Solutions Manual for Friendly Introduction To Numerical Analysis 1st Edition by Bradie
- 1. Floating Point Number Systems 1 1.3 Floating Point Number Systems 1. Provide the floating point equivalent for each of the following numbers from the floating point number system F(10, 4, 0, 4). Consider both chopping and round- ing. Compute the absolute and relative error in each floating point equivalent. (a) π (b) e (c) √ 2 (d) 1/7 (e) cos 22◦ (f) ln 10 (g) 3 √ 9 In the following table, δ denotes the absolute error and ǫ the relative error. Chopping Rounding y fl(y) error fl(y) error π 3.141 δ = 5.927 × 10−4 3.142 δ = 4.073 × 10−4 ǫ = 1.886 × 10−4 ǫ = 1.297 × 10−4 e 2.718 δ = 2.818 × 10−4 2.718 δ = 2.818 × 10−4 ǫ = 1.037 × 10−4 ǫ = 1.037 × 10−4 √ 2 1.414 δ = 2.136 × 10−4 1.414 δ = 2.136 × 10−4 ǫ = 1.510 × 10−4 ǫ = 1.510 × 10−4 1/7 0.1428 δ = 5.714 × 10−4 0.1429 δ = 4.286 × 10−4 ǫ = 4.000 × 10−4 ǫ = 3.000 × 10−4 cos 22◦ 0.9271 δ = 8.385 × 10−5 0.9272 δ = 1.615 × 10−5 ǫ = 9.044 × 10−5 ǫ = 1.741 × 10−5 ln 10 2.302 δ = 5.851 × 10−4 2.303 δ = 4.149 × 10−4 ǫ = 2.541 × 10−4 ǫ = 1.802 × 10−4 3 √ 9 2.080 δ = 8.382 × 10−5 2.080 δ = 8.382 × 10−5 ǫ = 4.030 × 10−5 ǫ = 4.030 × 10−5 2. Prove the bounds on the absolute and relative roundoff error associated with rounding: |flround(y) − y| ≤ 1 2 βe−k and |flround(y) − y| |y| ≤ 1 2 β1−k . Consider the floating point system F(β, k, m, M) with rounding. Let y be a real number whose expansion is given by y = ±(0.d1d2d3 · · · dkdk+1 · · ·)β × βe Solutions Manual for Friendly Introduction To Numerical Analysis 1st Edition by Bradie Full Download: https://downloadlink.org/p/solutions-manual-for-friendly-introduction-to-numerical-analysis-1st-edition-by-bradie/ Full download all chapters instantly please go to Solutions Manual, Test Bank site: TestBankLive.com
- 2. 2 Section 1.3 with d1 = 0 and m ≤ e ≤ M. If we let d denote β/2, then a bound on the absolute size of the roundoff error is |flround(y) − y| ≤ (0.d)β × βe−k = 1 2 βe−k . Provided y = 0, given the restriction on d1, |y| = (0.d1d2d3 · · ·)β × βe ≥ (0.1)β × βe = βe−1 . Therefore, the relative error in flround(y) is bounded by |flround(y) − y| |y| ≤ 1 2 βe−k βe−1 = 1 2 β1−k . 3. Show that machine precision is the smallest floating point number, v, such that fl(1 + v) > 1. First consider the floating point number system F(β, k, m, M) with chopping. The number one is represented by the expansion (0.1 00 · · · 00 k−1 zeros )β × β1 . If we let v = u = β1−k = (0.1 00 · · · 00 k−1 zeros )β × β2−k = (0. 00 · · · 00 k−1 zeros 1 00 · · · 00 k−1 zeros )β × β1 , then 1 + v = (0.1 00 · · · 00 k−2 zeros 1 00 · · · 00 k−1 zeros )β × β1 and flchop(1 + v) = 1.00 · · · 001 > 1. If we assign to v any value smaller than u, then the kth digit in the mantissa of 1 + v is zero and flchop(1 + v) = 1. Thus, with chopping, machine precision is the smallest floating point number, v, such that fl(1 + v) > 1. Now, consider the floating point number system F(β, k, m, M) with rounding. For notational convenience, let d denote β/2. If we take v = u = 1 2 β1−k = (0.d 00 · · · 00 k−1 zeros )β × β1−k
- 3. Floating Point Number Systems 3 = (0. 00 · · · 00 k zeros d 00 · · · 00 k−1 zeros )β × β1 , then 1 + v = (0.1 00 · · · 00 k−1 zeros d 00 · · · 00 k−1 zeros )β × β1 and flround(1 + v) = 1.00 · · · 001 > 1. If we assign to v any value smaller than u, then the (k + 1)st digit in the mantissa of 1+v is smaller than β/2 and flround(1+v) = 1. Thus, with rounding, machine precision is the smallest floating point number, v, such that fl(1 + v) > 1. 