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Template:Sequence of the Day for November 15

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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 ) = qneiωt
(
i
being the complex unit) is
qn  + 1 − 2  (1x) 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  (1x)     −1
1     0
,

with (
 T
for matrix transpose)
[qn +1, qn ] T  =  R (x) [qn , qn  − 1] T.

Iteration yields

[qn +1, qn ] T  =  Mn(x) [q1, q0 ] T,
with
Mn(x) = (R (x))n
, and the two arbitrary inputs
q1
and
q0
. It follows that
     
Mn(x)  = 
S (n, 2  (1x))     S (n − 1, 2  (1x))
S (n − 1, 2  (1x))     − S (n − 2, 2  (1x))

due to the recurrence for the
S
-polynomials:
S (−1, x)  =  0; S (0, x)  =  1; S (n, x)  =  xS (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)) q1S (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 Mn(x) = 1
, identically, therefore on has the so called Cassini–Simson identity
(S (n − 1, y)) 2S (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