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Pillwein’s Proof of
Schoberl’s Conjecture
Manuel Kauers
currently at INRIA-Rocquencourt
usually at RISC-Linz
Polynomial Inequalities
Polynomial Inequalities
. . . arise regularly as Monthly problems . . .
Polynomial Inequalities
. . . arise regularly as Monthly problems . . .
. . . are often considered difficult . . .
Polynomial Inequalities
. . . arise regularly as Monthly problems . . .
. . . are often considered difficult . . .
. . . but are in fact trivial . . .
Polynomial Inequalities
. . . arise regularly as Monthly problems . . .
. . . are often considered difficult . . .
. . . but are in fact trivial . . .
. . . if we have a fast computer.
Polynomial Inequalities
I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847):Let a, b, c be positive real numbers, two of which are ≤ 1,satisfying ab + ac + bc = 3. Show that
1
(a + b)2+
1
(a + c)2+
1
(b + c)2− 3
4≥ 3(a − 1)(b − 1)(c − 1)
2(a + b)(a + c)(b + c)
Polynomial Inequalities
I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847):Let a, b, c be positive real numbers, two of which are ≤ 1,satisfying ab + ac + bc = 3. Show that
1
(a + b)2+
1
(a + c)2+
1
(b + c)2− 3
4≥ 3(a − 1)(b − 1)(c − 1)
2(a + b)(a + c)(b + c)
I Problem 11301 (Finbarr Holland; vol. 114(10), 2007, p. 547):Find the least number M such that for all a, b, c,
|ab(a2 − b2) + bc(b2 − c2) + ca(c2 − a2)| ≤ M(a2 + b2 + c2)2.
Polynomial Inequalities
A Tarski formula is a formula composed of
Polynomial Inequalities
A Tarski formula is a formula composed of
I rational numbers (1, 2,−3117 , . . . )
Polynomial Inequalities
A Tarski formula is a formula composed of
I rational numbers (1, 2,−3117 , . . . )
I variables (x, y, . . . )
Polynomial Inequalities
A Tarski formula is a formula composed of
I rational numbers (1, 2,−3117 , . . . )
I variables (x, y, . . . )
I arithmetic operations (+,−, ·, /)
Polynomial Inequalities
A Tarski formula is a formula composed of
I rational numbers (1, 2,−3117 , . . . )
I variables (x, y, . . . )
I arithmetic operations (+,−, ·, /)
I comparison predicates (=, 6=, <, >,≤,≥)
Polynomial Inequalities
A Tarski formula is a formula composed of
I rational numbers (1, 2,−3117 , . . . )
I variables (x, y, . . . )
I arithmetic operations (+,−, ·, /)
I comparison predicates (=, 6=, <, >,≤,≥)
I boolean operations (∧,∨, . . . )
Polynomial Inequalities
A Tarski formula is a formula composed of
I rational numbers (1, 2,−3117 , . . . )
I variables (x, y, . . . )
I arithmetic operations (+,−, ·, /)
I comparison predicates (=, 6=, <, >,≤,≥)
I boolean operations (∧,∨, . . . )
I quantifiers (∀,∃)
Polynomial Inequalities
I Problem 11251:
∀a, b, c :(
a > 0 ∧ 1 ≥ b > 0 ∧ 1 ≥ c > 0 ∧ ab + ac + bc = 3
⇒ 1(a+b)2
+ 1(a+c)2
+ 1(b+c)2
− 34 ≥ 3(a−1)(b−1)(c−1)
2(a+b)(a+c)(b+c)
)
Polynomial Inequalities
I Problem 11251:
∀a, b, c :(
a > 0 ∧ 1 ≥ b > 0 ∧ 1 ≥ c > 0 ∧ ab + ac + bc = 3
⇒ 1(a+b)2
+ 1(a+c)2
+ 1(b+c)2
− 34 ≥ 3(a−1)(b−1)(c−1)
2(a+b)(a+c)(b+c)
)
I Problem 11301:
∀a, b, c :(∣∣ab(a2 − b2) + bc(b2 − c2) + ca(c2 − a2)
∣∣
≤ M(a2 + b2 + c2)2)
Polynomial Inequalities
Theorem. (Tarski, 1948) Every Tarski formula is, as a statementabout real numbers, equivalent to a Tarski formula without anyquantifiers.
Polynomial Inequalities
Theorem. (Tarski, 1948) Every Tarski formula is, as a statementabout real numbers, equivalent to a Tarski formula without anyquantifiers.
There are Quantifier Elimination algorithms which take arbitraryTarski formulas as input and compute an equivalent quantifier freeformula.
Polynomial Inequalities
Theorem. (Tarski, 1948) Every Tarski formula is, as a statementabout real numbers, equivalent to a Tarski formula without anyquantifiers.
There are Quantifier Elimination algorithms which take arbitraryTarski formulas as input and compute an equivalent quantifier freeformula.
One such algorithm is due to Collins (Cylindrical AlgebraicDecomposition, CAD, 1975).
Polynomial Inequalities
I Problem 11251:
∀a, b, c :(
a > 0 ∧ 1 ≥ b > 0 ∧ 1 ≥ c > 0 ∧ ab + ac + bc = 3
⇒ 1(a+b)2
+ 1(a+c)2
+ 1(b+c)2
− 34 ≥ 3(a−1)(b−1)(c−1)
2(a+b)(a+c)(b+c)
)
Polynomial Inequalities
I Problem 11251:
∀a, b, c :(
a > 0 ∧ 1 ≥ b > 0 ∧ 1 ≥ c > 0 ∧ ab + ac + bc = 3
⇒ 1(a+b)2
+ 1(a+c)2
+ 1(b+c)2
− 34 ≥ 3(a−1)(b−1)(c−1)
2(a+b)(a+c)(b+c)
)
CAD
−−−→ true
Polynomial Inequalities
I Problem 11251:
∀a, b, c :(
a > 0 ∧ 1 ≥ b > 0 ∧ 1 ≥ c > 0 ∧ ab + ac + bc = 3
⇒ 1(a+b)2
+ 1(a+c)2
+ 1(b+c)2
− 34 ≥ 3(a−1)(b−1)(c−1)
2(a+b)(a+c)(b+c)
)
CAD
−−−→ true
I Problem 11301:
∀a, b, c :(∣∣ab(a2 − b2) + bc(b2 − c2) + ca(c2 − a2)
∣∣
≤ M(a2 + b2 + c2)2)
Polynomial Inequalities
I Problem 11251:
∀a, b, c :(
a > 0 ∧ 1 ≥ b > 0 ∧ 1 ≥ c > 0 ∧ ab + ac + bc = 3
⇒ 1(a+b)2
+ 1(a+c)2
+ 1(b+c)2
− 34 ≥ 3(a−1)(b−1)(c−1)
2(a+b)(a+c)(b+c)
)
CAD
−−−→ true
I Problem 11301:
∀a, b, c :(∣∣ab(a2 − b2) + bc(b2 − c2) + ca(c2 − a2)
∣∣
≤ M(a2 + b2 + c2)2)
CAD
−−−→ M ≥ 932
√2
Polynomial Inequalities
Message:
Polynomial inequalities can be proven by CAD
without thinking.
Polynomial Inequalities
Message:
Polynomial inequalities can be proven by CAD
without thinking.
The rest of this talk is aboutinequalities that can be proven by CAD with thinking only.
Bernoulli’s Inequality
∀ n ∈ N ∀ x ≥ −1 : (x + 1)n ≥ 1 + nx.
Bernoulli’s Inequality
∀ n ∈ N ∀ x ≥ −1 : (x + 1)n − (1 + nx) ≥ 0.
Bernoulli’s Inequality
∀ n ∈ N ∀ x ≥ −1 : (x + 1)n − (1 + nx) ≥ 0.
What exactly does (x + 1)n − (1 + nx) mean?
Bernoulli’s Inequality
∀ n ∈ N ∀ x ≥ −1 : (x + 1)n − (1 + nx) ≥ 0.
What exactly does (x + 1)n − (1 + nx) mean?
I For any specific integer n, it is a polynomial in x.
Bernoulli’s Inequality
∀ n ∈ N ∀ x ≥ −1 : (x + 1)n − (1 + nx) ≥ 0.
What exactly does (x + 1)n − (1 + nx) mean?
I For any specific integer n, it is a polynomial in x.
I View (x + 1)n − (1 + nx) as a sequence of polynomials.
Bernoulli’s Inequality
∀ n ∈ N ∀ x ≥ −1 : (x + 1)n − (1 + nx) ≥ 0.
What exactly does (x + 1)n − (1 + nx) mean?
I For any specific integer n, it is a polynomial in x.
I View (x + 1)n − (1 + nx) as a sequence of polynomials.
I View Bernoulli’s inequality as a sequence of polynomialinequalities.
Bernoulli’s Inequality
∀ n ∈ N ∀ x ≥ −1 : (x + 1)n − (1 + nx) ≥ 0.
-4 -3 -2 -1 1 2 3 4
-20
-15
-10
-5
5
10
15
20
Bernoulli’s Inequality
∀ n ∈ N ∀ x ≥ −1 : (x + 1)n − (1 + nx) ≥ 0.
-4 -3 -2 -1 1 2 3 4
-20
-15
-10
-5
5
10
15
20
Bernoulli’s Inequality
∀ n ∈ N ∀ x ≥ −1 : (x + 1)n − (1 + nx) ≥ 0.
-4 -3 -2 -1 1 2 3 4
-20
-15
-10
-5
5
10
15
20
Bernoulli’s Inequality
∀ n ∈ N ∀ x ≥ −1 : (x + 1)n − (1 + nx) ≥ 0.
-4 -3 -2 -1 1 2 3 4
-20
-15
-10
-5
5
10
15
20
Bernoulli’s Inequality
∀ n ∈ N ∀ x ≥ −1 : (x + 1)n − (1 + nx) ≥ 0.
-4 -3 -2 -1 1 2 3 4
-20
-15
-10
-5
5
10
15
20
Bernoulli’s Inequality
∀ n ∈ N ∀ x ≥ −1 : (x + 1)n − (1 + nx) ≥ 0.
-4 -3 -2 -1 1 2 3 4
-20
-15
-10
-5
5
10
15
20
Bernoulli’s Inequality
∀ n ∈ N ∀ x ≥ −1 : (x + 1)n − (1 + nx) ≥ 0.
-4 -3 -2 -1 1 2 3 4
-20
-15
-10
-5
5
10
15
20
Bernoulli’s Inequality
∀ n ∈ N ∀ x ≥ −1 : (x + 1)n − (1 + nx) ≥ 0.
-4 -3 -2 -1 1 2 3 4
-20
-15
-10
-5
5
10
15
20
Bernoulli’s Inequality
∀ n ∈ N ∀ x ≥ −1 : (x + 1)n − (1 + nx) ≥ 0.
-4 -3 -2 -1 1 2 3 4
-20
-15
-10
-5
5
10
15
20
Bernoulli’s Inequality
∀ n ∈ N ∀ x ≥ −1 : (x + 1)n − (1 + nx) ≥ 0.
-4 -3 -2 -1 1 2 3 4
-20
-15
-10
-5
5
10
15
20
Bernoulli’s Inequality
∀ n ∈ N ∀ x ≥ −1 : (x + 1)n − (1 + nx) ≥ 0.
-4 -3 -2 -1 1 2 3 4
-20
-15
-10
-5
5
10
15
20
Bernoulli’s Inequality
∀ n ∈ N ∀ x ≥ −1 : (x + 1)n − (1 + nx) ≥ 0.
I Can we show Bernoulli’s inequality with CAD?
Bernoulli’s Inequality
∀ n ∈ N ∀ x ≥ −1 : (x + 1)n − (1 + nx) ≥ 0.
I Can we show Bernoulli’s inequality with CAD?
I Can CAD be used to do induction on n?
Bernoulli’s Inequality
∀ n ∈ N ∀ x ≥ −1 : (x + 1)n − (1 + nx) ≥ 0.
I Can we show Bernoulli’s inequality with CAD?
I Can CAD be used to do induction on n?
I Let fn(x) := (x + 1)n − (1 + nx).
Bernoulli’s Inequality
∀ n ∈ N ∀ x ≥ −1 : (x + 1)n − (1 + nx) ≥ 0.
I Can we show Bernoulli’s inequality with CAD?
I Can CAD be used to do induction on n?
I Let fn(x) := (x + 1)n − (1 + nx).
I Induction step:
∀n ∈ N ∀x ≥ −1 : fn(x) ≥ 0 ⇒ fn+1(x) ≥ 0
Bernoulli’s Inequality
∀ n ∈ N ∀ x ≥ −1 : (x + 1)n − (1 + nx) ≥ 0.
I Can we show Bernoulli’s inequality with CAD?
I Can CAD be used to do induction on n?
I Let fn(x) := (x + 1)n − (1 + nx).
I Induction step:
∀n ∈ N ∀x ≥ −1 : fn(x) ≥ 0 ⇒ fn+1(x) ≥ 0
I Exploit the recurrence fn+1(x) = (x + 1)fn(x) + nx2
Bernoulli’s Inequality
∀ n ∈ N ∀ x ≥ −1 : (x + 1)n − (1 + nx) ≥ 0.
I Can we show Bernoulli’s inequality with CAD?
I Can CAD be used to do induction on n?
I Let fn(x) := (x + 1)n − (1 + nx).