4. (a) Construct an algorithm to determine machine precision and another algo- rithm to determine the smallest positive number of a floating point number system. (b) Implement the algorithms from part (a) to determine machine precision and the smallest positive number on your computing system. Consider both single and double precision. (c) Assuming that your computing system uses β = 2 and rounding, use the results from part (b) to determine the values for k and m. (a) Assuming the floating point system uses rounding, here is an algorithm to determine machine precision. Multiplication by β is performed in the output step because the while loop terminates when one too many divisions by β have been carried out. GIVEN: base β STEP 1: initialize u = 1/2 STEP 2: while (1 + u > 1) replace u by u/β OUTPUT: β · u Here is an algorithm to determine the smallest positive number, assuming that underflow is handled by setting the value to zero. GIVEN: base β STEP 1: initialize temp = 1 STEP 2: while (temp > 0) STEP 3: set sm = temp replace temp by temp/β OUTPUT: sm (b) Answers will of course vary. On a SunBlade 100, machine precision in both single and double precision is 2.22045×10−16 . The smallest positive number in single precision is 1.4013 × 10−45 and in double precision is 4.94066 × 10−324 .
- 4. 4 Section 1.3 (c) In general, machine precision with rounding is 1 2 β1−k and the smallest positive number is (0.1)β × βm = βm−1 . Assuming β = 2, we solve 2−k = 2.22045 × 10−16 to find k = 52 in both single and double precision on the SunBlade 100. In single precision, we solve 2m−1 = 1.4013 × 10−45 to find m = −148; in double precision, the equation 2m−1 = 4.94066 × 10−324 yields m = −1073. 5. Determine machine precision, the smallest positive number and the largest pos- itive number for the floating point number system used by your calculator. Assuming the calculator uses β = 10, determine the values for k, m and M. Answers will of course vary. On a Casio fx − 300SA, machine precision is 5 × 10−10 , the smallest positive number is 10−99 and the largest positive number is 9.999999999 × 1099 . In general, machine precision with rounding is 1 2 β1−k , the smallest positive number is (0.1)β ×βm = βm−1 and the largest positive number is (1 − β−k ) × βM . Assuming that the calculator uses β = 10, the values for machine precision, the smallest positive number and the largest positive number on the Casio fx − 300SA determine k = 10, m = −98 and M = 100. 6. Determine the number of significant decimal digits and the number of significant binary digits to which each of the following pairs of numbers agree. (a) 355/113 and π (b) 685/252 and e (c) √ 10002 and √ 10001 (d) 103/280 and 1/e (a) Because 355 113 − π π = 8.491 × 10−8 and 10−8 < 8.491 × 10−8 ≤ 10−7 , it follows that 355 113 and π agree to at least 7 and at most 8 decimal digits. Since 2−24 = 5.960 × 10−8 < 8.491 × 10−8 < 1.192 × 10−7 = 2−23 , we see that 355 113 and π agree to at least 23 and at most 24 binary digits. (b) Because 685 252 − e e = 1.025 × 10−5 and 10−5 < 1.025 × 10−5 ≤ 10−4 ,
- 5. Floating Point Number Systems 5 it follows that 685 252 and e agree to at least 4 and at most 5 decimal digits. Since 2−17 = 7.629 × 10−6 < 1.025 × 10−5 < 1.526 × 10−5 = 2−16 , we see that 685 252 and e agree to at least 16 and at most 17 binary digits. (c) Because √ 10002 − √ 10001 √ 10001 = 4.999 × 10−5 and 10−5 < 4.999 × 10−5 ≤ 10−4 , it follows that √ 10002 and √ 10001 agree to at least 4 and at most 5 decimal digits. Since 2−15 = 3.052 × 10−5 < 4.999 × 10−5 < 6.103 × 10−5 = 2−14 , we see that √ 10002 and √ 10001 agree to at least 14 and at most 15 binary digits. (d) Because 103 280 − 1/e 1/e = 6.061 × 10−5 and 10−5 < 6.061 × 10−5 ≤ 10−4 , it follows that 103 280 and 1/e agree to at least 4 and at most 5 decimal digits. Since 2−15 = 3.052 × 10−5 < 6.061 × 10−5 < 6.103 × 10−5 = 2−14 , we see that 103 280 and 1/e agree to at least 14 and at most 15 binary digits. 7. The ideal gas law states that PV = nRT, where P is the pressure of the gas, V is the volume, n is the number of moles, T is the temperature and R = 0.08206 atm·m3 /moles·K is the universal gas constant. (a) Experimentally, it has been determined that P = 0.750 atm, V = 1.15 m3 and T = 294.1K. Assuming that all values have been rounded to the digits shown, in what range of values does n fall? (b) Experimentally, it has been determined that V = 0.331 m3 , n = 0.00712 moles and T = 264.7K. Assuming that all values have been rounded to the digits shown, in what range of values does P fall?
