**Sol. Non-linear Fun.**

^{1}

**Solutions of Non-linear Equations** **in One Variable**

**NTNU**

**Tsung-Min Hwang**
**November 16, 2003**

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

**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**

**Sol. Non-linear Fun.**

^{3}

**1 – Preliminaries**

**Definition 1 Let**

### {x

^{n}

### } → x

^{∗}

*. We say that*

*the rate of convergence*

*is*

*1.* *linear* *if*

### ∃

^{a constant}### 0 < c < 1

*and an integer*

### N > 0

^{such that}### |x

^{n+1}

### − x

^{∗}

### | ≤ c|x

^{n}

### − x

^{∗}

### |, ∀ n ≥ N;

*2.* *superlinear* *if*

### ∃ {c

^{n}

### }

^{,}### c

_{n}

### → 0

^{as}### n → ∞

*, and an integer*

### N > 0

^{such that}### |x

^{n+1}

### − x

^{∗}

### | ≤ c

^{n}

### |x

^{n}

### − x

^{∗}

### |, ∀ n ≥ N,

*or, equivalently,*

n→∞

### lim

### |x

^{n+1}

### − x

^{∗}

### |

### |x

^{n}

### − x

^{∗}

### | = 0;

*3.* *quadratic* *if*

### ∃

^{a constant}### c > 0

*(not necessarily less than 1) and an integer*

### N > 0

*such that*

### |x

^{n+1}

### − x

^{∗}

### | ≤ c|x

^{n}

### − x

^{∗}

### |

^{2}

### , ∀ n ≥ N.

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

**Sol. Non-linear Fun.**

^{3}

**1 – Preliminaries**

**Definition 1 Let**

### {x

^{n}

### } → x

^{∗}

*. We say that*

*the rate of convergence*

*is*

*1.*

*linear*

*if*

### ∃

^{a constant}### 0 < c < 1

*and an integer*

### N > 0

^{such that}### |x

^{n+1}

### − x

^{∗}

### | ≤ c|x

^{n}

### − x

^{∗}

### |, ∀ n ≥ N;

*2.* *superlinear* *if*

### ∃ {c

^{n}

### }

^{,}### c

_{n}

### → 0

^{as}### n → ∞

*, and an integer*

### N > 0

^{such that}### |x

^{n+1}

### − x

^{∗}

### | ≤ c

^{n}

### |x

^{n}

### − x

^{∗}

### |, ∀ n ≥ N,

*or, equivalently,*

n→∞

### lim

### |x

^{n+1}

### − x

^{∗}

### |

### |x

^{n}

### − x

^{∗}

### | = 0;

*3.* *quadratic* *if*

### ∃

^{a constant}### c > 0

*(not necessarily less than 1) and an integer*

### N > 0

*such that*

### |x

^{n+1}

### − x

^{∗}

### | ≤ c|x

^{n}

### − x

^{∗}

### |

^{2}

### , ∀ n ≥ N.

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

**Sol. Non-linear Fun.**

^{3}

**1 – Preliminaries**

**Definition 1 Let**

### {x

^{n}

### } → x

^{∗}

*. We say that*

*the rate of convergence*

*is*

*1.*

*linear*

*if*

### ∃

^{a constant}### 0 < c < 1

*and an integer*

### N > 0

^{such that}### |x

^{n+1}

### − x

^{∗}

### | ≤ c|x

^{n}

### − x

^{∗}

### |, ∀ n ≥ N;

*2.* *superlinear* *if*

### ∃ {c

^{n}

### }

^{,}### c

_{n}

### → 0

^{as}### n → ∞

*, and an integer*

### N > 0

^{such that}### |x

^{n+1}

### − x

^{∗}

### | ≤ c

^{n}

### |x

^{n}

### − x

^{∗}

### |, ∀ n ≥ N,

*or, equivalently,*

n→∞

### lim

### |x

^{n+1}

### − x

^{∗}

### |

### |x

^{n}

### − x

^{∗}

### | = 0;

*3.* *quadratic* *if*

### ∃

^{a constant}### c > 0

*(not necessarily less than 1) and an integer*

### N > 0

*such that*

### |x

^{n+1}

### − x

^{∗}

### | ≤ c|x

^{n}

### − x

^{∗}

### |

^{2}

### , ∀ n ≥ N.

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

**Sol. Non-linear Fun.**

^{3}

**1 – Preliminaries**

**Definition 1 Let**

### {x

^{n}

### } → x

^{∗}

*. We say that*

*the rate of convergence*

*is*

*1.*

*linear*

*if*

### ∃

^{a constant}### 0 < c < 1

*and an integer*

### N > 0

^{such that}### |x

^{n+1}

### − x

^{∗}

### | ≤ c|x

^{n}

### − x

^{∗}

### |, ∀ n ≥ N;

*2.* *superlinear* *if*

### ∃ {c

^{n}

### }

^{,}### c

_{n}

### → 0

^{as}### n → ∞

*, and an integer*

### N > 0

^{such that}### |x

^{n+1}

### − x

^{∗}

### | ≤ c

^{n}

### |x

^{n}

### − x

^{∗}

### |, ∀ n ≥ N,

*or, equivalently,*

n→∞

### lim

### |x

^{n+1}

### − x

^{∗}

### |

### |x

^{n}

### − x

^{∗}

### | = 0;

*3.* *quadratic* *if*

### ∃

^{a constant}### c > 0

*(not necessarily less than 1) and an integer*

### N > 0

*such that*

