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Chemnitz, CMS2013, September of 2013 – p. 1

Testing the remote control. . .

Chemnitz, CMS2013, September of 2013 – p. 2

The “(xn)” sequence

Paul Nevai

paul@nevai.us

Chemnitz, CMS2013, September of 2013 – p. 3

Introduction

Linear difference equations are the discrete version of linear differential equations, and have a theory that parallels the latter. Even here, the problems can be quite difficult. E.g., going from Fibonacci’s

Fn+1 = Fn + Fn−1

to, say,

Fn+1 =

• 1 + sin 1

n

• Fn + Fn−1

is highly nontrivial, c.f. Poincaré’s theorem on linear difference equations. Destroying linearity by, say,

Fn+1 = F 2

n + Fn−1

can lead to very interesting problems.

Chemnitz, CMS2013, September of 2013 – p. 4

OPs

def

= Orthogonal Polynomials

Let α be a positive Borel measure with infinite support in R and let

pn(α, x) = γn(α)xn + l.d.t., n ∈ N0

def

= {0, 1, 2, . . . },

denote the OPs w.r.t. α. (l.d.t. def

= lower degree terms)

Orthogonality:

• R

pmpn dα = δm,n.

The three-term recurrence:

xpn = an+1pn+1 + bnpn + anpn−1

where an = γn−1/γn > 0 and bn ∈ R “describe” the symmetry

• f the measure w.r.t. a vertical line.

Chemnitz, CMS2013, September of 2013 – p. 5

Favard’s theorem

• THEOREM. (Favard, 1935) Given (an > 0) and (bn ∈ R), if

(pn) satisfy the three-term recurrence xpn = an+1pn+1 + bnpn + anpn−1

then they are OPs w.r.t. some, not necessarily unique, α in R.

Chemnitz, CMS2013, September of 2013 – p. 6

Hermite polynomials

The Hermite polynomials, well, the Hermite functions, are the eigenfunctions of the Fourier transform. So there can be no doubt that they are worthy of study.

Chemnitz, CMS2013, September of 2013 – p. 7

Hermite polynomials

The Hermite polynomials, well, the Hermite functions, are the eigenfunctions of the Fourier transform. So there can be no doubt that they are worthy of study.

• PUZZLE. What is the ugliest word in mathematical English?

Chemnitz, CMS2013, September of 2013 – p. 7

Hermite polynomials

The Hermite polynomials, well, the Hermite functions, are the eigenfunctions of the Fourier transform. So there can be no doubt that they are worthy of study.

• PUZZLE. What is the ugliest word in mathematical English?
• ANSWER. IMHO, it’s a tie between eigenfunction and eigen-

value.

• Note. IMHO means “in my humble opinion”.

Chemnitz, CMS2013, September of 2013 – p. 7

Hermite polynomials

The Hermite polynomials, well, the Hermite functions, are the eigenfunctions of the Fourier transform. So there can be no doubt that they are worthy of study.

• PUZZLE. What is the ugliest word in mathematical English?
• ANSWER. IMHO, it’s a tie between eigenfunction and eigen-

value.

• Note. IMHO means “in my humble opinion”.

In the next few slides I will show how to obtain certain proper- ties of Hermite polynomials using “nothing” but orthogonality. The very same ideas can be applied to much more “sophis- ticated” weight functions.

Chemnitz, CMS2013, September of 2013 – p. 7

Hermite polynomials (hn)

dα(x) = w(x)dx

where

w(x) def = exp

• −x2

, x ∈ R.

Orthogonality:

• R

hmhn w = δm,n.

Chemnitz, CMS2013, September of 2013 – p. 8

Hermite polynomials (hn)

dα(x) = w(x)dx

where

w(x) def = exp

• −x2

, x ∈ R.

Orthogonality:

• R

hmhn w = δm,n.

The three-term recurrence:

xhn =

• n + 1

2 hn+1 +

• n

2 hn−1, n = 0, 1, 2, . . . ,

that is,

an =

• n

2 (n = 2, 3, . . . ) & bn = 0 (n = 0, 1, 2, . . . )

and

a2

0 = 0

& a2

1 =

• R x2 exp
• −x2

dx

• R exp (−x2) dx

= Γ (3/2) Γ (1/2) = 1 2 .

Chemnitz, CMS2013, September of 2013 – p. 8

an for hn

The following tricky computation of (an) probably goes back to Shohat and then Freud, probably innocently, repeated the same argument; it is based on the the property of w that

w(x)′ = −2 x w(x)

i.e.

y′ = −2 x y.

Chemnitz, CMS2013, September of 2013 – p. 9

an for hn

The following tricky computation of (an) probably goes back to Shohat and then Freud, probably innocently, repeated the same argument; it is based on the the property of w that

w(x)′ = −2 x w(x)

i.e.

y′ = −2 x y.

On one hand,

• R

(hnhn−1)′ w =

• R
• h′

nhn−1 + hnh′ n−1

• w orth

=

• R

h′

nhn−1w =

Chemnitz, CMS2013, September of 2013 – p. 9

an for hn

The following tricky computation of (an) probably goes back to Shohat and then Freud, probably innocently, repeated the same argument; it is based on the the property of w that

w(x)′ = −2 x w(x)

i.e.

y′ = −2 x y.

