NONLINEAR EQUATION.pdf

NONLINEAR EQUATION
July 7, 2023
NONLINEAR EQUATION July 7, 2023 1 / 32

1 MOTIVATION
2 THE BISECTION METHOD
3 FIXED POINT ITERATION METHOD
4 NEWTON’S METHOD

MOTIVATION
MOTIVATION
The equation f (x) = ax2
+bx +c = 0 can be solved directly.

MOTIVATION
MOTIVATION
EXAMPLE 1.1
The growth of population in America is described by model:
N(t) = 1000eλ
+
435
λ
(eλ
−1), where N(t) is the number of population
in time t and λ is the constant birthrate of population. Suppose that
in 2001, population contains 1000,000 people, find the birthrate in
this year.

MOTIVATION
MOTIVATION
EXAMPLE 1.1
The growth of population in America is described by model:
N(t) = 1000eλ
+
435
λ
(eλ
−1), where N(t) is the number of population
in time t and λ is the constant birthrate of population. Suppose that
in 2001, population contains 1000,000 people, find the birthrate in
this year.
The aim of this chapter is to solve approximately nonlinear
equations that are impossible to solve exactly.

MOTIVATION
ISOLATED INTERVAL
DEFINITION 1.1
If on [a,b], the equation f (x) = 0 has only one real root, then we say
that [a,b] is an isolated interval of f (x) = 0.

MOTIVATION
ISOLATED INTERVAL
DEFINITION 1.1
THEOREM 1.1
Given an equation f (x) = 0 and an interval [a,b]. If f (x) is continous
on [a,b],f (a).f (b) < 0 and f is monotone on [a,b], then [a,b] is an
isolated interval of f (x) = 0.

MOTIVATION
ISOLATED INTERVAL
DEFINITION 1.1
THEOREM 1.1
There are 2 methods to determine the number of intervals
containing root:

MOTIVATION
ISOLATED INTERVAL
DEFINITION 1.1
THEOREM 1.1
containing root:
Analytical method

MOTIVATION
ISOLATED INTERVAL
DEFINITION 1.1
THEOREM 1.1
containing root:
Analytical method
Geometrical method

MOTIVATION
EXAMPLE 1.2
Determine all isolated intervals containing root of the following
equations:
x3
−6x +2 = 0
ex
− x2
= 0

MOTIVATION
ERROR
THEOREM 1.2 (GENERAL FORMULA OF ERROR OF ROOT EQUATION)
Suppose that f (x) is continuous and differentiable on [a,b] and
∀x ∈ [a,b],|f ′
(x)| Ê m > 0. Let x∗
, x̄ be approx. and exact roots of
f (x) = 0 on [a,b] respectively, then the absolute error ∆x∗ of x∗
is:
|x∗
− x| É
|f (x∗
)|
m
= ∆x∗ .
Proof

MOTIVATION
ERROR
THEOREM 1.2 (GENERAL FORMULA OF ERROR OF ROOT EQUATION)
Suppose that f (x) is continuous and differentiable on [a,b] and
∀x ∈ [a,b],|f ′
(x)| Ê m > 0. Let x∗
, x̄ be approx. and exact roots of
f (x) = 0 on [a,b] respectively, then the absolute error ∆x∗ of x∗
is:
|x∗
− x| É
|f (x∗
)|
m
= ∆x∗ .
Proof
EXAMPLE 1.3
Suppose that the equation f (x) = x3
−5x2
+12 = 0 has approximate
solution x∗
= −1.37 on [−2,−1]. Estimate the error of x∗
, rounding to
4 decimal digits.

THE BISECTION METHOD
Let f (x) = 0 have one real root on [a,b].

Set a0 = a,b0 = b,x0 =
a0 +b0
2
, d0 = b0 − a0

Set a0 = a,b0 = b,x0 =
a0 +b0
2
, d0 = b0 − a0
If f (x0) = 0, we’ve done. If not, there are 2 cases: if
f (a0)f (x0) < 0, then let a1 = a0,b1 = x0, otherwise, set
a1 = x0,b1 = b0 and x1 = (a1 +b1)/2

Set a0 = a,b0 = b,x0 =
a0 +b0
2
, d0 = b0 − a0
If f (x0) = 0, we’ve done. If not, there are 2 cases: if
f (a0)f (x0) < 0, then let a1 = a0,b1 = x0, otherwise, set
a1 = x0,b1 = b0 and x1 = (a1 +b1)/2
reapply the process on [a1,b1],[a2,b2],.....,[an,bn]
At the nth
step, we get:





an É x É bn, an É xn =
an +bn
2
É bn
f (an).f (bn) < 0, dn = bn − an =
b − a
2n
And xn is the approximate solution of equation.

