MCSE-004 Dec 2013 Solutions Manual IGNOUUSER
1 1. (a) Verify the distributive property of floating point numbers i.e. prove : a(b-c) ab ac a=.5555e1, b=.4545e1, c=.4535e1 Define : Truncation error, Absolute Error and Relative Error. { } Addition/Subtraction of floating point numbers To add/subtract 2 floating point numbers, the exponents of the two numbers must be equal. If not the mantissa is shifted appropriately. Then the Mantissas are added or subtracted. After that the result is normalized and the exponent is appropriately adjusted. Multiplication of floating point numbers To multiply 2 floating point numbers, the mantissas are multiplied and exponents are added. After that the resulting mantissa is normalized and the exponent is appropriately adjusted. LHS (b-c) =.0010E1 =.1000E-1 a(b-c) =.5555E1 *.1000E-1 =.05555E0 =.5555E-1 RHS ab =.5555E1 *.4545E1 =.2524E2 ac =.5555E1 *.4535E1 =.2519E2 ab ac =.0005E0 =.5000E-3 Clearly LHS RHS For definitions, please refer block. 1. (b)find the real root of the equation x = e -x using Newton-Raphson Method. List the cases where Newton s Method fail. Given x=e -x multiplying both sides by e x and arranging, we get xe x 1 = 0 f(x) = xe x 1 f (x) = (1 + x)e x
2 Let x 0 = 1 By Newton Raphson method we have, x n+1 = x n - x 1 = 1 - = 0.6839397 f(x 1 ) = 0.3553424, f (x 1 ) = 3.337012 x 2 = 0.6839397 - = 0.5774545 f(x 2 ) = 0.028733951, f (x 2 ) = 2.810232 x 3 = 0.5774545 - = 0.5672297 f(x 3 ) = 0.000238786, f (x 3 ) = 2.763614 x 4 = 0.5672297 - = 0.567143297 Since x3 and x4 are same upto 3 decimal places, 0.567 is the required root. Method fails in the cases listed below i) Division by zero occurs in some cases. let f(x) = x 3 2x 2 + 3 = 0 f (x) = 3x 2 4x By Newton Raphson method we have x n+1 = x n - If we choose initial value x 0 = 0 then we get division by zero error. ii) Infinite loop case In some cases, say x 0 is the initial value and 1 st approximation gives x 1 which inturn gives 2 nd approximation as x 0 again and the loop continues.
3 1. (c) Solve by Gauss-Seidel method. 2x 1 x 2 + x 3 = -1 x 1 + 2x 2 x 3 = 6 x 1 x 2 + 2x 3 = -3 correct to 3 decimal places The given system of equations can we written as x 1 = (1/2)(-1 + x 2 - x 3 ) x 2 = (1/2)(6 - x 1 + x 3 ) x 3 = (1/2)(-3 - x 1 + x 2 ) Let initial approximation x 0, y 0, z 0 is 0,0,0 First approximation x 1 (1) = (1/2)(-1 + 0 + 0) = - 0.5 x 2 (1) = (1/2)(6 + 0.5 + 0) = 3.25 x 3 (1) = (1/2)(-3 + 0.5 + 3.25) = 0.375 Second approximation x 1 (2) = (1/2)(-1 + 3.25 0.375) = 0.9375 x 2 (2) = (1/2)(6 0.9375 + 0.375) = 2.71875 x 3 (2) = (1/2)(-2-0.9375 + 2.71875) = -0.60938 Similarly, we have [x 1 (3), x 2 (3), x 3 (3) ] = [1.164063, 2.113281, -1.02539] [x 1 (4), x 2 (4), x 3 (4) ] = [1.069336, 1.952637, -1.05835] [x 1 (5), x 2 (5), x 3 (5) ] = [1.005493, 1.968079, -1.01871] [x 1 (6), x 2 (6), x 3 (6) ] = [0.993393, 2.001721, -0.99972] [x 1 (7), x 2 (7), x 3 (7) ] = [0.996836, 2.001721, -0.99756] [x 1 (8), x 2 (8), x 3 (8) ] = [0.999639, 2.001402, -0.99912] [x 1 (9), x 2 (9), x 3 (9) ] = [1.00026, 2.00031, -0.99997] [x 1 (10), x 2 (10), x 3 (10) ] = [1.000143, 1.999941, -1.0001] [x 1 (11), x 2 (11), x 3 (11) ] = [1.000021, 1.999939, -1.00004] Since last 2 approximations are same upto 3 decimal places. Hence the required solution is x 1 = 1.000 x 2 = 1.999 x 3 = -1.000