On one hand,

• R

(hnhn−1)′ w =

• R
• h′

nhn−1 + hnh′ n−1

• w orth

=

• R

h′

nhn−1w =

• R
• n γn xn−1 + l.d.t.
• hn−1w orth

= n γn

• R

xn−1hn−1w orth =

Chemnitz, CMS2013, September of 2013 – p. 9

an for hn

The following tricky computation of (an) probably goes back to Shohat and then Freud, probably innocently, repeated the same argument; it is based on the the property of w that

w(x)′ = −2 x w(x)

i.e.

y′ = −2 x y.

On one hand,

• R

(hnhn−1)′ w =

• R
• h′

nhn−1 + hnh′ n−1

• w orth

=

• R

h′

nhn−1w =

• R
• n γn xn−1 + l.d.t.
• hn−1w orth

= n γn

• R

xn−1hn−1w orth = n γn γn−1

• R
• γn−1 xn−1 + l.d.t.
• hn−1w orth

= n γn γn−1

• R

h2

n−1w = n

an .

Chemnitz, CMS2013, September of 2013 – p. 9

an for hn

On the other hand, by integration by parts,

• R

(hnhn−1)′ w

ibp

= −

• R

hnhn−1w′ = 2

• R

hnhn−1x w orth = 2an

Chemnitz, CMS2013, September of 2013 – p. 10

an for hn

On the other hand, by integration by parts,

• R

(hnhn−1)′ w

ibp

= −

• R

hnhn−1w′ = 2

• R

hnhn−1x w orth = 2an

because, in general,

• R

x pnpn−1 dα orth =

• R

x pn

• γn−1 xn−1 + l.d.t.
• dα orth

= γn−1 γn

• R

pn (γn xn + l.d.t.) dα = γn−1 γn

• R

p2

n dα = γn−1

γn = an.

Chemnitz, CMS2013, September of 2013 – p. 10

an for hn

On the other hand, by integration by parts,

• R

(hnhn−1)′ w

ibp

= −

• R

hnhn−1w′ = 2

• R

hnhn−1x w orth = 2an

because, in general,

• R

x pnpn−1 dα orth =

• R

x pn

• γn−1 xn−1 + l.d.t.
• dα orth

= γn−1 γn

• R

pn (γn xn + l.d.t.) dα = γn−1 γn

• R

p2

n dα = γn−1

γn = an.

So

n an =

• R

(hnhn−1)′ w = 2an.

Chemnitz, CMS2013, September of 2013 – p. 10

bn for hn

The coefficient bn = 0 because w is even. In general, from the recurrence formula,

bn =

• R

x p2

n dα.

From the so-to-speak Hermitian point of view,

bn =

• R

x h2

n w = −1

2

• R

h2

n w′ ibp

= 1 2

• R
• h2

n

′ w =

• R

hn h′

n w orth

= 0.

Chemnitz, CMS2013, September of 2013 – p. 11

bn for hn

The coefficient bn = 0 because w is even. In general, from the recurrence formula,

bn =

• R

x p2

n dα.

From the so-to-speak Hermitian point of view,

bn =

• R

x h2

n w = −1

2

• R

h2

n w′ ibp

= 1 2

• R
• h2

n

′ w =

• R

hn h′

n w orth

= 0.

• HOMEWORK. Find the an’s and the bn’s for the weight function

w(x) def = exp

• −β2 x2 + β1 x + β0
• with x ∈ R where β2 > 0.

Chemnitz, CMS2013, September of 2013 – p. 11

bn for hn

The coefficient bn = 0 because w is even. In general, from the recurrence formula,

bn =

• R

x p2

n dα.

From the so-to-speak Hermitian point of view,

bn =

• R

x h2

n w = −1

2

• R

h2

n w′ ibp

= 1 2

• R
• h2

n

′ w =

• R

hn h′

n w orth

= 0.

• HOMEWORK. Find the an’s and the bn’s for the weight function

w(x) def = exp

• −β2 x2 + β1 x + β0
• with x ∈ R where β2 > 0.
• HOMEWORK. What if x ∈ R+ for w above instead of x ∈ R?

Chemnitz, CMS2013, September of 2013 – p. 11

DE for hn

Let k < n − 1 & let rk be a polynomial of degree at most k. Then

• R

h′

n rk w orth

=

• R

(hn rk)′ w

ibp

= −2

• R

(hn rk) x w = −2

• R

hn (rk x) w orth = 0

Chemnitz, CMS2013, September of 2013 – p. 12

DE for hn

Let k < n − 1 & let rk be a polynomial of degree at most k. Then

• R

h′

n rk w orth

=

• R

(hn rk)′ w

ibp

= −2

• R

(hn rk) x w = −2

• R

hn (rk x) w orth = 0

so that

h′

n = const ×hn−1.

Thus, we have an Appell sequence, and by comparing leading coefficients

h′

n =

√ 2 n hn−1

and, therefore,

h′′

n =

√ 2 n h′

n−1 =

√ 2 n

• 2 (n − 1) hn−2 = 2
• n(n − 1) hn−2.

Chemnitz, CMS2013, September of 2013 – p. 12

DE for hn

Replacing hn−1 and hn−2 by

h′

n =

√ 2 n hn−1 & h′′

n = 2

• n(n − 1) hn−2

in the recurrence formula

xhn−1 =

• n

2 hn +

• n − 1

2 hn−2, n = 0, 1, 2, . . . ,

we get the differential equation for the Hermite polynomials

2 x y′ = 2 n y + y′′, y = hn.