THE ERROR OF ROOT
THEOREM 2.1
Let a function f (x) be continuous on [a,b] and f (a)f (b) < 0. Then the
bisection method generates a sequence {xn} that converges to the exact
solution x of the equation f (x) = 0 with the error:
|xn − x| =
¯
¯
¯
¯
an +bn
2
− x
¯
¯
¯
¯ É
1
2
(bn − an) =
b − a
2n+1
.

EXAMPLE 2.1
Let the equation f (x) = 5x3
−cos3x = 0 have a root on [0,1]. Using the
bisection method to find x4 and estimate its error by two formulas.

EXAMPLE 2.1
Let the equation f (x) = 5x3
−cos3x = 0 have a root on [0,1]. Using the
bisection method to find x4 and estimate its error by two formulas.
x4 = 0.40625; ∆x4 = 0.0313.

EXAMPLE 2.2
Let the equation f (x) =
p
x −cosx = 0 have a root on [0,1]. Using the
bisection method to find x4 and find its error.

EXAMPLE 2.2
Let the equation f (x) =
p
x −cosx = 0 have a root on [0,1]. Using the
bisection method to find x4 and find its error.
x4 = 0.56625;∆x4 = 0.0313.

EXAMPLE 2.3
Let the equation f (x) = x −tanx = 0 have a root on [4,4.5]. Using
bisection method to find approximate solution which has error less
than 10−2
.

EXAMPLE 2.3
than 10−2
.
Find the least value n ∈ N such that ∆xn =
4.5−4
2n+1
< 10−2
.

EXAMPLE 2.3
than 10−2
.
4.5−4
2n+1
< 10−2
.
n = 5.

EXAMPLE 2.3
than 10−2
.
4.5−4
2n+1
< 10−2
.
n = 5.
x5 = 4.4921875.

FIXED POINT ITERATION METHOD
FIXED POINT METHOD
DEFINITION 3.1
The point x is called fixed point of function g if g(x) = x.

FIXED POINT METHOD
DEFINITION 3.1
From a given equation f (x) = 0, we always can define the equivalent
equation x = g(x) in many ways.

FIXED POINT METHOD
DEFINITION 3.1
EXAMPLE 3.1
Given the equation x3
− x −1 = 0 , we can write:
x = x3
−1

FIXED POINT METHOD
DEFINITION 3.1
EXAMPLE 3.1
x = x3
−1
x =
3
p
1+ x

FIXED POINT METHOD
DEFINITION 3.1
EXAMPLE 3.1
x = x3
−1
x =
3
p
1+ x
x =
1
x
+
1
x2

THEOREM 3.1
If g(x) is differentiable on [a,b], moreover
max
[a,b]
|g′
(x)| < 1
then g has exactly one fixed point on [a,b]. Function g is called a
contraction mapping.

THEOREM 3.1
If g(x) is differentiable on [a,b], moreover
max
[a,b]
|g′
(x)| < 1
then g has exactly one fixed point on [a,b]. Function g is called a
contraction mapping.
EXAMPLE 3.2
Show that g(x) = (x2
−1)/3 has a unique fixed point on the interval
[−1,1].

FIXED-POINT ITERATION
Given the equation x = g(x).
Choose an initial value x0 ∈ [a,b].
Construct the sequence by xn = g(xn−1).

FIXED-POINT ITERATION
Given the equation x = g(x).
Choose an initial value x0 ∈ [a,b].
Construct the sequence by xn = g(xn−1).
THEOREM 3.2
Condition for convergence Function g is differentiable on [a,b] and
q = |g′
(x)| < 1,∀x ∈ [a,b], then for any x0 ∈ [a,b], the sequence defined
by xn = g(xn−1) converges to the unique fixed point x of g(x) in [a,b].

CONDITION FOR CONVERGENCE
Proof
THEOREM 3.3 (MEAN VALUE THEOREM)
If the function f (x) is differentiable on (a,b), then there exists a point
c in (a,b) such that:
f (b)− f (a) = f ′
(c)(b − a).

