Question

QUESTION 1 (a) Apart from 1 = 0 the equation f(1) = x2 - 4 sin 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) = 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]

this is numerical analysis

Answer 1

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f(x) = x² -usina Givent one root is 2 O the other root [1,2.5] bisection method in using Solution o between [1, 2.5] f(x) = x2- 4 sinx and f(25) 6.25 - usin (25) 1st iterations fol 1 - 4 Sin (1) = 3.856170 - 2.3659 LO I and 9.5 Root lies between & to = 1.75 1+ 2.5 2 and iteration & f (xo) f (1-75) = (1.75) ?- 4 Sin (1-75) -0.8734 20 and f (0.5) = 3:856170 f (1.75) -0.8734 <0 1.75 and 2.5 root lies between 1.75 + 2.5 2.125 2 1.1143 70 - f(x) = (0.125)2 - 4 Sin (2.125) -0.8734 Ko the root lies 3rd iteration 6 f (1.75) = & f ( 2.125) = 4.75€ 1.1143 70 botween 2.125 1.75 + 2.125 1.9375 - f(xs) = (1.9375)2 - 4 SM (1.9375) 0.019870 1C Scanned with CamScanner

and +(1.9375) = 0.019870 Here f(175) = -0.8734 <0 1.9375 and lies 700+ 1.75 between 1.8438 1.75 + 1.9375 23 1.8438 3 iteration = ooot from bisection method after error = 2= 1.433 75 1.93375 - 1.8438 .xlvo 1.93375 actual root root by bisection = x=18438 method after 3 iteration error = 4.66% Newton's method & f(x)= 22- 4 Sinx = f'(x) = O 1- 4 C8x => x= 10.5 also 1.5 ho = Given Ist iteration & f (no) = f (1.5) = (1.5). 4 Sin (-5) = -|.74 f' (20) = f'(1.5) = 2 (1.5) - 4 Gos (1,5) = 2.7171 1.5 -1.74 2.1404 2,5 = Xo - f(xo) f (xo) 2.7171 (8.1404)- 4 Sin (201404) Qud iteration e f (x) = f (8.1404) = - 1.2128 f'(x) f' (2.1404) = 2X2.1404-4. Cos (2-1404) 6.438 Danned with-CamScanner

2 = x, f(x) 2.1404 7.2128 6.438 f (x) 1.952 0.0975 3rd iteration & f(X21 = (1.982) ? 4 Siu (1.952) = f'(22) = 2 (1.952/- 4 col (1.952) = 5.3922 0.0975 1.952 - x3 = x2 f(x2) f (x2) 5. 3922 X3 1.9339 Terror = 1.93375 - 1.9339 1.93375 -0.008%. x=0 f(x)= x2 - 4 Sinx=0 one root is is and 1.93375 the the another lost of equation New x= 1.8438 the root after 3 iterations using bisection method x= 1.9339 root after 3 iterations using Needter's method 5 & Scanned with CamScanner