Solutions of Equations in One Variable

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Tsung-Ming Huang

Department of Mathematics National Taiwan Normal University, Taiwan

October 13, 2014

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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

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Bisection Method

Idea

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

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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

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

a = a₀≤ a₁ ≤ a₂ ≤ · · · ≤ b₀= b

⇒ {a_n} and {bn} are bounded

⇒ lim

n→∞an and lim

n→∞bn exist

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2 |c_k− c_k−1| < δ, or

3 |f (c_k)| < ε.

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

a = a₀≤ a₁ ≤ a₂ ≤ · · · ≤ b₀= b

⇒ lim

n→∞an and lim

n→∞bn exist

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2 |c_k− c_k−1| < δ, or

3 |f (c_k)| < ε.

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

a = a₀≤ a₁ ≤ a₂ ≤ · · · ≤ b₀= b

⇒ lim

n→∞an and lim

n→∞bn exist

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Since

b1− a₁ = 1

2(b0− a₀) b2− a₂ = 1

2(b1− a₁) = 1

4(b0− a₀) ...

b_n− a_n = 1

2ⁿ(b₀− a₀) hence

n→∞lim b_n− lim

n→∞a_n= lim

n→∞(b_n− a_n) = 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).

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Since

b1− a₁ = 1

2(b0− a₀) b2− a₂ = 1

2(b1− a₁) = 1

4(b0− a₀) ...

b_n− a_n = 1

n→∞a_n= lim

n→∞

1

2ⁿ(b₀− a₀) = 0.

Therefore

n→∞lim an= lim

n→∞bn≡ z.

n→∞bn) = f (z).

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Since

b1− a₁ = 1

2(b0− a₀) b2− a₂ = 1

2(b1− a₁) = 1

4(b0− a₀) ...

b_n− a_n = 1

n→∞a_n= lim

n→∞

1

2ⁿ(b₀− a₀) = 0.

Therefore

n→∞lim an= lim

n→∞bn≡ z.

n→∞bn) = f (z).

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Since

b1− a₁ = 1

2(b0− a₀) b2− a₂ = 1

2(b1− a₁) = 1

4(b0− a₀) ...

b_n− a_n = 1

n→∞a_n= lim

n→∞

1

2ⁿ(b₀− a₀) = 0.

Therefore

n→∞lim an= lim

n→∞bn≡ z.

n→∞bn) = f (z).

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On the other hand,

f (a_n)f (b_n) < 0

⇒ lim

n→∞f (a_n)f (b_n) = f²(z) ≤ 0

⇒ f (z) = 0

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

|z − c_n| = lim

n→∞a_n−1

2(a_n+ b_n)

= 1 2

h

n→∞lim a_n− b_ni + 1

2 h

n→∞lim a_n− a_ni

≤ max lim

n→∞an− b_n ,

lim

n→∞an− a_n

≤ |b_n− a_n| = 1

2ⁿ|b₀− a₀|.

This proves the following theorem.

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On the other hand,

f (a_n)f (b_n) < 0

⇒ lim

n→∞f (a_n)f (b_n) = f²(z) ≤ 0

⇒ f (z) = 0

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

|z − c_n| = lim

n→∞a_n−1

2(a_n+ b_n)

= 1 2

h

2 h

≤ max lim

n→∞an− b_n ,

lim

n→∞an− a_n

≤ |b_n− a_n| = 1

2ⁿ|b₀− a₀|.

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On the other hand,

f (a_n)f (b_n) < 0

⇒ lim

n→∞f (a_n)f (b_n) = f²(z) ≤ 0

⇒ f (z) = 0

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

|z − c_n| = lim

n→∞a_n−1

2(a_n+ b_n)

= 1 2

h

2 h

≤ max lim

n→∞an− b_n ,

lim

n→∞an− a_n

≤ |b_n− a_n| = 1

2ⁿ|b₀− a₀|.

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On the other hand,

f (a_n)f (b_n) < 0

⇒ lim

n→∞f (a_n)f (b_n) = f²(z) ≤ 0

⇒ f (z) = 0

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

|z − c_n| = lim

n→∞a_n−1

2(a_n+ b_n)

= 1 2

h

2 h

≤ max lim

n→∞an− b_n ,

lim

n→∞an− a_n

≤ |b_n− a_n| = 1

2ⁿ|b₀− a₀|.

