Solution manual for numerical methods 4th edition by faires

9.* Use the error term of a Taylor polynomial to estimate the error involved in using sin x ≈ x... Perform the following computations i exactly, ii using three-digit chopping arithmetic,

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Mathematical Preliminaries

Review of Calculus

Exercise Set 1.2 (page 14)

1 Show that the following equations have at least one solution in the given intervals.

(a) x cos x − 2x2+ 3x − 1 = 0, [0.2, 0.3] and [1.2, 1.3]

(b) (x − 2)2− ln x = 0, [1, 2] and [e, 4]

(c) 2x cos(2x) − (x − 2)2= 0, [2, 3] and [3, 4]

(d∗ ) x − (ln x) x = 0, [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) = 0.

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 0 = f (x) = x − (ln x) x

2 Find intervals containing solutions to the following equations.

(a) x − 3 −x= 0

(b) 4x2− e x= 0(c∗ ) x3− 2x2− 4x + 3 = 0

that is, when x = −2

3 and x = 2 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(−2

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SOLUTION: 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

SOLUTION: (a) First note that f ′ (x) = ( −e x + 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) 0.8SOLUTION: (c) 5.164000SOLUTION: (d) 1.582572

5 Let f (x) = x3

(a) Find the second Taylor polynomial P2(x) about x0= 0

(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= 1

(d) Repeat part (b) for the polynomial found in part (c)

SOLUTION: (a) P2(x) = 0 SOLUTION: (b) R2(0.5) = 0.125; actual error = 0.125 SOLUTION: (c) P2(x) = 1 + 3(x − 1) + 3(x − 1)2

SOLUTION: (d) R2(0.5) = −0.125; actual error = −0.125

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7.* Find the second Taylor polynomial for f (x) = e x cos x about x0= 0.

(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

∫ 1 0

f (x) dx using

∫ 1 0

≤13(0.5)2 max

ξ ∈[0,0.5] |e ξ (sin ξ + cos ξ) |.

To maximize this quantity on [0, 0.5], ﬁrst note that D x e x (sin x + cos x) = 2e x cos x > 0, for all x in [0, 0.5] This implies that the maximum and minimum values of e x (sin x + cos x) on [0, 0.5] occur at the endpoints of the interval, and

e0(sin 0 + cos 0) = 1 < e 0.5 (sin 0.5 + cos 0.5) ≈ 2.24.

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4 Exercise Set 1.2

(c)

∫ 1 0

f (x) dx ≈

∫ 1 0

= 3

2.

(d) From (b),

∫ 1 0

|R2(x) | dx ≤

∫ 1 0

1

3e

1(cos 1 + sin 1)x3 dx =

∫ 1 0

1.252x3dx = 0.313.

Since

∫ 1 0

so the actual error is|1.3780 − 1.5| ≈ 0.12.

8 Find the third Taylor polynomial P3(x) for the function f (x) = (x − 1) ln x about x0= 1

(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].

(c) Approximate∫1.5

0.5 f (x) dx using∫1.5

0.5 P3(x) dx.

(d) Find an upper bound for the error in (c) using∫1.5

0.5 |R3(x) dx |, and compare the bound

to the actual error

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

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Mathematical Preliminaries 5

If we use the bound| cos ξ| ≤ 1, then

sin(180π )− π

180

= R

2

( π180

) = − cos ξ

3!

( π180

)3

≤ 8.86 × 10 −7 .

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 x

For|Rn(7π30)| < 10 −6 , it suﬃces to take n = 3 To 7 digits,

cos 42◦ = 0.7431448 and P

3(42◦ ) = P

3(7π

30) = 0.7431446,

so the actual error is 2× 10 −7.

11.* Let f (x) = e x/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 tocompute the ﬁrst three terms of the Taylor polynomial for f(x)

taylor(f(x),’Order’,3)MATLAB responds with

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diff(f(x),4)

and MATLAB responds with

ans =(5*cos(x/3)*exp(x/2))/54 - (119*sin(x/3)*exp(x/2))/1296Now deﬁne

f4=inline(’(5*cos(x/3)*exp(x/2))/54 - (119*sin(x/3)*exp(x/2))/1296’,’x’)

which results inf4 = Inline function: f4(x) = (5*cos(x/3)*exp(x/2))/54 - (119*sin(x/3)*exp(x/2))/1296Then 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))/2592For the evaluations we will use the long format of MATLAB, implemented withformat 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 = 0.604738907555866Deﬁne this value so that we can evaluate the fourth derivative at this point

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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= 1

(b) The maximum error|f(x) − P3(x) | for 0 ≤ x ≤ 1.

(c) The Maclaurin polynomial ˜P3(x) for f

(d) The maximum error|f(x) − ˜ P3(x) | for 0 ≤ x ≤ 1.

(e) Does P3(0) approximate f (0) better than ˜ P3(1) approximates f (1)?

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= 1 if the

tangent line at x0= 1 has equation y = 4x − 1, and if f ′′(1) = 6.

SOLUTION: (a) P n (k) (x0) = f (k) (x0) for k = 0, 1, , n The shapes of P n and f are the same at x0

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 0 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 −t2

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(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 inpart (b)

(e) Explain why diﬃculties occur using the series in part (b) to approximate erf(x).

SOLUTION: (a) Use the series

2

√ π

√

π e

−x2∑∞ k=0

2k x 2k+1

1· 3 · · · (2k + 1) =

2

√ π

We have no such bound for the positive term series in part (b)

16.* In Example 3 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 ≥ 0 the function f(x) = x − sin x is non-decreasing, which implies

that sin x ≤ x with equality only when x = 0.

