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.
Taylor Series centered about \(k\) is given by the formula:
\[\begin{gather*}
f(x) = f(k) + f'(k)(x-k) + \frac{f''(k)}{2!} + \cdot\cdot\cdot \\
= \sum_{n=0}^{\infty}\frac{f^{(n)}(k)}{n!}x^n
\end{gather*}\]
If k=0 (centered about 0) it forms a Maclaurin Series:
\[\begin{gather*} 1 + x + x^2 + x^3 + \cdot\cdot\cdot \\ = \sum_{n = 0}^{\infty}x^n \ \ \ x \in (-1, 1) \end{gather*}\]
This forms a geometric series.
Any derivative of \(f(x) = e^x\) is \(e^x\)
Hence:
\(f^{(n)}(x)=e^x\)
When centered about 0:
\(f^{(n)}(0)=1\)
Using Maclaurin Series:
\[\begin{gather*} 1 + \frac{1}{1!}x^1 + \frac{1}{2!}x^2 + \frac{1}{3!}x^3 + \frac{1}{4!}x^4 + \cdot\cdot\cdot \\ = \sum_{n = 0}^{\infty}(-1)^{n+1}\frac{x^n}{n} = \sum_{n = 0}^{\infty}(-1)^{n-1}\frac{x^n}{n} \ \ \ x \in (-1, 1] \end{gather*}\]
\[\begin{gather*} 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdot\cdot\cdot \\ = \sum_{n = 0}^{\infty} \frac{x^n}{n!} \ \ \ x \in \mathbb{R} \end{gather*}\]
The first four derivatives of at \(x=0\):
\(f(x) = ln(x)\) →\(f(0) = 0\)
\(f^{(1)}(x) = \frac{1}{1+x}\) →\(f^{(1)}(0) = 1\)
\(f^{(2)}(x) = \frac{-1}{(1+x)^2}\) →\(f^{(2)}(0) = -1\)
\(f^{(3)}(x) = \frac{2}{(1+x)^3}\) →\(f^{(3)}(0) = 2\)
\(f^{(4)}(x) = \frac{-6}{(1+x)^4}\) →\(f^{(4)}(0) = -6\)
At \(x=0\) Taylor’s Series becomes Maclaurin Series:
\[\begin{gather*} 0 + x - \frac{1}{2!}x^2 + \frac{2}{3!}x^3 - \frac{6}{4!}x^4 + \cdot\cdot\cdot \\ = \sum_{n = 0}^{\infty}(-1)^{n+1}\frac{x^n}{n} = \sum_{n = 0}^{\infty}(-1)^{n-1}\frac{x^n}{n} \ \ \ x \in (-1, 1] \end{gather*}\]