《数值分析》黄仿伦改编英文版课后习题答案.doc

Answers for ExercisesNumerical methods using MatlabChapter 1P10 2. Solution (a) produces an equation . Solving it gives the roots and . Since and , thus, both and are fixed points of . (b) (d) The iterative rule using is. The results for part (b)-(d) with starting value and are listed in Table 1.Table 101.90000.10000.050003.8000.20000.050011.79500.20500.102513.98000.02000.005021.56900.43100.215523.99980.00020.000131.04510.95490.477534.00000.00000.0000 (e) Calculate values of at and ., and . Since is continuous, there exists a number such that for all . There also exists a number such that for all . Therefore, is an attractive fixed point. The sequence generated by with starting value converges to . is a repelling fixed point. The sequence generated by with starting value does not converge to .P11 4. Find the fixed point for : gives. Find the derivative: . Evaluate and : , . Both and gives . There is no reason to find the solution(s) using the fixed-point iteration. P11 6. Proof P21 4. False position method: Assume thatcontains the root. The equation of the secand line through and is . It intersects x-axise at (Eq. 1.36, p18), ;Since , then .Similarly, we have, , 10. Bisection method: Assume that contains the root. Then .(a) , then .Since , then .Similarly, we can obtain . The results are listed in Table 3. Table 3nanf(an)bnf(bn)cnf(cn)03-0.1425 41.15783.50.374613-0.14253.50.37463.250.108823-0.14253.250.10883.125-0.016633.125-0.01663.250.10883.18750.045943.125-0.01663.18750.04593.15630.014753.125-0.04793.15630.01473.1406-9.4265e-004 The values of tan(x) at midpoints are going to zero while the sequence converges (b) Since , , there exist a root in .The results using Bisection method are listed in Table 4.Table 4nanf(an)bnf(bn)cnf(cn)011.55743-0.14252-2.1850111.55742-2.18501.514.101421.514.10142-2.18501.75-5.520431.514.10141.75-5.52041.625 -18.430941.514.10141.625-18.43091.5625120.532551.5625120.53251.625-18.43091.5938-43.5584 Although the sequence converges, the values of tan (x) at midpoints are not going to zero. P36 2. has two zeros . ()The first derivative of is.The Newton-Raphson iterative function is .The Newton-Raphson formula is , . The results are listed in Table 5 with starting value p0=1.6 and p0=0.0 respectively.Table 5p0p1p2p3p41.62.52732.31522.30282.30280.0-3.0000-1.9615-1.1472-0.0066Obviously, the sequence generated by the starting value p0=0.0 does not converge.11. Use Newton-raphson method to solve .The derivative of is .Newton-Raphoson formula is , .Since is a zero of and ,The sequence generated by the recursive formula will converge to for any starting value , where .·Answers for ExercisesNumerical methods using MatlabChapter 2P44 2. Solution The 4th equation yields . Substituting to the 3rd equation gives . Substituting both and to the 2nd equation produces . is obtained by sustituting all , and to the 1st equation. The value of the determinant of the coefficient matrix is .4. Proof (a) Calculating the product of the two given upper-triangular matrices gives. It is also an upper-triangular matrix. (b) Let and where and when . Let . According to the definition of product of the two matrices, we have for all . when because and when . That means that the product of the two upper-triangular matrices is also upper triangular. 5. Solution From the first equation we have . Substituting to the second equation gives . is obtained from the third equation and is attained from the last equation.The value of the determinant of the coefficient is 7. Proof The formula of the back substitution for an upper-triangular system is and for .The process requires divisions, multiplications, and additions or subtractions. P53 1. Solution Using elementary transformations for the augmented matrix gives That means that is equivalent to The set of solutions is 11. Solution Using the algorithm of Gaussian Elimination gives The set of solutions of the system is obtained by the back substitutions and (Chasing method for solving tridiagonal linear systems)14. (a) (i) Solution Applying Gaussian elimination with partial pivoting to the augment matrix results inThe set of solutions is and 15. Solution The Hilbert matrix is defined by where for .