Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which

Pillwein’s Proof of

Schoberl’s Conjecture

Manuel Kauers

currently at INRIA-Rocquencourt

usually at RISC-Linz

Polynomial Inequalities


. . . arise regularly as Monthly problems . . .



. . . are often considered difficult . . .




. . . but are in fact trivial . . .




. . . but are in fact trivial . . .

. . . if we have a fast computer.


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 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.


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 boolean operations (∧,∨, . . . )

I quantifiers (∀,∃)


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 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)


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.



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).


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 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 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:


∣∣

≤ M(a2 + b2 + c2)2)


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:


∣∣

≤ M(a2 + b2 + c2)2)

CAD

−−−→ M ≥ 932

√2


Message:

Polynomial inequalities can be proven by CAD

without thinking.


Message:

Polynomial inequalities can be proven by CAD

without thinking.

The rest of this talk is aboutinequalities that can be proven by CAD with thinking only.

Bernoulli’s Inequality

∀ n ∈ N ∀ x ≥ −1 : (x + 1)n ≥ 1 + nx.


∀ n ∈ N ∀ x ≥ −1 : (x + 1)n − (1 + nx) ≥ 0.


∀ n ∈ N ∀ x ≥ −1 : (x + 1)n − (1 + nx) ≥ 0.

What exactly does (x + 1)n − (1 + nx) mean?


∀ n ∈ N ∀ x ≥ −1 : (x + 1)n − (1 + nx) ≥ 0.


I For any specific integer n, it is a polynomial in x.


∀ n ∈ N ∀ x ≥ −1 : (x + 1)n − (1 + nx) ≥ 0.



I View (x + 1)n − (1 + nx) as a sequence of polynomials.


∀ n ∈ N ∀ x ≥ −1 : (x + 1)n − (1 + nx) ≥ 0.



I View (x + 1)n − (1 + nx) as a sequence of polynomials.

I View Bernoulli’s inequality as a sequence of polynomialinequalities.


∀ 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


∀ 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


∀ 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


∀ 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


∀ 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


∀ 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


∀ 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


∀ 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


∀ 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


∀ 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


∀ 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


∀ n ∈ N ∀ x ≥ −1 : (x + 1)n − (1 + nx) ≥ 0.

I Can we show Bernoulli’s inequality with CAD?


∀ n ∈ N ∀ x ≥ −1 : (x + 1)n − (1 + nx) ≥ 0.


I Can CAD be used to do induction on n?


∀ n ∈ N ∀ x ≥ −1 : (x + 1)n − (1 + nx) ≥ 0.



I Let fn(x) := (x + 1)n − (1 + nx).


∀ n ∈ N ∀ x ≥ −1 : (x + 1)n − (1 + nx) ≥ 0.



I Let fn(x) := (x + 1)n − (1 + nx).

I Induction step:

∀n ∈ N ∀x ≥ −1 : fn(x) ≥ 0 ⇒ fn+1(x) ≥ 0


∀ n ∈ N ∀ x ≥ −1 : (x + 1)n − (1 + nx) ≥ 0.



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


∀ n ∈ N ∀ x ≥ −1 : (x + 1)n − (1 + nx) ≥ 0.



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



∀ n ∈ N ∀ x ≥ −1 : (x + 1)n − (1 + nx) ≥ 0.



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 Generalize fn(x) to y and n ∈ N to n ≥ 0


∀ n ∈ N ∀ x ≥ −1 : (x + 1)n − (1 + nx) ≥ 0.



I Let fn(x) := (x + 1)n − (1 + nx).

I Induction step:

∀n ≥ 0 ∀y ∀x ≥ −1 : y ≥ 0 ⇒ (x + 1)y + nx2 ≥ 0




∀ n ∈ N ∀ x ≥ −1 : (x + 1)n − (1 + nx) ≥ 0.



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 The resulting formula is indeed true.


∀ n ∈ N ∀ x ≥ −1 : (x + 1)n − (1 + nx) ≥ 0.



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.