CONDITION FOR CONVERGENCE
Proof
THEOREM 3.3 (MEAN VALUE THEOREM)
If the function f (x) is differentiable on (a,b), then there exists a point
c in (a,b) such that:
f (b)− f (a) = f ′
(c)(b − a).
Applying the Mean value theorem
|xn − x| = |g(xn−1)− g(x)| = |g′
(ζ1)||xn−1 − x|
= |g′
(ζ1)||g(xn−2)− g(x)|
= |g′
(ζ1)||g′
(ζ2)||xn−2 − x|
≤ q2
|xn−2 − x|.... ≤ qn
|x0 − x|.
Then lim
n→∞
xn = x. The proof is finished.

ERROR ESTIMATION
Priori error
|xn − x| ≤
qn
1− q
|x1 − x0|.

ERROR ESTIMATION
Priori error
|xn − x| ≤
qn
1− q
|x1 − x0|.
Posteriori error
|xn − x| ≤
q
1− q
|xn − xn−1|.

EXAMPLE 3.3
Let the equation x =
5x3
+3
20
= g(x) have a root on (0,1). Find
approximate solution x3 by the fixed point method and its error by
two formulas, x0 is the midpoint on (0,1).
SOLUTION

EXAMPLE 3.3
5x3
+3
20
SOLUTION
1 x0 = 0.5 → x1 = 29/160 → x2 = 0.151489(A) → x3 = 0.150869(B)

EXAMPLE 3.3
5x3
+3
20
SOLUTION
1 x0 = 0.5 → x1 = 29/160 → x2 = 0.151489(A) → x3 = 0.150869(B)
2 q = max
x∈[0,1]
¯
¯3
4
x2
¯
¯ = 3
4
(M)

EXAMPLE 3.3
5x3
+3
20
SOLUTION
1 x0 = 0.5 → x1 = 29/160 → x2 = 0.151489(A) → x3 = 0.150869(B)
2 q = max
x∈[0,1]
¯
¯3
4
x2
¯
¯ = 3
4
(M)
3 Priori error: ∆x3 =
q3
1−q |x1 − x0| ≈ 0.5379

EXAMPLE 3.3
5x3
+3
20
SOLUTION
1 x0 = 0.5 → x1 = 29/160 → x2 = 0.151489(A) → x3 = 0.150869(B)
2 q = max
x∈[0,1]
¯
¯3
4
x2
¯
¯ = 3
4
(M)
3 Priori error: ∆x3 =
q3
1−q |x1 − x0| ≈ 0.5379
4 Posterrior error: ∆x3 =
q
1−q
|x3 − x2| =
q
1−q
|B − A| ≈ 0.0019

EXAMPLE 3.4
Using the fixed-point iteration method, find approximate solution x3
of the equation x = 3+
5
x2
= g(x) on [3,4] with x0 = 3.5 and its priori
error .

EXAMPLE 3.4
Using the fixed-point iteration method, find approximate solution x3
of the equation x = 3+
5
x2
= g(x) on [3,4] with x0 = 3.5 and its priori
error .
ANS
x3 = 3.4248799, ∆x3 = 0.0074103

EXAMPLE 3.5
Using the fixed point method, find an approximate root x3 for
equation x =
3
p
x +1 = g(x) on [1,2] with x0 = 1.5 and compute its
posteriori error. Find the least number of steps of calculation to get the
solution whose priori error is less than 10−3
.

EXAMPLE 3.5
Using the fixed point method, find an approximate root x3 for
equation x =
3
p
x +1 = g(x) on [1,2] with x0 = 1.5 and compute its
posteriori error. Find the least number of steps of calculation to get the
solution whose priori error is less than 10−3
.
Priori error ∆xn ≈
qn
1−q |x1 − x0| < 10−3
⇒ qn
<
10−3
(1− q)
|x1 − x0|
⇒
n > logq
10−3
(1− q)
|x1 − x0|
.

EXAMPLE 3.6
+2x2
− x −3 = 0. Transform it equivalently to
x = g(x) in the two cases below
x =
4
p
3+ x −2x2
x =
3x4
+2x2
+3
4x3 +4x −1
.
With the initial value x0 = 1, which case gives us the better
approximate solution?