Chemnitz, CMS2013, September of 2013 – p. 13

DE for hn

Replacing hn−1 and hn−2 by

h′

n =

√ 2 n hn−1 & h′′

n = 2

• n(n − 1) hn−2

in the recurrence formula

xhn−1 =

• n

2 hn +

• n − 1

2 hn−2, n = 0, 1, 2, . . . ,

we get the differential equation for the Hermite polynomials

2 x y′ = 2 n y + y′′, y = hn.

• HOMEWORK. Using “nothing” but orthogonality, find the Ro-

drigues’ formula for hn. Well, first google “rodrigues formula”.

Chemnitz, CMS2013, September of 2013 – p. 13

exp

• −c/4 x4 − K/2 x2

Let

w(x) def = exp

• −c

4 x4 − K 2 x2

• ,

x ∈ R,

where c > 0 & K ∈ R (interesting case) or c = 0 & K ∈ R+. Then the weight w satisfies a DE similar to the Hermite weight:

w(x)′ =

• −c x3 − K x
• w(x)

i.e.

y′ = −

• c x3 + K x
• y.

Chemnitz, CMS2013, September of 2013 – p. 14

exp

• −c/4 x4 − K/2 x2

Let

w(x) def = exp

• −c

4 x4 − K 2 x2

• ,

x ∈ R,

where c > 0 & K ∈ R (interesting case) or c = 0 & K ∈ R+. Then the weight w satisfies a DE similar to the Hermite weight:

w(x)′ =

• −c x3 − K x
• w(x)

i.e.

y′ = −

• c x3 + K x
• y.

Note.

(− log w)′′ = 3cx2 + 2K

so that

− log w is convex ⇐ ⇒ K ≥ 0.

It is unknown if convexity matters but the case K < 0 is “orders of magnitude” harder than K ≥ 0.

Chemnitz, CMS2013, September of 2013 – p. 14

The recurrence coefficients

The bn’s are all 0 (even w).

Chemnitz, CMS2013, September of 2013 – p. 15

The recurrence coefficients

The bn’s are all 0 (even w). The an’s can be computed using exactly the same idea as for the Hermite case, albeit it is a little more complicated because the recurrence formula must be used three times on the right-hand side.

Chemnitz, CMS2013, September of 2013 – p. 15

The recurrence coefficients

The bn’s are all 0 (even w). The an’s can be computed using exactly the same idea as for the Hermite case, albeit it is a little more complicated because the recurrence formula must be used three times on the right-hand side. We get

n an = c an

• a2

n+1 + a2 n + a2 n−1

• + K an,

n ∈ N,

where

a2

0 = 0

& a2

1 =

• R x2 exp
• − c

4 x4 − K 2 x2

dx

• R exp
• − c

4 x4 − K 2 x2

dx .

For K = 0,

a2

1 = Γ (3/4)

Γ (1/4) .

• Note. By their very nature, all the an’s are positive for n ∈ N.

Chemnitz, CMS2013, September of 2013 – p. 15

The (an)’s

The case K = 0 was found by Shohat in 1939 and by Freud in the early 1970s, and the general case by Stan Bonan and PN in 1984. Neither Shohat nor Freud studied it except that Fred proved the following in 1976:

w(x) = exp

• −x4/4
• =

⇒ lim

n→∞

an n1/4 = 1 31/4

It turned out that his completely elementary ad hoc tech- nique became a successful tool for proving limits in similar situations.

Chemnitz, CMS2013, September of 2013 – p. 16

The Freud Kunstgriff

We have

n = a2

n

• a2

n+1 + a2 n + a2 n−1

• =

⇒ a4

n ≤ n

and then

0 ≤ ℓ def = lim inf

n→∞

a2

n

√n ≤ L def = lim sup

n→∞

a2

n

√n < ∞

Chemnitz, CMS2013, September of 2013 – p. 17

The Freud Kunstgriff

We have

n = a2

n

• a2

n+1 + a2 n + a2 n−1

• =

⇒ a4

n ≤ n

and then

0 ≤ ℓ def = lim inf

n→∞

a2

n

√n ≤ L def = lim sup

n→∞

a2

n

√n < ∞

so that dividing both sides by n and going to lim inf with a2

n/√n

1 ≤ ℓ(L + ℓ + L)

Chemnitz, CMS2013, September of 2013 – p. 17

The Freud Kunstgriff

We have

n = a2

n

• a2

n+1 + a2 n + a2 n−1

• =

⇒ a4

n ≤ n

and then

0 ≤ ℓ def = lim inf

n→∞

a2

n

√n ≤ L def = lim sup

n→∞

a2

n

√n < ∞

so that dividing both sides by n and going to lim inf with a2

n/√n

1 ≤ ℓ(L + ℓ + L)

and going to lim sup with a2

n/√n

1 ≥ L(ℓ + L + ℓ)

Chemnitz, CMS2013, September of 2013 – p. 17

The Freud Kunstgriff

We have

n = a2

n

• a2

n+1 + a2 n + a2 n−1

• =

⇒ a4

n ≤ n

and then

0 ≤ ℓ def = lim inf

n→∞

a2

n

√n ≤ L def = lim sup

n→∞

a2

n

√n < ∞

so that dividing both sides by n and going to lim inf with a2

n/√n

1 ≤ ℓ(L + ℓ + L)

and going to lim sup with a2

n/√n

1 ≥ L(ℓ + L + ℓ)

from which

L2 + 2Lℓ ≤ ℓ2 + 2Lℓ = ⇒ L ≤ ℓ = ⇒ L = ℓ.