∀ n ∈ N ∀ x ≥ −1 : (x + 1)n − (1 + nx) ≥ 0.



I Let fn(x) := (x + 1)n − (1 + nx).

I Induction step:

∀n ∈ N ∀x ≥ −1 : fn(x) ≥ 0 ⇒ fn+1(x) ≥ 0


I The induction base 0 ≥ 0 is trivial.


∀ n ∈ N ∀ x ≥ −1 : (x + 1)n − (1 + nx) ≥ 0.



I Let fn(x) := (x + 1)n − (1 + nx).

I Induction step:

∀n ∈ N ∀x ≥ −1 : fn(x) ≥ 0 ⇒ fn+1(x) ≥ 0


I The induction base 0 ≥ 0 is trivial.

I This completes the proof.


Message:

We may use CAD to construct an induction proof

for the positivity of a quantity satisfying a

recurrence.

Legendre Polynomials

There are other interesting sequences of polynomials.



For example, Legendre Polynomials Pn(x).




I P0(x) = 1

-1 -0.5 0.5 1

-2

-1

1

2




I P0(x) = 1

I P1(x) = x-1 -0.5 0.5 1

-2

-1

1

2




I P0(x) = 1

I P1(x) = x

I P2(x) = 32x2 − 1

2

-1 -0.5 0.5 1

-2

-1

1

2




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




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




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




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




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




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




-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.




-1 -0.5 0.5 1

-2

-1

1

2


∫ 1

−1Pn(x)Pm(x)dx =

2

2n + 1δn,m


They also satisfy a recurrence:

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




-1 -0.5 0.5 1

-2

-1

1

2


∫ 1

−1Pn(x)Pm(x)dx =

2

2n + 1δn,m



(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.




-1 -0.5 0.5 1

-2

-1

1

2


∫ 1

−1Pn(x)Pm(x)dx =

2

2n + 1δn,m



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

and various interesting inequalities, e.g.,

∀n ∈ N ∀x : −1 ≤ x ≤ 1 ⇒ |Pn(x)| ≤ 1.

Legendre Polynomials: Turan’s Inequality

Here is an inequality about Pn(x) that can be shown with CAD:

∀n ∈ N ∀x : −1 ≤ x ≤ 1 ⇒ P 2n+1(x) − Pn(x)Pn+2(x)

︸︷︷︸

=:∆n(x)

≥ 0



∀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



∀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



∀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



∀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



∀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



∀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



∀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



∀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



∀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



∀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.



∀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


inequality.

I For specific n, it is just apolynomial inequality.



∀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


inequality.


I For general n, it is not easy.(Try it.)



∀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


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.



∀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.



∀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



∀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 ∈ N ∀x : −1 ≤ x ≤ 1 ⇒ P 2n+1(x) − Pn(x)Pn+2(x)

︸︷︷︸

=:∆n(x)

≥ 0


∀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

)



∀n ∈ N ∀x : −1 ≤ x ≤ 1 ⇒ P 2n+1(x) − Pn(x)Pn+2(x)

︸︷︷︸

=:∆n(x)

≥ 0


∀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.



∀n ∈ N ∀x : −1 ≤ x ≤ 1 ⇒ P 2n+1(x) − Pn(x)Pn+2(x)

︸︷︷︸

=:∆n(x)

≥ 0


∀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.



∀n ∈ N ∀x : −1 ≤ x ≤ 1 ⇒ P 2n+1(x) − Pn(x)Pn+2(x)

︸︷︷︸

=:∆n(x)

≥ 0


∀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

),


The induction base ∆0(x) ≥ 0 is trivial.



∀n ∈ N ∀x : −1 ≤ x ≤ 1 ⇒ P 2n+1(x) − Pn(x)Pn+2(x)

︸︷︷︸

=:∆n(x)

≥ 0


∀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

),


The induction base ∆0(x) ≥ 0 is trivial. This completes the proof.


Message:

A “deep” special function inequality may be just an

immediate consequence of a polynomial 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


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).



