Numerical Analysis By Shanker Rao

36 NUMERICAL ANALYSIS
The second approximation of the root is
x
f x1
= x −
f ′ x
2 1
bg
bg 1
∴ x 2 = 0. 73250699 ≈ 0. 7321 (correct to four decimal places).
0.
0019805
= 07316 . +
4.
39428
The root of the equation = 0.7321 (approximately).
Example 2.11 By applying Newton’s method twice, find the real root near 2 of the equation x 4 – 12x + 7 = 0.
Solution
Let
4
fbg x1
= x − 12x
+ 7
3
∴ f ′ bg x = 4x
−12
Here x 0
= 2
4
bg 0 bg .
3
b 0g bg bg
∴ f x = f 2 = 2 − 122 + 7 = −1
f ′ x = f ′ 2 = 4 2 − 12 = 20
∴ x = x −
1 0
bg
f x1
and x2 = x1
−
f ′ x
∴ The root of the equation is 2.6706.
= 205 . −
b g
.
f x0
f ′ bg x
= − − 1 41
2 = = 205
0 20 20
bg
bg 1
b
4
g b g
2
2. 05 − 12 2.
05 + 7
= 26706 .
b g
4 2.
05 − 12
Example 2.12 Find the Newton’s method, the root of the e x = 4x, which is approximately 2, correct to three places
of decimals.
Solution Here
f(x) = e x – 4x
f(2) = e 2 – 8 = 7.389056 – 8 = – 0.610944 = – ve
f(3) = e 3 – 12 = 20.085537 – 12 = 8.085537 = + ve
∴ f(2) f(3) < 0
∴ f(x) = 0 has a root between 2 and 3.
Let x 0
= 2.1 be the approximate value of the root
f(x) = e x – 4x
bg 0
21 .
bg 4
bg
21 .
bg 0
⇒ f ′ x = e x −
∴ f x = e − 4 21 . = 816617 . − 8. 4 = −0.
23383
f ′ x = e − 4 = 416617 . .