Research Article

Further Development of Jarratt Method for Solving Nonlinear Equations

Table 1

Test functions and their roots.

FunctionsRoots

𝑓 1 ( 𝑥 ) = e x p ( 𝑥 ) s i n ( 𝑥 ) + l n ( 1 + 𝑥 2 ) 𝛼 = 0
𝑓 2 ( 𝑥 ) = ( 𝑥 2 1 ) 1 1 𝛼 = 2
𝑓 3 ( 𝑥 ) = ( 𝑥 2 ) ( 𝑥 1 0 + 𝑥 + 1 ) e x p ( 𝑥 1 ) 𝛼 = 2
𝑓 4 ( 𝑥 ) = ( 𝑥 + 1 ) e x p ( s i n ( 𝑥 ) ) 𝑥 2 e x p ( c o s ( 𝑥 ) ) 1 𝛼 = 0
𝑓 5 ( 𝑥 ) = s i n ( 𝑥 ) 2 𝑥 2 + 1 𝛼 = 1 . 4 0 4 4 9 1 6 5
𝑓 6 ( 𝑥 ) = e x p ( 𝑥 ) c o s ( 𝑥 ) , 𝛼 = 0 . 6 6 6 2 7 3 1 2 6
𝑓 7 ( 𝑥 ) = l n ( 𝑥 2 + 𝑥 + 2 ) 𝑥 + 1 𝛼 = 4 . 1 5 2 5 9 0 7 4
𝑓 8 ( 𝑥 ) = 𝑥 1 0 2 𝑥 3 𝑥 + 1 𝛼 = 0 . 5 9 1 4 4 8 0 9 3
𝑓 9 ( 𝑥 ) = c o s ( 𝑥 ) 2 5 1 𝑥 𝛼 = 1 . 0 8 5 9 8 2 6 8
𝑓 1 0 ( 𝑥 ) = s i n ( 𝑥 ) 2 1 𝑥 𝛼 = 0