F06 - Numerical Methods - Tutorial (Solutions)

National Junior College Mathematics Department 2016 Numerical Methods Page 1 of 22 National Junior College 2016 – 2017 H2 Further Mathematics Numerical Methods Tutorial Basic Mastery Questions Interval Bisection, Linear Interpolation and Newton-Raphson’s Method 1 Estimate the positive root of the equation  ln 2x x  , giving your answer to four significant figures. J73/P1/14 Solution:               2   there exists a Let f ln 2 1f ' 1 . 2 f ' 0 1 1f '' 0 2 1 gives a minimum poin root in 1, .t x x x x x x x x x x x                           f 0 ln 2 0 f 1 1 ln 3 0 f 2 2 ln 4 0         There exists a positive root in 1 2 x  . Estimate of root f ( ) Interval containing root 1 < 0 2 > 0  1, 2 1.5 > 0  1,1.5 1.25 > 0  1,1.25 1.125 < 0  1.125,1.25 1.1875 > 0  1.125,1.1875 1.15625 > 0  1.125,1.15625 1.140625 < 0  1.140625,1.15625 1.1484375 > 0  1.140625,1.1484375 1.14453125 < 0  1.14453125,1.1484375 1.146484375 > 0  1.14453125,1.146484375 1.145507813 < 0  ,11. .11455 464807813 4375 1.145996094 < 0  ,1.145507813 1.145996094 www.KiasuExamPaper.com 368

National Junior College Mathematics Department 2016 Numerical Methods Page 2 of 22 Since there exists a positive root in ,1.145507813 1.145996094 , the positive root is 1.146(to 4 significant figures). 2 Use linear interpolation to find an approximation to the root of equation  ln 4sin 0x x  which lies between 1 and 2. Give your answer correct to three decimal places. Solution:    Let f ln 4sinx x x  and be the root. Estimate of root f ( ) Interval containing 1 < 0 2 > 0  1, 2 1.5 > 0  1,1.5         1.5 f 1 1 f 1.5 1.32387 (5 .d.p)f 1 f 1.5   < 0  1.32387,1.5         1.5 f 1.32387 1.32387 f 1.5 1.36155 (5 .d.p)f 1.32387 f 1.5   < 0  1.36155,1.5         1.5 f 1.36155 1.36155 f 1.5 1.36468 (5 .d.p)f 1.36155 f 1.5   < 0  1.36468,1.5         1.5 f 1.36468 1.36468 f 1.5 1.36494 (5 .d.p)f 1.36468 f 1.5   < 0  1.36494,1.5 Check if 1.365  ,       f 1.3645 0 1.3645,1.3654 f 1.3654 0       1.365 to 3.d.p  3 Show that the equation e 2 0 x x   has one and only one real root. Use the Newton - Raphson process to find this root correct to three decimal places. J75/P1/14 Solution: Let  f e 2 xx x    f ' e 1 xx   For all real values of x, e 0x  .    f ' e 1 f ' 0xx x      f x is strictly increasing function and thus cuts the x-axis only once. e 2 0 x x    has one and only one real root. Let 1 0x  and  be the root of  f 0x  . By Newton – Raphson’s method: www.KiasuExamPaper.com 369

National Junior College Mathematics Department 2016 Numerical Methods Page 3 of 22                 