1. Not defined when x = 1 as denom will be 0 Evaluation at x = 0

\(F(x) = \frac{1}{(1-x)}\)

\(F(x) = \frac{1}{(1-x)} , F(0) = 1\) \(F'(x) = \frac{1}{(1-x)^2} , F'(0) = 1\) \(F''(x) = \frac{2}{(1-x)^3} , F''(0) = 2\) \(F'''(x) = \frac{6}{(1-x)^4} , F'''(0) = 6\) \(F''''(x) = \frac{24}{(1-x)^5} , F''''(0) = 24\)

Series formula:

\(1 + \frac{1}{1!}x^1 + \frac{2}{2!}x^2 + \frac{6}{3!}x^3 + \frac{24}{4!}x^4 ....\)

Simplifying to :

\(1 + x + x^2 + x^3 + x^4....x^n\)

  1. Again x = 0

\(F(x) = e^x\)

\(F(x) = e^x , F(0) = 1\) \(F'(x) = e^x , F'(0) = 1\) \(F''(x) = e^x , F''(0) = 1\) \(F'''(x) = e^x , F'''(0) = 1\) \(F''''(x) = e^x , F''''(0) = 1\) \(F^5(x) = e^x , F^5(0) = 1\)

Series formula:

\(1 + \frac{1}{1!}x^1 + \frac{1}{2!}x^2 + \frac{1}{3!}x^3 ....\)

Simplifying to :

\(1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24}....\frac{x^n}{n!}\)

  1. X = 0

\(F(x) = ln(1+x)\)

\(F(x) = ln(1+x) , F(0) = 0\) \(F'(x) = \frac{1}{(x+1)} , F'(0) = 1\) \(F''(x) = \frac{-1}{(x+1)^2} , F'(0) = -1\) \(F'''(x) = \frac{2}{(x+1)^3} , F'''(0) = 2\) \(F''''(x) = \frac{-6}{(x+1)^4} , F''''(0) = -6\) \(F^5(x) = \frac{24}{(x+1)^5} , F^5(0) = -24\)

Series formula:

\(0 + \frac{1}{1!}x^1 - \frac{1}{2!}x^2 + \frac{2}{3!}x^3 - \frac{6}{4!}x^4....\)

Simplifying to :

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