4 1. (d)let f(x) = ln(1+x), x 0 = 1 and x 1 = 1.1. Use linear interpolation to calculate an approximate value of f(1.04) and obtain a bound on the truncation error. We have x 0 = 1, x 1 = 1.1 f(x 0 ) = ln(1+x 0 ) = 0.693147 f(x 1 ) = ln(1+x 1 ) = 0.741937 Lets compute Lagranges fundamental polynomials, l 0 (x) = (x-x 1 ) / (x 0 x 1 ) = (x 1.1)/(1 1.1) = (x 1.1)/(-0.1) l 1 (x) = (x-x 0 ) / (x 1 x 0 ) = (x 1)/(1.1 1) = (x 1)/( 0.1) Hence Lagranges Interpolating polynomial is given by P 1 (x) = l 0 (x)* f(x 0 ) + l 1 (x)* f(x 1 ) =- 10(x 1.1)* 0.693147 + 10(x 1)* 0.741937 f(1.04) = P 1 (1.04) = -10(1.04 1.1)* 0.693147 + 10(1.04 1)* 0.741937 = 0.712663 The truncate error in linear interpolation is given by E 1 (f;x) = (1/2)(x-x 0 )(x-x 1 )f ( ) where lies between 1 and 1.1 = (1/2)(x 1)(x 1.1)(-1/(1+x) 2 ) The max value of is 0.892857 and maximum value of (x-1)(x-1.1) occurs at (1+1.1)/2=1.05 * (0.892857) = 0.0011160 1. (e) Consider initial value problem = x+y; y(0) = 1 Find y(0.2) using Runge-Kutta Method of fourth order. Also compare it with exact solution y = -(1+x) + 2e x to find the error. We have, x 0 =0, y 0 =1, h=0.2 Then we get, k 1 =hf(x 0,y 0 ) = 0.2(0+1) = 0.2 k 2 = hf(x 0 + h/2,y 0 + k 1 /2) = 0.2(0.1+ 1.1) = 0.24 k 3 = hf(x 0 + h/2,y 0 + k 2 /2) = 0.2(0.1+1.12) =0.244 k 4 = hf(x 0 + h,y 0 + k 3 ) = 0.2(0.2+1.244) = 0.2888 Therefore y = y 0 + 1/6(k 1 +2 k 2 + 2 k 3 + k 4 ) = 1 + 0.2428= 1.2428
5 Given, y = -(1+x) + 2e x Putting x = 0.2, we get y = -1.2 + 2.442806 = 1.242806 Error = 1.242806 1.2428 = 0.000006 2. (a) Find the interval in which the smallest positive root of the following equation lies using Bisection Method x 3 x 4 = 0 Given f(x) = x 3 x 4 f(0) = 0 0 4 = -4 f(1) = 1 1 4 = -4 f(2) = 8 2 4 = 2 Clearly f(1)*f(2)<0, hence root lies between 1 and 2 First approximation of root, x 1 = (1+2)/2 = 1.5 f(1.5) = -2.125 Clearly f(1.5)*f(2)<0, hence root lies between 1.5 and 2. Second approximation of root, x 2 = (1.5+2)/2 = 1.75 f(1.75) = -0.39063 Clearly f(1.75)*f(2)<0, hence root lies between 1.75 and 2. Third approximation of root, x 3 = 1.875 f(1.875) = 0.716797 Now, f(1.875)*f(2) > 0, Hence the interval in which smallest root lie is [1.75,2] and the root is 1.875. 2. (b) Solve the following linear system of equations using Gauss Elimination Method. x 1 + x 2 + x 3 = 3 4x 1 + 3x 2 + 4x 3 = 8 9x 1 + 3x 2 + 4x 3 = 7
6 Given system of equations x 1 + x 2 + x 3 = 3 4x 1 + 3x 2 + 4x 3 = 8 9x 1 + 3x 2 + 4x 3 = 7...(1) Eliminate x 1 from second equation of (1), multiply first equation by 4 and subtract it from second equation (3 4) x 2 + (4 4) x 3 = 8 4*3 - x 2 = 8 12 x 2 = 4 Eliminate x 1 from third equation of (1), multiply first equation by 9 and subtract it from third equation. (3 9) x 2 + (4 9) x 3 = 7 9*3-6 x 2 + (-5 x 3 ) = 7-27 6 x 2 + 5 x 3 = 20 Hence given system of equations become x 1 + x 2 + x 3 = 3 x 2 = 4 6 x 2 + 5 x 3 = 20...(2) Substituting value of x 2 in third equation we get, x 3 = -4 /5 = -0.8 Substituting value of x 2, x 3 in the first equation we get, x 1 = 3 4 (-0.8) = -0.2 Hence the required solution is x 1 = - 0.2 x 2 = 4 x 3 = -0.8 2. (c) Give properties of polynomial equations. i) A polynomial equation of degree n has n solutions. ii) Non-real complex solution with real coefficient, if exist, occur in conjugate pairs. iii) Descarte s rule of signs The number of positive real solutions of the given polynomial equation is equal to the number of variations in signs of the polynomial.