∆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




∆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




∆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




∆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




∆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




∆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




∆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




∆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


Can we show this also by induction?



∆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



We have the recurrence

(n + 3)(n + 4)αn+2

= (2n + 5)αn+1 + (n + 1)(n + 2)αn.



∆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



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. . .



∆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).



∆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.



∆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.



∆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:



∆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.



∆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 For x = 0 there is nothing to show.



∆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 For x > 0, it suffices to show that∆n(x)/(1 − x2) is increasing.



∆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 For x > 0, it suffices to show that∆n(x)/(1 − x2) is increasing.

New idea: Show thatd

dx

∆n(x)

1 − x2≥ 0



∆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))



∆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.


Message:

A special function inequality may require some

non-obvious manipulation before an induction proof

via CAD succeeds.



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)q(x)dx = q(0) for all q with deg q ≤ n.

I

∫ 1

−1

|fn(x)|dx ≤ C for some constant C.


I The Legendre kernel polynomials

kn(x, y) :=n + 1

2(x − y)(Pn+1(x)Pn(y) − Pn+1(y)Pn(x))



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.



kn(x, y) :=n + 1

2(x − y)(Pn+1(x)Pn(y) − Pn+1(y)Pn(x))

have the property

∫ 1



I So fn(x) := kn(x, 0) satisfies the first condition.



kn(x, y) :=n + 1

2(x − y)(Pn+1(x)Pn(y) − Pn+1(y)Pn(x))

have the property

∫ 1



I So fn(x) := kn(x, 0) satisfies the first condition.

I But not the second.


I Schoberl next considered the “gliding averages”

fn(x) :=1

n + 1

2n∑

i=n

ki(x, 0).



fn(x) :=1

n + 1

2n∑

i=n

ki(x, 0).

I He could show that this family does the job



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.



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.


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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.


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 For specific n ∈ N, it can beshown without thinking.


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 It can be also be shown forx = −1, x = 0, x = +1.


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 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.


Message:

Special function inequalities arise

in real world applications.

Similar Inequalities

I Fejer’s inequality:

n∑

k=0

Pk(x) ≥ 0 (x ∈ [−1, 1], n ∈ N)



n∑

k=0

Pk(x) ≥ 0 (x ∈ [−1, 1], n ∈ N)

-1 -0.5 0.5 1

-1

1

2

3

4

5



n∑

k=0

Pk(x) ≥ 0 (x ∈ [−1, 1], n ∈ N)

-1 -0.5 0.5 1

-1

1

2

3

4

5



n∑

k=0

Pk(x) ≥ 0 (x ∈ [−1, 1], n ∈ N)

-1 -0.5 0.5 1

-1

1

2

3

4

5



n∑

k=0

Pk(x) ≥ 0 (x ∈ [−1, 1], n ∈ N)

-1 -0.5 0.5 1

-1

1

2

3

4

5



n∑

k=0

Pk(x) ≥ 0 (x ∈ [−1, 1], n ∈ N)

-1 -0.5 0.5 1

-1

1

2

3

4

5



n∑

k=0

Pk(x) ≥ 0 (x ∈ [−1, 1], n ∈ N)

-1 -0.5 0.5 1

-1

1

2

3

4

5



n∑

k=0

Pk(x) ≥ 0 (x ∈ [−1, 1], n ∈ N)

-1 -0.5 0.5 1

-1

1

2

3

4

5



n∑

k=0

Pk(x) ≥ 0 (x ∈ [−1, 1], n ∈ N)

-1 -0.5 0.5 1

-1

1

2

3

4

5



n∑

k=0

Pk(x) ≥ 0 (x ∈ [−1, 1], n ∈ N)

-1 -0.5 0.5 1

-1

1

2

3

4

5



n∑

k=0

Pk(x) ≥ 0 (x ∈ [−1, 1], n ∈ N)

-1 -0.5 0.5 1

-1

1

2

3

4

5



n∑

k=0

Pk(x) ≥ 0 (x ∈ [−1, 1], n ∈ N)

