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1.2 Answers for Numerical Methods 633
ANSWERS FOR NUMERICAL METHODS
Exercise Set 1.2 (Page 000)
- For each part, f ∈ C[a, b] on the given interval. Since f (a) and f (b) are of opposite sign, the Intermediate Value Theorem implies a number c exists with f (c) = 0.
- For each part, f ∈ C[a, b], f ′^ exists on (a, b), and f (a) = f (b) = 0. Rolle’s Theorem implies that a number c exists in (a, b) with f ′(c) = 0. For part (d), we can use [a, b] = [− 1 , 0] or [a, b] = [0, 2].
- a. P 2 (x) = 0 b. R 2 (0.5) = 0.125; actual error = 0. 125 c. P 2 (x) = 1 + 3(x − 1) + 3(x − 1)^2 d. R 2 (0.5) = − 0 .125; actual error = − 0. 125
- Since P 2 (x) = 1 + x and R 2 (x) = −^2 e
ξ (^) (sin ξ + cos ξ) 6 x
3
for some number ξ between x and 0, we have the following: a. P 2 (0.5) = 1.5 and f (0.5) = 1.446889. An error bound is 0.093222 and |f (0.5) − P 2 (0.5)| ≤ 0. 0532 b. |f (x) − P 2 (x)| ≤ 1. 252 c. ∫^01 f (x) dx ≈ 1. 5 d. | ∫^01 f (x) dx−∫^01 P 2 (x) dx| ≤ ∫^01 |R 2 (x)| dx ≤ 0 .313, and the actual error is 0.122.
- The error is approximately 8. 86 × 10 −^7.
634 CHAPTER 1 Answers for Numerical Methods
- a. P 3 (x) = 13 x + 16 x^2 + 64823 x^3
b. We have f (4)(x) = − 2592199 ex/^2 sin x 3 + 388861 ex/^2 cos x 3 , so |f (4)(x)| ≤ |f (4)(0.60473891)| ≤ 0. 09787176 for 0 ≤ x ≤ 1, and |f (x) − P 3 (x)| ≤ |f^
(4)(ξ)| 4! |x|
4 ≤ 0.^09787176
- A bound for the maximum error is 0.0026.
- a. e−t^2 =
∑^ ∞
k=
(−1)kt^2 k k! Use this series to integrate √^2 π
∫ (^) x 0 e−t^2 dt and obtain the result. b. √^2 π e
−x^2 ∑^ ∞ k=
2 kx^2 k+ 1 · 3 · · · (2k + 1) =
√^2
π
[
1 − x^2 +^12 x^4 − 16 x^7 + 241 x^8 + · · ·
]
[
x +^23 x^3 + 154 x^5 + 1058 x^7 + 94516 x^9 + · · ·
]
= √^2 π
[
x − 13 x^3 + 101 x^5 − 421 x^7 + 2161 x^9 + · · ·
]
= erf (x)
c. 0. 8427008 d. 0. 8427069
636 CHAPTER 1 Answers for Numerical Methods
Approximation Absolute Error Relative Error a. 3. 14557613 3. 983 × 10 −^3 1. 268 × 10 −^3 b. 3. 14162103 2. 838 × 10 −^5 9. 032 × 10 −^6
- b. The first formula gives − 0 .00658 and the second formula gives − 0 .0100. The true three-digit value is − 0. 0116.
- a. 39. 375 ≤ volume ≤ 86. 625 b. 71. 5 ≤ surface area ≤ 119. 5
Exercise Set 1.4 (Page 000)
- x 1 Absolute Error^ Relative Error^ x 2 Absolute Error^ Relative Error a. 92. 26 0. 01542 1. 672 × 10 −^4 0. 005419 6. 273 × 10 −^7 1. 157 × 10 −^4 b. 0. 005421 1. 264 × 10 −^6 2. 333 × 10 −^4 − 92. 26 4. 580 × 10 −^3 4. 965 × 10 −^5 c. 10. 98 6. 875 × 10 −^3 6. 257 × 10 −^4 0. 001149 7. 566 × 10 −^8 6. 584 × 10 −^5 d. − 0. 001149 7. 566 × 10 −^8 6. 584 × 10 −^5 − 10. 98 6. 875 × 10 −^3 6. 257 × 10 −^4
- a. − 0. 1000 b. − 0. 1010 c. Absolute error for part (a) is 2. 331 × 10 −^3 with relative error 2. 387 × 10 −^2. Absolute error for part (b) is 3. 331 × 10 −^3 with relative error 3. 411 × 10 −^2.
