Solution manual for numerical methods 4th edition by faires

Solution Manual for Numerical Methods 4th Edition by Faires Mathematical Preliminaries Review of Calculus Exercise Set 1.2 (page 14) Show that the following equations have at least one solution in the given intervals (a) x cos x − 2x2 + 3x − = 0, (b) (x − 2) − ln x = 0, [1, 2] and [e, 4] (c) 2x cos(2x) − (x − 2) = 0, ∗ (d ) x − (ln x) = 0, x [0.2, 0.3] and [1.2, 1.3] [2, 3] and [3, 4] [4, 5] SOLUTION: 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 that a number c exists with f (c) = For part (d∗ ): It is not possible to algebraically solve for the solution x, but this is not required in the problem, we must show only that a solution exists Let f (x) = x − (ln x)x = x − exp(x(ln(ln x))) Since f is continuous on [4, 5] with f (4) ≈ 0.3066 and f (5) ≈ −5.799, the Intermediate Value Theorem implies that a number x must exist in (4, 5) with = f (x) = x − (ln x)x Find intervals containing solutions to the following equations (a) x − 3−x = (b) 4x2 − ex = (c∗ ) x3 − 2x2 − 4x + = (d) x3 + 4.001x2 + 4.002x + 1.101 = SOLUTION: (a) [0, 1] SOLUTION: (b) [0, 1], [4, 5], [−1, 0] SOLUTION: (c∗ ) Let f (x) = x3 − 2x2 − 4x + The critical points of f occur when = f ′ (x) = 3x2 − 4x − = (3x + 2)(x − 2); that is, when x = − 23 and x = Relative maximum and minimum values of f can occur only at these values There are at most three( solutions to f (x) = 0, because f (x) is a polynomial of ) degree three Since f (−2) = −5 and f − 23 ≈ 4.48; f (0) = and f (1) = −2; and f (2) = −5 and f (4) = 19; solutions lie in the intervals [−2, −2/3], [0, 1], and [2, 4] SOLUTION: (d) [−3, −2], [−1, −0.5], and [−0.5, 0] Full file at https://TestbankDirect.eu/Solution-Manual-for-Numerical-Methods-4th-Edition-by-Fai Solution Manual for Numerical Methods 4th Edition by Faires Exercise Set 1.2 Show that the first derivatives of the following functions are zero at least once in the given intervals (a) f (x) = − ex + (e − 1) sin((π/2)x), (b) f (x) = (x − 1) tan x + x sin πx, [0, 1] [0, 1] (c) f (x) = x sin πx − (x − 2) ln x, [1, 2] (d) f (x) = (x − 2) sin x ln(x + 2), [−1, 3] SOLUTION: For each part, f ∈ C[a, b], f ′ exists on (a, b) and f (a) = f (b) = Rolle’s Theorem implies that a number c exists in (a, b) with f ′ (c) = For part (d), we can use [a, b] = [−1, 0] or [a, b] = [0, 2] Find maxa≤x≤b |f (x)| for the following functions and intervals (a) f (x) = (2 − ex + 2x)/3, [0, 1] (b) f (x) = (4x − 3)/(x − 2x), [0.5, 1] (c) f (x) = 2x cos(2x) − (x − 2) , (d) f (x) = + e− cos(x−1) , [2, 4] [1, 2] SOLUTION: (a) First note that f ′ (x) = (−ex + 2) /3, so the only critical point of f occurs at x = ln 2, which lies in the interval [0, 1] The maximum for |f (x)| must consequently be max{|f (0)|, |f (ln 2)|, |f (1)|} = max{1/3, (2 ln 2)/3, (4 − e)/3} = (2 ln 2)/3 SOLUTION: (b) 0.8 SOLUTION: (c) 5.164000 SOLUTION: (d) 1.582572 Let f (x) = x3 (a) Find the second Taylor polynomial P2 (x) about x0 = (b) Find R2 (0.5) and the actual error when using P2 (0.5) to approximate f (0.5) (c) Repeat part (a) with x0 = (d) Repeat part (b) for the polynomial found in part (c) SOLUTION: (a) P2 (x) = SOLUTION: (b) R2 (0.5) = 0.125; actual error = 0.125 SOLUTION: (c) P2 (x) = + 3(x − 1) + 3(x − 1)2 SOLUTION: (d) R2 (0.5) = −0.125; actual error = −0.125 √ Let f (x) = x + (a) Find the third Taylor polynomial P3 (x) about x0 = √ √ √ √ (b) Use P3 (x) to approximate 0.5 , 0.75, 1.25, and 1.5 (c) Determine the actual error of the approximations in part (b) Full file at https://TestbankDirect.eu/Solution-Manual-for-Numerical-Methods-4th-Edition-by-Fai Solution Manual for Numerical Methods 4th Edition by Faires Mathematical Preliminaries SOLUTION: (a∗ ) The third Taylor polynomial is 1 P3 (x) = + x − x2 + x3 16 SOLUTION: ((b), (c)) The approximations and errors are shown in the following table x −0.5 −0.25 0.25 0.5 √P3 (x) x+1 √ | x + − P3 (x)| −0.5 0.7071068 0.003830719 −0.25 0.8660254 0.0001855 0.25 1.1180340 0.0001301 0.5 1.2247449 0.0018176 7.* Find the second Taylor polynomial for f (x) = ex cos x about x0 = (a) Use P2 (0.5) to approximate f (0.5), find an upper bound for |f (0.5) − P2 (0.5)|, and compare this to the actual error (b) Find a bound for the error |f (x) − P2 (x)|, for x in [0, 1] ∫ ∫ (c) Approximate f (x) dx using P2 (x) dx 0 (d) Find an upper bound for the error in part (c), and compare this bound to the actual error SOLUTION: Since f ′ (x) = ex (cos x − sin x), f ′′ (x) = −2ex (sin x), and f ′′′ (x) = −2ex (sin x + cos x), we have f (0) = 1, f ′ (0) = 1, and f ′′ (0) = So P2 (x) = + x and R2 (x) = −2eξ (sin ξ + cos ξ) x 3! (a) We have P2 (0.5) = + 0.5 = 1.5 and |f (0.5) − P2 (0.5)| ≤ max ξ∈[0.0.5] −2eξ (sin ξ + cos ξ) (0.5)2 ≤ (0.5)2 max |eξ (sin ξ + cos ξ)| 3! ξ∈[0,0.5] To maximize this quantity on [0, 0.5], first note that