[PPT] - Symbolic local Fourier Analysis Veronika Pillwein RISC Challenges PowerPoint Presentation

SLIDE 1

Symbolic local Fourier Analysis

Veronika Pillwein

RISC

Challenges in 21st Century Experimental Mathematical Computation, July 21–25, 2014, ICERM, Brown University

SLIDE 2

Symbolic local Fourier Analysis (and some Special Functions Inequalities)

Veronika Pillwein

RISC

Challenges in 21st Century Experimental Mathematical Computation, July 21–25, 2014, ICERM, Brown University

SLIDE 3

Cylindrical Algebraic Decomposition

SLIDE 4

Input: Polynomial Formulas

◮ Everything that can be formed from variables, rational (or real

algebraic) numbers, +, −, ·, /, =, =, ≤, <, ≥, >, ¬, ∧, ∨, ⇒, ⇔, True, False according to the usual syntactic rules.

SLIDE 5

Input: Polynomial Formulas

◮ Everything that can be formed from variables, rational (or real

algebraic) numbers, +, −, ·, /, =, =, ≤, <, ≥, >, ¬, ∧, ∨, ⇒, ⇔, True, False according to the usual syntactic rules.

◮ Examples:

◮ x2 − (x − 1)(y − 1) > 0 ∧ x ≥ y 2 ≥ 1 ⇒ y 2 − 3x < 0 ◮ xz2 −

√ 2y + z3 > 1 ∧ x > y > z > 1 ⇒ xyz > 1

2

SLIDE 6

Input: Polynomial Formulas

◮ Everything that can be formed from variables, rational (or real

algebraic) numbers, +, −, ·, /, =, =, ≤, <, ≥, >, ¬, ∧, ∨, ⇒, ⇔, True, False according to the usual syntactic rules.

◮ Examples:

◮ x2 − (x − 1)(y − 1) > 0 ∧ x ≥ y 2 ≥ 1 ⇒ y 2 − 3x < 0 ◮ xz2 −

√ 2y + z3 > 1 ∧ x > y > z > 1 ⇒ xyz > 1

2

◮ Counterexamples:

◮ x cos(x) − y sin(y) > π

4 ⇒ cos(x) sin(x) < sin2(x)

◮ x exp(y) − y 2 < xy ∨ log(y) < x ⇒ x2 + y 2 < 4

SLIDE 7

Cylindrical Algebraic Decomposition (CAD)

❘ ❘ ❘ ❘

SLIDE 8

Cylindrical Algebraic Decomposition (CAD)

CAD is an algorithm which can ...

◮ decide whether a given polynomial formula is consistent ◮ decide whether a given polynomial formula is universal ◮ decide whether a given polynomial formula implies another one ◮ determine a certificate point for a given satisfiable formula ◮ determine the semialgebraic set of points

(x1, . . . , xn−1) ∈ ❘n−1 such that there exists a number xn ∈ ❘ where a given formula is true at (x1, . . . , xn−1, xn)

◮ determine the semialgebraic set of all points

(x1, . . . , xn−1) ∈ ❘n−1 such that for all numbers xn ∈ ❘ a given formula is true at (x1 . . . , xn−1, xn)

SLIDE 9

Quantifier Elimination

In particular, given any polynomial formula Φ involving quantifiers (∀, ∃) CAD can compute a polynomial formula Ψ without quantifiers that is equivalent (over ❘) to Φ.

SLIDE 10

Quantifier Elimination

In particular, given any polynomial formula Φ involving quantifiers (∀, ∃) CAD can compute a polynomial formula Ψ without quantifiers that is equivalent (over ❘) to Φ. Examples:

∀ x : 0 < x < 1 ⇒ x2 − y + 1 ≤ B(xy + 1)

CAD

− → y > −1 ∧ B ≥ 2 − y y + 1

SLIDE 11

Quantifier Elimination

In particular, given any polynomial formula Φ involving quantifiers (∀, ∃) CAD can compute a polynomial formula Ψ without quantifiers that is equivalent (over ❘) to Φ. Examples:

∀ x : 0 < x < 1 ⇒ x2 − y + 1 ≤ B(xy + 1)

CAD

− → y > −1 ∧ B ≥ 2 − y y + 1 ∀ x ∃ y : 0 < x < 1 ∧ 0 < y < x2 ⇒ x2 − y + 1 ≤ B(xy + 1)

CAD

− → B ≥ 1

SLIDE 12

Quantifier Elimination

In particular, given any polynomial formula Φ involving quantifiers (∀, ∃) CAD can compute a polynomial formula Ψ without quantifiers that is equivalent (over ❘) to Φ. Examples:

∀ x : 0 < x < 1 ⇒ x2 − y + 1 ≤ B(xy + 1)

CAD

− → y > −1 ∧ B ≥ 2 − y y + 1 ∀ x ∃ y : 0 < x < 1 ∧ 0 < y < x2 ⇒ x2 − y + 1 ≤ B(xy + 1)

CAD

− → B ≥ 1 ∀ x ∃ y : x2 + y 2 − 4 > 0 ⇔ (x − 1)(y − 1) − 1 > 0

CAD

− → True

SLIDE 13

Quantifier Elimination

In particular, given any polynomial formula Φ involving quantifiers (∀, ∃) CAD can compute a polynomial formula Ψ without quantifiers that is equivalent (over ❘) to Φ. Examples:

∀ x : 0 < x < 1 ⇒ x2 − y + 1 ≤ B(xy + 1)

CAD

− → y > −1 ∧ B ≥ 2 − y y + 1 ∀ x ∃ y : 0 < x < 1 ∧ 0 < y < x2 ⇒ x2 − y + 1 ≤ B(xy + 1)

CAD

− → B ≥ 1 ∀ x ∃ y : x2 + y 2 − 4 > 0 ⇔ (x − 1)(y − 1) − 1 > 0

CAD

− → True

Note: The execution of CAD may be computationally expensive!

SLIDE 14

Quantifier Elimination

In particular, given any polynomial formula Φ involving quantifiers (∀, ∃) CAD can compute a polynomial formula Ψ without quantifiers that is equivalent (over ❘) to Φ. Examples:

∀ x : 0 < x < 1 ⇒ x2 − y + 1 ≤ B(xy + 1)

CAD

− → y > −1 ∧ B ≥ 2 − y y + 1 ∀ x ∃ y : 0 < x < 1 ∧ 0 < y < x2 ⇒ x2 − y + 1 ≤ B(xy + 1)

CAD

− → B ≥ 1 ∀ x ∃ y : x2 + y 2 − 4 > 0 ⇔ (x − 1)(y − 1) − 1 > 0

CAD

− → True

Note: The execution of CAD may be computationally expensive! Implementations: Mathematica, Maple, QEPCAD, Redlog, etc.

SLIDE 15

Symbolic Local Fourier Analysis

joint work with Stefan Takacs

SLIDE 16

Local Fourier analysis

◮ standard tool to analyze the convergence behaviour of a

numerical method

SLIDE 17

Local Fourier analysis

◮ standard tool to analyze the convergence behaviour of a

numerical method

◮ Model problem: optimization problem constrained to a partial

differential equation Find a state y and a control u that minimize J(y, u) := 1

2y − yD2 L2(Ω) + α 2 u2 L2(Ω),

subject to −∆y = u in the domain Ω.

SLIDE 18

Local Fourier analysis

◮ standard tool to analyze the convergence behaviour of a

numerical method

◮ Model problem: optimization problem constrained to a partial

differential equation Find a state y and a control u that minimize J(y, u) := 1

2y − yD2 L2(Ω) + α 2 u2 L2(Ω),

subject to −∆y = u in the domain Ω.

◮ solution method: Finite Element Method (FEM) with a

multigrid-solver (which ultimately means solving a large scale linear system)

SLIDE 19

Local Fourier analysis

◮ standard tool to analyze the convergence behaviour of a

numerical method

◮ Model problem: optimization problem constrained to a partial

differential equation Find a state y and a control u that minimize J(y, u) := 1

2y − yD2 L2(Ω) + α 2 u2 L2(Ω),

subject to −∆y = u in the domain Ω.

