Gonzalo Garcia

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14 years, 218 days

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Realted to this post I am trying to solve the following inequalities in the variable "t"




Cosenos := proc (m, x) local C, x1, x2, y, x0, i, j, j0, k, K0, Kaux, K1, L, R, S, t1, t2, t, r1, r2, d; C := NULL; K0 := NULL; Kaux := NULL; K1 := NULL; L := NULL; R := NULL; S := NULL; d := nops(x); for k to d-1 do if evalf(x[k]) = 1 then j0 := m else j0 := 1+floor(x[k]*m) end if; K0 := K0, [(j0-1)/m, j0/m] end do; K0 := [K0]; S := K0[1][1] <= t, t <= K0[1][2]; for k from 2 to d-1 do S := S, K0[k][1] <= 1/2-(1/2)*cos(Pi*m^(k-1)*t), 1/2-(1/2)*cos(Pi*m^(k-1)*t) <= K0[k][2] end do; K1 := solve({S}, t) end proc


However, this run very very slow....For instance, for Cosenos(50, [.3225, .25877, .325, 1])


we interrupt the opertion after more than 10 minutes...I am doing something wrong?

Many thanks for your comments and suggestions. 





Consider, fixed an integer m>1, the mapping given by the following procedure:


G := proc (t) local k, C; C := NULL; C := t; for k from 2 to d do C := C, 1/2-(1/2)*cos(Pi*m^(k-1)*t) end do; return [C] end proc

Then, it can be proved that given x in the cube [0,1]^{d} there is t in [0,1] such that the norm of x-G(t) is less, or equal, than sqrt(d-1)/m. Indeed, dividing the cube [0,1]^{d} into m^{d-1} subcubes of side-length 1/m x ... x 1/m x 1, the point x belongs to some of these subcubes, say J. As, by the properties of the cosines function, the curve G(t) lies in J whenever t in certain subinterval of [0,1], the result follows.

In other words, computing all the solutions of the equation

1/2*(1-cos(Pi*m^(d-1)*t)) = x[d], (j-1)/m <= t and t <= j/m

for some of these solutions the desired t is obtained, where j is such that x1 in [(j-1)/m,j/m] (x1 is the first coordinate of the point x). However, for large values of m and d, the above equation have many solutions, I have tried find all of them and the process is extremely slow....Other way to find such a t can be the following: find a t satisfying the following system of inequalities

EQ := abs(t-x[1]) <= 1/m; for k from 2 to d do EQ := EQ, abs(1/2*(1-cos(Pi*m^(k-1)*t))-x[k]) <= 1/m end do


and then, a solution of this system is a such t. I do not know how to find, efficiently, a t such that of x-G(t) is less, or equal, than sqrt(d-1)/m   :(

Some idea?

Many thanks for your comments in advance.



Looking the Maple's help, I see that the command "isolve"  tries to solve an equations   over the integers. Then, given m>1 and t in the interval [0,1], How can used this command to find an integer j>=1 such that (j-1)/m<=t<j/m. That is, fin j such that t belongs to the interval [(j-1)/m,j/m].


Thanks in advance for your comments and help.


Consider the problem min{|a+2*i| : i integer}, where a is a number of the form 2k*t for a fixed integer k>1 and t in [0,1]. A simple checking shows that i must belongs to the interval [-(1+a)/2,(1-a)/2] and as the legth of this interval is 1, there is two integers in this interval.

How can I compute this integer i with Maple? The commad "ceil" not seem very suitable. Some idea?
Many thanks in advance for your comments.


Let C a square in the n-diemnsional Euclidean space. Somebody know how to divide C into 2^{n} congruent subsquares? 

For instance, for n=2 and  say C:=[0,1]x[0,1], the unit closed square, we will obtain the 2^{2}=4 subsquares [0,1/4]x[0,1/4], [0,1/4]x[1/2], [1/2,1]x[0,1/4] and [1/2,1]x[1/2,1].  

Many thanks in advance for your comments!!

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