Taylor Series

Taylor’s Theorem states that any function that is infinitely differentiable can be represented as a polynomial of the following form:

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

Where \(f^{(n)}(a)\) represents the nth derivative of \(f(x)\) evaluated at \(x = a\)

Throughout this exercise, we will assume that the series will be evaluated at 0, i.e. a = 0, which is within the exclusive bounds provided in the class material of \(x \space \epsilon\space (-1,1)\)

Using the above definition on the following expansions:

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

First we will calculate the respective derivatives and their value at x = 0. Using Chain Rule

  • \(f^{(0)} = \frac{1}{1-x} \rightarrow 1\) for \(x = 0\)

  • \(f^{(1)} = \frac{1}{(1-x)^2} \rightarrow 1\) for \(x = 0\)

  • \(f^{(2)} = \frac{2}{(1-x)^3} \rightarrow 2\) for \(x = 0\)

  • \(f^{(3)} = \frac{6}{(1-x)^4} \rightarrow 2\) for \(x = 0\)

A pattern is emerging of \(f^{(n)} = \frac{n!}{(1-x)^{n+1}}\)

Using this in the Taylor Series definition for x = 0:

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

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

Any sequential derivative of \(e^x\) will always be \(e^x\), so we know that \(f^{(n)} = e^x\) and that \(e^0\) evaluates to 1.

Using this in the Taylor Series definition for x = 0:

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

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

First we will calculate the respective derivatives and their value at x = 0. Using Chain Rule

  • \(f^{(0)} = ln(1+x) \rightarrow 0\) for \(x = 0\)

  • \(f^{(1)} = \frac{1}{1+x} \rightarrow 1\) for \(x = 0\)

  • \(f^{(2)} = \frac{-1}{(1+x)^2} \rightarrow -1\) for \(x = 0\)

  • \(f^{(3)} = \frac{2}{(1+x)^3} \rightarrow 1\) for \(x = 0\)

With these results above, we see that further expansions have some similarities to \(\frac{1}{1-x}\) except that \(n\) will lag by 1 and that the derivative will be negative for even values of \(n\), i.e. \((-1)^{n+1}\).

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

We can rewrite \((n-1)!\) as \(\frac{n!}{n}\) resulting in \(f^{(n)} = \frac{(-1)^{n+1}n!}{n(1+x)^n}\)

Using this in the Taylor Series definition for x = 0:

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

It is important to note that this is undefined for n = 0 using the Taylor Series approximation. The correct form should be

\[f(x) = \left\{\begin{matrix} \sum_{n=1}^{\infty} \frac{(-1)^{n+1}\space x^n}{n} & n \neq 0\\ 0 & n=0 \end{matrix}\right.\]