202010022249 Problem 2.1.10

In $C[0,1]$ , determine the values of $d_{\infty}(x,y)$ and $d_{1}(x,y)$ , when

(a) $x(t)=t^3+t+1$ and $y(t)=t^3+t^2+\frac{1}{2}t+1$ ;

(b) $x(t)=\sin t$ and $y(t)=t$ ;
(c) $x(t)=\sin t$ and $y(t)=t-\displaystyle{\frac{t^3}{6}}$ ;
(d) $x(t)=\mathrm{exp}(t)$ and $y(t)=\displaystyle{\sum_{m=0}^n}\frac{t^m}{m!}$ .

Recall.

Definition. (uniform metric) Let $C[a,b]$ be the set of all real-valued continuous functions defined on $[a,b]$ . For any $x,y\in C[a,b]$ , define the uniform metric $d_{\infty}$ :

$d_{\infty}(x,y)=\displaystyle{\max_{t\in [a,b]}}|x(t)-y(t)|$ .

N.b. If we let $B[a,b]$ be the set of all real-valued functions defined and bounded on $[a,b]$ , the uniform metric is then defined

$d_{\infty}(x,y)=\displaystyle{\sup_{t\in [a,b]}}|x(t)-y(t)|$ .

_{(cited from Examples 14 and 15, pg. 13, Pawan K. Jain and Khalil Ahmad’s Metric Spaces (2e) on Introductory Concepts)}

Definition. For any $x, y\in C[a,b]$ , define

$d_1(x,y)=\displaystyle{\int_a^b}|x(t)-y(t)|\,\mathrm{d}t$

N.b. $d_1(x,y)$ represents the absolute area between the functions $x$ and $y$ as a measure of the distance between these two functions.

_{(cited from Example 16, pg. 14, Pawan K. Jain and Khalil Ahmad’s Metric Spaces (2e) on Introductory Concepts)}

Solution.

(a)

$\begin{aligned} & \quad\, d_{\infty} (x,y) \\ &= \max_{t\in [0,1]} |x(t)-y(t)| \\ \end{aligned}$

Roughwork.

Approach.

To know the maximum value of $|x(t)-y(t)|$ , apply differentiation to

$f(t)=-t^2+\displaystyle{\frac{1}{2}}t$

and attain

$\begin{aligned} f(t) & = -t^2 + \frac{1}{2}t \\ f'(t) & = -2t+\frac{1}{2} \\ f'(t) & = 0 \Leftrightarrow t=\frac{1}{4} \\ \end{aligned}$

If the quadratic function $f(t)$ is plotted in a graph, a parabola admits of no inflexion points, needless to check on $f''(x)=0$ . So,

$\begin{array}{c|c|c|c|c|c} & t=0 & 0<t<\frac{1}{4} & t=\frac{1}{4} & \frac{1}{4}<t<1 & t=1 \\ &&&&&\\ \hline &&&&&\\ f(t) & 0 & \dots & \displaystyle{\frac{1}{16}} & \dots & -\displaystyle{\frac{1}{2}} \\ &&&&&\\ \hline &&&&&\\ f'(t) & \displaystyle{\frac{1}{2}}\enspace (>0) & \dots & 0\enspace (=0) & \dots & -\displaystyle{\frac{3}{2}}\enspace (<0) \\ &&&&&\\ \hline &&&&&\\ \textrm{plot} & \diagup & \dots & --- & \dots & \diagdown \\ &&&&&\\ \end{array}$

The continuous function $f(t)$ in the closed interval $[0,1]$ attains its maximum value $f(\frac{1}{4})=\frac{1}{16}$ when $t=\frac{1}{4}$ .