I Induction step:
∀n ∈ N ∀x ≥ −1 : fn(x) ≥ 0 ⇒ (x + 1)fn(x) + nx2 ≥ 0
I Exploit the recurrence fn+1(x) = (x + 1)fn(x) + nx2
Bernoulli’s Inequality
∀ n ∈ N ∀ x ≥ −1 : (x + 1)n − (1 + nx) ≥ 0.
I Can we show Bernoulli’s inequality with CAD?
I Can CAD be used to do induction on n?
I Let fn(x) := (x + 1)n − (1 + nx).
I Induction step:
∀n ∈ N ∀x ≥ −1 : fn(x) ≥ 0 ⇒ (x + 1)fn(x) + nx2 ≥ 0
I Exploit the recurrence fn+1(x) = (x + 1)fn(x) + nx2
I Generalize fn(x) to y and n ∈ N to n ≥ 0
Bernoulli’s Inequality
∀ n ∈ N ∀ x ≥ −1 : (x + 1)n − (1 + nx) ≥ 0.
I Can we show Bernoulli’s inequality with CAD?
I Can CAD be used to do induction on n?
I Let fn(x) := (x + 1)n − (1 + nx).
I Induction step:
∀n ≥ 0 ∀y ∀x ≥ −1 : y ≥ 0 ⇒ (x + 1)y + nx2 ≥ 0
I Exploit the recurrence fn+1(x) = (x + 1)fn(x) + nx2
I Generalize fn(x) to y and n ∈ N to n ≥ 0
Bernoulli’s Inequality
∀ n ∈ N ∀ x ≥ −1 : (x + 1)n − (1 + nx) ≥ 0.
I Can we show Bernoulli’s inequality with CAD?
I Can CAD be used to do induction on n?
I Let fn(x) := (x + 1)n − (1 + nx).
I Induction step:
∀n ≥ 0 ∀y ∀x ≥ −1 : y ≥ 0 ⇒ (x + 1)y + nx2 ≥ 0
I Exploit the recurrence fn+1(x) = (x + 1)fn(x) + nx2
I Generalize fn(x) to y and n ∈ N to n ≥ 0
I The resulting formula is indeed true.
Bernoulli’s Inequality
∀ n ∈ N ∀ x ≥ −1 : (x + 1)n − (1 + nx) ≥ 0.
I Can we show Bernoulli’s inequality with CAD?
I Can CAD be used to do induction on n?
I Let fn(x) := (x + 1)n − (1 + nx).
I Induction step:
∀n ∈ N ∀x ≥ −1 : fn(x) ≥ 0 ⇒ fn+1(x) ≥ 0
I This proves the induction step.
Bernoulli’s Inequality
∀ n ∈ N ∀ x ≥ −1 : (x + 1)n − (1 + nx) ≥ 0.
I Can we show Bernoulli’s inequality with CAD?
I Can CAD be used to do induction on n?
I Let fn(x) := (x + 1)n − (1 + nx).
I Induction step:
∀n ∈ N ∀x ≥ −1 : fn(x) ≥ 0 ⇒ fn+1(x) ≥ 0
I This proves the induction step.
I The induction base 0 ≥ 0 is trivial.
Bernoulli’s Inequality
∀ n ∈ N ∀ x ≥ −1 : (x + 1)n − (1 + nx) ≥ 0.
I Can we show Bernoulli’s inequality with CAD?
I Can CAD be used to do induction on n?
I Let fn(x) := (x + 1)n − (1 + nx).
I Induction step:
∀n ∈ N ∀x ≥ −1 : fn(x) ≥ 0 ⇒ fn+1(x) ≥ 0
I This proves the induction step.
I The induction base 0 ≥ 0 is trivial.
I This completes the proof.
Bernoulli’s Inequality
Message:
We may use CAD to construct an induction proof
for the positivity of a quantity satisfying a
recurrence.
Legendre Polynomials
There are other interesting sequences of polynomials.
Legendre Polynomials
There are other interesting sequences of polynomials.
For example, Legendre Polynomials Pn(x).
Legendre Polynomials
There are other interesting sequences of polynomials.
For example, Legendre Polynomials Pn(x).
I P0(x) = 1
-1 -0.5 0.5 1
-2
-1
1
2
Legendre Polynomials
There are other interesting sequences of polynomials.
For example, Legendre Polynomials Pn(x).
I P0(x) = 1
I P1(x) = x-1 -0.5 0.5 1
-2
-1
1
2
Legendre Polynomials
There are other interesting sequences of polynomials.
For example, Legendre Polynomials Pn(x).
I P0(x) = 1
I P1(x) = x
I P2(x) = 32x2 − 1
2
-1 -0.5 0.5 1
-2
-1
1
2
Legendre Polynomials
There are other interesting sequences of polynomials.
For example, Legendre Polynomials Pn(x).
I P0(x) = 1
I P1(x) = x
I P2(x) = 32x2 − 1
2
I P3(x) = 52x3 − 3
2x
-1 -0.5 0.5 1
-2
-1
1
2
Legendre Polynomials
There are other interesting sequences of polynomials.
For example, Legendre Polynomials Pn(x).
I P0(x) = 1
I P1(x) = x
I P2(x) = 32x2 − 1
2
I P3(x) = 52x3 − 3
2x
I P4(x) = 358 x4 − 15
4 x2 + 38
-1 -0.5 0.5 1
-2
-1
1
2
Legendre Polynomials
There are other interesting sequences of polynomials.
For example, Legendre Polynomials Pn(x).
I P0(x) = 1
I P1(x) = x
I P2(x) = 32x2 − 1
2
I P3(x) = 52x3 − 3
2x
I P4(x) = 358 x4 − 15
4 x2 + 38
I P5(x) = 638 x5 − 35
4 x3 + 158 x
-1 -0.5 0.5 1
-2
-1
1
2
Legendre Polynomials
There are other interesting sequences of polynomials.
For example, Legendre Polynomials Pn(x).
I P0(x) = 1
I P1(x) = x
I P2(x) = 32x2 − 1
2
I P3(x) = 52x3 − 3
2x
I P4(x) = 358 x4 − 15
4 x2 + 38
I P5(x) = 638 x5 − 35
4 x3 + 158 x
I P6(x) = 23116 x6 − 315
16 x4 + 10516 x2 − 5
16
-1 -0.5 0.5 1
-2
-1
1
2
Legendre Polynomials
There are other interesting sequences of polynomials.
For example, Legendre Polynomials Pn(x).
I P0(x) = 1
I P1(x) = x
I P2(x) = 32x2 − 1
2
I P3(x) = 52x3 − 3
2x
I P4(x) = 358 x4 − 15
4 x2 + 38
I P5(x) = 638 x5 − 35
4 x3 + 158 x
I P6(x) = 23116 x6 − 315
16 x4 + 10516 x2 − 5
16
I P7(x) = 42916 x7 − 693
16 x5 + 31516 x3 − 35
16x
-1 -0.5 0.5 1
-2
-1
1
2
Legendre Polynomials
There are other interesting sequences of polynomials.
For example, Legendre Polynomials Pn(x).
I P0(x) = 1
I P1(x) = x
I P2(x) = 32x2 − 1
2
I P3(x) = 52x3 − 3
2x
I P4(x) = 358 x4 − 15
4 x2 + 38
I P5(x) = 638 x5 − 35
4 x3 + 158 x
I P6(x) = 23116 x6 − 315
16 x4 + 10516 x2 − 5
16
I P7(x) = 42916 x7 − 693
16 x5 + 31516 x3 − 35
16x
I P8(x) = 6435128 x8 − 3003
32 x6 + 346564 x4 − 315
32 x2 + 35128
-1 -0.5 0.5 1
-2
-1
1
2
Legendre Polynomials
There are other interesting sequences of polynomials.
For example, Legendre Polynomials Pn(x).
-1 -0.5 0.5 1
-2
-1
1
2
These polynomials satisfy
∫ 1
−1Pn(x)Pm(x)dx =
2
2n + 1δn,m
so they are orthogonal to each other.
Legendre Polynomials
There are other interesting sequences of polynomials.
For example, Legendre Polynomials Pn(x).
-1 -0.5 0.5 1
-2
-1
1
2
These polynomials satisfy
∫ 1
−1Pn(x)Pm(x)dx =
2
2n + 1δn,m
so they are orthogonal to each other.
They also satisfy a recurrence:
(n + 2)Pn+2(x) − (2n + 3)xPn+1(x) + (n + 1)Pn(x) = 0
Legendre Polynomials
There are other interesting sequences of polynomials.
For example, Legendre Polynomials Pn(x).
-1 -0.5 0.5 1
-2
-1
1
2
These polynomials satisfy
∫ 1
−1Pn(x)Pm(x)dx =
2
2n + 1δn,m
so they are orthogonal to each other.
They also satisfy a recurrence:
(n + 2)Pn+2(x) − (2n + 3)xPn+1(x) + (n + 1)Pn(x) = 0
and various interesting inequalities, e.g.,
∀n ∈ N ∀x : −1 ≤ x ≤ 1 ⇒ |Pn(x)| ≤ 1.
Legendre Polynomials
There are other interesting sequences of polynomials.
For example, Legendre Polynomials Pn(x).
-1 -0.5 0.5 1
-2
-1
1
2
These polynomials satisfy
∫ 1
−1Pn(x)Pm(x)dx =
2
2n + 1δn,m
so they are orthogonal to each other.
They also satisfy a recurrence:
(n + 2)Pn+2(x) − (2n + 3)xPn+1(x) + (n + 1)Pn(x) = 0
and various interesting inequalities, e.g.,
∀n ∈ N ∀x : −1 ≤ x ≤ 1 ⇒ |Pn(x)| ≤ 1.
Legendre Polynomials: Turan’s Inequality
Here is an inequality about Pn(x) that can be shown with CAD:
∀n ∈ N ∀x : −1 ≤ x ≤ 1 ⇒ P 2n+1(x) − Pn(x)Pn+2(x)
︸ ︷︷ ︸
=:∆n(x)
≥ 0
Legendre Polynomials: Turan’s Inequality
Here is an inequality about Pn(x) that can be shown with CAD:
∀n ∈ N ∀x : −1 ≤ x ≤ 1 ⇒ P 2n+1(x) − Pn(x)Pn+2(x)
︸ ︷︷ ︸
=:∆n(x)
≥ 0
-1 -0.5 0.5 1
-0.4
-0.2
0.2
0.4
Legendre Polynomials: Turan’s Inequality
Here is an inequality about Pn(x) that can be shown with CAD:
∀n ∈ N ∀x : −1 ≤ x ≤ 1 ⇒ P 2n+1(x) − Pn(x)Pn+2(x)
︸ ︷︷ ︸
=:∆n(x)
≥ 0
-1 -0.5 0.5 1
-0.4
-0.2
0.2
0.4
Legendre Polynomials: Turan’s Inequality
Here is an inequality about Pn(x) that can be shown with CAD:
∀n ∈ N ∀x : −1 ≤ x ≤ 1 ⇒ P 2n+1(x) − Pn(x)Pn+2(x)
︸ ︷︷ ︸
=:∆n(x)
≥ 0
-1 -0.5 0.5 1
-0.4
-0.2
0.2
0.4
Legendre Polynomials: Turan’s Inequality
Here is an inequality about Pn(x) that can be shown with CAD:
∀n ∈ N ∀x : −1 ≤ x ≤ 1 ⇒ P 2n+1(x) − Pn(x)Pn+2(x)
︸ ︷︷ ︸
=:∆n(x)
≥ 0
-1 -0.5 0.5 1
-0.4
-0.2
0.2
0.4
Legendre Polynomials: Turan’s Inequality
Here is an inequality about Pn(x) that can be shown with CAD:
∀n ∈ N ∀x : −1 ≤ x ≤ 1 ⇒ P 2n+1(x) − Pn(x)Pn+2(x)
︸ ︷︷ ︸
=:∆n(x)
≥ 0
-1 -0.5 0.5 1
-0.4
-0.2
0.2
0.4
Legendre Polynomials: Turan’s Inequality
Here is an inequality about Pn(x) that can be shown with CAD:
∀n ∈ N ∀x : −1 ≤ x ≤ 1 ⇒ P 2n+1(x) − Pn(x)Pn+2(x)
︸ ︷︷ ︸
=:∆n(x)
≥ 0
-1 -0.5 0.5 1
-0.4
-0.2
0.2
0.4
Legendre Polynomials: Turan’s Inequality
Here is an inequality about Pn(x) that can be shown with CAD:
∀n ∈ N ∀x : −1 ≤ x ≤ 1 ⇒ P 2n+1(x) − Pn(x)Pn+2(x)
︸ ︷︷ ︸
=:∆n(x)
≥ 0
-1 -0.5 0.5 1
-0.4
-0.2
0.2
0.4
Legendre Polynomials: Turan’s Inequality
Here is an inequality about Pn(x) that can be shown with CAD:
∀n ∈ N ∀x : −1 ≤ x ≤ 1 ⇒ P 2n+1(x) − Pn(x)Pn+2(x)
︸ ︷︷ ︸
=:∆n(x)
≥ 0
-1 -0.5 0.5 1
-0.4
-0.2
0.2
0.4
Legendre Polynomials: Turan’s Inequality
Here is an inequality about Pn(x) that can be shown with CAD:
∀n ∈ N ∀x : −1 ≤ x ≤ 1 ⇒ P 2n+1(x) − Pn(x)Pn+2(x)
︸ ︷︷ ︸
=:∆n(x)
≥ 0
-1 -0.5 0.5 1
-0.4
-0.2
0.2
0.4
Legendre Polynomials: Turan’s Inequality
Here is an inequality about Pn(x) that can be shown with CAD:
∀n ∈ N ∀x : −1 ≤ x ≤ 1 ⇒ P 2n+1(x) − Pn(x)Pn+2(x)
︸ ︷︷ ︸
=:∆n(x)
≥ 0
-1 -0.5 0.5 1
-0.4
-0.2
0.2
0.4
I This is known as Turan’s
inequality.
