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:

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:

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:

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:

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

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.

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.

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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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

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

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

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

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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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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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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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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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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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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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 − 4x
1/2

x³ = 10 − 4x² ⇒ x² = 10

x − 4x ⇒ x = ± 10 x − 4x

1/2

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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 − 4x
1/2

x³ = 10 − 4x² ⇒ x² = 10

x − 4x ⇒ x = ± 10 x − 4x

1/2

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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 − 4x
1/2

x³ = 10 − 4x² ⇒ x² = 10

x − 4x ⇒ x = ± 10 x − 4x

1/2

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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 − 4x
1/2

x³ = 10 − 4x² ⇒ x² = 10

x − 4x ⇒ x = ± 10 x − 4x

1/2

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(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 −^x3_3x^+4x2+8x²⁻¹⁰

x = g5(x) ≡ x − f (x) f⁰(x)

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(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 −^x3_3x^+4x2+8x²⁻¹⁰

x = g5(x) ≡ x − f (x) f⁰(x)

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(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 −^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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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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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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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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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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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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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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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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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

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

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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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

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

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

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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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].

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].

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].

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].

Hence, the sequence is convergent to the fixed point.

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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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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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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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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).

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).

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).

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).

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^∗.

師大

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^∗.

師大

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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This sets the stage for theNewton-Rapbson’smethod, which starts with an initial approximation x₀ and generates the sequence {x_n}^∞_n=0defined by

xn+1 = xn− f (xn) f⁰(xn).

Since the Taylor’s expansion of f (x) at x_nis given by f (x) = f (xn) + f⁰(xn)(x − xn) +1

2f⁰⁰(xn)(x − xn)²+ · · · . At xn, one uses thetangent line

y = `(x) = f (xn) + f⁰(xn)(x − xn)

toapproximate the curveof f (x) and uses the zero of the tangent line to approximate the zero of f (x).

師大

This sets the stage for theNewton-Rapbson’smethod, which starts with an initial approximation x₀ and generates the sequence {x_n}^∞_n=0defined by

xn+1 = xn− f (xn) f⁰(xn).

Since the Taylor’s expansion of f (x) at x_nis given by f (x) = f (xn) + f⁰(xn)(x − xn) +1

2f⁰⁰(xn)(x − xn)²+ · · · . At xn, one uses thetangent line

y = `(x) = f (xn) + f⁰(xn)(x − xn)

toapproximate the curveof f (x) and uses the zero of the tangent line to approximate the zero of f (x).

師大

This sets the stage for theNewton-Rapbson’smethod, which starts with an initial approximation x₀ and generates the sequence {x_n}^∞_n=0defined by

xn+1 = xn− f (xn) f⁰(xn).

Since the Taylor’s expansion of f (x) at x_nis given by f (x) = f (xn) + f⁰(xn)(x − xn) +1

2f⁰⁰(xn)(x − xn)²+ · · · . At xn, one uses thetangent line

y = `(x) = f (xn) + f⁰(xn)(x − xn)

toapproximate the curveof f (x) and uses the zero of the tangent line to approximate the zero of f (x).

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Newton’s Method

Given x₀, tolerance T OL, maximum number of iteration M . Set n = 1 and x = x₀− f (x₀)/f⁰(x₀).

While n ≤ M and |x − x₀| ≥ T OL

Set n = n + 1, x₀= xand x = x₀− f (x₀)/f⁰(x₀).

End While

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Three stopping-technique inequalities

(a). |x_n− x_n−1| < ε, (b). |x_n− x_n−1|

|x_n| < ε, xn6= 0, (c). |f (x_n)| < ε.

Note that Newton’s method for solving f (x) = 0 xn+1= xn− f (x_n)

f⁰(xn), for n ≥ 1 is just a special case of functional iteration in which

g(x) = x − f (x) f⁰(x).

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Three stopping-technique inequalities

(a). |x_n− x_n−1| < ε, (b). |x_n− x_n−1|

|x_n| < ε, xn6= 0, (c). |f (x_n)| < ε.

Note that Newton’s method for solving f (x) = 0 xn+1= xn− f (x_n)

f⁰(xn), for n ≥ 1 is just a special case of functional iteration in which

g(x) = x − f (x) f⁰(x).

