Chapter 2 Homework Each of the following equations has a root on the

2.4 Newton’s Method

1. Each of the following equations has a root on the interval (0,1). Perform New-

ton’s method to determine p4, the fourth approximation to the location of the

root.

(a) ln(1 + x)−cos x= 0 (b) x5+ 2x−1 = 0

(c) e−x−x= 0 (d) cos x−x= 0

(a) Let f(x) = ln(1 + x)−cos x. Then f′(x) = 1

1+x+ sin x. With p0= 0, four

iterations of Newton’s method yield

(b) Let f(x) = x5+ 2x−1. Then f′(x) = 5x4+ 2. With p0= 0, four iterations

of Newton’s method yield

−e−p0−1= 0.5000000000;

2Section 2.4

2. Construct an algorithm for Newton’s method. Is it necessary to save all calcu-

lated terms in the sequence {pn}?

Because convergence is quadratic, iteration is terminated when |pn−pn−1|falls

In Exercises 3 – 6, an equation, an interval on which the equation has a root,

and the exact value of the root are specified.

(a) Perform five (5) iterations of Newton’s method.

(b) For n≥1, compare |pn−pn−1|with |pn−1−p|and |pn−p|.

(c) For n≥1, compute the ratio |pn−p|/|pn−1−p|2and show that this value

approaches |f′′ (p)/2f′(p)|.

Newton’s Method 3

3. The equation x3+x2−3x−3 = 0 has a root on the interval (1,2), namely

x=√3.

npn|pn−pn−1| |pn−p| |pn−1−p| |pn−p|/|pn−1−p|2

1 3.00000000 2.00000000 1.26794919 0.73205081 2.366

3 1.83015075 0.36984925 0.09809995 0.46794919 0.448

Note that for n≥3,|pn−pn−1|provides an excellent estimate for |pn−1−p|and is

4. The equation x7= 3 has a root on the interval (1,2), namely x=7

√3.

npn|pn−pn−1| |pn−p| |pn−1−p| |pn−p|/|pn−1−p|2

0 1.00000000

2 1.19691682 0.08879746 0.02698601 0.11578347 2.013

Note that for all n,|pn−pn−1|provides a reasonable estimate for |pn−1−p|and is

5. The equation x3−13 = 0 has a root on the interval (2,3), namely 3

npn|pn−pn−1| |pn−p| |pn−1−p| |pn−p|/|pn−1−p|2

0 3.00000000

2 2.35804119 0.12344029 6.707 ×10−30.13014679 0.396

Note that for all n,|pn−pn−1|provides a reasonable estimate for |pn−1−p|and is

6. The equation 1/x−37 = 0 has a zero on the interval (0.01,0.1), namely x= 1/37.

npn|pn−pn−1| |pn−p| |pn−1−p| |pn−p|/|pn−1−p|2

1 0.01630000 0.00630000 0.010727027 0.017027027 37.000

3 0.02635634 0.00358687 0.000670691 0.004257557 37.000

Note that for all n,|pn−pn−1|provides a reasonable estimate for |pn−1−p|and

7. Show that when Newton’s method is applied to the equation x2−a= 0, the

resulting iteration function is g(x) = 1

2x+a

x.

Newton’s Method 5

8. Show that when Newton’s method is applied to the equation 1/x −a= 0, the

resulting iteration function is g(x) = x(2 −ax).

9. The function f(x) = sin xhas a zero on the interval (3,4), namely x=π.

Perform three iterations of Newton’s method to approximate this zero, using

p0= 4. Determine the absolute error in each of the computed approximations.

What is the apparent order of convergence? What explanation can you provide

for this behavior? (NOTE: If you have access to MAPLE, perform five iterations

with the Digits parameter set to at least 100.)

n|pn−p| |pn−p|/|pn−1−p|3

2 9.280 ×10−30.34577

Because the ratio in the third column of the table appears to be approaching a

6Section 2.4

10. (a) Verify that the equation x4−18x2+45 = 0 has a root on the interval (1,2).

Next, perform three iterations of Newton’s method, with p0= 1. Given

that the exact value of the root is x=√3, compute the absolute error in the

approximations just obtained. What is the apparent order of convergence?

What explanation can you provide for this behavior? (NOTE: If you have

access to MAPLE, perform five iterations with the Digits parameter set

to at least 100.)

(b) Verify that the equation x4−18x2+ 45 = 0 also has a root on the interval

(3,4). Perform five iterations of Newton’s method, and compute the abso-

lute error in each approximation. The exact value of the root is x=√15.

What is the apparent order of convergence in this case?