- Approximation Absolute Error Relative Error a. and b. 3. 743 1. 011 × 10 −^3 2. 694 × 10 −^3 c. and d, 3. 755 1. 889 × 10 −^4 5. 033 × 10 −^4
- a. The approximate sums are 1.53 and 1.54, respectively. The actual value is 1.549. Significant round-off error occurs earlier with the first method.
1.4 Answers for Numerical Methods 637
- Approximation Absolute Error Relative Error a. 2. 715 3. 282 × 10 −^3 1. 207 × 10 −^3 b. 2. 716 2. 282 × 10 −^3 8. 394 × 10 −^4 c. 2. 716 2. 282 × 10 −^3 8. 394 × 10 −^4 d. 2. 718 2. 818 × 10 −^4 1. 037 × 10 −^4
- The rates of convergence are as follows.
a. O(h^2 ) b. O(h) c. O(h^2 ) d. O(h)
- Since limn→∞ xn = limn→∞ xn+1 = x and xn+1 = 1 + (^) x^1 n , we have x = 1 + (^) x^1. This implies that x = (1 + √5)/2. This number is called the golden ratio. It appears frequently in mathematics and the sciences.
- a. n = 50 b. n = 500 c. An accuracy of 10−^4 cannot be obtained with Digits set to 10 in some earlier versions of Maple. However, in Release 7 we get n = 5001.
2.3 Answers for Numerical Methods 639
Thus, a 1 = a, f (a 1 ) < 0, b 1 = b, and f (b 1 ) > 0. a. Since a + b < 2, we have p 1 = a+ 2 b and 1/ 2 < p 1 < 1. Thus, f (p 1 ) > 0. Hence, a 2 = a 1 = a and b 2 = p 1. The only zero of f in [a 2 , b 2 ] is p = 0, so the convergence will be to 0. b. Since a + b > 2, we have p 1 = a+ 2 b and 1 < p 1 < 3 /2. Thus, f (p 1 ) < 0. Hence, a 2 = p 1 and b 2 = b 1 = b. The only zero of f in [a 2 , b 2 ] is p = 2, so the convergence will be to 2. c. Since a + b = 2, we have p 1 = a+ 2 b = 1 and f (p 1 ) = 0. Thus, a zero of f has been found on the first iteration. The convergence is to p = 1.
Exercise Set 2.3 (Page 000)
- a. p 3 = 2. 45454 b. p 3 = 2. 44444
- Using the endpoints of the intervals as p 0 and p 1 , we have the following. a. p 11 = 2. 69065 b. p 7 = − 2. 87939 c. p 6 = 0. 73909 d. p 5 = 0. 96433
- Using the endpoints of the intervals as p 0 and p 1 , we have the following. a. p 16 = 2. 69060 b. p 6 = − 2. 87938 c. p 7 = 0. 73908 d. p 6 = 0. 96433
- For p 0 = 0.1 and p 1 = 3 we have p 7 = 2. 363171. For p 0 = 3 and p 1 = 4 we have p 7 = 3. 817926. For p 0 = 5 and p 1 = 6 we have p 6 = 5. 839252. For p 0 = 6 and p 1 = 7 we have p 9 = 6. 603085.
- For p 0 = 1 and p 1 = 2, we have p 5 = 1.73205068, which compares to 14 iterations of the Bisection method.
640 CHAPTER 2 Answers for Numerical Methods
For p 0 = 0 and p 1 = 1, the Secant method gives p 7 = 0.589755. The closest point on the graph is (0. 589755 , 0 .347811).
a. For p 0 = −1 and p 1 = 0, we have p 17 = − 0 .04065850, and for p 0 = 0 and p 1 = 1, we have p 9 = 0.9623984. b. For p 0 = −1 and p 1 = 0, we have p 5 = − 0 .04065929, and for p 0 = 0 and p 1 = 1, we have p 12 = − 0 .04065929. The Secant method fails to find the zero in [0, 1].