Legendre Polynomials: Turan’s Inequality
Here is an inequality about Pn(x) that can be shown with CAD:
∀n ∈ N ∀x : −1 ≤ x ≤ 1 ⇒ P 2n+1(x) − Pn(x)Pn+2(x)
︸ ︷︷ ︸
=:∆n(x)
≥ 0
-1 -0.5 0.5 1
-0.4
-0.2
0.2
0.4
I This is known as Turan’s
inequality.
I For specific n, it is just apolynomial inequality.
Legendre Polynomials: Turan’s Inequality
Here is an inequality about Pn(x) that can be shown with CAD:
∀n ∈ N ∀x : −1 ≤ x ≤ 1 ⇒ P 2n+1(x) − Pn(x)Pn+2(x)
︸ ︷︷ ︸
=:∆n(x)
≥ 0
-1 -0.5 0.5 1
-0.4
-0.2
0.2
0.4
I This is known as Turan’s
inequality.
I For specific n, it is just apolynomial inequality.
I For general n, it is not easy.(Try it.)
Legendre Polynomials: Turan’s Inequality
Here is an inequality about Pn(x) that can be shown with CAD:
∀n ∈ N ∀x : −1 ≤ x ≤ 1 ⇒ P 2n+1(x) − Pn(x)Pn+2(x)
︸ ︷︷ ︸
=:∆n(x)
≥ 0
-1 -0.5 0.5 1
-0.4
-0.2
0.2
0.4
I This is known as Turan’s
inequality.
I For specific n, it is just apolynomial inequality.
I For general n, it is not easy.(Try it.)
A proof for general n can be obtained in the same way as forBernoulli’s inequality.
Legendre Polynomials: Turan’s Inequality
Here is an inequality about Pn(x) that can be shown with CAD:
∀n ∈ N ∀x : −1 ≤ x ≤ 1 ⇒ P 2n+1(x) − Pn(x)Pn+2(x)
︸ ︷︷ ︸
=:∆n(x)
≥ 0
Induction step:
∀n ∈ N ∀x :(−1 ≤ x ≤ 1 ∧ ∆n(x) ≥ 0
)⇒ ∆n+1(x) ≥ 0.
Legendre Polynomials: Turan’s Inequality
Here is an inequality about Pn(x) that can be shown with CAD:
∀n ∈ N ∀x : −1 ≤ x ≤ 1 ⇒ P 2n+1(x) − Pn(x)Pn+2(x)
︸ ︷︷ ︸
=:∆n(x)
≥ 0
Induction step:
∀n ∈ N ∀x :(−1 ≤ x ≤ 1 ∧ ∆n(x) ≥ 0
)⇒ ∆n+1(x) ≥ 0.
Use the recurrence for Pn(x) to obtain
∆n(x) = (n+1)n+2 Pn(x)2 − 2n+3
n+2 xPn+1(x)Pn(x) + Pn+1(x)2
∆n+1(x) = (n+1)2
(n+2)2Pn(x)2 − (n+1)(2n2+9n+8)x
(n+2)2(n+3)Pn+1(x)Pn(x)
+ (n+2)3−(2n+3)x2
(n+2)2(n+3)Pn+1(x)2
Legendre Polynomials: Turan’s Inequality
Here is an inequality about Pn(x) that can be shown with CAD:
∀n ∈ N ∀x : −1 ≤ x ≤ 1 ⇒ P 2n+1(x) − Pn(x)Pn+2(x)
︸ ︷︷ ︸
=:∆n(x)
≥ 0
Relaxing Pn(x) to y, and Pn+1(x) to z, and n ∈ N to n ≥ 0 leadsto the formula
Legendre Polynomials: Turan’s Inequality
Here is an inequality about Pn(x) that can be shown with CAD:
∀n ∈ N ∀x : −1 ≤ x ≤ 1 ⇒ P 2n+1(x) − Pn(x)Pn+2(x)
︸ ︷︷ ︸
=:∆n(x)
≥ 0
Relaxing Pn(x) to y, and Pn+1(x) to z, and n ∈ N to n ≥ 0 leadsto the formula
∀n ∀x∀y ∀z :(n ≥ 0 ∧ x2 ≤ 1 ∧ n+1
n+2y2 − 2n+3n+2 xyz + z2 ≥ 0
)
⇒( (n+1)2
(n+2)2y2 − (n+1)(2n2+9n+8)x
(n+2)2(n+3)yz + (n+2)3−(2n+3)x2
(n+2)2(n+3)z2 ≥ 0
)
Legendre Polynomials: Turan’s Inequality
Here is an inequality about Pn(x) that can be shown with CAD:
∀n ∈ N ∀x : −1 ≤ x ≤ 1 ⇒ P 2n+1(x) − Pn(x)Pn+2(x)
︸ ︷︷ ︸
=:∆n(x)
≥ 0
Relaxing Pn(x) to y, and Pn+1(x) to z, and n ∈ N to n ≥ 0 leadsto the formula
∀n ∀x∀y ∀z :(n ≥ 0 ∧ x2 ≤ 1 ∧ n+1
n+2y2 − 2n+3n+2 xyz + z2 ≥ 0
)
⇒( (n+1)2
(n+2)2y2 − (n+1)(2n2+9n+8)x
(n+2)2(n+3)yz + (n+2)3−(2n+3)x2
(n+2)2(n+3)z2 ≥ 0
),
which is indeed true.
Legendre Polynomials: Turan’s Inequality
Here is an inequality about Pn(x) that can be shown with CAD:
∀n ∈ N ∀x : −1 ≤ x ≤ 1 ⇒ P 2n+1(x) − Pn(x)Pn+2(x)
︸ ︷︷ ︸
=:∆n(x)
≥ 0
Relaxing Pn(x) to y, and Pn+1(x) to z, and n ∈ N to n ≥ 0 leadsto the formula
∀n ∀x∀y ∀z :(n ≥ 0 ∧ x2 ≤ 1 ∧ n+1
n+2y2 − 2n+3n+2 xyz + z2 ≥ 0
)
⇒( (n+1)2
(n+2)2y2 − (n+1)(2n2+9n+8)x
(n+2)2(n+3)yz + (n+2)3−(2n+3)x2
(n+2)2(n+3)z2 ≥ 0
),
which is indeed true. This proves the induction step.
Legendre Polynomials: Turan’s Inequality
Here is an inequality about Pn(x) that can be shown with CAD:
∀n ∈ N ∀x : −1 ≤ x ≤ 1 ⇒ P 2n+1(x) − Pn(x)Pn+2(x)
︸ ︷︷ ︸
=:∆n(x)
≥ 0
Relaxing Pn(x) to y, and Pn+1(x) to z, and n ∈ N to n ≥ 0 leadsto the formula
∀n ∀x∀y ∀z :(n ≥ 0 ∧ x2 ≤ 1 ∧ n+1
n+2y2 − 2n+3n+2 xyz + z2 ≥ 0
)
⇒( (n+1)2
(n+2)2y2 − (n+1)(2n2+9n+8)x
(n+2)2(n+3)yz + (n+2)3−(2n+3)x2
(n+2)2(n+3)z2 ≥ 0
),
which is indeed true. This proves the induction step.
The induction base ∆0(x) ≥ 0 is trivial.
Legendre Polynomials: Turan’s Inequality
Here is an inequality about Pn(x) that can be shown with CAD:
∀n ∈ N ∀x : −1 ≤ x ≤ 1 ⇒ P 2n+1(x) − Pn(x)Pn+2(x)
︸ ︷︷ ︸
=:∆n(x)
≥ 0
Relaxing Pn(x) to y, and Pn+1(x) to z, and n ∈ N to n ≥ 0 leadsto the formula
∀n ∀x∀y ∀z :(n ≥ 0 ∧ x2 ≤ 1 ∧ n+1
n+2y2 − 2n+3n+2 xyz + z2 ≥ 0
)
⇒( (n+1)2
(n+2)2y2 − (n+1)(2n2+9n+8)x
(n+2)2(n+3)yz + (n+2)3−(2n+3)x2
(n+2)2(n+3)z2 ≥ 0
),
which is indeed true. This proves the induction step.
The induction base ∆0(x) ≥ 0 is trivial. This completes the proof.
Legendre Polynomials: Turan’s Inequality
Message:
A “deep” special function inequality may be just an
immediate consequence of a polynomial inequality.
Legendre Polynomials: Turan’s Inequality
Turan’s inequality
∆n(x) = Pn+1(x)2 − Pn(x)Pn+2(x) ≥ 0
-1 -0.5 0.5 1
-0.4
-0.2
0.2
0.4
Legendre Polynomials: Turan’s Inequality
Turan’s inequality can be improved to
∆n(x) = Pn+1(x)2 − Pn(x)Pn+2(x) ≥ αn(1 − x2)
-1 -0.5 0.5 1
-0.4
-0.2
0.2
0.4
where αn = ∆n(0).
Legendre Polynomials: Turan’s Inequality
Turan’s inequality can be improved to
∆n(x) = Pn+1(x)2 − Pn(x)Pn+2(x) ≥ αn(1 − x2)
-1 -0.5 0.5 1
-0.4
-0.2
0.2
0.4
where αn = ∆n(0).
Legendre Polynomials: Turan’s Inequality
Turan’s inequality can be improved to
∆n(x) = Pn+1(x)2 − Pn(x)Pn+2(x) ≥ αn(1 − x2)
-1 -0.5 0.5 1
-0.4
-0.2
0.2
0.4
where αn = ∆n(0).
Legendre Polynomials: Turan’s Inequality
Turan’s inequality can be improved to
∆n(x) = Pn+1(x)2 − Pn(x)Pn+2(x) ≥ αn(1 − x2)
-1 -0.5 0.5 1
-0.4
-0.2
0.2
0.4
where αn = ∆n(0).
Legendre Polynomials: Turan’s Inequality
Turan’s inequality can be improved to
∆n(x) = Pn+1(x)2 − Pn(x)Pn+2(x) ≥ αn(1 − x2)
-1 -0.5 0.5 1
-0.4
-0.2
0.2
0.4
where αn = ∆n(0).
Legendre Polynomials: Turan’s Inequality
Turan’s inequality can be improved to
∆n(x) = Pn+1(x)2 − Pn(x)Pn+2(x) ≥ αn(1 − x2)
-1 -0.5 0.5 1
-0.4
-0.2
0.2
0.4
where αn = ∆n(0).
Legendre Polynomials: Turan’s Inequality
Turan’s inequality can be improved to
∆n(x) = Pn+1(x)2 − Pn(x)Pn+2(x) ≥ αn(1 − x2)
-1 -0.5 0.5 1
-0.4
-0.2
0.2
0.4
where αn = ∆n(0).
Legendre Polynomials: Turan’s Inequality
Turan’s inequality can be improved to
∆n(x) = Pn+1(x)2 − Pn(x)Pn+2(x) ≥ αn(1 − x2)
-1 -0.5 0.5 1
-0.4
-0.2
0.2
0.4
where αn = ∆n(0).
Legendre Polynomials: Turan’s Inequality
Turan’s inequality can be improved to
∆n(x) = Pn+1(x)2 − Pn(x)Pn+2(x) ≥ αn(1 − x2)
-1 -0.5 0.5 1
-0.4
-0.2
0.2
0.4
where αn = ∆n(0).
Can we show this also by induction?
Legendre Polynomials: Turan’s Inequality
Turan’s inequality can be improved to
∆n(x) = Pn+1(x)2 − Pn(x)Pn+2(x) ≥ αn(1 − x2)
-1 -0.5 0.5 1
-0.4
-0.2
0.2
0.4
where αn = ∆n(0).
Can we show this also by induction?
We have the recurrence
(n + 3)(n + 4)αn+2
= (2n + 5)αn+1 + (n + 1)(n + 2)αn.
Legendre Polynomials: Turan’s Inequality
Turan’s inequality can be improved to
∆n(x) = Pn+1(x)2 − Pn(x)Pn+2(x) ≥ αn(1 − x2)
-1 -0.5 0.5 1
-0.4
-0.2
0.2
0.4
where αn = ∆n(0).
Can we show this also by induction?
We have the recurrence
(n + 3)(n + 4)αn+2
= (2n + 5)αn+1 + (n + 1)(n + 2)αn.
A Tarski formula encoding the induction step would be. . .
Legendre Polynomials: Turan’s Inequality
Turan’s inequality can be improved to
∆n(x) = Pn+1(x)2 − Pn(x)Pn+2(x) ≥ αn(1 − x2)
∀n, x, y, z, a, b :(n ≥ 0 ∧ x2 ≤ 1 ∧ n+1
n+2y2 − 2n+3n+2 xyz + z2 ≥ a(1 − x2)
∧ (n+1)2
(n+2)2 y2 − (n+1)(2n2+9n+8)x
(n+2)2(n+3) yz + (n+2)3−(2n+3)x2
(n+2)2(n+3) z2 ≥ b(1 − x2))
⇒( (n+1)2((n+3)3−(2n+5)x2)
(n+4)(n+3)2(n+2)2 y2
+ (n+1)(2(2n+3)(2n+5)x2−(2n
4+21n3+83n
2+142n+86))(n+4)(n+3)2(n+2)2 xyz
+ ((n+4)(n+2)4−(2n+3)2(2n+5)x4+(n+1)(2n+3)(2n+5)x2)(n+4)(n+3)(n+2) z2
≥ (n+1)(n+2)(n+3)(n+4)a + (2n+5)
(n+3)(n+4)b).