(c) What explanation can you provide for the different convergence behavior

between parts (a) and (b)?

n|pn−p| |pn−p|/|pn−1−p|3

2 1.014 ×10−30.34730

Because the ratio in the third column of the table appears to be approaching a

(b) Let f(x) = x4−18x2+ 45. Then f(3) = −36 <0and f(4) = 13 >0,

npn|pn−p| |pn−p|/|pn−1−p|2

Newton’s Method 7

11. The function f(x) = 27x4+ 162x3−180x2+ 62x−7 has a zero at x= 1/3.

Perform ten iterations of Newton’s method on this function, starting with p0=

0. What is the apparent order of convergence of the sequence of approximations?

What is the multiplicity of the zero at x= 1/3? Would the sequence generated

by the bisection method converge faster?

npn|pn−p| |pn−p|/|pn−1−p|

2 0.1871468695 0.1461864638 0.663187

4 0.2687288261 0.0646045072 0.665169

6 0.3046911326 0.0286422007 0.666010

8 0.3206173081 0.0127160252 0.666378

Convergence is clearly linear with an asymptotic error constant of λ= 2/3 = 1−1/3;

12. Repeat Exercise 11 for the function

f(x) = x

1 + x2−500

841 1−21x

125 ,

which has a zero at x= 2.5. Start Newton’s method with p0= 2.

Let

8Section 2.4

npn|pn−p| |pn−p|/|pn−1−p|

1 2.2600472810 0.2399527190

3 2.4401402930 0.0598597070 0.500415

5 2.4850136160 0.0149863840 0.500274

7 2.4962520120 0.0037479880 0.500041

13. The function f(x) = x3+ 2x2−3x−1 has a zero on the interval (−1,0).

Approximate this zero to within an absolute tolerance of 5 ×10−5.

Let f(x) = x3+ 2x2−3x−1. With an initial approximation of p0= 0 and a

14. For each of the functions given below, use Newton’s method to approximate all

real roots. Use an absolute tolerance of 10−6as a stopping condition.

(a) f(x) = ex+x2−x−4

(b) f(x) = x3−x2−10x+ 7

(c) f(x) = 1.05 −1.04x+ ln x

Newton’s Method 9

np0=−2p0= 1

1−1.5610519106 1.3447071068

(b) Let f(x) = x3−x2−10x+ 7. By trial and error, we find that f(−4) <0,

np0=−4p0= 0 p0= 3

1−3.2826086957 0.7000000000 3.4545454545

10 Section 2.4

np0= 0.80 p0= 1.10

1 0.8244931015 1.1100083179

15. An equation of state relates the volume Voccupied by one mole of a gas to

the instantaneous pressure Pand the Kelvin absolute temperature Tof the gas.

The Redlich-Kwong equation of state is given by

P=RT

V−b−a

V(V+b)√T,

where aand bare related to the critical temperature Tcand the critical pressure

Pcby the equations

a= 0.42747 R2T5/2

c

Pc!and b= 0.08664 RTc

Pc.

The coeﬃcient Ris a universal constant equal to 0.08206.

(a) Determine the volume of one mole of carbon dioxide at a temperature of

T= 323.15K and a pressure of one atmosphere. For carbon dioxide, Tc=

304.2K and Pc= 72.9 atmospheres.

(b) Determine the volume of one mole of ammonia at a temperature of T=

450K and a pressure of 56 atmospheres. For ammonia, Tc= 405.5K and

Pc= 111.3 atmospheres.

Newton’s Method 11

(a) For carbon dioxide, Tc= 304.2K and Pc= 72.9atmospheres, so

With a convergence tolerance of 5×10−7, Newton’s method yields

(b) For ammonia, Tc= 405.5K and Pc= 111.3atmospheres, so

With a convergence tolerance of 5×10−7, Newton’s method yields

nVn

16. In determining the minimum cushion pressure needed to break a given thickness

of ice using an air cushion vehicle, Muller (“Ice Breaking with an Air Cushion

Vehicle,” in Mathematical Modeling: Classroom Notes in Applied Mathematics,

M.S. Klamkin, editor, SIAM, 1987) derived the equation

p3(1 −β2) + 0.4hβ2−σh2

r2p2+σ2h4

3r4p−σh2

3r23

= 0,

where pdenotes the cushion pressure, hthe thickness of the ice field, rthe size

of the air cushion, σthe tensile strength of the ice, and βis related to the width

of the ice wedge. Take β= 0.5, r= 40 feet and σ= 150 pounds per square inch

(psi). Determine pfor h= 0.6, 1.2, 1.8, 2.4, 3.0, 3.6 and 4.2 feet.

Let

17. A frame structure is composed of two vertical columns and one horizontal beam,

as shown below. The vertical columns are of length Land have modulus of elas-

ticity Eand moment of inertia I. The horizontal beam connecting the tops of

the columns is of length L1with modulus of elasticity Eand moment of inertia

I1. The structure is pinned at the bottom and free to displace laterally at the

top. The buckling load, P, for the structure is given by

P

P= (kL)2EI

L2,

where kL is the smallest positive solution of

kL tan kL = 6 I1L

IL1

.

Suppose E= 30 ×106lb/in2,I= 15.2 in4,

L= 144 in, I1= 9.7 in4and L1= 120 in.

Determine the buckling load of the structure.

Newton’s Method 13

Let x=kL and define the function

nxn

1 1.4472486988

4 1.2981292481