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Lee Davidson science forum beginner
Joined: 28 Sep 2005
Posts: 4

Posted: Mon Jul 17, 2006 4:34 pm Post subject:
why are the polynomials in this series all solvable by radicals?



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
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.
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? 

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Daniel Lichtblau science forum beginner
Joined: 12 May 2005
Posts: 48

Posted: Mon Jul 17, 2006 9:58 pm Post subject:
Re: why are the polynomials in this series all solvable by radicals?



Lee Davidson wrote:
Quote:  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
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.
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?

Offhand I do not know if there is something special to the physics
which would call for a family of radical solutions. But it certainly
can be shown that the recurrence has this property. Using Mathematica
it would be solved as below.
In[113]:= InputForm[pp[n_] = p[n] /.
First[RSolve[{p[n]==x*p[n1]p[n2],p[1]==x,p[0]==1}, p[n], n]]]
Out[113]//InputForm=
(2^(1  n)*((x*(x  Sqrt[4 + x^2])^n) +
Sqrt[4 + x^2]*(x  Sqrt[4 + x^2])^n + x*(x + Sqrt[4 + x^2])^n +
Sqrt[4 + x^2]*(x + Sqrt[4 + x^2])^n))/Sqrt[4 + x^2]
We'll rewrite in a way that might be more amenable to solving.
In[117]:= InputForm[expr[n_] = Simplify[
Collect[pp[n], {(x+Sqrt[x^24])^n,(xSqrt[x^24])^n}]]]
Out[117]//InputForm=
(2^(1  n)*((x  Sqrt[4 + x^2])^(1 + n) + (x + Sqrt[4 + x^2])^(1 +
n)))/
Sqrt[4 + x^2]
We have a numerator of the form constant * (a^(n+1)b^(n+1)) and a
denominator
Sqrt[4 + x^2]. Also we know a priori we have a polynomial for positive
integer n, hence the numerator will have Sqrt[4 + x^2] as a factor for
such values of n. So solutions will be of the form a =
Exp[2*Pi*I*j/(n+1)]*b where j takes on integer values from 0 to n, a is
(x+Sqrt[x^24]), b is (xSqrt[x^24]), and we discard parasite
solutions x = +/2.
In[118]:= InputForm[solns[n_] = x /.
Solve[(x+Sqrt[x^24])==Exp[2*Pi*I*j/(n+1)]*(xSqrt[x^24]), x]]
Out[118]//InputForm=
{(Sqrt[1 + 2*E^(((2*I)*j*Pi)/(1 + n)) + E^(((4*I)*j*Pi)/(1 + n))]/
E^((I*j*Pi)/(1 + n))),
Sqrt[1 + 2*E^(((2*I)*j*Pi)/(1 + n)) + E^(((4*I)*j*Pi)/(1 + n))]/
E^((I*j*Pi)/(1 + n))}
This is clearly going to give radicals for positive integer n and
integers 0<=j<=n. We'll indicate a few such families.
In[119]:= sol[n_] := Complement[Simplify[Flatten[Table[solns[n],
{j,0,n1}]]], {2,2}]
In[120]:= InputForm[Table[sol[n], {n,1,8}]]
Out[120]//InputForm=
{{}, {1, 1}, {0, Sqrt[2], Sqrt[2]}, {(1)^(2/5)*(1 + (1)^(1/5)),
(1)^(4/5)*(1 + (1)^(2/5)), (1)^(2/5)  (1)^(3/5),
(1)^(1/5)  (1)^(4/5), ((1)^(3/5)*Sqrt[1  (1)^(3/5) +
2*(1)^(4/5)]),
(1)^(3/5)*Sqrt[1  (1)^(3/5) + 2*(1)^(4/5)]},
{1, 0, 1, Sqrt[3], Sqrt[3]}, {(1)^(3/7)*(1 + (1)^(1/7)),
(1)^(6/7)*(1 + (1)^(2/7)), (1)^(2/7)*(1 + (1)^(3/7)),
(1)^(3/7)  (1)^(4/7), ((1)^(5/7)*Sqrt[1  (1)^(1/7) +
2*(1)^(4/7)]),
(1)^(5/7)*Sqrt[1  (1)^(1/7) + 2*(1)^(4/7)], (1)^(2/7) 
(1)^(5/7),
(1)^(1/7)  (1)^(6/7), ((1)^(4/7)*Sqrt[1  (1)^(5/7) +
2*(1)^(6/7)]),
(1)^(4/7)*Sqrt[1  (1)^(5/7) + 2*(1)^(6/7)]},
{0, Sqrt[2], Sqrt[2], ((1)^(3/*Sqrt[(1 + I)  2*(1)^(1/4)]),
(1)^(3/*Sqrt[(1 + I)  2*(1)^(1/4)],
((1)^(5/*Sqrt[(1 + I)*(1 + Sqrt[2])]),
(1)^(5/*Sqrt[(1 + I)*(1 + Sqrt[2])],
((1)^(7/*Sqrt[(1 + I)*(1 + Sqrt[2])]),
(1)^(7/*Sqrt[(1 + I)*(1 + Sqrt[2])]},
{1, 1, (1)^(4/9)*(1 + (1)^(1/9)), (1)^(8/9)*(1 + (1)^(2/9)),
(1)^(7/9)*(1 + (1)^(4/9)),
((1)^(2/9)*Sqrt[1  (1)^(1/9)  2*(1)^(5/9)]),
(1)^(2/9)*Sqrt[1  (1)^(1/9)  2*(1)^(5/9)], (1)^(4/9) 
(1)^(5/9),
(1)^(2/9)  (1)^(7/9), (1)^(1/9)  (1)^(8/9),
((1)^(5/9)*Sqrt[1  (1)^(7/9) + 2*(1)^(8/9)]),
(1)^(5/9)*Sqrt[1  (1)^(7/9) + 2*(1)^(8/9)]}}
I guess a few observations are in order. One is that solutions are
symmetric about the origin (so for odd n 0 is a solution). Another is
that they interleave with solutions for p[n+1] between those for p[n]
except the outermost two. The positive of these lies between the
largest solution of p[n1] and 2. This also lets us know that there is
never a case where +/2 are legitimate (as opposed to parasite)
solutions.
To see that solutions are always bounded by 2 one can show, by
induction, say, that derivatives for x>=2 are always positive and p[n]
at x=2 is equal to n+1, hence there are no roots to the right. This
seems to involve a few steps e.q showing p[n]>p[n1] for x>2. I have a
feeling there is a nicer way to go about this but I have not stumbled
over it.
Daniel Lichtblau
Wolfram Research 

