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.

Taylor and Maclaurin Series Definition

Let \(f(x)\) have derivatives of all orders at \(x = c\).

  1. The Taylor Series of \(f(x)\), centered at \(c\) is

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

  1. Setting \(c = 0\) gives the Maclaurin Series of \(f(x)\):

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

\(f(x) = \frac{1}{(1−x)}\)

First we find the first few derivates of \(f(x)\) and evaluate them at 0 to look for a pattern.

\[ \begin{align} f(x) = \frac{1}{(1−x)} = (1-x)^{-1} \qquad &\Rightarrow \qquad f(0) = 1 \\ f'(x) = -1(1-x)^{-2} (-1)= (1-x)^{-2} \qquad &\Rightarrow \qquad f'(0) = 1 \\ f''(x) = -2(1-x)^{-3} (-1)= 2(1-x)^{-3} \qquad &\Rightarrow \qquad f''(0) = 2 \\ f'''(x) = (2)(-3)(1-x)^{-4} (-1)= (2)(3)(1-x)^{-4} \qquad &\Rightarrow \qquad f'''(0) = 6 \\ f^{(4)}(x) = (2)(3)(4)(1-x)^{-5}\qquad &\Rightarrow \qquad f^{(4)}(0) = 24 \\ f^{(5)}(x) = (2)(3)(4)(5)(1-x)^{-6} \qquad &\Rightarrow \qquad f^{(5)}(0) = 120 \\ f^{(6)}(x) = (2)(3)(4)(5)(6)(1-x)^{-7} \qquad &\Rightarrow \qquad f^{(6)}(0) = 720\\ \end{align} \]

Note: Since we could already see the pattern emerging I skipped the chain rule step in the last three higher order derivatives.

We can see from the pattern above that…

\[ f^{(n)}(x) = n!(1-x)^{(-n-1)} = \frac{n!}{(1-x)^{n+1}} = \qquad \Rightarrow \qquad f^{(n)}(0) = n! \]

Plugging this into our Maclaurin Series definition we get:

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

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

\[ \begin{align} f(x) = e^x \qquad &\Rightarrow \qquad f(0) = 1 \\ f'(x) = e^x \qquad &\Rightarrow \qquad f'(0) = 1 \\ f''(x) = e^x \qquad &\Rightarrow \qquad f''(0) = 1 \\ f'''(x) = e^x \qquad &\Rightarrow \qquad f'''(0) = 1 \\ f^{(4)}(x) = e^x \qquad &\Rightarrow \qquad f^{(4)}(0) = 1 \\ \\ f^{(n)}(x) = e^x \qquad &\Rightarrow \qquad f^{(n)}(0) = 1 \\ \end{align} \]

All higher order derivatives of \(e^x\) are \(e^x\) which evaluated at zero gives us 1. Plugging this directly into our Maclaurin Series definition gives us:

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

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

\[ \begin{align} f(x) = \ln(1+x)\qquad &\Rightarrow \qquad f(0) = 0 \\ f'(x) = \frac{1}{1+x} (1) = (1 + x)^{-1} \qquad &\Rightarrow \qquad f'(0) = 1 \\ f''(x) = (-1)(1+x)^{-2}(1) = -(1+x)^{-2}\qquad &\Rightarrow \qquad f''(0) = -1 \\ f'''(x) = (-1)(-2)(1+x)^{-3}(1) = 2(1+x)^{-3} \qquad &\Rightarrow \qquad f'''(0) = 2 \\ f^{(4)}(x) = (-1)(-2)(-3)(1+x)^{-4}(1) = -6(1+x)^{-4} \qquad &\Rightarrow \qquad f^{(4)}(0) = -6 \\ f^{(5)}(x) = (-1)(-2)(-3)(-4)(1+x)^{-5} = 24(1+x)^{-5}\qquad &\Rightarrow \qquad f^{(5)}(0) = 24 \\ \\ f^{(n)}(x) = (-1)^{n+1} (n-1)!(1+x)^{-n} = (-1)^{n+1} \frac{(n-1)!}{(1+x)^n} \qquad &\Rightarrow \qquad f^{(n)}(0) = (-1)^{n+1} (n-1)!\\ \end{align} \]

Plugging the equation above into our Maclaurin Series definition gives us:

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

At \(n=0\) this would be undefined, so we will start at \(n=1\)…

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