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                     1 2 1 1 3 4 5 f f 0 0 0.5f ' f ' 0 f 0.50.5 0.44385 (5.d.p)f ' 0.5 f 0.443850.44385 0.44285 (5.d.p)f ' 0.44385 f 0.442850.44285 0.44285 (5.d.p)f ' 0.44285 xx x x x x x               Check if 0.443  ,       f 0.4425 0 0.4425,0.4434 f 0.4434 0      0.443 (3.d.p)  4 Sketch the following graphs on a single diagram, stating the x-coordinates of all intersections with the x-axis and the equations of any asymptotes. (i)      2 4 2 2y x x x x x     (ii)   2 1 1 xy x   Use linear interpolation once on the interval  1,0 to obtain an approximation to a root of the equation     2 2 14 . 1 xx x x    The Newton-Raphson method is to be used to find an approximation to another root of the equation. Use the method, with 2x as a first approximation, to obtain a second approximation to this root, giving your answer correct to 2 places of decimal. N2000/P2/12 Solution: 1x is a vertical asymptote. www.KiasuExamPaper.com 370

National Junior College Mathematics Department 2016 Numerical Methods Page 4 of 22 Let       2 2 1f 4 . 1 xx x x x     By linear interpolation, approximation to root =         1 f 0 1 1 .1 3 4f 0 f 1            2 4 1 3f ' 3 4 1 x xx x x      By Newton-Raphson’s method, 2nd approximation to root     f 2 32 2 2.23 (2.d.p)f ' 2 13      Iterations involving recurrence relations of the form  1 Fn nx x  5 The equation 3 12 1 0x x   has two positive roots,  and  ,    and one negative root. (i) Prove that 0 1   and 3 4   . (ii) Use the iterative formula   1 31 12 1n nx x   , 1n , with 3.5 as a starting value to approximate  correct to two decimal places. (iii) With the aid of a graph, show that the iterative formula in (ii) will converge to  for some starting values more than  . Solution: (i)   3f 12 1x x x       f 0 1 0 f 1 10 0     There exists a root in  0,1 .   3f 12 1x x x       f 3 8 0 f 4 17 0     There exists a root in  3, 4 . Given that   , 0 1   and 3 4   . (ii)                               1 1 1 3 32 1 1 1 3 33 2 1 1 3 34 3 1 1 3 35 4 1 1 3 36 5 3.5 12 1 41 3.4482 4.d.p) 12 1 40.3784 3.4307 4.d.p) 12 1 40.1684 3.4247 4.d.p) 12 1 40.0964 3.4227 4.d.p) 12 1 40.0724 3.4220 4.d.p) x x x x x x x x x x x                      www.KiasuExamPaper.com 371

National Junior College Mathematics Department 2016 Numerical Methods Page 5 of 22 Check if 3.42      f 3.415 0.1535 0 f 3.424 0.0542 0      3.42 to 2.d.p  (iii) Trapezium Rule and Simpson’s Rule 6 Estimate the values of the following definite integrals, taking the number of ordinates in each case, using (a) the trapezium rule, (b) Simpson’s Rule. (i) 2 0 1 d1 cos xx   , 3 ordinates (ii) 0.4 2 0 1 dx x , 5 ordinates Solution: (i) 02 3 1 4h     Let   1f 1 cosx x   f 0 1/ 2 , 1 2f 4 1 1/ 2 2 1         , f 1 2      Using Trapezium Rule, 02 3 1 4h     Let   1f 1 cosx x   f 0 1/ 2 , 1 2f 4 1 1/ 2 2 1         , f 1 2      Using Simpson’s Rule, x y O   1/3 12 1y x  y x   3x 2x 1x 4x www.KiasuExamPaper.com 372

National Junior College Mathematics Department 2016 Numerical Methods Page 6 of 22   2 0 21/ 2 1 22 4 1 2 1.04912 1.05 3 significant 1 d1 co fi r s 1 gu es xx                          2 0 1 d1 cos 1 21/ 2 4 13 4 2 1 1.00613 1.01 3 significant figures xx                       (ii) 0.4 0 0.15 1h    Let   2f 1 x x   f 0 1  ,  f 0.1 0.99 ,  f 0.2 0.96 ,  f 0.3 0.91 ,  f 0.4 0.84 , Using Trapezium Rule,     0.4 2 0 1 0.84 2 2 0.99 0.96 0.91 0.38870 0.389 3 significant fi 1 d 0 gures .1 x x               0.4 0 0.15 1h    Let   2f 1 x x   f 0 1  ,  f 0.1 0.99 ,  f 0.2 0.96 ,  f 0.3 0.91 ,  f 0.4 0.84 , Using Simpson’s Rule,       0.4 2 0 1 d 1 4 0.99 0.910.1 3 2 0.96 0.84 0.38906 0.389 3 significant figures x x             Euler’s Method and Improved Euler’s Method 7 Apply Euler’s method with step size of 0.1 to obtain an approximation to the given initial value problems at y(0.5). (a) 2d 1d y yt   ,  0 0.5y  (b) d d y t tyt   ,  0 1y  Solution: (a) f(t, y) = 1 – y2, thus by Euler’s method,   2 1 1n ny y t y     . n tn yn, approx 0 0.0 0.5000 1 0.1 0.5750 2 0.2 0.6419 www.KiasuExamPaper.com 373

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