◮ solution method: Finite Element Method (FEM) with a

multigrid-solver (which ultimately means solving a large scale linear system)

◮ robust with respect to parameters such as mesh-size and

regularization parameters

SLIDE 20

Local Fourier Analysis

◮ Given: Iterative procedure of the form

x(k+1) = Ax(k) the convergence rate is related to the matrix norm of A which can be estimated by the spectral radius, i.e., the largest eigenvalue in absolute value

SLIDE 21

Local Fourier Analysis

◮ Given: Iterative procedure of the form

x(k+1) = Ax(k) the convergence rate is related to the matrix norm of A which can be estimated by the spectral radius, i.e., the largest eigenvalue in absolute value

◮ typically rates are obtained only by numerical interpolation

SLIDE 22

Local Fourier Analysis

◮ Given: Iterative procedure of the form

x(k+1) = Ax(k) the convergence rate is related to the matrix norm of A which can be estimated by the spectral radius, i.e., the largest eigenvalue in absolute value

◮ typically rates are obtained only by numerical interpolation ◮ symbolic local Fourier analysis: exact bounds

SLIDE 23

FEM and multigrid

◮ in FEM the given domain is subdivided in

simple geometric objects (triangles, rectangles, tetrahedra, prisms, etc.)

SLIDE 24

FEM and multigrid

◮ in FEM the given domain is subdivided in

simple geometric objects (triangles, rectangles, tetrahedra, prisms, etc.)

◮ Multigrid methods operate on two (or

more) grids

SLIDE 25

FEM and multigrid

◮ in FEM the given domain is subdivided in

simple geometric objects (triangles, rectangles, tetrahedra, prisms, etc.)

◮ Multigrid methods operate on two (or

more) grids

◮ One step in a multigrid method consists

f

◮ (pre)smoothing steps

↓ restriction ↓

◮ coarse grid correction

↓ prolongation ↓

◮ (post)smoothing steps

SLIDE 26

FEM and multigrid

◮ in FEM the given domain is subdivided in

simple geometric objects (triangles, rectangles, tetrahedra, prisms, etc.)

◮ Multigrid methods operate on two (or

more) grids

◮ One step in a multigrid method consists

f

◮ (pre)smoothing steps

↓ restriction ↓

◮ coarse grid correction

↓ prolongation ↓

◮ (post)smoothing steps

SLIDE 27

FEM and multigrid

◮ in FEM the given domain is subdivided in

simple geometric objects (triangles, rectangles, tetrahedra, prisms, etc.)

◮ Multigrid methods operate on two (or

more) grids

◮ One step in a multigrid method consists

f

◮ (pre)smoothing steps

↓ restriction ↓

◮ coarse grid correction

↓ prolongation ↓

◮ (post)smoothing steps

SLIDE 28

All-at-once analysis for 1D

In 1D the largest eigenvalue of the iteration matrix can be computed explicitely for the whole twogrid-step: σ(τ, η, γ) = P1(τ, η, γ)P2(τ, η, γ) 64(9 + η)2(9(−1 + γ)2 + η(1 + 2γ)2) with

P1(τ, η, γ) = η

γ2τ 2 + γ
4 − 3τ 2

+ 4(τ − 1)2 + 36

γ2τ 2 + γ
6τ 2 − 6τ + 1
+ (τ − 1)2

P2(τ, η, γ) = η2 γ3τ 2 + γ2 4 + 16τ − 7τ 2 + γ

8τ 2 − 56τ + 52
+16(τ − 1)2

+ 36η

2γ3τ 2 + γ2

28τ 2 − 22τ + 5

+γ
34τ 2 − 34τ + 5
+ 8(τ − 1)2

+ 1296(γ − 1)2 (γ + 1)τ 2 − 2τ + 1

.

SLIDE 29

Convergence rate in 1D

◮ The convergence rate for the twogrid method can be

computed as q2

TG(τ) = sup η>0

sup

0≤γ≤1

σ(τ, η, γ)

SLIDE 30

Convergence rate in 1D

◮ The convergence rate for the twogrid method can be

computed as q2

TG(τ) = sup η>0

sup

0≤γ≤1

σ(τ, η, γ)

◮ i.e., as the least upper bound λ = λ(τ) satisfying

∀ η > 0, 0 ≤ γ ≤ 1: σ(τ, η, γ) ≤ λ

SLIDE 31

Convergence rate in 1D

◮ The convergence rate for the twogrid method can be

computed as q2

TG(τ) = sup η>0

sup

0≤γ≤1

σ(τ, η, γ)

◮ i.e., as the least upper bound λ = λ(τ) satisfying

∀ η > 0, 0 ≤ γ ≤ 1: σ(τ, η, γ) ≤ λ

◮ Direct approach using CAD to determine λ fails in reasonable

amount of time, so we guess the bound by considering the limiting cases η → 0 and η → ∞

SLIDE 32

Convergence rate in 1D

◮ The convergence rate for the twogrid method can be

computed as q2

TG(τ) = sup η>0

sup

0≤γ≤1

σ(τ, η, γ)

◮ i.e., as the least upper bound λ = λ(τ) satisfying

∀ η > 0, 0 ≤ γ ≤ 1: σ(τ, η, γ) ≤ λ

◮ Direct approach using CAD to determine λ fails in reasonable

amount of time, so we guess the bound by considering the limiting cases η → 0 and η → ∞

◮ Using CAD we can compute the bounds

q2

0(τ) = sup 0≤γ≤1

σ0(τ, γ) and q2

∞(τ) = sup 0≤γ≤1

σ∞(τ, γ).

SLIDE 33

Convergence rate in 1D

◮ The convergence rate for the twogrid method can be

computed as q2

TG(τ) = sup η>0

sup

0≤γ≤1

σ(τ, η, γ)

◮ i.e., as the least upper bound λ = λ(τ) satisfying

∀ η > 0, 0 ≤ γ ≤ 1: σ(τ, η, γ) ≤ λ

◮ Direct approach using CAD to determine λ fails in reasonable

amount of time, so we guess the bound by considering the limiting cases η → 0 and η → ∞

◮ Using CAD we can compute the bounds

q2

0(τ) = sup 0≤γ≤1

σ0(τ, γ) and q2

∞(τ) = sup 0≤γ≤1

σ∞(τ, γ).

◮ CAD can also verify that this guess yields the true bound

∀ τ, γ ∈ [0, 1], η > 0: σ(τ, η, γ) ≤ max{q0(τ)2, q∞(τ)2}

SLIDE 34

The convergence rate in 1D

qTG(τ) = max

|1 − 2τ|
2 + 4(τ − 1)τ, 1

4(τ − 2)2

0.2 0.4 0.6 0.8 1.0 Τ 0.2 0.4 0.6 0.8 1.0 1.2 1.4 q TG

SLIDE 35

The Problem in 2D

Given: matrix A(q, c1, c2, η) ∈ ❘8×8 with 0 < q < 1 and (c1, c2, η) ∈ Ω = {(c1, c2, η) | 0 ≤ c1 ≤ c2 < 1 ∧ η > 0}, where ci = cos(θi) for some frequencies θi and η = h4/α with mesh-size h and regularization parameter α. Find: bound B(q) for the maximal eigenvalue λmax(q, c1, c2, η)

f A over (0, 1) × Ω.

SLIDE 36

The Matrix A

            A1,1 A1,3 A1,4 A1,5 A1,6 A1,7 A1,8 A2,2 A2,3 A2,4 A2,5 A2,6 A2,7 A2,8 A3,1 A3,2 A3,3 A3,5 A3,6 A3,7 A3,8 A4,1 A4,2 A4,4 A4,5 A4,6 A4,7 A4,8 A5,1 A5,2 A5,3 A5,4 A5,5 A5,7 A5,8 A6,1 A6,2 A6,3 A6,4 A6,6 A6,7 A6,8 A7,1 A7,2 A7,3 A7,4 A7,5 A7,6 A7,7 A8,1 A8,2 A8,3 A8,4 A8,5 A8,6 A8,8             Common denominator of the matrix entries:

D =256

16c4

2c4 1η + 16c2 2c4 1η + 4c4 1η + 16c4 2c2 1η + 16c2 2c2 1η + 4c2 1η

+ 4c4

2η + 4c2 2η + 144c4 2c4 1 − 72c2 2c4 1 + 9c4 1 − 72c4 2c2 1 − 126c2 2c2 1

+36c2

1 + 9c4 2 + 36c2 2 + η + 36

SLIDE 37

The Matrix A

            A1,1 A1,3 A1,4 A1,5 A1,6 A1,7 A1,8 A2,2 A2,3 A2,4 A2,5 A2,6 A2,7 A2,8 A3,1 A3,2 A3,3 A3,5 A3,6 A3,7 A3,8 A4,1 A4,2 A4,4 A4,5 A4,6 A4,7 A4,8 A5,1 A5,2 A5,3 A5,4 A5,5 A5,7 A5,8 A6,1 A6,2 A6,3 A6,4 A6,6 A6,7 A6,8 A7,1 A7,2 A7,3 A7,4 A7,5 A7,6 A7,7 A8,1 A8,2 A8,3 A8,4 A8,5 A8,6 A8,8             Common denominator of the matrix entries:

D =256

16c4

2c4 1η + 16c2 2c4 1η + 4c4 1η + 16c4 2c2 1η + 16c2 2c2 1η + 4c2 1η

+ 4c4

2η + 4c2 2η + 144c4 2c4 1 − 72c2 2c4 1 + 9c4 1 − 72c4 2c2 1 − 126c2 2c2 1

+36c2

1 + 9c4 2 + 36c2 2 + η + 36

SLIDE 38

The Numerator of A1,1

432q2c6

2 c6 1 + 3q2ηc6 2 c6 1 + ηc6 2 c6 1 + 144c6 2 c6 1 + 1008q2c5 2 c6 1 + 16q2ηc5 2 c6 1 + 8ηc5 2 c6 1 + 720c5 2 c6 1 + 972q2c4 2 c6 1 +

30q2ηc4

2 c6 1 + 26ηc4 2 c6 1 + 1476c4 2 c6 1 + 1008q2c3 2 c6 1 + 28q2ηc3 2 c6 1 + 44ηc3 2 c6 1 + 1584c3 2 c6 1 + 108q2c6 1 + 1080q2c2 2 c6 1 +

27q2ηc2

2 c6 1 + 41ηc2 2 c6 1 + 936c2 2 c6 1 + 12q2ηc6 1 + 4ηc6 1 + 576q2c2c6 1 + 28q2ηc2c6 1 + 20ηc2c6 1 + 288c2c6 1 + 36c6 1 +

1008q2c6

2 c5 1 +16q2ηc6 2 c5 1 +8ηc6 2 c5 1 +720c6 2 c5 1 −360q2c5 2 c5 1 +80q2ηc5 2 c5 1 −64ηc5 2 c5 1 −1656c5 2 c5 1 −3600q2c4 2 c5 1 +

128q2ηc4

2 c5 1 − 304ηc4 2 c5 1 − 7056c4 2 c5 1 − 2736q2c3 2 c5 1 + 80q2ηc3 2 c5 1 − 352ηc3 2 c5 1 − 5328c3 2 c5 1 − 720q2c5 1 −

1872q2c2

2 c5 1 + 80q2ηc2 2 c5 1 − 184ηc2 2 c5 1 + 720c2 2 c5 1 + 64q2ηc5 1 − 96ηc5 1 − 2088q2c2c5 1 + 128q2ηc2c5 1 − 160ηc2c5 1 +

1800c2c5

1 + 432c5 1 + 972q2c6 2 c4 1 + 30q2ηc6 2 c4 1 + 26ηc6 2 c4 1 + 1476c6 2 c4 1 − 3600q2c5 2 c4 1 + 128q2ηc5 2 c4 1 − 304ηc5 2 c4 1 −

7056c5

2 c4 1 − 5616q2c4 2 c4 1 + 108q2ηc4 2 c4 1 + 2724ηc4 2 c4 1 + 21168c4 2 c4 1 + 1584q2c3 2 c4 1 − 136q2ηc3 2 c4 1 − 1672ηc3 2 c4 1 −

3600c3

2 c4 1 +648q2c4 1 +1404q2c2 2 c4 1 −114q2ηc2 2 c4 1 +3114ηc2 2 c4 1 −15660c2 2 c4 1 +120q2ηc4 1 +616ηc4 1 −576q2c2c4 1 +

152q2ηc2c4

1 − 760ηc2c4 1 − 2304c2c4 1 + 792c4 1 + 1008q2c6 2 c3 1 + 28q2ηc6 2 c3 1 + 44ηc6 2 c3 1 + 1584c6 2 c3 1 − 2736q2c5 2 c3 1 +

80q2ηc5

2 c3 1 − 352ηc5 2 c3 1 − 5328c5 2 c3 1 + 1584q2c4 2 c3 1 − 136q2ηc4 2 c3 1 − 1672ηc4 2 c3 1 − 3600c4 2 c3 1 + 12384q2c3 2 c3 1 −

640q2ηc3

2 c3 1 − 1936ηc3 2 c3 1 + 17568c3 2 c3 1 + 1872q2c3 1 + 5904q2c2 2 c3 1 − 580q2ηc2 2 c3 1 − 1012ηc2 2 c3 1 + 14544c2 2 c3 1 +

112q2ηc3

1 − 528ηc3 1 + 720q2c2c3 1 − 16q2ηc2c3 1 − 880ηc2c3 1 − 1872c2c3 1 − 2160c3 1 + 1080q2c6 2 c2 1 + 27q2ηc6 2 c2 1 +

41ηc6

2 c2 1 + 936c6 2 c2 1 − 1872q2c5 2 c2 1 + 80q2ηc5 2 c2 1 − 184ηc5 2 c2 1 + 720c5 2 c2 1 + 1404q2c4 2 c2 1 − 114q2ηc4 2 c2 1 +

3114ηc4

2 c2 1 − 15660c4 2 c2 1 + 5904q2c3 2 c2 1 − 580q2ηc3 2 c2 1 − 1012ηc3 2 c2 1 + 14544c3 2 c2 1 + 108q2c2 1 − 5184q2c2 2 c2 1 −

525q2ηc2

2 c2 1 + 3729ηc2 2 c2 1 − 15552c2 2 c2 1 + 108q2ηc2 1 + 676ηc2 1 − 6624q2c2c2 1 − 4q2ηc2c2 1 − 460ηc2c2 1 +

2880c2c2

1 + 6948c2 1 + 576q2c6 2 c1 + 28q2ηc6 2 c1 + 20ηc6 2 c1 + 288c6 2 c1 − 2088q2c5 2 c1 + 128q2ηc5 2 c1 − 160ηc5 2 c1 +

1800c5

2 c1 − 576q2c4 2 c1 + 152q2ηc4 2 c1 − 760ηc4 2 c1 − 2304c4 2 c1 + 720q2c3 2 c1 − 16q2ηc3 2 c1 − 880ηc3 2 c1 −

1872c3

2 c1 + 1440q2c1 − 6624q2c2 2 c1 − 4q2ηc2 2 c1 − 460ηc2 2 c1 + 2880c2 2 c1 + 112q2ηc1 − 240ηc1 − 3816q2c2c1 +

176q2ηc2c1 − 400ηc2c1 − 5112c2c1 − 6048c1 + 108q2c6

2 + 12q2ηc6 2 + 4ηc6 2 + 36c6 2 − 720q2c5 2 + 64q2ηc5 2 −

96ηc5

2 + 432c5 2 + 648q2c4 2 + 120q2ηc4 2 + 616ηc4 2 + 792c4 2 + 1872q2c3 2 + 112q2ηc3 2 − 528ηc3 2 − 2160c3 2 + 1728q2 +

108q2c2

2 + 108q2ηc2 2 + 676ηc2 2 + 6948c2 2 + 48q2η + 144η + 1440q2c2 + 112q2ηc2 − 240ηc2 − 6048c2 + 5184

SLIDE 39

Computational Issues

1. How to find the eigenvalues?
2. How to find the bound?
3. How to prove the bound?

SLIDE 40

Computational Issues

1. How to find the eigenvalues?

◮ symbolic interpolation in q

2. How to find the bound?
3. How to prove the bound?

SLIDE 41

Computational Issues

1. How to find the eigenvalues?

◮ symbolic interpolation in q

2. How to find the bound?

◮ consider the boundary of the domain to obtain a plausible

guess for the bound using CAD

3. How to prove the bound?

SLIDE 42

Computational Issues

1. How to find the eigenvalues?

◮ symbolic interpolation in q

2. How to find the bound?

◮ consider the boundary of the domain to obtain a plausible

guess for the bound using CAD

3. How to prove the bound?

◮ use CAD

SLIDE 43

Computational Issues

1. How to find the eigenvalues?

◮ symbolic interpolation in q

2. How to find the bound?

◮ consider the boundary of the domain to obtain a plausible

guess for the bound using CAD

3. How to prove the bound?

◮ use CAD ◮ necessary to split into subtasks and consider sufficient

conditions

SLIDE 44

The Eigenvalues

Using interpolation on the characteristic polynomial of the matrix we find that the eigenvalues are λ1 = 0, λ2 = q4,

3λ4 = 1

D

e(q2) ±
d(q2)
,

each of multiplicity 2 with q2 = q2 and e, d polynomials in c1, c2, η and q2.