Dx ex (sin x + cos x) = 2ex cos x > 0, for all x in [0, 0.5] This implies that the maximum and minimum values of ex (sin x + cos x) on [0, 0.5] occur at the endpoints of the interval, and e0 (sin + cos 0) = < e0.5 (sin 0.5 + cos 0.5) ≈ 2.24 Hence |f (0.5) − P2 (0.5)| ≤ (0.5)3 (2.24) ≈ 0.0933 (b) A similar analysis to that in (a) gives, for all x ∈ [0, 1], |f (x) − P2 (x)| ≤ (1.0)3 e1 (sin + cos 1) ≈ 1.252 Full file at https://TestbankDirect.eu/Solution-Manual-for-Numerical-Methods-4th-Edition-by-Fai Solution Manual for Numerical Methods 4th Edition by Faires Exercise Set 1.2 (c) ∫ ∫ 1 f (x) dx ≈ 0 [ x2 + x dx = x + ]1 = (d) From (b), ∫ ∫ 1 |R2 (x)| dx ≤ 0 1 e (cos + sin 1)x3 dx = ∫ 1.252x3 dx = 0.313 Since ∫ [ ex cos x dx = ex (cos x + sin x) ]1 = e (cos + sin 1) − (1 + 0) ≈ 1.378 2 Using Integration by Parts twice gives ∫ ex cos(x) dx = x (e cos(x) + sin(x)) ≈ 1.378, so the actual error is |1.3780 − 1.5| ≈ 0.12 Find the third Taylor polynomial P3 (x) for the function f (x) = (x − 1) ln x about x0 = (a) Use P3 (0.5) to approximate f (0.5) Find an upper bound for error |f (0.5) − P3 (0.5)| using the error formula, and compare it to the actual error SOLUTION: (P3 (0.5) = 0.312500, f (0.5) = 0.346574 An error bound is 0.2916, and the actual error is 0.034074.) (b) Find a bound for the error |f (x) − P3 (x)| in using P3 (x) to approximate f (x) on the interval [0.5, 1.5] ∫ 1.5 ∫ 1.5 (c) Approximate 0.5 f (x) dx using 0.5 P3 (x) dx ∫ 1.5 (d) Find an upper bound for the error in (c) using 0.5 |R3 (x) dx|, and compare the bound to the actual error First we have P3 (x) = (x − 1)2 − 12 (x − 1)3 Then SOLUTION: (a) |f (x) − P3 (x)| ≤ 0.2916 on [0.5, 1.5] ∫ 1.5 ∫ 1.5 SOLUTION: (b) 0.5 P3 (x) dx = 0.083, 0.5 (x − 1) ln x dx = 0.088020 SOLUTION: (c) An error bound is 0.0583, and the actual error is 4.687 × 10−3 9.* Use the error term of a Taylor polynomial to estimate the error involved in using sin x ≈ x to approximate sin 1◦ SOLUTION: First we need to convert the degree measure for the sine function to radians π We have 180◦ = π radians, so 1◦ = 180 radians Since f (x) = sin x, f ′ (x) = cos x, f ′′ (x) = − sin x, and f ′′′ (x) = − cos x, we have f (0) = 0, f ′ (0) = 1, and f ′′ (0) = The approximation sin x ≈ x is given by f (x) ≈ P2 (x) and R2 (x) = − cos ξ x 3! Full file at https://TestbankDirect.eu/Solution-Manual-for-Numerical-Methods-4th-Edition-by-Fai Solution Manual for Numerical Methods 4th Edition by Faires Mathematical Preliminaries If we use the bound | cos ξ| ≤ 1, then sin ( π ) ( π ) π − cos ξ ( π )3 − = R2 = ≤ 8.86 × 10−7 180 180 180 3! 180 10 Use a Taylor polynomial about π/4 to approximate cos 42◦ to an accuracy of 10−6 SOLUTION: Since 42◦ = 7π/30 radians, use x0 = π/4 Then ( Rn 7π 30 (π ) ≤ )n+1 − 7π (0.053)n+1 30 < (n + 1)! (n + 1)! −6 For |Rn ( 7π , it suffices to take n = To digits, 30 )| < 10 cos 42◦ = 0.7431448 and P3 (42◦ ) = P3 ( 7π ) = 0.7431446, 30 so the actual error is × 10−7 11.* Let f (x) = ex/2 sin(x/3) Use MATLAB to determine the following (a) The third Maclaurin polynomial P3 (x) (b) A bound for the error |f (x) − P3 (x)| on [0, 1] SOLUTION: We will use the symbolic toolbox in MATLAB to solve this problem (a) Define f (x) by syms x f =inline(’exp(x/2)*sin(x/3)’,’x’) MATLAB responds with f = Inline function: f(x) = exp(x/2)*sin(x/3) In order to use the symbolic toolbox of MATLAB we need to define x as the independent variable in the problem This requires the syms command We use the taylor command to compute the first three terms of the Taylor polynomial for f(x) taylor(f(x),’Order’,3) MATLAB responds with ans = x2 /6 + x/3 We define the third Maclaurin polynomial with the command p3=inline(’x^2/6 + x/3’,’x’) MATLAB responds with p3(x) = x /6 + x/3 (b) To determine the fourth derivative using the symbolic toolbox of MATLAB we first use the command Full file at https://TestbankDirect.eu/Solution-Manual-for-Numerical-Methods-4th-Edition-by-Fai Solution Manual for Numerical Methods 4th Edition by Faires Exercise Set 1.2 diff(f(x),4) and MATLAB responds with ans =(5*cos(x/3)*exp(x/2))/54 - (119*sin(x/3)*exp(x/2))/1296 Now define f4=inline(’(5*cos(x/3)*exp(x/2))/54 - (119*sin(x/3)*exp(x/2))/1296’,’x’) which results in f4 = Inline function: f4(x) = (5*cos(x/3)*exp(x/2))/54 - (119*sin(x/3)*exp(x/2))/1296 Then find the fifth derivative with diff(f(x),5) which MATLAB gives as ans = (61*cos(x/3)*exp(x/2))/3888 - (199*sin(x/3)*exp(x/2))/2592 For the evaluations we will use the long format of MATLAB, implemented with format long to determine if the fourth derivative has any critical points in [0, 1] we use fzero(inline(’(61*cos(x/3)*exp(x/2))/3888 - (199*sin(x/3)*exp(x/2))/2592’,’x’),[0,1]) and MATLAB responds with ans = 0.604738907555866 Define this value so that we can evaluate the fourth derivative at this point p= 0.604738907555866 