Legendre Polynomials: Turan’s Inequality
Turan’s inequality can be improved to
∆n(x) = Pn+1(x)2 − Pn(x)Pn+2(x) ≥ αn(1 − x2)
∀n, x, y, z, a, b :(n ≥ 0 ∧ x2 ≤ 1 ∧ n+1
n+2y2 − 2n+3n+2 xyz + z2 ≥ a(1 − x2)
∧ (n+1)2
(n+2)2 y2 − (n+1)(2n2+9n+8)x
(n+2)2(n+3) yz + (n+2)3−(2n+3)x2
(n+2)2(n+3) z2 ≥ b(1 − x2))
⇒( (n+1)2((n+3)3−(2n+5)x2)
(n+4)(n+3)2(n+2)2 y2
+ (n+1)(2(2n+3)(2n+5)x2−(2n
4+21n3+83n
2+142n+86))(n+4)(n+3)2(n+2)2 xyz
+ ((n+4)(n+2)4−(2n+3)2(2n+5)x4+(n+1)(2n+3)(2n+5)x2)(n+4)(n+3)(n+2) z2
≥ (n+1)(n+2)(n+3)(n+4)a + (2n+5)
(n+3)(n+4)b).
Unfortunately, this is false.
Legendre Polynomials: Turan’s Inequality
Turan’s inequality can be improved to
∆n(x) = Pn+1(x)2 − Pn(x)Pn+2(x) ≥ αn(1 − x2)
∀n, x, y, z, a, b :(n ≥ 0 ∧ x2 ≤ 1 ∧ n+1
n+2y2 − 2n+3n+2 xyz + z2 ≥ a(1 − x2)
∧ (n+1)2
(n+2)2 y2 − (n+1)(2n2+9n+8)x
(n+2)2(n+3) yz + (n+2)3−(2n+3)x2
(n+2)2(n+3) z2 ≥ b(1 − x2))
⇒( (n+1)2((n+3)3−(2n+5)x2)
(n+4)(n+3)2(n+2)2 y2
+ (n+1)(2(2n+3)(2n+5)x2−(2n
4+21n3+83n
2+142n+86))(n+4)(n+3)2(n+2)2 xyz
+ ((n+4)(n+2)4−(2n+3)2(2n+5)x4+(n+1)(2n+3)(2n+5)x2)(n+4)(n+3)(n+2) z2
≥ (n+1)(n+2)(n+3)(n+4)a + (2n+5)
(n+3)(n+4)b).
Unfortunately, this is false. We must be more careful.
Legendre Polynomials: Turan’s Inequality
Turan’s inequality can be improved to
∆n(x) = Pn+1(x)2 − Pn(x)Pn+2(x) ≥ αn(1 − x2)
-1 -0.5 0.5 1
-0.4
-0.2
0.2
0.4
Observations:
Legendre Polynomials: Turan’s Inequality
Turan’s inequality can be improved to
∆n(x) = Pn+1(x)2 − Pn(x)Pn+2(x) ≥ αn(1 − x2)
-1 -0.5 0.5 1
-0.4
-0.2
0.2
0.4
Observations:
I By symmetry, it suffices to considerx ≥ 0.
Legendre Polynomials: Turan’s Inequality
Turan’s inequality can be improved to
∆n(x) = Pn+1(x)2 − Pn(x)Pn+2(x) ≥ αn(1 − x2)
-1 -0.5 0.5 1
-0.4
-0.2
0.2
0.4
Observations:
I By symmetry, it suffices to considerx ≥ 0.
I For x = 0 there is nothing to show.
Legendre Polynomials: Turan’s Inequality
Turan’s inequality can be improved to
∆n(x) = Pn+1(x)2 − Pn(x)Pn+2(x) ≥ αn(1 − x2)
-1 -0.5 0.5 1
-0.4
-0.2
0.2
0.4
Observations:
I By symmetry, it suffices to considerx ≥ 0.
I For x = 0 there is nothing to show.
I For x > 0, it suffices to show that∆n(x)/(1 − x2) is increasing.
Legendre Polynomials: Turan’s Inequality
Turan’s inequality can be improved to
∆n(x) = Pn+1(x)2 − Pn(x)Pn+2(x) ≥ αn(1 − x2)
-1 -0.5 0.5 1
-0.4
-0.2
0.2
0.4
Observations:
I By symmetry, it suffices to considerx ≥ 0.
I For x = 0 there is nothing to show.
I For x > 0, it suffices to show that∆n(x)/(1 − x2) is increasing.
New idea: Show thatd
dx
∆n(x)
1 − x2≥ 0
Legendre Polynomials: Turan’s Inequality
Turan’s inequality can be improved to
∆n(x) = Pn+1(x)2 − Pn(x)Pn+2(x) ≥ αn(1 − x2)
We have
d
dx
∆n(x)
1 − x2=
(
(n + 1)Pn(x)2 − ((2n + 3)x + 1)Pn+1(x)Pn(x)
+ (n + x + 2)Pn+1(x)2)
/(
(n + 2)(x − 1)2(x + 1))
Legendre Polynomials: Turan’s Inequality
Turan’s inequality can be improved to
∆n(x) = Pn+1(x)2 − Pn(x)Pn+2(x) ≥ αn(1 − x2)
We have
d
dx
∆n(x)
1 − x2=
(
(n + 1)Pn(x)2 − ((2n + 3)x + 1)Pn+1(x)Pn(x)
+ (n + x + 2)Pn+1(x)2)
/(
(n + 2)(x − 1)2(x + 1))
A positivity proof for the latter expression by CAD and inductionon n succeeds.
Legendre Polynomials: Turan’s Inequality
Message:
A special function inequality may require some
non-obvious manipulation before an induction proof
via CAD succeeds.
Schoberl’s Conjecture
Schoberl’s Conjecture
I In the higher order finite elementmethod (FEM), solutions of PDEs arelocally approximated by polynomials.
Schoberl’s Conjecture
I In the higher order finite elementmethod (FEM), solutions of PDEs arelocally approximated by polynomials.
I Some basis polynomials lead to betternumerical performance than thestandard basis 1, x, x2, x3, . . . .
Schoberl’s Conjecture
I In the higher order finite elementmethod (FEM), solutions of PDEs arelocally approximated by polynomials.
I Some basis polynomials lead to betternumerical performance than thestandard basis 1, x, x2, x3, . . . .
I In a certain application, a basis f0(x), f1(x), f2(x), . . . wasneeded which satisfies
Schoberl’s Conjecture
I In the higher order finite elementmethod (FEM), solutions of PDEs arelocally approximated by polynomials.
I Some basis polynomials lead to betternumerical performance than thestandard basis 1, x, x2, x3, . . . .
I In a certain application, a basis f0(x), f1(x), f2(x), . . . wasneeded which satisfies
I
∫ 1
−1
fn(x)q(x)dx = q(0) for all q with deg q ≤ n.
Schoberl’s Conjecture
I In the higher order finite elementmethod (FEM), solutions of PDEs arelocally approximated by polynomials.
I Some basis polynomials lead to betternumerical performance than thestandard basis 1, x, x2, x3, . . . .
I In a certain application, a basis f0(x), f1(x), f2(x), . . . wasneeded which satisfies
I
∫ 1
−1
fn(x)q(x)dx = q(0) for all q with deg q ≤ n.
I
∫ 1
−1
|fn(x)|dx ≤ C for some constant C.
Schoberl’s Conjecture
I The Legendre kernel polynomials
kn(x, y) :=n + 1
2(x − y)(Pn+1(x)Pn(y) − Pn+1(y)Pn(x))
Schoberl’s Conjecture
I The Legendre kernel polynomials
kn(x, y) :=n + 1
2(x − y)(Pn+1(x)Pn(y) − Pn+1(y)Pn(x))
have the property
∫ 1
−1kn(x, y)q(x)dx = q(y),
for all q with deg q ≤ n.
Schoberl’s Conjecture
I The Legendre kernel polynomials
kn(x, y) :=n + 1
2(x − y)(Pn+1(x)Pn(y) − Pn+1(y)Pn(x))
have the property
∫ 1
−1kn(x, y)q(x)dx = q(y),
for all q with deg q ≤ n.
I So fn(x) := kn(x, 0) satisfies the first condition.
Schoberl’s Conjecture
I The Legendre kernel polynomials
kn(x, y) :=n + 1
2(x − y)(Pn+1(x)Pn(y) − Pn+1(y)Pn(x))
have the property
∫ 1
−1kn(x, y)q(x)dx = q(y),
for all q with deg q ≤ n.
I So fn(x) := kn(x, 0) satisfies the first condition.
I But not the second.
Schoberl’s Conjecture
I Schoberl next considered the “gliding averages”
fn(x) :=1
n + 1
2n∑
i=n
ki(x, 0).
Schoberl’s Conjecture
I Schoberl next considered the “gliding averages”
fn(x) :=1
n + 1
2n∑
i=n
ki(x, 0).
I He could show that this family does the job
Schoberl’s Conjecture
I Schoberl next considered the “gliding averages”
fn(x) :=1
n + 1
2n∑
i=n
ki(x, 0).
I He could show that this family does the job if and only if. . .
n∑
k=0
(4k + 1)(2n − 2k + 1)P2k(0)P2k(x) ≥ 0
for all x ∈ [−1, 1] and all n ∈ N.
Schoberl’s Conjecture
I Schoberl next considered the “gliding averages”
fn(x) :=1
n + 1
2n∑
i=n
ki(x, 0).
I He could show that this family does the job if and only if. . .
n∑
k=0
(4k + 1)(2n − 2k + 1)P2k(0)P2k(x) ≥ 0
for all x ∈ [−1, 1] and all n ∈ N.
I Hence was born the Schoberl conjecture.
Schoberl’s Conjecture
Consider
Sn(x) :=n∑
k=0
(4k + 1)(2n − 2k + 1)P2k(0)P2k(x)
for n = 0, 1, . . . , 20.
-1 -0.5 0.5 1
-5
-2.5
2.5
5
7.5
10
12.5
15
Schoberl’s Conjecture
Consider
Sn(x) :=n∑
k=0
(4k + 1)(2n − 2k + 1)P2k(0)P2k(x)
for n = 0, 1, . . . , 20.
-1 -0.5 0.5 1
-5
-2.5
2.5
5
7.5
10
12.5
15
Schoberl’s Conjecture
Consider
Sn(x) :=n∑
k=0
(4k + 1)(2n − 2k + 1)P2k(0)P2k(x)
for n = 0, 1, . . . , 20.
-1 -0.5 0.5 1
-5
-2.5
2.5
5
7.5
10
12.5
15
Schoberl’s Conjecture
Consider
Sn(x) :=n∑
k=0
(4k + 1)(2n − 2k + 1)P2k(0)P2k(x)
for n = 0, 1, . . . , 20.
-1 -0.5 0.5 1
-5
-2.5
2.5
5
7.5
10
12.5
15
Schoberl’s Conjecture
Consider
Sn(x) :=n∑
k=0
(4k + 1)(2n − 2k + 1)P2k(0)P2k(x)
for n = 0, 1, . . . , 20.
-1 -0.5 0.5 1
-5
-2.5
2.5
5
7.5
10
12.5
15
Schoberl’s Conjecture
Consider
Sn(x) :=n∑
k=0
(4k + 1)(2n − 2k + 1)P2k(0)P2k(x)
for n = 0, 1, . . . , 20.
-1 -0.5 0.5 1
-5
-2.5
2.5
5
7.5
10
12.5
15
Schoberl’s Conjecture
Consider
Sn(x) :=n∑
k=0
(4k + 1)(2n − 2k + 1)P2k(0)P2k(x)
for n = 0, 1, . . . , 20.
-1 -0.5 0.5 1
-5
-2.5
2.5
5
7.5
10
12.5
15
Schoberl’s Conjecture
Consider
Sn(x) :=n∑
k=0
(4k + 1)(2n − 2k + 1)P2k(0)P2k(x)
for n = 0, 1, . . . , 20.
-1 -0.5 0.5 1
-5
-2.5
2.5
5
7.5
10
12.5
15
Schoberl’s Conjecture
Consider
Sn(x) :=n∑
k=0
(4k + 1)(2n − 2k + 1)P2k(0)P2k(x)
for n = 0, 1, . . . , 20.
-1 -0.5 0.5 1
-5
-2.5
2.5
5
7.5
10
12.5
15
Schoberl’s Conjecture
Consider
Sn(x) :=n∑
k=0
(4k + 1)(2n − 2k + 1)P2k(0)P2k(x)
for n = 0, 1, . . . , 20.
-1 -0.5 0.5 1
-5
-2.5
2.5
5
7.5
10
12.5
15
Schoberl’s Conjecture
Consider
Sn(x) :=n∑
k=0
(4k + 1)(2n − 2k + 1)P2k(0)P2k(x)
for n = 0, 1, . . . , 20.
-1 -0.5 0.5 1
-5
-2.5
2.5
5
7.5
10
12.5
15
Schoberl’s Conjecture
Consider
Sn(x) :=n∑
k=0
(4k + 1)(2n − 2k + 1)P2k(0)P2k(x)
for n = 0, 1, . . . , 20.
