《数值分析》黄仿伦改编英文版课后习题答案

Answers for Exercises—Numerical methods using Matlab Chapter 1 P10 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 1 0 1.9000 0.1000 0.0500 0 3.800 0.2000 0.0500 1 1.7950 0.2050 0.1025 1 3.9800 0.0200 0.0050 2 1.5690 0.4310 0.2155 2 3.9998 0.0002 0.0001 3 1.0451 0.9549 0.4775 3 4.0000 0.0000 0.0000 (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 that contains 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 3 n an f(an) bn f(bn) cn f(cn) 0 3 -0.1425 4 1.1578 3.5 0.3746 1 3 -0.1425 3.5 0.3746 3.25 0.1088 2 3 -0.1425 3.25 0.1088 3.125 -0.0166 3 3.125 -0.0166 3.25 0.1088 3.1875 0.0459 4 3.125 -0.0166 3.1875 0.0459 3.1563 0.0147 5 3.125 -0.0479 3.1563 0.0147 3.1406 -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 4 n an f(an) bn f(bn) cn f(cn) 0 1 1.5574 3 -0.1425 2 -2.1850 1 1 1.5574 2 -2.1850 1.5 14.1014 2 1.5 14.1014 2 -2.1850 1.75 -5.5204 3 1.5 14.1014 1.75 -5.5204 1.625 -18.4309 4 1.5 14.1014 1.625 -18.4309 1.5625 120.5325 5 1.5625 120.5325 1.625 -18.4309 1.5938 -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 5 p0 p1 p2 p3 p4 1.6 2.5273 2.3152 2.3028 2.3028 0.0 -3.0000 -1.9615 -1.1472 -0.0066 Obviously, 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 Exercises—Numerical methods using Matlab Chapter 2 P44 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. 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 Solution Using the algorithm of Gaussian Elimination gives EMBED Equation.3 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 in The set of solutions is and 15. Solution The Hilbert matrix is defined by where for . The inverse of the Hilbert matrix is The exact solution is . 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 2.1 0 (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 2.2 0 (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 matrices Split the coefficient matrix into three matrices . The iterative matrix of Jacobi iteration is The 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 2.3 k 0 0 0 0 1 3.2500 1.6000 0.3333 2 2.9333 2.1833 1.1500 3 2.9917 1.9567 0.9472 4 2.9976 2.0089 1.0044 5 2.9989 1.9986 0.9977 6 2.9998 2.0002 0.9999 7 2.9999 2.0000 0.9999 8 3.0000 2.0000 1.0000 9 3.0000 2.0000 1.0000 Jacobi iteration converges to the solution (3, 2, 1)’ Gauss-Seidel iterative formula is for ’ for are listed in Table 2.4 with starting value Table 2.4 k 0 0 0 0 1 3.2500 2.2500 1.0417 2 2.9479 1.9813 0.9858 3 3.0011 2.0031 0.9999 4 2.9992 1.9999 0.9998 5 3.0000 2.0000 1.0000 6 3.0000 2.0000 1.0000 7 3.0000 2.0000 1.0000 8 3.0000 2.0000 1.0000 9 3.0000 2.0000 1.0000 Gauss-Seidel iteration converges to the solution (3, 2, 1)’ Answers for Exercises—Numerical methods using Matlab Chapter 3 P99 1. Solution (a) The nth order derivative of is . Therefore, , and . Estimating the remainder term gives for . Substituting to gives and . By using Taylor polynomial we have P108 1. (a) Using the Horner’s method to find gives 4 -0.02 0.1 -0.2 1.66 -0.08 0.08 -0.48 -0.02 0.02 -0.12 1.18 So =1.18. From part (a) we have . can be also obtained by using Horner’s method. 4 -0.02 0.02 -0.12 -0.08 -0.24 -0.02 -0.06 -0.36 So =-0.36 Another method: 4 -0.02*3 0.1*2 -0.2 -0.24 -0.16 -0.06 -0.04 -0.36 Hence, P(4)=-0.36. Find and firstly. 4 -0.02/4 0.1/3 -0.2/2 1.66 0 -0.02 0.0532 -0.1872 5.8912 -0.005 0.0133 -0.0468 1.4728 5.8912 1 -0.02/4 0.1/3 -0.2/2 1.66 0 -0.005 0.0283 -0.0717 1.5883 -0.005 0.0283 -0.0717 1.5883 1.5883 Then 4.3029. (d) Use Horner’s method to evaluate P(5.5) 5.5 -0.02 0.1 -0.2 1.66 -0.1100 -0.0550 -1.4025 -0.02 -0.0100 -0.2550 0.2575 Hence, P(5.5)=0.2575. 