MCSE004. June 2014 Solutions Manual IGNOUUSER


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1 MCSE004 June 2014 Solutions Manual IGNOUUSER
2 1 1. (a) Determine the root of the equation 2x=cosx+3 correct to three decimal places. { } We will use the simplest method called Bisection Method to solve this problem. Bisection Method states that if F is a continuous function between a and b and F(a) and F(b) are of opposite signs i.e., F(a)*F(b) < 0 then a root lies between a and b and first approximation of the root is (a+b)/2 Given f(x) = 2x  cosx 3 We have f(0) = 2*0 cos(0) 3 = 4 f(1) = 2*1 cos(1) 3 = f(2) = 2*2 cos(2) 3 = 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*1.5 cos(1.5) 3 = 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) = 2*1.75 cos(1.75) 3 = Clearly f(1.75)*f(2)<0, hence root lies between 1.75 and 2. Third approximation of root, x 3 = (1.75+2)/2 = Continuing like this, we get the following result after 11 th approximation Since x10 and x11 are same upto 3 decimal places, hence is the real approximate root of the given equation.
3 2 1. (b)solve the following system of equations using Gauss Elimination method. 2x + y + z = 10 3x + 2y + 3z = 18 x + 4y + 9z = 16 Given system of equations 2x + y + z = 10 3x + 2y + 3z = 18 x + 4y + 9z = 16...(1) Eliminate x from second equation of (1), multiply first equation by 3/2 and subtract it from second equation (2 3/2)y + (3 3/2)z = 18 (3/2)*10 (1/2)y + (3/2)z = y + 3z = 6 Eliminate x from third equation of (1), multiply first equation by 1/2 and subtract it from third equation. (4 1/2)y + (9 1/2)z = 16 (1/2)*10 (7/2)y + (17/2)z = 11 7y + 17z = 22 Hence given system of equations become 2x + y + z = 10 y + 3z = 6 7y + 17z = 22...(2) Eliminate y from third equation of (2), multiply second equation by 7 and subtract it from third equation (17 7*3)z = 22 7*64z = 20 z = 5 Hence given system of equation becomes 2x + y + z = 10 y + 3z = 6 z = 5...(3) using back substitution, i.e substituting value of z in second equation we get, y=6 15 = 9 and substituting z and y in first equation we get, x = (10 (9) 5) / 2 = 7 Hence x=7,y=9 and z=5 is the required solution.
4 3 1. (c) Using Lagrange Interpolation, determine the value of log , from the tabulated data given below. X Log 10 X We have, x 0 = 300, f(x 0 ) = x 1 = 304, f(x 1 ) = x 2 = 305, f(x 2 ) = x 3 = 307, f(x 3 ) = Lagrange s formula is f(x) = + f(x) = + f(x) = + Putting x=301 we get f(301) = + = =
5 4 1. (d)ten coins are thrown simultaneously. Find the probability of getting atleast seven Heads. Let p = probability of getting success i.e Head in 1 toss = 1/2 q = probability of getting failure = 1 p = 1 (1/2) = 1/2 Probability of getting x heads in 10 coins = 10 C x p x q 10x = 10 C x (1/2) x (1/2) 10x = 10 C x (1/2) 10 Hence, probability of getting atleast 7 Heads P(X>=7) = P(7) + P(8) + P(9) + P(10) = 10 C 7 (1/2) C 8 (1/2) C 9 (1/2) C 10 (1/2) 10 = ( 10 C C C C 10 )(1/2) 10 = ( )(1/1024) = 176/1024 = 11/64 1. (e) What do you mean by Goodness to fit Test? What for the said test is required? Please refer Block. 1. (f) Calculate the value of the integral using Simpsons 3/8 th rule. { } Simpsons 3/8 th rule [(y 0 + y n ) + 3(y 1 + y 2 + y 4 + y ) + 2(y 3 + y 6 + y )] Divide the interval [4,5.2] into 6 equal subintervals, each of width = = 0.2 Values of y = logx are tabulated below. x y=logx By Simpson s 3/8 th rule, we get = [( ) + 3( ) + 2(1.5261)] = [ ] =
