For each function, only consider its valid ranges as indicated in the notes when you are computing the Taylor Series expansion. Please submit your assignment as an R- Markdown document.

“Taylor series of a real or complex-valued function f(x) that is infinitely differentiable at a real or complex number a is the power series” -wikipedia

\(f(a) + \frac{f'(a)}{1!}(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + ...\)

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

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

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

a = 0

converges for \(|x| < 1\)

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

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

\(f'(x) = e^x\)

\(f''(x) = e^x\)

\(f'''(x) = e^x\)

a = 0

converges for all values of x

\(f(a) + \frac{e^0}{1!}(x-0) + \frac{x^2}{2!} + \frac{e^0}{3!}(x-0)^3 + ...\)

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

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

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

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

a = 0

converges for \(|x| < 1\)

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

\(0 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + ...\)

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

\(f''(x) = \frac{-1}{4}x^{\frac{-3}{2}}\)

\(f'''(x) = \frac{3}{8}x^{\frac{-5}{2}}\)

a = 1

converges for \(|x - 1| < 1\)

\(f(1) + \frac{\frac{1}{2}1^{\frac{-1}{2}}}{1!}(x-1) + \frac{\frac{-1}{4}1^{\frac{-3}{2}}}{2!}(x-1)^2 + \frac{\frac{3}{8}1^{\frac{-5}{2}}}{3!}(x-1)^3 + ...\)

\(1 + \frac{\frac{1}{2}}{1!}(x-1) + \frac{\frac{-1}{4}}{2!}(x-1)^2 + \frac{\frac{3}{8}}{3!}(x-1)^3 + ...\)

\(1 + \frac{1}{2}(x-1) + \frac{-1}{8}(x-1)^2 + \frac{1}{16}(x-1)^3 + ...\)