(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) = 1 − cos x ≥ 0, because −1 ≤ cos x ≤ 1

for all values of x Also, the statement clearly holds when |x| ≥ π, because | sin x| ≤ 1.

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SOLUTION: (a) The observation implies that f (x) is non-decreasing for all values of x, and

in particular that f (x) > f (0) = 0 when x > 0 Hence for x ≥ 0, we have x ≥ sin x, and when

0≤ x ≤ π, we have | sin x| = sin x ≤ x = |x|.

SOLUTION: (b) When −π < x < 0, we have π ≥ −x > 0 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

1 Compute the absolute error and relative error in approximations of p by p ∗.

2 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)

(a) 4

5 +

13(b) 4

5 ·1

3(c)

(1

3 − 3

11

)+ 320(d)

(1

3 +

311

)

− 3

20

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SOLUTION: (a) (i) 17/15 (ii) 1.13 (iii) 1.13 (iv) both 3× 10 −3

SOLUTION: (b) (i) 4/15 (ii) 0.266 (iii) 0.266 (iv) both 2.5 × 10 −3

SOLUTION: (c) (i) 139/660 (ii) 0.211 (iii) 0.210 (iv) 2× 10 −3, 3× 10 −3

SOLUTION: (d) (i) 301/660 (ii) 0.455 (iii) 0.456 (iv) 2× 10 −3, 1× 10 −4

3 Use three-digit rounding arithmetic to perform the following calculations Compute the

abso-lute error and relative error with the exact value determined to at least ﬁve digits

2e − 5.4

(f) −10π + 6e − 3

62(g)

(29

)

·

(97)

(h) π −22

7 1 17

2e − 5.4 =

0.0720 0.0400 = 1.80.

The correct value is approximately 1.954, so the absolute and relative errors to three digits are

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2e − 5.4 =

0.0710 0.0200 = 3.55.

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|

1.954 = 0.817,

respectively The results in Exercise 3(e) were considerably better

5 Repeat Exercise 3 using four-digit rounding arithmetic.

6 Repeat Exercise 3 using four-digit chopping arithmetic.

Full file at https://TestbankDirect.eu/Solution-Manual-for-Numerical-Methods-4th-Edition-by-Faires

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)+ arctan

(13

)]

(b) 16 arctan

(15

)

− 4 arctan

(1239)

SOLUTION: (a) We have P(1

2 + arctan

13

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9 Suppose the points (x0, y0) and (x1, y1) are on a straight line with y1̸= y0 Two formulas are

available to ﬁnd the x-intercept of the line:

(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.

10 The Taylor polynomial of degree n for f (x) = e xis ∑n

i=0 x i /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

9

∑

i=0

(−1) i5i i!

(b) e −5= 1

e5 ≈∑9 1

i=05i /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

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11.* A rectangular parallelepiped has sides 3 cm, 4 cm, and 5 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 = 3 cm, y = 4 cm, and z = 5 cm to the nearest cm This

12 The following MATLAB M-ﬁle rounds or chops a number x to t digits where rnd = 1 for

rounding and rnd = 0 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

(a) x = 124.031, t = 5 (b) x = 124.036, t = 5 (c) x = −0.00653, t = 2

(d) x = −0.00656, t = 2

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.

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13 The binomial coeﬃcient (

m k

)

k! (m − k)!

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

for all k by the deﬁnition without causing overﬂow?

(b) Show that(m

k

)can also be computed by(

m k

by the formula in part (b) without causing overﬂow?

(d) Use the equation in (b) and four-digit chopping arithmetic to compute the number ofpossible 5-card hands in a 52-card deck Compute the actual and relative errors

SOLUTION: (a) m = 17

SOLUTION: (b)

(

m k

SOLUTION: (c) m = 181707 SOLUTION: (d) 2,597,000; actual error 1960; relative error 7.541 × 10 −4

Errors in Scientiﬁc Computation

1 (i) Use four-digit rounding arithmetic and the formulas of Example 1 to ﬁnd the most accurate

approximations to the roots of the following quadratic equations (ii) Compute the absoluteerrors and relative errors for these approximations

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The actual root is 10.98687488, so the absolute error is|10.98687488−10.98| = 6.87488×10 −3,

and the relative error is

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

0.001149076 = 6.5844 × 10 −5 .

2 Repeat Exercise 1 using four-digit chopping arithmetic.

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The actual root is 10.98687488, so the absolute error is

|10.98687488 − 10.98| = 6.87488 × 10 −3, and the relative error is

The absolute errors and relative errors for these approximations are:

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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 .

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

(a) Approximate e −0.98using ˆP5(0.49) and four-digit rounding arithmetic.

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(b) Compute the absolute and relative error for the approximations in part (a)

(c) Approximate e −0.98 using 1/P

5(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 ˆP5(0.49) gives

P5(0.49) =

1(((((−0.2667)(0.49) + 0.6667)(0.49) + 1.333)(0.49) + 2)(0.49) + 2)(0.49) + 1

6.* (a) Show that the polynomial nesting technique can be used to evaluate

f (x) = 1.01e 4x − 4.62e 3x − 3.11e 2x + 12.2e x − 1.99.

(b) Use three-digit rounding arithmetic and the formula given in the statement of part (a)

to evaluate f (1.53).

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- Faires< /h3>... absolute errors and relative errors for these approximations are:

Full file at https://TestbankDirect.eu /Solution- Manual- for- Numerical- Methods- 4th- Edition- by- Faires< /h3>

Trang... 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- Faires< /h3>

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