Question

QUESTION 1 (a) Apart from = 0 the equation f(t) = 12 - 4sin r = 0 has another root in (1, 2.5). Perform three (10) 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 (10) solution. Compare the error from the Newton's approximation with that incurred for the same number of iterations using the bisection method. [20]

Answer 1

Q- f(a)= x- & sinazo root lies between [1, 2.5] Bisection Method :- let a=1 and b=2.5 ist iteration = 1.75 u=2.5 <= (2+). = 2.5+! first checking values of f(a) at x= 1 and f(1) = (2-4 sen(I) = -2,3658 f(2-5) = (2-5)= 4 sn (2.5) = 3.8561 Now value of f(x) at XF 1.75 fav=f(1-75) = (1.75)2 454m (1-75) -0,8734 so f (1-15) <0 and f (2.5) 70 so root lies between 0.75) and 2.5 and Iteration a=1-75 b=2.5 - 2 U. 20125 - H₂ = 175+2.5 2 f(u) at Ug = 2.125 $(2-195) = (2.125)3_454(1.125) = 1.1143 f(*) 70 so and f(175) <0 so

root lies between 1.758 2125 3rd = 1.9375 Iteration d=1-75 b=2.125 atb f= 1.75 +2-125 2 2 Value of f(x) at U₂ = 1.9375 f(1) = f(1.9375) = (19375)2 4 SM(1.9379 = 0.0198 ; flus)>0; f(4)<o so root dies between 1075 8 1.9375 So After third iteration We get apprexi mate root is = 1.9375 Exact root of x² - 4 sense is 1.93375 so errore exact groot approximate root error-11.93375-1.9375) = 0.00375

(b) Using Newton raphson with initial value X=105 Newton Raphson Iteration formular 2,- 7.-H. f (*) f(x) = x² 4 sint t(a) = 2u - 4 cosa Yo=105 First Iteration Clus)? 4 sen(1-5)] [211.5) – 4 Lostos)] 26 = 1.5- = 2.1403 1,= 2.1403 2,= 1.5+ 117399 2: 7170 and steration H₂ = X- f(240) f (a) U2= [(201403) 4 sen (2.1403)] [2(2.1403) – 4 608(2.1963) 2.1403 2. 1403 Ag - १२१२२ 1.4374 = 1.95190

ra 3 Iteration X= 1.9519 H -f(A) 43= H₂ f (2) ts= 1.9519 – [(1.9519) 4-sum(119519)) [2(1.9519) — 4603 (1.9519)] 13=119519 O. 0968 1.9339 5.3915 so using Newton raphson after 3rd iteration we get approximate root is &z= 109339 using Bisection method we get 4₂= 1.9375 erol = 1.9375 -1.9339 =0,00 36 so error to calculate approximate rest using bisection and newton raphson is 0.0036