• 沒有找到結果。

# Solutions of Equations in One Variable

N/A
N/A
Protected

Academic year: 2022

Share "Solutions of Equations in One Variable"

Copied!
238
0
0

(1)

Tsung-Ming Huang

Department of Mathematics National Taiwan Normal University, Taiwan

October 13, 2014

(2)

### Outline

1 Bisection Method

2 Fixed-Point Iteration

3 Newton’s method

4 Error analysis for iterative methods

5 Accelerating convergence

6 Zeros of polynomials and M ¨uller’s method

(3)

### Outline

1 Bisection Method

2 Fixed-Point Iteration

3 Newton’s method

4 Error analysis for iterative methods

5 Accelerating convergence

6 Zeros of polynomials and M ¨uller’s method

(4)

### Outline

1 Bisection Method

2 Fixed-Point Iteration

3 Newton’s method

4 Error analysis for iterative methods

5 Accelerating convergence

6 Zeros of polynomials and M ¨uller’s method

(5)

### Outline

1 Bisection Method

2 Fixed-Point Iteration

3 Newton’s method

4 Error analysis for iterative methods

5 Accelerating convergence

6 Zeros of polynomials and M ¨uller’s method

(6)

### Outline

1 Bisection Method

2 Fixed-Point Iteration

3 Newton’s method

4 Error analysis for iterative methods

5 Accelerating convergence

6 Zeros of polynomials and M ¨uller’s method

(7)

### Outline

1 Bisection Method

2 Fixed-Point Iteration

3 Newton’s method

4 Error analysis for iterative methods

5 Accelerating convergence

6 Zeros of polynomials and M ¨uller’s method

(8)

### Bisection Method

Idea

Iff (x) ∈ C[a, b]andf (a)f (b) < 0, then∃ c ∈ (a, b)such that f (c) = 0.

(9)

Bisection method algorithm

Given f (x) defined on (a, b), the maximal number of iterations M, and stop criteria δ and ε, this algorithm tries to locate one root of f (x).

Compute u = f (a), v = f (b), and e = b − a If sign(u) = sign(v), then stop

For k = 1, 2, . . . , M

e = e/2, c = a + e, w = f (c) If |e| < δ or |w| < ε, then stop If sign(w) 6= sign(u)

b = c, v = w Else

a = c, u = w End If

End For

(10)

Let {cn} be the sequence of numbers produced. The algorithm should stop if one of the following conditions is satisfied.

1 the iteration number k > M ,

2 |ck− ck−1| < δ, or

3 |f (ck)| < ε.

Let [a0, b0], [a1, b1], . . .denote the successive intervals produced by the bisection algorithm. Then

a = a0≤ a1 ≤ a2 ≤ · · · ≤ b0= b

⇒ {an} and {bn} are bounded

⇒ lim

n→∞an and lim

n→∞bn exist

(11)

Let {cn} be the sequence of numbers produced. The algorithm should stop if one of the following conditions is satisfied.

1 the iteration number k > M ,

2 |ck− ck−1| < δ, or

3 |f (ck)| < ε.

Let [a0, b0], [a1, b1], . . .denote the successive intervals produced by the bisection algorithm.Then

a = a0≤ a1 ≤ a2 ≤ · · · ≤ b0= b

⇒ {an} and {bn} are bounded

⇒ lim

n→∞an and lim

n→∞bn exist

(12)

Let {cn} be the sequence of numbers produced. The algorithm should stop if one of the following conditions is satisfied.

1 the iteration number k > M ,

2 |ck− ck−1| < δ, or

3 |f (ck)| < ε.

Let [a0, b0], [a1, b1], . . .denote the successive intervals produced by the bisection algorithm. Then

a = a0≤ a1 ≤ a2 ≤ · · · ≤ b0= b

⇒ {an} and {bn} are bounded

⇒ lim

n→∞an and lim

n→∞bn exist

(13)

Since

b1− a1 = 1

2(b0− a0) b2− a2 = 1

2(b1− a1) = 1

4(b0− a0) ...

bn− an = 1

2n(b0− a0) hence

n→∞lim bn− lim

n→∞an= lim

n→∞(bn− an) = lim

n→∞

1

2n(b0− a0) = 0.

Therefore

n→∞lim an= lim

n→∞bn≡ z.

Since f is a continuous function, we have that

n→∞lim f (an) = f ( lim

n→∞an) = f (z) and lim

n→∞f (bn) = f ( lim

n→∞bn) = f (z).

(14)

Since

b1− a1 = 1

2(b0− a0) b2− a2 = 1

2(b1− a1) = 1

4(b0− a0) ...

bn− an = 1

2n(b0− a0) hence

n→∞lim bn− lim

n→∞an= lim

n→∞(bn− an) = lim

n→∞

1

2n(b0− a0) = 0.

Therefore

n→∞lim an= lim

n→∞bn≡ z.

Since f is a continuous function, we have that

n→∞lim f (an) = f ( lim

n→∞an) = f (z) and lim

n→∞f (bn) = f ( lim

n→∞bn) = f (z).

(15)

Since

b1− a1 = 1

2(b0− a0) b2− a2 = 1

2(b1− a1) = 1

4(b0− a0) ...

bn− an = 1

2n(b0− a0) hence

n→∞lim bn− lim

n→∞an= lim

n→∞(bn− an) = lim

n→∞

1

2n(b0− a0) = 0.