-1 -0.5 0.5 1

-1

1

2

3

4

5



n∑

k=0

Pk(x) ≥ 0 (x ∈ [−1, 1], n ∈ N)

-1 -0.5 0.5 1

-1

1

2

3

4

5



n∑

k=0

Pk(x) ≥ 0 (x ∈ [−1, 1], n ∈ N)

-1 -0.5 0.5 1

-1

1

2

3

4

5



n∑

k=0

Pk(x) ≥ 0 (x ∈ [−1, 1], n ∈ N)

-1 -0.5 0.5 1

-1

1

2

3

4

5



n∑

k=0

Pk(x) ≥ 0 (x ∈ [−1, 1], n ∈ N)

-1 -0.5 0.5 1

-1

1

2

3

4

5



n∑

k=0

Pk(x) ≥ 0 (x ∈ [−1, 1], n ∈ N)

-1 -0.5 0.5 1

-1

1

2

3

4

5



n∑

k=0

Pk(x) ≥ 0 (x ∈ [−1, 1], n ∈ N)

-1 -0.5 0.5 1

-1

1

2

3

4

5



n∑

k=0

Pk(x) ≥ 0 (x ∈ [−1, 1], n ∈ N)

-1 -0.5 0.5 1

-1

1

2

3

4

5



n∑

k=0

Pk(x) ≥ 0 (x ∈ [−1, 1], n ∈ N)

-1 -0.5 0.5 1

-1

1

2

3

4

5



n∑

k=0

Pk(x) ≥ 0 (x ∈ [−1, 1], n ∈ N)

-1 -0.5 0.5 1

-1

1

2

3

4

5



n∑

k=0

Pk(x) ≥ 0 (x ∈ [−1, 1], n ∈ N)

-1 -0.5 0.5 1

-1

1

2

3

4

5



n∑

k=0

Pk(x) ≥ 0 (x ∈ [−1, 1], n ∈ N)

-1 -0.5 0.5 1

-1

1

2

3

4

5



n∑

k=0

Pk(x) ≥ 0 (x ∈ [−1, 1], n ∈ N)

-1 -0.5 0.5 1

-1

1

2

3

4

5



n∑

k=0

Pk(x) ≥ 0 (x ∈ [−1, 1], n ∈ N)

-1 -0.5 0.5 1

-1

1

2

3

4

5



n∑

k=0

Pk(x) ≥ 0 (x ∈ [−1, 1], n ∈ N)

-1 -0.5 0.5 1

-1

1

2

3

4

5



n∑

k=0

Pk(x) ≥ 0 (x ∈ [−1, 1], n ∈ N)

-1 -0.5 0.5 1

-1

1

2

3

4

5



n∑

k=0

Pk(x) ≥ 0 (x ∈ [−1, 1], n ∈ N)

-1 -0.5 0.5 1

-1

1

2

3

4

5



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.



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.


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 This is work in progress.

I Now back to Schoberl’s conjecture. . .

Pillwein’s Proof

Pillwein has been able to bring this conjecture into a form forwhich a proof with CAD and induction succeeds.

Pillwein’s Proof


Her transformation is not trivial. It consists of

Pillwein’s Proof



I Generalizing the inequality to Jacobi polynomials P(α,α)n (x)

Pillwein’s Proof




I Proving the inequality at the boundary for α

Pillwein’s Proof





I Finding a decomposition for general α into two parts

Pillwein’s Proof






I Proving estimates for each part by hand

Pillwein’s Proof






I Proving estimates for each part by hand

I Combining the estimates for both components with CAD andinduction

Schoberl’s conjecture is not sharp

Consider the graph of

Sn(x) :=n∑

k=0

(4k + 1)(2n − 2k + 1)P2k(0)P2k(x)

for n = 15

-1 -0.5 0.5 1

-5

-2.5

2.5

5

7.5

10

12.5

15



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



Sn(x) :=n∑

k=0

(4k + 1)(2n − 2k + 1)P2k(0)P2k(x)


-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



Sn(x) :=n∑

k=0

(4k + 1)(2n − 2k + 1)P2k(0)P2k(x)


-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?