Chemnitz, CMS2013, September of 2013 – p. 17

The Freud conjecture

Based on this, Freud made the following conjecture in 1976 that fundamentally shaped the theory of orthogonal polyno- mials for the next 30+ years.

w(x) = exp (−|x|α) = ⇒ ∃ lim

n→∞

an n1/α ∈ (0, ∞)

Chemnitz, CMS2013, September of 2013 – p. 18

The Freud conjecture

Based on this, Freud made the following conjecture in 1976 that fundamentally shaped the theory of orthogonal polyno- mials for the next 30+ years.

w(x) = exp (−|x|α) = ⇒ ∃ lim

n→∞

an n1/α ∈ (0, ∞)

The conjecture has been solved but that’s another story; google “freud conjecture”.

Chemnitz, CMS2013, September of 2013 – p. 18

exp

• −x4

Returning to w(x) = exp

• −x4

, Shohat (1939) & Bonan (1983) proved that (pn) form a generalized Appel sequence:

p′

n = n

an pn−1 + 4 an an−1 an−2 pn−3,

and Shohat (1939) & I (1984) found the DE

z′′ + fnz = 0

where

z def = pn

w

ϕn

with

ϕn(x) def = a2

n+1 + a2 n + x2

and

fn(x) def = 4 a2

n

• 4 ϕn−1(x)ϕn(x) + 1 − 4 a2

n x2 − 4 x4 − 2 x2 ϕ−1 n (x)

• + 6 x2 − 4 x6 + (1 − 4 x4)ϕ−1

n (x) − 3 x2 ϕ−2 n (x).

Chemnitz, CMS2013, September of 2013 – p. 19

exp

• −x4

Returning to w(x) = exp

• −x4

, Shohat (1939) & Bonan (1983) proved that (pn) form a generalized Appel sequence:

p′

n = n

an pn−1 + 4 an an−1 an−2 pn−3,

and Shohat (1939) & I (1984) found the DE

z′′ + fnz = 0

where

z def = pn

w

ϕn

with

ϕn(x) def = a2

n+1 + a2 n + x2

and

fn(x) def = 4 a2

n

• 4 ϕn−1(x)ϕn(x) + 1 − 4 a2

n x2 − 4 x4 − 2 x2 ϕ−1 n (x)

• + 6 x2 − 4 x6 + (1 − 4 x4)ϕ−1

n (x) − 3 x2 ϕ−2 n (x).

Ich entschuldige mich für der häßlicher Formel

Chemnitz, CMS2013, September of 2013 – p. 19

exp

• −x4

Returning to w(x) = exp

• −x4

, Shohat (1939) & Bonan (1983) proved that (pn) form a generalized Appel sequence:

p′

n = n

an pn−1 + 4 an an−1 an−2 pn−3,

and Shohat (1939) & I (1984) found the DE

z′′ + fnz = 0

where

z def = pn

w

ϕn

with

ϕn(x) def = a2

n+1 + a2 n + x2

and

fn(x) def = 4 a2

n

• 4 ϕn−1(x)ϕn(x) + 1 − 4 a2

n x2 − 4 x4 − 2 x2 ϕ−1 n (x)

• + 6 x2 − 4 x6 + (1 − 4 x4)ϕ−1

n (x) − 3 x2 ϕ−2 n (x).

Ich entschuldige mich für der häßlicher Formel and for my terrible German.

Chemnitz, CMS2013, September of 2013 – p. 19

The (xn)’s

Let xn

def

= a2

• n. Then

n c = xn (xn+1 + xn + xn−1) + K c xn, x0 = 0, n ∈ N,

where

x1 =

• R x2 exp
• − c

4 x4 − K 2 x2

dx

• R exp
• − c

4 x4 − K 2 x2

dx .

Chemnitz, CMS2013, September of 2013 – p. 20

The (xn)’s

Let xn

def

= a2

• n. Then

n c = xn (xn+1 + xn + xn−1) + K c xn, x0 = 0, n ∈ N,

where

x1 =

• R x2 exp
• − c

4 x4 − K 2 x2

dx

• R exp
• − c

4 x4 − K 2 x2

dx .

From another point of view and with a different interpretation

• f facts: ∃x1 > 0 such that if (xn) is defined by

n c = xn (xn+1 + xn + xn−1) + K c xn, x0 = 0, n ∈ N,

(∗)

then

xn > 0, ∀n ∈ N.

• Note. (∗) is no longer associated with the original OPs.

Chemnitz, CMS2013, September of 2013 – p. 20

The (xn)’s; WLOG c = 1

Given K ∈ R and x0 ∈ R, we define (xn)n∈N by

n = xn (xn+1 + xn + xn−1) + K xn, n ∈ N,

(ℵ)

and then the question is under what conditions on x1 > 0 we have

xn > 0, ∀n ∈ N.

(∗)

Chemnitz, CMS2013, September of 2013 – p. 21

The (xn)’s; WLOG c = 1

Given K ∈ R and x0 ∈ R, we define (xn)n∈N by

n = xn (xn+1 + xn + xn−1) + K xn, n ∈ N,

(ℵ)

and then the question is under what conditions on x1 > 0 we have

xn > 0, ∀n ∈ N.

(∗)

• Theorem. ∀K ∈ R, if x0 = 0, then ∃!x1 > 0 such that (∗) holds.