Therefore

n→∞lim an= lim

n→∞bn≡ z.

Since f is a continuous function, we have that

n→∞lim f (an) = f ( lim

n→∞an) = f (z) and lim

n→∞f (bn) = f ( lim

n→∞bn) = f (z).

(16)

Since

b1− a1 = 1

2(b0− a0) b2− a2 = 1

2(b1− a1) = 1

4(b0− a0) ...

bn− an = 1

2n(b0− a0) hence

n→∞lim bn− lim

n→∞an= lim

n→∞(bn− an) = lim

n→∞

1

2n(b0− a0) = 0.

Therefore

n→∞lim an= lim

n→∞bn≡ z.

Since f is a continuous function, we have that

n→∞lim f (an) = f ( lim

n→∞an) = f (z) and lim

n→∞f (bn) = f ( lim

n→∞bn) = f (z).

(17)

On the other hand,

f (an)f (bn) < 0

⇒ lim

n→∞f (an)f (bn) = f2(z) ≤ 0

⇒ f (z) = 0

Therefore, the limit of the sequences {an} and {bn} is a zero of f in [a, b]. Let cn= 12(an+ bn).Then

|z − cn| = lim

n→∞an−1

2(an+ bn)

= 1 2

h

n→∞lim an− bni + 1

2 h

n→∞lim an− ani

≤ max lim

n→∞an− bn ,

lim

n→∞an− an

≤ |bn− an| = 1

2n|b0− a0|.

This proves the following theorem.

(18)

On the other hand,

f (an)f (bn) < 0

⇒ lim

n→∞f (an)f (bn) = f2(z) ≤ 0

⇒ f (z) = 0

Therefore, the limit of the sequences {an} and {bn} is a zero of f in [a, b].Let cn= 12(an+ bn). Then

|z − cn| = lim

n→∞an−1

2(an+ bn)

= 1 2

h

n→∞lim an− bni + 1

2 h

n→∞lim an− ani

≤ max lim

n→∞an− bn ,

lim

n→∞an− an

≤ |bn− an| = 1

2n|b0− a0|.

This proves the following theorem.

(19)

On the other hand,

f (an)f (bn) < 0

⇒ lim

n→∞f (an)f (bn) = f2(z) ≤ 0

⇒ f (z) = 0

Therefore, the limit of the sequences {an} and {bn} is a zero of f in [a, b]. Let cn= 12(an+ bn).Then

|z − cn| = lim

n→∞an−1

2(an+ bn)

= 1 2

h

n→∞lim an− bni + 1

2 h

n→∞lim an− ani

≤ max lim

n→∞an− bn ,

lim

n→∞an− an

≤ |bn− an| = 1

2n|b0− a0|.

This proves the following theorem.

(20)

On the other hand,

f (an)f (bn) < 0

⇒ lim

n→∞f (an)f (bn) = f2(z) ≤ 0

⇒ f (z) = 0

Therefore, the limit of the sequences {an} and {bn} is a zero of f in [a, b]. Let cn= 12(an+ bn). Then

|z − cn| = lim

n→∞an−1

2(an+ bn)

= 1 2

h

n→∞lim an− bni + 1

2 h

n→∞lim an− ani

≤ max lim

n→∞an− bn ,

lim

n→∞an− an

≤ |bn− an| = 1

2n|b0− a0|.

This proves the following theorem.

(21)

Theorem 1

Let{[an, bn]}denote the intervals produced by the bisection algorithm. Then lim

n→∞anand lim

n→∞bnexist, areequal, and represent azerooff (x). If

z = lim

n→∞an= lim

n→∞bn and cn= 1

2(an+ bn), then

|z − cn| ≤ 1

2n(b0− a0) . Remark

{cn}convergestozwith therateofO(2−n).

(22)

Theorem 1

Let{[an, bn]}denote the intervals produced by the bisection algorithm. Then lim

n→∞anand lim

n→∞bnexist, areequal, and represent azerooff (x). If

z = lim

n→∞an= lim

n→∞bn and cn= 1

2(an+ bn), then

|z − cn| ≤ 1

2n(b0− a0) . Remark

{cn}convergestozwith therateofO(2−n).

(23)

Example 2

How many steps should be taken to compute a root of f (x) = x3+ 4x2− 10 = 0 on [1, 2] with relative error 10−3? solution: Seek an n such that

|z − cn|

|z| ≤ 10−3 ⇒ |z − cn| ≤ |z| × 10−3. Since z ∈ [1, 2], it is sufficient to show

|z − cn| ≤ 10−3. That is, we solve

2−n(2 − 1) ≤ 10−3 ⇒ −n log102 ≤ −3 which gives n ≥ 10.

(24)

Example 2

How many steps should be taken to compute a root of f (x) = x3+ 4x2− 10 = 0 on [1, 2] with relative error 10−3? solution: Seek an n such that

|z − cn|

|z| ≤ 10−3 ⇒ |z − cn| ≤ |z| × 10−3. Since z ∈ [1, 2], it is sufficient to show

|z − cn| ≤ 10−3. That is, we solve

2−n(2 − 1) ≤ 10−3 ⇒ −n log102 ≤ −3 which gives n ≥ 10.

(25)

Exercise

Page 54: 1, 13, 14, 16, 17

(26)

### Fixed-Point Iteration

Definition 3

xis called afixed pointof a given functiongifg(x) = x.

Root-finding problems and fixed-point problems Find x such that f (x) = 0.