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)



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



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



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



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



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



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



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



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



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



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



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



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



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



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



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



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



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



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)


-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



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)


-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



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)


-1 -0.5 0.5 1

-5

-2.5

2.5

5

7.5

10

12.5

15

0.8 0.8250.850.875 0.9 0.9250.950.975

-0.5

0.5

1

1.5

2

Conjecture: Sαn (x) ≥ 0 for α ∈ [−1

2 , 12 ], x ∈ [−1, 1], n ∈ N

Note: Sn(x) = S0n(x).

Situation at the boundary

For α = 1/2, the sum Sαn (x) can be written in closed form.



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)



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

)



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.



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

)


=4

πx2

{(Un+1(x) − xUn(x))2 if n is even



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

)


=4

πx2

{(Un+1(x) − xUn(x))2 if n is even(1 − x2)Un(x)2 if n is odd


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.





Idea: WriteSα

n (x) = gαn(x) − fα

n (x)

where fαn (x) is a sum expression that vanishes for α = ±1/2, and

gαn(x) is a closed form expression.

This can be done in many ways.

Split

A good choice turned out to be

fαn (x) =

2n∑

k=0

4−α(1−4α2)(2α+2k−1)(2α+2k+3)

(2α+k

α

)

(α+k

α

) P(α,α)k (0)P

(α,α)k (x)

gαn(x) = 2−2α−1(2n + 1)

(2α+2n+1

α

)

(α+2n

α

) P(α,α)2n (0)

×(

xP(α,α)2n+1 (x) − 2(α+2n+1)

2α+4n+3 P(α,α)2n (x)

)

Split


fαn (x) =

2n∑

k=0

4−α(1−4α2)(2α+2k−1)(2α+2k+3)

(2α+k

α

)

(α+k

α

) P(α,α)k (0)P

(α,α)k (x)

gαn(x) = 2−2α−1(2n + 1)

(2α+2n+1

α

)

(α+2n

α

) P(α,α)2n (0)

×(

xP(α,α)2n+1 (x) − 2(α+2n+1)

2α+4n+3 P(α,α)2n (x)

)

Indeed, for this choice we have

Sαn (x) = 1

x2 (gαn(x) − fα

n (x)) and f±1/2n (x) = 0.

Split


fαn (x) =

2n∑

k=0

4−α(1−4α2)(2α+2k−1)(2α+2k+3)

(2α+k

α

)

(α+k

α

) P(α,α)k (0)P

(α,α)k (x)

gαn(x) = 2−2α−1(2n + 1)

(2α+2n+1

α

)

(α+2n

α

) P(α,α)2n (0)

×(

xP(α,α)2n+1 (x) − 2(α+2n+1)

2α+4n+3 P(α,α)2n (x)

)

Indeed, for this choice we have

Sαn (x) = 1

x2 (gαn(x) − fα

n (x)) and f±1/2n (x) = 0.

This can be verified (but not discovered!) by symbolic summation.

Split

Now Sαn (x) ≥ 0 is equivalent to gα

n(x) ≥ fαn (x).

Split


n(x) ≥ fαn (x).

Consider g0n(x) and f0

n(x) for n =0, . . . , 15.

-1 -0.5 0.5 1

-0.4

-0.2

0.2

0.4

Split


n(x) ≥ fαn (x).


n(x) for n =0, . . . , 15.

-1 -0.5 0.5 1

-0.4

-0.2

0.2

0.4

Split


n(x) ≥ fαn (x).


n(x) for n =0, . . . , 15.

-1 -0.5 0.5 1

-0.4

-0.2

0.2

0.4

Split


n(x) ≥ fαn (x).


n(x) for n =0, . . . , 15.

-1 -0.5 0.5 1

-0.4

-0.2

0.2

0.4

Split


n(x) ≥ fαn (x).


n(x) for n =0, . . . , 15.