The extreme values of the fourth derivative will occur at 0, p, or c1=f4(0),c2=f4(p),c3=f4(1) MATLAB responds with c1 = 0.092592592592593, c2 = 0.097871762181106, c3 =0.094723444582014 So the maximum absolute value of the fourth derivative, c2, occurs at p and the bound error is given by Full file at https://TestbankDirect.eu/Solution-Manual-for-Numerical-Methods-4th-Edition-by-Fai Solution Manual for Numerical Methods 4th Edition by Faires Mathematical Preliminaries error1=c2/24 MATLAB gives this as error1 =0.004077990090879 12 Let f (x) = ln(x2 + 2) Use MATLAB to determine the following (a) The Taylor polynomial P3 (x) for f expanded about x0 = (b) The maximum error |f (x) − P3 (x)| for ≤ x ≤ (c) The Maclaurin polynomial P˜3 (x) for f (d) The maximum error |f (x) − P˜3 (x)| for ≤ x ≤ (e) Does P3 (0) approximate f (0) better than P˜3 (1) approximates f (1)? SOLUTION: (a) P3 (x) = ln(3) + 23 (x − 1) + 19 (x − 1)2 − 10 81 (x − 1)3 SOLUTION: (b) max0≤x≤1 |f (x) − P3 (x)| = |f (0) − P3 (0)| = 0.02663366 SOLUTION: (c) P˜3 (x) = ln(2) + 12 x2 SOLUTION: (d) max0≤x≤1 |f (x) − P˜3 (x)| = |f (1) − P˜3 (1)| = 0.09453489 SOLUTION: (e) P3 (0) approximates f (0) better than P˜3 (1) approximates f (1) 13 The polynomial P2 (x) = − 12 x2 is to be used to approximate f (x) = cos x in [− 12 , 21 ] Find a bound for the maximum error SOLUTION: A bound for the maximum error is 0.0026 14 The nth Taylor polynomial for a function f at x0 is sometimes referred to as the polynomial of degree at most n that “best” approximates f near x0 (a) Explain why this description is accurate (b) Find the quadratic polynomial that best approximates a function f near x0 = if the tangent line at x0 = has equation y = 4x − 1, and if f ′′ (1) = (k) SOLUTION: (a) Pn (x0 ) = f (k) (x0 ) for k = 0, 1, , n The shapes of Pn and f are the same at x0 SOLUTION: (b) P2 (x) = + 4(x − 1) + 3(x − 1)2 15 The error function defined by erf(x) = √ π ∫ x e−t dt gives the probability that any one of a series of trials will lie within x units of the mean, assuming that the trials have a normal distribution with mean and standard deviation √ 2/2 This integral cannot be evaluated in terms of elementary functions, so an approximating technique must be used (a) Integrate the Maclaurin series for e−t to show that ∞ ∑ (−1)k x2k+1 erf(x) = √ (2k + 1)k! π k=0 Full file at https://TestbankDirect.eu/Solution-Manual-for-Numerical-Methods-4th-Edition-by-Fai Solution Manual for Numerical Methods 4th Edition by Faires Exercise Set 1.2 (b) The error function can also be expressed in the form ∞ ∑ 2k x2k+1 erf(x) = √ e−x · · · · · (2k + 1) π k=0 Verify that the two series agree for k = 1, 2, 3, and [Hint: Use the Maclaurin series for e−x ] (c) Use the series in part (a) to approximate erf(1) to within 10−7 (d) Use the same number of terms used in part (c) to approximate erf(1) with the series in part (b) (e) Explain why difficulties occur using the series in part (b) to approximate erf(x) SOLUTION: (a) Use the series e −t2 = ∞ ∑ (−1)k t2k k=0 k! to integrate √ π ∫ x e−t dt, and obtain the result SOLUTION: (b) We have [ ] ∞ ∑ 2k x2k+1 √ e−x √ = − x + x − x + x + ··· · · · · (2k + 1) 24 π π k=0 [ ] 16 · x + x3 + x5 + x + x + ··· 15 105 945 [ ] 1 = √ x − x3 + x5 − x7 + x + · · · = erf (x) 10 42 216 π SOLUTION: (c) 0.8427008 SOLUTION: (d) 0.8427069 SOLUTION: (e) The series in part (a) is alternating, so for any positive integer n and positive x we have the bound n ∑ (−1)k x2k+1 x2n+3 erf(x) − √ < (2k + 1)k! (2n + 3)(n + 1)! π k=0 We have no such bound for the positive term series in part (b) 16.* In Example it is stated that for all x we have | sin x| ≤ |x| Use the following to verify this statement (a) Show that for all x ≥ the function f (x) = x − sin x is non-decreasing, which implies that sin x ≤ x with equality only when x = (b) Use the fact that the sine function is odd to reach the conclusion First observe that for f (x) = x − sin x we have f ′ (x) = − cos x ≥ 0, because −1 ≤ cos x ≤ for all values of x Also, the statement clearly holds when |x| ≥ π, because | sin x| ≤ Full file at https://TestbankDirect.eu/Solution-Manual-for-Numerical-Methods-4th-Edition-by-Fai Solution Manual for Numerical Methods 4th Edition by Faires Mathematical Preliminaries SOLUTION: (a) The observation implies that f (x) is non-decreasing for all values of x, and in particular that f (x) > f (0) = when x > Hence for x ≥ 0, we have x ≥ sin x, and when ≤ x ≤ π, we have | sin x| = sin x ≤ x = |x| SOLUTION: (b) When −π < x < 0, we have π ≥ −x > Since sin x is an odd function, the fact (from part (a)) that sin(−x) ≤ (−x) implies that | sin x| = − sin x ≤ −x = |x| As a consequence, for all real numbers x we have | sin x| ≤ |x| Round-Off Error and Computer Arithmetic Exercise Set 1.3 (page 20) Compute the absolute error and relative