Let g(x) = x − f (x). Then g(x) = x− f (x) = x.

⇒ xis a fixed point for g(x).

Find x such that g(x) = x. Define f (x) = x − g(x) so that f (x) = x− g(x) = x− x= 0

⇒ xis a zero of f (x).

(27)

### Fixed-Point Iteration

Definition 3

xis called afixed pointof a given functiongifg(x) = x.

Root-finding problems and fixed-point problems Find x such that f (x) = 0.

Let g(x) = x − f (x). Then g(x) = x− f (x) = x.

⇒ xis a fixed point for g(x).

Find x such that g(x) = x. Define f (x) = x − g(x) so that f (x) = x− g(x) = x− x= 0

⇒ xis a zero of f (x).

(28)

### Fixed-Point Iteration

Definition 3

xis called afixed pointof a given functiongifg(x) = x.

Root-finding problems and fixed-point problems Find x such that f (x) = 0.

Let g(x) = x − f (x). Then g(x) = x− f (x) = x.

⇒ xis a fixed point for g(x).

Find x such that g(x) = x. Define f (x) = x − g(x) so that f (x) = x− g(x) = x− x= 0

⇒ xis a zero of f (x).

(29)

Example 4

The function g(x) = x2− 2, for −2 ≤ x ≤ 3, has fixed points at x = −1and x = 2 since

0 = g(x) − x = x2− x − 2 = (x + 1)(x − 2).

(30)

Theorem 5 (Existence and uniqueness)

1 If g ∈ C[a, b] such thata ≤ g(x) ≤ bfor all x ∈ [a, b], then g hasa fixed point in [a, b].

2 If, in addition, g0(x)exists in (a, b) and there exists a positive constantM < 1such that|g0(x)| ≤ M < 1for all x ∈ (a, b). Then the fixed point isunique.

(31)

Theorem 5 (Existence and uniqueness)

1 If g ∈ C[a, b] such thata ≤ g(x) ≤ bfor all x ∈ [a, b], then g hasa fixed point in [a, b].

2 If, in addition, g0(x)exists in (a, b) and there exists a positive constantM < 1such that|g0(x)| ≤ M < 1for all x ∈ (a, b). Then the fixed point isunique.

(32)

### Proof

Existence:

If g(a) = a or g(b) = b, then a or b is a fixed point of g and we are done.

Otherwise, it must be g(a) > a and g(b) < b. The function h(x) = g(x) − xis continuous on [a, b], with

h(a) = g(a) − a > 0 and h(b) = g(b) − b < 0.

By the Intermediate Value Theorem, ∃ x ∈ [a, b] such that h(x) = 0. That is

g(x) − x = 0 ⇒ g(x) = x. Hence g has a fixed point xin [a, b].

(33)

### Proof

Existence:

If g(a) = a or g(b) = b, then a or b is a fixed point of g and we are done.

Otherwise, it must be g(a) > a and g(b) < b. The function h(x) = g(x) − xis continuous on [a, b], with

h(a) = g(a) − a > 0 and h(b) = g(b) − b < 0.

By the Intermediate Value Theorem, ∃ x ∈ [a, b] such that h(x) = 0.That is

g(x) − x = 0 ⇒ g(x) = x. Hence g has a fixed point xin [a, b].

(34)

### Proof

Existence:

If g(a) = a or g(b) = b, then a or b is a fixed point of g and we are done.

Otherwise, it must be g(a) > a and g(b) < b. The function h(x) = g(x) − xis continuous on [a, b], with

h(a) = g(a) − a > 0 and h(b) = g(b) − b < 0.

By the Intermediate Value Theorem, ∃ x ∈ [a, b] such that h(x) = 0. That is

g(x) − x = 0 ⇒ g(x) = x. Hence g has a fixed point xin [a, b].

(35)

### Proof

Existence:

If g(a) = a or g(b) = b, then a or b is a fixed point of g and we are done.

Otherwise, it must be g(a) > a and g(b) < b. The function h(x) = g(x) − xis continuous on [a, b], with

h(a) = g(a) − a > 0 and h(b) = g(b) − b < 0.

By the Intermediate Value Theorem, ∃ x ∈ [a, b] such that h(x) = 0.That is

g(x) − x = 0 ⇒ g(x) = x. Hence g has a fixed point xin [a, b].

(36)

### Proof

Existence:

If g(a) = a or g(b) = b, then a or b is a fixed point of g and we are done.

Otherwise, it must be g(a) > a and g(b) < b. The function h(x) = g(x) − xis continuous on [a, b], with

h(a) = g(a) − a > 0 and h(b) = g(b) − b < 0.

By the Intermediate Value Theorem, ∃ x ∈ [a, b] such that h(x) = 0. That is

g(x) − x = 0 ⇒ g(x) = x. Hence g has a fixed point xin [a, b].

(37)

### Proof

Uniqueness:

Suppose that p 6= q are both fixed points of g in [a, b].By the Mean-Value theorem, there exists ξ between p and q such that

g0(ξ) = g(p) − g(q)

p − q = p − q p − q = 1.

However, this contradicts to the assumption that

|g0(x)| ≤ M < 1for all x in [a, b]. Therefore the fixed point of g is unique.

(38)

### Proof

Uniqueness:

Suppose that p 6= q are both fixed points of g in [a, b]. By the Mean-Value theorem, there exists ξ between p and q such that

g0(ξ) = g(p) − g(q)

p − q = p − q p − q = 1.

