Chapter 5.4, Problem 1E

(a)

To determine

To calculate: The length of each subinterval of the numbers $\left\{0,1.1,2.6,3.7,4.1,5\right\}$ which determines the partition P of an interval.

Answer to Problem 1E

The length of each subinterval of the numbers $\left\{0,1.1,2.6,3.7,4.1,5\right\}$ is $\Delta x_1=1.1,\Delta x_2=1.5,\Delta x_3=1.1,\Delta x_4=0.4,\Delta x_5=0.9$ .

Explanation of Solution

Given information:

The numbers $\left\{0,1.1,2.6,3.7,4.1,5\right\}$ which determines the partition P of an interval.

Formula used:

For a closed interval of the form $\left[c,d\right]$ , a partition P is decomposition of the interval $\left[c,d\right]$ into subintervals like $\left[x_0,x_1\right],\left[x_1,x_2\right],\dots ,\left[x_{n-1},x_n\right]$ , where n is positive integer such that

  $c=x_0<x_1<x_2<\dots <x_{n-1}<x_n=d$

Consider a $k^{th}$ subinterval $\left[x_{k-1},x_k\right]$ , then length of this interval is denoted by $\Delta x_k$ and is calculated as $\Delta x_k=x_k-x_{k-1}$ .

Calculation:

Consider the numbers $\left\{0,1.1,2.6,3.7,4.1,5\right\}$ which determines the partition P of an interval.

Recall that for a closed interval of the form $\left[c,d\right]$ , a partition P is decomposition of the interval $\left[c,d\right]$ into subintervals like $\left[x_0,x_1\right],\left[x_1,x_2\right],\dots ,\left[x_{n-1},x_n\right]$ , where n is positive integer such that

  $c=x_0<x_1<x_2<\dots <x_{n-1}<x_n=d$

Consider a $k^{th}$ subinterval $\left[x_{k-1},x_k\right]$ , then length of this interval is denoted by $\Delta x_k$ and is calculated as $\Delta x_k=x_k-x_{k-1}$ .

Denote the numbers $\left\{0,1.1,2.6,3.7,4.1,5\right\}$ as $\left\{x_1,x_2,x_3,x_4,x_5,x_6\right\}$ .

Now, the length of the subinterval $\Delta x_1$ is,

  $\begin{array}{c}\Delta x_1=x_2-x_1\\ =1.1-0\\ =1.1\end{array}$

Therefore, the length of the subinterval $\Delta x_1=1.1$ .

Next, the length of the subinterval $\Delta x_2$ is,

  $\begin{array}{c}\Delta x_2=x_3-x_2\\ =2.6-1.1\\ =1.5\end{array}$

Therefore, the length of the subinterval $\Delta x_2=1.5$ .

Next, the length of the subinterval $\Delta x_3$ is,

  $\begin{array}{c}\Delta x_3=x_4-x_3\\ =3.7-2.6\\ =1.1\end{array}$

Therefore, the length of the subinterval $\Delta x_3=1.1$ .

Next, the length of the subinterval $\Delta x_4$ is,

  $\begin{array}{c}\Delta x_4=x_5-x_4\\ =4.1-3.7\\ =0.4\end{array}$

Therefore, the length of the subinterval $\Delta x_4=0.4$ .

Last, the length of the subinterval $\Delta x_5$ is,

  $\begin{array}{c}\Delta x_5=x_5-x_4\\ =5-4.1\\ =0.9\end{array}$

Therefore, the length of the subinterval $\Delta x_5=0.9$ .

Thus, the length of each subinterval of the numbers $\left\{0,1.1,2.6,3.7,4.1,5\right\}$ is $\Delta x_1=1.1,\Delta x_2=1.5,\Delta x_3=1.1,\Delta x_4=0.4,\Delta x_5=0.9$ .

(b)

To determine

To calculate: The norm $\Vert P\Vert$ of the partition where numbers $\left\{0,1.1,2.6,3.7,4.1,5\right\}$ determines the partition P of an interval.