For p 0 = 12 , p 1 = π 4 , and tolerance of 10−^100 , the Secant method required 11 iterations, giving the 100-digit answer p 11 =. 73908513321516064165531208767387340401341175890075746496568063577328
For p 0 = 0.1 and p 1 = 0.2, the Secant method gives p 3 = 0.16616, so the depth of the water is 1 − p 3 = 0.83385 ft.
Exercise Set 2.4 (Page 000)
- p 2 = 2. 60714
- a. For p 0 = 2, we have p 5 = 2.69065. b. For p 0 = −3, we have p 3 = − 2 .87939. c. For p 0 = 0, we have p 4 = 0.73909. d. For p 0 = 0, we have p 3 = 0.96434.
- Newton’s method gives the following approximations: With p 0 = 1. 5 , p 6 = 2.363171; with p 0 = 3. 5 , p 5 = 3.817926; With p 0 = 5. 5 , p 4 = 5.839252; with p 0 = 7, p 5 = 6.603085.
642 CHAPTER 2 Answers for Numerical Methods
a. b. c. d. q 0 0. 258684 0. 907859 0. 548101 0. 731385 q 1 0. 257613 0. 909568 0. 547915 0. 736087 q 2 0. 257536 0. 909917 0. 547847 0. 737653 q 3 0. 257531 0. 909989 0. 547823 0. 738469 q 4 0. 257530 0. 910004 0. 547814 0. 738798 q 5 0. 257530 0. 910007 0. 547810 0. 738958
- Newton’s Method gives p 6 = − 0 .1828876, and the improved value is q 6 = − 0. 183387.
- a. (i) Since |pn+1 − 0 | = (^) n+1^1 < (^) n^1 = |pn − 0 |, the sequence {^1 n^ }^ converges linearly to
- (ii) We need (^) n^1 ≤ 0 .05 or n ≥ 20. (iii) Aitken’s ∆^2 method gives q 10 = 0.045. b. (i) Since |pn+1 − 0 | = (^) (n+1)^12 < (^) n^12 = |pn − 0 |, the sequence {^ n^12 }^ converges linearly to 0. (ii) We need (^) n^12 ≤ 0 .05 or n ≥ 5. (iii) Aitken’s ∆^2 method gives q 2 = 0.0363.
- a. Since |pn+1 − 0 | |pn − 0 |^2 =
10 −^2 n+ (10−^2 n^ )^2 =
10 −^2 n+ 10 −^2 n+1^ = 1, the sequence is quadratically convergent. b. Since |pn+1 − 0 | |pn − 0 |^2 =
10 −(n+1)k (10−nk^ )^2 =
10 −(n+1)k 10 −^2 nk^ = 10
2 nk^ −(n+1)k
diverges, the sequence pn = 10−nk does not converge quadratically.
Exercise Set 2.6 (Page 000)
- a. For p 0 = 1, we have p 22 = 2.69065. b. For p 0 = 1, we have p 5 = 0.53209; for p 0 = −1, we have p 3 = − 0 .65270, and for p 0 = −3, we have p 3 = − 2 .87939.
2.6 Answers for Numerical Methods 643
c. For p 0 = 1, we have p 4 = 1.12412; and for p 0 = 0, we have p 8 = − 0 .87605. d. For p 0 = 0, we have p 10 = 1.49819.
- The following table lists the initial approximation and the roots. p 0 p 1 p 2 Approximated Roots Complex Conjugate Roots a. − 1 0 1 p 7 = − 0. 34532 − 1. 31873 i − 0 .34532 + 1. 31873 i 0 1 2 p 6 = 2. 69065 b. 0 1 2 p 6 = 0. 53209 1 2 3 p 9 = − 0. 65270 − 2 − 3 − 2. 5 p 4 = − 2. 87939 c. 0 1 2 p 5 = 1. 12412 2 3 4 p 12 = − 0 .12403 + 1. 74096 i − 0. 12403 − 1. 74096 i − 2 0 − 1 p 5 = − 0. 87605 d. 0 1 2 p 6 = 1. 49819 − 1 − 2 − 3 p 10 = − 0. 51363 − 1. 09156 i − 0 .51363 + 1. 09156 i 1 0 − 1 p 8 = 0. 26454 − 1. 32837 i 0 .26454 + 1. 32837 i
- a. The roots are 1. 244 , 8 .847, and − 1 .091. The critical points are 0 and 6.
