DHS Numerical Approximation of Roots (9649) (Revision Solutions)

Numerical Methods – Approximation of Roots H2 Double Math (TOPICAL REVISION – SUGGESTED SOLUTIONS) NUMERICAL METHODS (PART 1) APPROXIMATION OF ROOTS OF EQUATIONS 1 Show that the equation 3 5 1 0,xx− + = has exactly one root in (0,1). Use two iterations of linear interpolation between 0x= and 1x= to yield a fractional approximation of this root. [4] Three possible rearrangements of the given equation in the form ()x F x= are 3 5 1,xx=− 31 ( 1),5xx=+ 3 4 1.x x x= − + Only one of these rearrangements will provide an iterative method, of the form 1 ()nnx F x+ = , 0 2,x = which converges to the root between 2 and 3 . Use this rearrangement to find this root correct to 3 significant figures. [4] It is known the correct rearrangement above provides a convergent iterative method by analysing the derivative of F . For this purpose, show that 0 '( ) 1,Fx whenever 2x . [2] [EJC/FM/2018/P1/Q9] [Solution] With 3g( ) 5 1,x x x= − + .x Note that g(0) 1 0,g(1) 3 0,=  =−  thus g(0) g(1) 0 Since g is continuous on [0,1] , the equation g( ) 0x = has at least 1 root in [0,1]. (1) 2g ( ) 3 5 0, [0,1].x x x= −   Thus g( ) 0x = has at most 1 root in [0,1]. Therefore, g( ) 0x = has exactly 1 real root in [0,1]. Using linear interpolation on [0,1] , we have 1x = (0)g(1) (1)g(0) g(1) g(0) − − = 1 31 − −− = 1 4 Since 1 1 1 1 5 16 64 15g 5 1 ,4 64 4 64 64 64 64    = − + = − + =−       < 0 and g(0) = 1 > 0  lies in (0, 1/4).

Numerical Methods – Approximation of Roots 2x = 11(0)g g(0)44 1g g(0)4    −        − = 11 1644 79 715 64 91 64    −      ==−  − 3 5 1,xx=− (This fixed point iterations based on this rearrangement converges to 2.13) 0 2x = 1 2.08x  2 2.11x  3 2.12x  4 2.13x  5 2.13x  Using GC, we can see for the other 2 schemes based on: 31 ( 1),5xx=+ the fixed point iteration does not converge to a value between 2 and 3, but to 0.205. 3 4 1,x x x= − + the fixed point iteration diverges. ( ) ( ) 3 1 2 (1) 3 2 (1) 3 3/2 3/2 ( ) 5 1 51( ) 5 1 0,35 5( ) 5 1 1, 3 5whenever 5 1 , 3 15. ., 1 0.630,53 which is satisfied as ( ) increases, when ever 2>0.630. F x x F x x x F x x x i e x F x x − − =− = −   = −  −    +   

Numerical Methods – Approximation of Roots 2 (a) The curve with equation 2 2e 3xyx= + − has exactly one stationary point in the interval  1, 0− . Use the Newton-Raphson method to find the x-coordinate of the stationary point, correct to 4 decimal places. [5] (b) (i) Show that the equation 3 7 2 0xx − + = has a root,  , in the interval [0, 1]. [1] (ii) Student A uses the recurrence relation ( ) 3 1 1 14 27 nnnx x x+ =− − for finding  . Explain why Student A will fail to find  . [2] (iii) Student B uses the recurrence relation ( ) 3 1 1 27 nnxx+ =+ for finding  . Find the approximate value of  to 5 decimal places. [2] NJC/FM/2019/P1/8 [Solution] 2(a) 2d 4 e 1 0d xy xx = + = The Newton-Raphson formula as follows: 2 221 2 4 e 1 8 e 4e n nn x n nn xx n xxx x + +=− + 1 3 4 0 2 0.5 0.29647 0.23906 0.23642 0.23641 x x x x x =− =− =− =− =− Hence, 0.2364x=− . Checking, ( ) ( ) ( ) ( ) 2 2 0.23635 4 0.23645 4 4 0.23635 e 1 2.9 10 0 4 0.23645 e 1 1.8 10 0 − − − − − + =   − + =−   (b) (i) ( ) ( ) ( ) 3 72 f 0 2 f 1 4 f xxx= −+ = =− Since ( ) ( )f 0 f 1 0  and f is continuous, there is a root in [0, 1]. (b) (ii) Either use graph or display a series of values to show convergence fail.

