Gonzalo Garcia

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11 years, 144 days

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I want to plot the approximation of a surface by polynomials. The surface is given by (x,y,f(x,y)) where f(x,y) is given by the following expression

proc (x) options operator, arrow; (sum(i*cos((i+1)*(-2+4*x[1])+i), i = 1 .. 5))*(sum(i*cos((i+1)*(-2+4*x[2])+i), i = 1 .. 5)) end proc

with both variables varying in the interval [0,1]. Then, by using the Bernstein polynomials of two variables (see, for instance, this paper for details  https://www.sciencedirect.com/science/article/pii/0021904589900956), the graph of the resulting (plot3d) surface (x,y,p(x,y))  it is not even like to the original surfaces.

Please, see this PDF of what I have done:  plots.pdf

Some idea or suggestion?




Assume we have the following "intervals" (I am not sure what is its formal name in Maple)


C :=[0,1/11],[1/11,1/9],[1/9, 1/7],[1/3,1/2],[1/2,1]


How can we get the "union" of these intervals? That is to say, obtain  [0,1/7],[1/3,1] 


Many thanks in advance for your comments and suggestions.




Assume we have the first N positive integres, 1,..,N, and we assing to these numbers a (discrete) probability distribution p1,...,pN. Of course, p1+...+pN=1.

Then, How can we select a number in {1,..,N} according to the given probability distribution? That is, the number 1 can be chosen with probability p1, 2 with a probability p2, etc.

Many thanks in advance for your comments.


I am interested in plot the so-called "free space diagram", please see https://en.wikipedia.org/wiki/Fr%C3%A9chet_distance for a formal definition. Essentially, given two curves (i.e., two polygonal from [0,1] to the plane) the "free space" is the points (s,t) in the square [0,1]x[0,1] such that the distance from P(s) to Q(t) (if P and Q are the parametrizations of the curves) is less or equal than a prefixed epsilon>0. An image can be the following (the free space is, in the image, the area not colored in black):



I have found nothing about this topic in Maple, any suggestion will be welcome!





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. 




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