However, this contradicts to the assumption that

|g0(x)| ≤ M < 1for all x in [a, b]. Therefore the fixed point of g is unique.

(39)

### Proof

Uniqueness:

Suppose that p 6= q are both fixed points of g in [a, b]. By the Mean-Value theorem, there exists ξ between p and q such that

g0(ξ) = g(p) − g(q)

p − q = p − q p − q = 1.

However, this contradicts to the assumption that

|g0(x)| ≤ M < 1for all x in [a, b]. Therefore the fixed point of g is unique.

(40)

Example 6

Show that the following function has a unique fixed point.

g(x) = (x2− 1)/3, x ∈ [−1, 1].

Solution: The Extreme Value Theorem implies that min

x∈[−1,1]g(x) = g(0) = −1 3, max

x∈[−1,1]g(x) = g(±1) = 0.

That is g(x) ∈ [−1, 1], ∀ x ∈ [−1, 1].

Moreover, g is continuous and

|g0(x)| =

2x 3

≤ 2

3, ∀ x ∈ (−1, 1).

By above theorem, g has a unique fixed point in [−1, 1].

(41)

Example 6

Show that the following function has a unique fixed point.

g(x) = (x2− 1)/3, x ∈ [−1, 1].

Solution: The Extreme Value Theorem implies that min

x∈[−1,1]g(x) = g(0) = −1 3, max

x∈[−1,1]g(x) = g(±1) = 0.

That is g(x) ∈ [−1, 1], ∀ x ∈ [−1, 1].

Moreover, g is continuous and

|g0(x)| =

2x 3

≤ 2

3, ∀ x ∈ (−1, 1).

By above theorem, g has a unique fixed point in [−1, 1].

(42)

Example 6

Show that the following function has a unique fixed point.

g(x) = (x2− 1)/3, x ∈ [−1, 1].

Solution: The Extreme Value Theorem implies that min

x∈[−1,1]g(x) = g(0) = −1 3, max

x∈[−1,1]g(x) = g(±1) = 0.

That is g(x) ∈ [−1, 1], ∀ x ∈ [−1, 1].

Moreover, g is continuous and

|g0(x)| =

2x 3

≤ 2

3, ∀ x ∈ (−1, 1).

By above theorem, g has a unique fixed point in [−1, 1].

(43)

Example 6

Show that the following function has a unique fixed point.

g(x) = (x2− 1)/3, x ∈ [−1, 1].

Solution: The Extreme Value Theorem implies that min

x∈[−1,1]g(x) = g(0) = −1 3, max

x∈[−1,1]g(x) = g(±1) = 0.

That is g(x) ∈ [−1, 1], ∀ x ∈ [−1, 1].

Moreover, g is continuous and

|g0(x)| =

2x 3

≤ 2

3, ∀ x ∈ (−1, 1).

By above theorem, g has a unique fixed point in [−1, 1].

(44)

Example 6

Show that the following function has a unique fixed point.

g(x) = (x2− 1)/3, x ∈ [−1, 1].

Solution: The Extreme Value Theorem implies that min

x∈[−1,1]g(x) = g(0) = −1 3, max

x∈[−1,1]g(x) = g(±1) = 0.

That is g(x) ∈ [−1, 1], ∀ x ∈ [−1, 1].

Moreover, g is continuous and

|g0(x)| =

2x 3

≤ 2

3, ∀ x ∈ (−1, 1).

By above theorem, g has a unique fixed point in [−1, 1].

(45)

Let p be such unique fixed point of g. Then p = g(p) = p2− 1

3 ⇒ p2− 3p − 1 = 0

⇒ p = 1 2(3 −√

13).

(46)

Fixed-point iteration or functional iteration

Given a continuous function g, choose an initial point x0 and generate {xk}k=0by

xk+1 = g(xk), k ≥ 0.

{xk} may not converge, e.g., g(x) = 3x. However, when the sequence converges, say,

k→∞lim xk= x, then, since g is continuous,

g(x) = g( lim

k→∞xk) = lim

k→∞g(xk) = lim

k→∞xk+1= x. That is, xis a fixed point of g.

(47)

Fixed-point iteration or functional iteration

Given a continuous function g, choose an initial point x0 and generate {xk}k=0by

xk+1 = g(xk), k ≥ 0.

{xk} may not converge, e.g., g(x) = 3x.However, when the sequence converges, say,

k→∞lim xk= x, then, since g is continuous,

g(x) = g( lim

k→∞xk) = lim

k→∞g(xk) = lim

k→∞xk+1= x. That is, xis a fixed point of g.

(48)

Fixed-point iteration or functional iteration

Given a continuous function g, choose an initial point x0 and generate {xk}k=0by

xk+1 = g(xk), k ≥ 0.

{xk} may not converge, e.g., g(x) = 3x. However, when the sequence converges, say,

k→∞lim xk= x, then, since g is continuous,

g(x) = g( lim

k→∞xk) = lim

k→∞g(xk) = lim

k→∞xk+1= x. That is, xis a fixed point of g.

(49)

Fixed-point iteration or functional iteration

Given a continuous function g, choose an initial point x0 and generate {xk}k=0by

xk+1 = g(xk), k ≥ 0.

{xk} may not converge, e.g., g(x) = 3x. However, when the sequence converges, say,

k→∞lim xk= x, then, since g is continuous,

g(x) = g( lim

k→∞xk) = lim

k→∞g(xk) = lim

k→∞xk+1= x. That is, xis a fixed point of g.