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Theorem 1

Let{[a_n, 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 − c_n| ≤ 1

2ⁿ(b₀− a₀) . Remark

{c_n}convergestozwith therateofO(2⁻ⁿ).

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Theorem 1

Let{[a_n, 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 − c_n| ≤ 1

2ⁿ(b₀− a₀) . Remark

{c_n}convergestozwith therateofO(2⁻ⁿ).

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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 − c_n|

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

|z − c_n| ≤ 10⁻³. That is, we solve

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

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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 − c_n|

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

|z − c_n| ≤ 10⁻³. That is, we solve

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

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Exercise

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

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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^∗.

⇒ x^∗is 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

⇒ x^∗is a zero of f (x).

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Fixed-Point Iteration

Definition 3

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Fixed-Point Iteration

Definition 3

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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).

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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.

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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.

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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 x^∗in [a, b].

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Proof

Existence:

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

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Proof

Existence:

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

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Proof

Existence:

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

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Proof

Existence:

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

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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.

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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.

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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.

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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].

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Example 6

g(x) = (x²− 1)/3, x ∈ [−1, 1].

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].

|g⁰(x)| =

2x 3

≤ 2

3, ∀ x ∈ (−1, 1).

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Example 6

g(x) = (x²− 1)/3, x ∈ [−1, 1].

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].

|g⁰(x)| =

2x 3

≤ 2

3, ∀ x ∈ (−1, 1).

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Example 6

g(x) = (x²− 1)/3, x ∈ [−1, 1].

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].

|g⁰(x)| =

2x 3

≤ 2

3, ∀ x ∈ (−1, 1).

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Example 6

g(x) = (x²− 1)/3, x ∈ [−1, 1].

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].

|g⁰(x)| =

2x 3

≤ 2

3, ∀ x ∈ (−1, 1).

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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).

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Fixed-point iteration or functional iteration

Given a continuous function g, choose an initial point x₀ and generate {x_k}^∞_k=0by

x_k+1 = g(x_k), k ≥ 0.

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

k→∞lim x_k= x^∗, then, since g is continuous,

g(x^∗) = g( lim

k→∞x_k) = lim

k→∞g(x_k) = lim

k→∞x_k+1= x^∗. That is, x^∗is a fixed point of g.

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x_k+1 = g(x_k), k ≥ 0.

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

g(x^∗) = g( lim

k→∞x_k) = lim

k→∞g(x_k) = lim

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x_k+1 = g(x_k), k ≥ 0.

g(x^∗) = g( lim

k→∞x_k) = lim

k→∞g(x_k) = lim

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x_k+1 = g(x_k), k ≥ 0.

g(x^∗) = g( lim

k→∞x_k) = lim

k→∞g(x_k) = lim

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x_k+1 = g(x_k), k ≥ 0.

g(x^∗) = g( lim

k→∞x_k) = lim

k→∞g(x_k) = lim

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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

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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) = ¹⁰_x − 4x1/2

x³ = 10 − 4x² ⇒ x² = 10

x − 4x ⇒ x = ± 10 x − 4x

1/2

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x³+ 4x²− 10 = 0

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

(b) x = g₂(x) = ¹⁰_x − 4x1/2

x³ = 10 − 4x² ⇒ x² = 10

x − 4x ⇒ x = ± 10 x − 4x

1/2

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x³+ 4x²− 10 = 0

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

(b) x = g₂(x) = ¹⁰_x − 4x1/2

x³ = 10 − 4x² ⇒ x² = 10

x − 4x ⇒ x = ± 10 x − 4x

1/2

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x³+ 4x²− 10 = 0

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

(b) x = g₂(x) = ¹⁰_x − 4x1/2

x³ = 10 − 4x² ⇒ x² = 10

x − 4x ⇒ x = ± 10 x − 4x

1/2

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(c) x = g3(x) = ¹₂ 10 − x³1/2

4x² = 10 − x³ ⇒ x = ±1

2 10 − x³1/2

(d) x = g₄(x) =

10 4+x

1/2

x²(x + 4) = 10 ⇒ x = ±

10 4 + x

1/2

(e) x = g₅(x) = x −^x³_3x^+4x2+8x²⁻¹⁰

x = g5(x) ≡ x − f (x) f⁰(x)

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(c) x = g3(x) = ¹₂ 10 − x³1/2