error in approximations of p by p∗ (a) p = π, p∗ = 22 (b) p = π, p∗ = 3.1416 (c) p = e, p∗ = 2.718 √ (d) p = 2, p∗ = 1.414 (e) p = e10 , p∗ = 22000 (f) p = 10π , p∗ = 1400 (g) p = 8!, p∗ = 39900 √ (h) p = 9!, p∗ = 18π (9/e) SOLUTION: We have (a) (b) (c) (d) (e) (f) (g) (h) Absolute error Relative error 0.001264 7.346 × 10−6 2.818 × 10−4 2.136 × 10−4 2.647 × 101 1.454 × 101 420 3.343 × 103 4.025 × 10−4 2.338 × 10−6 1.037 × 10−4 1.510 × 10−4 1.202 × 10−3 1.050 × 10−2 1.042 × 10−2 9.213 × 10−3 Perform the following computations (i) exactly, (ii) using three-digit chopping arithmetic, and (iii) using three-digit rounding arithmetic (iv) Compute the relative errors in parts (ii) and (iii) + (b) · ) ( − + (c) 11 ( ) + − (d) 11 (a) 20 20 Full file at https://TestbankDirect.eu/Solution-Manual-for-Numerical-Methods-4th-Edition-by-Fai Solution Manual for Numerical Methods 4th Edition by Faires 10 Exercise Set 1.3 Full file at https://TestbankDirect.eu/Solution-Manual-for-Numerical-Methods-4th-Edition-by-Fai SOLUTION: (a) (i) 17/15 (ii) 1.13 (iii) 1.13 SOLUTION: (b) (i) 4/15 (ii) 0.266 (iii) 0.266 (iv) both × 10−3 (iv) both 2.5 × 10−3 SOLUTION: (c) (i) 139/660 (ii) 0.211 (iii) 0.210 (iv) × 10−3 , × 10−3 SOLUTION: (d) (i) 301/660 (ii) 0.455 (iv) × 10−3 , × 10−4 (iii) 0.456 Use three-digit rounding arithmetic to perform the following calculations Compute the absolute error and relative error with the exact value determined to at least five digits (a) 133 + 0.921 (b) 133 − 0.499 (c) (121 − 0.327) − 119 (d) (121 − 119) − 0.327 (e) − 67 2e − 5.4 13 14 (f) −10π + 6e − (g) (h) ( ) ( ) · π− 62 22 17 SOLUTION: We have (a) (b) (c) (d) (e*) (f) (g) (h) Approximation Absolute error Relative error 134 133 2.00 1.67 1.80 −15.1 0.286 0.00 0.079 0.499 0.327 0.003 0.154 0.0546 2.86 × 10−4 0.0215 5.90 × 10−4 3.77 × 10−3 0.195 1.79 × 10−3 0.0786 3.60 × 10−3 10−3 1.00 SOLUTION: (e∗ ) Using three-digit rounding arithmetic gives 13 14 = 0.929, e = 2.72 So 13 − = 0.0720 and 2e − 5.4 = 5.44 − 5.40 = 0.0400 14 Hence 13 0.0720 14 − = = 1.80 2e − 5.4 0.0400 = 0.857, and The correct value is approximately 1.954, so the absolute and relative errors to three digits are |1.80 − 1.954| = 0.154 and |1.80 − 1.954| = 0.0788, 1.954 respectively Full file at https://TestbankDirect.eu/Solution-Manual-for-Numerical-Methods-4th-Edition-by-Fai Solution Manual for Numerical Methods 4th Edition by Faires Mathematical Preliminaries 11 Full file at https://TestbankDirect.eu/Solution-Manual-for-Numerical-Methods-4th-Edition-by-Fai Repeat Exercise using three-digit chopping arithmetic SOLUTION: We have (a) (b) (c) (d) (e*) (f) (g) (h) Approximation Absolute error Relative error 133 132 1.00 1.67 3.55 −15.2 0.284 0.921 0.501 0.673 0.003 1.60 0.0454 0.00171 0.02150 6.88 × 10−3 3.78 × 10−3 0.402 1.79 × 10−3 0.817 0.00299 0.00600 SOLUTION: (e∗ ) Using three-digit chopping arithmetic gives 13 14 = 0.928, e = 2.71 So 13 − = 0.0710 and 2e − 5.4 = 5.42 − 5.40 = 0.0200 14 7 = 0.857, and Hence − 67 0.0710 = = 3.55 2e − 5.4 0.0200 13 14 The correct value is approximately 1.954, so the absolute and relative errors to three digits are |3.55 − 1.954| = 1.60, and |3.55 − 1.954| = 0.817, 1.954 respectively The results in Exercise 3(e) were considerably better Repeat Exercise using four-digit rounding arithmetic SOLUTION: We have (a) (b) (c) (d) (e) (f) (g) (h) Approximation Absolute error Relative error 133.9 132.5 1.700 1.673 1.986 −15.16 0.2857 −0.01700 0.021 0.001 0.027 0.03246 0.005377 1.429 × 10−5 0.0045 1.568 × 10−4 7.55 × 10−6 0.01614 0.01662 3.548 × 10−4 × 10−5 0.2092 Repeat Exercise using four-digit chopping arithmetic Full file at https://TestbankDirect.eu/Solution-Manual-for-Numerical-Methods-4th-Edition-by-Fai Solution Manual for Numerical Methods 4th Edition by Faires 12 Exercise Set 1.3 Full file at https://TestbankDirect.eu/Solution-Manual-for-Numerical-Methods-4th-Edition-by-Fai SOLUTION: We have (a) (b) (c) (d) (e) (f) (g) (h) Approximation Absolute error Relative error 133.9 132.5 1.600 1.673 1.983 −15.15 0.2855 −0.01700 0.021 0.001 0.073 0.02945 0.004622 2.143 × 10−4 0.0045 1.568 × 10−4 7.55 × 10−6 0.04363 0.01508 3.050 × 10−4 7.5 × 10−4 0.2092 The first three nonzero terms of the Maclaurin series for the arctangent function are 1 P3 (x) = x − x3 + x5 [ ( ) ( ) ( )] ( ) 1 1 (a) arctan (b) 16 arctan + arctan − arctan 239 (1) (1) SOLUTION: (a) We have P = 0.4645833 and P = 0.3218107, so [ ] 1 π = arctan + arctan ≈ 3.145576 The absolute and relative errors are, respectively, |π − 3.145576| ≈ 3.983 × 10−3 SOLUTION: (b) We have P (1) |π − 3.145576| ≈ 1.268 × 10−3 |π| and = 0.1973973 and P ( 239 ) = 0.004184, so [ ] 1 π = 16 arctan − arctan ≈ 3.1416208 239 The absolute and relative errors are, respectively, |π − 3.1416208| ≈ 2.8 × 10−5 and |π − 3.1416208| ≈ 8.9 × 10−6 |π| The two-by-two linear