1. LECTURE 4 OF 4 7.2 SOLUTIONS OF NON-LINEAR EQUATIONS OBJECTIVES
Use the Iteration and Newton-Raphson methods to find the approximate root of an equation.

2. OBJECTIVES
Use the Iteration and Newton-Raphson methods to find the approximate root of an equation.

3. EXAMPLE 1
Show that the equations 2 sin x – x = 0
has a root between x = 1 rad and x= 2 rad.
Find the root of the equation by using iteration method and Newton Raphson method, giving your answer to two decimal places.

4. Since f(1) > 0 and f(2) < 0, f(x) has a root between x=1 rad and
SOLUTION
f(x) = 2 sin x – x
f(1) = 2 sin 1 – 1 = > 0
f(2) = 2 sin 2 – 2 = < 0
Since f(1) > 0 and f(2) < 0,
f(x) has a root between x=1 rad and
x=2 rad

5. Iteration method:
2 sin x – x = 0
x = 2 sin x
g(x) = 2 sin x
g(x) = 2 cos x
= | 2 cos 1.5 | = (<1 )
So,the iteration function : g(x) = 2 sinx

6. g(x) = 2 sin x
x1 = 1.5
x2 = 2sin =
x3 = 2 sin =
x4 = 2 sin =
x5 = 2 sin =
x6 =
x7 =
x8 =
x9 =
x10 =
x11 =
x12 =

7. x13 =
x14 =
x15 =
Thus, the root is ( two decimal places ).

8. Newton Raphson method:
f(x) = 2 sin x - x
f’(x) = 2 cos x – 1
x1 = 1.5 =

9. Thus, the root is
1.90 ( 2d.p).

10. EXAMPLE 2
Sketch the graph of y = ex and y = 2 – x
on the same axes. Get the first
approximation, x0 for the equation ex = 2 – x
where 0 < xo < 1. Hence, by using Newton-
Raphson method , solve the equation of
e-x =
to three decimal places.

11. y x 2 1 2
0<x0<2
Let x0= 0.4

12. Newton-Raphson method:
e-x =
= 2 – x
ex = 2 – x  ex – 2 + x = 0
f(x) = ex - 2+x
f ’(x) = ex + 1
x0 = ( from graph )

13. x0 = 0.4 = =
Thus, x=0.443 ( 3d.p )

14. EXAMPLE 3 (PAST YEAR 2006)
Use the trapezoidal rule with n = 4 to
approximate Using definite
Integration, find the value of
Compare the answers and give a reason for
the difference [7 MARKS]
(b)Approximate by using Newton-Raphson method and initial value 2, up to the 2nd iteration [3 MARKS]

15. (a) h =

16. (b) Let x =
Thus, x = 1.91 ( 3s.f )