Data 605 - Assignment 14

Hazal Gunduz

ASSIGNMENT 14 - TAYLOR SERIES

This week, we’ll work out some Taylor Series expansions of popular functions.

• \(f(x) = \frac{1}{(1-x)}\)

• \(f(x) = {e^x}\)

• \(f(x) = ln(1 + x)\)

• \(f(x) = x^{(1/2)}\)

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.

• \(f(x) = \frac{1}{(1-x)}, x = 0\)

\[f(x) = \sum_{n=0}^{∞}\frac {f^{(n)}(a)}{n!} (x - a) ^ {n}\]

f(x) = f(a) + f′(a)(x−a) + \(\frac{f''(a)}{2!} (x - a)^2 + \frac{f'''(a)}{3!} (x - a)^3 + \frac{f''''(a)}{4!} (x - a)^4 + ....\)

\(f(x) = \frac{1}{1-x}, f(0) = \frac{1}{(1-0)} = 1\)

\(f'(x) = \frac{1}{(1-x)^2}, f'(0) = \frac{1}{(1-0)^2} = 1\)

\(f''(x) = \frac{2}{(1-x)^3}, f''(0) = \frac{2}{(1-0)^3} = 2\)

\(f'''(x) = \frac{6}{(1-x)^4}, f'''(0) = \frac{6}{(1-0)^4} = 6\)

\(f''''(x) = \frac{24}{(1-x)^5}, f''''(0) = \frac{24}{(1-0)^5} = 24\)

\(f^{n}(x) =\frac{n!}{(1-x)^{n+1}}\)

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

\[f(x) = \sum_{n=0}^{∞} x^n\]

• \(f(x) = {e^x}, x = 0\)

\(f'(x) = {e^x}, f'(0) = {e^0} = 1\)

\(f''(x) = {e^x}, f''(0) = {e^0} = 1\)

\(f'''(x) = {e^x}, f'''(0) = {e^0} = 1\)

      .....

\(f^{(n)}(x) = {e^x}, f^{(n)}(0) = {e^0} = 1\)

\(f(x) = e^0 + e^0(x-0) + e^0(x-0)^2 + e^0(x-0)^3 + ...... + e^0(x-0)^n\)

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

\[f(x) = \sum_{n=0}^{∞}\frac {x^n}{n!}\]

• \(f(x) = ln(1 + x), x = 0\)

\(f(x) = ln(1 + x), f(0) = 0\)

\(f'(x) = \frac {1}{1+x}, f'(0) = 1\)

\(f''(x) = \frac {-1}{(1+x)^2}, f''(0) = -1\)

\(f'''(x) = \frac {2}{(1+x)^3}, f'''(0) = 2\)

\(f''''(x) = \frac {-6}{(1+x)^4}, f''''(0) = -6\)

      ......
      

\(f^n(x) = \frac {(-1)^{(n-1)}(n-1)!}{(x+1)^n}, f^n(0) = (-1)^{(n-1)}(n-1)!\)

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

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

\[f(x) = \sum_{n=1}^{\infty} \frac {(-1) ^{(n-1)}(n-1)!} {n!} (x)^n\]

• \(f(x) = x^{(1/2)}, x = 0\)

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

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

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