-1 -0.5 0.5 1
-5
-2.5
2.5
5
7.5
10
12.5
15
Schoberl’s Conjecture
Consider
Sn(x) :=n∑
k=0
(4k + 1)(2n − 2k + 1)P2k(0)P2k(x)
for n = 0, 1, . . . , 20.
-1 -0.5 0.5 1
-5
-2.5
2.5
5
7.5
10
12.5
15
Schoberl’s Conjecture
Consider
Sn(x) :=n∑
k=0
(4k + 1)(2n − 2k + 1)P2k(0)P2k(x)
for n = 0, 1, . . . , 20.
-1 -0.5 0.5 1
-5
-2.5
2.5
5
7.5
10
12.5
15
Schoberl’s Conjecture
Consider
Sn(x) :=n∑
k=0
(4k + 1)(2n − 2k + 1)P2k(0)P2k(x)
for n = 0, 1, . . . , 20.
-1 -0.5 0.5 1
-5
-2.5
2.5
5
7.5
10
12.5
15
Schoberl’s Conjecture
Consider
Sn(x) :=n∑
k=0
(4k + 1)(2n − 2k + 1)P2k(0)P2k(x)
for n = 0, 1, . . . , 20.
-1 -0.5 0.5 1
-5
-2.5
2.5
5
7.5
10
12.5
15
Schoberl’s Conjecture
Consider
Sn(x) :=n∑
k=0
(4k + 1)(2n − 2k + 1)P2k(0)P2k(x)
for n = 0, 1, . . . , 20.
-1 -0.5 0.5 1
-5
-2.5
2.5
5
7.5
10
12.5
15
Schoberl’s Conjecture
Consider
Sn(x) :=n∑
k=0
(4k + 1)(2n − 2k + 1)P2k(0)P2k(x)
for n = 0, 1, . . . , 20.
-1 -0.5 0.5 1
-5
-2.5
2.5
5
7.5
10
12.5
15
Schoberl’s Conjecture
Consider
Sn(x) :=n∑
k=0
(4k + 1)(2n − 2k + 1)P2k(0)P2k(x)
for n = 0, 1, . . . , 20.
-1 -0.5 0.5 1
-5
-2.5
2.5
5
7.5
10
12.5
15
Schoberl’s Conjecture
Consider
Sn(x) :=n∑
k=0
(4k + 1)(2n − 2k + 1)P2k(0)P2k(x)
for n = 0, 1, . . . , 20.
-1 -0.5 0.5 1
-5
-2.5
2.5
5
7.5
10
12.5
15
Schoberl’s Conjecture
Consider
Sn(x) :=n∑
k=0
(4k + 1)(2n − 2k + 1)P2k(0)P2k(x)
for n = 0, 1, . . . , 20.
-1 -0.5 0.5 1
-5
-2.5
2.5
5
7.5
10
12.5
15
Schoberl’s Conjecture
Consider
Sn(x) :=n∑
k=0
(4k + 1)(2n − 2k + 1)P2k(0)P2k(x)
for n = 0, 1, . . . , 20.
-1 -0.5 0.5 1
-5
-2.5
2.5
5
7.5
10
12.5
15
Schoberl’s Conjecture
Consider
Sn(x) :=n∑
k=0
(4k + 1)(2n − 2k + 1)P2k(0)P2k(x)
for n = 0, 1, . . . , 20.
-1 -0.5 0.5 1
-5
-2.5
2.5
5
7.5
10
12.5
15
I The conjecture seems to betrue.
Schoberl’s Conjecture
Consider
Sn(x) :=n∑
k=0
(4k + 1)(2n − 2k + 1)P2k(0)P2k(x)
for n = 0, 1, . . . , 20.
-1 -0.5 0.5 1
-5
-2.5
2.5
5
7.5
10
12.5
15
I The conjecture seems to betrue.
I For specific n ∈ N, it can beshown without thinking.
Schoberl’s Conjecture
Consider
Sn(x) :=n∑
k=0
(4k + 1)(2n − 2k + 1)P2k(0)P2k(x)
for n = 0, 1, . . . , 20.
-1 -0.5 0.5 1
-5
-2.5
2.5
5
7.5
10
12.5
15
I The conjecture seems to betrue.
I For specific n ∈ N, it can beshown without thinking.
I It can be also be shown forx = −1, x = 0, x = +1.
Schoberl’s Conjecture
Consider
Sn(x) :=n∑
k=0
(4k + 1)(2n − 2k + 1)P2k(0)P2k(x)
for n = 0, 1, . . . , 20.
-1 -0.5 0.5 1
-5
-2.5
2.5
5
7.5
10
12.5
15
I The conjecture seems to betrue.
I For specific n ∈ N, it can beshown without thinking.
I It can be also be shown forx = −1, x = 0, x = +1.
I But a proof for general x, ncould not be found for someyears.
Schoberl’s Conjecture
Message:
Special function inequalities arise
in real world applications.
Similar Inequalities
I Fejer’s inequality:
n∑
k=0
Pk(x) ≥ 0 (x ∈ [−1, 1], n ∈ N)
Similar Inequalities
I Fejer’s inequality:
n∑
k=0
Pk(x) ≥ 0 (x ∈ [−1, 1], n ∈ N)
-1 -0.5 0.5 1
-1
1
2
3
4
5
Similar Inequalities
I Fejer’s inequality:
n∑
k=0
Pk(x) ≥ 0 (x ∈ [−1, 1], n ∈ N)
-1 -0.5 0.5 1
-1
1
2
3
4
5
Similar Inequalities
I Fejer’s inequality:
n∑
k=0
Pk(x) ≥ 0 (x ∈ [−1, 1], n ∈ N)
-1 -0.5 0.5 1
-1
1
2
3
4
5
Similar Inequalities
I Fejer’s inequality:
n∑
k=0
Pk(x) ≥ 0 (x ∈ [−1, 1], n ∈ N)
-1 -0.5 0.5 1
-1
1
2
3
4
5
Similar Inequalities
I Fejer’s inequality:
n∑
k=0
Pk(x) ≥ 0 (x ∈ [−1, 1], n ∈ N)
-1 -0.5 0.5 1
-1
1
2
3
4
5
Similar Inequalities
I Fejer’s inequality:
n∑
k=0
Pk(x) ≥ 0 (x ∈ [−1, 1], n ∈ N)
-1 -0.5 0.5 1
-1
1
2
3
4
5
Similar Inequalities
I Fejer’s inequality:
n∑
k=0
Pk(x) ≥ 0 (x ∈ [−1, 1], n ∈ N)
-1 -0.5 0.5 1
-1
1
2
3
4
5
Similar Inequalities
I Fejer’s inequality:
n∑
k=0
Pk(x) ≥ 0 (x ∈ [−1, 1], n ∈ N)
-1 -0.5 0.5 1
-1
1
2
3
4
5
Similar Inequalities
I Fejer’s inequality:
n∑
k=0
Pk(x) ≥ 0 (x ∈ [−1, 1], n ∈ N)
-1 -0.5 0.5 1
-1
1
2
3
4
5
Similar Inequalities
I Fejer’s inequality:
n∑
k=0
Pk(x) ≥ 0 (x ∈ [−1, 1], n ∈ N)
-1 -0.5 0.5 1
-1
1
2
3
4
5
Similar Inequalities
I Fejer’s inequality:
n∑
k=0
Pk(x) ≥ 0 (x ∈ [−1, 1], n ∈ N)
-1 -0.5 0.5 1
-1
1
2
3
4
5
Similar Inequalities
I Fejer’s inequality:
n∑
k=0
Pk(x) ≥ 0 (x ∈ [−1, 1], n ∈ N)
-1 -0.5 0.5 1
-1
1
2
3
4
5
Similar Inequalities
I Fejer’s inequality:
n∑
k=0
Pk(x) ≥ 0 (x ∈ [−1, 1], n ∈ N)
-1 -0.5 0.5 1
-1
1
2
3
4
5
Similar Inequalities
I Fejer’s inequality:
n∑
k=0
Pk(x) ≥ 0 (x ∈ [−1, 1], n ∈ N)
-1 -0.5 0.5 1
-1
1
2
3
4
5
Similar Inequalities
I Fejer’s inequality:
n∑
k=0
Pk(x) ≥ 0 (x ∈ [−1, 1], n ∈ N)
-1 -0.5 0.5 1
-1
1
2
3
4
5
Similar Inequalities
I Fejer’s inequality:
n∑
k=0
Pk(x) ≥ 0 (x ∈ [−1, 1], n ∈ N)
-1 -0.5 0.5 1
-1
1
2
3
4
5
Similar Inequalities
I Fejer’s inequality:
n∑
k=0
Pk(x) ≥ 0 (x ∈ [−1, 1], n ∈ N)
-1 -0.5 0.5 1
-1
1
2
3
4
5
Similar Inequalities
I Fejer’s inequality:
n∑
k=0
Pk(x) ≥ 0 (x ∈ [−1, 1], n ∈ N)
-1 -0.5 0.5 1
-1
1
2
3
4
5
Similar Inequalities
I Fejer’s inequality:
n∑
k=0
Pk(x) ≥ 0 (x ∈ [−1, 1], n ∈ N)
-1 -0.5 0.5 1
-1
1
2
3
4
5
Similar Inequalities
I Fejer’s inequality:
n∑
k=0
Pk(x) ≥ 0 (x ∈ [−1, 1], n ∈ N)
-1 -0.5 0.5 1
-1
1
2
3
4
5
Similar Inequalities
I Fejer’s inequality:
n∑
k=0
Pk(x) ≥ 0 (x ∈ [−1, 1], n ∈ N)
-1 -0.5 0.5 1
-1
1
2
3
4
5
Similar Inequalities
I Fejer’s inequality:
n∑
k=0
Pk(x) ≥ 0 (x ∈ [−1, 1], n ∈ N)
-1 -0.5 0.5 1
-1
1
2
3
4
5
Similar Inequalities
I Fejer’s inequality:
n∑
k=0
Pk(x) ≥ 0 (x ∈ [−1, 1], n ∈ N)
-1 -0.5 0.5 1
-1
1
2
3
4
5
Similar Inequalities
I Fejer’s inequality:
n∑
k=0
Pk(x) ≥ 0 (x ∈ [−1, 1], n ∈ N)
-1 -0.5 0.5 1
-1
1
2
3
4
5
Similar Inequalities
I Fejer’s inequality:
n∑
k=0
Pk(x) ≥ 0 (x ∈ [−1, 1], n ∈ N)
-1 -0.5 0.5 1
-1
1
2
3
4
5
Similar Inequalities
I Fejer’s inequality:
n∑
k=0
Pk(x) ≥ 0 (x ∈ [−1, 1], n ∈ N)
-1 -0.5 0.5 1
-1
1
2
3
4
5
Similar Inequalities
I Fejer’s inequality:
n∑
k=0
Pk(x) ≥ 0 (x ∈ [−1, 1], n ∈ N)
-1 -0.5 0.5 1
-1
1
2
3
4
5
Similar Inequalities
I Fejer’s inequality:
n∑
k=0
Pk(x) ≥ 0 (x ∈ [−1, 1], n ∈ N)
I The Askey-Gasper inequality:
n∑
k=0
P(α,0)k ≥ 0 (x ∈ [−1, 1], α ≥ −2, n ∈ N)
where P(α,β)k (x) refers to the Jacobi polynomials.
Similar Inequalities
I Fejer’s inequality:
n∑
k=0
Pk(x) ≥ 0 (x ∈ [−1, 1], n ∈ N)
I The Askey-Gasper inequality:
n∑
k=0
P(α,0)k ≥ 0 (x ∈ [−1, 1], α ≥ −2, n ∈ N)
where P(α,β)k (x) refers to the Jacobi polynomials.
As Pk(x) = P(0,0)k (x), it includes Fejer’s inequality.
Similar Inequalities
I These inequalities are pretty deep.
Similar Inequalities
I These inequalities are pretty deep.
I Their classical proofs depend on rewriting the sums in termsof squares of other special functions.
Similar Inequalities
I These inequalities are pretty deep.
I Their classical proofs depend on rewriting the sums in termsof squares of other special functions.
I A computer proof would be highly interesting.
Similar Inequalities
I These inequalities are pretty deep.
I Their classical proofs depend on rewriting the sums in termsof squares of other special functions.
I A computer proof would be highly interesting.
I But all attempts to prove them directly by CAD and inductionhave failed so far.
Similar Inequalities
I These inequalities are pretty deep.
I Their classical proofs depend on rewriting the sums in termsof squares of other special functions.
I A computer proof would be highly interesting.
I But all attempts to prove them directly by CAD and inductionhave failed so far.
I It is not clear how the inequalities could be reformulated suchas to make the proof go through.
Similar Inequalities
I These inequalities are pretty deep.
I Their classical proofs depend on rewriting the sums in termsof squares of other special functions.
I A computer proof would be highly interesting.
I But all attempts to prove them directly by CAD and inductionhave failed so far.
I It is not clear how the inequalities could be reformulated suchas to make the proof go through.
I This is work in progress.
Similar Inequalities
I These inequalities are pretty deep.
I Their classical proofs depend on rewriting the sums in termsof squares of other special functions.
I A computer proof would be highly interesting.
I But all attempts to prove them directly by CAD and inductionhave failed so far.
I It is not clear how the inequalities could be reformulated suchas to make the proof go through.
I This is work in progress.
I Now back to Schoberl’s conjecture. . .
Pillwein’s Proof
Pillwein has been able to bring this conjecture into a form forwhich a proof with CAD and induction succeeds.
