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## Sol. Non-linear Fun.

1

### NTNU

Tsung-Min Hwang November 16, 2003

Department of Mathematics – NTNU Tsung-Min Hwang November 16, 2003

(2)

## Sol. Non-linear Fun.

2

1 – Preliminaries . . . 3

2 – Bisection Method . . . 6

3 – Newton’s Method . . . 12

3.1 – Derivation of Newton’s Method . . . 12

3.2 – Convergence Analysis . . . 16

3.3 – Examples . . . 24

4 – Quasi-Newton’s Method (Secant Method) . . . 27

4.1 – The Secant Method . . . 27

4.2 – Error Analysis of Secant Method . . . 31

5 – Fixed Point and Functional Iteration . . . 36

5.1 – Functional Iteration . . . 37

5.2 – Convergence Analysis . . . 43

Department of Mathematics – NTNU Tsung-Min Hwang November 16, 2003

(3)

## Sol. Non-linear Fun.

3

Definition 1 Let

n

### } → x

. We say that the rate of convergence is

1. linear if

a constant

and an integer

such that

n+1

n

### |, ∀ n ≥ N;

2. superlinear if

n

,

n

as

, and an integer

such that

n+1

n

n

### |, ∀ n ≥ N,

or, equivalently,

n→∞

n+1

n

a constant

### c > 0

(not necessarily less than 1) and an integer

such that

n+1

n

2

### , ∀ n ≥ N.

Department of Mathematics – NTNU Tsung-Min Hwang November 16, 2003

(4)

## Sol. Non-linear Fun.

3

Definition 1 Let

n

### } → x

. We say that the rate of convergence is 1. linear if

a constant

and an integer

such that

n+1

n

### |, ∀ n ≥ N;

2. superlinear if

n

,

n

as

, and an integer

such that

n+1

n

n

### |, ∀ n ≥ N,

or, equivalently,

n→∞

n+1

n

a constant

### c > 0

(not necessarily less than 1) and an integer

such that

n+1

n

2

### , ∀ n ≥ N.

Department of Mathematics – NTNU Tsung-Min Hwang November 16, 2003

(5)

## Sol. Non-linear Fun.

3

Definition 1 Let

n

### } → x

. We say that the rate of convergence is 1. linear if

a constant

and an integer

such that

n+1

n

### |, ∀ n ≥ N;

2. superlinear if

n

,

n

as

, and an integer

such that

n+1

n

n

### |, ∀ n ≥ N,

or, equivalently,

n→∞

n+1

n

a constant

### c > 0

(not necessarily less than 1) and an integer

such that

n+1

n

2

### , ∀ n ≥ N.

Department of Mathematics – NTNU Tsung-Min Hwang November 16, 2003

(6)

## Sol. Non-linear Fun.

3

Definition 1 Let

n

### } → x

. We say that the rate of convergence is 1. linear if

a constant

and an integer

such that

n+1

n

### |, ∀ n ≥ N;

2. superlinear if

n

,

n

as

, and an integer

such that

n+1

n

n

### |, ∀ n ≥ N,

or, equivalently,

n→∞

n+1

n

a constant

### c > 0

(not necessarily less than 1) and an integer

such that

n+1

n

2

### , ∀ n ≥ N.

Department of Mathematics – NTNU Tsung-Min Hwang November 16, 2003

(7)

## Sol. Non-linear Fun.

4

In general, if there are positive constants

and

and an integer

such that

n+1

n

α

### , ∀ n ≥ N,

then we say the rate of convergence is of order

### α

.

Definition 2 Suppose

n

and

n

. If

and an integer

such that

n

n

then we say

n

converges to

### x

with rate of convergence

n

, and write

n

n

### )

.

Example 1 Compare the convergence behavior of

n

and

n

, where

n

2

and

n

3

### .

Department of Mathematics – NTNU Tsung-Min Hwang November 16, 2003

(8)

## Sol. Non-linear Fun.

4

In general, if there are positive constants

and

and an integer

such that

n+1

n

α

### , ∀ n ≥ N,

then we say the rate of convergence is of order

### α

.

Definition 2 Suppose

n

and

n

. If

and an integer

such that

n

n

then we say

n

converges to

### x

with rate of convergence

n

, and write

n

n

### )

.

