Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian...
Transcript of Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian...
![Page 1: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/1.jpg)
Pillwein’s Proof of
Schoberl’s Conjecture
Manuel Kauers
currently at INRIA-Rocquencourt
usually at RISC-Linz
![Page 2: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/2.jpg)
Polynomial Inequalities
![Page 3: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/3.jpg)
Polynomial Inequalities
. . . arise regularly as Monthly problems . . .
![Page 4: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/4.jpg)
Polynomial Inequalities
. . . arise regularly as Monthly problems . . .
. . . are often considered difficult . . .
![Page 5: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/5.jpg)
Polynomial Inequalities
. . . arise regularly as Monthly problems . . .
. . . are often considered difficult . . .
. . . but are in fact trivial . . .
![Page 6: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/6.jpg)
Polynomial Inequalities
. . . arise regularly as Monthly problems . . .
. . . are often considered difficult . . .
. . . but are in fact trivial . . .
. . . if we have a fast computer.
![Page 7: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/7.jpg)
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)
![Page 8: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/8.jpg)
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.
![Page 9: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/9.jpg)
Polynomial Inequalities
A Tarski formula is a formula composed of
![Page 10: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/10.jpg)
Polynomial Inequalities
A Tarski formula is a formula composed of
I rational numbers (1, 2,−3117 , . . . )
![Page 11: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/11.jpg)
Polynomial Inequalities
A Tarski formula is a formula composed of
I rational numbers (1, 2,−3117 , . . . )
I variables (x, y, . . . )
![Page 12: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/12.jpg)
Polynomial Inequalities
A Tarski formula is a formula composed of
I rational numbers (1, 2,−3117 , . . . )
I variables (x, y, . . . )
I arithmetic operations (+,−, ·, /)
![Page 13: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/13.jpg)
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=, <, >,≤,≥)
![Page 14: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/14.jpg)
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 (∧,∨, . . . )
![Page 15: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/15.jpg)
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 (∀,∃)
![Page 16: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/16.jpg)
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)
)
![Page 17: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/17.jpg)
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)
![Page 18: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/18.jpg)
Polynomial Inequalities
Theorem. (Tarski, 1948) Every Tarski formula is, as a statementabout real numbers, equivalent to a Tarski formula without anyquantifiers.
![Page 19: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/19.jpg)
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.
![Page 20: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/20.jpg)
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).
![Page 21: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/21.jpg)
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)
)
![Page 22: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/22.jpg)
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
![Page 23: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/23.jpg)
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)
![Page 24: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/24.jpg)
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
![Page 25: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/25.jpg)
Polynomial Inequalities
Message:
Polynomial inequalities can be proven by CAD
without thinking.
![Page 26: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/26.jpg)
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.
![Page 27: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/27.jpg)
Bernoulli’s Inequality
∀ n ∈ N ∀ x ≥ −1 : (x + 1)n ≥ 1 + nx.
![Page 28: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/28.jpg)
Bernoulli’s Inequality
∀ n ∈ N ∀ x ≥ −1 : (x + 1)n − (1 + nx) ≥ 0.
![Page 29: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/29.jpg)
Bernoulli’s Inequality
∀ n ∈ N ∀ x ≥ −1 : (x + 1)n − (1 + nx) ≥ 0.
What exactly does (x + 1)n − (1 + nx) mean?
![Page 30: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/30.jpg)
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.
![Page 31: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/31.jpg)
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.
![Page 32: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/32.jpg)
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.
![Page 33: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/33.jpg)
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
![Page 34: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/34.jpg)
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
![Page 35: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/35.jpg)
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
![Page 36: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/36.jpg)
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
![Page 37: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/37.jpg)
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
![Page 38: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/38.jpg)
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
![Page 39: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/39.jpg)
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
![Page 40: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/40.jpg)
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
![Page 41: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/41.jpg)
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
![Page 42: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/42.jpg)
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
![Page 43: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/43.jpg)
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
![Page 44: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/44.jpg)
Bernoulli’s Inequality
∀ n ∈ N ∀ x ≥ −1 : (x + 1)n − (1 + nx) ≥ 0.
I Can we show Bernoulli’s inequality with CAD?
![Page 45: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/45.jpg)
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?
![Page 46: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/46.jpg)
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).
![Page 47: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/47.jpg)
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
![Page 48: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/48.jpg)
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
![Page 49: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/49.jpg)
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
![Page 50: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/50.jpg)
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
![Page 51: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/51.jpg)
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
![Page 52: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/52.jpg)
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.