Pillwein’s Proof
Pillwein has been able to bring this conjecture into a form forwhich a proof with CAD and induction succeeds.
Her transformation is not trivial. It consists of
Pillwein’s Proof
Pillwein has been able to bring this conjecture into a form forwhich a proof with CAD and induction succeeds.
Her transformation is not trivial. It consists of
I Generalizing the inequality to Jacobi polynomials P(α,α)n (x)
Pillwein’s Proof
Pillwein has been able to bring this conjecture into a form forwhich a proof with CAD and induction succeeds.
Her transformation is not trivial. It consists of
I Generalizing the inequality to Jacobi polynomials P(α,α)n (x)
I Proving the inequality at the boundary for α
Pillwein’s Proof
Pillwein has been able to bring this conjecture into a form forwhich a proof with CAD and induction succeeds.
Her transformation is not trivial. It consists of
I Generalizing the inequality to Jacobi polynomials P(α,α)n (x)
I Proving the inequality at the boundary for α
I Finding a decomposition for general α into two parts
Pillwein’s Proof
Pillwein has been able to bring this conjecture into a form forwhich a proof with CAD and induction succeeds.
Her transformation is not trivial. It consists of
I Generalizing the inequality to Jacobi polynomials P(α,α)n (x)
I Proving the inequality at the boundary for α
I Finding a decomposition for general α into two parts
I Proving estimates for each part by hand
Pillwein’s Proof
Pillwein has been able to bring this conjecture into a form forwhich a proof with CAD and induction succeeds.
Her transformation is not trivial. It consists of
I Generalizing the inequality to Jacobi polynomials P(α,α)n (x)
I Proving the inequality at the boundary for α
I Finding a decomposition for general α into two parts
I Proving estimates for each part by hand
I Combining the estimates for both components with CAD andinduction
Schoberl’s conjecture is not sharp
Consider the graph of
Sn(x) :=n∑
k=0
(4k + 1)(2n − 2k + 1)P2k(0)P2k(x)
for n = 15
-1 -0.5 0.5 1
-5
-2.5
2.5
5
7.5
10
12.5
15
Schoberl’s conjecture is not sharp
Consider the graph of
Sn(x) :=n∑
k=0
(4k + 1)(2n − 2k + 1)P2k(0)P2k(x)
for n = 15 near x = 1
-1 -0.5 0.5 1
-5
-2.5
2.5
5
7.5
10
12.5
15
0.8 0.8250.850.875 0.9 0.9250.950.975
-0.5
0.5
1
1.5
2
Schoberl’s conjecture is not sharp
Consider the graph of
Sn(x) :=n∑
k=0
(4k + 1)(2n − 2k + 1)P2k(0)P2k(x)
for n = 15 near x = 1
-1 -0.5 0.5 1
-5
-2.5
2.5
5
7.5
10
12.5
15
0.8 0.8250.850.875 0.9 0.9250.950.975
-0.5
0.5
1
1.5
2
Schoberl’s conjecture is not sharp
Consider the graph of
Sn(x) :=n∑
k=0
(4k + 1)(2n − 2k + 1)P2k(0)P2k(x)
for n = 15 near x = 1
-1 -0.5 0.5 1
-5
-2.5
2.5
5
7.5
10
12.5
15
0.8 0.8250.850.875 0.9 0.9250.950.975
-0.5
0.5
1
1.5
2
What is the reason for this gap?
Schoberl’s conjecture is not sharp
Consider, more generally, the graph of
Sαn (x) :=
n∑
k=0
(2α+4k+1)(2n−2k+1)
(2k+2α
α
)
4α(2k+α
α
)P(α,α)2k (0)P
(α,α)2k (x)
Schoberl’s conjecture is not sharp
Consider, more generally, the graph of
Sαn (x) :=
n∑
k=0
(2α+4k+1)(2n−2k+1)
(2k+2α
α
)
4α(2k+α
α
)P(α,α)2k (0)P
(α,α)2k (x)
for α = 1/2
-1 -0.5 0.5 1
-5
-2.5
2.5
5
7.5
10
12.5
15
Schoberl’s conjecture is not sharp
Consider, more generally, the graph of
Sαn (x) :=
n∑
k=0
(2α+4k+1)(2n−2k+1)
(2k+2α
α
)
4α(2k+α
α
)P(α,α)2k (0)P
(α,α)2k (x)
for α = 1/2
-1 -0.5 0.5 1
-5
-2.5
2.5
5
7.5
10
12.5
15
Schoberl’s conjecture is not sharp
Consider, more generally, the graph of
Sαn (x) :=
n∑
k=0
(2α+4k+1)(2n−2k+1)
(2k+2α
α
)
4α(2k+α
α
)P(α,α)2k (0)P
(α,α)2k (x)
for α = 1/2
-1 -0.5 0.5 1
-5
-2.5
2.5
5
7.5
10
12.5
15
Schoberl’s conjecture is not sharp
Consider, more generally, the graph of
Sαn (x) :=
n∑
k=0
(2α+4k+1)(2n−2k+1)
(2k+2α
α
)
4α(2k+α
α
)P(α,α)2k (0)P
(α,α)2k (x)
for α = 1/2
-1 -0.5 0.5 1
-5
-2.5
2.5
5
7.5
10
12.5
15
Schoberl’s conjecture is not sharp
Consider, more generally, the graph of
Sαn (x) :=
n∑
k=0
(2α+4k+1)(2n−2k+1)
(2k+2α
α
)
4α(2k+α
α
)P(α,α)2k (0)P
(α,α)2k (x)
for α = 1/2
-1 -0.5 0.5 1
-5
-2.5
2.5
5
7.5
10
12.5
15
Schoberl’s conjecture is not sharp
Consider, more generally, the graph of
Sαn (x) :=
n∑
k=0
(2α+4k+1)(2n−2k+1)
(2k+2α
α
)
4α(2k+α
α
)P(α,α)2k (0)P
(α,α)2k (x)
for α = 1/2
-1 -0.5 0.5 1
-5
-2.5
2.5
5
7.5
10
12.5
15
Schoberl’s conjecture is not sharp
Consider, more generally, the graph of
Sαn (x) :=
n∑
k=0
(2α+4k+1)(2n−2k+1)
(2k+2α
α
)
4α(2k+α
α
)P(α,α)2k (0)P
(α,α)2k (x)
for α = 1/2
-1 -0.5 0.5 1
-5
-2.5
2.5
5
7.5
10
12.5
15
Schoberl’s conjecture is not sharp
Consider, more generally, the graph of
Sαn (x) :=
n∑
k=0
(2α+4k+1)(2n−2k+1)
(2k+2α
α
)
4α(2k+α
α
)P(α,α)2k (0)P
(α,α)2k (x)
for α = 1/2
-1 -0.5 0.5 1
-5
-2.5
2.5
5
7.5
10
12.5
15
Schoberl’s conjecture is not sharp
Consider, more generally, the graph of
Sαn (x) :=
n∑
k=0
(2α+4k+1)(2n−2k+1)
(2k+2α
α
)
4α(2k+α
α
)P(α,α)2k (0)P
(α,α)2k (x)
for α = 1/2
-1 -0.5 0.5 1
-5
-2.5
2.5
5
7.5
10
12.5
15
Schoberl’s conjecture is not sharp
Consider, more generally, the graph of
Sαn (x) :=
n∑
k=0
(2α+4k+1)(2n−2k+1)
(2k+2α
α
)
4α(2k+α
α
)P(α,α)2k (0)P
(α,α)2k (x)
for α = 1/2
-1 -0.5 0.5 1
-5
-2.5
2.5
5
7.5
10
12.5
15
Schoberl’s conjecture is not sharp
Consider, more generally, the graph of
Sαn (x) :=
n∑
k=0
(2α+4k+1)(2n−2k+1)
(2k+2α
α
)
4α(2k+α
α
)P(α,α)2k (0)P
(α,α)2k (x)
for α = 1/2
-1 -0.5 0.5 1
-5
-2.5
2.5
5
7.5
10
12.5
15
Schoberl’s conjecture is not sharp
Consider, more generally, the graph of
Sαn (x) :=
n∑
k=0
(2α+4k+1)(2n−2k+1)
(2k+2α
α
)
4α(2k+α
α
)P(α,α)2k (0)P
(α,α)2k (x)
for α = 1/2
-1 -0.5 0.5 1
-5
-2.5
2.5
5
7.5
10
12.5
15
Schoberl’s conjecture is not sharp
Consider, more generally, the graph of
Sαn (x) :=
n∑
k=0
(2α+4k+1)(2n−2k+1)
(2k+2α
α
)
4α(2k+α
α
)P(α,α)2k (0)P
(α,α)2k (x)
for α = 1/2
-1 -0.5 0.5 1
-5
-2.5
2.5
5
7.5
10
12.5
15
Schoberl’s conjecture is not sharp
Consider, more generally, the graph of
Sαn (x) :=
n∑
k=0
(2α+4k+1)(2n−2k+1)
(2k+2α
α
)
4α(2k+α
α
)P(α,α)2k (0)P
(α,α)2k (x)
for α = 1/2
-1 -0.5 0.5 1
-5
-2.5
2.5
5
7.5
10
12.5
15
Schoberl’s conjecture is not sharp
Consider, more generally, the graph of
Sαn (x) :=
n∑
k=0
(2α+4k+1)(2n−2k+1)
(2k+2α
α
)
4α(2k+α
α
)P(α,α)2k (0)P
(α,α)2k (x)
for α = 1/2
-1 -0.5 0.5 1
-5
-2.5
2.5
5
7.5
10
12.5
15
Schoberl’s conjecture is not sharp
Consider, more generally, the graph of
Sαn (x) :=
n∑
k=0
(2α+4k+1)(2n−2k+1)
(2k+2α
α
)
4α(2k+α
α
)P(α,α)2k (0)P
(α,α)2k (x)
for α = 1/2
-1 -0.5 0.5 1
-5
-2.5
2.5
5
7.5
10
12.5
15
Schoberl’s conjecture is not sharp
Consider, more generally, the graph of
Sαn (x) :=
n∑
k=0
(2α+4k+1)(2n−2k+1)
(2k+2α
α
)
4α(2k+α
α
)P(α,α)2k (0)P
(α,α)2k (x)
for α = 1/2 near x = 1
-1 -0.5 0.5 1
-5
-2.5
2.5
5
7.5
10
12.5
15
0.8 0.8250.850.875 0.9 0.9250.950.975
-0.5
0.5
1
1.5
2
Schoberl’s conjecture is not sharp
Consider, more generally, the graph of
Sαn (x) :=
n∑
k=0
(2α+4k+1)(2n−2k+1)
(2k+2α
α
)
4α(2k+α
α
)P(α,α)2k (0)P
(α,α)2k (x)
for α = 1/2 near x = 1
-1 -0.5 0.5 1
-5
-2.5
2.5
5
7.5
10
12.5
15
0.8 0.8250.850.875 0.9 0.9250.950.975
-0.5
0.5
1
1.5
2
Schoberl’s conjecture is not sharp
Consider, more generally, the graph of
Sαn (x) :=
n∑
k=0
(2α+4k+1)(2n−2k+1)
(2k+2α
α
)
4α(2k+α
α
)P(α,α)2k (0)P
(α,α)2k (x)
for α = 1/2 near x = 1
-1 -0.5 0.5 1
-5
-2.5
2.5
5
7.5
10
12.5
15
0.8 0.8250.850.875 0.9 0.9250.950.975
-0.5
0.5
1
1.5
2
Conjecture: Sαn (x) ≥ 0 for α ∈ [−1
2 , 12 ], x ∈ [−1, 1], n ∈ N
Schoberl’s conjecture is not sharp
Consider, more generally, the graph of
Sαn (x) :=
n∑
k=0
(2α+4k+1)(2n−2k+1)
(2k+2α
α
)
4α(2k+α
α
)P(α,α)2k (0)P
(α,α)2k (x)
for α = 1/2 near x = 1
-1 -0.5 0.5 1
-5
-2.5
2.5
5
7.5
10
12.5
15
0.8 0.8250.850.875 0.9 0.9250.950.975
-0.5
0.5
1
1.5
2
Conjecture: Sαn (x) ≥ 0 for α ∈ [−1
2 , 12 ], x ∈ [−1, 1], n ∈ N
Note: Sn(x) = S0n(x).
Situation at the boundary
For α = 1/2, the sum Sαn (x) can be written in closed form.
Situation at the boundary
For α = 1/2, the sum Sαn (x) can be written in closed form.
With Uk(x) := π4
(k+11/2
)P
(1/2,1/2)k (x), we have
S1/2n (x) =
4
π
n∑
k=0
(2n − 2k + 1)U2k(0)U2k(x)
Situation at the boundary
For α = 1/2, the sum Sαn (x) can be written in closed form.
With Uk(x) := π4
(k+11/2
)P
(1/2,1/2)k (x), we have
S1/2n (x) =
4
π
n∑
k=0
(2n − 2k + 1)U2k(0)U2k(x)
=2
πx2
(1 + (−1)n − 2(−1)n(1 − x2)Un(x)2
)
Situation at the boundary
For α = 1/2, the sum Sαn (x) can be written in closed form.
With Uk(x) := π4
(k+11/2
)P
(1/2,1/2)k (x), we have
S1/2n (x) =
4
π
n∑
k=0
(2n − 2k + 1)U2k(0)U2k(x)
=2
πx2
(1 + (−1)n − 2(−1)n(1 − x2)Un(x)2
)
This identity was found with symbolic summation.
