Question

QUESTION 1 = = (a) Apart from x = 0 the equation f(x) 22 – 4 sin r 0 has another root in (1, 2.5). Perform three iterations of the bisection method to approximate the root. State the accuracy of the root after the three iterations. (b) Perform three iterations of Newton's method for the function in (a) above, using x(0) = 1.5 as the initial solution. Compare the error from the Newton's approximation with that incurred for the same number of iterations using the bisection method.

Answer 1

(a) f(x) = 7? - 48lux = 0 [1.8.5] got iteration: f(1) = -0.3659 <0 $(2.5) = 3.856170 Now Moot lies blo 1 & 2.5 20 1+2.5 2 = 1115 ond guel 2 f(xo) - 7(175) = 1752 -4 8in (175) = -0.873430 711.75) = -0.8434 <0 $ $(2.5) = 3-8561 >0 NOW, root lies b/w 1.75 & 2.5 24 = 1.75 +2.5 = 3.125 f(x) = f(p.125) = 01.414370 141.45) = -0.8134<0 $ $(2.135) = 1.1143 30 Now , Mools lies b/w 1.75 ¢ 2.125 1.75 +2.125 1.9375 $(30) = f(1.9375) 0.0198 > $(1.75)= -0,8734 <0 $ $(1.9375)=0.0198 80 NOW, 2001 lies b/w 1.75 & 1.9375 23= 1:7571.9375 1.8438 71X3) = f(18438) = -0.4525<o accuracy f(x) 1.75 -0.8734 2.125 01:1143 0.0198 -0.4525 eth 2 m 1 3 1.9375 1.8438 4 (6) 22-4 sina f(x) = f'(x) = px - COS X 20 e 1.5 jot f(x0) = 4(1-5) = -1-75 f'(xo) = f'(15) = 2.7171

X20- f(xo) f'(xo) 1.5 - (-1.94) - 2014ou 2.717) f(x1) = 12.1404) = 1.2128 ť (x1) 6.438 2g = x - f(x1) o nel f'(x) 39 320 1.952 f(x) = $11.952) = 0.09 +5 f'(43) = f(11952) = 5:39 22 &z= x2 (06) 7 (22) X3 = 1.9339 f(x3) = f(119339) - 0.0009 f (x₂) 5.2887 f (23) t'(43) 24 = 2z - 24 = 1.9338 apps.oximate most of method is the equetion geo-Lisin (a) = 0 1.9338 Newton's wing 1.5 f(x) 1.14 2.1404 7.952 1-9339 3 4 f'(xo) 9.4171 6.438 5.3922 5.0864 1.128 0.09 15 0.0009 9.1404 1.952 1.9339 1.9338 euror of newton's : I ca) : 1.9338 -1.9339 1.9338 0.005 1.1.4.3 fose bisection! -> Log I erunt log(2) + 3), log () log2 @: 1.875