system ax + by = e, cx + dy = f, Full file at https://TestbankDirect.eu/Solution-Manual-for-Numerical-Methods-4th-Edition-by-Fai Solution Manual for Numerical Methods 4th Edition by Faires Mathematical Preliminaries 13 Full file at https://TestbankDirect.eu/Solution-Manual-for-Numerical-Methods-4th-Edition-by-Fai where a, b, c, d, e, f are given, can be solved for x and y as follows: c , provided a ̸= 0; a = d − mb; set m = d1 = f − me; f1 ; y = d1 (e − by) x = a f1 Solve the following linear systems using four-digit rounding arithmetic (a) 1.130x − 6.990y 8.110x + 12.20y (b) = 14.20 = −0.1370 1.013x − 6.099y −18.11x + 112.2y = 14.22 = −0.1376 SOLUTION: (a) x = 2.451, y = −1.635 SOLUTION: (b) x = 507.7, y = 82.00 Suppose the points (x0 , y0 ) and (x1 , y1 ) are on a straight line with y1 ̸= y0 Two formulas are available to find the x-intercept of the line: x= x0 y1 − x1 y0 y1 − y0 and x = x0 − (x1 − x0 )y0 y1 − y0 (a) Show that both formulas are algebraically correct (b) Use the data (x0 , y0 ) = (1.31, 3.24) and (x1 , y1 ) = (1.93, 4.76) and three-digit rounding arithmetic to compute the x-intercept both ways Which method is better, and why? SOLUTION: The first formula gives −0.00658, and the second formula gives −0.0100 The true three-digit value is −0.0116 ∑n 10 The Taylor polynomial of degree n for f (x) = ex is i=0 xi /i! Use the Taylor polynomial of degree nine and three-digit chopping arithmetic to find an approximation to e−5 by each of the following methods (a) e−5 ≈ ∑ (−5)i i=0 (b) e−5 = i! = ∑ (−1)i 5i i=0 i! 1 ≈ ∑9 i e5 i=0 /i! An approximate value of e−5 correct to three digits is 6.74 × 10−3 Which formula, (a) or (b), gives the most accuracy, and why? SOLUTION: (a) −1.82 SOLUTION: (b) 7.09 × 10−3 The formula in (b) is more accurate since subtraction is not involved Full file at https://TestbankDirect.eu/Solution-Manual-for-Numerical-Methods-4th-Edition-by-Fai Solution Manual for Numerical Methods 4th Edition by Faires 14 Exercise Set 1.3 Full file at https://TestbankDirect.eu/Solution-Manual-for-Numerical-Methods-4th-Edition-by-Fai 11.* A rectangular parallelepiped has sides cm, cm, and cm, measured to the nearest centimeter (a) What are the best upper and lower bounds for the volume of this parallelepiped? (b) What are the best upper and lower bounds for the surface area? SOLUTION: The sides are x = cm, y = cm, and z = cm to the nearest cm This means that 2.5 ≤ x < 3.5, 3.5 ≤ y < 4.5 and 4.5 ≤ z < 5.5 (a) The volume is V = xyz so 2.5(2.5 · 3.5) ≤ V < 5.5(3.5 · 4.5) and 39.375 ≤ V < 86.625 (b) The surface area is S = 2xy + 2xz + 2y so 2(2.5 · 3.5 + 2.5 · 4.5 + 3.5 · 4.5) ≤ S < 2(3.5 · 4.5 + 3.5 · 5.5 + 4.5 · 6.5) that is 71.50 ≤ S < 128.50 12 The following MATLAB M-file rounds or chops a number x to t digits where rnd = for rounding and rnd = for chopping function [res] = CHIP(rnd,t,x) % This program is used to round or chop a number x to a specific number t of digits if x == w = 0; else ee = fix(log10(abs(x))); if abs(x) > ee = ee + 1; end; if rnd == w = round(x*10^(t-ee))*10^(ee-t); else w = fix(x*10^(t-ee))*10^(ee-t); end; end; res = w ; Verify that the procedure works for the following values (a) x = 124.031, t = (b) x = 124.036, t = (c) x = −0.00653, t = (d) x = −0.00656, t = SOLUTION: (a) We have 124.03 for rounding and 124.03 for chopping SOLUTION: (b) We have 124.04 for rounding and 124.03 for chopping SOLUTION: (c) We have −0.0065 for rounding and −0.0065 for chopping SOLUTION: (d) We have −0.0065 for rounding and −0.0065 for chopping Full file at https://TestbankDirect.eu/Solution-Manual-for-Numerical-Methods-4th-Edition-by-Fai Solution Manual for Numerical Methods 4th Edition by Faires Mathematical Preliminaries 15 Full file at https://TestbankDirect.eu/Solution-Manual-for-Numerical-Methods-4th-Edition-by-Fai ( ) m! m = k k! (m − k)! 13 The binomial coefficient describes the number of ways of choosing a subset of k objects from a set of m elements (a) Suppose decimal machine numbers are of the form ±0.d1 d2 d3 d4 × 10n , with ≤ d1 ≤ 9, ≤ di ≤ 9, if i = 2, 3, and |n| ≤ 15 What is the largest value of m for which the binomial coefficient for all k by the definition without causing overflow? ( ) (b) Show that m k can also be computed by (m) k can be computed ) ( ) ( ) ( )( m−1 m m m−k+1 = ··· k k−1 k (c) What is the largest value of m for which the binomial coefficient by the formula in part (b) without causing overflow? ( m) can be computed (d) Use the equation in (b) and four-digit chopping arithmetic to compute the number of possible 5-card hands in a 52-card deck Compute the actual and relative errors SOLUTION: (a) m = 17 SOLUTION: (b) ( ) m m! m(m − 1) · · · (m − k − 1)(m − k)! = = k k!(m − k)! k!