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 3.1 x -1 0 1 2 f(x) -1 0 1 8 (a) Find the Lagrange coefficient polynomials EMBED Equation.3 The interpolating polynomial is . (b) EMBED Equation.3 . (c) (d) EMBED Equation.3 . (e) EMBED Equation.3 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: k xk f[ ] f[,] f[, ,] f[, , ,] f[, , , ,] 0 1.0 3.60 1 2.0 1.80 -1.80 2 3.0 1.20 -0.6 0.6 3 4.0 0.90 -0.3 0.15 -0.15 4 5.0 0.72 -0.18 0.06 -0.03 0.03 (b) Find the Newton polynomials with order 1, 2, 3 and 4. , , , . (c)–(d) The results are listed in Table 3.2 Table 3.2 xk f(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)| 2.5 1.4400 0.9000 0.5400 1.3500 0.0900 1.4062 0.0338 1.3444 0.0956 3.5 1.0286 -0.9000 1.9286 1.3500 0.3214 1.0688 0.0402 1.3219 0.2933 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 Exercises—Numerical methods using Matlab Chapter 4 P157 1(a). Solution The sums for obtaining Normal equations are listed in Table 4.1 Table 4.1 -2 1 1.2 4 -2 0.04 -1 2 1.9 1 -2 0.01 0 3 2.6 0 0 0.16 1 3 3.3 1 3 0.09 2 4 4.0 4 8 0 0 13 10 7 0.3 The 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 that the 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 4.2 2.0 5.9 4.0 16.0 64 23.6 47.2 2.3 8.3 5.29 27.9841 148 43.907 100.9861 2.6 10.7 6.76 45.6976 308.9 72.332 188.0632 2.9 13.7 8.41 70.7281 594.8 115.217 334.1293 3.2 17.0 10.24 104.8576 1073.7 174.08 557.056 13 55.6 34.7 265.3 2189.5 429.136 1227.4 Hence, and . , . fits the given data better. P171 2(c). The sums for normal equations are listed in Table 4.3. -2 10 4 -8 16 -20 40 -1 1 1 -1 1 -1 1 0 0 0 0 0 0 0 1 2 1 1 1 2 2 2 9 4 8 16 18 36 0 22 10 0 34 -1 79 Using the formula produces the system with unkowns A, B, and C Solving 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 -1 6.62 -1 1.8901 1 -1.8901 0 3.94 0 1.3712 0 0 1 2.17 1 0.7747 1 0.7747 2 1.35 2 0.3001 4 0.6002 3 0.89 3 -0.1165 9 -0.3496 5 4.2196 15 -0.8648 The 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 -1 6.62 -1 0.1511 1 -1.8901 0 3.94 0 0.2538 0 0 1 2.17 1 0.4608 1 0.7747 2 1.35 2 0.7407 4 0.6002 3 0.89 3 1.1236 9 -0.3496 5 2.7300 15 5.1620 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 polynomial 4. 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 gives Solve the linear equation to obtain In addition, Use formula (4. 65) to find the spline coefficients: , ; , ; , . Therefore, , for ; , for for . Answers for Exercises—Numerical methods using Matlab Chapter 5 P209 1(b). Solution Let The result of using the trapezoidal rule with h=1 is Using Simpson’s rule with h=1/2, we have For Simpson’s 3/8 rule with h=1/3, we obtain The result of using the Boole’s rule with h=1/4 is 4. Proof Integrate over . = . The Quadrature formula is called the trapezoidal rule. 6. Solution The Simpson’s rule is . It will suffice to apply Simpson’s rule over the interval [0, 2] with the test functions and . For the first four functions, since , , , , the Simpson’s rule is exact. But for , . . Therefore, the degree of precision of Simpson’s rule is n=3. The Simpson’s rule and the Simpson’s 3/8 rule have the same degree of precision n=3. P220 3(a) Solution When for , . The values of at 11 sample points (M=10) are listed in the Table 5.1: Table 5.1 x 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 g(x) 0 0.0010 0.0081 0.0280 0.0710 0.1563 0.3179 0.6098 1.1084 1.9156 3.1623 (i) Using the composite Trapezoidal rule , the computation is = =0.1576+0.4216=0.5792. (ii) Using the composite Simposon’s rule , the computation is = =0.5672. 