- 6. 6 Section 1.3 (a) With 0.7495 atm < P < 0.7505 atm 1.145 m3 < V < 1.155 m3 294.05 K < T < 294.15 K it follows from the ideal gas law that (0.7495)(1.145) (0.08206)(294.15) < n < (0.7505)(1.155) (0.08206)(294.05) or 0.03555 moles < n < 0.03592 moles. (b) With 0.3305 m3 < V < 0.3315 m3 0.007115 moles < n < 0.007125 moles 264.65 K < T < 264.75 K it follows from the ideal gas law that (0.007115)(0.08206)(264.65) 0.3315 < P < (0.007125)(0.08206)(264.75) 0.3305 or 0.46612 atm < P < 0.46836 atm. 8. In a physics laboratory, students measure the mass of a rectangular block to be 243.27 ± 0.005 grams. The length, width and depth of the block are measured to be 7.8 ± 0.05 cm, 3.1 ± 0.05 cm and 4.2 ± 0.05 cm, respectively. (a) In what range of values does the volume of the block fall? (b) In what range of values does the density of the block fall? Density is mass per unit volume. (a) With 7.75 cm < length < 7.85 cm 3.05 cm < width < 3.15 cm 4.15 cm < depth < 4.25 cm it follows that 98.095625 cm3 = (7.75)(3.05)(4.15) < volume < (7.85)(3.15)(4.25) = 105.091875 cm3 . (b) Density is defined as mass per unit volume. It is given that 243.265 grams < mass < 243.275 grams, and we determined in part (a) that 98.095625 cm3 < volume < 105.091875 cm3 , so 2.32 grams cm3 = 243.625 105.091875 < density < 243.275 98.095625 = 2.48 grams cm3 .
- 7. Floating Point Number Systems 7 9. Students are using a pendulum to experimentally determine the acceleration due to gravity, g. They measure the period, T, of the pendulum to be 2.2 seconds, and the length, l, of the pendulum to be 1.15 meters. Assuming that all values are correct to the digits shown, in what range of values does g fall? The variables in this problem are related by the formula T = 2π l/g. Solving T = 2π l/g for g yields g = 4π2 l/T2 . With 2.15 s < T < 2.25 s and 1.145 m < l < 1.155 m, it follows that 4π2 1.145 2.252 < g < 4π2 1.155 2.152 , or 8.929 m/s 2 < g < 9.864 m/s 2 . 10. Determine machine precision, the smallest positive number and the largest posi- tive number in the IEEE standard double precision system. Approximately how many significant decimal digits does the double precision standard supply? With β = 2 and k = 53, machine precision with rounding is u = 1 2 21−53 = 2−53 ≈ 1.11 × 10−16 . Accordingly, there are between 15 and 16 significant decimal digits available in IEEE standard double precision. The smallest positive number in double precision is (0.1)2 × 2−1021 = 2−1022 ≈ 2.23 × 10−308 , while the largest positive number is (0.111 · · · 1)2 × 21024 = (1 − 2−53 )21024 ≈ 1.80 × 10308 . 11. In addition to the standard single and double precision floating point systems, Intel microprocessors also have an extended precision system F(2, 64, −16381, 16384). Determine machine precision, the smallest positive number and the largest pos- itive number for this extended precision system. With β = 2 and k = 64, machine precision with rounding is u = 1 2 21−64 = 2−64 ≈ 5.42 × 10−20 . Accordingly, there are between 19 and 20 significant decimal digits available in this extended precision format. The smallest positive number is (0.1)2 × 2−16381 = 2−16382 ≈ 3.36 × 10−4932 ,
- 8. 8 Section 1.3 while the largest positive number is (0.111 · · · 1)2 × 216384 = (1 − 2−64 )216384 ≈ 1.19 × 104932 . 12. IBM System/390 mainframes provide three floating point number systems: short precision F(16, 6, −64, 63), long precision F(16, 14, −64, 63) and extended preci- sion F(16, 28, −64, 63). Compare machine precision, the smallest positive num- ber and the largest positive number for each of these number systems. In the short precision system F(16, 6, −64, 63), machine precision with rounding is u = 1 2 161−6 = 2−21 ≈ 4.77 × 10−7 , while machine precision with rounding in the long precision system F(16, 14, −64, 63) is u = 1 2 161−14 = 2−53 ≈ 1.11 × 10−16 . In the extended precision system F(16, 28, −64, 63), machine precision with round- ing is u = 1 2 161−28 = 2−109 ≈ 1.54 × 10−33 . Accordingly, the short precision system provides between 6 and 7 significant decimal digits, the long precision system provides between 15 and 16 significant decimal digits and the extended precision system provides between 32 and 33 significant decimal digits. In all three systems, the smallest positive number is (0.1)16 × 16−64 = 16−65 ≈ 5.40 × 10−79 . The largest positive number in short, long and extended precision is (1 − 16−6 ) · 1663 , (1 − 16−14 ) · 1663 , and (1 − 16−28 ) · 1663 , respectively. 