Question

If I: = [0,4], calculate the norms of the following partitions:

a) P1: = (0,1,2,4)

b) P2: = (0,2,3,4)

c) P3: = (0,1,1.5,2,3.4,4)

d) P4: = (0,.5,2.5,3.5,4)

Answer 1

Solution: The norm of a partition is merely the length of the largest subinterval into which the partition divides [a,b]. Clearly many partition have the same norm, so partition is not a function of the norm.

for $i=[a,b]$ where $a<x_1<x_2<....<x_n<b,$ partition $p={x_1,x_2,....,x_n}$ for above condition $norm=max\{x_2-x_1,x_3-x_2,...,x_n-x_{n-1}\}$

a)

$\therefore x_2-x_1=1-0=1 \\x_3-x_2=2-1=1 \\x_4-x_3=4-2=2 \\Hence ~ norm=max\{x_2-x_1,x_3-x_2, x_4-x_3\} \\norm=max=\{1,1,2\} \\norm=2$

Hence norm of the partition P1: = (0,1,2,4) is 2.

b)

$\therefore x_2-x_1=2-0=2 \\x_3-x_2=3-2=1 \\x_4-x_3=4-3=1 \\Hence ~ norm=max\{x_2-x_1,x_3-x_2, x_4-x_3\} \\norm=max=\{2,1,1\} \\norm=2$

Hence norm of the partition  P2: = (0,2,3,4) is 2.

c)

 $\therefore x_2-x_1=1-0=1 \\x_3-x_2=1.5-1=0.5 \\x_4-x_3=2-1.5=0.5 \\x_5-x_4=3.4-2=1.4 \\x_6-x_5=4-3.4=0.6 \\Hence ~ norm=max\{x_2-x_1,x_3-x_2, x_4-x_3,x_5-x_4,x_6-x_5\} \\norm=max=\{1,0.5,0.5,1.4,0.6\} \\norm=1.4$

Hence norm of the partition  P3: = (0,1,1.5,2,3.4,4) is 1.4.

d)

 $\therefore x_2-x_1=0.5-0=0.5 \\x_3-x_2=2.5-0.5=2 \\x_4-x_3=3.5-2.5=1 \\x_5-x_4=4-3.5=0.5 \\Hence ~ norm=max\{x_2-x_1,x_3-x_2, x_4-x_3,x_5-x_4\} \\norm=max=\{0.5,2,1,0.5\} \\norm=2$

Hence norm of the partition  P4: = (0,0.5,2.5,3.5,4) is 2.