(50)

Fixed-point iteration or functional iteration

Given a continuous function g, choose an initial point x0 and generate {xk}k=0by

xk+1 = g(xk), k ≥ 0.

{xk} may not converge, e.g., g(x) = 3x. However, when the sequence converges, say,

k→∞lim xk= x, then, since g is continuous,

g(x) = g( lim

k→∞xk) = lim

k→∞g(xk) = lim

k→∞xk+1= x. That is, xis a fixed point of g.

(51)

Fixed-point iteration

Given x0, tolerance T OL, maximum number of iteration M . Set i = 1 and x = g(x0).

While i ≤ M and |x − x0| ≥ T OL Set i = i + 1, x0 = xand x = g(x0).

End While

(52)

Example 7 The equation

x3+ 4x2− 10 = 0

has a unique root in [1, 2]. Change the equation to the fixed-point form x = g(x).

(a) x = g1(x) ≡ x − f (x) = x − x3− 4x2+ 10

(b) x = g2(x) = 10x − 4x1/2

x3 = 10 − 4x2 ⇒ x2 = 10

x − 4x ⇒ x = ± 10 x − 4x

1/2

(53)

Example 7 The equation

x3+ 4x2− 10 = 0

has a unique root in [1, 2]. Change the equation to the fixed-point form x = g(x).

(a) x = g1(x) ≡ x − f (x) = x − x3− 4x2+ 10

(b) x = g2(x) = 10x − 4x1/2

x3 = 10 − 4x2 ⇒ x2 = 10

x − 4x ⇒ x = ± 10 x − 4x

1/2

(54)

Example 7 The equation

x3+ 4x2− 10 = 0

has a unique root in [1, 2]. Change the equation to the fixed-point form x = g(x).

(a) x = g1(x) ≡ x − f (x) = x − x3− 4x2+ 10

(b) x = g2(x) = 10x − 4x1/2

x3 = 10 − 4x2 ⇒ x2 = 10

x − 4x ⇒ x = ± 10 x − 4x

1/2

(55)

Example 7 The equation

x3+ 4x2− 10 = 0

has a unique root in [1, 2]. Change the equation to the fixed-point form x = g(x).

(a) x = g1(x) ≡ x − f (x) = x − x3− 4x2+ 10

(b) x = g2(x) = 10x − 4x1/2

x3 = 10 − 4x2 ⇒ x2 = 10

x − 4x ⇒ x = ± 10 x − 4x

1/2

(56)

(c) x = g3(x) = 12 10 − x31/2

4x2 = 10 − x3 ⇒ x = ±1

2 10 − x31/2

(d) x = g4(x) =

10 4+x

1/2

x2(x + 4) = 10 ⇒ x = ±

 10 4 + x

1/2

(e) x = g5(x) = x −x33x+4x2+8x2−10

x = g5(x) ≡ x − f (x) f0(x)

(57)

(c) x = g3(x) = 12 10 − x31/2

4x2 = 10 − x3 ⇒ x = ±1

2 10 − x31/2

(d) x = g4(x) =

10 4+x

1/2

x2(x + 4) = 10 ⇒ x = ±

 10 4 + x

1/2

(e) x = g5(x) = x −x33x+4x2+8x2−10

x = g5(x) ≡ x − f (x) f0(x)

(58)

(c) x = g3(x) = 12 10 − x31/2

4x2 = 10 − x3 ⇒ x = ±1

2 10 − x31/2

(d) x = g4(x) =

10 4+x

1/2

x2(x + 4) = 10 ⇒ x = ±

 10 4 + x

1/2

(e) x = g5(x) = x −x33x+4x2+8x2−10

x = g5(x) ≡ x − f (x) f0(x)

(59)

Results of the fixed-point iteration with initial point x0 = 1.5

(60)

Theorem 8 (Fixed-point Theorem)

Let g ∈ C[a, b] be such thatg(x) ∈ [a, b]for all x ∈ [a, b].

Suppose that g0 exists on (a, b) and that ∃ k with0 < k < 1such that

|g0(x)| ≤ k, ∀ x ∈ (a, b).

Then, for any number x0 in [a, b],

xn= g(xn−1), n ≥ 1, converges to the unique fixed point x in [a, b].

(61)

Theorem 8 (Fixed-point Theorem)

Let g ∈ C[a, b] be such thatg(x) ∈ [a, b]for all x ∈ [a, b].

Suppose that g0 exists on (a, b) and that ∃ k with0 < k < 1such that

|g0(x)| ≤ k, ∀ x ∈ (a, b).

Then, for any number x0 in [a, b],

xn= g(xn−1), n ≥ 1, converges to the unique fixed point x in [a, b].

(62)

Theorem 8 (Fixed-point Theorem)

Let g ∈ C[a, b] be such thatg(x) ∈ [a, b]for all x ∈ [a, b].

Suppose that g0 exists on (a, b) and that ∃ k with0 < k < 1such that

|g0(x)| ≤ k, ∀ x ∈ (a, b).

Then, for any number x0 in [a, b],

xn= g(xn−1), n ≥ 1, converges to the unique fixed point x in [a, b].

(63)

Proof: By the assumptions, a unique fixed point exists in [a, b].