-1 -0.5 0.5 1

-0.4

-0.2

0.2

0.4

Split


n(x) ≥ fαn (x).


n(x) for n =0, . . . , 15.

-1 -0.5 0.5 1

-0.4

-0.2

0.2

0.4

Split


n(x) ≥ fαn (x).


n(x) for n =0, . . . , 15.

-1 -0.5 0.5 1

-0.4

-0.2

0.2

0.4

Split


n(x) ≥ fαn (x).


n(x) for n =0, . . . , 15.

-1 -0.5 0.5 1

-0.4

-0.2

0.2

0.4

Split


n(x) ≥ fαn (x).


n(x) for n =0, . . . , 15.

-1 -0.5 0.5 1

-0.4

-0.2

0.2

0.4

Split


n(x) ≥ fαn (x).


n(x) for n =0, . . . , 15.

-1 -0.5 0.5 1

-0.4

-0.2

0.2

0.4

Split


n(x) ≥ fαn (x).


n(x) for n =0, . . . , 15.

-1 -0.5 0.5 1

-0.4

-0.2

0.2

0.4

Split


n(x) ≥ fαn (x).


n(x) for n =0, . . . , 15.

-1 -0.5 0.5 1

-0.4

-0.2

0.2

0.4

Split


n(x) ≥ fαn (x).


n(x) for n =0, . . . , 15.

-1 -0.5 0.5 1

-0.4

-0.2

0.2

0.4

Split


n(x) ≥ fαn (x).


n(x) for n =0, . . . , 15.

-1 -0.5 0.5 1

-0.4

-0.2

0.2

0.4

Split


n(x) ≥ fαn (x).


n(x) for n =0, . . . , 15.

-1 -0.5 0.5 1

-0.4

-0.2

0.2

0.4

Split


n(x) ≥ fαn (x).


n(x) for n =0, . . . , 15.

-1 -0.5 0.5 1

-0.4

-0.2

0.2

0.4

Split


n(x) ≥ fαn (x).


n(x) for n =0, . . . , 15.

The closed form part gαn(x) con-

tains the main oscillations.-1 -0.5 0.5 1

-0.4

-0.2

0.2

0.4

Split


n(x) ≥ fαn (x).


n(x) for n =0, . . . , 15.


tains the main oscillations.

So maybe the sum part fαn (x) is

now easier to handle.

-1 -0.5 0.5 1

-0.4

-0.2

0.2

0.4

Split


n(x) ≥ fαn (x).


n(x) for n =0, . . . , 15.


tains the main oscillations.

So maybe the sum part fαn (x) is

now easier to handle.

-1 -0.5 0.5 1

-0.4

-0.2

0.2

0.4

Next goal: Find eαn(x) in closed form such that

fαn (x) ≤ eα

n(x).

Bound the sum

Consider

fαn (x, y) =

2n∑

k=0

4−α(1−4α2)(2α+2k−1)(2α+2k+3)

(2α+k

α

)

(α+k

α

) P(α,α)k (x)P

(α,α)k (y)

Bound the sum

Consider

fαn (x, y) =

2n∑

k=0

4−α(1−4α2)(2α+2k−1)(2α+2k+3)

(2α+k

α

)

(α+k

α

) P(α,α)k (x)P

(α,α)k (y)

Then fαn (x) = fα

n (x, 0).

Bound the sum

Consider

fαn (x, y) =

2n∑

k=0

4−α(1−4α2)(2α+2k−1)(2α+2k+3)

(2α+k

α

)

(α+k

α

) P(α,α)k (x)P

(α,α)k (y)

Then fαn (x) = fα

n (x, 0).

It can be shown without too much effort that

fαn (x, y) ≤ 1

2

(fα

n (x, x) + fαn (y, y)

)(α ∈ [−1

2 , 12 ])

Bound the sum

Consider

fαn (x, y) =

2n∑

k=0

4−α(1−4α2)(2α+2k−1)(2α+2k+3)

(2α+k

α

)

(α+k

α

) P(α,α)k (x)P

(α,α)k (y)

Then fαn (x) = fα

n (x, 0).


fαn (x, y) ≤ 1

2

(fα

n (x, x) + fαn (y, y)

)(α ∈ [−1

2 , 12 ])

There are closed forms for the sums fαn (x, x) and fα

n (y, y).