$\begin{aligned} & d_{\infty} (x,y) \\ = & \max_{t\in [0,1]} |x(t)-y(t)| \\ = & \max_{t\in [0,1]} |f(t)| \\ = & \frac{1}{16} \qquad\qquad\qquad \checkmark\\ \end{aligned}$

and should you think of what follows as quite right

$\begin{aligned} d_1(x,y) & = \int_0^1 |x(t)-y(t)| \,\mathrm{d}t \\ & = \int_0^1 f(t)\,\mathrm{d}t \\ & = \int_0^1 \bigg| -t^2 + \frac{1}{2}t \bigg| \, \mathrm{d}t \\ & = \bigg[ -\frac{t^3}{3} + \frac{t^2}{4} \bigg] \bigg|_0^1 \\ & = -\frac{1}{12}\qquad\qquad\qquad \times \\ \end{aligned}$

you might have rather mistaken calculus.

Correction.

Get back to the basics,

$\begin{aligned} f(t) & = 0 \\ \bigg| -t^2+\frac{1}{2}t \bigg| & = 0 \\ t^2-\frac{1}{2}t &= 0 \\ (t-\frac{1}{2})t & = 0 \\ t & = 0\quad \textrm{\scriptsize{OR}}\quad \frac{1}{2} \\ \end{aligned}$

From the previous graph of $C[0,1]$ , $f(t)$ is found to be positive when $t\in (0,0.5)$ , zero when $t\in \{ 0\} \cup\{ 0.5\}$ , and negative when $t\in (0.5,1]$ .

Doing it step-by-step,

$\begin{aligned} d_1(x,y) & = \int_0^1 \bigg| -t^2+\frac{1}{2}t \bigg| \,\mathrm{d}t \\ & = \int_{0}^{0.5}\mathrm{d}t\enspace \bigg| -t^2+\frac{1}{2}t \bigg| + \int_{0.5}^{1}\mathrm{d}t \enspace \bigg| -t^2+\frac{1}{2}t \bigg| \\ & = \int_{0}^{0.5}\mathrm{d}t\enspace \bigg( - t^2 + \frac{1}{2}t \bigg) + \int_{0.5}^{1}\mathrm{d}t \enspace \bigg( t^2-\frac{1}{2}t\bigg) \\ \end{aligned}$

Evaluating term-by-term, the first term being

$\begin{aligned} & \int_{0}^{0.5} \bigg( -t^2 + \frac{1}{2}t \bigg) \,\mathrm{d}t \\ = & \bigg[ -\frac{t^3}{3} + \frac{t^2}{4} \bigg] \bigg|_{0}^{0.5} \\ = & \bigg[ -\frac{(0.5)^3}{3} + \frac{(0.5)^2}{4} \bigg] - \bigg[ -\frac{(0)^3}{3} + \frac{(0)^2}{4} \bigg] \\ \dots & \enspace \textrm{by arithmetic}\enspace \dots \\ = & \frac{1}{48} \\ \end{aligned}$

and the second term being

$\begin{aligned} & \int_{0.5}^{1}\bigg( t^2-\frac{1}{2}t\bigg) \,\mathrm{d}t \\ = & \bigg[ \frac{t^3}{3} - \frac{t^2}{4} \bigg]\bigg|_{0.5}^{1} \\ = & \bigg[ \frac{(1)^3}{3} - \frac{(1)^2}{4} \bigg] - \bigg[ \frac{(0.5)^3}{3} - \frac{(0.5)^2}{4} \bigg] \\ \dots & \enspace \textrm{by arithmetic}\enspace \dots \\ = & \bigg( \frac{1}{12} \bigg) - \bigg( -\frac{1}{48} \bigg) \\ = & \frac{5}{48} \\ \end{aligned}$

In sum,

$d_{1}(x,y)=\displaystyle{\frac{1}{48}+\frac{5}{48}=\frac{6}{48}=\frac{1}{8}}=0.125$ .

Part (b), (c), and (d) are not chosen.

M	T	W	T	F	S	S
			1	2	3	4
5	6	7	8	9	10	11
12	13	14	15	16	17	18
19	20	21	22	23	24	25
26	27	28	29	30	31

Leave a Reply Cancel reply