SLIDE 45

The Eigenvalues

Using interpolation on the characteristic polynomial of the matrix we find that the eigenvalues are λ1 = 0, λ2 = q4,

3λ4 = 1

D

e(q2) ±
d(q2)
,

each of multiplicity 2 with q2 = q2 and e, d polynomials in c1, c2, η and q2. The largest eigenvalue is λmax = 1 D

e(q2) +
d(q2)
.

SLIDE 46

The Eigenvalues

Using interpolation on the characteristic polynomial of the matrix we find that the eigenvalues are λ1 = 0, λ2 = q4,

3λ4 = 1

D

e(q2) ±
d(q2)
,

each of multiplicity 2 with q2 = q2 and e, d polynomials in c1, c2, η and q2. The largest eigenvalue is λmax = 1 D

e(q2) +
d(q2)
.

SLIDE 47

The Polynomial e(q2)

ηc6

2 c6 1 + 144c6 2 c6 1 + 8ηc5 2 c6 1 + 720c5 2 c6 1 + 26ηc4 2 c6 1 + 1476c4 2 c6 1 + 44ηc3 2 c6 1 + 1584c3 2 c6 1 + 41ηc2 2 c6 1 + 936c2 2 c6 1 +

9ηc6

2 q2 2c6 1 +1296c6 2 q2 2c6 1 −24ηc5 2 q2 2c6 1 −2160c5 2 q2 2c6 1 +138ηc4 2 q2 2c6 1 +6372c4 2 q2 2c6 1 −132ηc3 2 q2 2c6 1 −4752c3 2 q2 2c6 1 +

177ηc2

2 q2 2c6 1 + 4968c2 2 q2 2c6 1 + 36ηq2 2c6 1 − 60ηc2q2 2c6 1 − 864c2q2 2c6 1 + 324q2 2c6 1 + 4ηc6 1 + 20ηc2c6 1 + 288c2c6 1 +

6ηc6

2 q2c6 1 + 864c6 2 q2c6 1 + 16ηc5 2 q2c6 1 + 1440c5 2 q2c6 1 + 60ηc4 2 q2c6 1 + 1944c4 2 q2c6 1 + 88ηc3 2 q2c6 1 + 3168c3 2 q2c6 1 +

54ηc2

2 q2c6 1 +2160c2 2 q2c6 1 +24ηq2c6 1 +40ηc2q2c6 1 +576c2q2c6 1 +216q2c6 1 +36c6 1 +8ηc6 2 c5 1 +720c6 2 c5 1 −64ηc5 2 c5 1 −

1656c5

2 c5 1 −304ηc4 2 c5 1 −7056c4 2 c5 1 −352ηc3 2 c5 1 −5328c3 2 c5 1 −184ηc2 2 c5 1 +720c2 2 c5 1 −24ηc6 2 q2 2c5 1 −2160c6 2 q2 2c5 1 +

32ηc5

2 q2 2c5 1 +2664c5 2 q2 2c5 1 +272ηc4 2 q2 2c5 1 +10224c4 2 q2 2c5 1 +416ηc3 2 q2 2c5 1 +3312c3 2 q2 2c5 1 +296ηc2 2 q2 2c5 1 −2736c2 2 q2 2c5 1 +

32ηq2

2c5 1 + 128ηc2q2 2c5 1 − 792c2q2 2c5 1 − 144q2 2c5 1 − 96ηc5 1 − 160ηc2c5 1 + 1800c2c5 1 + 16ηc6 2 q2c5 1 + 1440c6 2 q2c5 1 +

32ηc5

2 q2c5 1 −1008c5 2 q2c5 1 +32ηc4 2 q2c5 1 −3168c4 2 q2c5 1 −64ηc3 2 q2c5 1 +2016c3 2 q2c5 1 −112ηc2 2 q2c5 1 +2016c2 2 q2c5 1 +

64ηq2c5

1 + 32ηc2q2c5 1 − 1008c2q2c5 1 − 288q2c5 1 + 432c5 1 + 26ηc6 2 c4 1 + 1476c6 2 c4 1 − 304ηc5 2 c4 1 − 7056c5 2 c4 1 +

2724ηc4

2 c4 1 + 21168c4 2 c4 1 − 1672ηc3 2 c4 1 − 3600c3 2 c4 1 + 3114ηc2 2 c4 1 − 15660c2 2 c4 1 + 138ηc6 2 q2 2c4 1 + 6372c6 2 q2 2c4 1 +

272ηc5

2 q2 2c4 1 + 10224c5 2 q2 2c4 1 + 4292ηc4 2 q2 2c4 1 + 15408c4 2 q2 2c4 1 + 1496ηc3 2 q2 2c4 1 + 8496c3 2 q2 2c4 1 + 5018ηc2 2 q2 2c4 1 + . . .

SLIDE 48

The Polynomial e(q2)

· · · − 25740c2

2 q2 2c4 1 + 1064ηq2 2c4 1 + 680ηc2q2 2c4 1 − 576c2q2 2c4 1 + 1368q2 2c4 1 + 616ηc4 1 − 760ηc2c4 1 − 2304c2c4 1 +

60ηc6

2 q2c4 1 + 1944c6 2 q2c4 1 + 32ηc5 2 q2c4 1 − 3168c5 2 q2c4 1 + 216ηc4 2 q2c4 1 − 11232c4 2 q2c4 1 + 176ηc3 2 q2c4 1 −

4896c3

2 q2c4 1 − 228ηc2 2 q2c4 1 + 2808c2 2 q2c4 1 + 240ηq2c4 1 + 80ηc2q2c4 1 + 2880c2q2c4 1 + 1296q2c4 1 + 792c4 1 +

44ηc6

2 c3 1 + 1584c6 2 c3 1 − 352ηc5 2 c3 1 − 5328c5 2 c3 1 − 1672ηc4 2 c3 1 − 3600c4 2 c3 1 − 1936ηc3 2 c3 1 + 17568c3 2 c3 1 −

1012ηc2

2 c3 1 + 14544c2 2 c3 1 − 132ηc6 2 q2 2c3 1 − 4752c6 2 q2 2c3 1 + 416ηc5 2 q2 2c3 1 + 3312c5 2 q2 2c3 1 + 1496ηc4 2 q2 2c3 1 +

8496c4

2 q2 2c3 1 + 1808ηc3 2 q2 2c3 1 − 13536c3 2 q2 2c3 1 + 1628ηc2 2 q2 2c3 1 − 14832c2 2 q2 2c3 1 + 176ηq2 2c3 1 + 944ηc2q2 2c3 1 −

144c2q2

2c3 1 +720q2 2c3 1 −528ηc3 1 −880ηc2c3 1 −1872c2c3 1 +88ηc6 2 q2c3 1 +3168c6 2 q2c3 1 −64ηc5 2 q2c3 1 +2016c5 2 q2c3 1 +

176ηc4

2 q2c3 1 − 4896c4 2 q2c3 1 + 128ηc3 2 q2c3 1 − 4032c3 2 q2c3 1 − 616ηc2 2 q2c3 1 + 288c2 2 q2c3 1 + 352ηq2c3 1 − 64ηc2q2c3 1 +

2016c2q2c3

1 + 1440q2c3 1 − 2160c3 1 + 41ηc6 2 c2 1 + 936c6 2 c2 1 − 184ηc5 2 c2 1 + 720c5 2 c2 1 + 3114ηc4 2 c2 1 − 15660c4 2 c2 1 −

1012ηc3

2 c2 1 + 14544c3 2 c2 1 + 3729ηc2 2 c2 1 − 15552c2 2 c2 1 + 177ηc6 2 q2 2c2 1 + 4968c6 2 q2 2c2 1 + 296ηc5 2 q2 2c2 1 − 2736c5 2 q2 2c2 1 +

5018ηc4

2 q2 2c2 1 − 25740c4 2 q2 2c2 1 + 1628ηc3 2 q2 2c2 1 − 14832c3 2 q2 2c2 1 + 6041ηc2 2 q2 2c2 1 − 24768c2 2 q2 2c2 1 + 1220ηq2 2c2 1 + . . .