(m − k)! ( ) (m) (m − 1) m−k−1 = ··· k k−1 SOLUTION: (c) m = 181707 SOLUTION: (d) 2,597,000; actual error 1960; relative error 7.541 × 10−4 Errors in Scientific Computation Exercise Set 1.4 (page 28) (i) Use four-digit rounding arithmetic and the formulas of Example to find the most accurate approximations to the roots of the following quadratic equations (ii) Compute the absolute errors and relative errors for these approximations 1 123 x − x+ =0 123 (b) x + x− =0 (c) 1.002x − 11.01x + 0.01265 = (a) (d) 1.002x2 + 11.01x + 0.01265 = Full file at https://TestbankDirect.eu/Solution-Manual-for-Numerical-Methods-4th-Edition-by-Fai Solution Manual for Numerical Methods 4th Edition by Faires 16 Exercise Set 1.4 Full file at https://TestbankDirect.eu/Solution-Manual-for-Numerical-Methods-4th-Edition-by-Fai SOLUTION: We have x1 (a) (b) (c) (d) Absolute error 92.26 0.005421 10.98 −0.001149 0.01542 1.264 × 10−6 6.875 × 10−3 7.566 × 10−8 Relative error −4 1.672 × 10 2.333 × 10−4 6.257 × 10−4 6.584 × 10−5 x2 Absolute error 0.005419 −92.26 0.001149 −10.98 −7 6.273 × 10 4.580 × 10−3 7.566 × 10−8 6.875 × 10−3 Relative error 1.157 × 10−4 4.965 × 10−5 6.584 × 10−5 6.257 × 10−4 SOLUTION: (c∗ ) For the four-digit rounding arithmetic and the formulas in Eqs (1.1) and (1.2) we have: ((i)) Since b = −11.01 is negative, we use the formulas x1 = −b + √ b2 − 4ac 2a and x2 = −2c √ b − b2 − 4ac to avoid the subtraction of nearly equal numbers Using four-digit rounding arithmetic gives √ √ −(−11.01) + (−11.01)2 − 4(1.002)(0.01265) −b + b2 − 4ac x1 = = 2a 2a √ √ 121.2 − 4.008(0.01265) 11.01 + 121.2 − 0.05070 = = 2.004 2.004 √ 11.01 + 121.1 11.01 + 11.00 22.01 = = = = 10.98 2.004 2.004 2.004 11.01 + The actual root is 10.98687488, so the absolute error is |10.98687488−10.98| = 6.87488×10−3 , and the relative error is 6.87488 × 10−3 = 6.25736 × 10−4 10.98687488 Also, −2c −2(0.01265) √ √ = b − b − 4ac −11.01 − (−11.01)2 − 4(1.002)(0.01265) −0.02530 −0.02530 = = = 0.001149 −11.01 − 11.00 −22.01 x2 = The absolute errors and relative errors for these approximations are: ((ii)) The actual root is 0.001149076 so the absolute error of this approximation is |0.001149076 − 0.001149| = 7.566 × 10−8 , and the relative error is 7.566 × 10−8 = 6.5844 × 10−5 0.001149076 Repeat Exercise using four-digit chopping arithmetic Full file at https://TestbankDirect.eu/Solution-Manual-for-Numerical-Methods-4th-Edition-by-Fai Solution Manual for Numerical Methods 4th Edition by Faires Mathematical Preliminaries 17 Full file at https://TestbankDirect.eu/Solution-Manual-for-Numerical-Methods-4th-Edition-by-Fai SOLUTION: Approximation for x1 Absolute error Relative error 92.24 0.005417 10.98 −0.001149 0.004580 2.736 × 10−6 6.875 × 10−3 7.566 × 10−8 4.965 × 10−5 5.048 × 10−4 6.257 × 10−4 6.584 × 10−5 Approximation for x2 Absolute error Relative error 0.005418 −92.25 0.001149 −10.98 2.373 × 10−6 5.420 × 10−3 7.566 × 10−8 6.875 × 10−3 4.377 × 10−4 5.875 × 10−5 6.584 × 10−5 6.257 × 10−4 (a) (b) (c) (d) (a) (b) (c) (d) SOLUTION: (c∗ ) For the four-digit chopping arithmetic and the formulas in Eqs (1.1) and (1.2) we have: (i) Since b = −11.01 is negative, we use the formulas √ −b + b2 − 4ac −2c √ x1 = and x2 = 2a b − b2 − 4ac to avoid the subtraction of nearly equal numbers Using four-digit chopping arithmetic gives √ √ −(−11.01) + (−11.01)2 − 4(1.002)(0.01265) −b + b2 − 4ac x1 = = 2a 2a √ √ 121.2 − 4.008(0.01265) 11.01 + 121.2 − 0.05070 = = 2.004 2.004 √ 11.01 + 121.1 11.01 + 11.00 22.01 = = = = 10.98 2.004 2.004 2.004 11.01 + The actual root is 10.98687488, so the absolute error is |10.98687488 − 10.98| = 6.87488 × 10−3 , and the relative error is 6.87488 × 10−3 = 6.25736 × 10−4 10.98687488 Also, −2c −2(0.01265) √ √ = b − b2 − 4ac −11.01 − (−11.01)2 − 4(1.002)(0.01265) −0.02530 −0.02530 = = 0.001149 = −11.01 − 11.00 −22.01 x2 = The absolute errors and relative errors for these approximations are: Full file at https://TestbankDirect.eu/Solution-Manual-for-Numerical-Methods-4th-Edition-by-Fai Solution Manual for Numerical Methods 4th Edition by Faires 18 Exercise Set 1.4 Full file at https://TestbankDirect.eu/Solution-Manual-for-Numerical-Methods-4th-Edition-by-Fai (ii) The actual root is 0.001149076 so the absolute error of this approximation is |0.001149076 − 0.001149| = 7.566 × 10−8 , and the relative error is 7.566 × 10−8 = 6.5844 × 10−5 0.001149076 Let f (x) = 1.013x5 − 5.262x3 − 0.01732x2 + 0.8389x − 1.912 (a) Evaluate f (2.279) by first calculating (2.279)2 , (2.279)3 , (2.279)4 , and (2.279)5 using four-digit rounding arithmetic (b) Evaluate f (2.279) using the formula f (x) = (((1.013x2 − 5.262)x − 0.01732)x + 0.8389)x − 1.912 and four-digit rounding arithmetic (c) Compute the absolute and relative errors in parts (a) and (b) SOLUTION: (a) −0.1000 SOLUTION: (b) −0.1010 SOLUTION: (c) Absolute error for part (a) 2.331 × 10−3 with relative error 2.387 × 10−2 Absolute error for part (b) 3.331 × 10−3 with relative error 3.411 × 10−2 Let f (x) = 1.013x5 − 5.262x3 − 0.01732x2 + 0.8389x − 1.912 (a) Evaluate f (2.279) by first calculating (2.279)2 , (2.279)3 , (2.279)4 , and (2.279)5 using four-digit chopping arithmetic (b) Evaluate f (2.279) using the formula f (x) = (((1.013x2 − 5.262)x − 0.01732)x + 0.8389)x − 1.912 and four-digit chopping arithmetic (c) Compute the absolute and relative errors in parts (a) and (b) SOLUTION: (a) −0.09000 SOLUTION: (b) −0.1140 SOLUTION: (c) The absolute error for (a) is 7.669 × 10−3 , and the relative error for (a) is 7.852 × 10−2 The absolute error for (b) is 1.633 × 10−2 , and the relative error for (b) is 1.672 × 10−1 5.