
Re: why are the polynomials in this series all solvable by radicals?
Posted:
Jul 18, 2006 12:11 AM


In article <1153154093.967916.305090@i42g2000cwa.googlegroups.com>, "Lee Davidson" <lmd@meta5.com> wrote:
> Consider n+1 identical masses m in linear series with n identical > springs of spring constant k connecting adjacent masses. Taking the > lengths of the springs as n variables, you can set up n independent > second order linear differential equations and analyze the solutions. > > What you want are eigenvalues, to determine characteristic vibrational > frequencies. After some simplification and change of variable, you get > polynomials according to the following recurrence relation: > > p(1) = x > p(2) = x^21 > p(n) = x * p(n1)  p(n2) for n >= 3
Actually, you have determined the polynomials for a system of n identical masses m in linear series with n+1 identical springs of spring constant k, where each end spring has one end fixed  which is a (slightly) different physical problem.
The associated matrix is tridiagonal with 2 along the diagonal and 1 along the super and subdiagonals. For n = 5, the matrix reads
2 1 0 0 0
1 2 1 0 0
0 1 2 1 0
0 0 1 2 1
0 0 0 1 2
and the characteristic polynomial is
 (1 + x) (2 + x) (3 + x) (1 + 4 x + x^2)
which, after a change of variables, x > 2 x, agrees with your p(5),
p(5) = (x  1) x (x + 1) (x^2  3) > What is remarkable about this series is that every polynomial I've > checked up to degree fifty is factorable, and every factor I've checked > is solvable by radicals. (Some factors in some high degrees were too > computationally difficult.) > > Furthermore, many of these solutions were expressible as linear > combinarions of pth roots of 1, or had such linear combinarions under > radicals. > > I wonder why this recurrence relation invariably (at least as far as > I've checked) leads to such polynomials.
Determining the closedform eigenvalues is straightforward; employing the recurrence relations for the Chebyshev polynomials, one obtains
p(n) = ChebyshevU[n, x/2] == Sin[(n + 1) ArcCos[x/2]]/Sqrt[1  x^2/4]
Alternatively, the eigenvalues of the matrix above are
2 (1 + Cos[Pi k / (n+1)]), k = 1, 2, ..., n > I tried varying the recurrence relation, and obtained some others with > similar behavior. However, other variations produce >= 5th degree > factors which are invariably not solvable by radicals, or produce > irreducible polynomials whose solution I did not attempt. > > So there is some quality of recurrence relations, rather than some > feature of the original physical problem, which explains this?
If you return to the original problem as you stated it, the associated matrix is still tridiagonal but the {1,1} and {n,n} elements are modified because the first and last masses are only connect to one spring, not two. For n = 5, the matrix reads
1 1 0 0 0
1 2 1 0 0
0 1 2 1 0
0 0 1 2 1
0 0 0 1 1
Now, explicit eigenvalues (and eigenvectors) of such tridiagonal matrices are also wellknown. See, e.g.,
L. Losonczi, Eigenvalues and eigenvectors of some tridiagonal matrices, Acta Math. Hungar. 60 (1992), no. 34, 309322.
Explicitly, the eigenvalues are
2 (1 + Cos[Pi k/n]), k = 1, 2, ..., n
Cheers, Paul
_______________________________________________________________________ Paul Abbott Phone: 61 8 6488 2734 School of Physics, M013 Fax: +61 8 6488 1014 The University of Western Australia (CRICOS Provider No 00126G) AUSTRALIA http://physics.uwa.edu.au/~paul