Since g([a, b]) ⊆ [a, b], {xn}n=0 is defined and xn∈ [a, b] for all n ≥ 0. Using the Mean Values Theorem and the fact that

|g0(x)| ≤ k, we have

|x − xn| = |g(xn−1) − g(x)| = |g0n)||x − xn−1| ≤ k|x − xn−1|, where ξn∈ (a, b).It follows that

|xn− x| ≤ k|xn−1− x| ≤ k2|xn−2− x| ≤ · · · ≤ kn|x0− x|. (1) Since 0 < k < 1, we have

n→∞lim kn= 0 and

n→∞lim |xn− x| ≤ lim

n→∞kn|x0− x| = 0.

Hence, {xn} converges to x.

(64)

Proof: By the assumptions, a unique fixed point exists in [a, b].

Since g([a, b]) ⊆ [a, b], {xn}n=0 is defined and xn∈ [a, b] for all n ≥ 0. Using the Mean Values Theorem and the fact that

|g0(x)| ≤ k, we have

|x − xn| = |g(xn−1) − g(x)| = |g0n)||x − xn−1| ≤ k|x − xn−1|, where ξn∈ (a, b). It follows that

|xn− x| ≤ k|xn−1− x| ≤ k2|xn−2− x| ≤ · · · ≤ kn|x0− x|. (1) Since 0 < k < 1, we have

n→∞lim kn= 0 and

n→∞lim |xn− x| ≤ lim

n→∞kn|x0− x| = 0.

Hence, {xn} converges to x.

(65)

Proof: By the assumptions, a unique fixed point exists in [a, b].

Since g([a, b]) ⊆ [a, b], {xn}n=0 is defined and xn∈ [a, b] for all n ≥ 0. Using the Mean Values Theorem and the fact that

|g0(x)| ≤ k, we have

|x − xn| = |g(xn−1) − g(x)| = |g0n)||x − xn−1| ≤ k|x − xn−1|, where ξn∈ (a, b).It follows that

|xn− x| ≤ k|xn−1− x| ≤ k2|xn−2− x| ≤ · · · ≤ kn|x0− x|. (1) Since 0 < k < 1, we have

n→∞lim kn= 0 and

n→∞lim |xn− x| ≤ lim

n→∞kn|x0− x| = 0.

Hence, {xn} converges to x.

(66)

Proof: By the assumptions, a unique fixed point exists in [a, b].

Since g([a, b]) ⊆ [a, b], {xn}n=0 is defined and xn∈ [a, b] for all n ≥ 0. Using the Mean Values Theorem and the fact that

|g0(x)| ≤ k, we have

|x − xn| = |g(xn−1) − g(x)| = |g0n)||x − xn−1| ≤ k|x − xn−1|, where ξn∈ (a, b). It follows that

|xn− x| ≤ k|xn−1− x| ≤ k2|xn−2− x| ≤ · · · ≤ kn|x0− x|. (1) Since 0 < k < 1, we have

n→∞lim kn= 0 and

n→∞lim |xn− x| ≤ lim

n→∞kn|x0− x| = 0.

Hence, {xn} converges to x.

(67)

Proof: By the assumptions, a unique fixed point exists in [a, b].

Since g([a, b]) ⊆ [a, b], {xn}n=0 is defined and xn∈ [a, b] for all n ≥ 0. Using the Mean Values Theorem and the fact that

|g0(x)| ≤ k, we have

|x − xn| = |g(xn−1) − g(x)| = |g0n)||x − xn−1| ≤ k|x − xn−1|, where ξn∈ (a, b). It follows that

|xn− x| ≤ k|xn−1− x| ≤ k2|xn−2− x| ≤ · · · ≤ kn|x0− x|. (1) Since 0 < k < 1, we have

n→∞lim kn= 0 and

n→∞lim |xn− x| ≤ lim

n→∞kn|x0− x| = 0.

Hence, {xn} converges to x.

(68)

Corollary 9

If g satisfies the hypotheses of above theorem, then

|x − xn| ≤ knmax{x0− a, b − x0} and

|xn− x| ≤ kn

1 − k|x1− x0|, ∀ n ≥ 1.

Proof: From (1),

|xn− x| ≤ kn|x0− x| ≤ knmax{x0− a, b − x0}.

For n ≥ 1, using the Mean Values Theorem,

|xn+1− xn| = |g(xn) − g(xn−1)| ≤ k|xn− xn−1| ≤ · · · ≤ kn|x1− x0|.

(69)

Corollary 9

If g satisfies the hypotheses of above theorem, then

|x − xn| ≤ knmax{x0− a, b − x0} and

|xn− x| ≤ kn

1 − k|x1− x0|, ∀ n ≥ 1.

Proof: From (1),

|xn− x| ≤ kn|x0− x| ≤ knmax{x0− a, b − x0}.

For n ≥ 1, using the Mean Values Theorem,

|xn+1− xn| = |g(xn) − g(xn−1)| ≤ k|xn− xn−1| ≤ · · · ≤ kn|x1− x0|.

(70)

Corollary 9

If g satisfies the hypotheses of above theorem, then

|x − xn| ≤ knmax{x0− a, b − x0} and

|xn− x| ≤ kn

1 − k|x1− x0|, ∀ n ≥ 1.

Proof: From (1),

|xn− x| ≤ kn|x0− x| ≤ knmax{x0− a, b − x0}.

For n ≥ 1, using the Mean Values Theorem,

|xn+1− xn| = |g(xn) − g(xn−1)| ≤ k|xn− xn−1| ≤ · · · ≤ kn|x1− x0|.