SLIDE 49

The Polynomial e(q2)

· · · + 740ηc2q2

2c2 1 + 4608c2q2 2c2 1 + 11844q2 2c2 1 + 676ηc2 1 − 460ηc2c2 1 + 2880c2c2 1 + 54ηc6 2 q2c2 1 + 2160c6 2 q2c2 1 −

112ηc5

2 q2c2 1 + 2016c5 2 q2c2 1 − 228ηc4 2 q2c2 1 + 2808c4 2 q2c2 1 − 616ηc3 2 q2c2 1 + 288c3 2 q2c2 1 − 1050ηc2 2 q2c2 1 −

10368c2

2 q2c2 1 + 216ηq2c2 1 − 280ηc2q2c2 1 − 7488c2q2c2 1 + 216q2c2 1 + 6948c2 1 + 20ηc6 2 c1 + 288c6 2 c1 −

160ηc5

2 c1 + 1800c5 2 c1 − 760ηc4 2 c1 − 2304c4 2 c1 − 880ηc3 2 c1 − 1872c3 2 c1 − 460ηc2 2 c1 + 2880c2 2 c1 − 60ηc6 2 q2 2c1 −

864c6

2 q2 2c1 + 128ηc5 2 q2 2c1 − 792c5 2 q2 2c1 + 680ηc4 2 q2 2c1 − 576c4 2 q2 2c1 + 944ηc3 2 q2 2c1 − 144c3 2 q2 2c1 + 740ηc2 2 q2 2c1 +

4608c2

2 q2 2c1 + 80ηq2 2c1 + 368ηc2q2 2c1 + 6120c2q2 2c1 + 2016q2 2c1 − 240ηc1 − 400ηc2c1 − 5112c2c1 +

40ηc6

2 q2c1 + 576c6 2 q2c1 + 32ηc5 2 q2c1 − 1008c5 2 q2c1 + 80ηc4 2 q2c1 + 2880c4 2 q2c1 − 64ηc3 2 q2c1 + 2016c3 2 q2c1 −

280ηc2

2 q2c1 − 7488c2 2 q2c1 + 160ηq2c1 + 32ηc2q2c1 − 1008c2q2c1 + 4032q2c1 − 6048c1 + 4ηc6 2 + 36c6 2 −

96ηc5

2 + 432c5 2 + 616ηc4 2 + 792c4 2 − 528ηc3 2 − 2160c3 2 + 676ηc2 2 + 6948c2 2 + 36ηc6 2 q2 2 + 324c6 2 q2 2 + 32ηc5 2 q2 2 −

144c5

2 q2 2 + 1064ηc4 2 q2 2 + 1368c4 2 q2 2 + 176ηc3 2 q2 2 + 720c3 2 q2 2 + 1220ηc2 2 q2 2 + 11844c2 2 q2 2 + 272ηq2 2 + 80ηc2q2 2 +

2016c2q2

2 + 9792q2 2 + 144η − 240ηc2 − 6048c2 + 24ηc6 2 q2 + 216c6 2 q2 + 64ηc5 2 q2 − 288c5 2 q2 + 240ηc4 2 q2 +

1296c4

2 q2 + 352ηc3 2 q2 + 1440c3 2 q2 + 216ηc2 2 q2 + 216c2 2 q2 + 96ηq2 + 160ηc2q2 + 4032c2q2 + 3456q2 + 5184

SLIDE 50

The Bound

◮ Consider extreme cases for

(c1, c2)

◮ Consider the limits η → 0

and η → ∞

0.5 1 0.5 1 0.5 1 0.5 1 c1 c2

SLIDE 51

The Bound

◮ Consider extreme cases for

(c1, c2)

◮ Consider the limits η → 0

and η → ∞

0.5 1 0.5 1 0.5 1 0.5 1 c1 c2

This yields the following guess: B(q2) =

lb(q2) =
q2+3

4

2 , 0 < q2 < Q2, rb(q2) = q2(q2 + 1), Q2 ≤ q2 < 1. , with Q2 = 1

15(4

√ 10 − 5).

SLIDE 52

Bound for the Convergence Rate

0.0 0.2 0.4 0.6 0.8 1.0 Τ 0.2 0.4 0.6 0.8 1.0 1.2 1.4 q TG

Two-grid convergence factor depending on τ for ν = νpre + νpost = 2 + 2 smoothing steps

SLIDE 53

Experimental all-at-once analysis for 2D

◮ Set up the full matrix for the all-at-once approach

SLIDE 54

Experimental all-at-once analysis for 2D

◮ Set up the full matrix for the all-at-once approach ◮ Consider only the limiting cases for c1, c2, η

SLIDE 55

Experimental all-at-once analysis for 2D

◮ Set up the full matrix for the all-at-once approach ◮ Consider only the limiting cases for c1, c2, η ◮ qTG(τ) = max

1

16(τ − 4)2, 1 4|3τ − 2|

9τ 2 − 12τ + 8
0.0

0.2 0.4 0.6 0.8 1.0 1.2 1.4 0.2 0.4 0.6 0.8 1.0 1.2 1.4

SLIDE 56

Experimental all-at-once analysis for 2D

◮ Set up the full matrix for the all-at-once approach ◮ Consider only the limiting cases for c1, c2, η ◮ qTG(τ) = max

1

16(τ − 4)2, 1 4|3τ − 2|

9τ 2 − 12τ + 8
0.0

0.2 0.4 0.6 0.8 1.0 1.2 1.4 0.2 0.4 0.6 0.8 1.0 1.2 1.4

http://www.risc.jku.at/people/vpillwei/sLFA/

SLIDE 57

The Gerhold-Kauers method

SLIDE 58

The Gerhold-Kauers method

◮ Method to prove inequalities involving expressions depending

n a discrete parameter

SLIDE 59

The Gerhold-Kauers method

◮ Method to prove inequalities involving expressions depending

n a discrete parameter

◮ Proof by induction: reformulating the proof goal as

polynomial formula

SLIDE 60

The Gerhold-Kauers method

◮ Method to prove inequalities involving expressions depending

n a discrete parameter

◮ Proof by induction: reformulating the proof goal as

polynomial formula

◮ Quantifier elimination using CAD applicable to give an

automatic proof

SLIDE 61

The Gerhold-Kauers method

◮ Method to prove inequalities involving expressions depending

n a discrete parameter

◮ Proof by induction: reformulating the proof goal as

polynomial formula

◮ Quantifier elimination using CAD applicable to give an

automatic proof

◮ Actually proving something more general: may be false, even

if the original statement is true!

SLIDE 62

The Gerhold-Kauers method

◮ Method to prove inequalities involving expressions depending

n a discrete parameter

◮ Proof by induction: reformulating the proof goal as

polynomial formula

◮ Quantifier elimination using CAD applicable to give an

automatic proof

◮ Actually proving something more general: may be false, even

if the original statement is true!

◮ Method rather than an algorithm (termination!)

SLIDE 63

The Gerhold-Kauers method

◮ Method to prove inequalities involving expressions depending

n a discrete parameter

◮ Proof by induction: reformulating the proof goal as

polynomial formula

◮ Quantifier elimination using CAD applicable to give an

automatic proof

◮ Actually proving something more general: may be false, even

if the original statement is true!

◮ Method rather than an algorithm (termination!) ◮ Implemented in the Mathematica package SumCracker

[Kauers]

SLIDE 64

The Gerhold-Kauers method

◮ Method to prove inequalities involving expressions depending

n a discrete parameter

◮ Proof by induction: reformulating the proof goal as

polynomial formula

◮ Quantifier elimination using CAD applicable to give an

automatic proof

◮ Actually proving something more general: may be false, even

if the original statement is true!

◮ Method rather than an algorithm (termination!) ◮ Implemented in the Mathematica package SumCracker

[Kauers] http://www.risc.jku.at/research/combinat/software

SLIDE 65

An extension of Tur´ an’s inequality

joint work with Geno Nikolov

SLIDE 66

Tur´ an’s inequality and a variation

◮ For all n ∈ ◆ and all x ∈ [−1, 1]:

Pn(x)2 − Pn−1(x)Pn+1(x) ≥ 0

1.0 0.5 0.5 1.0 0.02 0.02 0.04 0.06 0.08 0.10

◆

SLIDE 67

Tur´ an’s inequality and a variation

◮ For all n ∈ ◆ and all x ∈ [−1, 1]:

Pn(x)2 − Pn−1(x)Pn+1(x) ≥ 0

1.0 0.5 0.5 1.0 0.02 0.02 0.04 0.06 0.08 0.10 1.0 0.5 0.5 1.0 0.05 0.10 0.15

◮ (Gerhold+Kauers) For all n ∈ ◆ and all x ∈ [−1, 1]:

|x|Pn(x)2 − Pn−1(x)Pn+1(x) ≥ 0

SLIDE 68

Orthogonal Polynomials: Legendre polynomials

◮ The nth Legendre polynomial Pn(x) is defined on [−1, 1] and

Legendre polynomials are orthogonal w.r.t. the inner product f , g = 1

−1

f (x)g(x) dx.