Bound the sum

Consider

fαn (x, y) =

2n∑

k=0

4−α(1−4α2)(2α+2k−1)(2α+2k+3)

(2α+k

α

)

(α+k

α

) P(α,α)k (x)P

(α,α)k (y)

Then fαn (x) = fα

n (x, 0).


fαn (x, y) ≤ 1

2

(fα

n (x, x) + fαn (y, y)

)(α ∈ [−1

2 , 12 ])

There are closed forms for the sums fαn (x, x) and fα

n (y, y).

So we may set

eαn(x) := 1

2

(fα

n (x, x) + fαn (0, 0)

).

Putting things together. . .

I We want to show gαn(x) ≥ fα

n (x).



n (x).

I We already know that eαn(x) ≥ fα

n (x).



n (x).


n (x).

I Maybe we also have gαn(x) ≥ eα

n(x)?



n (x).


n (x).


n(x)?

-1 -0.5 0.5 1

-0.4

-0.2

0.2

0.4



n (x).


n (x).


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



n (x).


n (x).


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



n (x).


n (x).


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. . .


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

)


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).


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 .



(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.



(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.)



(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 .


But CAD and induction on n is applicable here.



(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 .


But CAD and induction on n is applicable here.

A Tarski formula for the induction step is. . .

. . . and leaving the rest to the computer

∀n, α, x, y, z, w`

(n ≥ 0 ∧ −1 ≤ x ≤ 1 ∧ −1 ≤ 2α ≤ 1 ∧ (2α + 4n + 1)(y2 + z2)(α + 2n + 1)2 −

(2α + 4n + 1)(2α + 4n + 3)wxz(α + 2n + 1) + (2n + 1)(2α + 2n + 1)(2α + 4n + 3)w2 ≥ 0) ⇒

(2n + 3)(α + 2n + 1)2(α + 2n + 3)2(2α + 2n + 3)(2α + 4n + 5)y2(α + 2n + 2)2 + (α + 2n +