Example 1 Compare the convergence behavior of

n

and

n

, where

n

2

and

n

3

### .

Department of Mathematics – NTNU Tsung-Min Hwang November 16, 2003

(9)

## Sol. Non-linear Fun.

4

In general, if there are positive constants

and

and an integer

such that

n+1

n

α

### , ∀ n ≥ N,

then we say the rate of convergence is of order

### α

.

Definition 2 Suppose

n

and

n

. If

and an integer

such that

n

n

then we say

n

converges to

### x

with rate of convergence

n

, and write

n

n

### )

.

Example 1 Compare the convergence behavior of

n

and

n

, where

n

2

and

n

3

### .

Department of Mathematics – NTNU Tsung-Min Hwang November 16, 2003

(10)

## Sol. Non-linear Fun.

5

Solution: Note that both

n→∞

n

and

n→∞

n

Let

n

n1 and

n

n12. Then

n

2

2

n

n

3

3

2

n

Hence

n

and

n

2

This shows that

n

### }

converges to 0 much faster than

n

### }

.

Department of Mathematics – NTNU Tsung-Min Hwang November 16, 2003

(11)

## Sol. Non-linear Fun.

5

Solution: Note that both

n→∞

n

and

n→∞

n

Let

n

n1 and

n

n12.

Then

n

2

2

n

n

3

3

2

n

Hence

n

and

n

2

This shows that

n

### }

converges to 0 much faster than

n

### }

.

Department of Mathematics – NTNU Tsung-Min Hwang November 16, 2003

(12)

## Sol. Non-linear Fun.

5

Solution: Note that both

n→∞

n

and

n→∞

n

Let

n

n1 and

n

n12. Then

n

2

2

n

n

3

3

2

n

Hence

n

and

n

2

This shows that

n

### }

converges to 0 much faster than

n

### }

.

Department of Mathematics – NTNU Tsung-Min Hwang November 16, 2003

(13)

## Sol. Non-linear Fun.

5

Solution: Note that both

n→∞

n

and

n→∞

n

Let

n

n1 and

n

n12. Then

n

2

2

n

n

3

3

2

n

Hence

n

and

n

2

This shows that

n

### }

converges to 0 much faster than

n

### }

.

Department of Mathematics – NTNU Tsung-Min Hwang November 16, 2003

(14)

## Sol. Non-linear Fun.

5

Solution: Note that both

n→∞

n

and

n→∞

n

Let

n

n1 and

n

n12. Then

n

2

2

n

n

3

3

2

n

Hence

n

and

n

2

This shows that

n

### }

converges to 0 much faster than

n

### }

.

Department of Mathematics – NTNU Tsung-Min Hwang November 16, 2003

(15)

## Sol. Non-linear Fun.

6

Idea: if

and

, then

such that

### f (c) = 0

.

Algorithm 1 (Bisection Method) Given

defined on

### (a, b)

, the maximal number of iterations

### M

, and stop criteria

and

### ε

, this algorithm tries to locate one root of

.

compute

,

, and

.

if

, then stop for

do

,

,

.

if

or

, then stop if

then

,

.

else

,

### u = w

end if end for

Department of Mathematics – NTNU Tsung-Min Hwang November 16, 2003

(16)

## Sol. Non-linear Fun.

7

Let

n

### }

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

1. the iteration number

,

2.

k

k−1

, or

3.

k

.

Let

0

0

1

1

### ], . . .

denote the successive intervals produced by the bisection algorithm. Then

0

1

2

0

n

and

n

are bounded.

n→∞

n and

n→∞

### b

n exist.

Department of Mathematics – NTNU Tsung-Min Hwang November 16, 2003

(17)

## Sol. Non-linear Fun.

7

Let

n

### }

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

1. the iteration number

,

2.

k

k−1

, or

3.

k

.

Let

0

0

1

1

### ], . . .

denote the successive intervals produced by the bisection algorithm.

Then

0

1

2

0

n

and

n

are bounded.

n→∞

n and

n→∞

### b

n exist.

Department of Mathematics – NTNU Tsung-Min Hwang November 16, 2003

(18)

## Sol. Non-linear Fun.

7

Let

n

### }

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

1. the iteration number

,

2.

k

k−1

, or

3.

k

.

