Exercises · 9. Differentiation

Easy

1Easy

Let $f$ be continuous on $[a,b]$ and differentiable on $(a,b)$ , with $\lvert f'(x) \rvert \le M$ for all $x \in (a,b)$ and some constant $M \ge 0$ . Prove that $\lvert f(x) - f(y) \rvert \le M\lvert x - y \rvert$ for all $x, y \in [a,b]$ . (Hence $f$ is uniformly continuous on $[a,b]$ .)

Show solution

The case $x = y$ is trivial, so let $x < y$ . Since $f$ is continuous on $[x,y] \subset [a,b]$ and differentiable on $(x,y) \subset (a,b)$ , the Mean Value Theorem applies on $[x,y]$ : for some $c \in (x,y)$ , $f(y) - f(x) = f'(c)(y - x)$ .
Taking absolute values, $\lvert f(y) - f(x) \rvert = \lvert f'(c) \rvert \cdot \lvert y - x \rvert \le M\lvert y - x \rvert$ . The case $x > y$ is the same with the roles swapped, so the inequality holds for all $x, y \in [a,b]$ . This is the Lipschitz condition, and uniform continuity follows with $\delta = \varepsilon / M$ (any $\delta > 0$ if $M = 0$ ).

2Easy

Start Newton's iteration for $x^2 - 2 = 0$ , $x_{n+1} = x_n - \dfrac{x_n^2 - 2}{2x_n} = \dfrac{x_n + 2/x_n}{2}$ , at $x_0 = 1$ . (a) Show $x_n > 0$ for all $n$ . (b) Use AM-GM to show $x_{n+1} \ge \sqrt{2}$ for all $n$ . (c) Show $x_1 > \sqrt{2}$ and that $(x_n)_{n \ge 1}$ is decreasing. (d) Conclude $x_n \to \sqrt{2}$ .

x_{n+1} = \frac{x_n + 2/x_n}{2}, \qquad x_0 = 1, \quad x_1 = \frac32

Riding Tangents to the Root

Starting at

x_0 = 1

: the first tangent overshoots

\sqrt{2}

to

x_1 = 3/2

, then the iterates descend monotonically to the root.

Show solution

(a) $x_0 = 1 > 0$ , and if $x_n > 0$ then $x_{n+1}$ is the average of two positive numbers, hence positive — by induction $x_n > 0$ for all $n$ . (b) By AM-GM, $x_{n+1} = \dfrac{x_n + 2/x_n}{2} \ge \sqrt{x_n \cdot \dfrac{2}{x_n}} = \sqrt{2}$ .
(c) $x_1 = \dfrac{1 + 2}{2} = \dfrac32$ , and $\left(\dfrac32\right)^2 = \dfrac94 > 2$ , so $x_1 > \sqrt{2}$ — the first step overshoots the root. For $n \ge 1$ , $x_{n+1} - x_n = \dfrac{2/x_n - x_n}{2} = \dfrac{2 - x_n^2}{2x_n}$ ; by (b), $x_n \ge \sqrt{2}$ , so the numerator is $\le 0$ . Hence $(x_n)_{n \ge 1}$ is decreasing.
(d) $(x_n)_{n \ge 1}$ is decreasing and bounded below by $\sqrt{2}$ , so it converges to some $L \ge \sqrt{2}$ . Passing to the limit in the recurrence gives $L = \dfrac{L + 2/L}{2}$ , i.e. $2L = L + 2/L$ , so $L = 2/L$ . Since $L > 0$ , $L^2 = 2$ , i.e. $L = \sqrt{2}$ .

3Easy

Prove via the triangle inequality that $f(x) = \lvert x \rvert$ is convex on $\mathbb{R}$ , and cite the continuity theorem for convex functions to conclude $f$ is continuous. Show $f$ is not differentiable at $0$ , confirming that convexity does not imply differentiability.

\lvert tx + (1-t)y \rvert \le t\lvert x \rvert + (1-t)\lvert y \rvert \qquad (0 \le t \le 1)

Convex Functions

Show solution

For $0 \le t \le 1$ and $x, y \in \mathbb{R}$ , the triangle inequality and homogeneity of the absolute value give $\lvert tx + (1-t)y \rvert \le \lvert tx \rvert + \lvert (1-t)y \rvert = t\lvert x \rvert + (1-t)\lvert y \rvert$ . This is exactly the definition of convexity, so $f$ is convex on $\mathbb{R}$ . A convex function is continuous on every open interval (the theorem of the convex scene), so $f$ is continuous everywhere.
But the difference quotient at $0$ is $\dfrac{f(h) - f(0)}{h} = \dfrac{\lvert h \rvert}{h} = \begin{cases} 1 & h > 0 \\ -1 & h < 0 \end{cases}$ , which has no limit. Hence $f$ is not differentiable at $0$ — the simplest example of a function that is convex and continuous but not differentiable.

