Nonlinear equation handout

DEFINITION 2.1 The closed interval [a, b] or the open interval a, b in which there is a unique root of equation 1, is called initial estimates... To find the approximation to the roots

SOLUTIONS OF EQUATIONS IN ONE VARIABLE

Dr Lê Xuân Đại

HoChiMinh City University of Technology Faculty of Applied Science, Department of Applied Mathematics

Email: ytkadai@hcmut.edu.vn

HCMC — 2019.

Some difficulties when solving the

equation(1)

f (x) = a n x n + a n−1 x n−1 + + a1x + a0 = 0,

(a n6= 0), wheren = 1,2we have explicit

formula for the root In addition, for

n = 3,4the formula becomes

complicated However, for n Ê 5 there is

no general formula

When f (x) = 0 is a transcendental

equation, for instance, cos x − 5x = 0, there

is no general formula

I NITIAL E STIMATES

We will consider an initial guess for the root

or a bracketing interval containing a root

DEFINITION 2.1

The closed interval [a, b] (or the open interval

(a, b) ) in which there is a unique root of

equation (1), is called initial estimates

To find the approximation to the roots ofequation (1), we have to follow the

following steps:

1 Determine initial estimates for the roots

of equation (1) It means that, we have tofind the interval containing a unique

root

2 In each interval, find the approximation

to the root using some given methods

I NITIAL ESTIMATES

THEOREM 2.1

If function f (x) is continuous on the interval

(a, b) and f (a).f (b) < 0, f0(x) exists and has the same sign in (a, b), then in (a, b) there exists unique real root x of equation (1).

The equation has roots in the intervals

[−3,−2];[0,1];[2,3]. Since the cubic functionhas maximum 3 roots, so in each interval,there is 1 unique root Therefore, they arethe initial estimates

X3− 6 ∗ X + 2

- Calc X = −3,−2, ,3

The equation has 1 root x = 0 and 1 root inthe interval · 1

2, 1

¸ Therefore, the initialestimates of the equation f (x) = 0 are

£− 1

2 ,12¤ ;£ 1

2 , 1¤

EXAMPLE 2.5

If the equation f (x) = x3− 5x2+ 12 = 0 on the interval [−2,−1]has the approximation to the root x∗= −1.37, then

|f0(x)| = |3x2− 10x| Ê 13 = m > 0, ∀x ∈ [−2, −1].

Therefore,

|x∗− x| É |f (−1.37)|

Therefore, the initial estimates for the roots

of the equation are [0, 1]and [1, 2].

M ULTIPLE CHOICE EXERCISES

EXERCISE 2.3

If the equation f (x) = 5x3+ 12x − 5 = 0 has in the interval [0, 1]the approximation to the root x∗= 0.40, the minimum error of x∗ is

B ISECTION T ECHNIQUE

Suppose that (a, b) is the initial estimates, inwhich the equation (1) has unique root Themethod calls for a repeated halving of

subinterval of [a, b]:

Suppose f is a continuous function

defined on the interval [a, b] and

f (a).f (b) < 0. Then there is a number x in

(a, b) such that f (x) = 0. Seta0= a, b0= b,

d0= b0− a0= b − aand x0 is the midpoint

of [a, b].

If f (x0) = 0, then x0 is the exact root.

Inversely, if f (x0).f (a0) < 0 then set

(

a n É x É b n , a nÉx n = a n +b n

2 É b n

f (a n ).f (b n ) < 0, d n = b n − a n = b−a2n

T HE CONVERGENCE OF THE BISECTION METHOD

Since the sequence (a n) is increasingandbounded above by b, and the sequence (b n)

is decreasing and bounded below by a,so as

E RROR ANALYSIS FOR THE BISECTION METHOD

EXAMPLE 3.1

If the equation f (x) = 5x3− cos 3x = 0 has a unique root in [0, 1], then use the bisection method to determine an approximation x5

to the root and estimate the error of this

method.

SOLUTION We have f (0) < 0and f (1) > 0

∆x5= 1−026 = 641 ≈ 0.0157.

EXERCISE 3.1

Use the bisection method to find an

approximation x5 to the root of the equation

f (x) =px − cosx = 0, if this equation has a

unique root in [0, 1]. Estimate the error of the root and the error of the bisection method.