7 3. (a) The table below gives the values of tanx for 0.10 x 0.30. x 0.10 0.15 0.20 0.25 0.30 y=tanx 0.1003 0.1511 0.2027 0.2553 0.3093 Find i) tan0.12 ii) tan0.26 First lets construct the difference table x y=tanx Δ Δ2 Δ3 Δ4 0.1 0.1003 0.0508 0.15 0.1511 0.0008 0.0516 0.0002 0.2 0.2027 0.001 0.0002 0.0526 0.0004 0.25 0.2553 0.0014 0.054 0.3 0.3093 i) tan0.12 here, h=0.05, a=0.10, x=0.12 u=(0.12-0.10)/0.05 = 0.4 By Newton s Forward Interpolation formula we have tan(0.12) = f(a) + u Δ f(a) + Δ 2 f(a) + Δ 3 f(a) +... = 0.1003 + 0.4*0.0508 + 0.0008 + 0.0002 + 0.0002 = 0.1003 + 0.02032 0.000096 + 0.0000128 0.00000832 = 0.1205 ii) tan0.26 here, h=0.05, a=0.10, x=0.26 u=(0.26-0.10)/0.05 = 3.2 By Newton s Forward Interpolation formula we have tan(0.26) = f(a) + u Δ f(a) + Δ 2 f(a) + Δ 3 f(a) +... = 0.1003 + 3.2*0.0508 + 0.0008 + 0.0002 + 0.0002 = 0.1003 + 0.16256 + 0.002816 + 0.0002816 + 0.00001408 = 0.2659
8 3. (b) Evaluate I = dx, correct to three decimal places using i) Trapezoidal, and ii) Simpson s rule with h=0.5 and h=0.25 Here, f(x) = The value of f(x) at x=0,0.25,0.5,0.75,1 are tabulated below. x 0 0.25 0.50 0.75 1 f(x) 1 0.8 0.67 0.57 0.5 f 0 f 1 f 2 f 3 f 4 i) Trapezoidal rule Taking h=0.25, we have I = (0.25/2)[ f 0 + f 4 + 2(f 1 + f 2 + f 3 )] = 0.125*(1.5 + 2*2.04) = 0.6975 ii) Simpsons rule h = 0.25, I = (0.25/3)[ f 0 + f 4 + 4(f 1 + f 3 ) + 2 f 2 ] = 0.0833 * (1.5 + 4*1.37 + 2*0.67) = 0.6930 h=0.5 we have to consider only f 0, f 2, f 4 I = (0.5/3)[ f 0 + f 4 + 4* f 2 ] = 0.16667*(1.5 + 4*0.67) = 0.6967 3. (c) Determine the value of y when x=0.1 given that y(0) = 1 and y = x 2 + y. let h = 0.1/5 = 0.02 x 0 = 0, y 0 = 1 By Euler s method we have y 1 = y 0 + hf(x 0,y 0 ) = 1 + 0.1*(0+1) = 1.1 y 2 = y 1 + hf(x 1,y 1 ) = 1.1 + 0.1*(0.02*0.02 + 1.1) = 1.21004 y 3 = y 2 + hf(x 2,y 2 ) = 1.21004 + 0.1*(0.04*0.04 + 1.21004) = 1.331204
9 y 4 = y 3 + hf(x 3,y 3 ) = 1.331204 + 0.1*(0.06*0.06 + 1.331204) = 1.464684 y 5 = y 5 + hf(x 4,y 4 ) = 1.464684 + 0.1*(0.08*0.08 + 1.464684) = 1.611792 Hence the value of y at 0.1 is 1.611792 4. (a) A problem in statistics is given to the three students A,B and C whose chances of solving it are 1/2, 3/4 and 1/4 respectively. What is the probability that the problem will be solved. Given, probability that A will solve the problem P(A) = 1/2 probability that B will solve the problem P(B) = 3/4 probability that C will solve the problem P(C) = 1/4 Hence, probability that A will not solve the problem P( ) = 1 1/2 = 1/2 probability that B will not solve the problem P( ) = 1 3/4 = 1/4 probability that C will not solve the problem P( ) = 1 1/4 = 3/4 Since all events are independent, so P(no one will solve the problem) = P( )* P( )* P( ) = (1/2)*(1/4)*(3/4) = (3/32) P(that prob will be solved) = 1 (3/32) = 29/32. 