Chemnitz, CMS2013, September of 2013 – p. 21

The (xn)’s; WLOG c = 1

Given K ∈ R and x0 ∈ R, we define (xn)n∈N by

n = xn (xn+1 + xn + xn−1) + K xn, n ∈ N,

(ℵ)

and then the question is under what conditions on x1 > 0 we have

xn > 0, ∀n ∈ N.

(∗)

• Theorem. ∀K ∈ R, if x0 = 0, then ∃!x1 > 0 such that (∗) holds.
• Theorem. ∀K ≥ 0, if x0 ∈ R, then ∃!x1 > 0 such that (∗) holds.

Chemnitz, CMS2013, September of 2013 – p. 21

The (xn)’s; WLOG c = 1

Given K ∈ R and x0 ∈ R, we define (xn)n∈N by

n = xn (xn+1 + xn + xn−1) + K xn, n ∈ N,

(ℵ)

and then the question is under what conditions on x1 > 0 we have

xn > 0, ∀n ∈ N.

(∗)

• Theorem. ∀K ∈ R, if x0 = 0, then ∃!x1 > 0 such that (∗) holds.
• Theorem. ∀K ≥ 0, if x0 ∈ R, then ∃!x1 > 0 such that (∗) holds.
• Conjecture. ∀K ∈ R, if x0 ∈ R, then ∃!x1 > 0 such that (∗) holds.

Chemnitz, CMS2013, September of 2013 – p. 21

The (xn)’s; WLOG c = 1

Given K ∈ R and x0 ∈ R, we define (xn)n∈N by

n = xn (xn+1 + xn + xn−1) + K xn, n ∈ N,

(ℵ)

and then the question is under what conditions on x1 > 0 we have

xn > 0, ∀n ∈ N.

(∗)

• Theorem. ∀K ∈ R, if x0 = 0, then ∃!x1 > 0 such that (∗) holds.
• Theorem. ∀K ≥ 0, if x0 ∈ R, then ∃!x1 > 0 such that (∗) holds.
• Conjecture. ∀K ∈ R, if x0 ∈ R, then ∃!x1 > 0 such that (∗) holds.
• Note. Of course, (ℵ) is d-PI.

Chemnitz, CMS2013, September of 2013 – p. 21

The (xn)’s; WLOG c = 1

Given K ∈ R and x0 ∈ R, we define (xn)n∈N by

n = xn (xn+1 + xn + xn−1) + K xn, n ∈ N,

(ℵ)

and then the question is under what conditions on x1 > 0 we have

xn > 0, ∀n ∈ N.

(∗)

• Theorem. ∀K ∈ R, if x0 = 0, then ∃!x1 > 0 such that (∗) holds.
• Theorem. ∀K ≥ 0, if x0 ∈ R, then ∃!x1 > 0 such that (∗) holds.
• Conjecture. ∀K ∈ R, if x0 ∈ R, then ∃!x1 > 0 such that (∗) holds.
• Note. Of course, (ℵ) is d-PI.

What???

Chemnitz, CMS2013, September of 2013 – p. 21

The (xn)’s; WLOG c = 1

Given K ∈ R and x0 ∈ R, we define (xn)n∈N by

n = xn (xn+1 + xn + xn−1) + K xn, n ∈ N,

(ℵ)

and then the question is under what conditions on x1 > 0 we have

xn > 0, ∀n ∈ N.

(∗)

• Theorem. ∀K ∈ R, if x0 = 0, then ∃!x1 > 0 such that (∗) holds.
• Theorem. ∀K ≥ 0, if x0 ∈ R, then ∃!x1 > 0 such that (∗) holds.
• Conjecture. ∀K ∈ R, if x0 ∈ R, then ∃!x1 > 0 such that (∗) holds.
• Note. Of course, (ℵ) is d-PI.

What??? What???

Chemnitz, CMS2013, September of 2013 – p. 21

The (xn)’s; WLOG c = 1

Given K ∈ R and x0 ∈ R, we define (xn)n∈N by

n = xn (xn+1 + xn + xn−1) + K xn, n ∈ N,

(ℵ)

and then the question is under what conditions on x1 > 0 we have

xn > 0, ∀n ∈ N.

(∗)

• Theorem. ∀K ∈ R, if x0 = 0, then ∃!x1 > 0 such that (∗) holds.
• Theorem. ∀K ≥ 0, if x0 ∈ R, then ∃!x1 > 0 such that (∗) holds.
• Conjecture. ∀K ∈ R, if x0 ∈ R, then ∃!x1 > 0 such that (∗) holds.
• Note. Of course, (ℵ) is d-PI.

What??? What??? What???

Chemnitz, CMS2013, September of 2013 – p. 21

The (xn)’s; WLOG c = 1

Given K ∈ R and x0 ∈ R, we define (xn)n∈N by

n = xn (xn+1 + xn + xn−1) + K xn, n ∈ N,

(ℵ)

and then the question is under what conditions on x1 > 0 we have

xn > 0, ∀n ∈ N.

(∗)

• Theorem. ∀K ∈ R, if x0 = 0, then ∃!x1 > 0 such that (∗) holds.
• Theorem. ∀K ≥ 0, if x0 ∈ R, then ∃!x1 > 0 such that (∗) holds.
• Conjecture. ∀K ∈ R, if x0 ∈ R, then ∃!x1 > 0 such that (∗) holds.
• Note. Of course, (ℵ) is d-PI.

What??? What??? What??? What???