(71)

Thus, for m > n ≥ 1,

|xm− xn| = |xm− xm−1+ xm−1− · · · + xn+1− xn|

≤ |xm− xm−1| + |xm−1− xm−2| + · · · + |xn+1− xn|

≤ km−1|x1− x0| + km−2|x1− x0| + · · · + kn|x1− x0|

= kn|x1− x0| 1 + k + k2+ · · · + km−n−1 . It implies that

|x − xn| = lim

m→∞|xm− xn| ≤ lim

m→∞kn|x1− x0|

m−n−1

X

j=0

kj

≤ kn|x1− x0|

X

j=0

kj = kn

1 − k|x1− x0|.

(72)

Thus, for m > n ≥ 1,

|xm− xn| = |xm− xm−1+ xm−1− · · · + xn+1− xn|

≤ |xm− xm−1| + |xm−1− xm−2| + · · · + |xn+1− xn|

≤ km−1|x1− x0| + km−2|x1− x0| + · · · + kn|x1− x0|

= kn|x1− x0| 1 + k + k2+ · · · + km−n−1 . It implies that

|x − xn| = lim

m→∞|xm− xn| ≤ lim

m→∞kn|x1− x0|

m−n−1

X

j=0

kj

≤ kn|x1− x0|

X

j=0

kj = kn

1 − k|x1− x0|.

(73)

Example 10

For previous example,

f (x) = x3+ 4x2− 10 = 0.

For g1(x) = x − x3− 4x2+ 10, we have g1(1) = 6 and g1(2) = −12, so g1([1, 2]) * [1, 2].Moreover,

g10(x) = 1 − 3x2− 8x ⇒ |g10(x)| ≥ 1 ∀ x ∈ [1, 2]

• DOES NOT guarantee to converge or not

(74)

Example 10

For previous example,

f (x) = x3+ 4x2− 10 = 0.

For g1(x) = x − x3− 4x2+ 10, we have g1(1) = 6 and g1(2) = −12, so g1([1, 2]) * [1, 2].Moreover,

g10(x) = 1 − 3x2− 8x ⇒ |g10(x)| ≥ 1 ∀ x ∈ [1, 2]

• DOES NOT guarantee to converge or not

(75)

Example 10

For previous example,

f (x) = x3+ 4x2− 10 = 0.

For g1(x) = x − x3− 4x2+ 10, we have g1(1) = 6 and g1(2) = −12, so g1([1, 2]) * [1, 2]. Moreover,

g10(x) = 1 − 3x2− 8x ⇒ |g10(x)| ≥ 1 ∀ x ∈ [1, 2]

• DOES NOT guarantee to converge or not

(76)

Example 10

For previous example,

f (x) = x3+ 4x2− 10 = 0.

For g1(x) = x − x3− 4x2+ 10, we have g1(1) = 6 and g1(2) = −12, so g1([1, 2]) * [1, 2]. Moreover,

g10(x) = 1 − 3x2− 8x ⇒ |g10(x)| ≥ 1 ∀ x ∈ [1, 2]

• DOES NOT guarantee to converge or not

(77)

For g3(x) = 12(10 − x3)1/2, ∀ x ∈ [1, 1.5], g30(x) = −3

4x2(10 − x3)−1/2< 0, ∀ x ∈ [1, 1.5], so g3 is strictly decreasing on [1, 1.5] and

1 < 1.28 ≈ g3(1.5) ≤ g3(x) ≤ g3(1) = 1.5, ∀ x ∈ [1, 1.5].

On the other hand,

|g03(x)| ≤ |g03(1.5)| ≈ 0.66, ∀ x ∈ [1, 1.5].

Hence, the sequence is convergent to the fixed point.

(78)

For g3(x) = 12(10 − x3)1/2, ∀ x ∈ [1, 1.5], g30(x) = −3

4x2(10 − x3)−1/2< 0, ∀ x ∈ [1, 1.5], so g3 is strictly decreasing on [1, 1.5]and

1 < 1.28 ≈ g3(1.5) ≤ g3(x) ≤ g3(1) = 1.5, ∀ x ∈ [1, 1.5].

On the other hand,

|g03(x)| ≤ |g03(1.5)| ≈ 0.66, ∀ x ∈ [1, 1.5].

Hence, the sequence is convergent to the fixed point.

(79)

For g3(x) = 12(10 − x3)1/2, ∀ x ∈ [1, 1.5], g30(x) = −3

4x2(10 − x3)−1/2< 0, ∀ x ∈ [1, 1.5], so g3 is strictly decreasing on [1, 1.5] and

1 < 1.28 ≈ g3(1.5) ≤ g3(x) ≤ g3(1) = 1.5, ∀ x ∈ [1, 1.5].

On the other hand,

|g03(x)| ≤ |g03(1.5)| ≈ 0.66, ∀ x ∈ [1, 1.5].

Hence, the sequence is convergent to the fixed point.

(80)

For g3(x) = 12(10 − x3)1/2, ∀ x ∈ [1, 1.5], g30(x) = −3

4x2(10 − x3)−1/2< 0, ∀ x ∈ [1, 1.5], so g3 is strictly decreasing on [1, 1.5] and

1 < 1.28 ≈ g3(1.5) ≤ g3(x) ≤ g3(1) = 1.5, ∀ x ∈ [1, 1.5].