4. (b) Calculate the correlation coefficient for the following heights (in inches) of fathers (x) and their sons (y) : x: 65 66 67 67 68 69 70 y: 67 68 65 68 72 72 69 The various calculations on the given data are tabulated below. x y x-x' y-y' (x-x')*(y-y') (x-x')*(x-x') (y-y')*(y-y') 65 67-2.42857-1.71429 4.16326531 4225 4489 66 68-1.42857-0.71429 1.02040816 4356 4624 67 65-0.42857-3.71429 1.59183673 4489 4225 67 68-0.42857-0.71429 0.30612245 4489 4624 68 72 0.571429 3.285714 1.87755102 4624 5184 69 72 1.571429 3.285714 5.16326531 4761 5184 70 69 2.571429 0.285714 0.73469388 4900 4761 14.8571429 31844 33091
10 where x = mean value of x = 67.42857 y = mean value of y = 68.71429 We have the correlation coefficient r = = = 0.000457 4. (c) Three identical bags have the following proportion of balls. First bag: 2 black 1 white Second bag: 1 black 2 white Third bad: 2 black 2 white One of the bag is selected and one ball is drawn. It turns out to be white. What is the probability of drawing a white ball again. The first one not been returned. Let E = event of selection of ith bag(i=1,2,3) A = event of drawing white ball Then P(E 1 ) = P(E 2 ) = P(E 3 ) = 1/3 Also, P(A E 1 ) = C(1,1)/C(3,1) = 1/3 P(A E 2 ) = C(2,1)/C(3,1) = 2/3 P(A E 3 ) = C(2,1)/C(4,1) = 1/2 Let C denote the further event of drawing another white ball. P(C E 1 A) = 0/2 = 0 P(C E 2 A) = 1/2 P(C E 3 A) = 1/3 Hence by Bayes theorem P(C A) = = = = 1/6
11 5. (a) Evaluate [ ( )] using Simpsons rule with 11 points. Here, h = (6-1)/11 = 5/11 x f(x) 1 2.0349 1+(5/11) 2.04209 1+(10/11) 2.04821 1+(15/11) 2.05364 1+(20/11) 2.05857 1+(25/11) 2.06311 1+(30/11) 2.06734 1+(35/11) 2.07132 1+(40/11) 2.07509 1+(45/11) 2.07868 1+(50/11) 2.08211 6 2.0854 By Simpsons one-third rule, [ ( )] = (5/33) [(2.0349 + 2.0854) + 4(2.04209 + 2.05364 + 2.06311 + 2.07132 + 2.07868) + 2(2.04821 + 2.05857 + 2.06734 + 2.07509 + 2.08211 )] = (5/33) [4.1203 + 4(10.30884) + 2(10.33132) ] = 10.003
12 5. (b) Estimate the sale of a particular quantity for 1966 using the following table. Year 1931 1941 1951 1961 1971 1981 Sale in thousands 12 15 20 27 39 52 First lets construct the difference table. Year Sale 2 3 4 5 1931 12 3 1941 15 2 5 0 1951 20 2 3 7 3-10 1961 27 5-7 12-4 1971 39 1 13 1981 52 here, h=1981-1971=10 a=1981 x=1966 hence, u=(1966 1981)/10 = - 1.5 By Newton s Backward Interpolation formula we have f(1966) = f(a) + u f(a) + 2 f(a) + 3 f(a) + 4 f(a) +... = 52 + (-1.5)*13 + 1 + + = 52-19.5 + 0.375-0.25 0.164 0.117 = 32.344 (-7) + (-10)
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