(4)• We have x(1

Roots of polynomials

Lesson 1-2

Example

• εx² − x + 1 = 0, with small ε > 0.

Example

• εx² − x + 1 = 0, with small ε > 0.

• Consider the iteration

x^(I) − 1 = ε(x^(I−1))²,

where the initial condition is defined as a root of the initial equation with ε = 0. So

x⁽⁰⁾ = 1.

• We have

x⁽¹⁾ = 1 + ε(x⁽⁰⁾)² = 1 + ε, x⁽²⁾ = 1 + ε(1 + ε)² = 1 + ε + 2ε² + ε³, x⁽³⁾ = 1 + ε(1 + ε + 2ε² + ε³)²

= 1 + ε + 2ε² + 5ε³ + 6ε⁴ + 6ε⁵ + 4ε⁶ + ε⁷.

• We have

x⁽¹⁾ = 1 + ε(x⁽⁰⁾)² = 1 + ε, x⁽²⁾ = 1 + ε(1 + ε)² = 1 + ε + 2ε² + ε³, x⁽³⁾ = 1 + ε(1 + ε + 2ε² + ε³)²

= 1 + ε + 2ε² + 5ε³ + 6ε⁴ + 6ε⁵ + 4ε⁶ + ε⁷.

• The problem is that the coefficient before ε^k is

changed from step to step. It would be nice to have

x = x + ε¹x + ε²x + ε³x + . . . .

Example

Substitute x = x₀ + ε¹x₁ + ε²x₂ + ε³x₃ + . . . into

εx² − x + 1 = 0. We get

ε(x₀ + ε¹x₁ + ε²x₂ + ε³x₃ + . . .)²

− (x₀ + ε¹x₁ + ε²x₂ + ε³x₃ + . . .) + 1 = 0

Since the right hand side is ₀, we get

ε⁰ : −x₀ + 1 = 0 ⇒ x₀ = 1,

1 2

Example

The following perturbation expansion for the solution of the equation εx² − x + 1 = 0 is deduced

x = 1 + ε¹ + 2ε² + 5ε³ + . . .

and it agrees with

x⁽³⁾ = 1 + ε + 2ε² + 5ε³ + 6ε⁴ + 6ε⁵ + 4ε⁶ + ε⁷

up to terms of order ε³.

Regular perturbation expansion

• Let us see on the exact solution of εx² − x + 1 = 0

x_1,2 = 1 ± √

1 − 4ε 2ε .

Regular perturbation expansion

• Let us see on the exact solution of εx² − x + 1 = 0

x_1,2 = 1 ± √

1 − 4ε 2ε .

• Using the formular (1 + δ)^µ = 1 + µδ + ^µ(µ_2!⁻¹⁾δ² . . .

we get

1 ± (1 − 2ε − 2ε² . . .)

2ε −→

Regular perturbation expansion

• Let us see on the exact solution of εx² − x + 1 = 0

x_1,2 = 1 ± √

1 − 4ε 2ε .

• Using the formular (1 + δ)^µ = 1 + µδ + ^µ(µ_2!⁻¹⁾δ² . . .

we get

1 ± (1 − 2ε − 2ε² . . .)

2ε −→

• x = 1 + ε + . . . → x = 1 as ε → 0,

Regular perturbation expansion

• Let us see on the exact solution of εx² − x + 1 = 0

x_1,2 = 1 ± √

1 − 4ε 2ε .

• Using the formular (1 + δ)^µ = 1 + µδ + ^µ(µ_2!⁻¹⁾δ² . . .

we get

1 ± (1 − 2ε − 2ε² . . .)

2ε −→

• x = 1 + ε + . . . → x = 1 as ε → 0,

• x = ¹ − 1 − ε + . . .

Singular perturbation expansion

• Consider εx³ + x² − 1 = 0. Substituting the standard expansion x = x₀ + ε¹x₁ + ε²x₂ + . . . into the

equation we get

Singular perturbation expansion

• Consider εx³ + x² − 1 = 0. Substituting the standard expansion x = x₀ + ε¹x₁ + ε²x₂ + . . . into the

equation we get

• ε⁰ : x²₀ − 1 = 0 −→ x₀ = ±1,

Singular perturbation expansion

• Consider εx³ + x² − 1 = 0. Substituting the standard expansion x = x₀ + ε¹x₁ + ε²x₂ + . . . into the

equation we get

• ε⁰ : x²₀ − 1 = 0 −→ x₀ = ±1,

• ε¹ : x³₀ + 2x₀x₁ = 0 −→ x₁ = −^x₂²⁰ = −¹₂,

Singular perturbation expansion

• Consider εx³ + x² − 1 = 0. Substituting the standard expansion x = x₀ + ε¹x₁ + ε²x₂ + . . . into the

equation we get

• ε⁰ : x²₀ − 1 = 0 −→ x₀ = ±1,

• ε¹ : x³₀ + 2x₀x₁ = 0 −→ x₁ = −^x₂²⁰ = −¹₂,

• ε² : 3x²₀x₁ + 2x₀x₂ + x²₁ = 0 −→ x₂ = _8x⁵

0 .