Figure 0 Placed Here
for Exercise 5a b. The roots are 0.5798, 1.521, 2.332, and − 2 .432, and the critical points are 1, 2 .001, and − 1 .5. Note: New Figures
3.2 Answers for Numerical Methods 645
Exercise Set 3.2 (Page 000)
- a. (i) P 1 (x) = − 0. 29110731 x + 1; P 1 (0.45) = 0.86900171; | cos 0. 45 − P 1 (0.45)| = 0 .03144539; (ii) P 2 (x) = − 0. 43108687 x^2 − 0. 03245519 x+1; P 2 (0.45) = 0.89810007; | cos 0. 45 − P 2 (0.45)| = 0. 0023470 b. (i) P 1 (x) = 0. 44151844 x + 1; P 1 (0.45) = 1.1986833; |√ 1. 45 − P 1 (0.45)| = 0 .00547616; (ii) P 2 (x) = − 0. 070228596 x^2 +0. 483655598 x+1; P 2 (0.45) = 1.20342373; |√ 1. 45 − P 2 (0.45)| = 0. 00073573 c. (i) P 1 (x) = 0. 78333938 x; P 1 (0.45) = 0 .35250272; | ln 1. 45 − P 1 (0.45)| = 0 .01906083; (ii) P 2 (x) = − 0. 23389466 x^2 + 0. 92367618 x; P 2 (0.45) = 0.36829061; | ln 1. 45 − P 2 (0.45)| = 0. 00327294 d. (i) P 1 (x) = 1. 14022801 x; P 1 (0.45) = 0. 0 .51310260; | tan 0. 45 − P 1 (0.45)| = 0 .03004754; (ii) P 2 (x) = 0. 86649261 x^2 + 0. 62033245 x; P 2 (0.45) = 0.45461436; | tan 0. 45 − P 2 (0.45)| = 0. 02844071
a. n x 0 , x 1 ,... , xn Pn(8.4) 1 8. 3 , 8. 6 17. 87833 2 8. 3 , 8. 6 , 8. 7 17. 87716 3 8. 3 , 8. 6 , 8. 7 , 8. 1 17. 87714 b. n x 0 , x 1 ,... , xn Pn(− 13 ) 1 − 0. 5 , − 0. 25 0. 21504167 2 − 0. 5 , − 0. 25 , 0. 0 0. 16988889 3 − 0. 5 , − 0. 25 , 0. 0 , − 0. 75 0. 17451852 c. n x 0 , x 1 ,... , xn Pn(0.25) 1 0. 2 , 0. 3 − 0. 13869287 2 0. 2 , 0. 3 , 0. 4 − 0. 13259734 3 0. 2 , 0. 3 , 0. 4 , 0. 1 − 0. 13277477 d. n x 0 , x 1 ,... , xn Pn(0.9) 1 0. 8 , 1. 0 0. 44086280 2 0. 8 , 1. 0 , 0. 7 0. 43841352 3 0. 8 , 1. 0 , 0. 7 , 0. 6 0. 44198500
646 CHAPTER 3 Answers for Numerical Methods
- √ 3 ≈ P 4 (^12 )^ = 1. 7083
a. n Actual Error Error Bound 1 0. 00118 0. 00120 2 1. 367 × 10 −^5 1. 452 × 10 −^5 b. n Actual Error Error Bound 1 4. 0523 × 10 −^2 4. 5153 × 10 −^2 2 4. 6296 × 10 −^3 4. 6296 × 10 −^3 c. n Actual Error Error Bound 1 5. 9210 × 10 −^3 6. 0971 × 10 −^3 2 1. 7455 × 10 −^4 1. 8128 × 10 −^4 d. n Actual Error Error Bound 1 2. 7296 × 10 −^3 1. 4080 × 10 −^2 2 5. 1789 × 10 −^3 9. 2215 × 10 −^3
- f (1.09) ≈ 0 .2826. The actual error is 4. 3 × 10 −^5 , and an error bound is 7. 4 × 10 −^6. The discrepancy is due to the fact that the data are given to only four decimal places and only four-digit arithmetic is used.