◮ Legendre polynomials satisfy a three term recurrence,

(n + 1)Pn(x) − (2n + 3)xPn+1(x) + (n + 2)Pn+2(x) = 0,

P−1(x) = 0,

P0(x) = 1.

◮ The first few are given by

P1(x) = x, P2(x) = 1 2

3x2 − 1
,

P3(x) = 1 2x

5x2 − 3
, . . .

SLIDE 69

Tur´ an’s inequality and a variation

◮ For all n ∈ ◆ and all x ∈ [−1, 1]:

Pn(x)2 − Pn−1(x)Pn+1(x) ≥ 0

1.0 0.5 0.5 1.0 0.02 0.02 0.04 0.06 0.08 0.10 1.0 0.5 0.5 1.0 0.05 0.10 0.15

◮ (Gerhold+Kauers) For all n ∈ ◆ and all x ∈ [−1, 1]:

|x|Pn(x)2 − Pn−1(x)Pn+1(x) ≥ 0

SLIDE 70

The modified Tur´ an inequality for Gegenbauer polynomials

◮ Let pn(x) = C λ n (x)/C λ n (1) and λ ∈ (−1/2, 1/2]. Then, for

every n ∈ N and x ∈ [−1, 1], |x|p2

n(x) − pn−1(x)pn+1(x) ≥ 0

1.0 0.5 0.5 1.0 0.05 0.10 0.15 0.20 0.25 1.0 0.5 0.5 1.0 0.05 0.10 0.15 0.20

◮ The normalized Gegenbauer polynomials satisfy the recurrence

npn−1(x) − 2(n + λ)xpn(x) + (n + 2λ)pn+1(x) = 0.

SLIDE 71

Proof using SumCracker (Kauers)

With C λ

n (x) = GegenbauerC[n, λ, x],

In[1]:= ProveInequality[

xCλ

n+1(x)2

(n + 1)! (2λ)n+1 2 − Cλ

n (x)Cλ n+2(x) n!(n + 2)!

(2λ)n(2λ)n+2 ≥ 0, Using → {− 1

2 < λ ≤ 1 2 && 0 ≤ x ≤ 1}, Variable → n] Out[1]= True

SLIDE 72

Proof using SumCracker (Kauers)

Or, providing the definition of the polynomials,

In[2]:= ProveInequality[xp[n + 1]2 − p[n]p[n + 2] ≥ 0,

Where → {p[2 + n]==2(n + λ + 1) n + 2λ + 1 xp[n + 1] − n + 1 n + 2λ + 1p[n], p[0]==1, p[1]==x}, Using → {− 1

2 < λ ≤ 1 2 && 0 ≤ x ≤ 1}, Variable → n] Out[2]= True

SLIDE 73

A Classical Proof

An extension of Tur´ an’s inequality for ultraspherical polynomials

Geno Nikolov

Faculty of Mathematics and Informatics, Sofia University “St. Kliment Ohridski” 5 James Bourchier Blvd., 1164 Sofia, Bulgaria

Veronika Pillwein

Research Institute for Symbolic Computation, Johannes Kepler University Altenberger Straße 69, A-4040 Linz, Austria Abstract Let pm(x) = P (λ)

m (x)/P (λ) m (1) be the m-th ultraspherical polynomial normalized by

pm(1) = 1. We prove the inequality |x|p2

n(x) − pn−1(x)pn+1(x) ≥ 0, x ∈ [−1, 1],

for −1/2 < λ ≤ 1/2. The equality holds only for x = ±1 and, if n is even, for x = 0. Further partial results on an extension of this inequality to normalized Jacobi polynomials are given. Key words: Tur´ an inequality, orthogonal polynomials, ultraspherical polynomials 1991 MSC: 41A17, 68W30 1 Introduction and statement of the result Let P (λ)

m , m = 0, 1, 2, . . . be the m-th ultraspherical polynomial, orthogonal in

[−1, 1] with respect to the weight function wλ(x) = (1 − x2)λ−1/2, λ > −1/2, and normalized by P (λ)

m (1) = m+2λ−1 m

. We shall need a different normalization

The first named author was supported by the National Science Foundation through Contract no. DDVU 02/30. The second named author was supported by the Austrian Science Fund (FWF) under grant P22748-N18. Email addresses: geno@fmi.uni-sofia.bg (Geno Nikolov), veronika.pillwein@risc.jku.at (Veronika Pillwein). Preprint submitted to Elsevier 26 April 2013

1 1

2 1 2

1

1 3 2 3

1 1

2 1 2

1

1 3 2 3

Fig. 1. Graphs of

∆n(x)/(1 − x2) (straight line) and ∆n(x)/(1 − x2) (dashed) for n = 7, 8 and λ = 1

4

for these polynomials, namely, we shall require that they take value 1 at x = 1, so we set pm(x) = p(λ)

m (x) := P (λ) m (x)/P (λ) m (1),

m = 0, 1, . . . , (1.1) where, for the sake of brevity, the superscript (λ) will be omitted hereafter. We prove the following extension of Tur´ an’s inequality: Theorem 1 Let pn be defined by (1.1), and λ ∈ (−1/2, 1/2]. Then, for every n ∈ N, |x|p2

n(x) − pn−1(x)pn+1(x) ≥ 0

for every x ∈ [−1, 1] . (1.2) The equality in (1.2) holds only for x = ±1 and, if n is even, for x = 0. Moreover, (1.2) fails for every λ > 1/2 and n ∈ N. This variation of Tur´ an’s inequality was introduced by Gerhold and Kauers [15] and proven in the limit case λ = 1/2, i.e., for the Legendre polynomials. 2 Classical analysis of Theorem 1 Assume first that {pm} is a general sequence of orthogonal polynomials, defined by the three term recurrence equation xpn(x) = γnpn+1(x) + αnpn−1(x), n = 0, 1, 2, . . . , (2.1) where p−1(x) := 0, p0(x) = 1, αn > 0, γn > 0, and αn + γn = 1 for every n ∈ N0. Clearly, pm(−x) = (−1)m and pm(1) = 1 for every m ∈ N0. By these properties is easy to see that |x|p2

n(x) − pn−1(x)pn+1(x) is an even function,

which vanishes at x = ±1 and, if n is even, also at x = 0. We therefore set

∆n(x) := xp2

n(x) − pn−1(x)pn+1(x) ,

n ∈ N , (2.2) and our goal is to examine what conditions guarantee that ∆n(x) > 0 for every x ∈ (0, 1). We start with some representations of ∆n(x). 2

SLIDE 74

A Classical Proof

Lemma 2 Assume that the sequence {pm} satisfies the three term recurrence relation (2.1). Then the following representations hold true: γn ∆n(x) = γnxp2

n(x) + αnp2 n−1(x) − xpn−1(x)pn(x) ,

(2.3) αn ∆n(x) = αnxp2

n(x) + γnp2 n+1(x) − xpn(x)pn+1(x) ,

(2.4) γn ∆n(x) = (γnx − γn−1)p2

n(x) + (αn − αn−1x)p2 n−1(x) + αn−1

∆n−1(x) , (2.5) γn ∆n(x) =x

αnpn−1(x) − γnpn(x)
pn−1(x) − pn(x)
+ αn(1 − x)p2

n−1(x) ,

(2.6)

Proof. Formulae (2.3) and (2.4) follow from rewriting γnpn+1(x) and αnpn−1(x)

in γn ∆n and αn ∆n, respectively, using the recurrence equation (2.1). Subtract- ing (2.4) (with n − 1 instead of n) from (2.3), we obtain (2.5). Formula (2.6) is deduced by multiplying xpn−1(x)pn(x) in the right-hand side of (2.3) by γn + αn (= 1), and then adding and subtracting αn x p2

n−1(x).

Our next lemma shows that the inequality

∆n(x) > 0 generally holds true in a subinterval of (0, 1) . Lemma 3 Assume that the sequence {pm} satisfies the three term recurrence relation (2.1). Then

∆n(x) > 0

for every x ∈ (0, 4αnγn) .

Proof. The right-hand side of (2.3) is equal to

√αnpn−1(x) − x 2√αn pn(x) 2 + x 4αn

4αnγn − x
p2

n(x) .

Both summands are non-negative if x ∈ (0, 4αnγn). Moreover, for x ∈ (0, 4αnγn) this expression would be equal to zero only if both pn(x) and pn−1(x) are equal to zero, which is impossible, since the zeros of pn and pn−1 interlace.