EXAMPLE 3.6
+2x2
− x −3 = 0. Transform it equivalently to
x = g(x) in the two cases below
x =
4
p
3+ x −2x2
x =
3x4
+2x2
+3
4x3 +4x −1
.
With the initial value x0 = 1, which case gives us the better
approximate solution?
n xn g1(xn) g2(xn)
1 x1 1.1892 1.1429
2 x2 1.0801 1.1245
3 x3 1.1497 1.1241
4 x4 1.1078 1.1241

NEWTON’S METHOD
GEOMETRIC IDEA

NEWTON’S METHOD
CONSTRUCT THE FORMULA
Suppose that equation f (x) = 0 has a root on [a,b].
Guess an initial value x0

NEWTON’S METHOD
The tangent line to f (x) at x0 is y = f (x0)+ f ′
(x0)(x − x0).

NEWTON’S METHOD
(x0)(x − x0).
The intersection point of tangent line with x-axis is root of
f (x0)+ f ′
(x0)(x − x0) = 0

NEWTON’S METHOD
(x0)(x − x0).
The intersection point of tangent line with x-axis is root of
f (x0)+ f ′
(x0)(x − x0) = 0
⇒ x = x0 −
f (x0)
f ′(x0)
, then x1 = x0 −
f (x0)
f ′(x0)
.
Repeat the process to find x2,x3,..,xn.
Generate the sequence xn = xn−1 −
f (xn−1)
f ′(xn−1)
.

NEWTON’S METHOD
EXAMPLE 4.1
Given the equation cosx − x = 0, using Newton’s method to find an
approximate root on [0,π/2] after 3 iterations with the initial guess
x0 = 0.

NEWTON’S METHOD
EXAMPLE 4.1
x0 = 0.
SOLUTION
f = cosx − x, f ′
= −sinx −1, set function
g(x) = x −
f (x)
f ′(x)
= x −
cosx − x
−sinx −1

NEWTON’S METHOD
EXAMPLE 4.1
x0 = 0.
SOLUTION
g(x) = x −
f (x)
f ′(x)
= x −
cosx − x
−sinx −1
First iteration: x1 = g(x0) = 1

NEWTON’S METHOD
EXAMPLE 4.1
x0 = 0.
SOLUTION
g(x) = x −
f (x)
f ′(x)
= x −
cosx − x
−sinx −1
Second iteration: x2 = g(x1) = 0.750363867840244

NEWTON’S METHOD
EXAMPLE 4.1
x0 = 0.
SOLUTION
g(x) = x −
f (x)
f ′(x)
= x −
cosx − x
−sinx −1
Second iteration: x2 = g(x1) = 0.750363867840244
Third iteration: x3 = g(x2) = 0.739112890911362

NEWTON’S METHOD

NEWTON’S METHOD
• Exactly root : A = 0.739085133
• B = x2 = 0.750363867840244
• C = x3 = 0.739112890911362

NEWTON’S METHOD
CONVERGENCE
THEOREM 4.1
The sequence xn generated by Newton’s method converges to exact root
x of function f (x) = 0 if the following conditions satisfy:
f ′
(x) ̸= 0,∀x ∈ [a,b].

NEWTON’S METHOD
CONVERGENCE
THEOREM 4.1
f ′
(x) ̸= 0,∀x ∈ [a,b].
f ′′
(x) is continuous on [a,b].

NEWTON’S METHOD
CONVERGENCE
THEOREM 4.1
f ′
(x) ̸= 0,∀x ∈ [a,b].
f ′′
(x) is continuous on [a,b].
x0(initial guest) is sufficiently close to the exact root x.

NEWTON’S METHOD
EXAMPLE 4.2
Given the equation f (x) = x3
−2x2
−5 on [−1,4], construct the
sequence by Newton’s method with x0 = −1 after 5 iterations.

NEWTON’S METHOD
EXAMPLE 4.2
−2x2
Exact root x = 2.690647448028614.
x1 = 0.142857142857143
x2 = −9.731428571428571
x3 = −6.276696653150196
x4 = −2.398586261805698
x5 = −1.270038382108753

NEWTON’S METHOD
EXAMPLE 4.2
−2x2
Exact root x = 2.690647448028614.
x1 = 0.142857142857143
x2 = −9.731428571428571
x3 = −6.276696653150196
x4 = −2.398586261805698
x5 = −1.270038382108753
⇒ Method failed

NEWTON’S METHOD
INITIAL GUESS
Consider the equation f (x) = 0. Let f (x) such that f ′
(x) > 0 or
f ′
(x) < 0 on [a,b], f (a)f (b) < 0. In addition, f ′′
(x) > 0 or f ′′
(x) < 0 on
[a,b]. We will choose x0 as below:
Taking any point p in (a,b)

NEWTON’S METHOD
INITIAL GUESS
(x) > 0 or
f ′
(x) > 0 or f ′′
(x) < 0 on
If f ′
(p)f ′′
(p) < 0, then choose x0 = a.

