Taylor's Theorem for Differentiable Functions

# Taylor's Theorem for Differentiable Functions

Recall from the Taylor Polynomials of a Function page that if $f : [a, b] \to \mathbb{R}$ is $n$-times differentiable at $x_0 \in A$ then the $n^{\mathrm{th}}$ Taylor polynomial of $f$ at $x_0$ is defined to be:

(1)
\begin{align} \quad P_n(x) = f(x_0) + f'(x_0)(x - x_0) + \frac{f^{(2)}(x_0)}{2!}(x - x_0)^2 + ... + \frac{f^{(n)}(x_0)}{n!}(x - x_0)^n \end{align}
 Theorem 1 (Taylor's Theorem): Let $f : [a, b] \to \mathbb{R}$ and let $x_0 \in I$. If $f$ is $n$ times differentiable on $[a, b]$,$f$, $f'$, …, $f^{(n)}$ are continuous on $[a, b]$, and $f^{(n+1)}$ exists on $(a, b)$ then for every $x \in [a, b]$ there exists a point $c$ between $x$ and $x_0$ such that $\displaystyle{f(x) = P_n(x) + \frac{f^{(n+1)}(c)}{(n+1)!} (x - x_0)^{n+1}}$ where $P_n$ is the $n^{\mathrm{th}}$ Taylor polynomial of $f$ at $x_0$.
• Proof: Let $x \in [a, b]$. Without loss of generality assume that $x < x_0$ and define a function $F : (x, x_0) \to \mathbb{R}$ by:
(2)
\begin{align} \quad F(t) = f(x) - f(t) - (x - t)f'(t) - \frac{(x - t)^2}{2!}f^{(2)}(t) - ... - \frac{(x - t)^n}{n!}f^{(n)}(t) \end{align}
• (If instead $x > x_0$ then we define $F$ the same way as above, but on the interval $(x_0, x)$ instead).
• Define another function, $G : (x, x_0) \to \mathbb{R}$ by:
(3)
\begin{align} \quad G(t) = F(t) - \left ( \frac{x - t}{x - x_0} \right )^{n+1} F(x_0) \end{align}
• Then $G$ is continuous on $[x, x_0]$ and differentiable on $(x, x_0)$. Furthermore by plugging in $t = x$ and $t = x_0$ we see that
(4)
\begin{align} \quad G(x) &= F(x) - \left ( \frac{x - x}{x - x_0} \right )^{n+1}F(x_0) = F(x) = f(x) - f(x) - (x - x)f'(x) - ... - \frac{(x - x)^n}{n!} f^{(n)}(x) = 0 \\ \quad G(x_0) &= F(x_0) - \left ( \frac{x - x_0}{x - x_0} \right )^{n+1}F(x_0) = F(x_0) - F(x_0) = 0 \end{align}
(5)
\begin{align} \quad G'(t) &= \frac{d}{dt} \left [ F(t) - \left ( \frac{x - t}{x - x_0} \right )^{n+1} F(x_0) \right ] \\ &= F'(t) - (n+1) \left ( \frac{x - t}{x - x_0} \right )^n F(x_0) \cdot \left ( - \frac{1}{(x - x_0)} \right ) \\ &= F'(t) + (n + 1) \frac{(x - t)^n}{(x - x_0)^{n+1}} F(x_0) \\ &= \frac{d}{dt} \left [ f(x) - f(t) - (x - t)f'(t) - \frac{(x - t)^2}{2!}f^{(2)}(t) - ... - \frac{(x - t)^n}{n!}f^{(n)}(t) \right ] + (n + 1) \frac{(x - t)^n}{(x - x_0)^{n+1}} F(x_0) \\ &= - \frac{(x - t)^n}{n!}f^{(n+1)}(t) + (n + 1) \frac{(x - t)^n}{(x - x_0)^{n+1}} F(x_0) \end{align}
• Plugging in $t = c$ gives us:
(6)
\begin{align} \quad 0 = G'(c) = - \frac{(x - c)^n}{n!}f^{(n+1)}(c) + (n + 1) \frac{(x - c)^n}{(x - x_0)^{n+1}} F(x_0) \end{align}
• Therefore:
(7)
\begin{align} \quad F(x_0) &= \frac{(x - x_0)^{n+1}}{(x - c)^n} \cdot \frac{(x - c)^n}{n!} f^{(n+1)}(c) \\ &= \frac{(x - x_0)^{n+1}}{(n+1)!} f^{(n+1)}(c) \\ &= \frac{f^{(n+1}(c)}{(n + 1)!}(x - x_0)^{n+1} \end{align}
• And:
(8)
\begin{align} \quad f(x) \underbrace{- f(x_0) - (x - x_0)f'(x_0) - ... - \frac{(x - x_0)^n}{n!} f^{(n)}(x_0)}_{= -P_n(x)} &= \frac{f^{(n+1}(c)}{(n + 1)!}(x - x_0)^{n+1} \end{align}
• Hence we conclude that:
(9)
\begin{align} \quad f(x) = P_n(x) + \frac{f^{(n+1)}(c)}{(n+1)!}(x - x_0)^{n+1} \quad \blacksquare \end{align}