Friday, February 8, 2013 Taylor's Theorem

1Taylor's Theorem¶

Theorem: (Taylor's) Let $n \in \mathbb{N}$, $I = [a,b]$ and $f : I \to \mathbb{R}$ be such that $f, f', f'', \ldots, f^{(n)}$ are continuous on $I$ and $f^{(n+1)}$ exists on $(a,b)$. If $x_0 \in I$, then, for $x \in I$ there exists a $c \in I$ between $x_0$ and $x$ such that

$$f(x) = f(x_0) + \frac{f'(x_0)}{1!}(x - x_0) + \frac{f''(x_0)}{2!}(x - x_0)^2 + \ldots + \frac{f^{(n)}(x_0)}{n!}(x - x_0)^n + \frac{f^{(n+1)}(c)}{(n+1)!}(x - x_0)^{n+1}$$

i.e., $P_n(x) + R_n(x)$. The final term is the remainder term $R_n(x)$, called the Lagrange form of the remainder.

Proof: Let $x$ and $x_0$ be given, and let $J$ be the closed interval with $x$ and $x_0$ as endpoints. We define

$$F(t) = f(x) - f(t) - (x-t)f'(t) - \ldots - \frac{(x-t)^n}{n!}f^{(n)}(t)$$

It follows that

$$F'(t) = -f'(t) + f'(t) - (x-t)f''(t) + (x-t)f''(t) - \ldots - \frac{(x-t)^n}{n!}f^{(n+1)}(t)$$

This is a sort of telescoping sum; notice all but the final term cancel, leaving us with $\displaystyle F'(t) = -\frac{(x-t)^n}{n!}f^{(n+1)}(t)$. Now define

$$G(t) = F(t) - \left(\frac{x-t}{x-x_0}\right)^{n+1}F(x_0)$$

Then $G(t)$ is differentiable on $J$. Note that $G(x_0) = 0$. $G(x) = F(x) = 0$. Apply Rolle's Theorem to $G$, getting a $c \in J$ such that $G'(c) = 0$. But

\begin{align*} G'(t) &= F'(t) - F(x_0)(n+1)\left(\frac{x-t}{x-x_0}\right)^n \cdot \left(\frac{-1}{x-x_0}\right) \\ &= -\frac{(x-t)^n}{n!}f^{(n+1)}(t) + (n+1)F(x_0)\frac{(x-t)^n}{(x-x_0)^{n+1}} \\ G'(c) &= 0 = -\frac{(x-c)^n}{n!}f^{(n+1)}(c) + (n+1)F(x_0)\frac{(x-c)^n}{(x-x_0)^{n+1}} \end{align*}

i.e. $\displaystyle F(x_0) = \frac{(x-x_0)^{n+1}}{(n+1)!}f^{(n+1)}(c)$. Thus

$$F(x_0) = f(x) - f(x_0) - (x-x_0)f'(x_0) - \ldots - \frac{(x-x_0)^n}{n!}f^{(n)}(x_0)$$

i.e.

$$f(x) = f(x_0) + \frac{f'(x_0)}{1!}(x - x_0) + \frac{f''(x_0)}{2!}(x - x_0)^2 + \ldots + \frac{f^{(n)}(x_0)}{n!}(x - x_0)^n + \frac{f^{(n+1)}(c)}{(n+1)!}(x - x_0)^{n+1}$$

Thus ends Analysis 1!