Medium

4Medium

Let $f$ be differentiable on $[a,b]$ with $f'(x) \ne 0$ for every $x \in [a,b]$ . Combining Darboux's theorem with the Mean Value Theorem, prove that $f$ is strictly monotone on $[a,b]$ .

Show solution

By Darboux's theorem, $f'$ has the intermediate value property. Suppose $f'$ changed sign: there exist $u, v \in [a,b]$ with $f'(u) > 0 > f'(v)$ . Applying Darboux's theorem on the interval between $u$ and $v$ would produce a point $c$ with $f'(c) = 0$ , contradicting the hypothesis. Hence $f'$ cannot change sign, and since it never vanishes, it is either everywhere positive or everywhere negative on $[a,b]$ .
Consider the case $f' > 0$ (the negative case differs only in sign). For $x < y$ , the Mean Value Theorem gives $c \in (x,y)$ with $f(y) - f(x) = f'(c)(y - x)$ ; since $f'(c) > 0$ , we get $f(y) > f(x)$ . So $f$ is strictly increasing. Likewise $f' < 0$ makes $f$ strictly decreasing.

5Medium

Let $f$ be $n+1$ times continuously differentiable on $[a,b]$ ( $n \ge 0$ ). For $a \le x \le b$ , prove Taylor's formula with integral remainder \[f(x) = \sum_{k=0}^{n} \frac{f^{(k)}(a)}{k!}(x-a)^k + R_n(x), \qquad R_n(x) = \frac{1}{n!}\int_a^x (x-t)^n f^{(n+1)}(t)\,dt.\] The case $n = 0$ is the Fundamental Theorem of Calculus; for the induction step apply integration by parts once.

Degree

n = 5

: the band trapping the remainder

R_5(x)

— the integral remainder is the exact formula measuring this error.

Show solution

For $n = 0$ the formula is $f(x) = f(a) + \int_a^x f'(t)\,dt$ , the Fundamental Theorem of Calculus. As the induction hypothesis, assume the formula for $n$ holds for $C^{n+1}$ functions. For $g \in C^{n+2}$ , applying it to $g$ gives $g(x) = T_n(x) + \frac{1}{n!}\int_a^x (x-t)^n g^{(n+1)}(t)\,dt$ , where $T_n$ is the degree- $n$ Taylor polynomial.
Apply integration by parts to the remainder integral: with $u = g^{(n+1)}(t)$ and $dv = (x-t)^n\,dt$ , we have $du = g^{(n+2)}(t)\,dt$ and $v = -\dfrac{(x-t)^{n+1}}{n+1}$ . Since $g^{(n+1)}$ and $(x-t)^{n+1}$ are continuous, parts is legitimate and \[\int_a^x (x-t)^n g^{(n+1)}(t)\,dt = \left[-\frac{(x-t)^{n+1}}{n+1}\,g^{(n+1)}(t)\right]_a^x + \frac{1}{n+1}\int_a^x (x-t)^{n+1} g^{(n+2)}(t)\,dt.\] The boundary term is $0$ at $t = x$ and gives $\dfrac{(x-a)^{n+1}}{n+1}\,g^{(n+1)}(a)$ at $t = a$ (the minus sign flips).
Multiplying by $\frac{1}{n!}$ , the remainder integral becomes $\dfrac{g^{(n+1)}(a)}{(n+1)!}(x-a)^{n+1} + \dfrac{1}{(n+1)!}\int_a^x (x-t)^{n+1} g^{(n+2)}(t)\,dt$ . The first term is exactly the degree- $(n+1)$ Taylor term, and the second is $R_{n+1}(x)$ . Hence $g(x) = T_{n+1}(x) + R_{n+1}(x)$ , which is the formula for $n+1$ . By induction the proof is complete for all $n \ge 0$ .

6Medium

Consider the map $T(x) = \dfrac{1}{2+x}$ on $[0,1]$ . (a) Show $T$ maps $[0,1]$ into itself and is a contraction with constant $c = 1/4$ . (b) Cite the Banach fixed point theorem to find the unique fixed point of $T$ (solve the quadratic), and explain why iteration from any starting point converges to it.