6 5 1. (g) Find the probability that an individual s IQ score is between 91 and 121 provided the individuals IQ score has a Normal Distribution N(100,15 2 ). We require P[91 < X < 121] Given, N(100,15 2 ) Mean=100 St dev = 15 Standardizing X, P[ < < P[0.6 < Z < 1.4] =F(1.4) F(0.6) =F(1.4) (1 F(0.6)) from table we have, = ( ) = (a) Determine the value of the expression x = + + accurate upto 4 significant digits, and also find the absolute and relative errors. value upto 4 significant digits of = = = x = = We know, if a number is correct to n significant digits then error = * 10 1n here n=4, error = * = * 103 = Absolute Error E a = = Also total absolute error shows that the sum x is correct to 3 significant digits only hence x = 6.61 and Relative Error E r = E a / x = / 6.61 =
7 6 2. (b) Determine the value of Y using Euler s method, when x=0.1 Given Y(0)=1 and Y = X 2 +Y { } Euler Method y n+1 = y n + h*f(x n,y n ) Let h=0.1 / 5 = 0.02 Given, x 0 =0,y 0 =1,f(x,y) = x 2 + y Taking n=0,1,2,3,4 in succession y 1 = y * f(x 0,y 0 ) = * (0 + 1) = 1.02 x 1 = x = 0.02 y 2 = y * f(x 1,y 1 ) = *( ) = x 2 = x = 0.04 y 3 = y *f(x 2,y 2 ) = *( ) = x3 = x = 0.06 y4 = y *f(x3,y3) = *( ) = x4 = x = 0.08 y5 = y *f(x4,y4) = *( ) = Hence the value of Y at X=0.1 is (c) Find the value of Δ tan 1 x, where Δ is the difference operator, with differencing step size h. Δtan1x = tan 1 (x+h) tan 1 (x) = tan 1 ( = tan 1 ( ) ) )
8 7 2. (d) Solve the following system of equations by using LU Decomposition method. x + y = 2, 2x + 3y = 5 The given system of equations in matrix form Ax=B is [ ] [ ] [ ] Decomposing A into LU form, A=LU [ ] [ ] [ ] [ ] [ ] Comparing both sides, we have u 11 =1 u 12 =1 l 21 u 11 =2 which gives l 21 =2 l 21 u 12 + u 22 =3 which gives u 22 =1 Substituting these values, we have L = [ ] [ ] Now, Ax = B (LU)x=B L(Ux)=B Ly=B where Ux=y We solve first Ly=B [ ] [ ] [ ] Using forward substitution, y 1 =2 2y 1 +y 2 =5 which gives y 2 =1 Now, Ux=y gives [ ] [ ] [ ] Using backward substitution, x 2 =1 x 1 +x 2 =2 which gives x 1 =1 Hence 1,1 is the solution of the given equation.
9 8 3. (a) Solve the initial value problem given below by using RungeKutta Method. =y x with y(0)=2 and h=0.1 Also find y(0.1) and y(0.2) correct to four decimal places. We have, x 0 =0, y 0 =2, h=0.1 Then we get, k 1 =hf(x 0,y 0 ) = 0.1(20) = 0.2 k 2 = hf(x 0 + h/2,y 0 + k 1 /2) = 0.1( ) = k 3 = hf(x 0 + h/2,y 0 + k 2 /2) = 0.1( ) = k 4 = hf(x 0 + h,y 0 + k 3 ) = 0.1( ) = Therefore y = y 0 + 1/6(k 1 +2 k k 3 + k 4 ) = = y(0.1) = upto 4 decimal places. For y(0.2) we have x 0 =0.1, y 0 = k 1 =hf(x 0,y 0 ) = 0.1( ) = k 2 = hf(x 0 + h/2,y 0 + k 1 /2) = 0.1( ) = k 3 = hf(x 0 + h/2,y 0 + k 2 /2) = 0.1( ) = k 4 = hf(x 0 + h,y 0 + k 3 ) = 0.1( ) = y(0.2) = /6(k 1 +2 k k 3 + k 4 ) = = upto 4 decimal places. 3. (b) Determine the goodness to fit parameter R for the data given below. X Y Analyse the results and comment on whether the predicted line fits well into the data or not. We have