SOLUTION We have f (0) < 0and f (1) > 0

EXERCISE 3.2

Use the bisection method to find root

accurate to within 10−2 for the equation

x = tanx, if the equation has a unique root in

[4, 4.5].

SOLUTION The error of the bisection

method ∆x n= 4.5 − 4

2n+1 < 10−2⇒ 2n > 25. Thenumber n of iterations necessary to solvethe equation with accuracy to within 10−2

will satisfy 2n > 25 ⇒ n = 5. Let f (x) = x − tanx.

We have f (4) > 0,f (4.5) < 0

EXERCISE 3.3

Use the bisection method to find root

accurate to within 10−2 for the equation

2 + cos(e x − 2) − e x= 0, if this equation has a unique root in [0.5, 1.5].

SOLUTION. The error of the bisection

method is ∆x n = 1.5 − 0.5

2n+1 < 10−2⇒ 2n> 50.

Therefore, n = 6 iterations will ensure an

approximation accurate to within 10−2. Let

f (x) = 2 + cos(e x − 2) − e x ⇒ f (0.5) > 0, f (1.5) < 0

M ULTIPLE CHOICE EXERCISES

If the equation f (x) = 2x3− 6x2+ 6x − 13 = 0

has a unique root in [2, 3]. Use the bisectionmethod to find approximation x5 to the rootfor the equation

F IXED -P OINT I TERATION

DEFINITION 4.1

The number x is a fixed point for a given

function g if g(x) = x.

Suppose that the equation f (x) = 0 has a

unique root in (a, b). Given a root-findingproblem f (x) = 0, we can define function g

with a fixed point x in a number of ways

For example, for the equation x3− x − 1 = 0

We choose an initial approximation

x0∈ [a, b] and generate the sequence(x n) bysubstituting x = x0 into the right side of (2),then x1= g(x0 ). Reapply this process by

substituting x = x1 into the right side of (2),

we get x2= g(x1 ).

In the end, we have the sequence (x n),

which is defined by

Question: How can we find a fixed-point

problem that produces a sequence (x n) thatreliably and rapidly converges to a root togiven root-finding problem?

S UFFICIENT CONDITIONS FOR THE EXISTENCE AND UNIQUENESS OF A FIXED POINT

T HEOREM 4.1

1 If g(x) is continuous on [a, b] and

g(x) ∈ [a,b],∀x ∈ [a,b], then g has at least one fixed point in [a, b].

2 If, in addition, g0(x) exists on (a, b) and a positive constant q ∈ (0,1) exists with

|g0(x)| É q,∀x ∈ (a,b), (4)

then there is exactly one fixed point in [a, b].

THEOREM 4.2

Suppose that g(x) is continuous on [a, b] and

g : [a, b] → [a,b]. In addition, g0 exists on (a, b)

and a constant q ∈ (0,1) exists with

|g0(x)| É q,∀x ∈ (a,b). Then for any number

x0∈ [a, b], the sequence defined by

x n = g(x n−1 ), n Ê 1, converges to the unique

fixed point x ∈ [a,b]. Error analysis

Note The rate of convergence depends on the factorq n The smaller the value ofq, the faster the convergence, which may

be very slow ifqis close to 1.

EXAMPLE 4.2

Use a fixed-point iteration method to

determine a root accurate to within 10−4 for the equation f (x) = 5x3− 20x + 3 = 0 on (0, 1).

Therefore, for g3(x) we have

Use x0= 0.75 ∈ [0, 1]. Evaluatex n , n = 1,2, bythe formula x n = 5x

Perform four iterations for each function

g k (x), k = 1,2,3,4 using x0= 1. Which function

do you think gives the best approximation

to the root?

SOLUTION Usingx0= 1 we have

EXERCISE 4.2

Use a fixed-point iteration method to

determine a root accurate to within 10−3 for the equation x3− 3x2− 5 = 0 on [3, 4]. Use

= 0.0017

Use x0= 3.5 ∈ [3, 4]. Evaluate x n , n = 1,2, bythe formula x n = 3 + 5

x n−12

EXERCISE 4.3

Use a fixed-point iteration method to

determine the root accurate to within 10−3

for the equation x = x2−e3x+2 on [0, 1]. Use

= 0.002.