13. A common floating point number system used on modern calculators is F(10, 10, −98, 100). Determine machine precision, the smallest positive num- ber and the largest positive number for this extended precision system. With β = 10 and k = 10, machine precision with rounding is u = 1 2 101−10 = 5 × 10−10 .
- 9. Floating Point Number Systems 9 Accordingly, there are between 9 and 10 significant decimal digits available in F(10, 10, −98, 100). The smallest positive number is (0.1)10 × 10−98 = 10−99 , while the largest positive number is (1 − 10−10 )10100 = 9.999999999 × 1099 . 14. (a) Show that the number of elements in the set F(β, k, m, M) is given by 1 + 2(β − 1)βk−1 (M − m + 1). (b) How many elements are in the IEEE standard single precision number system? (c) How many elements are in the IEEE standard double precision number system? (a) Let’s start by counting the number of positive elements in F(β, k, m, M). The only restriction on the mantissa is that the first digit cannot be zero; hence, there are (β − 1)βk−1 different mantissas that can be formed. With a minimum exponent of m and a maximum exponent of M, there are M − m + 1 different exponents. Consequently, there are (β − 1)βk−1 (M − m + 1) positive elements in F(β, k, m, M). Because the number system contains a zero element and equally many negative elements as positive elements, it follows that F(β, k, m, M) contains a total of 1 + 2(β − 1)βk−1 (M − m + 1) elements. (b) IEEE standard single precision is the system F(2, 24, −125, 128). Therefore, the IEEE standard single precision number system has 1 + 2(2 − 1)224−1 (128 − (−125) + 1) = 4, 261, 412, 865 elements. (c) IEEE standard double precision is the system F(2, 53, −1021, 1024). There- fore, the IEEE standard single precision number system has 1 + 2(2 − 1)253−1 (1024 − (−1021) + 1) = 18, 428, 729, 675, 200, 069, 633 ≈ 1.84 × 1019 elements. 15. Consider the function f(x) = x2 − 4x + 4. (a) What are the zeros of f?
- 10. 10 Section 1.3 (b) Suppose we were to change the constant term to 4 − 10−8 . What are the zeros of this new function? Relative to the size of the change in the constant term, how big is the change in the zeros of the function? (c) Now, suppose we were to change the constant term to 4 + 10−8 . What are the zeros of this new function? Relative to the size of the change in the constant term, how big is the change in the zeros of the function? (a) The function f(x) = x2 − 4x + 4 = (x − 2)2 has a (repeated) zero at x = 2. (b) Now consider the function f(x) = x2 − 4x + (4 − 10−8 ). By the quadratic formula, the zeros of this new function are x = 4 ± 16 − 4(4 − 10−8) 2 = 2 ± 10−4 = 1.9999, 2.0001 Observe that the change in the zeros (±0.0001) is 10,000 times larger than the change in the constant term in the function. (c) Finally, consider the function f(x) = x2 − 4x + (4 + 10−8 ). By the quadratic formula, the zeros of this new function are x = 4 ± 16 − 4(4 + 10−8) 2 = 2 ± 0.0001 · i Observe that the change in the zeros (±0.0001 · i) is 10,000 times larger than the change in the constant term in the function. 16. Consider the linear, first-order differential equation dx dt + 1 t x = sin t t . (a) Solve this equation subject to the initial condition x(π/2) = x0. (b) Solve this equation subject to the perturbed initial condition x(π/2) = x0 + ǫ. (c) By considering the difference between the solutions obtained in parts (a) and (b), comment on the conditioning of this problem. (a) Multiplying dx dt + 1 t x = sin t t by t yields dx dt + x = sin t.