Chemnitz, CMS2013, September of 2013 – p. 21

d-PI

d-PI stands for Painlevé’s discrete equation #1, and, since I know no more than you about either this or the origi- nal, nondiscrete versions, I recommend that you google “painleve equations” or “painleve transcendents”. You will find out that these are some of the most important nonlinear differential equations.

Chemnitz, CMS2013, September of 2013 – p. 22

d-PI

d-PI stands for Painlevé’s discrete equation #1, and, since I know no more than you about either this or the origi- nal, nondiscrete versions, I recommend that you google “painleve equations” or “painleve transcendents”. You will find out that these are some of the most important nonlinear differential equations. True story. A couple of years ago, I bet a French mathe- matician that Painlevé was also a prime minister of France whereas he stated he was not, only a minister of defense.

Chemnitz, CMS2013, September of 2013 – p. 22

d-PI

d-PI stands for Painlevé’s discrete equation #1, and, since I know no more than you about either this or the origi- nal, nondiscrete versions, I recommend that you google “painleve equations” or “painleve transcendents”. You will find out that these are some of the most important nonlinear differential equations. True story. A couple of years ago, I bet a French mathe- matician that Painlevé was also a prime minister of France whereas he stated he was not, only a minister of defense. I won: he was prime minister even twice

Chemnitz, CMS2013, September of 2013 – p. 22

d-PI

d-PI stands for Painlevé’s discrete equation #1, and, since I know no more than you about either this or the origi- nal, nondiscrete versions, I recommend that you google “painleve equations” or “painleve transcendents”. You will find out that these are some of the most important nonlinear differential equations. True story. A couple of years ago, I bet a French mathe- matician that Painlevé was also a prime minister of France whereas he stated he was not, only a minister of defense. I won: he was prime minister even twice (and Kriegsminister, that is, ministre de la Guerre once).

Chemnitz, CMS2013, September of 2013 – p. 22

d-PI

d-PI stands for Painlevé’s discrete equation #1, and, since I know no more than you about either this or the origi- nal, nondiscrete versions, I recommend that you google “painleve equations” or “painleve transcendents”. You will find out that these are some of the most important nonlinear differential equations. True story. A couple of years ago, I bet a French mathe- matician that Painlevé was also a prime minister of France whereas he stated he was not, only a minister of defense. I won: he was prime minister even twice (and Kriegsminister, that is, ministre de la Guerre once).

• Confession. Okay, I admit that this French mathematician

was, in fact, a German living in France and thinking like a Frenchman (Bernhard Beckermann).

Chemnitz, CMS2013, September of 2013 – p. 22

Generalizations

The equation

n = xn (xn+1 + xn + xn−1) + K xn

can be generalized in more than one direction.

Chemnitz, CMS2013, September of 2013 – p. 23

Generalizations

The equation

n = xn (xn+1 + xn + xn−1) + K xn

can be generalized in more than one direction. Examples.

• Tinker with the setup as in

n = xn (xn+1 + xn + xn−1) + K xn−1

Chemnitz, CMS2013, September of 2013 – p. 23

Generalizations

The equation

n = xn (xn+1 + xn + xn−1) + K xn

can be generalized in more than one direction. Examples.

• Tinker with the setup as in

n = xn (xn+1 + xn + xn−1) + K xn−1

• Introduce coefficients as in

ℓn = xn (σn,1 xn+1 + σn,0 xn + σn,−1 xn−1) + κn xn

Chemnitz, CMS2013, September of 2013 – p. 23

Generalizations

The equation

n = xn (xn+1 + xn + xn−1) + K xn

can be generalized in more than one direction. Examples.

• Tinker with the setup as in

n = xn (xn+1 + xn + xn−1) + K xn−1

• Introduce coefficients as in

ℓn = xn (σn,1 xn+1 + σn,0 xn + σn,−1 xn−1) + κn xn

• Increase the order as in

n = xn (xn+1 + xn + xn−1 + xn−2) + K xn

Chemnitz, CMS2013, September of 2013 – p. 23

Generalizations

The equation

n = xn (xn+1 + xn + xn−1) + K xn

can be generalized in more than one direction. Examples.

• Tinker with the setup as in

n = xn (xn+1 + xn + xn−1) + K xn−1

• Introduce coefficients as in

ℓn = xn (σn,1 xn+1 + σn,0 xn + σn,−1 xn−1) + κn xn

• Increase the order as in

n = xn (xn+1 + xn + xn−1 + xn−2) + K xn

• Go to several dimensions. . .

Chemnitz, CMS2013, September of 2013 – p. 23

Generalizations

The equation

n = xn (xn+1 + xn + xn−1) + K xn

can be generalized in more than one direction. Examples.

• Tinker with the setup as in

n = xn (xn+1 + xn + xn−1) + K xn−1

• Introduce coefficients as in

ℓn = xn (σn,1 xn+1 + σn,0 xn + σn,−1 xn−1) + κn xn

• Increase the order as in

n = xn (xn+1 + xn + xn−1 + xn−2) + K xn

• Go to several dimensions. . .

Each of them leads to a principal problem: no explicit solu- tions exist.

Chemnitz, CMS2013, September of 2013 – p. 23

A theorem and a proof

• Theorem. ∀K ≥ 0, if x0 ∈ R, then ∃!x1 > 0 such that (xn)n∈N

defined by

n = xn (xn+1 + xn + xn−1) + K xn, n ∈ N,

is a positive sequence.

