Differentiation

# Differentiation

A function $$f : [a, b] \rightarrow \mathbb{R}$$ is differentiable at $$x \in [a, b]$$ if there exists the limit $f'(x) = \lim\limits_{t \rightarrow x} \frac{f(t) - f(x)}{t - x}$ where $$t \in (a, b)$$ and $$t \neq x$$. The function $$f'$$ is call the derivative of $$f$$.

For example,

1. If $$f$$ is continuous on $$[a, b]$$, $$f$$ is not necessarily differentiable on $$[a, b]$$.
2. If $$f$$ is differentiable on $$[a, b]$$, $$f$$ is continuous on $$[a, b]$$.

Why? If $$t \rightarrow x$$ then $$\lim\limits_{t \rightarrow x} f(t) - f(x) = \lim\limits_{t \rightarrow x} \frac{f(t) - f(x)}{t - x}(t - x) = f'(x) \cdot \lim\limits_{t \rightarrow x} t - x = f'(x) \cdot 0 = 0$$.

3. If $$f$$ is differentiable on $$[a, b]$$, $$f'$$ is not necessarily continuous. However, $$f'$$ satisfies the intermediate value property but it does not have any simple discontinuities.

Consider a version of the topologist’s sine curve. \begin{align} f(x) \begin{cases} 0 & \text{if } x = 0 \\ x^\frac{4}{3}\sin\left(\frac{1}{x}\right) & \text{otherwise} \end{cases} \end{align}

Call a function $$f$$ a $$\mathcal{C}^1$$-function if $$f'$$ exists and is continuous. Similarly, call a function $$f$$ a $$\mathcal{C}^k$$-function if it has a $$k$$th derivative $$f^{(k)}$$ which is continuous. A function is a $$\mathcal{C}^0$$-function if it is continuous. A function is a $$\mathcal{C}^\infty$$-function if all of its derivatives exists and are continuous; alternatively, $$\mathcal{C}^\infty$$-functions are called smooth. As defined, $$\mathcal{C}^k$$-functions are also $$\mathcal{C}^{k-1}$$-functions.

If $$f'$$ is a limit then the sum, product, and quotient rules follow.

For example,

1. $$(f + g)' = f' + g'$$ because the sum of limits is the limit of the sums.
2. $$(f \cdot g)' = f' \cdot g + f \cdot g'$$.

Theorem. There exist functions that are continuous everywhere but are differentiable nowhere.

For example,

1. Let $$0 < b < 1$$ and $$a$$ be an odd integer, where $$ab > 1 + \frac{3 \pi}{2}$$. Consider the following function.

$f(x) = \sum\limits_{n = 1}^\infty b^n cos(a^n \pi x)$

# The Mean Value Theorem

Theorem (the mean value theorem). If $$f$$ is continuous on $$[a, b]$$ and differentiable on $$(a, b)$$ then there exists a point $$c \in (a, b)$$ such that $$f(b) - f(a) = (b - a)f'(c)$$.

$$f(b) - f(a) = (b - a)f'(c)$$ relates value of $$f$$ to value of $$f'$$ without involving limits.

For example,

1. If $$f'(x) > 0$$ for all $$x \in (a, b)$$ then $$f(b) > f(a)$$.

Proof. By the mean value theorem, $$f(b) - f(a) = (b - a)f'(c) > 0$$ for some $$c \in (a, b)$$, as desired. QED.

Theorem (the generalized mean value theorem). If $$f(x)$$ and $$g(x)$$ are continuous on $$[a, b]$$ and differentiable on $$(a, b)$$ then there exists a point $$c \in (a, b)$$ such that $$(f(b) - f(a))g'(c) = (g(b) - g(a))f'(c)$$.

If $$g(x) = x$$, we get the mean value theorem.

Consider $$h(x) = (f(b) - f(a))g'(c) - (g(b) - g(a))f'(c)$$. Notice $$h(a) = 0$$ and $$h(b) = 0$$.

# Taylor’s Theorem

Suppose we know about $$f(a)$$. We want to approximate $$f(b)$$. The mean value theorem says $$f(b) = f(a) + (b - a)f'(c)$$ for some $$c \in (a, b)$$. $$(b - a)f'(c)$$ is an error term, not precisely known. This suggests $$f(b) = f(a) + f'(a)(b - a) + e$$ with $$e$$ as an error term. In fact, $$e = \frac{f''(d)}{2}(b - a)^2$$, for some $$d \in (a, b)$$.

More generally, if $$P_{n - 1}(x) = f(a) + f'(a)(x - a) + \frac{f''(a)}{2!}(x - a)^2 + \dots + \frac{f^{(n - 1)}(a)}{(n - 1)!}(x - a)^{n - 1}$$, which is a polynomial with degree $$n - 1$$, then Taylor’s theorem states that $$P_{n - 1}(x)$$ approximates $$f(x)$$ with a small error term.

Theorem (due to Taylor). If $$f^{(n - 1)}$$ is continuous on $$[a, b]$$ and $$f^{(n)}$$ exists on $$(a, b)$$ then $$P_{n - 1}(x)$$ approximates $$f(x)$$ and $$f(x) = P_{n - 1}(x) + \frac{f^{(n)}(c)}{n!}(x - a)^n$$ for some $$c \in (a, x)$$.

• When $$n = 1$$, Taylor’s theorem is the mean value theorem.
• $$P_n(x)$$ is the “best” polynomial approximation of order $$n$$ at $$a$$.

Proof. Clearly, for some number $$\mathcal{M}$$, $$f(b) = P_{n-1}(b) + \mathcal{M}(b - a)^n$$. Let $$g(x) = f(x) - P_{n-1}(x) - \mathcal{M}(x - a)^n$$. Consider $$g^{(n)}(x) = f^{(n)}(x) - \mathcal{M}n!$$. So we have to show $$g^{(n)}(c) = 0$$ for some $$c \in (a, b)$$. Consider $$g(a) = 0$$, $$g'(a) = 0$$, $$g''(a) = 0$$, upto $$g^{(n-1)}(a) = 0$$. Also, $$g(b) = 0$$. Since $$g(a) = g(b) = 0$$, there exists $$c_1 \in (a, b)$$ such that $$g'(c_1) = 0$$. Since $$g'(a) = g'(c_1) = 0$$, there exists $$c_2 \in (a, b)$$ such that $$g''(c_2) = 0$$, and so on. So, there exists $$c_{n - 1} \in (a, b)$$ such that $$g^{(n-1)}(c_{n-1}) = 0$$. Since $$g^{(n-1)}(a) = g^{(n-1)}(c_{n - 1}) = 0$$, there exists $$c \in (a, b)$$ such that $$g^{(n)}(c) = 0$$. This shows $$\mathcal{M} = \frac{f^{(n)}(c)}{n!}(x - a)^n$$, as desired. QED.