}{(1-a)^{2}1!}(x-a)^{1} + \frac{2}{(1-a)^{3}2!}(x-a)^{2} + \frac{6}{(1-a)^{4}3!}(x-a)^{3} ...\] \[= \sum_{n=0}^{\infty} \frac{1}{(1-a)^{n+1}}(x-c)^{n}\]
The Maclaurin Series, a = 0:
\[ f(x) = \sum_{n=0}^{\infty}x^{n} = 1 + x + x^2 + x^3 + ... \] The series converges if absolute value of x < 1. Range should be (-1,1).
\[f^{0}(a) = e^{a}\] \[f'(a) = e^{a}\] \[f''(a) = e^{a}\] \[f'''(a) = e^{a}\] By definition,
\[ f(x) = \frac{e^{a}}{0!}(x-a)^{0} + \frac{e^{a}}{1!}(x-a)^{1} + \frac{e^{a}}{2!}(x-a)^{2}+... \] \[ = e^{a} \sum_{n=0}^{\infty}\frac{(x-a)^{n}}{n!} \] According to the Maclaurin Series, with a = 0: \[ f(x) = \sum_{n=0}^{\infty}\frac{x^n}{n!} = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + ... \] The series has valid range of \[(-\infty , \infty )\]
Few derivates,
\[f^{0}(a) = ln(1+a)\] \[f'(a) = \frac{1}{(a+1)}\] \[f''(a) = -\frac{1}{(a+1)^{2}}\] \[f'''(a) = \frac{2}{(a+1)^{3}}\] By definition,
\[ f(x) = \frac{ln(1+a)}{0!}(x-a)^{0} + \frac{1}{(a+1)1!}(x-a)^1 - \frac{1}{(c+1)^{2}2!(x-a)^2} + ... \] According to the Maclaurin Series, a=0: \[f(x)=\sum_{n=1}^{\infty}(-1)^{n+1}\frac{x^n}{n} = x - \frac{x^2}{2} + \frac{x^3}{3} + ...\] By ratio test, we know absolute value of x should be less than 1 for this case. Therefore, the valid range is (-1,1)