![Page 53: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/53.jpg)
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.
![Page 54: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/54.jpg)
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.
![Page 55: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/55.jpg)
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.
![Page 56: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/56.jpg)
Bernoulli’s Inequality
Message:
We may use CAD to construct an induction proof
for the positivity of a quantity satisfying a
recurrence.
![Page 57: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/57.jpg)
Legendre Polynomials
There are other interesting sequences of polynomials.
![Page 58: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/58.jpg)
Legendre Polynomials
There are other interesting sequences of polynomials.
For example, Legendre Polynomials Pn(x).
![Page 59: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/59.jpg)
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
![Page 60: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/60.jpg)
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
![Page 61: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/61.jpg)
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
![Page 62: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/62.jpg)
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
![Page 63: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/63.jpg)
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
![Page 64: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/64.jpg)
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
![Page 65: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/65.jpg)
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
![Page 66: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/66.jpg)
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
![Page 67: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/67.jpg)
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
![Page 68: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/68.jpg)
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.
![Page 69: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/69.jpg)
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
![Page 70: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/70.jpg)
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.
![Page 71: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/71.jpg)
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.
![Page 72: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/72.jpg)
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
![Page 73: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/73.jpg)
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
![Page 74: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/74.jpg)
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
![Page 75: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/75.jpg)
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
![Page 76: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/76.jpg)
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
![Page 77: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/77.jpg)
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
![Page 78: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/78.jpg)
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
![Page 79: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/79.jpg)
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
![Page 80: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/80.jpg)
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
![Page 81: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/81.jpg)
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
![Page 82: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/82.jpg)
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.
![Page 83: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/83.jpg)
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.
![Page 84: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/84.jpg)
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.)
![Page 85: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/85.jpg)
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.
![Page 86: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/86.jpg)
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.
![Page 87: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/87.jpg)
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
![Page 88: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/88.jpg)
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
![Page 89: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/89.jpg)
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
)
![Page 90: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/90.jpg)
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.
![Page 91: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/91.jpg)
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.
![Page 92: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/92.jpg)
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.
![Page 93: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/93.jpg)
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.
![Page 94: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/94.jpg)
Legendre Polynomials: Turan’s Inequality
Message:
A “deep” special function inequality may be just an
immediate consequence of a polynomial inequality.
![Page 95: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/95.jpg)
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
![Page 96: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/96.jpg)
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).
![Page 97: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/97.jpg)
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).
![Page 98: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/98.jpg)
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).
![Page 99: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/99.jpg)
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).
![Page 100: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/100.jpg)
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).
![Page 101: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/101.jpg)
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).
![Page 102: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/102.jpg)
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).
![Page 103: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/103.jpg)
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).
![Page 104: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/104.jpg)
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?
![Page 105: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/105.jpg)
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.
![Page 106: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/106.jpg)
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. . .
![Page 107: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/107.jpg)
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).
![Page 108: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/108.jpg)
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.
![Page 109: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/109.jpg)
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.
![Page 110: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/110.jpg)
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:
![Page 111: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/111.jpg)
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.
![Page 112: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/112.jpg)
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.
![Page 113: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/113.jpg)
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.
![Page 114: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/114.jpg)
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
![Page 115: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/115.jpg)
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))
![Page 116: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/116.jpg)
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.
![Page 117: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/117.jpg)
Legendre Polynomials: Turan’s Inequality
Message:
A special function inequality may require some
non-obvious manipulation before an induction proof
via CAD succeeds.
![Page 118: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/118.jpg)
Schoberl’s Conjecture
![Page 119: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/119.jpg)
Schoberl’s Conjecture
I In the higher order finite elementmethod (FEM), solutions of PDEs arelocally approximated by polynomials.
![Page 120: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/120.jpg)
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, . . . .
![Page 121: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/121.jpg)
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
![Page 122: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/122.jpg)
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.
![Page 123: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/123.jpg)
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.
![Page 124: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/124.jpg)
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))
![Page 125: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/125.jpg)
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.
![Page 126: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/126.jpg)
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.
![Page 127: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/127.jpg)
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.
![Page 128: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/128.jpg)
Schoberl’s Conjecture
I Schoberl next considered the “gliding averages”
fn(x) :=1
n + 1
2n∑
i=n
ki(x, 0).
![Page 129: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/129.jpg)
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
![Page 130: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/130.jpg)
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.
![Page 131: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/131.jpg)
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.