* The fifth Maclaurin polynomials for e2x and e−2x are (((( ) ) ) ) 4 P5 (x) = x+ x+ x+2 x+2 x+1 15 3 and (((( Pˆ5 (x) = ) ) ) ) 4 x− x+2 x−2 x+1 − x+ 15 3 (a) Approximate e−0.98 using Pˆ5 (0.49) and four-digit rounding arithmetic Full file at https://TestbankDirect.eu/Solution-Manual-for-Numerical-Methods-4th-Edition-by-Fai Solution Manual for Numerical Methods 4th Edition by Faires Mathematical Preliminaries 19 Full file at https://TestbankDirect.eu/Solution-Manual-for-Numerical-Methods-4th-Edition-by-Fai (b) Compute the absolute and relative error for the approximations in part (a) (c) Approximate e−0.98 using 1/P5 (0.49) and four-digit rounding arithmetic (d) Compute the absolute and relative errors for the approximations in part (c) SOLUTION: (a) Using four-digit rounding arithmetic to evaluate Pˆ5 (0.49) gives Pˆ5 (0.49) = (((((−0.2667)(0.49) + 0.6667)(0.49) − 1.333)(0.49) + 2)(0.49) − 2)(0.49) + = ((((−0.1307 + 0.6667)(0.49) − 1.333)(0.49) + 2)(0.49) − 2)(0.49) + = ((((0.5360)(0.49) − 1.333)(0.49) + 2)(0.49) − 2)(0.49) + = (((0.2626 − 1.333)(0.49) + 2)(0.49) − 2)(0.49) + = (((−1.070)(0.49) + 2)(0.49) − 2)(0.49) + = (((−0.5243) + 2)(0.49) − 2)(0.49) + = ((1.476)(0.49) − 2)(0.49) + = (0.7232 − 2)(0.49) + = (−1.277)(0.49) + = −0.6257 + = 0.3743 SOLUTION: (b) The absolute error is |e−0.98 − 0.3743| = 1.011 × 10−3 , and the relative error is 1.011 × 10−3 = 2.694 × 10−3 e−0.98 SOLUTION: (c) We have 1 = P5 (0.49) (((((−0.2667)(0.49) + 0.6667)(0.49) + 1.333)(0.49) + 2)(0.49) + 2)(0.49) + 1 = = 0.3755 2.663 SOLUTION: (d) The absolute error is |e−0.98 − 0.3755| = 1.889 × 10−4 , and the relative error is 6.* 1.889 × 10−4 = 5.033 × 10−4 e−0.98 (a) Show that the polynomial nesting technique can be used to evaluate f (x) = 1.01e4x − 4.62e3x − 3.11e2x + 12.2ex − 1.99 (b) Use three-digit rounding arithmetic and the formula given in the statement of part (a) to evaluate f (1.53) Full file at https://TestbankDirect.eu/Solution-Manual-for-Numerical-Methods-4th-Edition-by-Fai Solution Manual for Numerical Methods 4th Edition by Faires 20 Exercise Set 1.4 Full file at https://TestbankDirect.eu/Solution-Manual-for-Numerical-Methods-4th-Edition-by-Fai (c) Redo the calculations in part (b) using the nesting form of f (x) that was found in part (a) (d) Compare the approximations in parts (b) and (c) n SOLUTION: (a) Since enx = (ex ) , we can write f (x) = ((((1.01)ex − 4.62) ex − 3.11) ex + 12.2) ex − 1.99 SOLUTION: (b) Using e1.53 = 4.62 and three-digit rounding gives e2(1.53) = (4.62)2 = 21.3, e3(1.53) = (4.62)2 (4.62) = (21.3)(4.62) = 98.4, and e4(1.53) = (98.4)(4.62) = 455 So f (1.53) = 1.01(455) − 4.62(98.4) − 3.11(21.3) + 12.2(4.62) − 1.99 = 460 − 455 − 66.2 + 56.4 − 1.99 = 5.00 − 66.2 + 56.4 − 1.99 = −61.2 + 56.4 − 1.99 = −4.80 − 1.99 = −6.79 SOLUTION: (c) We have f (1.53) = (((1.01)4.62 − 4.62)4.62 − 3.11)4.62 + 12.2)4.62 − 1.99 = (((4.67 − 4.62)4.62 − 3.11)4.62 + 12.2)4.62 − 1.99 = ((0.231 − 3.11)4.62 + 12.2)4.62 − 1.99 = (−13.3 + 12.2)4.62 − 1.99 = −7.07 SOLUTION: (d) The exact result is 7.61, so the absolute errors in parts (b) and (c) are, respectively, | − 6.79 + 7.61| = 0.82 and | − 7.07 + 7.61| = 0.54 The relative errors are, respectively, 0.108 and 0.0710 ∑10 Use three-digit chopping arithmetic to compute the sum i=1 1/i2 first by 11 + 14 + · · · + 100 1 and then by 100 + 81 + · · · + Which method is more accurate, and why? SOLUTION: (We have 1 + + + = 1.53; 100 1 + + + = 1.54 100 81 and The actual value is 1.549 Significant round-off error occurs much earlier in the first method.) 8.