Numerical Methods – Approximation of Roots (b) (iii) 0 1 2 3 4 5 6 1 0.428571429 0.2969596 0.289455341 0.289178834 0.289168915 0.28916856 x x x x x x x = = = = = = = 0.28917 = Checking, 35 35 0.289165 7(0.289165) 2 2.4 10 0 0.289175 7(0.289175) 2 4.4 10 0 − − − + =   − + =−  

Numerical Methods – Approximation of Roots 3 Let 2f( ) sec 2 e xxx=− . The equation f( ) 0x = has a root  in the interval [0.1, 0.3]. (a) Use Newton-Raphson method with initial approximation 0 0.1x = to find the first four approximations 1 2 3 4, , , x x x x to  . Deduce the behaviour of nx for large n and use a graph of the function f to help you explain why the sequence is not converging to  . [5] (b) Determine, with explanation, which one of these iterative formula e is more suitable in approximating the root  . (I) ( )( ) 2 1 ln sec 2nnxx+ = (II) 1 2 1 1 cos e2 nx nx −− + =   Using an initial approximation of 0 0.3x = and the chosen iterative for mula, find the approximation to the root  , correct to three decimal places. [4] [RVHS/FM/2018/P1/Q10] [Solution] 2 2f '( ) 4sec 2 tan 2 e xx x x=− 1 2 3 4 f (0.1)0.1 f '(0.1) 0.1455040283 0.146 0.0417668268 0.0418 0.0048039104 0.00480 0.0000781946 0.00008 x x x x =− =− − =− − =− − =− − As , 0 nnx→ → The initial approximation 0x is too far from  . It is nearer to the root 0x= . Thus the sequence converges to this root instead of  . π 4x=

Numerical Methods – Approximation of Roots (b) Let ( )( ) 2F( ) ln sec 2xx= and 1 21G( ) cos e2 x x −− =   2 2 4sec 2 tan 2F'( ) sec 2 4 tan 2 xxx x x = = 2 2 2 2 1 21G '( ) 2 1 41 x x x x e x e e e − − − − − =− −  = − '(0.3) 2.737 1 '(0.3) 0.4227 1 F G = = Thus 1 2 1 1 cos e2 nx nx −− + =   is a more suitable iteration to use. 0 0.3x = 1 2 3 4 5 6 7 0.2670688194 0.2526829817 0.2460804385 0.2415055517 0.2408020568 0.2404652776 0.2403038578 x x x x x x x = = = = = = = 0.240 (to 3 d.p.)

Numerical Methods – Approximation of Roots 4 (i) The function f is such that f( a)f(b) < 0, where a < b. A stud ent concludes that the equation f(x) = 0 has exactly one root in the interval (a, b). Illustrate with a sketch, two possible scenarios in which the student could be wrong. [2] GIVEN NOW THAT T HE EQUATION 2 230x x − − = HAS EXACTLY ONE ROOT  IN THE INTERVAL (3,4). (ii) Derive the iterative formula 1 4 43n n x x + =+ using the fixed point iteration method. Using 3 as the initial value, apply this iterative formula to find an approximation for , correct to 3 decimal places. You are required to check the accuracy of your answer in this question. [3] (iii) By using linear interpolation once, obtain, correct to 3 decimal places, a first approximation 1 to  . Using 1 as the initial value, apply the Newton -Raphson method once to obtain 2 , leaving your answer to 3 decimal places. [3] (iv) Explain why the Newton-Raphson method in this case fails to give an approximation to . [1] (v) Illustrate, on a single diagram, how 1 and 2 are obtained. [2] [CJC/FM/2018/P1/Q5] [Solution] (i) Two possible scenarios with more than one root: The student failed to check if f is strictly increasing or decreasing function in the interval (a, b) or if the curve is continuous over the interval (a,b). (ii) 22 144 2233 44Squaring both sides, 3 3 n n xx xx xx xx + − −  − = − =  = + Let f(x) = 2 23x x −− Method 1: 1 4 43n n x x + =+ o = 3 x x