![Page 132: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/132.jpg)
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
![Page 133: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/133.jpg)
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
![Page 134: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/134.jpg)
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
![Page 135: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/135.jpg)
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
![Page 136: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/136.jpg)
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
![Page 137: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/137.jpg)
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
![Page 138: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/138.jpg)
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
![Page 139: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/139.jpg)
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
![Page 140: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/140.jpg)
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
![Page 141: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/141.jpg)
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
![Page 142: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/142.jpg)
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
![Page 143: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/143.jpg)
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
![Page 144: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/144.jpg)
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
![Page 145: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/145.jpg)
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
![Page 146: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/146.jpg)
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
![Page 147: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/147.jpg)
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
![Page 148: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/148.jpg)
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
![Page 149: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/149.jpg)
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
![Page 150: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/150.jpg)
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
![Page 151: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/151.jpg)
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
![Page 152: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/152.jpg)
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
![Page 153: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/153.jpg)
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
![Page 154: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/154.jpg)
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.
![Page 155: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/155.jpg)
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.
![Page 156: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/156.jpg)
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.
![Page 157: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/157.jpg)
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.
![Page 158: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/158.jpg)
Schoberl’s Conjecture
Message:
Special function inequalities arise
in real world applications.
![Page 159: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/159.jpg)
Similar Inequalities
I Fejer’s inequality:
n∑
k=0
Pk(x) ≥ 0 (x ∈ [−1, 1], n ∈ N)
![Page 160: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/160.jpg)
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
![Page 161: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/161.jpg)
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
![Page 162: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/162.jpg)
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
![Page 163: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/163.jpg)
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
![Page 164: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/164.jpg)
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
![Page 165: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/165.jpg)
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
![Page 166: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/166.jpg)
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
![Page 167: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/167.jpg)
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
![Page 168: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/168.jpg)
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
![Page 169: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/169.jpg)
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
![Page 170: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/170.jpg)
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
![Page 171: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/171.jpg)
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
![Page 172: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/172.jpg)
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
![Page 173: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/173.jpg)
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
![Page 174: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/174.jpg)
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
![Page 175: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/175.jpg)
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
![Page 176: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/176.jpg)
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
![Page 177: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/177.jpg)
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
![Page 178: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/178.jpg)
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
![Page 179: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/179.jpg)
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
![Page 180: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/180.jpg)
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
![Page 181: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/181.jpg)
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
![Page 182: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/182.jpg)
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
![Page 183: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/183.jpg)
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
![Page 184: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/184.jpg)
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
![Page 185: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/185.jpg)
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
![Page 186: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/186.jpg)
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
![Page 187: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/187.jpg)
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.
![Page 188: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/188.jpg)
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.
![Page 189: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/189.jpg)
Similar Inequalities
I These inequalities are pretty deep.
![Page 190: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/190.jpg)
Similar Inequalities
I These inequalities are pretty deep.
I Their classical proofs depend on rewriting the sums in termsof squares of other special functions.
![Page 191: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/191.jpg)
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.
![Page 192: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/192.jpg)
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.
![Page 193: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/193.jpg)
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.
![Page 194: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/194.jpg)
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.
![Page 195: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/195.jpg)
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. . .
![Page 196: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/196.jpg)
Pillwein’s Proof
Pillwein has been able to bring this conjecture into a form forwhich a proof with CAD and induction succeeds.
![Page 197: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/197.jpg)
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
![Page 198: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/198.jpg)
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)
![Page 199: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/199.jpg)
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
![Page 200: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/200.jpg)
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
![Page 201: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/201.jpg)
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
![Page 202: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/202.jpg)
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
![Page 203: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/203.jpg)
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
![Page 204: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/204.jpg)
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
![Page 205: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/205.jpg)
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
![Page 206: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/206.jpg)
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?
![Page 207: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/207.jpg)
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)
![Page 208: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/208.jpg)
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
![Page 209: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/209.jpg)
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
![Page 210: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/210.jpg)
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
![Page 211: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/211.jpg)
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
![Page 212: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/212.jpg)
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
![Page 213: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/213.jpg)
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
![Page 214: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/214.jpg)
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
![Page 215: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/215.jpg)
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
![Page 216: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/216.jpg)
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
![Page 217: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/217.jpg)
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
![Page 218: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/218.jpg)
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
![Page 219: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/219.jpg)
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
![Page 220: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/220.jpg)
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
![Page 221: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/221.jpg)
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
![Page 222: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/222.jpg)
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
![Page 223: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/223.jpg)
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
![Page 224: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/224.jpg)
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
![Page 225: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/225.jpg)
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
![Page 226: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/226.jpg)
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
![Page 227: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/227.jpg)
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).