On the other hand,

|g03(x)| ≤ |g03(1.5)| ≈ 0.66, ∀ x ∈ [1, 1.5].

Hence, the sequence is convergent to the fixed point.

(81)

For g3(x) = 12(10 − x3)1/2, ∀ x ∈ [1, 1.5], g30(x) = −3

4x2(10 − x3)−1/2< 0, ∀ x ∈ [1, 1.5], so g3 is strictly decreasing on [1, 1.5] and

1 < 1.28 ≈ g3(1.5) ≤ g3(x) ≤ g3(1) = 1.5, ∀ x ∈ [1, 1.5].

On the other hand,

|g03(x)| ≤ |g03(1.5)| ≈ 0.66, ∀ x ∈ [1, 1.5].

Hence, the sequence is convergent to the fixed point.

(82)

For g4(x) =p10/(4 + x), we have r10

6 ≤ g4(x) ≤ r10

5 , ∀ x ∈ [1, 2] ⇒ g4([1, 2]) ⊆ [1, 2]

Moreover,

|g04(x)| =

√ −5

10(4 + x)3/2

≤ 5

√10(5)3/2 < 0.15, ∀ x ∈ [1, 2].

The bound of |g04(x)|is much smaller than the bound of |g30(x)|, which explains the more rapid convergence using g4.

(83)

For g4(x) =p10/(4 + x), we have r10

6 ≤ g4(x) ≤ r10

5 , ∀ x ∈ [1, 2] ⇒ g4([1, 2]) ⊆ [1, 2]

Moreover,

|g04(x)| =

√ −5

10(4 + x)3/2

≤ 5

√10(5)3/2 < 0.15, ∀ x ∈ [1, 2].

The bound of |g04(x)|is much smaller than the bound of |g30(x)|, which explains the more rapid convergence using g4.

(84)

For g4(x) =p10/(4 + x), we have r10

6 ≤ g4(x) ≤ r10

5 , ∀ x ∈ [1, 2] ⇒ g4([1, 2]) ⊆ [1, 2]

Moreover,

|g04(x)| =

√ −5

10(4 + x)3/2

≤ 5

√10(5)3/2 < 0.15, ∀ x ∈ [1, 2].

The bound of |g04(x)|is much smaller than the bound of |g30(x)|, which explains the more rapid convergence using g4.

(85)

Exercise

Page 64: 1, 3, 7, 11, 13

(86)

Suppose that f : R → R and f ∈ C2[a, b], i.e., f00exists and is continuous.If f (x) = 0and x= x + hwhere h is small, then byTaylor’stheorem

0 = f (x) = f (x + h)

= f (x) + f0(x)h + 1

2f00(x)h2+ 1

3!f000(x)h3+ · · ·

= f (x) + f0(x)h + O(h2).

Sincehissmall, O(h2)is negligible. It is reasonable to drop O(h2)terms. This implies

f (x) + f0(x)h ≈ 0 and h ≈ −f (x)

f0(x), if f0(x) 6= 0.

Hence

x + h = x − f (x) f0(x) is a better approximation to x.

(87)

Suppose that f : R → R and f ∈ C2[a, b], i.e., f00exists and is continuous. If f (x) = 0and x= x + hwhere h is small,then byTaylor’stheorem

0 = f (x) = f (x + h)

= f (x) + f0(x)h + 1

2f00(x)h2+ 1

3!f000(x)h3+ · · ·

= f (x) + f0(x)h + O(h2).

Sincehissmall, O(h2)is negligible. It is reasonable to drop O(h2)terms. This implies

f (x) + f0(x)h ≈ 0 and h ≈ −f (x)

f0(x), if f0(x) 6= 0.

Hence

x + h = x − f (x) f0(x) is a better approximation to x.

(88)

Suppose that f : R → R and f ∈ C2[a, b], i.e., f00exists and is continuous. If f (x) = 0and x= x + hwhere h is small, then byTaylor’stheorem

0 = f (x) = f (x + h)

= f (x) + f0(x)h + 1

2f00(x)h2+ 1

3!f000(x)h3+ · · ·

= f (x) + f0(x)h + O(h2).

Sincehissmall, O(h2)is negligible. It is reasonable to drop O(h2)terms. This implies

f (x) + f0(x)h ≈ 0 and h ≈ −f (x)

f0(x), if f0(x) 6= 0.

Hence

x + h = x − f (x) f0(x) is a better approximation to x.

We would like to point out that unlike the pure potential case considered in [RW19], here, in order to guarantee the bulk decay of ˜u, we also need the boundary decay of ∇u due to

After the Opium War, Britain occupied Hong Kong and began its colonial administration. Hong Kong has also developed into an important commercial and trading port. In a society

Real Schur and Hessenberg-triangular forms The doubly shifted QZ algorithm.. Above algorithm is locally

Quadratically convergent sequences generally converge much more quickly thank those that converge only linearly.

Developing Students’ Multimodal Literacy in the Secondary English Language Classroom is a resource package produced by the English Language Education Section,

In this chapter we develop the Lanczos method, a technique that is applicable to large sparse, symmetric eigenproblems.. The method involves tridiagonalizing the given

volume suppressed mass: (TeV) 2 /M P ∼ 10 −4 eV → mm range can be experimentally tested for any number of extra dimensions - Light U(1) gauge bosons: no derivative couplings. =&gt;

• Formation of massive primordial stars as origin of objects in the early universe. • Supernova explosions might be visible to the most