(a) The inverse of the Hilbert matrix isThe exact solution is .(b) The solution is . >>1 H is ill-conditioned.A miss is as good as a mile. (失之毫厘，谬以千里)P62 5 (a) Solving gives . From we have . The product of and is . That means (b) Similarly to the part (a), we have , , and . 6. , P72 7. (a) Jacobi Iterative formula is for Results for, are listed in Table 2.1 with starting value . Table 2.10(0, 0, 0)1(-8.0000, 13.0000, 0.3333)2(57.3333, 45.3333, -4.5000)3(214.1667, -220.8333, 11.8889)The numerical results show that Jacobi iteration does not converge.(b) Gauss-Seidel Iterative formula is for Results, are listed in Table 2.2 with starting value Table 2.20(0, 0, 0)1(-8, 45, -10)2(207, -825, 207)3(-3929, 15934, -3965) Gauss-Seidel iteration does not converge as well.Reasons: Conside the eigenvalues of iterative matricesSplit the coefficient matrix into three matrices.The iterative matrix of Jacobi iteration isThe spectral raduis of is. So Jacobi method doesnot converge.Similarly, the iterative matrix of Gauss-Seidel iteration is .The spectral radius of is >1. So Gauss-Seidel method does not converge. 8. (a) Jacobi Iterative formula is for for are listed in Table 2.3 with starting value . Table 2.3k000013.25001.60000.333322.93332.18331.150032.99171.95670.947242.99762.00891.004452.99891.99860.997762.99982.00020.999972.99992.00000.999983.00002.00001.000093.00002.00001.0000Jacobi iteration converges to the solution (3, 2, 1)(b) Gauss-Seidel iterative formula is for for are listed in Table 2.4 with starting value Table 2.4k000013.25002.25001.041722.94791.98130.985833.00112.00310.999942.99921.99990.999853.00002.00001.000063.00002.00001.000073.00002.00001.000083.00002.00001.000093.00002.00001.0000 Gauss-Seidel iteration converges to the solution (3, 2, 1)Answers for ExercisesNumerical methods using MatlabChapter 3P99 1. Solution (a) The nth order derivative of is . Therefore, , and.(b) Estimating the remainder term gives for .(c) Substituting to gives and .By using Taylor polynomial we haveP108 1. (a) Using the Horners method to find gives4-0.020.1-0.21.66-0.080.08-0.48-0.020.02-0.121.18 So =1.18.(b) From part (a) we have . can be also obtained by using Horners method. 4-0.020.02-0.12-0.08-0.24-0.02-0.06-0.36 So =-0.36Another method: 4-0.02*30.1*2-0.2-0.24-0.16-0.06-0.04-0.36 Hence, P(4)=-0.36.(c) Find and firstly. 4-0.02/40.1/3-0.2/21.660-0.020.0532-0.18725.8912-0.0050.0133-0.04681.47285.8912 1-0.02/40.1/3-0.2/21.660-0.0050.0283-0.07171.5883-0.0050.0283-0.07171.58831.5883Then 4.3029.(d) Use Horners method to evaluate P(5.5)5.5-0.020.1-0.21.66-0.1100-0.0550-1.4025-0.02-0.0100-0.25500.2575 Hence, P(5.5)=0.2575.(d) Let . There are 4 coefficients needed to found. Substituting four known point , i=1, 2, 3, 4, into gives four linear equations with unknown , i=1, 2, 3, 4. The coefficients can be found by solving this linear system: P120 1. The values of f(x) at the given points are listed in Table 3.1: Table 3.1x-1012f(x)-1018(a) Find the Lagrange coefficient polynomials The interpolating polynomial is .(b) .(c) (d) . (e) 7. (a) Note that each Lagrange polynomial is of degree at most 2 and is a combination of . Hence is also a polynomial of degree at most 2.(b) For each , , the Lagrange coefficient polynomial , and for , . Therefore, .(c) is a polynomial of degree and has zeroes. According to the fundamental theorem of algebra, for all x.9. Let . is a polynomial of degree . is degree with at N+1 points implies that has N+1 zeroes. Therefore, for all x, that is, for all x.P131 6. (a) Find the divided-difference table: kxkf f,f, ,f, , ,f, , , ,01.03.6012.01.80-1.8023.01.20-0.60.634.00.90-0.30.15-0.1545.00.72-0.180.06-0.030.03(b) Find the Newton polynomials with order 1, 2, 3 and 4., , ,.