Gonzalo Garcia

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14 years, 55 days

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@vv Many thanks for your anwer. F(34234) returns -17117, but |34234-17117 | returns 17117 wich is not in [0,1].

@Carl Love  many thanks!! It is seem works!

@vv

The appendix of the attached PDF file shows how to associate to a rectangle a number....How can we implement this in Maple?

 

Again, thanks!!

 

lip_algorithm.pdf 

@vv First at all, many thanks for your time and yoru patience....It is seem to return an error for n,m different from 2:

Cubes(3, 3, [a, b], h);
Error, (in Cubes) adding lists of different length

 

@vv Many thanks for your answer, 

Can we modify the procedure to divide an input cube?? That is, one of the argument of the procedure be the cube to divide.

 

Again, thanks.

@Kitonum A lot of thanks!!!

@vv What do you mean by the "extremal points"? Also, Why there cannot be 3 points with the same abscisa? 

@Kitonum , many thanks for your suggestion. However, my knowledge of maple are very limited and I do not understand the Maple's help guide ... you could exemplify your answer takin X:=[[1,2],[1,3],[1,4],[2,1],[2,2],[3,1]], please?

 

Again, thanks.

@Carl Love very thanks for your useful help!

Many thanks again! 

I will continue looking for some algorithm for the Penao or Hilbert space-filling curves.

Regards,

G. García.

 

Thansk vv 1739!

I know these space-filling curves, SFC, from the book "Space-filling curves", by H. Sagan. However, Lebesgue (or Shoenberg) have not some "good" properties, as the Peano or Hilbert SFC, such as Hölder continuity or measure preserving. For this, usually these curves are selected in optimization problems or integral quadrature instead the Lebesgue or Shoenberg one.

On the other hand, do you know some Maple code to compute the Peano curve for higher dimension? say, n=3,4,5....I have seen something here, but I am not sure if that is correct code. Moreover, I do not know as programming as convert the code (in C, C#) to Maple  :-(  The Sagan book only works with 2 or 3 dimensions.

I reiterate my gratitude for your time.

Regards,

G. García.

 

Dear Prebem,

Firstly, many thanks for your answer.However, I need to make operations as fsolve(f(peano(t,5)=0,t=0..1), but it not works. Neither  fsolve('f(peano(t,5)=0',t=0..1), and if I remove "t::numeric" form the peano procedure, the fsolve command returns the error:

Error, invalid input: f uses a 2nd argument, y, which is missing

For the other procecure: Error, (in fsolve) peano[t, 5][2] is in the equation, and is not solved for

Some hint?  

I retiterate my gratitude for your time.

 

Many thanks!!  The option Digits=45 works.

 

 

@Preben Alsholm 

Oppssss!!! Sorry, I have patesd images intead text! The other procedures are:

CreaC := proc (n, c1, c2) local i, H; H := NULL; for i to n do H := H, [c1, c2] end do; return [H] end proc;

Rdensf := proc (C, V0, n, m) local k, k0, R, raiz, V, t1, alpha; alpha := evalf(sqrt(n-1)/m); t1 := -10; V := NULL; for k to m do if evalf((k-1)/m) <= evalf(V0[1]) and evalf(V0[1]) <= evalf(k/m) then k0 := k; break end if end do; R := sort([Analytic(CreaCos(C, n, m, x)[n]-V0[n], x, re = (k0-1)/m .. k0/m, im = 0 .. .1)]); for raiz in R do if evalf(Im(raiz)) <= 1/100000000 then V := evalf(`<,>`(CreaCos(C, n, m, raiz))); if evalf(Norm(V-V0)) <= alpha then t1 := raiz; break end if end if end do; return t1 end proc;

 

And the instruction:

t0 := Rdensf(CreaC(3, 0, 1), `<,>`(1/18, 1/9, 1/9), 3, 250);

I retiterate my gratitude for your time!

 

 

 

@Carl Love Thanks very much for your time and help. Many of the error syntaxis as "f:=f(..." are the product of my confusions, but certainly many of them are by my ignorance of the program.

 

I reiterate my gratitude for your time and help.

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