### |x

^{n+1}

### − x

^{∗}

### | ≤ c|x

^{n}

### − x

^{∗}

### |

^{2}

### , ∀ n ≥ N.

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

**Sol. Non-linear Fun.**

^{4}

*In general, if there are positive constants*

### c

^{and}### α

*and an integer*

### N > 0

^{such that}### |x

^{n+1}

### − x

^{∗}

### | ≤ c|x

^{n}

### − x

^{∗}

### |

^{α}

### , ∀ n ≥ N,

*then we say the* *rate of convergence* *is of* *order*

### α

^{.}**Definition 2 Suppose**

### {β

^{n}

### } → 0

^{and}### {x

^{n}

### } → x

^{∗}

^{. If}### ∃ c > 0

*and an integer*

### N > 0

*such that*

### |x

^{n}

### − x

^{∗}

### | ≤ c|β

^{n}

### |, ∀ n ≥ N,

*then we say*

### {x

^{n}

### }

^{converges}

^{to}### x

^{∗}

^{with}*rate of convergence*

### O(β

_{n}

### )

*, and write*

### x

_{n}

### = x

^{∗}

### + O(β

_{n}

### )

^{.}**Example 1 Compare the convergence behavior of**

### {x

^{n}

### }

^{and}### {y

^{n}

### }

^{, where}### x

_{n}

### = n + 1

### n

^{2}

### ,

^{and}### y

_{n}

### = n + 3 n

^{3}

### .

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

**Sol. Non-linear Fun.**

^{4}

*In general, if there are positive constants*

### c

^{and}### α

*and an integer*

### N > 0

^{such that}### |x

^{n+1}

### − x

^{∗}

### | ≤ c|x

^{n}

### − x

^{∗}

### |

^{α}

### , ∀ n ≥ N,

*then we say the* *rate of convergence* *is of* *order*

### α

^{.}**Definition 2 Suppose**

### {β

^{n}

### } → 0

^{and}### {x

^{n}

### } → x

^{∗}

^{. If}### ∃ c > 0

*and an integer*

### N > 0

*such that*

### |x

^{n}

### − x

^{∗}

### | ≤ c|β

^{n}

### |, ∀ n ≥ N,

*then we say*

### {x

^{n}

### }

^{converges}

^{to}### x

^{∗}

^{with}*rate of convergence*

### O(β

_{n}

### )

*, and write*

### x

_{n}

### = x

^{∗}

### + O(β

_{n}

### )

^{.}**Example 1 Compare the convergence behavior of**

### {x

^{n}

### }

^{and}### {y

^{n}

### }

^{, where}### x

_{n}

### = n + 1

### n

^{2}

### ,

^{and}### y

_{n}

### = n + 3 n

^{3}

### .

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

**Sol. Non-linear Fun.**

^{4}

*In general, if there are positive constants*

### c

^{and}### α

*and an integer*

### N > 0

^{such that}### |x

^{n+1}

### − x

^{∗}

### | ≤ c|x

^{n}

### − x

^{∗}

### |

^{α}

### , ∀ n ≥ N,

*then we say the* *rate of convergence* *is of* *order*

### α

^{.}**Definition 2 Suppose**

### {β

^{n}

### } → 0

^{and}### {x

^{n}

### } → x

^{∗}

^{. If}### ∃ c > 0

*and an integer*

### N > 0

*such that*

### |x

^{n}

### − x

^{∗}

### | ≤ c|β

^{n}

### |, ∀ n ≥ N,

*then we say*

### {x

^{n}

### }

^{converges}

^{to}### x

^{∗}

^{with}*rate of convergence*

### O(β

_{n}

### )

*, and write*

### x

_{n}

### = x

^{∗}

### + O(β

_{n}

### )

^{.}**Example 1 Compare the convergence behavior of**

### {x

^{n}

### }

^{and}### {y

^{n}

### }

^{, where}### x

_{n}

### = n + 1

### n

^{2}

### ,

^{and}### y

_{n}

### = n + 3 n

^{3}

### .

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

**Sol. Non-linear Fun.**

^{5}

*Solution: Note that both*

n→∞

### lim x

_{n}

### = 0

^{and}

### lim

n→∞

### y

_{n}

### = 0.

Let

### α

_{n}

### =

_{n}

^{1}

^{and}

### β

_{n}

### =

_{n}

^{1}2. Then

### |x

^{n}

### − 0| = n + 1

### n

^{2}

### ≤ n + n

### n

^{2}

### = 2

### n = 2α

_{n}

### ,

### |y

^{n}

### − 0| = n + 3

### n

^{3}

### ≤ n + 3n

### n

^{3}

### = 4

### n

^{2}

### = 4β

_{n}

### .

Hence

### x

_{n}

### = 0 + O 1 n

and

### y

_{n}

### = 0 + O 1 n

^{2}

### .

This shows that

### {y

^{n}

### }

converges to 0 much faster than### {x

^{n}

### }

^{.}

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

**Sol. Non-linear Fun.**

^{5}

*Solution: Note that both*

n→∞

### lim x

_{n}

### = 0

^{and}

### lim

n→∞

### y

_{n}

### = 0.

Let

### α

_{n}

### =

_{n}

^{1}

^{and}

### β

_{n}

### =

_{n}

^{1}2.