- y = 4. 25
- The largest possible step size is 0.004291932, so 0.004 would be a reasonable choice.
- The difference between the actual value and the computed value is 23.
- a. x erf(x)
- 0 0
- 2 0. 2227
- 4 0. 4284
- 6 0. 6039
- 8 0. 7421
- 0 0. 8427
648 CHAPTER 3 Answers for Numerical Methods
- ∆^2 P (10) = 1140.
- The approximation to f (0.3) should be increased by 5.9375.
- f [x 0 ] = f (x 0 ) = 1, f [x 1 ] = f (x 1 ) = 3, f [x 0 , x 1 ] = 5
Exercise Set 3.4 (Page 000)
- The coefficients for the polynomials in divided-difference form are given in the follow- ing tables. For example, the polynomial in part (a) is H 3 (x) = 17.56492+3.116256(x− 8 .3)+0.05948(x− 8 .3)^2 − 0 .00202222(x− 8 .3)^2 (x− 8 .6). a. b. c. d.
- 56492 0. 022363362 − 0. 02475 − 0. 62049958
- 116256 2. 1691753 0. 751 3. 5850208
- 05948 0. 01558225 2. 751 − 2. 1989182 − 0. 00202222 − 3. 2177925 1 − 0. 490447 0 0. 037205 0 0. 040475 − 0. 0025277777
- 0029629628
- a. We have sin 0. 34 ≈ H 5 (0.34) = 0.33349. b. The formula gives an error bound of 3. 05 × 10 −^14 , but the actual error is 2. 91 × 10 −^6. The discrepancy is due to the fact that the data are given to only five decimal places. c. We have sin 0. 34 ≈ H 7 (0.34) = 0.33350. Although the error bound is now
- 4 × 10 −^20 , the accuracy of the given data dominates the calculations. This result is actually less accurate than the approximation in part (b), since sin 0.34 = 0 .333487.
- For 2(a) we have an error bound of 5. 9 × 10 −^8. The error bound for 2(c) is 0 since f (n)(x) ≡ 0 for n > 3.
3.5 Answers for Numerical Methods 649
- The Hermite polynomial generated from these data is H 9 (x) = 75x + 0. 222222 x^2 (x − 3) − 0. 0311111 x^2 (x − 3)^2 − 0. 00644444 x^2 (x − 3)^2 (x − 5) + 0. 00226389 x^2 (x − 3)^2 (x − 5)^2 − 0. 000913194 x^2 (x − 3)^2 (x − 5)^2 (x − 8) + 0. 000130527 x^2 (x − 3)^2 (x − 5)^2 (x − 8)^2 − 0. 0000202236 x^2 (x − 3)^2 (x − 5)^2 (x − 8)^2 (x − 13).
a. The Hermite polynomial predicts a position of H 9 (10) = 743 ft and a speed of H 9 ′(10) = 48 ft/s. Although the position approximation is reasonable, the low-speed prediction is suspect. b. To find the first time the speed exceeds 55 mi/h = 80.6 ft/s, we solve for the smallest value of t in the equation 80.6 = H 9 ′(x). This gives x ≈ 5. 6488092. c. The estimated maximum speed is H 9 ′(12.37187) = 119.423 ft/s ≈ 81 .425 mi/h.
