Taylor Series Expansions

Valid Ranges (See “Important Taylor Series Expansions”, Key Idea 8.8.1)

The Taylor polynomial of degree \(n\) of \(f\) at \(x=c\) is:

\(p_n(x) = f(c) + f'(c)(x-c) + \frac{f''(c)(x-c)^2}{2!} + \frac{f'''(c)(x-c)^3}{3!} + \dots + \frac{f^{(n)}(c)(x-c)^n}{n!}\)

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

Step 1: Find the first \(n\) derivatives of \(f(x)\).

  • \(f(x) = \frac{1}{1-x}\)
  • \(f'(x) = \frac{1}{(1-x)^2}\)
  • \(f''(x) = \frac{2}{(1-x)^3}\)
  • \(f'''(x) = \frac{6}{(1-x)^4}\)
  • \(f^{(n)}(x) = \frac{n!}{(1-x)^{n+1}}\)

Step 2: Evaluate these derivatives at \(x=0\).

  • \(f(0) = \frac{1}{1-0} = 1\)
  • \(f'(0) = \frac{1}{(1-0)^2} = 1\)
  • \(f''(0) = \frac{2}{(1-0)^3} = 2\)
  • \(f'''(0) = \frac{6}{(1-0)^4} = 6\)
  • \(f^{(n)}(0) = \frac{n!}{(1-0)^{n+1}} = n!\)

Step 3: Substitute the derivatives into the Taylor Polynomial.

\[\begin{align*} p_n(x) & = 1 + x + x^2 + x^3 + \dots \\ & = \sum_{n=0}^{\infty} x^n \end{align*}\]

which is centered at \(c=0\). The interval of convergence is \((-1,1)\).

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

Step 1: Find the first \(n\) derivatives of \(f(x)\).

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

Step 2: Evaluate these derivatives at \(x=0\).

  • \(f(0) = e^0 = 1\)
  • \(f'(0) = e^0 = 1\)
  • \(f''(0) = e^0 = 1\)
  • \(f'''(0) = e^0 = 1\)
  • \(f^{(n)}(0) = e^0 = 1\)

Step 3: Substitute the derivatives into the Taylor Polynomial.

\[\begin{align*} p_n(x) & = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots \\ & = \sum_{n=0}^{\infty} \frac{x^n}{n!} \end{align*}\]

which is centered at \(c=0\). The interval of convergence is \((-\infty, \infty)\).

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

This one can be done in a different way. From Key Idea 8.8.1,

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

The first few terms are,

\((x-1) - \frac{(x-1)^2}{2} + \frac{(x-1)^3}{3} - \dots\)

The interval of convergence is \((0,2]\) and the series is centered at \(x=1\). We can center the series for \(f(x) = ln(1+x)\) at

\(x=1\) as well.

Substitute \(1+x\) to obtain,

\[\begin{align*} ln(1+x) & = \sum_{n=1}^{\infty} (-1)^{n+1} \cdot \frac{([1+x]-1)^n}{n} \\ & = \sum_{n=1}^{\infty} (-1)^{n+1} \cdot \frac{x^n}{n} \end{align*}\]

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

We will center the series on \(x=1\).

Step 1: Find the first few derivatives of \(f(x)\).

  • \(f(x) = x^{\frac{1}{2}}\)
  • \(f'(x) = \frac{1}{2} \cdot x^{\frac{-1}{2}}\)
  • \(f''(x) = \frac{-1}{4} \cdot x^{\frac{-3}{2}}\)
  • \(f'''(x) = \frac{3}{8} \cdot x^{\frac{-5}{2}}\)

Step 2: Evaluate the derivatives at \(x=1\).

  • \(f(1) = 1^{\frac{1}{2}} = 1\)
  • \(f'(1) = \frac{1}{2} \cdot 1^{\frac{-1}{2}} = \frac{1}{2}\)
  • \(f''(1) = \frac{-1}{4} \cdot 1^{\frac{-3}{2}} = \frac{-1}{4}\)
  • \(f'''(1) = \frac{3}{8} \cdot 1^{\frac{-5}{2}} = \frac{3}{8}\)

Step 2: Substitute the derivatives.

\[\begin{align*} p_n(x) & = 1 + \frac{1}{2}(x-1) - \frac{1}{4}\frac{(x-1)^2}{2!} + \frac{3}{8}\frac{(x-1)^3}{3!} \dots \\ & = 1 + \frac{x-1}{2} - \frac{(x-1)^2}{8} + \frac{(x-1)^3}{16} - \dots \end{align*}\]