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Example 11

The following table shows the convergence behavior of

Newton’s method applied to solving f (x) = x²− 1 = 0. Observe the quadratic convergence rate.

n xn |e_n| ≡ |1 − x_n|

0 2.0 1

1 1.25 0.25

2 1.025 2.5e-2

3 1.0003048780488 3.048780488e-4 4 1.0000000464611 4.64611e-8

5 1.0 0

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Theorem 12

Assumef (x^∗) = 0, f⁰(x^∗) 6= 0andf (x),f⁰(x)andf⁰⁰(x)are continuouson N_ε(x^∗).Then if x₀is chosensufficiently closeto x^∗, then



xn+1 = xn− f (xn) f⁰(x_n)



→ x^∗.

Proof: Define

g(x) = x − f (x) f⁰(x). Find an interval [x^∗− δ, x^∗+ δ]such that

g([x^∗− δ, x^∗+ δ]) ⊆ [x^∗− δ, x^∗+ δ]

and

|g⁰(x)| ≤ k < 1, ∀ x ∈ (x^∗− δ, x^∗+ δ).

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Theorem 12

Assumef (x^∗) = 0, f⁰(x^∗) 6= 0andf (x),f⁰(x)andf⁰⁰(x)are continuouson N_ε(x^∗). Then if x₀is chosensufficiently closeto x^∗,then



xn+1 = xn− f (xn) f⁰(x_n)



→ x^∗.

Proof: Define

g(x) = x − f (x) f⁰(x). Find an interval [x^∗− δ, x^∗+ δ]such that

g([x^∗− δ, x^∗+ δ]) ⊆ [x^∗− δ, x^∗+ δ]

and

|g⁰(x)| ≤ k < 1, ∀ x ∈ (x^∗− δ, x^∗+ δ).

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Theorem 12

Assumef (x^∗) = 0, f⁰(x^∗) 6= 0andf (x),f⁰(x)andf⁰⁰(x)are continuouson N_ε(x^∗). Then if x₀is chosensufficiently closeto x^∗, then



xn+1 = xn− f (xn) f⁰(x_n)



→ x^∗.

Proof: Define

g(x) = x − f (x) f⁰(x). Find an interval [x^∗− δ, x^∗+ δ]such that

g([x^∗− δ, x^∗+ δ]) ⊆ [x^∗− δ, x^∗+ δ]

and

|g⁰(x)| ≤ k < 1, ∀ x ∈ (x^∗− δ, x^∗+ δ).

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Theorem 12

Assumef (x^∗) = 0, f⁰(x^∗) 6= 0andf (x),f⁰(x)andf⁰⁰(x)are continuouson N_ε(x^∗). Then if x₀is chosensufficiently closeto x^∗, then



xn+1 = xn− f (xn) f⁰(x_n)



→ x^∗.

Proof: Define

g(x) = x − f (x) f⁰(x). Find an interval [x^∗− δ, x^∗+ δ]such that

g([x^∗− δ, x^∗+ δ]) ⊆ [x^∗− δ, x^∗+ δ]

and

|g⁰(x)| ≤ k < 1, ∀ x ∈ (x^∗− δ, x^∗+ δ).

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Theorem 12

Assumef (x^∗) = 0, f⁰(x^∗) 6= 0andf (x),f⁰(x)andf⁰⁰(x)are continuouson N_ε(x^∗). Then if x₀is chosensufficiently closeto x^∗, then



xn+1 = xn− f (xn) f⁰(x_n)



→ x^∗.

Proof: Define

g(x) = x − f (x) f⁰(x). Find an interval [x^∗− δ, x^∗+ δ]such that

g([x^∗− δ, x^∗+ δ]) ⊆ [x^∗− δ, x^∗+ δ]

and

|g⁰(x)| ≤ k < 1, ∀ x ∈ (x^∗− δ, x^∗+ δ).

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Since f⁰ is continuous and f⁰(x^∗) 6= 0, it implies that ∃ δ₁ > 0 such that f⁰(x) 6= 0 ∀ x ∈ [x^∗− δ₁, x^∗+ δ₁] ⊆ [a, b].Thus, g is defined and continuous on [x^∗− δ₁, x^∗+ δ1]. Also

g⁰(x) = 1 − f⁰(x)f⁰(x) − f (x)f⁰⁰(x)

[f⁰(x)]² = f (x)f⁰⁰(x) [f⁰(x)]² , for x ∈ [x^∗− δ₁, x^∗+ δ1].Since f⁰⁰ is continuous on [a, b], we have g⁰ is continuous on [x^∗− δ₁, x^∗+ δ₁].

