DATA 605 - Discussion 14/15

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

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

\(f(x) = e^x\) –> \(f(0) = 1\)
\(f'(x) = e^x\) –> \(f'(0) = 1\)
\(f''(x) = e^x\) –> \(f''(0) = 1\)
\(f'''(x) = e^x\) –> \(f'''(0) = 1\)

Becomes:

\(1 + \frac{1}{1!} x^1 + \frac{1}{2!} x^2 + \frac{1}{3!}x^3...\)

Simplifies to:

\(1 + x +x^2 + x^3 + ... + x^n\)$

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

\(f(x) = e^x\) –> \(f(0) = 1\)
\(f'(x) = e^x\) –> \(f'(0) = 1\)
\(f''(x) = e^x\) –> \(f''(0) = 1\)
\(f'''(x) = e^x\) –> \(f'''(0) = 1\)

Becomes:

\(1 + \frac{1}{1!} x^1 + \frac{1}{2!} x^2 + \frac{1}{3!}x^3...\)

Simplifies to:

\(1 + x +x^2 + x^3 + ... + x^n\)

As a summation:

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

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

\(f(x) = ln(1+x)\) –> \(f(0) = 0\)
\(f'(x) = \frac{1}{(1+x)}\) –> \(f'(0) = 1\)
\(f''(x) = -\frac{2}{(1+x)}\) –> \(f''(0) = -1\)
\(f'''(x) = \frac{6}{(1+x)}\) –> \(f'''(0) = 2\)

Becomes:

\(0 + \frac{1}{1!} x^1 - \frac{1}{2!} x^2 + \frac{2}{3!}x^3 ...\)

Or:

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

Equivalent to:

\((-1)^{n+1}\)

Simplifies to:

\(x - \frac{1}{2} x^2 + \frac{1}{4}x^3 ... (-1)^{n+1}\frac{1}{n}x^n\)

As a summation:

\(\sum_{n=0}^{\infty} (-1)^{n+1}\frac{1}{n}x^n\)