Exercise Set 3.5 (Page 000)
- S(x) = x on [0, 2]
- The equations of the respective free cubic splines are given by S(x) = Si(x) = ai + bi(x − xi) + ci(x − xi)^2 + di(x − xi)^3 , for x in [xi, xi+1] and the coefficients in the following tables. a. i ai bi ci di 0 17. 564920 3. 13410000 0. 00000000 0. 00000000 b. i ai bi ci di 0 0. 22363362 2. 17229175 0. 00000000 0. 00000000 c. i ai bi ci di 0 − 0. 02475000 1. 03237500 0. 00000000 6. 50200000 1 0. 33493750 2. 25150000 4. 87650000 − 6. 50200000
3.5 Answers for Numerical Methods 651
- a. The equation of the spline is s(x) = si(x) = ai + bi(x − xi) + ci(x − xi)^2 + di(x − xi)^3 on the interval [xi, xi+1], where the coefficients are given in the following table. xi ai bi ci di 0 1. 0 0. 0 − 5. 193321 2. 028118
- 25 0. 7071068 − 2. 216388 − 3. 672233 4. 896310
- 5 0. 0 − 3. 134447 0. 0 4. 896310
- 75 − 0. 7071068 − 2. 216388 3. 672233 2. 028118
b.
0 s(x) dx = 0. 000000 c. s′(0.5) = − 3. 13445 , and s′′(0.5) = 0. 0.
a = 2, b = −1, c = −3, d = 1
B = 14 , D = 14 , b = − 12 , d = (^14)
Let f (x) = a + bx + cx^2 + dx^3. Clearly, f satisfies properties (a), (c), (d), (e) of the definition and f interpolates itself for any choice of x 0 ,... , xn. Since (ii) of (f) in the definition holds, f must be its own clamped cubic spline. However, f ′′(x) = 2c + 6dx can be zero only at x = −c/ 3 d. Thus, part (i) of (f) in the definition cannot hold at two values x 0 and xn, and f cannot be a natural cubic spline.
xi ai bi ci di 1940 132165 1651. 85 0. 00000 2. 64248 1950 151326 2444. 59 79. 2744 − 4. 37641 1960 179323 2717. 16 − 52. 0179 2. 00918 1970 203302 2279. 55 8. 25746 − 0. 381311 1980 226542 2330. 31 − 3. 18186 0. 106062 S(1930) = 113004, S(1965) = 191860, and S(2010) = 296451.
- a. S(x) = Si(x) = ai + bi(x − xi) + ci(x − xi)^2 + di(x − xi)^3 on [xi, xi+1], where
652 CHAPTER 3 Answers for Numerical Methods
xi ai bi ci di 0 0 88. 8 0 12. 8
- 25 22. 4 91. 2 9. 6 0
- 5 45. 8 96. 0 9. 6 − 4. 8
- 0 95. 6 102. 0 2. 4 − 3. 2
- 25 b. 1:10 (^1340) c. Starting speed ≈ 40 .54 mi/h. Ending speed ≈ 35 .09 mi/h.
Exercise Set 3.6 (Page 000)
- a. x(t) = − 10 t^3 + 14t^2 + t, y(t) = − 2 t^3 + 3t^2 + t b. x(t) = − 10 t^3 + 14. 5 t^2 + 0. 5 t, y(t) = − 3 t^3 + 4. 5 t^2 + 0. 5 t c. x(t) = − 10 t^3 + 14t^2 + t, y(t) = − 4 t^3 + 5t^2 + t d. x(t) = − 10 t^3 + 13t^2 + 2t, y(t) = 2t
- a. x(t) = − 11. 5 t^3 + 15t^2 + 1. 5 t + 1, y(t) = − 4. 25 t^3 + 4. 5 t^2 + 0. 75 t + 1 b. x(t) = − 6. 25 t^3 + 10. 5 t^2 + 0. 75 t + 1, y(t) = − 3. 5 t^3 + 3t^2 + 1. 5 t + 1 c. For t between (0, 0) and (4, 6) we have x(t) = − 5 t^3 + 7. 5 t^2 + 1. 5 t, y(t) = − 13. 5 t^3 + 18t^2 + 1. 5 t, and for t between (4, 6) and (6, 1) we have x(t) = − 5. 5 t^3 + 6t^2 + 1. 5 t + 4, y(t) = 4t^3 − 6 t^2 − 3 t + 6.
d. For t between (0, 0) and (2, 1) we have x(t) = − 5. 5 t^3 + 6t^2 + 1. 5 t, y(t) = − 0. 5 t^3 + 1. 5 t,