* (a) Determine the number n of terms of the series arctan x = lim Pn (x) = n→∞ ∞ ∑ x2i−1 (−1)i+1 (2i − 1) i=1 that are required to ensure that |4Pn (1) − π| < 10−3 (b) How many terms are required to ensure the 10−10 accuracy needed for an approximation to π? SOLUTION: (a) Since the terms of the series π = arctan = ∞ ∑ (−1)i+1 i=1 2i − Full file at https://TestbankDirect.eu/Solution-Manual-for-Numerical-Methods-4th-Edition-by-Fai Solution Manual for Numerical Methods 4th Edition by Faires Mathematical Preliminaries 21 Full file at https://TestbankDirect.eu/Solution-Manual-for-Numerical-Methods-4th-Edition-by-Fai alternate in sign, the error produced by truncating the series at any term is less than the magnitude of the next term To ensure significant accuracy, we need to choose n so that < 10−3 2(n + 1) − or 4000 < 2n + So n ≥ 2000 SOLUTION: (b) In this case, we need < 10−10 2(n + 1) − or n > 20,000,000,000 Clearly, a more rapidly convergent method is needed for this approximation ∑∞ The number e is defined by e = n=0 1/n!, where n! = n(n − 1) · · · · 1, for n ̸= and 0! = (i) Use four-digit chopping arithmetic to compute the following approximations to e (ii) Compute absolute and relative errors for these approximations (a) (b) ∑ n! n=0 ∑ j=0 (c) (d) (5 − j)! 10 ∑ n! n=0 10 ∑ j=0 (10 − j)! SOLUTION: We have (a) (b) (c) (d) Approximation Absolute Error Relative Error 2.715 2.716 2.716 2.718 3.282 × 10−3 2.282 × 10−3 2.282 × 10−3 2.818 × 10−4 1.207 × 10−3 8.394 × 10−4 8.394 × 10−4 1.037 × 10−4 10 Find the rates of convergence of the following sequences as n → ∞ ( ) =0 (a) lim sin n→∞ n ( ) =0 (b) lim sin n→∞ n2 ( ( ))2 (c) lim sin =0 n→∞ n Full file at https://TestbankDirect.eu/Solution-Manual-for-Numerical-Methods-4th-Edition-by-Fai Solution Manual for Numerical Methods 4th Edition by Faires 22 Exercise Set 1.4 Full file at https://TestbankDirect.eu/Solution-Manual-for-Numerical-Methods-4th-Edition-by-Fai (d) lim [ln(n + 1) − ln(n)] = n→∞ SOLUTION: (a) O SOLUTION: (b) SOLUTION: (c) SOLUTION: (d) (1) n ) O n12 ( ) O n12 ( ) O n1 ( 11 Find the rates of convergence of the following functions as h → sin h − h cos h =0 h − eh = −1 (b) lim h→0 h sin h (c) lim =1 h→0 h − cos h (d) lim =0 h→0 h (a) lim h→0 SOLUTION: (a) O(h2 ) SOLUTION: (b) O(h) SOLUTION: (c) O(h2 ) SOLUTION: (d) O(h) 12.* (a) How many calculations are needed to determine a sum of the form n ∑ i ∑ bj ? i=1 j=1 (b) Re-express the series in a way that will reduce the number of calculations needed to determine this sum ∑i SOLUTION: (a) For each i, the inner sum j=1 bj requires i multiplications and i − additions, for a total of n ∑ i=1 i= n(n + 1) n ∑ multiplications and i−1= i=1 n(n + 1) −n additions Once the n inner sums are computed, n − additions are required for the final sum The final total is: n(n + 1) multiplications and (n + 2)(n − 1) additions SOLUTION: (b) By rewriting the sum as n ∑ i ∑ i=1 j=1 bj = n ∑ i=1 i ∑ bj , j=1 Full file at https://TestbankDirect.eu/Solution-Manual-for-Numerical-Methods-4th-Edition-by-Fai Solution Manual for Numerical Methods 4th Edition by Faires Mathematical Preliminaries 23 Full file at https://TestbankDirect.eu/Solution-Manual-for-Numerical-Methods-4th-Edition-by-Fai we can significantly reduce the amount of calculation For each i, we now need i − additions to sum bj ’s for a total of n ∑ i−1= i=1 n(n + 1) −n additions Once the bj ’s are summed, we need n multiplications by the ’s, followed by n − additions of the products The total additions by this method is still 12 (n + 2)(n − 1), but the number of multiplications has been reduced from 12 n(n + 1) to n 13 The sequence {Fn } described by F0 = 1, F1 = 1, and Fn+2 = Fn + Fn+1 , if n ≥ 0, is called a Fibonacci sequence Its terms occur naturally in many botanical species, particularly those with petals or scales arranged in the form of a logarithmic spiral Consider the sequence {xn }, where √ xn = Fn+1 /Fn Assuming that limn→∞ xn = x exists, show that x is the golden ratio (1 + 5)/2 SOLUTION: Since lim xn = lim xn+1 = x n→∞ n→∞ we have x=1+ , x and xn+1 = + , xn so x2 − x − = The quadratic formula implies that x= √ ) 1( 1+ This number is called the golden ratio It appears frequently in mathematics and the sciences Full file at https://TestbankDirect.eu/Solution-Manual-for-Numerical-Methods-4th-Edition-by-Fai Solution Manual for Numerical Methods 4th Edition by Faires 24 Exercise Set 1.4 Full file at https://TestbankDirect.eu/Solution-Manual-for-Numerical-Methods-4th-Edition-by-Fai Full file at https://TestbankDirect.eu/Solution-Manual-for-Numerical-Methods-4th-Edition-by-Fai ... https://TestbankDirect.eu /Solution- Manual- for- Numerical- Methods- 4th- Edition- by- Fai Solution Manual for Numerical Methods 4th Edition by Faires 14 Exercise Set 1.3 Full file at https://TestbankDirect.eu /Solution- Manual- for- Numerical- Methods- 4th- Edition- by- Fai... https://TestbankDirect.eu /Solution- Manual- for- Numerical- Methods- 4th- Edition- by- Fai Solution Manual for Numerical Methods 4th Edition by Faires 24 Exercise Set 1.4 Full file at https://TestbankDirect.eu /Solution- Manual- for- Numerical- Methods- 4th- Edition- by- Fai... https://TestbankDirect.eu /Solution- Manual- for- Numerical- Methods- 4th- Edition- by- Fai Solution Manual for Numerical Methods 4th Edition by Faires 10 Exercise Set 1.3 Full file at https://TestbankDirect.eu /Solution- Manual- for- Numerical- Methods- 4th- Edition- by- Fai