Chemnitz, CMS2013, September of 2013 – p. 24

A theorem and a proof

• Theorem. ∀K ≥ 0, if x0 ∈ R, then ∃!x1 > 0 such that (xn)n∈N

defined by

n = xn (xn+1 + xn + xn−1) + K xn, n ∈ N,

is a positive sequence.

• Note. If x0 = 0, then the existence of a positive solution follows

from the explicit solution coming from exp

• −x4/4 − Kx2/2
• . Oth-

erwise, it can be proved using either some fixed point arguments

• r by constructing positive sequences (xn)N

n=1 satisfying the equa-

tion up to N and then letting N → ∞ and justifying that the latter is legitimate.

Chemnitz, CMS2013, September of 2013 – p. 24

A theorem and a proof

• Theorem. ∀K ≥ 0, if x0 ∈ R, then ∃!x1 > 0 such that (xn)n∈N

defined by

n = xn (xn+1 + xn + xn−1) + K xn, n ∈ N,

is a positive sequence.

• Note. If x0 = 0, then the existence of a positive solution follows

from the explicit solution coming from exp

• −x4/4 − Kx2/2
• . Oth-

erwise, it can be proved using either some fixed point arguments

• r by constructing positive sequences (xn)N

n=1 satisfying the equa-

tion up to N and then letting N → ∞ and justifying that the latter is legitimate. In what follows, I will prove the uniqueness part that I found back in 1983 and still think as one of the most beautiful proofs I ever dreamt up.

Chemnitz, CMS2013, September of 2013 – p. 24

Uniqueness

If K ≥ 0 and (xn) is positive, then from

n = xn (xn+1 + xn + xn−1) + K xn, n ∈ N,

(ℵ)

we have

n ≥ x2

n,

n ∈ N.

Let (αn) and (βn) be two positive solutions of (ℵ), that is, of

n xn = xn+1 + xn + xn−1 + K, n ∈ N,

and let εn

def

= βn − αn, Then, − n αnβn εn = n βn − n αn = εn+1 + εn + εn−1, n ∈ N,

(K is gone), that is,

εn+1 +

• 1 +

n αnβn

• εn + εn−1 = 0,

n ∈ N.

Chemnitz, CMS2013, September of 2013 – p. 25

Uniqueness

Since

n αnβn ≥ 1, we get

2 |εn| ≤ |εn+1| + |εn−1| , n ∈ N.

Chemnitz, CMS2013, September of 2013 – p. 26

Uniqueness

Since

n αnβn ≥ 1, we get

2 |εn| ≤ |εn+1| + |εn−1| , n ∈ N.

so that,

2

n

• n=M

|εn| ≤

N

• n=M

|εn+1| +

N

• n=M

|εn−1| , 1 ≤ M ≤ N,

Chemnitz, CMS2013, September of 2013 – p. 26

Uniqueness

Since

n αnβn ≥ 1, we get

2 |εn| ≤ |εn+1| + |εn−1| , n ∈ N.

so that,

2

n

• n=M

|εn| ≤

N

• n=M

|εn+1| +

N

• n=M

|εn−1| , 1 ≤ M ≤ N,

that is, for 1 ≤ M ≤ N,

2

N

• n=M

|εn| ≤ 2

N

• n=M

|εn| + |εN+1| − |εM| − |εN| + |εM−1| ,

Chemnitz, CMS2013, September of 2013 – p. 26

Uniqueness

Since

n αnβn ≥ 1, we get

2 |εn| ≤ |εn+1| + |εn−1| , n ∈ N.

so that,

2

n

• n=M

|εn| ≤

N

• n=M

|εn+1| +

N

• n=M

|εn−1| , 1 ≤ M ≤ N,

that is, for 1 ≤ M ≤ N,

2

N

• n=M

|εn| ≤ 2

N

• n=M

|εn| + |εN+1| − |εM| − |εN| + |εM−1| ,

i.e.,

|εM| − |εM−1| ≤ |εN+1| − |εN| , 1 ≤ M ≤ N.

We are getting there. . .

Chemnitz, CMS2013, September of 2013 – p. 26

Uniqueness

From

|εM| − |εM−1| ≤ |εN+1| − |εN| , 1 ≤ M ≤ N,

we get

P

• N=M

{|εM| − |εM−1|} ≤

P

• N=M

{|εN+1| − |εN|} , 1 ≤ M ≤ P,

Chemnitz, CMS2013, September of 2013 – p. 27

Uniqueness

From

|εM| − |εM−1| ≤ |εN+1| − |εN| , 1 ≤ M ≤ N,

we get

P

• N=M

{|εM| − |εM−1|} ≤

P

• N=M

{|εN+1| − |εN|} , 1 ≤ M ≤ P,

that is,

|εM| − |εM−1| ≤ |εP+1| − |εM| P − M + 1 ≤ 2 √ P + 1 P − M + 1, 1 ≤ M ≤ P,

Chemnitz, CMS2013, September of 2013 – p. 27

Uniqueness

From

|εM| − |εM−1| ≤ |εN+1| − |εN| , 1 ≤ M ≤ N,

we get

P

• N=M

{|εM| − |εM−1|} ≤

P

• N=M

{|εN+1| − |εN|} , 1 ≤ M ≤ P,

that is,

|εM| − |εM−1| ≤ |εP+1| − |εM| P − M + 1 ≤ 2 √ P + 1 P − M + 1, 1 ≤ M ≤ P,

so that, letting P → ∞,

|εM| − |εM−1| ≤ 0, 1 ≤ M.