(c)(d) The results are listed in Table 3.2Table 3.2xkf(xk)P1 (xk)|P1 (xk)- f(xk)|P2 (xk)|P2 (xk)- f(xk)|P3 (xk)|P3 (xk)- f(xk)|P4 (xk)|P4 (xk)- f(xk)|2.51.44000.90000.5400 1.3500 0.0900 1.4062 0.0338 1.3444 0.0956 3.51.0286-0.90001.92861.35000.32141.06880.04021.32190.2933 P143 6. , . The derivative of is . yields . Evaluating at and gives , , and . Therefore, , .10. When , the Chebyshev nodes are and .Calculating the Lagrange coefficient polynomials based on can produce the following results:The proof is finished.Answers for ExercisesNumerical methods using MatlabChapter 4P157 1(a). Solution The sums for obtaining Normal equations are listed in Table 4.1 Table 4.1-211.24-20.04-121.91-20.01032.6000.16133.3130.09244.04800131070.3The normal equations are . Then .The least-squares line is . P158 4. Proof Suppose the linear-squares line is where A and B satisfies the Normal equations and . meas thatthe point lies on the linear-squares line .5. First eliminating B on the Normal equations and gives where .Substituting A into the first equation gets .Note that .Simplifying B gives .8(b). The sums needed in the Normal equations are listed in Table 4.22.05.94.016.06423.647.22.38.35.2927.984114843.907100.98612.610.76.7645.6976308.972.332188.06322.913.78.4170.7281594.8115.217334.12933.217.010.24104.85761073.7174.08557.0561355.634.7265.32189.5429.1361227.4 Hence, and . , . fits the given data better.P171 2(c). The sums for normal equations are listed in Table 4.3. -2104-816-2040-111-11-1100000001211122294816183602210034-179Using the formula produces the system with unkowns A, B, and CSolving the obove system gives The fitting curve is P172 4. (a) Translate points in x-y plane into X-Y plane using . The results are listed in Table 4.4. Table 4.4-16.62-11.89011-1.890103.9401.37120012.1710.774710.774721.3520.300140.600230.893-0.11659-0.349654.219615-0.8648The Normal equations give the system Then , . Thus .The fitting curve is , and .(b) Translate points in x-y plane into X-Y plane using . The results are listed in Table 4.5. Table 4.5-16.62-10.15111-1.890103.9400.25380012.1710.460810.774721.3520.740740.600230.8931.12369-0.349652.7300155.1620 The Normal equations give the system Then , . The fitting curve is and .(c) It is easy to see that the exponential function is better comparing with errors in part (a) and part (b).P188 1. (a) Derivativing gives . Substituting the conditions into the derivative pruduces the system of equations (b) Solving the linear system of equations in (a) gives . The cubic polynomial is . Figure: Graph of the cubic polynomial4. Step 1 Find the quantities: , , , Step 2 Use to obtain the linear system.The solutions are .Step 3 Compute and using clamaped boundary. , Step 4 Find the spline coefficients , ; , ;, ;Step 5 The cubic spline is for , for , and for .5. Calculate the quantities: , , , . ( Same values as Ex. 4)Substituting , and into givesSolve the linear equation to obtain In addition, Use formula (4. 65) to find the spline coefficients:, ;, ;, .Therefore, , for ; , for for .Answers for ExercisesNumerical methods using MatlabChapter 5P209 1(b). Solution LetThe result of using the trapezoidal rule with h=1 is Using Simpsons rule with h=1/2, we haveFor Simpsons 3/8 rule with h=1/3, we obtainThe result of using the Booles rule with h=1/4 is4. Proof Integrate over .=.The Quadrature formula is called the trapezoidal rule.6. Solution The Simpsons rule is . It will suffice to apply Simpsons rule over the interval 0, 2 with the test functions and . For the first four functions, since, ,the Simpsons rule is exact. But for , .Therefore, the degree of precision of Simpsons rule is n=3.The Simpsons rule and the Simpsons 3/8 rule have the same degree of precision n=3.P220 3(a) Solution Whenfor , . The values of at 11 sample points (M=10) are listed in the Table 5.1:Table 5.1x