Numerical Analysis By Shanker Rao

30 NUMERICAL ANALYSIS
Similarly, x 5
= 1.4250
x 6
= 1.4340
x 7
= 1.4390
x 8
= 1.442
Correct to 2 decimal places, we get
x 7
= 1.44 and x 8
= 1.44
∴ The root of (1) is 1.44 (approximately).
2.4.1 Aitken’s ∆ 2 Method
Let x =α be a root of the equation
f(x) = 0 (1)
and let I be an interval containing the point x =α. The equation (1) can be written as
x =φbg x
such that φbg x and φ′
bg x are continuous in I and φ′ bg x < 1 for all x in I. Let
x i – 1
, x i
and x i + 1
be three successive approximations of the desired root α. Then we know that
d i
1 b xig
α − x = λ α − x −
i i 1
and α − x = λ α −
i+
( λ is a constant such that φ′ ≤ λ <
bg x i 1 for all i )
dividing we get
α − xi
α − x
i + 1
α − xi
=
α − x
⇒ α = x −
i + 1
−1
i
d
d
x
i+
1
− x
x − 2x − x
i+ 1 i i−1
i
i
2
i
(2)
Since ∆ x = x − x
i i+1
i
and ∆ 2 x −1
= E −1
x
(2) can be written as
i
b
d
2
g
2
i−1
= E − 2E + 1 x i −
= x − 2 x + x
α= x −
i+ 1 i i−1
i + 1
i −1
i
1
2
b∆
xig 2
(3)
∆ x