Then

### |x

^{n}

### − 0| = n + 1

### n

^{2}

### ≤ n + n

### n

^{2}

### = 2

### n = 2α

_{n}

### ,

### |y

^{n}

### − 0| = n + 3

### n

^{3}

### ≤ n + 3n

### n

^{3}

### = 4

### n

^{2}

### = 4β

_{n}

### .

Hence

### x

_{n}

### = 0 + O 1 n

and

### y

_{n}

### = 0 + O 1 n

^{2}

### .

This shows that

### {y

^{n}

### }

converges to 0 much faster than### {x

^{n}

### }

^{.}

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

**Sol. Non-linear Fun.**

^{5}

*Solution: Note that both*

n→∞

### lim x

_{n}

### = 0

^{and}

### lim

n→∞

### y

_{n}

### = 0.

Let

### α

_{n}

### =

_{n}

^{1}

^{and}

### β

_{n}

### =

_{n}

^{1}2. Then

### |x

^{n}

### − 0| = n + 1

### n

^{2}

### ≤ n + n

### n

^{2}

### = 2

### n = 2α

_{n}

### ,

### |y

^{n}

### − 0| = n + 3

### n

^{3}

### ≤ n + 3n

### n

^{3}

### = 4

### n

^{2}

### = 4β

_{n}

### .

Hence

### x

_{n}

### = 0 + O 1 n

and

### y

_{n}

### = 0 + O 1 n

^{2}

### .

This shows that

### {y

^{n}

### }

converges to 0 much faster than### {x

^{n}

### }

^{.}

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

**Sol. Non-linear Fun.**

^{5}

*Solution: Note that both*

n→∞

### lim x

_{n}

### = 0

^{and}

### lim

n→∞

### y

_{n}

### = 0.

Let

### α

_{n}

### =

_{n}

^{1}

^{and}

### β

_{n}

### =

_{n}

^{1}2. Then

### |x

^{n}

### − 0| = n + 1

### n

^{2}

### ≤ n + n

### n

^{2}

### = 2

### n = 2α

_{n}

### ,

### |y

^{n}

### − 0| = n + 3

### n

^{3}

### ≤ n + 3n

### n

^{3}

### = 4

### n

^{2}

### = 4β

_{n}

### .

Hence

### x

_{n}

### = 0 + O 1 n

and

### y

_{n}

### = 0 + O 1 n

^{2}

### .

This shows that

### {y

^{n}

### }

converges to 0 much faster than### {x

^{n}

### }

^{.}

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

**Sol. Non-linear Fun.**

^{5}

*Solution: Note that both*

n→∞

### lim x

_{n}

### = 0

^{and}

### lim

n→∞

### y

_{n}

### = 0.

Let

### α

_{n}

### =

_{n}

^{1}

^{and}

### β

_{n}

### =

_{n}

^{1}2. Then

### |x

^{n}

### − 0| = n + 1

### n

^{2}

### ≤ n + n

### n

^{2}

### = 2

### n = 2α

_{n}

### ,

### |y

^{n}

### − 0| = n + 3

### n

^{3}

### ≤ n + 3n

### n

^{3}

### = 4

### n

^{2}

### = 4β

_{n}

### .

Hence

### x

_{n}

### = 0 + O 1 n

and

### y

_{n}

### = 0 + O 1 n

^{2}

### .

This shows that

### {y

^{n}

### }

converges to 0 much faster than### {x

^{n}

### }

^{.}

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

**Sol. Non-linear Fun.**

^{6}

**2 – Bisection Method**

Idea: if

### f (x) ∈ C[a, b]

^{and}

### f (a)f (b) < 0

^{, then}

### ∃ c ∈ (a, b)

^{such that}

### f (c) = 0

^{.}

**Algorithm 1 (Bisection Method) 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

^{do}### e = e/2

^{,}

### c = a + e

^{,}

### w = f (c)

^{.}

**if**

### |e| < δ

^{or}

### |w| < ε

**, then stop**

**if**

### sign(w) 6= sign(u)

^{then}### b = c

^{,}

### v = w

^{.}

**else**

### a = c

^{,}

### u = w

**end if**
**end for**

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

**Sol. Non-linear Fun.**

^{7}

Let

### {c

^{n}

### }

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.

### |c

^{k}

### − c

^{k−1}

### | < δ

^{, or}

3.

### |f(c

^{k}

### )| < ε

^{.}

Let

### [a

_{0}

### , b

_{0}

### ], [a

_{1}

### , b

_{1}

### ], . . .

denote the successive intervals produced by the bisection algorithm. Then### a = a

_{0}

### ≤ a

^{1}

### ≤ a

^{2}

### ≤ · · · ≤ b

^{0}

### = b

### ⇒ {a

^{n}

### }

^{and}

### {b

^{n}

### }

are bounded.### ⇒ lim

n→∞

### a

_{n}

^{and}

### lim

n→∞

### b

_{n}

^{exist.}

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

**Sol. Non-linear Fun.**

^{7}

Let

### {c

^{n}

### }

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.

### |c

^{k}

### − c

^{k−1}

### | < δ

^{, or}

3.

### |f(c

^{k}

### )| < ε

^{.}

Let

### [a

_{0}

### , b

_{0}

### ], [a

_{1}

### , b

_{1}

### ], . . .

denote the successive intervals produced by the bisection algorithm.Then

### a = a

_{0}

### ≤ a

^{1}

### ≤ a

^{2}

### ≤ · · · ≤ b

^{0}

### = b

### ⇒ {a

^{n}

### }

^{and}

### {b

^{n}

### }

are bounded.### ⇒ lim

n→∞

### a

_{n}

^{and}

### lim

n→∞

### b

_{n}

^{exist.}

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

**Sol. Non-linear Fun.**

^{7}

Let

### {c

^{n}

### }

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.

### |c

^{k}

### − c

^{k−1}

### | < δ

^{, or}

3.

### |f(c

^{k}

### )| < ε

^{.}

Let

### [a

_{0}

### , b

_{0}

### ], [a

_{1}

### , b

_{1}

### ], . . .

denote the successive intervals produced by the bisection algorithm. Then### a = a

_{0}

### ≤ a

^{1}

### ≤ a

^{2}

### ≤ · · · ≤ b

^{0}

### = b

### ⇒ {a

^{n}

### }

^{and}

### {b

^{n}

### }

are bounded.### ⇒ lim

n→∞

### a

_{n}

^{and}

### lim

n→∞

### b

_{n}

^{exist.}

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

**Sol. Non-linear Fun.**

^{7}

Let

### {c

^{n}

### }

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.