Let

0

0

1

1

### ], . . .

denote the successive intervals produced by the bisection algorithm. Then

0

1

2

0

n

and

n

are bounded.

n→∞

n and

n→∞

### b

n exist.

Department of Mathematics – NTNU Tsung-Min Hwang November 16, 2003

(19)

## Sol. Non-linear Fun.

7

Let

n

### }

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

1. the iteration number

,

2.

k

k−1

, or

3.

k

.

Let

0

0

1

1

### ], . . .

denote the successive intervals produced by the bisection algorithm. Then

0

1

2

0

n

and

n

are bounded.

n→∞

n and

n→∞

### b

n exist.

Department of Mathematics – NTNU Tsung-Min Hwang November 16, 2003

(20)

## Sol. Non-linear Fun.

7

Let

n

### }

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

1. the iteration number

,

2.

k

k−1

, or

3.

k

.

Let

0

0

1

1

### ], . . .

denote the successive intervals produced by the bisection algorithm. Then

0

1

2

0

n

and

n

are bounded.

n→∞

n and

n→∞

### b

n exist.

Department of Mathematics – NTNU Tsung-Min Hwang November 16, 2003

(21)

## Sol. Non-linear Fun.

8

Since

1

1

0

0

2

2

1

1

0

0

.. .

n

n

n

0

0

hence

n→∞

n

n→∞

n

n→∞

n

n

n→∞

n

0

0

Therefore

n→∞

n

n→∞

n

Since

### f

is a continuous function

n→∞

n

n→∞

n

and

n→∞

n

n→∞

n

### ) = f (z).

Department of Mathematics – NTNU Tsung-Min Hwang November 16, 2003

(22)

## Sol. Non-linear Fun.

8

Since

1

1

0

0

2

2

1

1

0

0

.. .

n

n

n

0

0

hence

n→∞

n

n→∞

n

n→∞

n

n

n→∞

n

0

0

Therefore

n→∞

n

n→∞

n

Since

### f

is a continuous function

n→∞

n

n→∞

n

and

n→∞

n

n→∞

n

### ) = f (z).

Department of Mathematics – NTNU Tsung-Min Hwang November 16, 2003

(23)

## Sol. Non-linear Fun.

8

Since

1

1

0

0

2

2

1

1

0

0

.. .

n

n

n

0

0

hence

n→∞

n

n→∞

n

n→∞

n

n

n→∞

n

0

0

Therefore

n→∞

n

n→∞

n

Since

### f

is a continuous function

n→∞

n

n→∞

n

and

n→∞

n

n→∞

n

### ) = f (z).

Department of Mathematics – NTNU Tsung-Min Hwang November 16, 2003

(24)

## Sol. Non-linear Fun.

8

Since

1

1

0

0

2

2

1

1

0

0

.. .

n

n

n

0

0

hence

n→∞

n

n→∞

n

n→∞

n

n

n→∞

n

0

0

Therefore

n→∞

n

n→∞

n

Since

### f

is a continuous function

n→∞

n

n→∞

n

and

n→∞

n

n→∞

n

### ) = f (z).

Department of Mathematics – NTNU Tsung-Min Hwang November 16, 2003

(25)

## Sol. Non-linear Fun.

9

Since

n

n

.

n→∞

n

n

2

.

.

### ⇒

The limit of the sequences

n

and

n

is a zero of

in

.

Let

n

12

n

n

. Then

n

n→∞

n

n

n

n→∞

n

n

n→∞

n

n

n→∞

n

n

n→∞

n

n

n

n

n+1

0

0

### |.

This proves the following theorem. Department of Mathematics – NTNU Tsung-Min Hwang November 16, 2003

(26)

## Sol. Non-linear Fun.

9

Since

n

n

.

n→∞

n

n

2

.

.

### ⇒

The limit of the sequences

n

and

n

is a zero of

in

.

Let

n

12

n

n

. Then

n

n→∞

n

n

n

n→∞

n

n

n→∞

n

n

n→∞

n

n

n→∞

n

n

n

n

n+1

0

0

### |.

This proves the following theorem. Department of Mathematics – NTNU Tsung-Min Hwang November 16, 2003

(27)

## Sol. Non-linear Fun.

9

Since

n

n

.

n→∞

n

n

2

.

.

### ⇒

The limit of the sequences

n

and

n

is a zero of

in

.

