MaplePrimes - answers and comments on Question, Frequency with Maple

2011

Robert Israel — Thu, 21 Jul 2011 06:48:06 Z

Hint for (a): The general solution of the DE is

where the sum is over all 2011'th roots of -1. For this to be real,

For t -> infinity, the solution is dominated by the nonzero terms where Re(omega) is maximized. These have frequency given by Im(omega)/(2*Pi). For the largest possible frequency, take the roots closest to I and -I...

Retraction

Markiyan Hirnyk — Thu, 21 Jul 2011 22:13:52 Z

Unfortunately, my attempt to realize the idea by Robert Israel is not correct. I use that the frequency of the sum is not greater that the sum of the frequencies. This statement is not true without additional assumptions. The question a) remains open.

4022/log(2)

Markiyan Hirnyk — Sat, 23 Jul 2011 19:08:12 Z

This question originates from Math Dept of Moscow State University:http://en.wikipedia.org/wiki/Moscow_State_University . Not being a specialist on ODEs, I expected that the answer could be obtained through the Green function with Maple.
As professor Robert Israel pointed out, the general solution (of the ODE diff(x(t),t$2011)+x(t)=0 under consideration) x(t) equals the sum over omega s.t. omega^2011=-1 of C(omega)* exp((omega)*t). It is easy to find the frequency of each term, but it is difficult to find the frequency of x(t). The say on the dominant term does not work because we do not know the behavior of x(t) on the joints.
I succeeded to answer the part a). This is an interesting interplay of ODEs, complex analysis, and Maple (a little). The value distibution of meromorphic function theory is applied (see the book "Value distribution of meromorphic functions" by A. Goldberg and I. Ostrovskii , pp. 13-16). Applying the First Main Theorem by R. Nevanlinna, we obtain that int(nu(x,t)/t,t=0..2*s)<=T(2*s,1/x)=T(2*s,x)+O(1)=
1/(2*Pi)*int(max(log(abs(x(2*s*exp(I*phi)))),0),phi=0..2*Pi) +O(1),s->+infinity,
where T(r,f) is the Nevanlinna characteristic of a meromorphic function f. Using the properties of the Nevanlinna characteristic,
we continue the estimate: the integral
1/(2*Pi)*int(max(log(abs(x(2*s*exp(I*phi))),0),phi=0..2*Pi) is less than or equal to the sum over omega s.t. omega^2011=-1 of
the integrals int(max(Re(2*s*exp(I*omega*phi)),0),phi=0..2*Pi)+O(1),s->+infinity. The periodicity of the integrands implies
the equality of the integrals. Therefore, in view of
> 2*s*evalf(int(max(Re(exp(I*phi)), 0), phi = 0 .. 2*Pi))/(2*Pi);

2.000000000 *s/Pi ,
the integral int(nu(x,t)/t,t=0..2*s)<=4022.000*s/Pi+O(1), s->+infinity. Next, int(nu(x,t)/t,t=0..2*s)>=int(nu(x,t)/t,t=s..2*s)>=
nu(x,s)*log(2). Comparing the last two estimates, we conclude that omega(x)<= 4022/log(2).
Concerning b), one can obtain the upper estimate by the same way. However I believe in the existence of the limit, not the upper limit.

Edit. Vanising text.

PS. Because my text permanently vanishes, see it in estimates.txt

Simpler example

Markiyan Hirnyk — Thu, 21 Jul 2011 08:25:40 Z

As far as I understand it, a simple example doesn't confirm your suggestion (PS. concerning the value of the frequency):

Download four.mw

By the way, how to find the closest to I root of -1 of order 2011 ?

Good idea

Markiyan Hirnyk — Thu, 21 Jul 2011 17:05:55 Z

The idea by Robert Israel is good, he has understood merits of case, but the value of the frequency of a[omega]*exp(Re(omega)*t)*cos(Im(omega)*t) doesn't equal Im(omega)/(2*Pi).
In fact, nu(cos(Im(omega)*t),T)=T*Im(omega)/Pi+O(1), as T->+infinity, therefore the frequency equals Im(omega)<1. Because limsup(A+B)<=limsup(A)+limsup(B), this implies the frequency of x(t) is less than or equal to 2011. I tried to improve this by the max(map(evalf@evalc@Im,[solve(y^2011+1=0,y)])) command, but ran out of the limitations of my comp (1.8G, 4000s, kernel connection has been lost). The output of max(map(evalf@evalc@Im,[solve(y^100+1=0,y)])) is 0.9995065605. As far as I remember it, there are estimates of the trigonometrical polynomials by Weil and Vinogradov,
but I am not a specialist in this field. I vote up for the partial answer by Robert Israel. The part b) is still open.

Jensen formula

Markiyan Hirnyk — Sun, 24 Jul 2011 20:22:36 Z

Applying more elementary tools, namely Jensen's formula and the triangle inequality, one can prove a somewhat weaker estimate. Hint: = the Stiltjes integral over [0,r] from r/t by dnu(f,t) = .