Numerical Analysis By Shanker Rao

28 NUMERICAL ANALYSIS
F ⇒ ⋅H G πI fbg 0 K J <
f
F πI π π 3π
3 1
HG
2 K J = cos − + = − + <
2 2 2
2
0
1 0
∴ A root of f(x) = 0 lies between 0 and π 2 .
1
The given equation can be written as x = + x
3 1 cos .
Here
1
φbg x = + x
3 1 cos
Iteration method can be applied
be the initial approximation.
∴ φ′ bg x = − sin x
3
sin x
bg
3
F π
⇒ ′ = < H G I
φ x
in , K J
Let x 0
= 0
1 0 2
φb g cos .
∴ We get x1 = 1
x0
= 3 1 + 0 = 0 66667
x
x
x
x
x
x
φb g cos . . .
= 1
x = 1 0 66667 0 595295 0 59530
3 + = ≈
2 1
φb g cos . . .
= 1
x = 1 0 59530 0 6093267 0 60933
3 + = ≈
3 2
φb g cos . . .
= 1
x = 1 0 60933 0 6066772 0 60668
3 + = ≈
4 3
φb g cos . . .
= 1
x = 1 0 60668 0 6071818 0 60718
3 + = ≈
5 4
φb g cos . . .
= 1
x = 1 0 60718 0 6070867 0 60709
3 + = ≈
6 5
φb g cos . . .
= 1
x = 1 0 60709 0 6071039 0 60710
3 + = ≈
7 6
x
φb g cos . .
= 1
x = 1 0 60710 0 60710
3 + =
8 7
The correct root of the equation is 0.607 correct to three decimal places.