Let

n

12

n

n

. Then

n

n→∞

n

n

n

n→∞

n

n

n→∞

n

n

n→∞

n

n

n→∞

n

n

n

n

n+1

0

0

### |.

This proves the following theorem. Department of Mathematics – NTNU Tsung-Min Hwang November 16, 2003

(28)

## Sol. Non-linear Fun.

9

Since

n

n

.

n→∞

n

n

2

.

.

### ⇒

The limit of the sequences

n

and

n

is a zero of

in

.

Let

n

12

n

n

. Then

n

n→∞

n

n

n

n→∞

n

n

n→∞

n

n

n→∞

n

n

n→∞

n

n

n

n

n+1

0

0

### |.

This proves the following theorem. Department of Mathematics – NTNU Tsung-Min Hwang November 16, 2003

(29)

## Sol. Non-linear Fun.

9

Since

n

n

.

n→∞

n

n

2

.

.

### ⇒

The limit of the sequences

n

and

n

is a zero of

in

.

Let

n

12

n

n

.

Then

n

n→∞

n

n

n

n→∞

n

n

n→∞

n

n

n→∞

n

n

n→∞

n

n

n

n

n+1

0

0

### |.

This proves the following theorem. Department of Mathematics – NTNU Tsung-Min Hwang November 16, 2003

(30)

## Sol. Non-linear Fun.

9

Since

n

n

.

n→∞

n

n

2

.

.

### ⇒

The limit of the sequences

n

and

n

is a zero of

in

.

Let

n

12

n

n

. Then

n

n→∞

n

n

n

n→∞

n

n

n→∞

n

n

n→∞

n

n

n→∞

n

n

n

n

n+1

0

0

### |.

This proves the following theorem. Department of Mathematics – NTNU Tsung-Min Hwang November 16, 2003

(31)

## Sol. Non-linear Fun.

9

Since

n

n

.

n→∞

n

n

2

.

.

### ⇒

The limit of the sequences

n

and

n

is a zero of

in

.

Let

n

12

n

n

. Then

n

n→∞

n

n

n

n→∞

n

n

n→∞

n

n

n→∞

n

n

n→∞

n

n

n

n

n+1

0

0

### |.

This proves the following theorem.

Department of Mathematics – NTNU Tsung-Min Hwang November 16, 2003

(32)

## Sol. Non-linear Fun.

10

Theorem 1 Let

n

n

### ]}

denote the intervals produced by the bisection algorithm. Then

n→∞

n and

n→∞

### b

n exist, are equal, and represent a zero of

. If

n→∞

n

n→∞

n and

n

n

n

then

n

n+1

0

0

Remarks 1

n

converges to

with the rate of

n

### )

.

Example 2 If bisection method starts with interval

### [50, 75]

, then how many steps should be taken to compute a root with relative error that is less than

### 10

−12?

Department of Mathematics – NTNU Tsung-Min Hwang November 16, 2003

(33)

## Sol. Non-linear Fun.

10

Theorem 1 Let

n

n

### ]}

denote the intervals produced by the bisection algorithm. Then

n→∞

n and

n→∞

### b

n exist, are equal, and represent a zero of

. If

n→∞

n

n→∞

n and

n

n

n

then

n

n+1

0

0

Remarks 1

n

converges to

with the rate of

n

### )

.

Example 2 If bisection method starts with interval

### [50, 75]

, then how many steps should be taken to compute a root with relative error that is less than

### 10

−12?

Department of Mathematics – NTNU Tsung-Min Hwang November 16, 2003

(34)

## Sol. Non-linear Fun.

10

Theorem 1 Let

n

n

### ]}

denote the intervals produced by the bisection algorithm. Then

n→∞

n and

n→∞

### b

n exist, are equal, and represent a zero of

. If

n→∞

n

n→∞

n and

n

n

n

then

n

n+1

0

0

Remarks 1

n

converges to

with the rate of

n

### )

.

Example 2 If bisection method starts with interval

### [50, 75]

, then how many steps should be taken to compute a root with relative error that is less than

### 10

−12?