### |c

^{k}

### − c

^{k−1}

### | < δ

^{, or}

3.

### |f(c

^{k}

### )| < ε

^{.}

Let

### [a

_{0}

### , b

_{0}

### ], [a

_{1}

### , b

_{1}

### ], . . .

denote the successive intervals produced by the bisection algorithm. Then### a = a

_{0}

### ≤ a

^{1}

### ≤ a

^{2}

### ≤ · · · ≤ b

^{0}

### = b

### ⇒ {a

^{n}

### }

^{and}

### {b

^{n}

### }

are bounded.### ⇒ lim

n→∞

### a

_{n}

^{and}

### lim

n→∞

### b

_{n}

^{exist.}

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

**Sol. Non-linear Fun.**

^{7}

Let

### {c

^{n}

### }

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.

### |c

^{k}

### − c

^{k−1}

### | < δ

^{, or}

3.

### |f(c

^{k}

### )| < ε

^{.}

Let

### [a

_{0}

### , b

_{0}

### ], [a

_{1}

### , b

_{1}

### ], . . .

denote the successive intervals produced by the bisection algorithm. Then### a = a

_{0}

### ≤ a

^{1}

### ≤ a

^{2}

### ≤ · · · ≤ b

^{0}

### = b

### ⇒ {a

^{n}

### }

^{and}

### {b

^{n}

### }

are bounded.### ⇒ lim

n→∞

### a

_{n}

^{and}

### lim

n→∞

### b

_{n}

^{exist.}

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

**Sol. Non-linear Fun.**

^{8}

Since

### b

_{1}

### − a

^{1}

### = 1

### 2 (b

_{0}

### − a

^{0}

### ) b

_{2}

### − a

^{2}

### = 1

### 2 (b

_{1}

### − a

^{1}

### ) = 1

### 4 (b

_{0}

### − a

^{0}

### )

.. .

### b

_{n}

### − a

^{n}

### = 1

### 2

^{n}

### (b

_{0}

### − a

^{0}

### )

hence

n→∞

### lim b

_{n}

### − lim

n→∞

### a

_{n}

### = lim

n→∞

### (b

_{n}

### − a

^{n}

### ) = lim

n→∞

### 1

### 2

^{n}

### (b

_{0}

### − a

^{0}

### ) = 0.

Therefore

n→∞

### lim a

_{n}

### = lim

n→∞

### b

_{n}

### ≡ z.

Since

### f

is a continuous functionn→∞

### lim f (a

_{n}

### ) = f ( lim

n→∞

### a

_{n}

### ) = f (z)

^{and}

### lim

n→∞

### f (b

_{n}

### ) = f ( lim

n→∞

### b

_{n}

### ) = f (z).

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

**Sol. Non-linear Fun.**

^{8}

Since

### b

_{1}

### − a

^{1}

### = 1

### 2 (b

_{0}

### − a

^{0}

### ) b

_{2}

### − a

^{2}

### = 1

### 2 (b

_{1}

### − a

^{1}

### ) = 1

### 4 (b

_{0}

### − a

^{0}

### )

.. .

### b

_{n}

### − a

^{n}

### = 1

### 2

^{n}

### (b

_{0}

### − a

^{0}

### )

hence

n→∞

### lim b

_{n}

### − lim

n→∞

### a

_{n}

### = lim

n→∞

### (b

_{n}

### − a

^{n}

### ) = lim

n→∞

### 1

### 2

^{n}

### (b

_{0}

### − a

^{0}

### ) = 0.

Therefore

n→∞

### lim a

_{n}

### = lim

n→∞

### b

_{n}

### ≡ z.

Since

### f

is a continuous functionn→∞

### lim f (a

_{n}

### ) = f ( lim

n→∞

### a

_{n}

### ) = f (z)

^{and}

### lim

n→∞

### f (b

_{n}

### ) = f ( lim

n→∞

### b

_{n}

### ) = f (z).

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

**Sol. Non-linear Fun.**

^{8}

Since

### b

_{1}

### − a

^{1}

### = 1

### 2 (b

_{0}

### − a

^{0}

### ) b

_{2}

### − a

^{2}

### = 1

### 2 (b

_{1}

### − a

^{1}

### ) = 1

### 4 (b

_{0}

### − a

^{0}

### )

.. .

### b

_{n}

### − a

^{n}

### = 1

### 2

^{n}

### (b

_{0}

### − a

^{0}

### )

hence

n→∞

### lim b

_{n}

### − lim

n→∞

### a

_{n}

### = lim

n→∞

### (b

_{n}

### − a

^{n}

### ) = lim

n→∞

### 1

### 2

^{n}

### (b

_{0}

### − a

^{0}

### ) = 0.

Therefore

n→∞

### lim a

_{n}

### = lim

n→∞

### b

_{n}

### ≡ z.

Since

### f

is a continuous functionn→∞

### lim f (a

_{n}

### ) = f ( lim

n→∞

### a

_{n}

### ) = f (z)

^{and}

### lim

n→∞

### f (b

_{n}

### ) = f ( lim

n→∞

### b

_{n}

### ) = f (z).

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

**Sol. Non-linear Fun.**

^{8}

Since

### b

_{1}

### − a

^{1}

### = 1

### 2 (b

_{0}

### − a

^{0}

### ) b

_{2}

### − a

^{2}

### = 1

### 2 (b

_{1}

### − a

^{1}

### ) = 1

### 4 (b

_{0}

### − a

^{0}

### )

.. .

