1 Question 1

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.

1.1 Part (a)

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

Answer:

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

Compute the first four derivatives corresponding to the \(f(x)\) and their values at \(x=0\):

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

\(f'(0) = 1 = 1!\)

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

\(f''(0) = 2!\)

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

\(f'''(0) = 3!\)

\(f''''(x) = (3!)\cdot(-4)\cdot(1-x)^{-5}\cdot(-1) = 4!(1-x)^{-5}\)

\(f''''(0) = 4!\)

Therefore, we have the Taylor Series expansions:

\(f(x) = (1-x)^{-1}\)

\(= \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}x^{n}\)

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

\(= 1 + 1!\cdot x + 2!\cdot \frac{x^{2}}{2!} + 3!\cdot \frac{x^{3}}{3!} + 4!\cdot \frac{x^{4}}{4!} + \cdots\)

\(= 1 + x + x^{2} + x^{3} + x^{4} + \cdots\)

This function converges when \(|x|<1\).

1.2 Part (b)

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

Answer:

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

Compute the first four derivatives corresponding to the \(f(x)\) and their values at \(x=0\):

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

\(f'(0) = 1\)

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

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

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

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

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

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

Therefore, we have the Taylor Series expansions:

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

\(= \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}x^{n}\)

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

\(= 1 + x + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + \frac{x^{4}}{4!} + \cdots\)

\(= \sum_{n=0}^{\infty} \frac{x^{n}}{n!}\)

This function converges for all \(x \in R\).

1.3 Part (c)

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

Answer:

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

Compute the first four derivatives corresponding to the \(f(x)\) and their values at \(x=0\):

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

\(f'(0) = 1 = (-1)^{(1+1)} \cdot 0!\)

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

\(f''(0) = -1 = (-1)^{(2+1)} \cdot 1!\)

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

\(f'''(0) = 2 = (-1)^{(3+1)} \cdot 2!\)

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

\(f''''(0) = -6 = (-1)^{(4+1)} \cdot 3!\)

Therefore, we have the Taylor Series expansions:

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

\(= \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}x^{n}\)

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

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

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

\(= \sum_{n=1}^{\infty} \frac{(-1)^{n+1}\cdot x^{n}}{n}\)

This function converges for \(|x|<1\).