Re: Bound for quotient of eigenvalues

On 25-04-2009 13:28, achille wrote:

For each real-analytic function _f_ from [0,1] into itself, consider the
2x2 real matrix
a b
b c
where a = int(x^2 f(x),dx), b = int(x f(x),dx) and c= int(f(x),dx) (all
integrals are from 0 to 1). Unless _f_ is the null function, this matrix
is positive definite and therefore it has two positive eigenvalues.
Consider the quotient e_1/e_2 where e_1 is the greatest eigenvalue and
e_2 is the smallest one. My question is: is there an upper bound for
the quotients obtained by this method? My guess is that the answer is
negative, but I was unable to prove it.
Consider f(x) = (x(1-x))^(m-1),
a = B(m+2,m) = (m+1)/(2*(2m+1)) B(m,m)
b = B(m+1,m) = 1/2 B(m,m)
c = B(m,m)
where B(x,y) is the beta function. Notice
B(m,m)^(-1) [ a b ] = [ 1/4 +O(1/m) 1/2 ]
[ b c ] [ 1/2 1 ]
converges to a singular matrix as m -> \infinity.
It is easy to see for this f(x), e_1/e_2 ~ O(m)
for large m.
I wonder whether you (or someone else) can provide an example in which
there is a function _f_ from [0,1] into itself with finitely many zeros
such that, for each _n_, f <= f_n.
What is n, and what is f_n?
Achille (and you, in your reply to my first post) provided a sequence
(f_n)_n such that the sequence of quotients of eigenvalues obtained from
it is not bounded. His (and your sequence) has this property: the only
continuous function _f_ such that, for each _n_, f <= f_n is the null
function. My question is: is it possible to obtain a sequence such as
achille's our yours but which is bounded below by a continuous function
with only finitely many zeros?

Best regards,

Jose Carlos Santos

No, you cannot. Suppose f_n(x) >= f(x) on [0,1]
with f(x) has only finitely many zeros. Let

a_n, b_n, c_n be \int_0^1 x^i f_n(x) dx

for i = 2, 1,0 and M_n be the matrix.

1/c_n [ a_n b_n ]
[ b_n c_n ]

Let a, b, c and M be the corresponding entries for f(x).
It is easy to see
tr(M_n) <= 2
and det(M_n) = (1/c_n)^2 (a_n c_n - b_n^2)
= 1/c_n \int_0^1 (x - b_n/c_n)^2 f_n(x) dx
>= 1/c_n \int_0^1 (x - b_n/c_n)^2 f(x) dx
>= 1/c_n \int_0^1 (x - b/c)^2 f(x) dx
= c/c_n det(M)
>= c det(M) (since c_n >= 1 ).

From this, we can derive an upper bound of (e_1/e_2)_n
independent of n:

(e_1/e_2)_n <= ((e_1+e_2)^2/e_1 e_2)
= tr(M_n)^2/det(M_n)
<= 4/(c det(M)).

Thanks a lot. You were most helpful.

Best regards,

Jose Carlos Santos
.

Relevant Pages

  • Re: converge........
    ... Prove the the sequence of the logarithms has limit -oo. ... Best regards, ... Jose Carlos Santos ...
    (sci.math)
  • Re: A help with a sequence please
    ... Best regards, ... Jose Carlos Santos ... But I'm a bit confused, I didn't get the law of this sequence, how ...
    (sci.math)
  • Re: sequence of functionals in l^2
    ... Does converge poinwise to 0? ... Yes, it's easy to see that, if _x_ is a finite supported sequence, then ... Best regards, ... Jose Carlos Santos ...
    (sci.math)
  • Re: A help with a sequence please
    ... Best regards, ... Jose Carlos Santos ... But I'm a bit confused, I didn't get the law of this sequence, how ...
    (sci.math)
  • Re: Reducing Pseudorandomness
    ... I have a case where I need a pseudorandom number generated. ... the sequence can be generated very efficiently by a device called ... a linear shift register, by initializing ... the register to contain any seqeunce of n-bits (the all zeros being ...
    (comp.lang.pascal.delphi.misc)
Quantcast