\lvert T(x) - T(y) \rvert = \frac{\lvert x - y \rvert}{(2+x)(2+y)} \le \frac14\,\lvert x - y \rvert

x = \frac{1}{2+x} \iff x^2 + 2x - 1 = 0 \iff x = \sqrt{2} - 1 \approx 0.4142

The Contraction Cobweb

Show solution

(a) $T$ is decreasing on $[0,1]$ with $T(0) = 1/2$ and $T(1) = 1/3$ , so $T([0,1]) = [1/3, 1/2] \subset [0,1]$ . For $x, y \in [0,1]$ , $\lvert T(x) - T(y) \rvert = \left\lvert \dfrac{2+y-(2+x)}{(2+x)(2+y)} \right\rvert = \dfrac{\lvert x - y \rvert}{(2+x)(2+y)}$ . Since $(2+x)(2+y) \ge 2 \cdot 2 = 4$ , we get $\lvert T(x) - T(y) \rvert \le \dfrac14\lvert x - y \rvert$ — a contraction with constant $1/4$ .
(b) $[0,1]$ is complete, so the Banach fixed point theorem (the Banach scene) gives $T$ a unique fixed point $x^\ast \in [0,1]$ . The equation $x = \dfrac{1}{2+x}$ is equivalent to $x^2 + 2x - 1 = 0$ , with roots $x = -1 \pm \sqrt{2}$ . The only one in $[0,1]$ is $x^\ast = \sqrt{2} - 1 \approx 0.4142$ . By the same theorem, the iteration $x_{n+1} = T(x_n)$ from any $x_0 \in [0,1]$ converges to $x^\ast$ — the error shrinks by at least a factor of $1/4$ each step.

Hard

7Hard

Let $f(x) = e^{-1/x^2}$ for $x \ne 0$ and $f(0) = 0$ . Prove $f$ is infinitely differentiable on $\mathbb{R}$ with $f^{(n)}(0) = 0$ for every $n \ge 0$ . Deduce that the Maclaurin series of $f$ converges for every $x$ but equals $f(x)$ only at $x = 0$ — a Taylor series need not recover the function.

f^{(n)}(x) = P_n(1/x)\,e^{-1/x^2} \quad (x \ne 0),\qquad \lim_{\lvert u \rvert \to \infty} P(u)e^{-u^2} = 0

T(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}x^n \equiv 0 \quad\text{but}\quad f(x) > 0\ (x \ne 0)

Show solution

Lemma: for any polynomial $P$ , $\lim_{\lvert u \rvert \to \infty} P(u)e^{-u^2} = 0$ . Indeed, for $v = u^2 \ge 0$ , $e^{v} \ge \dfrac{v^m}{m!}$ (one term of the exponential's Taylor expansion), so with $2m > \deg P$ we get $\lvert P(u) \rvert e^{-u^2} \le m!\,\dfrac{\lvert P(u) \rvert}{u^{2m}} \to 0$ .
We claim $f^{(n)}(x) = P_n(1/x)e^{-1/x^2}$ for $x \ne 0$ , for some polynomial $P_n$ , by induction. For $n = 0$ , $P_0 \equiv 1$ . With $u = 1/x$ , \[\frac{d}{dx}\Bigl[P(u)e^{-1/x^2}\Bigr] = P'(u)\!\left(-\frac{1}{x^2}\right)e^{-1/x^2} + P(u)e^{-1/x^2}\!\left(\frac{2}{x^3}\right) = \bigl[-u^2P'(u) + 2u^3P(u)\bigr]e^{-u^2},\] and the bracket is a polynomial in $u$ , completing the induction step. Hence $f$ is infinitely differentiable at $x \ne 0$ .
Now $f^{(n)}(0) = 0$ by induction. The case $n = 0$ is the definition. $f'(0) = \lim_{h \to 0} \dfrac{e^{-1/h^2}}{h} = \lim_{\lvert u \rvert \to \infty} u e^{-u^2} = 0$ (lemma, with $u = 1/h$ ). Assuming $f^{(n)}(0) = 0$ , \[f^{(n+1)}(0) = \lim_{h \to 0}\frac{f^{(n)}(h) - f^{(n)}(0)}{h} = \lim_{h \to 0}\frac{P_n(1/h)e^{-1/h^2}}{h} = \lim_{\lvert u \rvert \to \infty} uP_n(u)e^{-u^2} = 0\] (lemma applied to $uP_n(u)$ ). Thus $f^{(n)}(0) = 0$ for all $n$ , and $f$ is infinitely differentiable at $0$ too.
The Maclaurin series is $T(x) = \sum_{n=0}^{\infty} \dfrac{f^{(n)}(0)}{n!}x^n \equiv 0$ — it converges (to $0$ ) at every $x$ . But for $x \ne 0$ , $f(x) = e^{-1/x^2} > 0$ . So the Taylor series converges everywhere yet matches $f$ only at $x = 0$ : even a smooth function need not be recovered by its Taylor series.