Template:Sequence of the Day for November 15

Intended for: November 15, 2012

Timetable

• First draft entered by Wolfdieter Lang on November 6, 2011 ✓
• Draft reviewed by Daniel Forgues on October 16, 2016 (presentation review) (help needed: may some editor please review the content) ✓
• Draft to be approved by October 15, 2012

Yesterday's SOTD * Tomorrow's SOTD

The line below marks the end of the <noinclude> ... </noinclude> section.

---

A049310: Triangle of coefficients of Chebyshev ’s

S (n, x) := U (n, x / 2)

polynomials* (exponents in increasing order).

{ 1, 0, 1, −1, 0, 1, 0, −2, 0, 1, ... }

This sequence appears in linear atomic chains with

N

uniformly harmonic interacting atoms of the same mass. The eigenmodes have scaled frequency squares

x

given by the zeros of

S (N, 2  (1  −  x))

.
The recurrence for the oscillations with frequency

ω

and displacement

qn

from the equilibrium position at site no.

n

,

qn(t ) = qn e i ω t

(

i

being the complex unit) is

qn  + 1 − 2  (1 − x) qn + qn  − 1  =  0.

Here we have

x := 1 2 (   ω ω0   ) 2

, the normalized frequency squared, with

ω0 := k  / m

, where

k

is the uniform spring constant and

m

is the atom’s mass. This leads to a so called

2  ×  2

transfer matrix
     

R (x)  :=  2  (1 − x)     −1 1     0 ,

with (

T

for matrix transpose)

[qn +1, qn ] T  =  R (x) [qn , qn  − 1] T.

Iteration yields

[qn +1, qn ] T  =  M n(x) [q1, q0 ] T,

with

M n(x) = (R (x)) n

, and the two arbitrary inputs

q1

and

q0

. It follows that
     

M n(x)  =  S (n, 2  (1 − x))     S (n − 1, 2  (1 − x)) − S (n − 1, 2  (1 − x))     − S (n − 2, 2  (1 − x))

due to the recurrence for the

S

-polynomials:

S (−1, x)  =  0; S (0, x)  =  1; S (n, x)  =  x S (n − 1, x) − S (n − 2, x), n ≥ 1.

Thus one obtains the general solution for the displacements

qn +1(x)  =  S (n, 2  (1 − x)) q1 − S (n − 1, 2  (1 − x)) q0 .

For finite

N

-chains, with fixed boundary conditions

q0 = 0 = qN +1

, one therefore has to solve

S (N, 2 (1  −  x)) = 0

, and thus obtains the

N

normalized eigenfrequency squares for the

N

-chain:

x ( N ) k  =  2 sin π  k 2  (N  + 1)   2, k  =  1, ..., N.

A side remark: because

Det R (x) = 1

, also

Det M n(x) = 1

, identically, therefore on has the so called Cassini–Simson identity

(S (n − 1, y)) 2 − S (n, y) S (n − 2, y)  =  1, n ≥ 0.

For this and another nine applications of this sequence and the row polynomials

S (n, x)

see the a link under A049310.

_______________

* Chebyshev polynomials