![Page 228: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/228.jpg)
Situation at the boundary
For α = 1/2, the sum Sαn (x) can be written in closed form.
![Page 229: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/229.jpg)
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)
![Page 230: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/230.jpg)
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
)
![Page 231: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/231.jpg)
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.
![Page 232: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/232.jpg)
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
![Page 233: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/233.jpg)
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
![Page 234: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/234.jpg)
Situation at the boundary
The case α = −1/2 can be handled similarly.
![Page 235: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/235.jpg)
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.
![Page 236: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/236.jpg)
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.
![Page 237: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/237.jpg)
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.
![Page 238: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/238.jpg)
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.
![Page 239: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/239.jpg)
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)
)
![Page 240: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/240.jpg)
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.
![Page 241: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/241.jpg)
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.
![Page 242: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/242.jpg)
Split
Now Sαn (x) ≥ 0 is equivalent to gα
n(x) ≥ fαn (x).
![Page 243: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/243.jpg)
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
![Page 244: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/244.jpg)
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
![Page 245: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/245.jpg)
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
![Page 246: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/246.jpg)
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
![Page 247: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/247.jpg)
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
![Page 248: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/248.jpg)
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
![Page 249: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/249.jpg)
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
![Page 250: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/250.jpg)
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
![Page 251: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/251.jpg)
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
![Page 252: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/252.jpg)
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
![Page 253: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/253.jpg)
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
![Page 254: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/254.jpg)
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
![Page 255: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/255.jpg)
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
![Page 256: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/256.jpg)
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
![Page 257: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/257.jpg)
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
![Page 258: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/258.jpg)
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
![Page 259: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/259.jpg)
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
![Page 260: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/260.jpg)
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
![Page 261: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/261.jpg)
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).
![Page 262: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/262.jpg)
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)
![Page 263: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/263.jpg)
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).
![Page 264: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/264.jpg)
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 ])
![Page 265: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/265.jpg)
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).
![Page 266: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/266.jpg)
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)
).
![Page 267: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/267.jpg)
Putting things together. . .
I We want to show gαn(x) ≥ fα
n (x).
![Page 268: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/268.jpg)
Putting things together. . .
I We want to show gαn(x) ≥ fα
n (x).
I We already know that eαn(x) ≥ fα
n (x).
![Page 269: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/269.jpg)
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)?
![Page 270: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/270.jpg)
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
![Page 271: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/271.jpg)
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
![Page 272: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/272.jpg)
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
![Page 273: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/273.jpg)
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. . .
![Page 274: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/274.jpg)
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
)
![Page 275: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/275.jpg)
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).
![Page 276: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/276.jpg)
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 .
![Page 277: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/277.jpg)
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.
![Page 278: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/278.jpg)
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.)
![Page 279: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/279.jpg)
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.
![Page 280: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/280.jpg)
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. . .
![Page 281: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/281.jpg)
. . . 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
´
![Page 282: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/282.jpg)
. . . and leaving the rest to the computer
Mathematica’s CAD asserts (after some hours) that this is true.
![Page 283: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/283.jpg)
. . . 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).
![Page 284: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/284.jpg)
. . . 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.
![Page 285: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/285.jpg)
. . . 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.
![Page 286: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/286.jpg)
. . . 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).
![Page 287: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/287.jpg)
. . . 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).
![Page 288: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/288.jpg)
. . . 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.
![Page 289: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/289.jpg)
Pillwein’s Proof
Message:
A special function inequality may require some very
non-obvious manipulation before an induction proof
via CAD succeeds.
![Page 290: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/290.jpg)
Summary
![Page 291: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/291.jpg)
Summary
I Polynomial inequalities can be proven without thinking.
![Page 292: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/292.jpg)
Summary
I Polynomial inequalities can be proven without thinking.
I We may use CAD to construct an induction proof for a specialfunction inequality.
![Page 293: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/293.jpg)
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.
![Page 294: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/294.jpg)
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.
![Page 295: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/295.jpg)
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.
![Page 296: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/296.jpg)
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).
![Page 297: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/297.jpg)
Conclusion
![Page 298: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/298.jpg)
Conclusion
I Classical inequality proofs proceed by reducing the claim to anobvious statement.
![Page 299: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/299.jpg)
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.
![Page 300: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/300.jpg)
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. . .
![Page 301: Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which](https://reader030.fdocuments.net/reader030/viewer/2022040617/5f2223688d8c5f4d117be13c/html5/thumbnails/301.jpg)
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.