Singular perturbation expansion

• Consider εx³ + x² − 1 = 0. Substituting the standard expansion x = x₀ + ε¹x₁ + ε²x₂ + . . . into the

equation we get

• ε⁰ : x²₀ − 1 = 0 −→ x₀ = ±1,

• ε¹ : x³₀ + 2x₀x₁ = 0 −→ x₁ = −^x₂²⁰ = −¹₂,

• ε² : 3x²₀x₁ + 2x₀x₂ + x²₁ = 0 −→ x₂ = _8x⁵

0 .

• x = 1 − ₂^ε + ^5ε₈² + . . . and x = −1 − ₂^ε − ^5ε₈² + . . .,

Singular perturbation expansion

• Consider εx³ + x² − 1 = 0. Substituting the standard expansion x = x₀ + ε¹x₁ + ε²x₂ + . . . into the

equation we get

• ε⁰ : x²₀ − 1 = 0 −→ x₀ = ±1,

• ε¹ : x³₀ + 2x₀x₁ = 0 −→ x₁ = −^x₂²⁰ = −¹₂,

• ε² : 3x²₀x₁ + 2x₀x₂ + x²₁ = 0 −→ x₂ = _8x⁵

0 .

• x = 1 − ₂^ε + ^5ε₈² + . . . and x = −1 − ₂^ε − ^5ε₈² + . . .,

• What about the third equation?

Singular perturbation expansion

• The third equation is associated with large values

x. So the principal equation is εx³ + x² = 0 for large

x. The nontrivial solution is x = −¹_ε. We look for the expansion

x = x₀

ε + x₁ + εx₂ + . . . , expecting x₀ = −1.

Singular perturbation expansion

• The third equation is associated with large values

x. So the principal equation is εx³ + x² = 0 for large

x. The nontrivial solution is x = −¹_ε. We look for the expansion

x = x₀

ε + x₁ + εx₂ + . . . , expecting x₀ = −1.

• Substitute the expansion into εx³ + x² − 1 = 0. Get

ε⁻² : x³₀ + x²₀ = 0 → x₀ = −1,

1 2

Singular perturbation expansion

• The third solution is x = ⁻_ε¹ + ε + . . .. We expect that powers of ε will increase by 2.

Singular perturbation expansion

• The third solution is x = ⁻_ε¹ + ε + . . .. We expect that powers of ε will increase by 2.

• Let us make a substitution x = ^z_ε and seek a

perturbation expansion for z³ + z² − ε² = 0. We take the form z = z₀ + ε²z₁ + ε⁴z₂ + . . ..

Singular perturbation expansion

• The third solution is x = ⁻_ε¹ + ε + . . .. We expect that powers of ε will increase by 2.

• Let us make a substitution x = ^z_ε and seek a

perturbation expansion for z³ + z² − ε² = 0. We take the form z = z₀ + ε²z₁ + ε⁴z₂ + . . ..

• Put the expansion into the equation and deduce

ε⁰ : z₀³ + z₀² = 0 → z₀ = −1, (we reject z₀ = 0) ε² : 3z²z + 2z z − 1 = 0 → z = 1,

Singular perturbation expansion

• The third solution is x = ⁻_ε¹ + ε + . . .. We expect that powers of ε will increase by 2.

• Let us make a substitution x = ^z_ε and seek a

perturbation expansion for z³ + z² − ε² = 0. We take the form z = z₀ + ε²z₁ + ε⁴z₂ + . . ..

• Put the expansion into the equation and deduce

ε⁰ : z₀³ + z₀² = 0 → z₀ = −1, (we reject z₀ = 0) ε² : 3z₀²z₁ + 2z₀z₁ − 1 = 0 → z₁ = 1,

Singular perturbation expansion

• Try to use for εx³ + x² − 1 = 0 the expansions

x = x₀

ε² + x₁ + εx₂ + . . . ,

or

x = x₀

ε^1/2 + x₁ + εx₂ + . . . ,

Singular perturbation expansion

• Try to use for εx³ + x² − 1 = 0 the expansions

x = x₀

ε² + x₁ + εx₂ + . . . ,

or

x = x₀

ε^1/2 + x₁ + εx₂ + . . . ,

• In both cases it leads to take x₀ = 0.

Singular perturbation expansion

• Try to use for εx³ + x² − 1 = 0 the expansions

x = x₀

ε² + x₁ + εx₂ + . . . ,

or

x = x₀

ε^1/2 + x₁ + εx₂ + . . . ,

• In both cases it leads to take x₀ = 0.

