1.2 Taylor Polynomials

Main Theorem

For some integer $n \geq 0$ , suppose $f (x)$ has $n + 1$ derivative on $(a, b)$ and $x_{0} \in (a, b)$ . Then for each $x \in (a, b)$ there exists a $ξ$ , depending on x, lying between $x$ and $x_{0}$ such that:

f (x) = f (x_{0}) + \sum_{j = 1}^{n} (\frac{f^{(j)} (x_{0})}{j!} (x - x_{0})^{j}) + \frac{f^{(n + 1) (ξ)}}{(n + 1)!} (x - x_{0})^{n + 1}

Proof

$x = x_{0}$ is trivial

assume $x \neq x_{0}$ :
$I$ is an open interval between $x$ and $x_{0}$ . $I \subset \bar{I} \subset (a, b)$ .
$f, f^{'}, f^{″}, \dots, f^{(n)}$ are all continuous on $\bar{I}$ and $f^{(n + 1)}$ exists on $I$ .

F (z) = f (x) - f (z) - \sum_{j = 1}^{n} \frac{f^{(j)} (z)}{j!} (x - z)^{j}

\begin{array}{rcl} F^{'} (z) & = & - f^{'} (z) - \sum_{j = 1}^{n} [\frac{f^{(j + 1)} (z)}{j!} (x - z)^{j} - \frac{f^{(j)} (z)}{(j - 1)!} (x - z)^{j - 1}] \\ = & - f^{'} (z) - [\frac{f^{(n + 1)} (z)}{n!} (x - z)^{n} - f^{'} (z)] \\ = & - \frac{f^{(n + 1)} (z)}{n!} (x - z)^{n} . \end{array}

Define $g (z) = F (z) - {(\frac{x - z}{x - x_{0}})}^{n + 1} F (x_{0}) .$ So we have $g (x_{0}) = g (x) = 0$ and since it is continuous and differentiable by Rolle's Theorem there exists and $ξ \in I$ such that $g^{'} (ξ) = 0 = F^{'} (ξ) + (n + 1) \frac{(x - ξ)^{n}}{(x - x_{0})^{n - 1}} F (x_{0})$

\begin{array}{rcl} F (x_{0}) & = & - F^{'} (ξ) \frac{(x - x_{0})^{n + 1}}{(n + 1) (x - ξ)^{n}} \\ = & \frac{f^{(n + 1)} (ξ)}{n! (n + 1)} (x - x_{0})^{n + 1} \\ = & \frac{f^{(n + 1)} (ξ)}{(n + 1)!} (x - x_{0})^{n + 1} . \end{array}

Taylor Polynomials

$n^{t h}$ Taylor polynomial of $f$ expanded about $x_{0}$ :

T_{n} (x) = f (x_{0}) + \sum_{j = 1}^{n} (\frac{f^{(j)} (x_{0})}{j!} (x - x_{0})^{j})

Remainder term:

R_{n} (x) = \frac{f^{(n + 1)} (ξ)}{(n + 1)!} (x - x_{0})^{n + 1}

Error

\begin{array}{rcl} | T_{n} (x) - f (x) | = | R_{n} (x) | & \leq & \max_{ξ} | \frac{f^{(n + 1)} (ξ)}{(n + 1)!} (x - x_{0})^{n + 1} | \\ = & \frac{| x - x_{0} |^{n + 1}}{(n + 1)!} \max_{ξ} | f^{(n + 1)} (ξ) | . \end{array}