Use x0= 0.5 ∈ [0, 1]. Evaluate x n , n = 1,2, bythe formula x n = x

2

n−1 − e x n−1+ 2

3

Use x0= 2.5 ⇒ x1 = 2.8 Error analysis

⇒ n Ê

ln

h

10−4.(1−0.64) 0.3

i

ln 0.64 ≈20.23 ⇒ n = 21

M ULTIPLE CHOICE EXERCISES

EXERCISE 4.5

Use a fixed-point method to determine a

root for the equation x =p3 6x + 14 on[3, 4].

Using x0= 3.2, find the approximation x2

EXERCISE 4.6

Use fixed-point method to determine the

root for the equation x =p3 6x + 14 on[3, 4].

Use x0= 3.2 estimate the minimum absolute error of the approximation x2

N EWTON ’ S M ETHOD

1 Suppose that the equation f (x) = 0has aunique root on[a, b]. Newton’s methodillustrates how the approximations areobtained using successive tangents

2 Starting with the initial approximation

x0, the approximationx1 is the

x−intercept of the tangent line to the

graph of f at (x0, f (x0)).

In order to evaluate x1, we will consider 2cases:

Case 1.f0(x).f00(x) > 0.We will consider 2 situations

1.f (a) < 0,f (b) > 0,f0(x) > 0,f00(x) > 0,∀x ∈ (a,b)

2.f (a) > 0,f (b) < 0,f0(x) < 0,f00(x) < 0,∀x ∈ (a,b)

x1∉ (a, b).So we choosex0= b.

The equation of the tangent line to the

graph of y = f (x) at (b, f (b))is

y − f (b) = f0(b)(x − b). Since x1 is x−intercept ofthe tangent line of the graph of f , so

Continuing this process, we generate thesequence defined, for n Ê 1,by

x n = x n−1− f (x n−1)

f0(x n−1) ·

Case 2.f0(x).f00(x) < 0.We will consider 2 situations

1.f (a) < 0,f (b) > 0,f0(x) > 0,f00(x) < 0,∀x ∈ (a,b)

2.f (a) > 0,f (b) < 0,f0(x) < 0,f00(x) > 0,∀x ∈ (a,b)

x1∉ (a, b).

The equation of the tangent line to the

graph of y = f (x) at (a, f (a))is

y − f (a) = f0(a)(x − a). Since x1 is x−intercept ofthe tangent line of the graph of f , so

0 − f (a) = f0(a)(x1− a) ⇔ x1= a − f (a)

Continuing this process, we generate thesequence defined, for n Ê 1,by

x n = x n−1− f (x n−1)

f0(x n−1) ·

F OURIER ’ S C ONDITION

If we choose x0 such that f (x0) and f00(x0)

have different sign, then Newton’s

method can not be used

Fourier’s Condition: Choose x0= b iff (b)

and f00(x) have the same sign Choose

x0= a iff (a) and f00(x) have the same sign

Suppose that the equationf (x) = 0has a unique root

on[a, b], f00(x) andf0(x)remain the sign unchanged in

(a, b).

defined by the formula

EXAMPLE 5.1

Suppose that the equation

f (x) = x3− 3x + 1 = 0 has a unique root on

[0, 0.5]. Find the approximation x3 to the root using Newton’s method and estimate its

We generate the sequence (x n) by the

CASIO Evaluate x n

x − x

3

− 3x + 1 3x2 − 3

We generate the sequence(x n) by the formula

estimated by the formula

EXERCISE 5.2

Use Newton’s method to find the root

accurate to within 10−5 for the equation

We generate the sequence(x n) by the formula

M ULTIPLE CHOICE EXERCISES

CALC X = 0.3 = ⇒ x1 ≈ 0.3202 ⇒ answer 3

EXERCISE 5.4

Suppose that the equation

f (x) = 2x3+ 6x2+ 7x + 5 = 0 has a unique root

on [−1.9,−1.8]. Use Newton’s method and

Fourier’s condition to choose x0, estimate the error of the approximation x1

THANK YOU FOR YOUR ATTENTION