By assumption f (x^∗) = 0, so

g⁰(x^∗) = f (x^∗)f⁰⁰(x^∗)

|f⁰(x^∗)|² = 0.

Since g⁰ is continuous on [x^∗− δ₁, x^∗+ δ1]and g⁰(x^∗) = 0, ∃ δ with 0 < δ < δ1 and k ∈ (0, 1) such that

|g⁰(x)| ≤ k, ∀ x ∈ [x^∗− δ, x^∗+ δ].

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Since f⁰ is continuous and f⁰(x^∗) 6= 0, it implies that ∃ δ₁ > 0 such that f⁰(x) 6= 0 ∀ x ∈ [x^∗− δ₁, x^∗+ δ₁] ⊆ [a, b]. Thus, g is defined and continuous on [x^∗− δ₁, x^∗+ δ1].Also

g⁰(x) = 1 − f⁰(x)f⁰(x) − f (x)f⁰⁰(x)

[f⁰(x)]² = f (x)f⁰⁰(x) [f⁰(x)]² , for x ∈ [x^∗− δ₁, x^∗+ δ1]. Since f⁰⁰ is continuous on [a, b], we have g⁰ is continuous on [x^∗− δ₁, x^∗+ δ₁].

By assumption f (x^∗) = 0, so

g⁰(x^∗) = f (x^∗)f⁰⁰(x^∗)

|f⁰(x^∗)|² = 0.

Since g⁰ is continuous on [x^∗− δ₁, x^∗+ δ1]and g⁰(x^∗) = 0, ∃ δ with 0 < δ < δ1 and k ∈ (0, 1) such that

|g⁰(x)| ≤ k, ∀ x ∈ [x^∗− δ, x^∗+ δ].

師大

Since f⁰ is continuous and f⁰(x^∗) 6= 0, it implies that ∃ δ₁ > 0 such that f⁰(x) 6= 0 ∀ x ∈ [x^∗− δ₁, x^∗+ δ₁] ⊆ [a, b]. Thus, g is defined and continuous on [x^∗− δ₁, x^∗+ δ1]. Also

g⁰(x) = 1 − f⁰(x)f⁰(x) − f (x)f⁰⁰(x)

[f⁰(x)]² = f (x)f⁰⁰(x) [f⁰(x)]² , for x ∈ [x^∗− δ₁, x^∗+ δ1].Since f⁰⁰ is continuous on [a, b], we have g⁰ is continuous on [x^∗− δ₁, x^∗+ δ₁].

By assumption f (x^∗) = 0, so

g⁰(x^∗) = f (x^∗)f⁰⁰(x^∗)

|f⁰(x^∗)|² = 0.

Since g⁰ is continuous on [x^∗− δ₁, x^∗+ δ1]and g⁰(x^∗) = 0, ∃ δ with 0 < δ < δ1 and k ∈ (0, 1) such that

|g⁰(x)| ≤ k, ∀ x ∈ [x^∗− δ, x^∗+ δ].

師大

Since f⁰ is continuous and f⁰(x^∗) 6= 0, it implies that ∃ δ₁ > 0 such that f⁰(x) 6= 0 ∀ x ∈ [x^∗− δ₁, x^∗+ δ₁] ⊆ [a, b]. Thus, g is defined and continuous on [x^∗− δ₁, x^∗+ δ1]. Also

g⁰(x) = 1 − f⁰(x)f⁰(x) − f (x)f⁰⁰(x)

[f⁰(x)]² = f (x)f⁰⁰(x) [f⁰(x)]² , for x ∈ [x^∗− δ₁, x^∗+ δ1]. Since f⁰⁰ is continuous on [a, b], we have g⁰ is continuous on [x^∗− δ₁, x^∗+ δ₁].

By assumption f (x^∗) = 0, so

g⁰(x^∗) = f (x^∗)f⁰⁰(x^∗)

|f⁰(x^∗)|² = 0.

Since g⁰ is continuous on [x^∗− δ₁, x^∗+ δ1]and g⁰(x^∗) = 0,∃ δ with 0 < δ < δ1 and k ∈ (0, 1) such that

|g⁰(x)| ≤ k, ∀ x ∈ [x^∗− δ, x^∗+ δ].