From now on, we restrict our considerations to the case of ultraspherical poly-

nomials, i.e., {pm} = {p(λ)

m }, as normalized by (1.1). The zeros of pm are

denoted henceforth by x1,m(λ) < x2,m(λ) < · · · < xm,m(λ). We collect in the next lemma some well-known properties of ultraspherical polynomials, which will be needed for the proof of Theorem 1. Lemma 4 (i) {pn} = {p(λ)

n } satisfy the recurrence relation (2.1) with

γn = n + 2λ 2(n + λ), αn = n 2(n + λ) . (2.7) 3 (ii) The positive zeros of p(λ)

n

are strictly monotone decreasing functions of λ in (−1/2, ∞). Moreover, for every n ≥ 2, xn,n(λ) ≤

(n − 1)(n + 2λ + 1)

(n + λ)2 + 3λ + 5/4 + 3(λ + 1/2)2/(n − 1) 1/2 . (2.8) (iii) The following relations hold true: p

n(x) := d

dx

p(λ)

n (x)

= n(n + 2λ)

2λ + 1 p(λ+1)

n−1 (x) ,

(2.9) pn−1(x) = 1 n(1 − x2)p

n(x) + xpn(x) .

(2.10) The above properties of p(λ)

n

are easily obtained from their analogues for P (λ)

n ,

given, e.g., in Szeg˝

’s monograph [34]. The recurrence relation (2.1) with the

coefficients γn and αn given in (2.7) follows from [34, Eqn. (4.7.17)]. Formulae (2.9) and (2.10) are consequences of [34, loc. cit. (4.7.14) and (4.7.27)]. The monotone dependence of the zeros of p(λ)

n

n λ follows from a well-known
bservation due to A. A. Markov, see e.g., [34, Theorem 6.12.1]. The upper

bound (2.8) for the extreme zeros of ultraspherical polynomials is proved in [26, Lemma 3.5] (for other bounds for the extreme zeros of classical orthogonal polynomials, see, e.g., [10] and the references therein). Set zn(λ) := 4αnγn = 1 − λ2 (n + λ)2 . The following is an immediate consequence of Lemma 3: Corollary 5 Let {pm} = {p(λ)

m }, λ > −1/2, be the sequence of ultraspherical

polynomials. Then for every n ∈ N,
∆n(x) > 0 ,

x ∈

0, zn(λ)
.

In view of Corollary 5, (1.2) is true for λ = 0, and to prove (1.2) for λ ∈ (−1/2, 0) ∪ (0, 1/2], we have to show that ∆n(x) > 0 when x ∈ [zn(λ), 1). The case n = 1, λ ∈ (−1/2, 1/2] is easily verified. Namely,

∆1(x) = x3 − 2(λ + 1)

2λ + 1 x2 + 1 2λ + 1 = x − 1 2λ + 1

(2λ + 1)x2 − x − 1
,

and the polynomial q(x) = (2λ+1)x2 −x−1 has a unique positive root. Since q(1) = 2λ − 1 ≤ 0, it follows that q(x) < 0, and consequently ∆1(x) > 0 for every x ∈ (0, 1). 4

SLIDE 75

A Classical Proof

We therefore assume in what follows that n ≥ 2. In our proof of the inequality

∆n(x) > 0, x ∈ [zn(λ), 1) we shall distinguish between the cases λ ∈ (−1/2, 0)

and λ ∈ (0, 1/2). Lemma 6 If λ ∈ (−1/2, 0), then ∆n(x) > 0 for every x ∈ [zn(λ), 1).

Proof. We use induction with respect to n. The case n = 1 was settled above,

and we assume that, for some n ≥ 2, ∆n−1(x) > 0 for every x ∈ [zn−1(λ), 1). Since zn−1(λ) < zn(λ), we have also ∆n−1(x) > 0 for every x ∈ [zn(λ), 1). By the interlacing property and monotonicity of the zeros of ultraspherical polynomials, we have 0 ≤ xn−1,n−1(λ + 1) ≤ xn−1,n−1(λ) < xn,n(λ) < xn+1,n+1(λ) < 1 , (2.11) where the first two inequalities are strict unless n = 2, in which case we have x1,1(λ + 1) = x1,1(λ) = 0. Next, we show that if λ ∈ (−1/2, 0), then the largest zero of p

n, which, in

view of (2.9), is xn−1,n−1(λ + 1), satisfies xn−1,n−1(λ + 1) < zn(λ) . (2.12) Indeed, by Lemma 4(ii) we readily get xn−1,n−1(λ+1) ≤ xn−1,n−1(1/2) ≤

1−

5 n2−n+3 1/2 < 1− 1 (2n−1)2 ≤ zn(λ) . As is seen from (2.3) and (2.4), the inequality ∆n(x) > 0 is true whenever pn−1(x)pn(x) ≤ 0 or pn(x)pn+1(x) ≤ 0, in particular, ∆n(x) > 0 in the interval [xn−1,n−1(λ), xn+1,n+1(λ)]. Set In := (xn−1,n−1(λ + 1), xn−1,n−1(λ)) . In view of (2.11) and (2.12), the induction step from n − 1 to n will be done if we manage to show that ∆n(x) > 0 for x ∈ In (this interval is void when n = 2) and for x ∈ (xn+1,n+1(λ), 1). Assume first that x ∈ (xn+1,n+1(λ), 1), then 0 < pn(x) < pn−1(x), since the zeros of pn − pn−1 interlace with the zeros of pn, and the rightmost zero of pn − pn−1 is at x = 1. Moreover, since αn > 1

2 > γn > 0 for λ ∈ (−1/2, 0), we

have αnpn−1(x) − γnpn(x) > 0. Then by (2.6) we conclude that γn ∆n(x) > x

αnpn−1(x)−γnpn(x)
pn−1(x)−pn(x)
> 0 , x ∈ (xn+1,n+1(λ), 1) .

5 Now assume that n ≥ 3 and x ∈ In ∩ [zn(λ), 1). By (2.5) and the inductional hypothesis, we have γn ∆n(x) > (αn−1x − αn) γnx − γn−1 αn−1x − αn p2

n(x) − p2 n−1(x)

,

(2.13) and it suffices to show that the right-hand side of the inequality (2.13) is positive in In ∩[zn(λ), 1). A straightforward calculation using (2.1) shows that if λ ∈ (−1/2, 0) and x ∈ [zn(λ), 1), then αn−1x−αn ≥ 4αn−1αnγn−αn = αn(4αn−1γn−1) = − λ(λ + 1)αn (n + λ)(n + λ − 1) > 0. Therefore, the right-hand side of inequality (2.13) is positive in In ∩ [zn(λ), 1) when γnx − γn−1 αn−1x − αn p2

n(x) − p2 n−1(x) > 0 ,

x ∈ In ∩ [zn(λ), 1) . (2.14) According to (2.10) we have pn−1(x) − xpn(x) = (1 − x2)p

n(x)/n, hence

pn−1(x) > xpn(x) for x ∈ In ; moreover, since both pn−1(x) and xpn(x) are negative in In, we get x2p2

n(x) > p2 n−1(x) ,

x ∈ In . (2.15) We shall show that ψ(x) := γnx − γn−1 αn−1x − αn > x2 , x ∈ [zn(λ), 1) , then obviously (2.14) is a consequence from (2.15). The function ψ is contin- uous in [zn(λ), 1]; moreover, from αn + γn = αn−1 + γn−1 = 1 we find that ψ(1) = 1 and ψ(x) = (αn − αn−1)(αn−1 + αn − 1) (αn−1x − αn)2 < 0 , x ∈ [zn(λ), 1) , since αn−1 > αn > 1/2. Thus, ψ(x) is decreasing function in [zn(λ), 1), and ψ(x) > 1 > x2 therein. Consequently, (2.14) is true, and therefore ∆n(x) > 0 for x ∈ In ∩ [zn(λ), 1). The proof of Lemma 6 is complete.

Next, we prove the analogue of Lemma 6 for the case λ ∈ (0, 1/2].

Lemma 7 If λ ∈ (0, 1/2], then ∆n(x) > 0 for every x ∈ [zn(λ), 1).