• We have to try to preserve the two leading terms

εx³ and x² (the principle of least degeneracy).

Two singular perturbation expansion

• εx³ − x + 1 = 0, 0 < ε ≪ 1.

Two singular perturbation expansion

• εx³ − x + 1 = 0, 0 < ε ≪ 1.

• To find the regular solution we take x₀ = 1 (Why?) and the standard expansion

x = x₀ + ε¹x₁ + ε²x₂ + . . . .

We get ε⁰ : −x₀ + 1 = 0 → x₀ = −1, ε¹ : x³ − x₁ = 0 → x₁ = 1,

Two singular perturbation expansion

• εx³ − x + 1 = 0, 0 < ε ≪ 1.

• To find the regular solution we take x₀ = 1 (Why?) and the standard expansion

x = x₀ + ε¹x₁ + ε²x₂ + . . . .

Two singular perturbation expansion

If we proceed as in the previous example, we take en expansion x = ^x_ε⁰ + x₁ + εx₂ + . . ., substitute into

equation εx³ − x + 1 and get

εx³₀

ε³ + 3x²₀x₁

ε² + . . .

− x₀

ε + . . . + 1 = 0.

What is bad?

Two singular perturbation expansion

If we proceed as in the previous example, we take en expansion x = ^x_ε⁰ + x₁ + εx₂ + . . ., substitute into

equation εx³ − x + 1 and get

εx³₀

ε³ + 3x²₀x₁

ε² + . . .

− x₀

ε + . . . + 1 = 0.

What is bad?

x₀ = 0. Let us try the dominate term ^x_εp⁰ . We get the dominate term after substitution

x³ x

Two singular perturbation expansion

The asymptotic sequence is _{ε^−1/2, ε⁰, ε^1/2, ε¹, ε^3/2} and the corresponding expansion

x = x₀

ε^1/2 + x₁ + ε^1/2x₂ + ε¹x₃ + . . . .

Two singular perturbation expansion

The asymptotic sequence is _{ε^−1/2, ε⁰, ε^1/2, ε¹, ε^3/2} and the corresponding expansion

x = x₀

ε^1/2 + x₁ + ε^1/2x₂ + ε¹x₃ + . . . .

Then ε^−1/2 : x³₀ − x₀ = 0 ⇒ x₀ = ±1,

ε⁰ : 3x²₀x₁ − x₁ + 1 = 0 ⇒ x₁ = −1/2,

ε^1/2 : 3x²₀x₂ + 3x₀x²₁ − x₂ = 0 → x₂ = ∓3/8, . . .

Two singular perturbation expansion

Two singular expansions are

x = 1

ε^1/2 − 1

2 − 3ε^1/2

8 + . . . , x = 1

ε^1/2 − 1

2 − 3ε^1/2

8 + . . . .

ε as a free term

• x³ − x + ε = 0, 0 < ε ≪ 1

ε as a free term

• x³ − x + ε = 0, 0 < ε ≪ 1

• x = x₀ + ε¹x₁ + ε²x₂ + . . .

ε as a free term

• x³ − x + ε = 0, 0 < ε ≪ 1

• x = x₀ + ε¹x₁ + ε²x₂ + . . .

•

ε⁰ : x³₀ − x₀ = 0 ⇒ x₀ = ±1, 0

ε¹ : 3x²₀x₁ − x₁ + 1 = 0 ⇒ x₁ = 1

1 − 3x²₀ , ε² : 3x²₀x₂ + 3x₀x²₁ − x₂ = 0 → x₂ = 3x₀x²₁

1 − 3x²₀ , . . .

ε as a free term

•

x₀ = 0, x₁ = 1, x₂ = 0,

x₀ = 1, x₁ = −1/2, x₂ = −3/8, x₀ = −1, x₁ = −1/2, x₂ = 3/8.

ε as a free term

•

x₀ = 0, x₁ = 1, x₂ = 0,

x₀ = 1, x₁ = −1/2, x₂ = −3/8, x₀ = −1, x₁ = −1/2, x₂ = 3/8.

•

x = 1 − ε/2 − 3ε²/8 + O(ε³), x = −1 − ε/2 + 3ε²/8 + O(ε³), x = ε + O(ε³).

ε as a free term

•

x₀ = 0, x₁ = 1, x₂ = 0,

x₀ = 1, x₁ = −1/2, x₂ = −3/8, x₀ = −1, x₁ = −1/2, x₂ = 3/8.

•

x = 1 − ε/2 − 3ε²/8 + O(ε³), x = −1 − ε/2 + 3ε²/8 + O(ε³),

ε in a middle term

• x⁴ + εx³ − 5x² + 4 = 0, 0 < ε ≪ 1

ε in a middle term

• x⁴ + εx³ − 5x² + 4 = 0, 0 < ε ≪ 1

• x = x₀ + ε¹x₁ + ε²x₂ + . . .

