@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.

@Carl Love   No, the file is not uploaded.....Mmm.....

For positive integer i and j=1,...,2^{i}-1 define:

f:=f(x,y,i,j)-->((x+j-1)/2^{i}, (y+2^{i}-j-1)/2^{i})

Then, defiene

CreaF:=proc(i)

local j,n,g,V;

n:=1;V:=NULL;

for j from 1 to 2 ^{i}-1 do

g[n]:=(x,y)-->[f(x,y,i,j)];

V:=V,g[n];

n:=n+1;

end do;

end proc:

@Carl Love Oppsss!!! We can not see the images, i think. I enclose the Maple file.

Thanks!!

@Carl Love thanks!!

@Carl Love thank you very much!

@Axel Vogt   Ok, thanks!!

@Carl Love 

Thanks!!!!

@Carl Love 

Thanks very much for your answer Carl Love, but what I want is that the "function" or "procedure" fun(curva(m,t)) (for a given "m") returns a function on the variable "t". The function above "fun", really is a tree variables function (because we intrate respect to "x" from 0 to 1), so the composition  "fun(curva(m,t))" (for a given "m")  should return a single one variable function. In the above example , for instance fun(curva(2,t)) returns the function

But I think that this return is wrong because, for instance, no "sin" appear. In fact, the function "fun" I have defined as

But fun(x,y,z) returns

which I think that is not correct, where is the term in "x"??

Thanks very much again!