Equation 1

\[\pmb{f(x) = \frac{1}{(1-x)}}\]

Function evaluated at \(0\):

\[f(0) = \frac{1}{1-0} = 1\]

Function re-written:

\[\frac{1}{(1-x)} = 1(1-x)^{-1}\]

Derivatives

  1. First Derivative:

\[f'(x) = -1(1-x)^{-2}.(-1)= 1(1-x)^{-2} = \frac{1}{(1-x)^{2}}\]

First Derivative evaluated: \(f'(0) = 1\)

  1. Second Derivative:

\[f''(x) = -2(1-x)^{-3}.(-1)= 2(1-x)^{-3} = \frac{2}{(1-x)^{3}}\]

Second Derivative evaluated: \(f''(0) = 2\)

  1. Third Derivative

\[f'''(x) = -6(1-x)^{-4}.(-1)= 6(1-x)^{-4} = \frac{6}{(1-x)^{4}}\]

Third Derivative evaluated: \(f'''(0) = 6\)

  1. Fourth Derivative

\[f^{(4)}(x) = -24(1-x)^{-5}.(-1)= 24(1-x)^{-5} = \frac{24}{(1-x)^{5}}\]

Fourth Derivative evaluated: \(f^{(4)}(0) = 24\)

  1. Fifth Derivative

\[f^{(5)}(x) = -120(1-x)^{-6}.(-1)= 120(1-x)^{-6} = \frac{120}{(1-x)^{6}}\]

Fifth Derivative evaluated: \(f^{(5)}(0) = 120\)

\[ f(x) \approx 1 + \frac{1}{1-x} + \frac{1}{(1-x)^{2}} + \frac{2}{(1-x)^{3}} + \frac{6}{(1-x)^{4}} + \frac{24}{(1-x)^{5}} + \frac{120}{(1-x)^{6}}+ ...+ \frac{n!}{(1-x)^{n+1}}\]

Taylor Series:

\[f(x) = f(c) + f'(c)(x-c) + \frac{f''(c)}{2!}(x-c)^2 + \frac{f'''(c)}{3!}(x-c)^3 + \frac{f^{(4)}(c)}{4!}(x-c)^4 + \frac{f^{(5)}(c)}{5!}(x-c)^5...+ \frac{f^n(c)}{n!}(x-c)^n\]

\[f\left(x\right)\approx\frac{1}{0!}x^{0}+\frac{1}{1!}x^{1}+\frac{2}{2!}x^{2}+\frac{6}{3!}x^{3}+\frac{24}{4!}x^{4}+\frac{120}{5!}x^{5}\]

\[ 1 + x + x^2 + x^3 + x^4 + x^5+...+x^n = \sum_{n=0}^\infty x^n\]

Equation 2

\[\pmb{f(x) = e^x}\]

\[f(0) = 1\]

Derivatives

First Derivative: \[f'(x) = e^x\]

Evaluated: \(f'(0) = 1\)

Second Derivative: \[f''(x) = e^x\]

Evaluated: \(f''(0) = 1\)

Third Derivative: \[f'''(x) = e^x\]

Evaluated: \(f'''(0) = 1\)

Fourth Derivative: \[f^{(4)}(x) = e^x\]

Evaluated: \(f^{(4)}(0) = 1\)

Fifth Derivative: \[f^{(5)}(x) = e^x\]

Evaluated: \(f^{(5)}(0) = 1\)

Taylor Series

\[1 + \frac{1}{1!}x + \frac{1}{2!}x^2 + \frac{1}{3!}x^3 + \frac{1}{4!}x^4 + \frac{1}{5!}x^5 + ...+\frac{1}{n!}\]

\[1 + x + \frac{1}{2}x^2 + \frac{1}{6}x^3 + \frac{1}{24}x^4 + \frac{1}{120}x^5 + ...+\frac{1}{n!}x^n = \sum_{n=0}^\infty \frac{x^n}{n!} \]

Equation 3

\[\pmb{f(x) = ln(1 + x)}\]

\[f(0) = ln(1) = 0\]

Derivatives

First Derivative:

\[f'(x) = \frac{1}{x + 1}\]

Second Derivative:

\[f''(x) = \frac{-1}{(x + 1)^2}\]

Third Derivative:

\[f'''(x) = \frac{2}{(x + 1)^3}\]

Fourth Derivative:

\[f^{(4)}(x) = \frac{-6}{(x + 1)^4}\]

Fifth Derivative:

\[f^{(4)}(x) = \frac{24}{(x + 1)^5}\]

Taylor Series

\[0 + \frac{1}{1!}x - \frac{1}{2!} x^2+ \frac{2}{3!}x^3 - \frac{6}{4!}x^4 + \frac{24}{5!}x^5 + ...\]

\[x - \frac{1}{2}x^2 + \frac{1}{3}x^3 - \frac{1}{4}x^4 + \frac{1}{5}x^5 + ...+ \frac{1}{n}x^n = \sum_{n=0}^\infty \frac{x^n}{n}\]