Numerical Analysis By Shanker Rao

32 NUMERICAL ANALYSIS
constructing the table, we have
x ∆x ∆ 2
x 1
= 0.667
x 2
= 0.5953
x 3
= 0.6093
−0.
0714
∆x
1
0.
014
∆ x
2
0.
0854
2
∆ x
1
2
2
b g b0014
. g
∆ x2
Hence x4 = x3
− = 06093 . −
2
∆ x
= 0.607
∴ The required root is 0.607.
1
b
g
00854 .
Exercise 2.3
1. Find the real root of the equation x 3 + x – 1 = 0 by the iteration method.
2. Solve the equation x 3 – 2x 2 – 5 = 0 by the method of iteration.
3. Find a real root of the equation, x 3 + x 2 – 100 = 0 by the method of successive approximations (the
iteration method).
4. Find the negative root of the equation x 3 – 2x + 5 = 0.
5. Find the real root of the equation x – sin x = 0.25 to three significant digits.
6. Find the root of x 2 = sin x, which lies between 0.5 and 1 correct to four decimals.
7. Find the real root of the equation x 3 – 5x – 11 = 0.
8. Find the real root of the equation
b g n
x 3 x 5 x 7 x 9 x
11
−1
x − + − + − + ... + −1
3 10 42 216 1320
2n
− 1
9. Find the smallest root of the equation by iteration method
2
3
x x x x
1 − x + − + − + ... = 0
2 2 2 2
( 2!) ( 3!) ( 4!) ( 5!)
4
5
x
+ ... = 0.
4431135
( n−1)!( 2n−1)
10. Use iteration method to find a root of the equations to four decimal places.
(i) x 3 + x 2 – 1 = 0
(ii) x = 1/2 + sin x and
(iii) e x – 3x = 0, lying between 0 and 1.
(iv) x 3 – 3x + 1 = 0
(v) 3x – log 10
x – 16 = 0