Chemnitz, CMS2013, September of 2013 – p. 27

Uniqueness

Therefore,

(|εM|)∞

M=1 is a decreasing nonnegative se-

quence, and, since ε0 = β0 − α0 = x0 − x0 = 0, we have

εM ≡ 0 for all M ∈ N, that is, βM = αM for all M ∈ N.

Chemnitz, CMS2013, September of 2013 – p. 28

The prize

I offer a cash prize of $100,

Chemnitz, CMS2013, September of 2013 – p. 29

The prize

I offer a cash prize of $100, or, let me be generous, of e 100, to the first person who succeeds in modifying the proof so it would also work for the case K < 0,

Chemnitz, CMS2013, September of 2013 – p. 29

The prize

I offer a cash prize of $100, or, let me be generous, of e 100, to the first person who succeeds in modifying the proof so it would also work for the case K < 0, with the additional condition that the proof is actually correct. If you have a proof, please contact me at paul@nevai.us.

Chemnitz, CMS2013, September of 2013 – p. 29

Explanation of difficulties

This theorem has several proofs; all are more or less elementary.

• Theorem. Let x0 = 0 and K ≥ 0. Then ∃!x1 > 0 such that (xn)n∈N

defined by

n = xn (xn+1 + xn + xn−1) + K xn, n ∈ N,

(ℵ)

is a positive sequence.

Chemnitz, CMS2013, September of 2013 – p. 30

Explanation of difficulties

This theorem has several proofs; all are more or less elementary.

• Theorem. Let x0 = 0 and K ≥ 0. Then ∃!x1 > 0 such that (xn)n∈N

defined by

n = xn (xn+1 + xn + xn−1) + K xn, n ∈ N,

(ℵ)

is a positive sequence. Currently, this one has only one proof; not at all elementary.

• Theorem. Let x0 = 0 and K < 0. Then ∃!x1 > 0 such that (xn)n∈N

defined by (ℵ) is a positive sequence.

Chemnitz, CMS2013, September of 2013 – p. 30

Explanation of difficulties

This theorem has several proofs; all are more or less elementary.

• Theorem. Let x0 = 0 and K ≥ 0. Then ∃!x1 > 0 such that (xn)n∈N

defined by

n = xn (xn+1 + xn + xn−1) + K xn, n ∈ N,

(ℵ)

is a positive sequence. Currently, this one has only one proof; not at all elementary.

• Theorem. Let x0 = 0 and K < 0. Then ∃!x1 > 0 such that (xn)n∈N

defined by (ℵ) is a positive sequence. This one has no proof. Conjecture Let x0 = 0 and K < 0. Then ∃!x1 > 0 such that

(xn)n∈N defined by (ℵ) is a positive sequence.

Chemnitz, CMS2013, September of 2013 – p. 30

Explanation of difficulties

This theorem has several proofs; all are more or less elementary.

• Theorem. Let x0 = 0 and K ≥ 0. Then ∃!x1 > 0 such that (xn)n∈N

defined by

n = xn (xn+1 + xn + xn−1) + K xn, n ∈ N,

(ℵ)

is a positive sequence. Currently, this one has only one proof; not at all elementary.

• Theorem. Let x0 = 0 and K < 0. Then ∃!x1 > 0 such that (xn)n∈N

defined by (ℵ) is a positive sequence. This one has no proof. Conjecture Let x0 = 0 and K < 0. Then ∃!x1 > 0 such that

(xn)n∈N defined by (ℵ) is a positive sequence.

Why? I have only intuitive reasons related to the convexity of

− log exp

• −x4/4 − Kx2/2
• .

Chemnitz, CMS2013, September of 2013 – p. 30

Progress report

Theorems for both the existence and uniqueness of positive solutions have been extended to

ℓn = xn (σn,1 xn+1 + σn,0 xn + σn,−1 xn−1) + κn xn

recently under rather mild conditions of the coefficients except that although κn can be negative, we must have

lim inf κn ≥ 0 in the uniqueness results. I am sure that this

limitation is due rather to the methods used and not to the nature of the equation.

Chemnitz, CMS2013, September of 2013 – p. 31

Progress report

Theorems for both the existence and uniqueness of positive solutions have been extended to

ℓn = xn (σn,1 xn+1 + σn,0 xn + σn,−1 xn−1) + κn xn

recently under rather mild conditions of the coefficients except that although κn can be negative, we must have

lim inf κn ≥ 0 in the uniqueness results. I am sure that this

limitation is due rather to the methods used and not to the nature of the equation. I conjecture that even eventually positive solutions of

n = xn (xn+1 + xn + xn−1) + K xn, n ∈ N,

are unique.

Chemnitz, CMS2013, September of 2013 – p. 31

Progress report

Theorems for both the existence and uniqueness of positive solutions have been extended to

ℓn = xn (σn,1 xn+1 + σn,0 xn + σn,−1 xn−1) + κn xn

recently under rather mild conditions of the coefficients except that although κn can be negative, we must have

lim inf κn ≥ 0 in the uniqueness results. I am sure that this

limitation is due rather to the methods used and not to the nature of the equation. I conjecture that even eventually positive solutions of

n = xn (xn+1 + xn + xn−1) + K xn, n ∈ N,

are unique. Even probably O (√n) solutions are unique.

Chemnitz, CMS2013, September of 2013 – p. 31

Das Ende

Chemnitz, CMS2013, September of 2013 – p. 32