Situation at the boundary
For α = 1/2, the sum Sαn (x) can be written in closed form.
With Uk(x) := π4
(k+11/2
)P
(1/2,1/2)k (x), we have
S1/2n (x) =
4
π
n∑
k=0
(2n − 2k + 1)U2k(0)U2k(x)
=2
πx2
(1 + (−1)n − 2(−1)n(1 − x2)Un(x)2
)
This identity was found with symbolic summation.
=4
πx2
{(Un+1(x) − xUn(x))2 if n is even
Situation at the boundary
For α = 1/2, the sum Sαn (x) can be written in closed form.
With Uk(x) := π4
(k+11/2
)P
(1/2,1/2)k (x), we have
S1/2n (x) =
4
π
n∑
k=0
(2n − 2k + 1)U2k(0)U2k(x)
=2
πx2
(1 + (−1)n − 2(−1)n(1 − x2)Un(x)2
)
This identity was found with symbolic summation.
=4
πx2
{(Un+1(x) − xUn(x))2 if n is even(1 − x2)Un(x)2 if n is odd
Situation at the boundary
The case α = −1/2 can be handled similarly.
Situation at the boundary
The case α = −1/2 can be handled similarly.
But for general α, the sum does not have a closed form.
Situation at the boundary
The case α = −1/2 can be handled similarly.
But for general α, the sum does not have a closed form.
Note: Symbolic summation can also assert the absense of closedforms.
Situation at the boundary
The case α = −1/2 can be handled similarly.
But for general α, the sum does not have a closed form.
Note: Symbolic summation can also assert the absense of closedforms.
Idea: WriteSα
n (x) = gαn(x) − fα
n (x)
where fαn (x) is a sum expression that vanishes for α = ±1/2, and
gαn(x) is a closed form expression.
Situation at the boundary
The case α = −1/2 can be handled similarly.
But for general α, the sum does not have a closed form.
Note: Symbolic summation can also assert the absense of closedforms.
Idea: WriteSα
n (x) = gαn(x) − fα
n (x)
where fαn (x) is a sum expression that vanishes for α = ±1/2, and
gαn(x) is a closed form expression.
This can be done in many ways.
Split
A good choice turned out to be
fαn (x) =
2n∑
k=0
4−α(1−4α2)(2α+2k−1)(2α+2k+3)
(2α+k
α
)
(α+k
α
) P(α,α)k (0)P
(α,α)k (x)
gαn(x) = 2−2α−1(2n + 1)
(2α+2n+1
α
)
(α+2n
α
) P(α,α)2n (0)
×(
xP(α,α)2n+1 (x) − 2(α+2n+1)
2α+4n+3 P(α,α)2n (x)
)
Split
A good choice turned out to be
fαn (x) =
2n∑
k=0
4−α(1−4α2)(2α+2k−1)(2α+2k+3)
(2α+k
α
)
(α+k
α
) P(α,α)k (0)P
(α,α)k (x)
gαn(x) = 2−2α−1(2n + 1)
(2α+2n+1
α
)
(α+2n
α
) P(α,α)2n (0)
×(
xP(α,α)2n+1 (x) − 2(α+2n+1)
2α+4n+3 P(α,α)2n (x)
)
Indeed, for this choice we have
Sαn (x) = 1
x2 (gαn(x) − fα
n (x)) and f±1/2n (x) = 0.
Split
A good choice turned out to be
fαn (x) =
2n∑
k=0
4−α(1−4α2)(2α+2k−1)(2α+2k+3)
(2α+k
α
)
(α+k
α
) P(α,α)k (0)P
(α,α)k (x)
gαn(x) = 2−2α−1(2n + 1)
(2α+2n+1
α
)
(α+2n
α
) P(α,α)2n (0)
×(
xP(α,α)2n+1 (x) − 2(α+2n+1)
2α+4n+3 P(α,α)2n (x)
)
Indeed, for this choice we have
Sαn (x) = 1
x2 (gαn(x) − fα
n (x)) and f±1/2n (x) = 0.
This can be verified (but not discovered!) by symbolic summation.
Split
Now Sαn (x) ≥ 0 is equivalent to gα
n(x) ≥ fαn (x).
Split
Now Sαn (x) ≥ 0 is equivalent to gα
n(x) ≥ fαn (x).
Consider g0n(x) and f0
n(x) for n =0, . . . , 15.
-1 -0.5 0.5 1
-0.4
-0.2
0.2
0.4
Split
Now Sαn (x) ≥ 0 is equivalent to gα
n(x) ≥ fαn (x).
Consider g0n(x) and f0
n(x) for n =0, . . . , 15.
-1 -0.5 0.5 1
-0.4
-0.2
0.2
0.4
Split
Now Sαn (x) ≥ 0 is equivalent to gα
n(x) ≥ fαn (x).
Consider g0n(x) and f0
n(x) for n =0, . . . , 15.
-1 -0.5 0.5 1
-0.4
-0.2
0.2
0.4
Split
Now Sαn (x) ≥ 0 is equivalent to gα
n(x) ≥ fαn (x).
Consider g0n(x) and f0
n(x) for n =0, . . . , 15.
-1 -0.5 0.5 1
-0.4
-0.2
0.2
0.4
Split
Now Sαn (x) ≥ 0 is equivalent to gα
n(x) ≥ fαn (x).
Consider g0n(x) and f0
n(x) for n =0, . . . , 15.
-1 -0.5 0.5 1
-0.4
-0.2
0.2
0.4
Split
Now Sαn (x) ≥ 0 is equivalent to gα
n(x) ≥ fαn (x).
Consider g0n(x) and f0
n(x) for n =0, . . . , 15.
-1 -0.5 0.5 1
-0.4
-0.2
0.2
0.4
Split
Now Sαn (x) ≥ 0 is equivalent to gα
n(x) ≥ fαn (x).
Consider g0n(x) and f0
n(x) for n =0, . . . , 15.
-1 -0.5 0.5 1
-0.4
-0.2
0.2
0.4
Split
Now Sαn (x) ≥ 0 is equivalent to gα
n(x) ≥ fαn (x).
Consider g0n(x) and f0
n(x) for n =0, . . . , 15.
-1 -0.5 0.5 1
-0.4
-0.2
0.2
0.4
Split
Now Sαn (x) ≥ 0 is equivalent to gα
n(x) ≥ fαn (x).
Consider g0n(x) and f0
n(x) for n =0, . . . , 15.
-1 -0.5 0.5 1
-0.4
-0.2
0.2
0.4
Split
Now Sαn (x) ≥ 0 is equivalent to gα
n(x) ≥ fαn (x).
Consider g0n(x) and f0
n(x) for n =0, . . . , 15.
-1 -0.5 0.5 1
-0.4
-0.2
0.2
0.4
Split
Now Sαn (x) ≥ 0 is equivalent to gα
n(x) ≥ fαn (x).
Consider g0n(x) and f0
n(x) for n =0, . . . , 15.
-1 -0.5 0.5 1
-0.4
-0.2
0.2
0.4
Split
Now Sαn (x) ≥ 0 is equivalent to gα
n(x) ≥ fαn (x).
Consider g0n(x) and f0
n(x) for n =0, . . . , 15.
-1 -0.5 0.5 1
-0.4
-0.2
0.2
0.4
Split
Now Sαn (x) ≥ 0 is equivalent to gα
n(x) ≥ fαn (x).
Consider g0n(x) and f0
n(x) for n =0, . . . , 15.
-1 -0.5 0.5 1
-0.4
-0.2
0.2
0.4
Split
Now Sαn (x) ≥ 0 is equivalent to gα
n(x) ≥ fαn (x).
Consider g0n(x) and f0
n(x) for n =0, . . . , 15.
-1 -0.5 0.5 1
-0.4
-0.2
0.2
0.4
Split
Now Sαn (x) ≥ 0 is equivalent to gα
n(x) ≥ fαn (x).
Consider g0n(x) and f0
n(x) for n =0, . . . , 15.
-1 -0.5 0.5 1
-0.4
-0.2
0.2
0.4
Split
Now Sαn (x) ≥ 0 is equivalent to gα
n(x) ≥ fαn (x).
Consider g0n(x) and f0
n(x) for n =0, . . . , 15.
-1 -0.5 0.5 1
-0.4
-0.2
0.2
0.4
Split
Now Sαn (x) ≥ 0 is equivalent to gα
n(x) ≥ fαn (x).
Consider g0n(x) and f0
n(x) for n =0, . . . , 15.
The closed form part gαn(x) con-
tains the main oscillations.-1 -0.5 0.5 1
-0.4
-0.2
0.2
0.4
Split
Now Sαn (x) ≥ 0 is equivalent to gα
n(x) ≥ fαn (x).
Consider g0n(x) and f0
n(x) for n =0, . . . , 15.
The closed form part gαn(x) con-
tains the main oscillations.
So maybe the sum part fαn (x) is
now easier to handle.
-1 -0.5 0.5 1
-0.4
-0.2
0.2
0.4
Split
Now Sαn (x) ≥ 0 is equivalent to gα
n(x) ≥ fαn (x).
Consider g0n(x) and f0
n(x) for n =0, . . . , 15.
The closed form part gαn(x) con-
tains the main oscillations.
So maybe the sum part fαn (x) is
now easier to handle.
-1 -0.5 0.5 1
-0.4
-0.2
0.2
0.4
Next goal: Find eαn(x) in closed form such that
fαn (x) ≤ eα
n(x).
Bound the sum
Consider
fαn (x, y) =
2n∑
k=0
4−α(1−4α2)(2α+2k−1)(2α+2k+3)
(2α+k
α
)
(α+k
α
) P(α,α)k (x)P
(α,α)k (y)
Bound the sum
Consider
fαn (x, y) =
2n∑
k=0
4−α(1−4α2)(2α+2k−1)(2α+2k+3)
(2α+k
α
)
(α+k
α
) P(α,α)k (x)P
(α,α)k (y)
Then fαn (x) = fα
n (x, 0).
Bound the sum
Consider
fαn (x, y) =
2n∑
k=0
4−α(1−4α2)(2α+2k−1)(2α+2k+3)
(2α+k
α
)
(α+k
α
) P(α,α)k (x)P
(α,α)k (y)
Then fαn (x) = fα
n (x, 0).
It can be shown without too much effort that
fαn (x, y) ≤ 1
2
(fα
n (x, x) + fαn (y, y)
)(α ∈ [−1
2 , 12 ])
Bound the sum
Consider
fαn (x, y) =
2n∑
k=0
4−α(1−4α2)(2α+2k−1)(2α+2k+3)
(2α+k
α
)
(α+k
α
) P(α,α)k (x)P
(α,α)k (y)
Then fαn (x) = fα
n (x, 0).
It can be shown without too much effort that
fαn (x, y) ≤ 1
2
(fα
n (x, x) + fαn (y, y)
)(α ∈ [−1
2 , 12 ])
There are closed forms for the sums fαn (x, x) and fα
n (y, y).
Bound the sum
Consider
fαn (x, y) =
2n∑
k=0
4−α(1−4α2)(2α+2k−1)(2α+2k+3)
(2α+k
α
)
(α+k
α
) P(α,α)k (x)P
(α,α)k (y)
Then fαn (x) = fα
n (x, 0).
It can be shown without too much effort that
fαn (x, y) ≤ 1
2
(fα
n (x, x) + fαn (y, y)
)(α ∈ [−1
2 , 12 ])
There are closed forms for the sums fαn (x, x) and fα
n (y, y).
So we may set
eαn(x) := 1
2
(fα
n (x, x) + fαn (0, 0)
).
Putting things together. . .
I We want to show gαn(x) ≥ fα
n (x).
Putting things together. . .
I We want to show gαn(x) ≥ fα
n (x).
I We already know that eαn(x) ≥ fα
n (x).
Putting things together. . .
I We want to show gαn(x) ≥ fα
n (x).
I We already know that eαn(x) ≥ fα
n (x).
I Maybe we also have gαn(x) ≥ eα
n(x)?
Putting things together. . .
I We want to show gαn(x) ≥ fα
n (x).
I We already know that eαn(x) ≥ fα
n (x).
I Maybe we also have gαn(x) ≥ eα
n(x)?
-1 -0.5 0.5 1
-0.4
-0.2
0.2
0.4
Putting things together. . .
I We want to show gαn(x) ≥ fα
n (x).
I We already know that eαn(x) ≥ fα
n (x).
I Maybe we also have gαn(x) ≥ eα
n(x)?
-1 -0.5 0.5 1
-0.4
-0.2
0.2
0.4
0.5 0.6 0.7 0.8 0.9
-0.25
-0.2
-0.15
-0.1
Putting things together. . .
I We want to show gαn(x) ≥ fα
n (x).
I We already know that eαn(x) ≥ fα
n (x).
I Maybe we also have gαn(x) ≥ eα
n(x)?
-1 -0.5 0.5 1
-0.4
-0.2
0.2
0.4
-0.2 -0.1 0.1 0.2
-0.35
-0.3
-0.25
-0.2
Putting things together. . .
I We want to show gαn(x) ≥ fα
n (x).
I We already know that eαn(x) ≥ fα
n (x).
I Maybe we also have gαn(x) ≥ eα
n(x)?
-1 -0.5 0.5 1
-0.4
-0.2
0.2
0.4
-0.2 -0.1 0.1 0.2
-0.35
-0.3
-0.25
-0.2
Looks promising. . .
Putting things together. . .