ε in a middle term

• x⁴ + εx³ − 5x² + 4 = 0, 0 < ε ≪ 1

• x = x₀ + ε¹x₁ + ε²x₂ + . . .

• Even the first equation is not linear, it is easy to solve

ε⁰ : x⁴₀ − 5x²₀ + 4 = (x²₀ − 4)(x²₀ − 1) = 0 ⇒ x₀ = ±1, ±2, ε¹ : 4x³₀x₁ + x³₀ − 10x₀x₁ = 0 ⇒ x₁ = 1

6, −2

3 , . . .

ε in a middle term

• x⁴ + εx³ − 5x² + 4 = 0, 0 < ε ≪ 1

• x = x₀ + ε¹x₁ + ε²x₂ + . . .

• Even the first equation is not linear, it is easy to solve

ε⁰ : x⁴₀ − 5x²₀ + 4 = (x²₀ − 4)(x²₀ − 1) = 0 ⇒ x₀ = ±1, ±2, ε¹ : 4x³₀x₁ + x³₀ − 10x₀x₁ = 0 ⇒ x₁ = 1

6, −2

3 , . . .

•

Repeated solutions

• x³ − 2x² + x(1 + ε) − 2ε = 0, 0 < ε ≪ 1

Repeated solutions

• x³ − 2x² + x(1 + ε) − 2ε = 0, 0 < ε ≪ 1

• x = x₀ + ε¹x₁ + ε²x₂ + . . .

Repeated solutions

• x³ − 2x² + x(1 + ε) − 2ε = 0, 0 < ε ≪ 1

• x = x₀ + ε¹x₁ + ε²x₂ + . . .

•

ε⁰ : x³₀ − 2x²₀ + x₀ = 0 ⇒ x₀ = 0, 1( order 2), ε¹ : 3x²₀x₁ − 4x₀x₁ + x₁ + x₀ = 0 ⇒ x₁ = 2

corresponding to x₀ = 0, . . .

Repeated solutions

• x³ − 2x² + x(1 + ε) − 2ε = 0, 0 < ε ≪ 1

• x = x₀ + ε¹x₁ + ε²x₂ + . . .

•

ε⁰ : x³₀ − 2x²₀ + x₀ = 0 ⇒ x₀ = 0, 1( order 2), ε¹ : 3x²₀x₁ − 4x₀x₁ + x₁ + x₀ = 0 ⇒ x₁ = 2

corresponding to x₀ = 0, . . .

• If x = 1 the second equation can not be satisfied.

Repeated solutions

• x³ − 2x² + x(1 + ε) − 2ε = 0, 0 < ε ≪ 1

• x = x₀ + ε¹x₁ + ε²x₂ + . . .

•

ε⁰ : x³₀ − 2x²₀ + x₀ = 0 ⇒ x₀ = 0, 1( order 2), ε¹ : 3x²₀x₁ − 4x₀x₁ + x₁ + x₀ = 0 ⇒ x₁ = 2

corresponding to x₀ = 0, . . .

• If x₀ = 1 the second equation can not be satisfied.

Repeated solutions

• x³ − 2x² + x(1 + ε) − 2ε = 0, 0 < ε ≪ 1

Repeated solutions

• x³ − 2x² + x(1 + ε) − 2ε = 0, 0 < ε ≪ 1

• Probably we need to change the asymptotic sequence. Let λ = x − x₀ = x − 1. Substitute

x = λ + 1 into the equation and get

λ² + λ³ − ε + ελ = 0. For small ε and λ the

dominated terms are ε and λ². Thus ε^1/2 ∼ λ and

Repeated solutions

• x³ − 2x² + x(1 + ε) − 2ε = 0, 0 < ε ≪ 1

• Probably we need to change the asymptotic sequence. Let λ = x − x₀ = x − 1. Substitute

x = λ + 1 into the equation and get

λ² + λ³ − ε + ελ = 0. For small ε and λ the

dominated terms are ε and λ². Thus ε^1/2 ∼ λ and

• x = 1 + ε^1/2x₁ + εx₂ + ε^3/2x₃ + . . .

Repeated solutions

•

ε¹ : x²₁ − 1 = 0 ⇒ x₁ = ±1,

ε^3/2 : x³₁ + 2x₁x₂ + x₁ = 0 ⇒ x₂ = −1

Repeated solutions

•

ε¹ : x²₁ − 1 = 0 ⇒ x₁ = ±1,

ε^3/2 : x³₁ + 2x₁x₂ + x₁ = 0 ⇒ x₂ = −1

• x = 1 + ε^1/2 − ε + O(ε^3/2), x = 1 − ε^1/2 − ε + O(ε^3/2).

The end