### b

_{n}

### − a

^{n}

### = 1

### 2

^{n}

### (b

_{0}

### − a

^{0}

### )

hence

n→∞

### lim b

_{n}

### − lim

n→∞

### a

_{n}

### = lim

n→∞

### (b

_{n}

### − a

^{n}

### ) = lim

n→∞

### 1

### 2

^{n}

### (b

_{0}

### − a

^{0}

### ) = 0.

Therefore

n→∞

### lim a

_{n}

### = lim

n→∞

### b

_{n}

### ≡ z.

Since

### f

is a continuous functionn→∞

### lim f (a

_{n}

### ) = f ( lim

n→∞

### a

_{n}

### ) = f (z)

^{and}

### lim

n→∞

### f (b

_{n}

### ) = f ( lim

n→∞

### b

_{n}

### ) = f (z).

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

**Sol. Non-linear Fun.**

^{9}

Since

### f (a

_{n}

### )f (b

_{n}

### ) ≤ 0

^{.}

### ⇒ lim

_{n→∞}

### f (a

_{n}

### )f (b

_{n}

### ) = f

^{2}

### (z) ≤ 0

^{.}

### ⇒ f(z) = 0

^{.}

### ⇒

The limit of the sequences### {a

^{n}

### }

^{and}

### {b

^{n}

### }

is a zero of### f

^{in}

### [a, b]

^{.}

Let

### c

_{n}

### =

^{1}

_{2}

### (a

_{n}

### + b

_{n}

### )

^{. Then}

### |z − c

^{n}

### | =

### lim

n→∞

### a

_{n}

### − 1

### 2 (a

_{n}

### + b

_{n}

### )

### =

### 1 2

### h lim

n→∞

### a

_{n}

### − b

^{n}

### i

### + 1 2

### h lim

n→∞

### a

_{n}

### − a

^{n}

### i

### ≤ 1

### 2 max

### lim

n→∞

### a

_{n}

### − b

^{n}

### ,

### lim

n→∞

### a

_{n}

### − a

^{n}

### ≤ 1

### 2 |b

^{n}

### − a

^{n}

### | = 1

### 2

^{n+1}

### |b

^{0}

### − a

^{0}

### |.

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

**Sol. Non-linear Fun.**

^{9}

Since

### f (a

_{n}

### )f (b

_{n}

### ) ≤ 0

^{.}

### ⇒ lim

_{n→∞}

### f (a

_{n}

### )f (b

_{n}

### ) = f

^{2}

### (z) ≤ 0

^{.}

### ⇒ f(z) = 0

^{.}

### ⇒

The limit of the sequences### {a

^{n}

### }

^{and}

### {b

^{n}

### }

is a zero of### f

^{in}

### [a, b]

^{.}

Let

### c

_{n}

### =

^{1}

_{2}

### (a

_{n}

### + b

_{n}

### )

^{. Then}

### |z − c

^{n}

### | =

### lim

n→∞

### a

_{n}

### − 1

### 2 (a

_{n}

### + b

_{n}

### )

### =

### 1 2

### h lim

n→∞

### a

_{n}

### − b

^{n}

### i

### + 1 2

### h lim

n→∞

### a

_{n}

### − a

^{n}

### i

### ≤ 1

### 2 max

### lim

n→∞

### a

_{n}

### − b

^{n}

### ,

### lim

n→∞

### a

_{n}

### − a

^{n}

### ≤ 1

### 2 |b

^{n}

### − a

^{n}

### | = 1

### 2

^{n+1}

### |b

^{0}

### − a

^{0}

### |.

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

**Sol. Non-linear Fun.**

^{9}

Since

### f (a

_{n}

### )f (b

_{n}

### ) ≤ 0

^{.}

### ⇒ lim

_{n→∞}

### f (a

_{n}

### )f (b

_{n}

### ) = f

^{2}

### (z) ≤ 0

^{.}

### ⇒ f(z) = 0

^{.}

### ⇒

The limit of the sequences### {a

^{n}

### }

^{and}

### {b

^{n}

### }

is a zero of### f

^{in}

### [a, b]

^{.}

Let

### c

_{n}

### =

^{1}

_{2}

### (a

_{n}

### + b

_{n}

### )

^{. Then}

### |z − c

^{n}

### | =

### lim

n→∞

### a

_{n}

### − 1

### 2 (a

_{n}

### + b

_{n}

### )

### =

### 1 2

### h lim

n→∞

### a

_{n}

### − b

^{n}

### i

### + 1 2

### h lim

n→∞

### a

_{n}

### − a

^{n}

### i

### ≤ 1

### 2 max

### lim

n→∞

### a

_{n}

### − b

^{n}

### ,

### lim

n→∞

### a

_{n}

### − a

^{n}

### ≤ 1

### 2 |b

^{n}

### − a

^{n}

### | = 1

### 2

^{n+1}

### |b

^{0}

### − a

^{0}

### |.

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

**Sol. Non-linear Fun.**

^{9}

Since

### f (a

_{n}

### )f (b

_{n}

### ) ≤ 0

^{.}

### ⇒ lim

_{n→∞}

### f (a

_{n}

### )f (b

_{n}