師大

Since f⁰ is continuous and f⁰(x^∗) 6= 0, it implies that ∃ δ₁ > 0 such that f⁰(x) 6= 0 ∀ x ∈ [x^∗− δ₁, x^∗+ δ₁] ⊆ [a, b]. Thus, g is defined and continuous on [x^∗− δ₁, x^∗+ δ1]. Also

g⁰(x) = 1 − f⁰(x)f⁰(x) − f (x)f⁰⁰(x)

[f⁰(x)]² = f (x)f⁰⁰(x) [f⁰(x)]² , for x ∈ [x^∗− δ₁, x^∗+ δ1]. Since f⁰⁰ is continuous on [a, b], we have g⁰ is continuous on [x^∗− δ₁, x^∗+ δ₁].

By assumption f (x^∗) = 0, so

g⁰(x^∗) = f (x^∗)f⁰⁰(x^∗)

|f⁰(x^∗)|² = 0.

Since g⁰ is continuous on [x^∗− δ₁, x^∗+ δ1]and g⁰(x^∗) = 0, ∃ δ with 0 < δ < δ1 and k ∈ (0, 1) such that

|g⁰(x)| ≤ k, ∀ x ∈ [x^∗− δ, x^∗+ δ].

師大

Since f⁰ is continuous and f⁰(x^∗) 6= 0, it implies that ∃ δ₁ > 0 such that f⁰(x) 6= 0 ∀ x ∈ [x^∗− δ₁, x^∗+ δ₁] ⊆ [a, b]. Thus, g is defined and continuous on [x^∗− δ₁, x^∗+ δ1]. Also

g⁰(x) = 1 − f⁰(x)f⁰(x) − f (x)f⁰⁰(x)

[f⁰(x)]² = f (x)f⁰⁰(x) [f⁰(x)]² , for x ∈ [x^∗− δ₁, x^∗+ δ1]. Since f⁰⁰ is continuous on [a, b], we have g⁰ is continuous on [x^∗− δ₁, x^∗+ δ₁].

By assumption f (x^∗) = 0, so

g⁰(x^∗) = f (x^∗)f⁰⁰(x^∗)

|f⁰(x^∗)|² = 0.

Since g⁰ is continuous on [x^∗− δ₁, x^∗+ δ1]and g⁰(x^∗) = 0,∃ δ with 0 < δ < δ1 and k ∈ (0, 1) such that

|g⁰(x)| ≤ k, ∀ x ∈ [x^∗− δ, x^∗+ δ].

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Since f⁰ is continuous and f⁰(x^∗) 6= 0, it implies that ∃ δ₁ > 0 such that f⁰(x) 6= 0 ∀ x ∈ [x^∗− δ₁, x^∗+ δ₁] ⊆ [a, b]. Thus, g is defined and continuous on [x^∗− δ₁, x^∗+ δ1]. Also

g⁰(x) = 1 − f⁰(x)f⁰(x) − f (x)f⁰⁰(x)

[f⁰(x)]² = f (x)f⁰⁰(x) [f⁰(x)]² , for x ∈ [x^∗− δ₁, x^∗+ δ1]. Since f⁰⁰ is continuous on [a, b], we have g⁰ is continuous on [x^∗− δ₁, x^∗+ δ₁].

By assumption f (x^∗) = 0, so

g⁰(x^∗) = f (x^∗)f⁰⁰(x^∗)

|f⁰(x^∗)|² = 0.

Since g⁰ is continuous on [x^∗− δ₁, x^∗+ δ1]and g⁰(x^∗) = 0, ∃ δ with 0 < δ < δ1 and k ∈ (0, 1) such that

|g⁰(x)| ≤ k, ∀ x ∈ [x^∗− δ, x^∗+ δ].

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Claim: g([x^∗− δ, x^∗+ δ]) ⊆ [x^∗− δ, x^∗+ δ].

If x ∈ [x^∗− δ, x^∗+ δ], then, by the Mean Value Theorem, ∃ ξ between x and x^∗ such that

|g(x) − g(x^∗)| = |g⁰(ξ)||x − x^∗|.

It implies that

|g(x) − x^∗| = |g(x) − g(x^∗)| = |g⁰(ξ)||x − x^∗|

≤ k|x − x^∗| < |x − x^∗| < δ.

Hence, g([x^∗− δ, x^∗+ δ]) ⊆ [x^∗− δ, x^∗+ δ].

By the Fixed-Point Theorem, the sequence {x_n}^∞_n=0defined by x_n= g(x_n−1) = x_n−1− f (x_n−1)

f⁰(xn−1), for n ≥ 1, converges to x^∗ for any x0 ∈ [x^∗− δ, x^∗+ δ].

