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Topic: why are the polynomials in this series all solvable by radicals?
Replies: 5   Last Post: Jul 18, 2006 11:04 AM

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 Paul Abbott Posts: 1,437 Registered: 12/7/04
Re: why are the polynomials in this series all solvable by radicals?
Posted: Jul 18, 2006 12:11 AM

"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^2-1
> p(n) = x * p(n-1) - p(n-2) 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 tri-diagonal with -2 along the diagonal and 1
along the super- and sub-diagonals. 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)

> 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
>
> I wonder why this recurrence relation invariably (at least as far as
> I've checked) leads to such polynomials.

Determining the closed-form 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 tri-diagonal 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 tri-diagonal
matrices are also well-known. See, e.g.,

L. Losonczi, Eigenvalues and eigenvectors of some tridiagonal matrices,
Acta Math. Hungar. 60 (1992), no. 3-4, 309-322.

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

Date Subject Author
7/17/06 Lee M Davidson
7/17/06 Daniel Lichtblau
7/17/06 Peter L. Montgomery
7/17/06 Robert Israel
7/18/06 Paul Abbott
7/18/06 Lee M Davidson