Taylor polynomial approximation

Theorem (Taylor's formula (with error))

Given a function $f$ whose $(n+1)th$ derivative exists for all $x$ $\in$ $\Omega,$ where $\Omega$ is an open interval such that $a \in \Omega.$ We have:

\begin{equation} f(x)=f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^2+\dots+\frac{f^{(n_{th} \space \text{derivative})}(a)}{n!}(x-a)^n+R_n(x) \end{equation}

where $R_n(x),$ the remainder/error is

\begin{equation} R_n(x)=\frac{f^{((n+1)th \space \text{derivative})}(\xi)}{\Gamma(n+2)}(x-a)^{n+1} \end{equation}

where $\Gamma$ is the gamma function (note $\Gamma$ ( $n+2)=(n+1)!)$ and $\xi \in [x,a].$

With this we can numerically approximate certain a function and its derivative at a single point.

Example

Given $f(x)=e^{2.7x^2}$ with $a=2,$ find a polynomial of degree $2$ using Definition 2.1 to evaluate $f(x)$ at $x=3$ and $x=5$ , (i.e. $e^{2.7(3)^2}$ and $e^{2.7(5)^2}.$ )

Solution

First observe what the data is. We are given a function $f(x)=e^{2.7x^2}$ and have been asked to evaluate it at $f(3)$ and $f(5)$ by finding a polynomial of degree $2$ . We’ll start by denoting the polynomial of degree 2 $P_2(x),$ and then find what $P_2(x)$ is. By Definition 2.1

\begin{equation} P_2(x) = f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^2 \\ \end{equation}

From a degree $2$ polynomial we need the first two derivatives of $f(a)$ to go any further, so we’ll do that now.

\begin{equation} f(a)=e^{2.7x^2} , f'(a)= 2e^{2.7x} , f''(a)=2e^{2.7}. \end{equation}

Plugging eqref(6) into eqref(5) with $a=2$ returns

P_2(x) = e^{2.7(2)^2}+2e^{2.7(2)}(x-2)+\frac{2e^{2.7}}{2!}(x-2)^2 \\ = 49020.8011364+442.812832408(x-2)+14.8797317249(x-2)^2

Notice that $P_2(x)\approx f(x),$ so evaluating $P_2(3)\approx f(3)$ and $P_2(5)\approx f(5)$ will give us our solution.

P_2(3) = 49020.8011364+442.812832408(3-2)+14.8797317249(3-2)^2 \\ = 49478.4937005

and

P_2(5) = 49020.8011364+442.812832408(5-2)+14.8797317249(5-2)^2 \\ = 50483.1572191

as desired.

Taylor Error

Lagrange polynomial approximation