1)2(64n5 −256x

2n4 +160αn

4 +464n4 +144α

2n3 −512αx

2n3 −1184x

2n3 +928αn

3 +1344n3 +

56α3n2 + 628α

2n2 − 384α

2x2n2 − 1776αx

2n2 − 1984x

2n2 + 2016αn

2 + 1944n2 + 8α

4n +

164α3n + 912α

2n − 128α

3x2n − 888α

2x2n − 1984αx

2n − 1434x

2n + 1944αn + 1404n + 12α

4 +120α

3 +441α2 −16α

4x2 −148α

3x2 −496α

2x2 −717αx

2 −378x2 +702α+405)z2(α+2n+2)2 −

w2(−256n

7+4096x4n6−3072x

2n6−896αn

6−1728n6+12288αx

4n5+25088x

4n5−1216α

2n5−

9216αx2n5 − 19968x

2n5 − 5184αn

5 − 4864n5 + 15360α

2x4n4 + 62720αx

4n4 + 62464x

4n4 −

800α3n4 − 5872α

2n4 − 11008α

2x2n4 − 49920αx

2n4 − 53120x

2n4 − 12160αn

4 − 7408n4 −

256α4n3+10240α

3x4n3+62720α

2x4n3+124928αx

4n3+81216x

4n3−3104α

3n3−11072α

2n3−

6656α3x2n3 − 47744α

2x2n3 − 106240αx

2n3 − 74176x

2n3 − 14816αn

3 − 6592n3 − 32α

5n2 −

752α4n2+3840α

4x4n2+31360α

3x4n2+93696α

2x4n2+121824αx

4n2+58320x

4n2−4448α

3n2−

10192α2n2 − 2112α

4x2n2 − 21696α

3x2n2 − 76416α

2x2n2 − 111264αx

2n2 − 57396x

2n2 −

9888αn2 − 3424n

2 − 64α5n− 736α

4n +768α

5x4n +7840α

4x4n +31232α

3x4n +60912α

2x4n +

58320αx4n + 21978x

4n − 2784α

3n − 4576α

2n − 320α

5x2n − 4608α

4x2n − 23296α

3x2n −

53568α2x2n − 57396αx

2n − 23340x

2n − 3424αn − 960n − 32α

5 − 240α4 + 64α

6x4 + 784α

5x4 +

3904α4x4+10152α

3x4+14580α

2x4+10989αx

4+3402x4−640α

3−800α2−16α

6x2−352α

5x2−

2504α4x2−8240α

3x2−13881α

2x2−11670αx

2−3897x2−480α−112)(α+2n+2)2−2(α+2n+

1)wx(128n6 − 1024x

2n5 + 384αn

5 + 1408n5 + 448α

2n4 − 2560αx

2n4 − 5504x

2n4 + 3520αn

4 +

5592n4+256α

3n3+3296α

2n3−2560α

2x2n3−11008αx

2n3−11488x

2n3+11184αn

3+10888n3+

72α4n2 + 1424α

3n2 + 7870α

2n2 − 1280α

3x2n2 − 8256α

2x2n2 − 17232αx

2n2 − 11688x

2n2 +

16332αn2 + 11258n

2 + 8α5n + 272α

4n + 2278α

3n + 7692α

2n − 320α

4x2n − 2752α

3x2n −

8616α2x2n − 11688αx

2n − 5814x

2n + 11258αn + 5940n + 16α

5 + 220α4 + 1124α

3 + 2669α2 −

32α5x2−344α

4x2−1436α

3x2−2922α

2x2−2907αx

2−1134x2+2970α+1257)z(α+2n+2)2 ≥ 0

´


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).





n(x).

Showing the induction base

gα0 (x) ≥ eα

0 (x)

is not a problem.





n(x).


(α + 1)(2αx2 + 3x2 − 2)

2(2α + 3)≥ α + 1

2α + 3

is not a problem.





n(x).


(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).





n(x).


(α + 1)(2αx2 + 3x2 − 2)

2(2α + 3)≥ α + 1

2α + 3

is not a problem.


n(x) (n ∈ N).

This completes the proof of gαn(x) ≥ fα

n (x) (n ∈ N).





n(x).


(α + 1)(2αx2 + 3x2 − 2)

2(2α + 3)≥ α + 1

2α + 3

is not a problem.


n(x) (n ∈ N).

This completes the proof of gαn(x) ≥ fα

n (x) (n ∈ N).

This completes the proof of Schoberl’s conjecture.

Pillwein’s Proof

Message:

A special function inequality may require some very

non-obvious manipulation before an induction proof

via CAD succeeds.

Summary

Summary

I Polynomial inequalities can be proven without thinking.

Summary


I We may use CAD to construct an induction proof for a specialfunction inequality.

Summary



I Special function inequalities arise in real world applications.

Summary




I Some “deep” special function inequalities are just animmediate consequence of a polynomial inequality.

Summary





I Some inequalities require human preprocessing.

Summary





I Some inequalities require human preprocessing.

I The preprocessing may be hard (if at all possible).

Conclusion

Conclusion

I Classical inequality proofs proceed by reducing the claim to anobvious statement.

Conclusion


I Modern inequality proofs proceed by reducing the claim tosomething that can be done with the computer.

Conclusion



I Stronger computer algebra methods for proving specialfunction inequalities would be highly appreciated. . .

Conclusion



I Stronger computer algebra methods for proving specialfunction inequalities would be highly appreciated. . .

I . . . because these inequalities are soo difficult.

Documents

Pillwein’s Proof of Schoberl’s Conjecture...Polynomial Inequalities I Problem 11251 (Marian Tetiva; vol. 113(10), 2006, p. 847): Let a,b,c be positive real numbers, two of which