### ) = f

^{2}

### (z) ≤ 0

^{.}

### ⇒ f(z) = 0

^{.}

### ⇒

The limit of the sequences### {a

^{n}

### }

^{and}

### {b

^{n}

### }

is a zero of### f

^{in}

### [a, b]

^{.}

Let

### c

_{n}

### =

^{1}

_{2}

### (a

_{n}

### + b

_{n}

### )

^{. Then}

### |z − c

^{n}

### | =

### lim

n→∞

### a

_{n}

### − 1

### 2 (a

_{n}

### + b

_{n}

### )

### =

### 1 2

### h lim

n→∞

### a

_{n}

### − b

^{n}

### i

### + 1 2

### h lim

n→∞

### a

_{n}

### − a

^{n}

### i

### ≤ 1

### 2 max

### lim

n→∞

### a

_{n}

### − b

^{n}

### ,

### lim

n→∞

### a

_{n}

### − a

^{n}

### ≤ 1

### 2 |b

^{n}

### − a

^{n}

### | = 1

### 2

^{n+1}

### |b

^{0}

### − a

^{0}

### |.

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

**Sol. Non-linear Fun.**

^{9}

Since

### f (a

_{n}

### )f (b

_{n}

### ) ≤ 0

^{.}

### ⇒ lim

_{n→∞}

### f (a

_{n}

### )f (b

_{n}

### ) = f

^{2}

### (z) ≤ 0

^{.}

### ⇒ f(z) = 0

^{.}

### ⇒

The limit of the sequences### {a

^{n}

### }

^{and}

### {b

^{n}

### }

is a zero of### f

^{in}

### [a, b]

^{.}

Let

### c

_{n}

### =

^{1}

_{2}

### (a

_{n}

### + b

_{n}

### )

^{.}

Then

### |z − c

^{n}

### | =

### lim

n→∞

### a

_{n}

### − 1

### 2 (a

_{n}

### + b

_{n}

### )

### =

### 1 2

### h lim

n→∞

### a

_{n}

### − b

^{n}

### i

### + 1 2

### h lim

n→∞

### a

_{n}

### − a

^{n}

### i

### ≤ 1

### 2 max

### lim

n→∞

### a

_{n}

### − b

^{n}

### ,

### lim

n→∞

### a

_{n}

### − a

^{n}

### ≤ 1

### 2 |b

^{n}

### − a

^{n}

### | = 1

### 2

^{n+1}

### |b

^{0}

### − a

^{0}

### |.

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

**Sol. Non-linear Fun.**

^{9}

Since

### f (a

_{n}

### )f (b

_{n}

### ) ≤ 0

^{.}

### ⇒ lim

_{n→∞}

### f (a

_{n}

### )f (b

_{n}

### ) = f

^{2}

### (z) ≤ 0

^{.}

### ⇒ f(z) = 0

^{.}

### ⇒

The limit of the sequences### {a

^{n}

### }

^{and}

### {b

^{n}

### }

is a zero of### f

^{in}

### [a, b]

^{.}

Let

### c

_{n}

### =

^{1}

_{2}

### (a

_{n}

### + b

_{n}

### )

^{. Then}

### |z − c

^{n}

### | =

### lim

n→∞

### a

_{n}

### − 1

### 2 (a

_{n}

### + b

_{n}

### )

### =

### 1 2

### h lim

n→∞

### a

_{n}

### − b

^{n}

### i

### + 1 2

### h lim

n→∞

### a

_{n}

### − a

^{n}

### i

### ≤ 1

### 2 max

### lim

n→∞

### a

_{n}

### − b

^{n}

### ,

### lim

n→∞

### a

_{n}

### − a

^{n}

### ≤ 1

### 2 |b

^{n}

### − a

^{n}

### | = 1

### 2

^{n+1}

### |b

^{0}

### − a

^{0}

### |.

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

**Sol. Non-linear Fun.**

^{9}

Since

### f (a

_{n}

### )f (b

_{n}

### ) ≤ 0

^{.}

### ⇒ lim

_{n→∞}

### f (a

_{n}

### )f (b

_{n}

### ) = f

^{2}

### (z) ≤ 0

^{.}

### ⇒ f(z) = 0

^{.}

### ⇒

The limit of the sequences### {a

^{n}

### }

^{and}

### {b

^{n}

### }

is a zero of### f

^{in}

### [a, b]

^{.}

Let

### c

_{n}

### =

^{1}

_{2}

### (a

_{n}

### + b

_{n}

### )

^{. Then}

### |z − c

^{n}

### | =

### lim

n→∞

### a

_{n}

### − 1

### 2 (a

_{n}

### + b

_{n}

### )

### =

### 1 2

### h lim

n→∞

### a

_{n}

### − b

^{n}

### i

### + 1 2

### h lim

n→∞

### a

_{n}

### − a

^{n}

### i

### ≤ 1

### 2 max

### lim

n→∞

### a

_{n}

### − b

^{n}

### ,

### lim

n→∞

### a

_{n}

### − a

^{n}

### ≤ 1

### 2 |b

^{n}

### − a

^{n}

### | = 1

### 2

^{n+1}

### |b

^{0}

### − a

^{0}

### |.

This proves the following theorem.

