Gonzalo Garcia

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14 years, 11 days

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Hi!

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.

 

Hi!

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.

Dears,

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.

Dears,

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!!

Hi!,

Assume that we hace a set points in the plane, put X:=[a1,a2,...,aN] where each ai is given by its coordinates [x,y]. The commnad "convexhull(X)" give us the points of the convex hull of X, but how I can find to "lower-right" of these points? Please, see the attached image. I need to findo the points A,C,E and F, marked with a solid circle.

Many thanks in advances for your comments.

 

 

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