- 11. Floating Point Number Systems 11 Note that the terms on the left-hand side of this latter equation are equal to the derivative of the product tx. Integrating both sides of this equation therefore produces tx = − cos t + C or x(t) = C − cos t t , where C is a constant of integration. Using the initial condition x(π/2) = x0, we determine x0 = C − 0 π/2 or C = πx0 2 . Hence, the solution of the initial value problem is x(t) = πx0 2t − cos t t . (b) The general solution to the differential equation remains x(t) = C − cos t t , where C is a constant of integration. Using the initial condition x(π/2) = x0 + ǫ, we determine x0 + ǫ = C − 0 π/2 or C = π(x0 + ǫ) 2 . Hence, the solution to the perturbed initial value problem is x(t) = π(x0 + ǫ) 2t − cos t t . (c) The difference between the solutions obtained in parts (a) and (b) is πǫ 2t . Because this difference decays to zero as t → ∞, this problem is not ill- conditioned. 17. Consider the linear, first-order differential equation dx dt − 1 t x = t sin t. (a) Solve this equation subject to the initial condition x(π/2) = x0. (b) Solve this equation subject to the perturbed initial condition x(π/2) = x0 + ǫ. (c) By considering the difference between the solutions obtained in parts (a) and (b), comment on the conditioning of this problem.
- 12. 12 Section 1.3 (a) Multiplying dx dt − 1 t x = t sin t by t−1 yields dx dt − 1 t2 x = sin t. Note that the terms on the left-hand side of this latter equation are equal to the derivative of the product t−1 x. Integrating both sides of this equation therefore produces x t = − cos t + C or x(t) = t(C − cos t), where C is a constant of integration. Using the initial condition x(π/2) = x0, we determine x0 = π 2 (C − 0) or C = 2x0 π . Hence, the solution of the initial value problem is x(t) = t 2x0 π − cos t . (b) The general solution to the differential equation remains x(t) = t(C − cos t), where C is a constant of integration. Using the initial condition x(π/2) = x0 + ǫ, we determine x0 + ǫ = π 2 (C − 0) or C = 2(x0 + ǫ) π . Hence, the solution to the perturbed initial value problem is x(t) = t 2(x0 + ǫ) π − cos t . (c) The difference between the solutions obtained in parts (a) and (b) is 2ǫt π . Because this difference tends toward infinity as t → ∞, meaning that a small change in input data can result in a large change in the output, this problem is ill conditioned. 18. Consider the linear system of equations 1.1 2.1 2 3.8 x y = b.
- 13. Floating Point Number Systems 13 (a) Solve the system for the right-hand side vector b = 3.2 5.8 T . (b) Solve the system for the right-hand side vector b = 3.21 5.79 T . (c) Solve the system for the right-hand side vector b = 3.1 5.7 T . (d) By considering the difference between the solutions obtained in parts (a), (b), and (c), comment on the conditioning of this problem. (a) The system of equations is 1.1x + 2.1y = 3.2 2x + 3.8y = 5.8 If we multiply the first equation by 2 and the second equation by−1.1 and then add, we obtain 0.02y = 0.02. Thus, y = 1. Back substituting into either of the original equations yields x = 1. (b) Working as we did in part (a), we find the solution corresponding to the right- hand side vector b = 3.21 5.79 T is x = −1.95 and y = 2.55. (c) Working as we did in part (a), we find the solution corresponding to the right- hand side vector b = 3.1 5.7 T is x = 9.5 and y = −3.5. (d) Given that small changes to the right-hand side vector resulted in relatively large changes to the solution vector, it appears that this problem is ill condi- tioned Solutions Manual for Friendly Introduction To Numerical Analysis 1st Edition by Bradie Full Download: https://downloadlink.org/p/solutions-manual-for-friendly-introduction-to-numerical-analysis-1st-edition-by-bradie/ Full download all chapters instantly please go to Solutions Manual, Test Bank site: TestBankLive.com