Please refer to the Assignment 14 Document.

1 Problem Set

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

  • \(f(x) = \frac{1}{(1-x)}\)
  • \(f(x) = e^{x}\)
  • \(f(x) = ln(1+x)\)

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 a R-Markdown document.

2 Taylor Series

3 Answers

3.1 Q1

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

Taking derivatives, we have

\[f'(x) = (-1)(1-x)^{-2}(-1)=(1-x)^{-2}\] \[f''(x) = (-2)(1-x)^{-3}(-1)=2(1-x)^{-3}\] \[f'''(x) = (2)(-3)(1-x)^{-4}(-1) = 6(1-x)^{-4}\] \[f''''(x) = (6)(-4)(1-x)^{-5}(-1) = 24(1-x)^{-5}\]

Evaluating the formula at \(x=0\), we get

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

\[= \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}x^{n}\] \[= f(0) + f'(0)x + f''(0)\frac{x^{2}}{2!} + f'''(0)\frac{x^{3}}{3!} + \cdots\] \[= 1 + x + \frac{2}{2!}x^{2} + \frac{6}{3!}x^{3} + \frac{24}{4!}x^{4} + \cdots \] \[= 1+x+x^{2}+x^{3}+x^{4}\cdots\]

3.2 Q2

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

Taking derivatives, we have

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

Evaluating the formula at \(x=0\), we get

\[f(x) = e^{x}\] \[= \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}x^{n}\] \[= f(0) + f'(0)x + f''(0)\frac{x^{2}}{2!} + f'''(0)\frac{x^{3}}{3!} + \cdots\] \[= 1 + x + \frac{1}{2!}x^{2} + \frac{1}{3!}x^{3} + \frac{1}{4!}x^{4} + \cdots \] \[= \sum_{n=0}^{\infty} \frac{x^{n}}{n!}\]

3.3 Q3

\[f(x) = ln(1+x)\]

Taking derivatives, we have

\[f'(x) = (1+x)^{-1}\] \[f''(x) = -(1+x)^{-2}\] \[f'''(x) = 2(1+x)^{-3}\] \[f''''(x) = -6(1+x)^{-4}\]

Evaluating the formula at \(x=0\), we get

\[f(x) = ln(1+x)\]

\[= \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}x^{n}\] \[= f(0) + f'(0)x + f''(0)\frac{x^{2}}{2!} + f'''(0)\frac{x^{3}}{3!} + \cdots\]

\[= 0 + (0!)x + \frac{-1!}{2!}x^{2} + \frac{2!}{3!}x^{3} + \frac{-3!}{4!}x^{4} + \cdots \] \[= 0 + \sum_{n=1}^{\infty} \frac{(-1)^{n-1}(n-1)!x^{n}}{n!}\] \[= \sum_{n=1}^{\infty} \frac{(-1)^{n-1}x^{n}}{n}\]