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Peter L. Montgomery science forum Guru Wannabe
Joined: 01 May 2005
Posts: 181

Posted: Tue Jul 18, 2006 12:46 am Post subject:
Re: why are the polynomials in this series all solvable by radicals?



In article <1153154093.967916.305090@i42g2000cwa.googlegroups.com>
"Lee Davidson" <lmd@meta5.com> writes:
Quote:  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
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.)

The recurrence for p(n) is linear with constant coefficients.
The differential equation (with x a constant, as in the recurrence)
d^2 y /dz^2 = x * dy/dz  y
has characteristic polynomial t^2  x*t + 1.
If alpha and beta are the roots of this polynomial, and alpha <> beta,
then the differential equation has solution
y(z) = c1 * exp(alpha*z) + c2 * exp(beta*z)
for some constants c1, c2.
Likewise your difference equation for p(n) has characteristic
polynomial t^2  x*t + 1. Letting alpha, beta be these roots,
then the recurrence solutions are
p(n) = c1 * alpha^2 + c2*beta^n
when c1 <> c2. After we look at the initial conditions
on p(1) and p(2), the solution is
alpha^(n+1)  beta^(n+1)
p(n) = 
alpha  beta
where alpha = (x + sqrt(x^2  4))/2 and beta = (x  sqrt(x^2  4))/2.
Given n, if you want to solve p(n)(x) = 0 for x,
you can solve alpha^(n+1) = beta^(n+1) for
alpha/beta. This and alpha*beta = 1 give alpha and beta.
Then x = alpha + beta. Two root extractions.
You can express x in terms of trigonometric functions.
Quote:  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.
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?


Why do storms (e.g., Katrina) have human names,
while earthquakes (e.g., Loma Prieta) have place names?
pmontgom@cwi.nl Microsoft Research and CWI Home: Bellevue, WA 

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Robert B. Israel science forum Guru
Joined: 24 Mar 2005
Posts: 2151

Posted: Tue Jul 18, 2006 1:40 am Post subject:
Re: why are the polynomials in this series all solvable by radicals?



In article <1153154093.967916.305090@i42g2000cwa.googlegroups.com>,
Lee Davidson <lmd@meta5.com> wrote:
Quote:  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

Your p(n) = i^n F_{n+1}(i x) where F_n are the Fibonacci polynomials.
They have a closedform formula
p(n) = (r_1(x)^(n+1)  r_2(x)^(n+1))/sqrt(x^24)
where r_1(x) = (x + sqrt(x^24))/2 and r_2(x) = (x  sqrt(x^24))/2
Quote:  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.)

r1(x)^(n+1) = r2(x)^(n+1) iff r2(x) = w r1(x) where w is an (n+1)'th
root of unity. We disregard the case w = 1 since (+/) 2 are not
roots of p(n). But that says (1+w) sqrt(x^24) = (1w) x
and then x = (+/) (sqrt(w) + 1/sqrt(w)).
Quote:  Furthermore, many of these solutions were expressible as linear
combinarions of pth roots of 1, or had such linear combinarions under
radicals.

They all are, as shown above.
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada 

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Paul Abbott science forum addict
Joined: 19 May 2005
Posts: 99

Posted: Tue Jul 18, 2006 4:11 am Post subject:
Re: why are the polynomials in this series all solvable by radicals?



In article <1153154093.967916.305090@i42g2000cwa.googlegroups.com>,
"Lee Davidson" <lmd@meta5.com> wrote:
Quote:  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)
Quote:  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
Quote:  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 

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Lee Davidson science forum beginner
Joined: 28 Sep 2005
Posts: 4

Posted: Tue Jul 18, 2006 3:04 pm Post subject:
Re: why are the polynomials in this series all solvable by radicals?



Paul Abbott wrote:
Quote:  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
Quite true. After posting, I switched from the "floating" n+1 body 
problem to the "anchored" n body problem, as they seem to be
mathematically equivalent.
I did realize that one can predict the degrees of the factors, with
little knowledge beyond one's picture of the physical situation. The
anchored n = 5 problem has symmetrical modes with mass 3 fixed. It
becomes two mirrorimage n = 2 problems, and so picks up the factors
for n = 2. Fixing masses 2 and 4 makes it 3 n = 1 problems  the x
factor. Two degrees of freedom remain, accounting for the remaining
degree 2 irreducible factor.
If n = 6, the midpoint of the middle spring is stationary, and so this
breaks up into 2 3body problems, but these are slightly different from
my n = 3 problem. That factors the n = 6 case into two irreducible
degree 3 polynomials.
Thanks all for the responses. 

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