## 140 Reputation

14 years, 55 days

## @Carl Love   No, the file is n...

@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...

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

Thanks!!

## Thanks...

@Carl Love thanks!!

## @Carl Love thank you very much!...

@Carl Love thank you very much!

## @Axel Vogt   Ok, thanks!!...

@Axel Vogt   Ok, thanks!!

## @Carl Love    Thanks!!!!  ...

Thanks!!!!

It seems that begins to work!    :-)

## @Carl Love    Thanks very much...

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!

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