12  Taylor Remainder Theorem

12.1 Single variable case

⚠ Intermediate Value Theorem

Suppose that f : [a,b] \to \mathbb R is continuous. Then, f is bounded on [a,b] and for all y \in [\min_{[a,b]} f, \max_{[a,b]} f ] there exists x\in[a,b] such that f(x) = y.

⚠ Mean Value Theorem

Suppose that f \colon [a,b] \to \mathbb R is continuous on [a,b] and differentiable on (a,b). Then, there exists c\in[a,b] for which

\begin{align*} f(b) - f(a) = f'(c) (b-a) \end{align*}

Proof: Consider g(x) := f(x) - \frac{f(b)-f(a)}{b-a} x, a function for which g(a) = g(b). You can show that there exists c\in [a,b] for which g'(c) = 0 (this is known as Rolle’s theorem). This follows from the intermediate value theorem - either the max or min is achieved at a and b (and the function is constant), or there is an extreme point in (a,b). In the latter case, you can show this point is stationary: i.e. g' vanishes here.

Cauchy’s Mean Value Theorem (need this for the proof of Taylor remainder theorem)

Suppose that f, g \colon [a,b] \to \mathbb R are continuous on [a,b] and differentiable on (a,b). Then, there exists c\in[a,b] for which

\begin{align*} \big( f(b) - f(a) \big) g'(c) = \big(g(b) - g(a) \big) f'(c) \end{align*}

Proof: h(x) = \big( g(b) - g(a) \big) f(x) - \big( f(b) - f(a) \big) g(x) is such that h(a) = h(b) and so there exists c \in [a,b] such that h'(c) = 0.

⚠ Taylor Remainder Theorem

Suppose that f\colon [a,b] \to \mathbb R is a function with n times continuously differentiable. Then, there exists c\in[a,b] for which

\begin{align*} f(x) = f(a) + f'(a) (x - a) + \dots + \frac{f^{(n-1)}(a)}{(n-1)!} (x - a)^{n-1} + \frac{f^{(n)}(c)}{n!} (x-a)^{n} \end{align*}

Proof: Define F(x) = f(x) - \sum_{k=0}^{n-1} \frac{f^{(k)}(a)}{k!} (x - a)^k and G(x) = (x-a)^n and note that F(a) = F'(a) = \cdots = F^{(n-1)}(a) = 0 and G(a) = \cdots = G^{(n-1)}(a) = 0 and G^{(n)}(x) = n!. Applying Cauchy’s Mean Value Theorem n times, we can conclude there exists c_1,\dots,c_n between a and x such that

\begin{align*} \frac{F(x)}{G(x)} &= \frac{F(x) - F(a)}{G(x) - G(a)} = \frac{F'(c_1)}{G'(c_1)} \nonumber\\ &= \frac{F'(c_1) - F'(a)}{G'(c_1) - G'(a)} = \frac{F''(c_2)}{G''(c_2)} \nonumber\\ &= \cdots = \frac{F^{(n)}(c_n)}{G^{(n)}(c_n)} % = \frac {f^{(n)}(c_n)} { n! } \end{align*}

12.2 Multivariable case

In the course, we used the fact that, for smooth f:[0,T] \times \mathbb R \to \mathbb R, we have

\begin{align} f(t,u) = f(t_0,u_0) &+ \frac{\partial f}{\partial t}(t_0,u_0) (t-t_0) % + \frac{\partial f}{\partial u}(t_0,u_0) (u-u_0) \nonumber\\ % &+ O( |t-t_0|^2 + |t-t_0|^2 ). \end{align}

This result follows from Taylor expansion in multiple variables. In the following, we will use the notation \bm x = (x_1,\dots,x_d) \in \mathbb R^d with d \in \mathbb N:

Theorem 12.1 (Multivariable Taylor Remainder Theorem) Suppose that f: \mathbb R^d \to \mathbb R is n+1 times continuously differentiable in each of the variables and fix \bm x_0 \in \mathbb R^d. Then, for all \bm x \in \mathbb R^d there exists \bm \xi = t \bm x + (1-t) \bm x_0 \in \mathbb R^d for some t\in [0,1] such that

\begin{align} f(\bm x) = f(\bm x_0) &+ \sum_{i=1}^d \frac{\partial f(\bm x_0)}{\partial x_i} (x-x_0)_i \nonumber\\ % &+ \frac{1}{2} \sum_{i,j=1}^d \frac{\partial^2 f(\bm x_0)}{\partial x_i \partial x_j} (x-x_0)_i (x-x_0)_j % + \cdots \nonumber\\ % &+ \frac{1}{n!} \sum_{i_1,\dots,i_n = 1}^d\frac{\partial^n f(\bm x_0)}{\partial x_{i_1} \cdots \partial x_{i_n}} (x-x_0)_{i_1} \cdots (x-x_0)_{i_n} \nonumber\\ % &+ \frac{1}{(n+1)!} \sum_{i_1,\dots,i_{n+1}=1}^d\frac{\partial^{n+1} f(\bm \xi)}{\partial x_{i_1} \cdots \partial x_{i_{n+1}}} (x-x_0)_{i_1} \cdots (x-x_0)_{i_{n+1}}. \end{align}

Proof.

We can define F(t) := f\big( t \bm x + (1-t) \bm x_0 \big), a real valued function of one variable, and apply the Taylor theorem that you know. Since F is n+1 times continuously differentiable, there exists \xi \in [0,1] such that

\begin{align} F(1) = F(0) + F'(0) + \frac{1}{2} F''(0) + \dots + \frac{1}{n!} F^{(n)}(0) + \frac{1}{(n+1)!} F^{(n+1)}(\xi). \end{align}

We conclude by noting that, by the chain rule

\begin{align} F'(t) &= \sum_{i=1}^d \frac{\partial f}{\partial x_i}\big( t \bm x + (1-t) \bm x_0 \big) (x-x_0)_i \nonumber\\ F''(t) &= \sum_{i,j=1}^d \frac{\partial^2 f}{\partial x_i \partial x_j}\big( t \bm x + (1-t) \bm x_0 \big) (x-x_0)_i (x-x_0)_j, \end{align}

etc.