Answer to Problem 1E

The norm $\Vert P\Vert$ of the partition is $\Delta x_2=1.5$ .

Explanation of Solution

Given information:

The numbers $\left\{0,1.1,2.6,3.7,4.1,5\right\}$ which determines the partition P of an interval.

Formula used:

For a closed interval of the form $\left[c,d\right]$ , a partition P is decomposition of the interval $\left[c,d\right]$ into subintervals like $\left[x_0,x_1\right],\left[x_1,x_2\right],\dots ,\left[x_{n-1},x_n\right]$ , where n is positive integer such that

  $c=x_0<x_1<x_2<\dots <x_{n-1}<x_n=d$

Consider a $k^{th}$ subinterval $\left[x_{k-1},x_k\right]$ , then length of this interval is denoted by $\Delta x_k$ and is calculated as $\Delta x_k=x_k-x_{k-1}$ .

The largest number among all $\Delta x_1,\Delta x_2,\dots \Delta x_n$ is the norm $\Vert P\Vert$ of the partition P .

Calculation:

Consider the numbers $\left\{0,1.1,2.6,3.7,4.1,5\right\}$ which determines the partition P of an interval.

Recall that for a closed interval of the form $\left[c,d\right]$ , a partition P is decomposition of the interval $\left[c,d\right]$ into subintervals like $\left[x_0,x_1\right],\left[x_1,x_2\right],\dots ,\left[x_{n-1},x_n\right]$ , where n is positive integer such that

  $c=x_0<x_1<x_2<\dots <x_{n-1}<x_n=d$

Consider a $k^{th}$ subinterval $\left[x_{k-1},x_k\right]$ , then length of this interval is denoted by $\Delta x_k$ and is calculated as $\Delta x_k=x_k-x_{k-1}$ .

Denote the numbers $\left\{0,1.1,2.6,3.7,4.1,5\right\}$ as $\left\{x_1,x_2,x_3,x_4,x_5,x_6\right\}$ .

Now, the length of the subinterval $\Delta x_1$ is,

  $\begin{array}{c}\Delta x_1=x_2-x_1\\ =1.1-0\\ =1.1\end{array}$

Therefore, the length of the subinterval $\Delta x_1=1.1$ .

Next, the length of the subinterval $\Delta x_2$ is,

  $\begin{array}{c}\Delta x_2=x_3-x_2\\ =2.6-1.1\\ =1.5\end{array}$

Therefore, the length of the subinterval $\Delta x_2=1.5$ .

Next, the length of the subinterval $\Delta x_3$ is,

  $\begin{array}{c}\Delta x_3=x_4-x_3\\ =3.7-2.6\\ =1.1\end{array}$

Therefore, the length of the subinterval $\Delta x_3=1.1$ .

Next, the length of the subinterval $\Delta x_4$ is,

  $\begin{array}{c}\Delta x_4=x_5-x_4\\ =4.1-3.7\\ =0.4\end{array}$

Therefore, the length of the subinterval $\Delta x_4=0.4$ .

Last, the length of the subinterval $\Delta x_5$ is,

  $\begin{array}{c}\Delta x_5=x_5-x_4\\ =5-4.1\\ =0.9\end{array}$

Therefore, the length of the subinterval $\Delta x_5=0.9$ .

Hence, the length of each subinterval of the numbers $\left\{0,1.1,2.6,3.7,4.1,5\right\}$ is $\Delta x_1=1.1,\Delta x_2=1.5,\Delta x_3=1.1,\Delta x_4=0.4,\Delta x_5=0.9$ .

Recall that the largest number among all $\Delta x_1,\Delta x_2,\dots \Delta x_n$ is the norm $\Vert P\Vert$ of the partition P .

Therefore, highest number is $\Delta x_2=1.5$ .

Thus, the norm $\Vert P\Vert$ of the partition is $\Delta x_2=1.5$ .