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Claim: g([x^∗− δ, x^∗+ δ]) ⊆ [x^∗− δ, x^∗+ δ].

If x ∈ [x^∗− δ, x^∗+ δ], then, by the Mean Value Theorem, ∃ ξ between x and x^∗ such that

|g(x) − g(x^∗)| = |g⁰(ξ)||x − x^∗|.

It implies that

|g(x) − x^∗| = |g(x) − g(x^∗)| = |g⁰(ξ)||x − x^∗|

≤ k|x − x^∗| < |x − x^∗| < δ.

Hence, g([x^∗− δ, x^∗+ δ]) ⊆ [x^∗− δ, x^∗+ δ].

By the Fixed-Point Theorem, the sequence {x_n}^∞_n=0defined by x_n= g(x_n−1) = x_n−1− f (x_n−1)

f⁰(xn−1), for n ≥ 1, converges to x^∗ for any x0 ∈ [x^∗− δ, x^∗+ δ].

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Claim: g([x^∗− δ, x^∗+ δ]) ⊆ [x^∗− δ, x^∗+ δ].

If x ∈ [x^∗− δ, x^∗+ δ], then, by the Mean Value Theorem, ∃ ξ between x and x^∗ such that

|g(x) − g(x^∗)| = |g⁰(ξ)||x − x^∗|.

It implies that

|g(x) − x^∗| = |g(x) − g(x^∗)| = |g⁰(ξ)||x − x^∗|

≤ k|x − x^∗| < |x − x^∗| < δ.

Hence, g([x^∗− δ, x^∗+ δ]) ⊆ [x^∗− δ, x^∗+ δ].

By the Fixed-Point Theorem, the sequence {x_n}^∞_n=0defined by x_n= g(x_n−1) = x_n−1− f (x_n−1)

f⁰(xn−1), for n ≥ 1, converges to x^∗ for any x0 ∈ [x^∗− δ, x^∗+ δ].

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Claim: g([x^∗− δ, x^∗+ δ]) ⊆ [x^∗− δ, x^∗+ δ].

If x ∈ [x^∗− δ, x^∗+ δ], then, by the Mean Value Theorem, ∃ ξ between x and x^∗ such that

|g(x) − g(x^∗)| = |g⁰(ξ)||x − x^∗|.

It implies that

|g(x) − x^∗| = |g(x) − g(x^∗)| = |g⁰(ξ)||x − x^∗|

≤ k|x − x^∗| < |x − x^∗| < δ.

Hence, g([x^∗− δ, x^∗+ δ]) ⊆ [x^∗− δ, x^∗+ δ].

By the Fixed-Point Theorem, the sequence {x_n}^∞_n=0defined by x_n= g(x_n−1) = x_n−1− f (x_n−1)

f⁰(xn−1), for n ≥ 1, converges to x^∗ for any x0 ∈ [x^∗− δ, x^∗+ δ].

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Claim: g([x^∗− δ, x^∗+ δ]) ⊆ [x^∗− δ, x^∗+ δ].

If x ∈ [x^∗− δ, x^∗+ δ], then, by the Mean Value Theorem, ∃ ξ between x and x^∗ such that

|g(x) − g(x^∗)| = |g⁰(ξ)||x − x^∗|.

It implies that

|g(x) − x^∗| = |g(x) − g(x^∗)| = |g⁰(ξ)||x − x^∗|

≤ k|x − x^∗| < |x − x^∗| < δ.

Hence, g([x^∗− δ, x^∗+ δ]) ⊆ [x^∗− δ, x^∗+ δ].

By the Fixed-Point Theorem, the sequence {x_n}^∞_n=0defined by x_n= g(x_n−1) = x_n−1− f (x_n−1)

f⁰(xn−1), for n ≥ 1, converges to x^∗ for any x0 ∈ [x^∗− δ, x^∗+ δ].

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Example 13

When Newton’s method applied to f (x) = cos x with starting point x₀ = 3, which is close to the root ^π₂ of f , it produces x₁ = −4.01525, x₂= −4.8526, · · · ,which converges to another root −^3π₂ .

5 4 3 2 1 0 1 2 3 4 5

1. 5 0 1.5

x₀ y = c os ( x)

x^*

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在文檔中 Solutions of Equations in One Variable (頁 32-118)