7. (a) Because the formula is exact for the three functions , , and , we obtain three equations with unkowns , , and : , , . Solving this linear system gives , and . Thus, Let and denote EMBED Equation.3 Then the change of variable translates into [0, 2] and converts the integral expresion into . Hence, . The formula is known as the Simpson’s rule over . 8(a). 9(a). P234 1(a) Let . The Romberg table with three rows for is given as follows: J Trapezoidal Rule Simpson’s Rule Boole’s Rule 0 -0.04191 1 0.04418 0.07288 2 0.3800 0.4919 0.5198 Where , , , , , , 2. Proof If , then and . 9. (a) Let . implies . Thus . (b) Let . implies . Thus . 10. (a) Do variable translation . Then . That means the two integrals and have the same numerical value. (b) Let and . Use means that the truncation error is approximately. Note that . It means . But for , is not true for all and any integer . Thus the Romberg sequence is faster for than for even though they have the same numerical value. P242 1 (a) Applying the change of variable to gives . Thus the two integrals are and equivalent. (b) =0.0809 +58.5857=58.6667 If using to approximate the integral, The result is 6. Analysis: The fact that the degree of precision of N-point Gauss-Legendra integration is 2N-1 implies that the error term can be represented in the form . (a) Since , and implies . Thus = . (b) and implies . Thus = . 7. The nth Legendre polynomial is defined by The first five polynomials are The roots of them are same as ones in Table 5.8. 11. The conditions that the relation is exact for the functions means the three equations: Sloving the system gives . is called three-point Gauss-Legendre rule. Answers for Exercises—Numerical methods using Matlab Chapter 6 P249 1. (a) Proof Differentiate . Substitute and into the right-hand side of the equation . (b) Solution Let . Then for any . So, the Lipschitz constant is . 12 . Integrate both side of over [a, b]: . Then, , where is the solution of the I. V. P , for with . That means that the definite integral can be computed using the two values and of the solution of the given I. V. P.. 14. Solution Separate the two variables of the equation into the form . Integrate and yeild the general solution . The initial-value condition means that . The solution for the I. V. P. is . P257 3. (a)-(c) The formula using Euler’s method to solve the I. V. P. , can be represented in the form . When and , the results are listed in Table 6.1. Table 6.1 h=0.1 h=0.2 h=0.1 h=0.2 0 1 1 1 0 0.1 1.0000 0.9950 0.0050 0.2 0.9850 1.0000 0.9802 0.0048 0.0198 0.3 0.9606 0.9560 0.0046 0.4 0.9273 0.9408 0.9231 0.0042 0.0177 (d) The F. G. E. does decrease half approximatelly as expacted when h is halved. 6. When , and , the Euler’s formula for is in the form . With , the missing entries can be filled in the table. Year 1900 0.0 76.1 76.1 1910 10.0 92.4 89.0 1920 20.0 106.5 103.6 1930 30.0 123.1 120.1 1940 40.0 132.6 138.2 1950 50.0 152.3 158.3 1960 60.0 180.7 180.0 1970 70.0 204.9 202.8 1980 80.0 226.5 227.1 7. Proof Using Euler’s method to calculate gives = =… …= = P265 3(a) Solution Heun’s scheme for is in the form , . With and , the results are listed in Table 6.2. Table 6.2 k h=0.1 h=0.2 h=0.1 h=0.2 0 0 1.0000 1.0000 1 0 0 1 0.1 0.9950 0.9950 0.0000 2 0.2 0.9802 0.9800 0.9802 0.0000 0.0002 3 0.3 0.9560 0.9560 0.0000 4 0.4 0.9231 0.9228 0.9231 0.0000 0.0003 (b) when and when . (c) The F. G. E is reduced aproximatelly by a factor of when is halved. 6. Note that in the given I. V. P. Then Heun formula can be represented in the following form: Therefore, =… … = 7. The missing entries can be filled using . 1 1.732422 1/2 1.682121 1.665354 1/4 1.672269 1.668985 1/8 1.670076 1.669345 1/16 1.669558 1.669385 1/32 1.6694