Department of Mathematics – NTNU Tsung-Min Hwang November 16, 2003

(35)

## Sol. Non-linear Fun.

11

Solution: Seek an

such that

n

12

### .

Since the bisection method starts with the interval

### [50, 75]

, this implies that

### z ≥ 50

, hence

it is sufficient to show

n

n

12

### .

That is, we solve

(n+1)

12

for

, which gives

### n ≥ 38

.

Department of Mathematics – NTNU Tsung-Min Hwang November 16, 2003

(36)

## Sol. Non-linear Fun.

11

Solution: Seek an

such that

n

12

### .

Since the bisection method starts with the interval

### [50, 75]

, this implies that

### z ≥ 50

,

hence it is sufficient to show

n

n

12

### .

That is, we solve

(n+1)

12

for

, which gives

### n ≥ 38

.

Department of Mathematics – NTNU Tsung-Min Hwang November 16, 2003

(37)

## Sol. Non-linear Fun.

11

Solution: Seek an

such that

n

12

### .

Since the bisection method starts with the interval

### [50, 75]

, this implies that

### z ≥ 50

, hence

it is sufficient to show

n

n

12

### .

That is, we solve

(n+1)

12

for

, which gives

### n ≥ 38

.

Department of Mathematics – NTNU Tsung-Min Hwang November 16, 2003

(38)

## Sol. Non-linear Fun.

11

Solution: Seek an

such that

n

12

### .

Since the bisection method starts with the interval

### [50, 75]

, this implies that

### z ≥ 50

, hence

it is sufficient to show

n

n

12

### .

That is, we solve

(n+1)

12

for

, which gives

### n ≥ 38

.

Department of Mathematics – NTNU Tsung-Min Hwang November 16, 2003

(39)

## Sol. Non-linear Fun.

12

Suppose that

and

2

, i.e.,

### f

00 exists and is continuous.

If

and

where

### h

is small, then by Taylor’s theorem

0

00

2

000

3

0

2

Since

is small,

2

### )

is negligible. It is reasonable to drop

2

### )

terms. This implies

0

and

0

if

0

### (x) 6= 0.

Department of Mathematics – NTNU Tsung-Min Hwang November 16, 2003

(40)

## Sol. Non-linear Fun.

12

Suppose that

and

2

, i.e.,

### f

00 exists and is continuous. If

and

where

### h

is small,

then by Taylor’s theorem

0

00

2

000

3

0

2

Since

is small,

2

### )

is negligible. It is reasonable to drop

2

### )

terms. This implies

0

and

0

if

0

### (x) 6= 0.

Department of Mathematics – NTNU Tsung-Min Hwang November 16, 2003

(41)

## Sol. Non-linear Fun.

12

Suppose that

and

2

, i.e.,

### f

00 exists and is continuous. If

and

where

### h

is small, then by Taylor’s theorem

0

00

2

000

3

0

2

Since

is small,

2

### )

is negligible. It is reasonable to drop

2

### )

terms. This implies

0

and

0

if

0

### (x) 6= 0.

Department of Mathematics – NTNU Tsung-Min Hwang November 16, 2003

(42)

## Sol. Non-linear Fun.

12

Suppose that

and

2

, i.e.,

### f

00 exists and is continuous. If

and

where

### h

is small, then by Taylor’s theorem

0

00

2

000

3

0

2

Since

is small,

2

### )

is negligible.

It is reasonable to drop

2

### )

terms. This implies

0

and

0

if

0

### (x) 6= 0.

Department of Mathematics – NTNU Tsung-Min Hwang November 16, 2003

(43)

## Sol. Non-linear Fun.

12

Suppose that

and

2

, i.e.,

### f

00 exists and is continuous. If

and

where

### h

is small, then by Taylor’s theorem

0

00

2

000

3

0

2

Since

is small,

2

### )

is negligible. It is reasonable to drop

2

terms.

This implies

0

and

0

if

0

### (x) 6= 0.

Department of Mathematics – NTNU Tsung-Min Hwang November 16, 2003

(44)

## Sol. Non-linear Fun.

12

Suppose that

and

2

, i.e.,

### f

00 exists and is continuous. If

and

where

### h

is small, then by Taylor’s theorem

0

00

2

000

3

0

2

Since

is small,

2

### )

is negligible. It is reasonable to drop

2

### )

terms. This implies

0

and

0

if

0

### (x) 6= 0.

Department of Mathematics – NTNU Tsung-Min Hwang November 16, 2003

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