LECTURE 4 OF SOLUTIONS OF NON-LINEAR EQUATIONS OBJECTIVES

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Presentation transcript:

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.

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

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.

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 = 0.6829 > 0 f(2) = 2 sin 2 – 2 = -0.1814 < 0 Since f(1) > 0 and f(2) < 0, f(x) has a root between x=1 rad and x=2 rad

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 | = 0.1415 (<1 ) So,the iteration function : g(x) = 2 sinx

g(x) = 2 sin x x1 = 1.5 x2 = 2sin 1.5 = 1.9950 x3 = 2 sin 1.9950 = 1.8227 x4 = 2 sin 1.8227 = 1.9369 x5 = 2 sin 1.9369 = 1.8675 x6 = 1.9126 x7 = 1.8843 x8 = 1.9025 x9 = 1.8910 x10 = 1.8984 x11 = 1.8937 x12 = 1.8967

x13 = 1.89474 x14 = 1.89597 x15 = 1.89519 Thus, the root is 1.90 ( two decimal places ).

Newton Raphson method: f(x) = 2 sin x - x f’(x) = 2 cos x – 1 x1 = 1.5 = 2.0766

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

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.

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

Newton-Raphson method: e-x = = 2 – x ex = 2 – x  ex – 2 + x = 0 f(x) = ex - 2+x f ’(x) = ex + 1 x0 = 0.4 ( from graph )

x0 = 0.4 =0.44341 =0.44285 Thus, x=0.443 ( 3d.p )

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]

(a) h =

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