Proof. Again, we apply induction with respect to n, and the base case n = 1

was already settled. As in the proof of Lemma 6, we assume that, for some n ≥ 2, ∆n−1(x) > 0 in [zn−1(λ, 1), then ∆n−1(x) > 0 in [zn(λ, 1), too. To 6

SLIDE 76

A Classical Proof

accomplish the induction step from n − 1 to n, we observe that if λ ∈ (0, 1/2], then zn(λ) > xn+1,n+1(λ) . (2.16) Indeed, by Lemma 4(ii) we have xn+1,n+1(λ) < xn+1,n+1(0) = cos

π 2n+2, while

zn(λ) ≥ zn(1/2) = 1 − 1/(2n + 1)2. Then (2.16) follows from the inequality sin2

π 4(n+1) > 1 2(2n+1)2, which is true since sin t > 2 πt for t ∈ (0, π/2).

In view of (2.6), the inductional hypothesis and (2.16), to prove the inequality

∆n(x) > 0 for x ∈ (zn(λ), 1), it suffices to show that

(γnx − γn−1)p2

n(x) + (αn − αn−1x)p2 n−1(x) > 0,

x ∈ [xn+1,n+1(λ), 1) . (2.17) For λ > 0 the sequences {γn} and {αn} defined by (2.7) satisfy γn 1

2 and αn 1 2

as n → ∞ . Therefore, γnx − γn−1 ≤ γn − γn−1 < 0 and αn − αn−1x ≥ αn − αn−1 > 0. Since pn(x) > 0 and pn−1(x) > 0 for x ∈ [xn+1,n+1(λ), 1), the inequality (2.17) is equivalent to ϕ(x) := pn−1(x) pn(x) ≥

γn−1 − γnx

αn − αn−1x, [xn+1,n+1(λ), 1) . (2.18) It is well-known that ϕ(x) is monotone decreasing and convex in (xn,n, ∞), where xn,n is the rightmost zero of pn, see e.g. [34, Theorem 3.3.5] for a general

result. For the sake of completeness, we propose a direct proof for the case

pn = p(λ)

n . By (2.10), we have

ϕ(x) = pn−1(x) pn(x) = x + 1 − x2 n p

n(x)

pn(x) = x + 1 − x2 n

n

k=1

1 x − xk,n , (2.19) where {xk,n} = {xk,n(λ)} are the zeros of pn. Differentiating the last expression, we obtain that ϕ(x) < 0 and ϕ(x) > 0 for x > xn,n(λ). Indeed, ϕ(x) = 1 − 2x n

n

k=1

1 x − xk,n − 1 − x2 n

n

k=1

1 (x − xk,n)2 = 1 n

n

k=1

(x − xk,n)2 − 2x(x − xk,n) − 1 + x2 (x − xk,n)2 = 1 n

n

k=1

x2

k,n − 1

(x − xk,n)2 < 0 , and ϕ(x) = 2 n

n

k=1

1 − x2

k,n

(x − xk,n)3 > 0 . 7 Since ϕ(1) = 1, it follows from the convexity of ϕ that ϕ(x) > 1 + ϕ(1)(x − 1) in (xn,n(λ), 1). We make use of (2.19) and (2.9) to calculate ϕ(1): ϕ(1) = 1 − 2 n p

n(1) = 2n + 2λ − 1

2λ + 1 . Therefore, we have ϕ(x) > 1 + 2n + 2λ − 1 2λ + 1 (1 − x) for x ∈ [xn+1,n+1(λ), 1) . (2.20) Now we estimate the right-hand side of (2.18). On using (2.7), we find γn−1 − γnx αn − αn−1x = 1+(1 − αn−1 − αn)(1 − x) αn − αn−1x = 1+ λ(2n + 2λ − 1)(1 − x)

n(n + λ − 1) − λ
(1 − x) + λ

. For n ≥ 1, λ > 0 and 0 < x < 1 we have

n(n + λ − 1) − λ
(1 − x) + λ ≥ λ,

therefore γn−1 − γnx αn − αn−1x ≤ 1 + (2n + 2λ − 1)(1 − x) . In view of this estimate and (2.20), the inequality (2.18) will be proved if we manage to show that 1+2n + 2λ − 1 2λ + 1 (1−x) ≥

1 + (2n + 2λ − 1)(1 − x) for x ∈ [xn+1,n+1(λ), 1) .

After squaring the both sides of this inequality, we find that a sufficient con- dition for its validity is 2/(2λ + 1) ≥ 1, i.e., λ ≤ 1/2. Thus, (2.17) is true, and therefore ∆n(x) > 0 for x ∈ [zn(λ), 1). The proof of Lemma 7 is complete. Summarizing, the inequality (2.1) follows from: 1) Corollary 5 for λ − 0; 2) Corollary 5 and Lemma 6 for λ ∈ (−1/2, 0); 3) Corollary 5 and Lemma 7 for λ ∈ (0, 1/2]. It remains to prove the last claim of Theorem 1, namely that if λ > 1/2, then (1.2) fails for every n ∈ N. On using pn−1(1) = pn(1) = 1, (2.9) and (2.1), we find γn ∆

n(1) = −αn + (2γn − 1)p n(1) + (2αn − 1)p n−1(1)

= (2λ − 1)(n + 2λ) 2(2λ + 1)(n + λ) . If λ > 1/2, then ∆

n(1) > 0, and hence

∆n(1−ε) < ∆n(1) = 0 for a sufficiently small ε > 0. This completes the proof of Theorem 1. 8

SLIDE 77

A conjectured Tur´ an inequality

Let pn(x) = P(α,β)

n

(x)/P(α,β)

n

(1). Then for all n ≥ 0 and −1 < α ≤ 0, −1 ≤ β ≤ 1 and all x ∈ [−1, 1], ∆n(x) = |x|p2

n(x) − pn−1(x)pn+1(x) ≥ 0.

1.0 0.5 0.5 1.0 2 4 6 8 10 12 1.0 0.5 0.5 1.0 2 4 6 8 10

SLIDE 78

A partial result

◮ It is easy to show that

cn∆n(x) = |x|cnpn(x)2 + anpn−1(x)2 − (x − bn)pn−1(x)pn(x).

SLIDE 79

A partial result

◮ It is easy to show that

cn∆n(x) = |x|cnpn(x)2 + anpn−1(x)2 − (x − bn)pn−1(x)pn(x).

◮ Using CAD we can determine ξ1, ξ2 ∈ [0, 1] depending on a, c

such that ∀ y0, y−1, a, c, x ∈ R: (0 ≤ c ≤ 1 ∧ 0 ≤ a ≤ 1 ∧ ξ1 ≤ x ≤ ξ2) = ⇒ xcy2

0 + ay2 −1 − (x − (1 − a − c))y−1y0 ≥ 0

SLIDE 80

A partial result

◮ It is easy to show that

cn∆n(x) = |x|cnpn(x)2 + anpn−1(x)2 − (x − bn)pn−1(x)pn(x).

◮ Using CAD we can determine ξ1, ξ2 ∈ [0, 1] depending on a, c

such that ∀ y0, y−1, a, c, x ∈ R: (0 ≤ c ≤ 1 ∧ 0 ≤ a ≤ 1 ∧ ξ1 ≤ x ≤ ξ2) = ⇒ xcy2

0 + ay2 −1 − (x − (1 − a − c))y−1y0 ≥ 0 ◮ For −1 < α ≤ 0, −1 ≤ β ≤ 1 we have

lim

n→∞ ξ1(n, α, β) = 0,

and lim

n→∞ ξ2(n, α, β) = 1.

SLIDE 81

A partial result

◮ It is easy to show that

cn∆n(x) = |x|cnpn(x)2 + anpn−1(x)2 − (x − bn)pn−1(x)pn(x).

◮ Using CAD we can determine ξ1, ξ2 ∈ [0, 1] depending on a, c

such that ∀ y0, y−1, a, c, x ∈ R: (0 ≤ c ≤ 1 ∧ 0 ≤ a ≤ 1 ∧ ξ1 ≤ x ≤ ξ2) = ⇒ xcy2

0 + ay2 −1 − (x − (1 − a − c))y−1y0 ≥ 0 ◮ For −1 < α ≤ 0, −1 ≤ β ≤ 1 we have

lim

n→∞ ξ1(n, α, β) = 0,

and lim

n→∞ ξ2(n, α, β) = 1. ◮ Analogously bounds for x ∈ [−1, 0] can be found.

SLIDE 82

A conjectured Tur´ an inequality

Let pn(x) = P(α,β)

n

(x)/P(α,β)

n

(1). Then for all n ≥ 0 and −1 < α ≤ 0, −1 ≤ β ≤ 1 and all x ∈ [−1, 1], ∆n(x) = |x|p2

n(x) − pn−1(x)pn+1(x) ≥ 0.

1.0 0.5 0.5 1.0 2 4 6 8 10 12 1.0 0.5 0.5 1.0 2 4 6 8 10