We have
gαn(x) = 2−2α−1(2n + 1)
(2α+2n+1
α
)
(α+2n
α
) P(α,α)2n (0)
×(
xP(α,α)2n+1 (x) − 2(α+2n+1)
2α+4n+3 P(α,α)2n (x)
)
eαn(x) = 2−2α−1(2n + 1)
(2α+2n+1
α
)
(α+2n
α
)
×(
xP(α,α)2n (x)P
(α,α)2n+1 (x) − α+2n+1
2α+4n+3P(α,α)2n (x)2
− α+2n+12α+4n+3P
(α,α)2n (0)2 − (2n+1)(2α+2n+1)
(α+2n+1)(2α+4n+1)P(α,α)2n+1 (x)2
)
Putting things together. . .
We have
gαn(x) = 2−2α−1(2n + 1)
(2α+2n+1
α
)
(α+2n
α
) P(α,α)2n (0)
×(
xP(α,α)2n+1 (x) − 2(α+2n+1)
2α+4n+3 P(α,α)2n (x)
)
eαn(x) = 2−2α−1(2n + 1)
(2α+2n+1
α
)
(α+2n
α
)
×(
xP(α,α)2n (x)P
(α,α)2n+1 (x) − α+2n+1
2α+4n+3P(α,α)2n (x)2
− α+2n+12α+4n+3P
(α,α)2n (0)2 − (2n+1)(2α+2n+1)
(α+2n+1)(2α+4n+1)P(α,α)2n+1 (x)2
)
It remains to show gαn(x) ≥ eα
n(x).
Putting things together. . .
After some simplifications, it remains to show
(α + 2n + 1)2(2α + 4n + 1)(P(α,α)2n (0)2 + P
(α,α)2n (x)2)
+ (2n + 1)(2α + 2n + 1)(2α + 4n + 3)P(α,α)2n+1 (x)2
− (α + 2n + 1)(2α + 4n + 1)
× (2α + 4n + 3)xP(α,α)2n (x)P
(α,α)2n+1 (x) ≥ 0
for −1 ≤ x ≤ 1 and −12 ≤ α ≤ 1
2 .
Putting things together. . .
After some simplifications, it remains to show
(α + 2n + 1)2(2α + 4n + 1)(P(α,α)2n (0)2 + P
(α,α)2n (x)2)
+ (2n + 1)(2α + 2n + 1)(2α + 4n + 3)P(α,α)2n+1 (x)2
− (α + 2n + 1)(2α + 4n + 1)
× (2α + 4n + 3)xP(α,α)2n (x)P
(α,α)2n+1 (x) ≥ 0
for −1 ≤ x ≤ 1 and −12 ≤ α ≤ 1
2 .
This would still be a hard thing to do by hand.
Putting things together. . .
After some simplifications, it remains to show
(α + 2n + 1)2(2α + 4n + 1)(P(α,α)2n (0)2 + P
(α,α)2n (x)2)
+ (2n + 1)(2α + 2n + 1)(2α + 4n + 3)P(α,α)2n+1 (x)2
− (α + 2n + 1)(2α + 4n + 1)
× (2α + 4n + 3)xP(α,α)2n (x)P
(α,α)2n+1 (x) ≥ 0
for −1 ≤ x ≤ 1 and −12 ≤ α ≤ 1
2 .
This would still be a hard thing to do by hand. (Try it.)
Putting things together. . .
After some simplifications, it remains to show
(α + 2n + 1)2(2α + 4n + 1)(P(α,α)2n (0)2 + P
(α,α)2n (x)2)
+ (2n + 1)(2α + 2n + 1)(2α + 4n + 3)P(α,α)2n+1 (x)2
− (α + 2n + 1)(2α + 4n + 1)
× (2α + 4n + 3)xP(α,α)2n (x)P
(α,α)2n+1 (x) ≥ 0
for −1 ≤ x ≤ 1 and −12 ≤ α ≤ 1
2 .
This would still be a hard thing to do by hand. (Try it.)
But CAD and induction on n is applicable here.
Putting things together. . .
After some simplifications, it remains to show
(α + 2n + 1)2(2α + 4n + 1)(P(α,α)2n (0)2 + P
(α,α)2n (x)2)
+ (2n + 1)(2α + 2n + 1)(2α + 4n + 3)P(α,α)2n+1 (x)2
− (α + 2n + 1)(2α + 4n + 1)
× (2α + 4n + 3)xP(α,α)2n (x)P
(α,α)2n+1 (x) ≥ 0
for −1 ≤ x ≤ 1 and −12 ≤ α ≤ 1
2 .
This would still be a hard thing to do by hand. (Try it.)
But CAD and induction on n is applicable here.
A Tarski formula for the induction step is. . .
. . . and leaving the rest to the computer
∀n, α, x, y, z, w`
(n ≥ 0 ∧ −1 ≤ x ≤ 1 ∧ −1 ≤ 2α ≤ 1 ∧ (2α + 4n + 1)(y2 + z2)(α + 2n + 1)2 −
(2α + 4n + 1)(2α + 4n + 3)wxz(α + 2n + 1) + (2n + 1)(2α + 2n + 1)(2α + 4n + 3)w2 ≥ 0) ⇒
(2n + 3)(α + 2n + 1)2(α + 2n + 3)2(2α + 2n + 3)(2α + 4n + 5)y2(α + 2n + 2)2 + (α + 2n +
1)2(64n5 −256x
2n4 +160αn
4 +464n4 +144α
2n3 −512αx
2n3 −1184x
2n3 +928αn
3 +1344n3 +
56α3n2 + 628α
2n2 − 384α
2x2n2 − 1776αx
2n2 − 1984x
2n2 + 2016αn
2 + 1944n2 + 8α
4n +
164α3n + 912α
2n − 128α
3x2n − 888α
2x2n − 1984αx
2n − 1434x
2n + 1944αn + 1404n + 12α
4 +120α
3 +441α2 −16α
4x2 −148α
3x2 −496α
2x2 −717αx
2 −378x2 +702α+405)z2(α+2n+2)2 −
w2(−256n
7+4096x4n6−3072x
2n6−896αn
6−1728n6+12288αx
4n5+25088x
4n5−1216α
2n5−
9216αx2n5 − 19968x
2n5 − 5184αn
5 − 4864n5 + 15360α
2x4n4 + 62720αx
4n4 + 62464x
4n4 −
800α3n4 − 5872α
2n4 − 11008α
2x2n4 − 49920αx
2n4 − 53120x
2n4 − 12160αn
4 − 7408n4 −
256α4n3+10240α
3x4n3+62720α
2x4n3+124928αx
4n3+81216x
4n3−3104α
3n3−11072α
2n3−
6656α3x2n3 − 47744α
2x2n3 − 106240αx
2n3 − 74176x
2n3 − 14816αn
3 − 6592n3 − 32α
5n2 −
752α4n2+3840α
4x4n2+31360α
3x4n2+93696α
2x4n2+121824αx
4n2+58320x
4n2−4448α
3n2−
10192α2n2 − 2112α
4x2n2 − 21696α
3x2n2 − 76416α
2x2n2 − 111264αx
2n2 − 57396x
2n2 −
9888αn2 − 3424n
2 − 64α5n− 736α
4n +768α
5x4n +7840α
4x4n +31232α
3x4n +60912α
2x4n +
58320αx4n + 21978x
4n − 2784α
3n − 4576α
2n − 320α
5x2n − 4608α
4x2n − 23296α
3x2n −
53568α2x2n − 57396αx
2n − 23340x
2n − 3424αn − 960n − 32α
5 − 240α4 + 64α
6x4 + 784α
5x4 +
3904α4x4+10152α
3x4+14580α
2x4+10989αx
4+3402x4−640α
3−800α2−16α
6x2−352α
5x2−
2504α4x2−8240α
3x2−13881α
2x2−11670αx
2−3897x2−480α−112)(α+2n+2)2−2(α+2n+
1)wx(128n6 − 1024x
2n5 + 384αn
5 + 1408n5 + 448α
2n4 − 2560αx
2n4 − 5504x
2n4 + 3520αn
4 +
5592n4+256α
3n3+3296α
2n3−2560α
2x2n3−11008αx
2n3−11488x
2n3+11184αn
3+10888n3+
72α4n2 + 1424α
3n2 + 7870α
2n2 − 1280α
3x2n2 − 8256α
2x2n2 − 17232αx
2n2 − 11688x
2n2 +
16332αn2 + 11258n
2 + 8α5n + 272α
4n + 2278α
3n + 7692α
2n − 320α
4x2n − 2752α
3x2n −
8616α2x2n − 11688αx
2n − 5814x
2n + 11258αn + 5940n + 16α
5 + 220α4 + 1124α
3 + 2669α2 −
32α5x2−344α
4x2−1436α
3x2−2922α
2x2−2907αx
2−1134x2+2970α+1257)z(α+2n+2)2 ≥ 0
´
. . . and leaving the rest to the computer
Mathematica’s CAD asserts (after some hours) that this is true.
. . . and leaving the rest to the computer
Mathematica’s CAD asserts (after some hours) that this is true.
This proves that gαn+1(x) ≥ eα
n+1(x) whenever gαn(x) ≥ eα
n(x).
. . . and leaving the rest to the computer
Mathematica’s CAD asserts (after some hours) that this is true.
This proves that gαn+1(x) ≥ eα
n+1(x) whenever gαn(x) ≥ eα
n(x).
Showing the induction base
gα0 (x) ≥ eα
0 (x)
is not a problem.
. . . and leaving the rest to the computer
Mathematica’s CAD asserts (after some hours) that this is true.
This proves that gαn+1(x) ≥ eα
n+1(x) whenever gαn(x) ≥ eα
n(x).
Showing the induction base
(α + 1)(2αx2 + 3x2 − 2)
2(2α + 3)≥ α + 1
2α + 3
is not a problem.
. . . and leaving the rest to the computer
Mathematica’s CAD asserts (after some hours) that this is true.
This proves that gαn+1(x) ≥ eα
n+1(x) whenever gαn(x) ≥ eα
n(x).
Showing the induction base
(α + 1)(2αx2 + 3x2 − 2)
2(2α + 3)≥ α + 1
2α + 3
is not a problem.
This completes the proof of gαn(x) ≥ eα
n(x) (n ∈ N).
. . . and leaving the rest to the computer
Mathematica’s CAD asserts (after some hours) that this is true.
This proves that gαn+1(x) ≥ eα
n+1(x) whenever gαn(x) ≥ eα
n(x).
Showing the induction base
(α + 1)(2αx2 + 3x2 − 2)
2(2α + 3)≥ α + 1
2α + 3
is not a problem.
This completes the proof of gαn(x) ≥ eα
n(x) (n ∈ N).
This completes the proof of gαn(x) ≥ fα
n (x) (n ∈ N).
. . . and leaving the rest to the computer
Mathematica’s CAD asserts (after some hours) that this is true.
This proves that gαn+1(x) ≥ eα
n+1(x) whenever gαn(x) ≥ eα
n(x).
Showing the induction base
(α + 1)(2αx2 + 3x2 − 2)
2(2α + 3)≥ α + 1
2α + 3
is not a problem.
This completes the proof of gαn(x) ≥ eα
n(x) (n ∈ N).
This completes the proof of gαn(x) ≥ fα
n (x) (n ∈ N).
This completes the proof of Schoberl’s conjecture.
Pillwein’s Proof
Message:
A special function inequality may require some very
non-obvious manipulation before an induction proof
via CAD succeeds.
Summary
Summary
I Polynomial inequalities can be proven without thinking.
Summary
I Polynomial inequalities can be proven without thinking.
I We may use CAD to construct an induction proof for a specialfunction inequality.
Summary
I Polynomial inequalities can be proven without thinking.
I We may use CAD to construct an induction proof for a specialfunction inequality.
I Special function inequalities arise in real world applications.
Summary
I Polynomial inequalities can be proven without thinking.
I We may use CAD to construct an induction proof for a specialfunction inequality.
I Special function inequalities arise in real world applications.
I Some “deep” special function inequalities are just animmediate consequence of a polynomial inequality.
Summary
I Polynomial inequalities can be proven without thinking.
I We may use CAD to construct an induction proof for a specialfunction inequality.
I Special function inequalities arise in real world applications.
I Some “deep” special function inequalities are just animmediate consequence of a polynomial inequality.
I Some inequalities require human preprocessing.
Summary
I Polynomial inequalities can be proven without thinking.
I We may use CAD to construct an induction proof for a specialfunction inequality.
I Special function inequalities arise in real world applications.
I Some “deep” special function inequalities are just animmediate consequence of a polynomial inequality.
I Some inequalities require human preprocessing.
I The preprocessing may be hard (if at all possible).
Conclusion
Conclusion
I Classical inequality proofs proceed by reducing the claim to anobvious statement.
Conclusion
I Classical inequality proofs proceed by reducing the claim to anobvious statement.
I Modern inequality proofs proceed by reducing the claim tosomething that can be done with the computer.
Conclusion
I Classical inequality proofs proceed by reducing the claim to anobvious statement.
I Modern inequality proofs proceed by reducing the claim tosomething that can be done with the computer.
I Stronger computer algebra methods for proving specialfunction inequalities would be highly appreciated. . .
Conclusion
I Classical inequality proofs proceed by reducing the claim to anobvious statement.
I Modern inequality proofs proceed by reducing the claim tosomething that can be done with the computer.
I Stronger computer algebra methods for proving specialfunction inequalities would be highly appreciated. . .
I . . . because these inequalities are soo difficult.
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