10 9 X Y XY X*X Y*Y Putting these values in the formula, we get r = / = Also, r 2 = As r and r 2 are close to 1, x and y have strong positive correlation, hence the predicted line fits given data very well. 4. (a) Develop the difference table for the data given below and use it to find the first and tenth term for the given data. X Y The forward difference table of above data is
11 10 X Y ΔY Δ 2 Y Δ 3 Y Δ 4 Y By Newtons forward interpolation formula, f(x) = f(0) + (x3) f(0) + + We have f(x) = (x3) * * = x x 2 8.4x x 3 1.2x x 6.0 = 0.1x 3 f(1) = 0.1 * 1 3 = 0.1 f(10) = 0.1 * 10 3 = (b) Find the smallest root of the equation f(x) = x 3 6x x 6 = 0 by using Newton Raphson method. Give two drawbacks of NewtonRaphson method. We have, f(x) = x 3 6x x 6 f (x) = 3x 2 12x + 11 By Newton Raphson method we have, x n+1 = x n  = x n  = Now, f(0) = 6 f(1) = 0, i.e 1 is root of the eqn.
12 11 Taking initial approximation as x 0 = 1.2 First approximation x 1 = = Second approximation x 2 = = Third approximation x 3 = = Fourth approximation x 4 = = 1 Fifth approximation x 5 = = 1 Since x4 = x5, 1 is the root of the given equation. Drawbacks: i) Division by zero occurs in some cases. let f(x) = x 3 2x = 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) Root jumping takes place in some cases. i.e, we move far away from the root in subsequent approximations.
13 12 5. (a) In a partially destroyed laboratory record of an analysis of correlation data, the following data are only legible. i) Variance of X = 9 ii) Regression equation: 8X 10Y + 66 = 0 40X 18Y = 214 Using this legible data determine the following. i) Mean value of X and Y ii) Correlation coefficient between Xand Y iii) Standard Deviation of Y Since both the lines of regression pass through the point, Solving (1) and (2) we get, Let 8X 10Y + 66 = 0 and 40X 18Y 214 = 0 be the lines of regression of y on x and x on y resp. rewriting the equations as, Y = 0.8X X = 0.45Y regression coefficient of y on x is = 0.8 regression coefficient of x on y is = 0.45 Hence, r2 = 0.8 * 0.45 = 0.36 r = 0.6 (+ sign with square root is taken as regression coefficients are +ve) Given, Variance of X = Also, regression coefficient of y on x = = 0.8 * 3 / 0.6 = 4
14 13 5. (b) Evaluate the integral I = intervals. by using composite Trapezoidal rule with 2 and 4 sub { } Composite Trapezoidal Rule I T [f] = [first + last + 2*(sum of remaining)] The table below shows the value of function f for x=0 to x=1 with h=1/4 x 0 1/4 2/4 3/4 1 f(x) f0 f1 f2 f3 f4 If N=2, number of subintervals, then h=1/2 and we have to use f0,f2,f4 Hence we get, I T [f] = [ (0.667)] = If N=4, number of subintervals, then h=1/4 and we have to use f0,f1,f2,f3,f4 Hence we get, I T [f] = [ ( )] = 0.697
15 14 5. (c) Find the approximate value of the root of the equation x 3 + x 1 = 0, near x = 1. using RegularFalsi method, twice. Given f(x) = x 3 + x 1 f(0) = 1 f(1) = 1 Since f(0)*f(1) < 0, hence a root lies between 0 and 1. First Approximation x 0 =0,x 1 =1 f(x 0 ) = 1, f(x 1 ) = 1 x2 = x0  f(x0) = 0  (1) = 0.5 f(0.5) = (0.5) = Since f(0.5)*f(1) < 0, hence a root lies between 0.5 and 1. Second Approximation x 0 =0.5,x 1 =1 f(x 0 ) = , f(x 1 ) = 1 x3 = x0  f(x0) = (0.375) = 0.63 Hence 0.63 is the approx. root of the given equation.
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