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

**Sol. Non-linear Fun.**

^{10}

**Theorem 1 Let**

### {[a

^{n}

### , b

_{n}

### ]}

*denote the intervals produced by the bisection algorithm. Then*

n→∞

### lim a

_{n}

^{and}### lim

n→∞

### b

_{n}

^{exist, are}*equal, and represent a*

*zero*

*of*

### f (x)

^{. If}### z = lim

n→∞

### a

_{n}

### = lim

n→∞

### b

_{n}

^{and}### c

_{n}

### = 1

### 2 (a

_{n}

### + b

_{n}

### ),

*then*

### |z − c

^{n}

### | ≤ 1

### 2

^{n+1}

### (b

_{0}

### − a

^{0}

### ) .

**Remarks 1**

### {c

^{n}

### }

^{converges}

^{to}### z

^{with the}

^{rate}

^{of}### O(2

^{−}

^{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**

**Sol. Non-linear Fun.**

^{10}

**Theorem 1 Let**

### {[a

^{n}

### , b

_{n}

### ]}

*denote the intervals produced by the bisection algorithm. Then*

n→∞

### lim a

_{n}

^{and}### lim

n→∞

### b

_{n}

^{exist, are}*equal, and represent a*

*zero*

*of*

### f (x)

^{. If}### z = lim

n→∞

### a

_{n}

### = lim

n→∞

### b

_{n}

^{and}### c

_{n}

### = 1

### 2 (a

_{n}

### + b

_{n}

### ),

*then*

### |z − c

^{n}

### | ≤ 1

### 2

^{n+1}

### (b

_{0}

### − a

^{0}

### ) .

**Remarks 1**

### {c

^{n}

### }

^{converges}

^{to}### z

^{with the}

^{rate}

^{of}### O(2

^{−}

^{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**

**Sol. Non-linear Fun.**

^{10}

**Theorem 1 Let**

### {[a

^{n}

### , b

_{n}

### ]}

*denote the intervals produced by the bisection algorithm. Then*

n→∞

### lim a

_{n}

^{and}### lim

n→∞

### b

_{n}

^{exist, are}*equal, and represent a*

*zero*

*of*

### f (x)

^{. If}### z = lim

n→∞

### a

_{n}

### = lim

n→∞

### b

_{n}

^{and}### c

_{n}

### = 1

### 2 (a

_{n}

### + b

_{n}

### ),

*then*

### |z − c

^{n}

### | ≤ 1

### 2

^{n+1}

### (b

_{0}

### − a

^{0}

### ) .

**Remarks 1**

### {c

^{n}

### }

^{converges}

^{to}### z

^{with the}

^{rate}

^{of}### O(2

^{−}

^{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**

**Sol. Non-linear Fun.**

^{11}

*Solution: Seek an*

### n

^{such that}

### |z − c

^{n}

### |

### |z| ≤ 10

^{−}

^{12}

### .

Since the bisection method starts with the interval

### [50, 75]

, this implies that### z ≥ 50

^{, hence}

it is sufficient to show

### |z − c

^{n}

### |

### |z| ≤ |z − c

^{n}

### |

### 50 ≤ 10

^{−}

^{12}

### .

That is, we solve

### 2

^{−}

^{(n+1)}

### (75 − 50) ≤ 50 × 10

^{−}

^{12}

for

### n

, which gives### n ≥ 38

^{.}

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

**Sol. Non-linear Fun.**

^{11}

*Solution: Seek an*

### n

^{such that}

### |z − c

^{n}

### |

### |z| ≤ 10

^{−}

^{12}

### .

Since the bisection method starts with the interval

### [50, 75]

, this implies that### z ≥ 50

^{,}

hence it is sufficient to show

### |z − c

^{n}

### |

### |z| ≤ |z − c

^{n}

### |

### 50 ≤ 10

^{−}

^{12}

### .

That is, we solve

### 2

^{−}

^{(n+1)}

### (75 − 50) ≤ 50 × 10

^{−}

^{12}

for

### n

, which gives### n ≥ 38

^{.}

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

**Sol. Non-linear Fun.**

^{11}

*Solution: Seek an*

### n

^{such that}

### |z − c

^{n}

### |

### |z| ≤ 10

^{−}

^{12}

### .

Since the bisection method starts with the interval

### [50, 75]

, this implies that### z ≥ 50

^{, hence}

it is sufficient to show

### |z − c

^{n}

### |

### |z| ≤ |z − c

^{n}

### |

### 50 ≤ 10

^{−}

^{12}

### .

That is, we solve

### 2

^{−}

^{(n+1)}

### (75 − 50) ≤ 50 × 10

^{−}

^{12}

for

### n

, which gives### n ≥ 38

^{.}

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

**Sol. Non-linear Fun.**

^{11}

*Solution: Seek an*

### n

^{such that}

### |z − c

^{n}

### |

### |z| ≤ 10

^{−}

^{12}

### .

Since the bisection method starts with the interval

### [50, 75]

, this implies that### z ≥ 50

^{, hence}

it is sufficient to show

### |z − c

^{n}

### |

### |z| ≤ |z − c

^{n}

### |

### 50 ≤ 10

^{−}

^{12}

### .

That is, we solve

### 2

^{−}

^{(n+1)}

### (75 − 50) ≤ 50 × 10

^{−}

^{12}

for

### n

, which gives### n ≥ 38

^{.}

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

**Sol. Non-linear Fun.**

^{12}

**3 – Newton’s Method**

**3.1 – Derivation of Newton’s Method**

Suppose that

### f : R → R

^{and}

### f ∈ C

^{2}

### [a, b]

^{, i.e.,}

### f

^{00}exists and is continuous.