4x² = 10 − x³ ⇒ x = ±1

2 10 − x³1/2

(d) x = g₄(x) =

10 4+x

1/2

x²(x + 4) = 10 ⇒ x = ±

10 4 + x

1/2

(e) x = g₅(x) = x −^x³_3x^+4x2+8x²⁻¹⁰

x = g5(x) ≡ x − f (x) f⁰(x)

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(c) x = g3(x) = ¹₂ 10 − x³1/2

4x² = 10 − x³ ⇒ x = ±1

2 10 − x³1/2

(d) x = g₄(x) =

10 4+x

1/2

x²(x + 4) = 10 ⇒ x = ±

10 4 + x

1/2

(e) x = g₅(x) = x −^x³_3x^+4x2+8x²⁻¹⁰

x = g5(x) ≡ x − f (x) f⁰(x)

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

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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],

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

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|g⁰(x)| ≤ k, ∀ x ∈ (a, b).

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|g⁰(x)| ≤ k, ∀ x ∈ (a, b).

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

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

|g⁰(x)| ≤ k, we have

|x − x_n| = |g(x_n−1) − g(x)| = |g⁰(ξn)||x − xn−1| ≤ k|x − x_n−1|, where ξn∈ (a, b).It follows that

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

n→∞lim kⁿ= 0 and

n→∞lim |x_n− x| ≤ lim

n→∞kⁿ|x₀− x| = 0.

Hence, {x_n}^∞ converges to x.

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|x − x_n| = |g(x_n−1) − g(x)| = |g⁰(ξn)||x − xn−1| ≤ k|x − x_n−1|, where ξn∈ (a, b). It follows that

n→∞kⁿ|x₀− x| = 0.

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|x − x_n| = |g(x_n−1) − g(x)| = |g⁰(ξn)||x − xn−1| ≤ k|x − x_n−1|, where ξn∈ (a, b).It follows that

n→∞kⁿ|x₀− x| = 0.

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n→∞kⁿ|x₀− x| = 0.

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n→∞kⁿ|x₀− x| = 0.

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Corollary 9

If g satisfies the hypotheses of above theorem, then

|x − x_n| ≤ kⁿmax{x₀− a, b − x₀} and

|x_n− x| ≤ kⁿ

1 − k|x₁− x₀|, ∀ n ≥ 1.

Proof: From (1),

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

For n ≥ 1, using the Mean Values Theorem,

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

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Corollary 9

|x_n− x| ≤ kⁿ

1 − k|x₁− x₀|, ∀ n ≥ 1.

Proof: From (1),

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Corollary 9

|x_n− x| ≤ kⁿ

1 − k|x₁− x₀|, ∀ n ≥ 1.

Proof: From (1),

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Thus, for m > n ≥ 1,

|x_m− x_n| = |x_m− x_m−1+ xm−1− · · · + x_n+1− x_n|

≤ |x_m− x_m−1| + |x_m−1− x_m−2| + · · · + |x_n+1− x_n|

≤ k^m−1|x₁− x₀| + k^m−2|x₁− x₀| + · · · + kⁿ|x₁− x₀|

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

|x − x_n| = lim

m→∞|x_m− x_n| ≤ lim

m→∞kⁿ|x₁− x₀|

m−n−1

X

j=0

k^j

≤ kⁿ|x₁− x₀|

∞

X

j=0

k^j = kⁿ

1 − k|x₁− x₀|.

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Thus, for m > n ≥ 1,

|x_m− x_n| = |x_m− x_m−1+ xm−1− · · · + x_n+1− x_n|

≤ |x_m− x_m−1| + |x_m−1− x_m−2| + · · · + |x_n+1− x_n|

≤ k^m−1|x₁− x₀| + k^m−2|x₁− x₀| + · · · + kⁿ|x₁− x₀|

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

|x − x_n| = lim

m→∞|x_m− x_n| ≤ lim

m→∞kⁿ|x₁− x₀|

m−n−1

X

j=0

k^j

≤ kⁿ|x₁− x₀|

∞

X

j=0

k^j = kⁿ

1 − k|x₁− x₀|.

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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

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Example 10

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]

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Example 10

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]

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Example 10

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]

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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.

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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].

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

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].

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

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].

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

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].

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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.

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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].

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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].

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Exercise

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

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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^∗.

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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).

f (x) + f⁰(x)h ≈ 0 and h ≈ −f (x)

f⁰(x), if f⁰(x) 6= 0.

Hence

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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).

f (x) + f⁰(x)h ≈ 0 and h ≈ −f (x)

f⁰(x), if f⁰(x) 6= 0.

Hence