Question

You are given that there exists a solution to the equation et = 4 sin (2) in the interval [0, 0.5]. a) Use the iterative method explained during the course to find this solution up to 3 decimal places. Make sure you include enough details of the calculation and that your cal- culator is set in radians for this problem. [3 marks] b) Now estimate the same solution using the Maclaurin expansion of the function f(x) = e" – 4 sin (2) up to second order. This means you should neglect terms with x and higher powers in the expansion and solve a quadratic equation. You are going to find two solutions in part b. Only take the one in the interval under consideration. I [4 marks)

Answer 1

(a é = 4 Singel Let f(x) = { _4 Sinb) = 0 We have to find the root in the interval [0, 0.5] let's use Newton-Raphson Method. By Newton-Raphson Method fcren f'(zn) 4 Sin Cen) Cos Can) Xnt = an un e f(en) f'(on) = ē" Un 4 enti an Un e Un 4 Sin Can) - 4 Coscan) е Let No put n=o ซอ e 1-0 0- no 21 = = 4 Sin() é*_4008(0) - 1 / 2 = 1/2 · 4 = 0.333 ut n=1 2 = 0.3333 e-4 Sin (20) 0.3333_ e - 4 Sin(0.3333) 21 0.3333 e - 4C08 (0.3333) =0.333 -0.0364 é - 4 cos (21) = 0.369 csscanned with CamScanner

put n = 2 X 2 X₂ X2 e 4 Sin(x2) e-4 cos (12) 22 0.369 = 0.369 - e 4 Sin(0.369) 0.369 e 4 cos(0.369) 0.369 +0.0016 = 0,371 put n=3 13 e 4 Sin (13) X4 Xz 13 e 4 cos (1₂) 0.371 = 0.371 e 4 Sin (0.371) -- 4 Cos Co. 37) 0.37) e 0.371 – 0.0005 - 0.371 Since we have same values in hoo successive élérations: 71-0.371 Answer: The solution 0.371 the of given equation e = 4 Sina is CSScanned with CamScanner

OC second order is (b) f(a) e_4 sinx Meclaurin's polynomial f(a)-f(0) + 앙 2 T a f"CO) frost 21 . f(a)=e_ 4 Sina f(a) = en f(0) = 1 f'(o) f(0) =/ 1 - 4 = 4 cosa f(n) e + 4 sina :; f(a) = H+H.(-5)+**.1 ie, e _ 4 sina I- 3x + 2 zel_8 sina 2-62+72 1 21 8 sina x²64 +2 ze f(a)=0 → a 61+2=0 6+ √36-8 6+528 = 2 67 257 2 3+57, 3-5 = 5.64, 0.35 Since the root lies in (0,0.5) ; 2 = 0. 35 Answer: solution in [0,0.5 0.35 csscanned with CamScanner