If

### f (x

^{∗}

### ) = 0

^{and}

### x

^{∗}

### = x + h

^{where}

### h

is small, then by Taylor’s theorem### 0 = f (x

^{∗}

### ) = f (x + h) = f (x) + f

^{0}

### (x)h + 1

### 2 f

^{00}

### (x)h

^{2}

### + 1

### 3! f

^{000}

### (x)h

^{3}

### + · · ·

### = f (x) + f

^{0}

### (x)h + O(h

^{2}

### ).

Since

### h

^{is}

^{small,}

### O(h

^{2}

### )

is negligible. It is reasonable to drop### O(h

^{2}

### )

terms. This implies### f (x) + f

^{0}

### (x)h ≈ 0

^{and}

### h ≈ − f (x)

### f

^{0}

### (x) ,

^{if}

### f

^{0}

### (x) 6= 0.

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

**Sol. Non-linear Fun.**

^{12}

**3 – Newton’s Method**

**3.1 – Derivation of Newton’s Method**

Suppose that

### f : R → R

^{and}

### f ∈ C

^{2}

### [a, b]

^{, i.e.,}

### f

^{00}exists and is continuous. If

### f (x

^{∗}

### ) = 0

^{and}

### x

^{∗}

### = x + h

^{where}

### h

^{is small,}

then by Taylor’s theorem

### 0 = f (x

^{∗}

### ) = f (x + h) = f (x) + f

^{0}

### (x)h + 1

### 2 f

^{00}

### (x)h

^{2}

### + 1

### 3! f

^{000}

### (x)h

^{3}

### + · · ·

### = f (x) + f

^{0}

### (x)h + O(h

^{2}

### ).

Since

### h

^{is}

^{small,}

### O(h

^{2}

### )

is negligible. It is reasonable to drop### O(h

^{2}

### )

terms. This implies### f (x) + f

^{0}

### (x)h ≈ 0

^{and}

### h ≈ − f (x)

### f

^{0}

### (x) ,

^{if}

### f

^{0}

### (x) 6= 0.

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

**Sol. Non-linear Fun.**

^{12}

**3 – Newton’s Method**

**3.1 – Derivation of Newton’s Method**

Suppose that

### f : R → R

^{and}

### f ∈ C

^{2}

### [a, b]

^{, i.e.,}

### f

^{00}exists and is continuous. If

### f (x

^{∗}

### ) = 0

^{and}

### x

^{∗}

### = x + h

^{where}

### h

is small, then by Taylor’s theorem### 0 = f (x

^{∗}

### ) = f (x + h) = f (x) + f

^{0}

### (x)h + 1

### 2 f

^{00}

### (x)h

^{2}

### + 1

### 3! f

^{000}

### (x)h

^{3}

### + · · ·

### = f (x) + f

^{0}

### (x)h + O(h

^{2}

### ).

Since

### h

^{is}

^{small,}

### O(h

^{2}

### )

is negligible. It is reasonable to drop### O(h

^{2}

### )

terms. This implies### f (x) + f

^{0}

### (x)h ≈ 0

^{and}

### h ≈ − f (x)

### f

^{0}

### (x) ,

^{if}

### f

^{0}

### (x) 6= 0.

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

**Sol. Non-linear Fun.**

^{12}

**3 – Newton’s Method**

**3.1 – Derivation of Newton’s Method**

Suppose that

### f : R → R

^{and}

### f ∈ C

^{2}

### [a, b]

^{, i.e.,}

### f

^{00}exists and is continuous. If

### f (x

^{∗}

### ) = 0

^{and}

### x

^{∗}

### = x + h

^{where}

### h

is small, then by Taylor’s theorem### 0 = f (x

^{∗}

### ) = f (x + h) = f (x) + f

^{0}

### (x)h + 1

### 2 f

^{00}

### (x)h

^{2}

### + 1

### 3! f

^{000}

### (x)h

^{3}

### + · · ·

### = f (x) + f

^{0}

### (x)h + O(h

^{2}

### ).

Since

### h

^{is}

^{small,}

### O(h

^{2}

### )

is negligible.It is reasonable to drop

### O(h

^{2}

### )

terms. This implies### f (x) + f

^{0}

### (x)h ≈ 0

^{and}

### h ≈ − f (x)

### f

^{0}

### (x) ,

^{if}

### f

^{0}

### (x) 6= 0.

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

**Sol. Non-linear Fun.**

^{12}

**3 – Newton’s Method**

**3.1 – Derivation of Newton’s Method**

Suppose that

### f : R → R

^{and}

### f ∈ C

^{2}

### [a, b]

^{, i.e.,}

### f

^{00}exists and is continuous. If

### f (x

^{∗}

### ) = 0

^{and}

### x

^{∗}

### = x + h

^{where}

### h

is small, then by Taylor’s theorem### 0 = f (x

^{∗}

### ) = f (x + h) = f (x) + f

^{0}

### (x)h + 1

### 2 f

^{00}

### (x)h

^{2}

### + 1

### 3! f

^{000}

### (x)h

^{3}

### + · · ·

### = f (x) + f

^{0}

### (x)h + O(h

^{2}

### ).

Since

### h

^{is}

^{small,}

### O(h

^{2}

### )

is negligible. It is reasonable to drop### O(h

^{2}

### )

^{terms.}

This implies

### f (x) + f

^{0}

### (x)h ≈ 0

^{and}

### h ≈ − f (x)

### f

^{0}

### (x) ,

^{if}

### f

^{0}

### (x) 6= 0.

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

**Sol. Non-linear Fun.**

^{12}

**3 – Newton’s Method**

**3.1 – Derivation of Newton’s Method**

Suppose that

### f : R → R

^{and}

### f ∈ C

^{2}

### [a, b]

^{, i.e.,}

### f

^{00}exists and is continuous. If

### f (x

^{∗}

### ) = 0

^{and}

### x

^{∗}

### = x + h

^{where}

### h

is small, then by Taylor’s theorem### 0 = f (x

^{∗}

### ) = f (x + h) = f (x) + f

^{0}

### (x)h + 1

### 2 f

^{00}

### (x)h

^{2}

### + 1

### 3! f

^{000}

### (x)h

^{3}

### + · · ·

### = f (x) + f

^{0}

### (x)h + O(h

^{2}

### ).

Since

### h

^{is}

^{small,}

### O(h

^{2}

### )

is negligible. It is reasonable to drop### O(h

^{2}

### )

terms. This implies### f (x) + f

^{0}

### (x)h ≈ 0

^{and}

### h ≈ − f (x)

### f

^{0}

### (x) ,

^{if}

### f

^{0}

### (x) 6= 0.

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