NEWTON’S METHOD
INITIAL GUESS
(x) > 0 or
f ′
(x) > 0 or f ′′
(x) < 0 on
If f ′
(p)f ′′
If f ′
(p)f ′′
(p) > 0, then choose x0 = b.
or

NEWTON’S METHOD
INITIAL GUESS
(x) > 0 or
f ′
(x) > 0 or f ′′
(x) < 0 on
If f ′
(p)f ′′
If f ′
(p)f ′′
or
If f (a)f ′′
(a) > 0 then choose x0 = a.

NEWTON’S METHOD
INITIAL GUESS
(x) > 0 or
f ′
(x) > 0 or f ′′
(x) < 0 on
If f ′
(p)f ′′
If f ′
(p)f ′′
or
If f (a)f ′′
(a) > 0 then choose x0 = a.
If f (b)f ′′
(b) > 0 then choose x0 = b.

NEWTON’S METHOD
ERROR
ERROR
We estimate error of the approximate root xn in Newton’s method by
using the general formula :
|xn − x| É ∆xn :=
|f (xn)|
m

NEWTON’S METHOD
EXAMPLE 4.3
Let the equation f (x) = x3
−3x +1 = 0 have a root on [0,0.5]. Using
the Newton method, find x3 and its error.

NEWTON’S METHOD
EXAMPLE 4.3
SOLUTION
f ′
= 3x2
−3, f ′′
(x) = 6x

NEWTON’S METHOD
EXAMPLE 4.3
SOLUTION
f ′
= 3x2
−3, f ′′
(x) = 6x
Taking p = 0.1 ∈ [0,0.5], check the sign of f ′
(x)f ′′
(x) at p,

NEWTON’S METHOD
EXAMPLE 4.3
SOLUTION
f ′
= 3x2
−3, f ′′
(x) = 6x
(x)f ′′
(x) at p,
f ′
(0,1)f ′′
(0.1) < 0, then choose x0 = 0.

NEWTON’S METHOD
EXAMPLE 4.3
SOLUTION
f ′
= 3x2
−3, f ′′
(x) = 6x
(x)f ′′
(x) at p,
f ′
(0,1)f ′′
(0.1) < 0, then choose x0 = 0.
Consider g(x) = x −
f
f ′
= x −
x3
−3x +1
3x2 −3
,

NEWTON’S METHOD
EXAMPLE 4.3
SOLUTION
f ′
= 3x2
−3, f ′′
(x) = 6x
(x)f ′′
(x) at p,
f ′
(0,1)f ′′
(0.1) < 0, then choose x0 = 0.
Consider g(x) = x −
f
f ′
= x −
x3
−3x +1
3x2 −3
, xn = g(xn−1).

NEWTON’S METHOD
x1 0.333333
x2 0.347222
x3 0.347296

NEWTON’S METHOD
x1 0.333333
x2 0.347222
x3 0.347296
Estimate of error
∆x3 ≤
|f (x3)|
m
≈ 2.5443·10−9
,
where
m = min
x∈[0,0.5]
|f ′
(x)| = |f ′
(0.5)| = 9/4

NEWTON’S METHOD
EXAMPLE 4.4
Given the equation sinx −e−x
= 0, using Newton’s method to
find approximate root x3 on [0,1] and estimate its error.
find approximate root xn such that ∆xn < 10−6
.

NEWTON’S METHOD
EXAMPLE 4.4
.
SOLUTION
f ′
= cosx +e−x
, f ′′
= −sinx −e−x

NEWTON’S METHOD
EXAMPLE 4.4
.
SOLUTION
f ′
= cosx +e−x
, f ′′
= −sinx −e−x
since f (0)f ′′
(0) > 0 so we choose x0 = 0

NEWTON’S METHOD
x1 0.5
x2 0.585644
x3 0.588529

NEWTON’S METHOD
x1 0.5
x2 0.585644
x3 0.588529
Estimate of error
∆x3 ≤
|f (x3)|
m
≈ 6·10−6
,
where m = min
x∈[0,1]
|f ′
(x)| = |f ′
(1)| =
cos1+e−1
.

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