Research Article

Approximation Algorithm for a System of Pantograph Equations

Table 3

Comparison of the absolute errors for Example 3.3.

tExact solution 𝑢 1
𝑢 1 = c o s ( 𝑡 ) 𝑛 = 1 𝑛 = 2 𝑛 = 3

0.2 9 . 8 0 0 6 5 8 𝐸 1 2 . 1 2 4 𝐸 2 1 . 5 2 5 𝐸 4 8 . 9 0 4 𝐸 5
0.4 9 . 2 1 0 6 1 0 𝐸 1 8 . 9 0 8 𝐸 2 2 . 4 7 4 𝐸 3 1 . 5 1 1 𝐸 3
0.6 8 . 2 5 3 3 5 6 𝐸 1 2 . 0 7 4 𝐸 1 3 . 2 6 7 𝐸 2 8 . 0 5 1 𝐸 3
0.8 6 . 9 6 7 0 6 7 𝐸 1 3 . 7 6 4 𝐸 1 4 . 0 4 2 𝐸 2 2 . 6 6 5 𝐸 2
1.0 5 . 4 0 3 0 2 3 𝐸 1 5 . 9 2 0 𝐸 1 1 . 9 3 4 𝐸 1 6 . 7 6 6 𝐸 2

𝑢 2 = 𝑡 c o s ( 𝑡 ) 𝑢 2
𝑛 = 1 𝑛 = 2 𝑛 = 3

0.2 1 . 9 6 0 1 3 3 𝐸 1 1 . 3 2 9 𝐸 3 1 . 9 3 5 𝐸 4 5 . 4 9 6 𝐸 6
0.4 3 . 6 8 4 2 4 4 𝐸 1 1 . 0 5 2 𝐸 2 5 . 8 2 4 𝐸 3 1 . 8 0 8 𝐸 4
0.6 4 . 9 5 2 0 1 4 𝐸 1 3 . 4 8 9 𝐸 2 1 . 1 3 9 𝐸 2 1 . 4 0 8 𝐸 3
0.8 5 . 5 7 3 6 5 4 𝐸 1 8 . 0 7 1 𝐸 2 2 . 3 1 2 𝐸 2 6 . 0 6 9 𝐸 3
1.0 5 . 4 0 3 0 2 3 𝐸 1 1 . 5 2 8 𝐸 1 1 . 0 7 8 𝐸 1 1 . 8 9 0 𝐸 2

𝑢 3 = s i n ( 𝑡 ) 𝑢 3
𝑛 = 1 𝑛 = 2 𝑛 = 3

0.2 1 . 9 8 6 6 9 3 𝐸 1 2 . 7 2 8 5 𝐸 3 1 . 4 6 2 9 𝐸 3 6 . 4 5 5 8 𝐸 5
0.4 3 . 8 9 4 1 8 3 𝐸 1 2 . 2 2 4 5 𝐸 2 1 . 2 6 6 6 𝐸 2 9 . 9 5 9 5 𝐸 4
0.6 5 . 6 4 6 4 2 5 𝐸 1 7 . 6 2 0 9 𝐸 2 4 . 5 6 9 1 𝐸 2 4 . 8 3 9 7 𝐸 3
0.8 7 . 1 7 3 5 6 1 𝐸 1 1 . 8 2 6 4 𝐸 1 1 . 1 4 4 0 𝐸 1 1 . 4 6 1 3 𝐸 2
1.0 8 . 4 1 4 7 1 0 𝐸 1 3 . 5 9 3 0 𝐸 1 2 . 3 3 4 0 𝐸 1 3 . 3 9 1 7 𝐸 2