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Question: Subprocedures in a main procedure

Question:Subprocedures in a main procedure

janhardo 997 Maple
array procedure programming module
December 17 2021
1

How to add any/many subprocedure in a mainprocedure?
Somehow the mainprocedure must deliver the input for the mainprocedure

Here is a example 

Note: probably can this old procedure be rewritten in modern Maple programming language ?  

> 

restart;

 

Code to Find the Max and Min Values in a List A

 

> 

maxmin:=proc(A,maxv::evaln,minv::evaln)
> 

   local i;
> 

   maxv:=A[1]; minv:=A[1];
> 

   for i from 2 to nops(A) do
> 

     if eval(maxv) < A[i] then maxv:=A[i] end if;
> 

     if eval(minv) > A[i] then minv:=A[i] end if;
> 

   end do;
> 

   RETURN()
> 

end proc:
> 

 

Now we can call the maxmin procedure from a procedure named optimize, which is designed to create the polygonal approximation, use maxmin to find the largest and smallest y-values from among the vertices, and produce some graphic output.

See the documentation in the book.

 

Code to Approximate the Max and Min Values of a Function  f :

 

> 

optimize:=proc(f,a,b,N,pic::evaln)
> 

  local X,Y,L,i,A,xmax,xmin;
> 

  X:=array(0..N);Y:=array(0..N);L:=array(0..N);
> 

  with(plots,display);
> 

  for i from 0 to N do
> 

    X[i]:=evalf(a+i*(b-a)/N);

> 

    Y[i]:=f(X[i]);
> 

    L[i]:=plot([[X[i],0],[X[i],Y[i]]],color=black):
> 

  end do;
> 

  A:=[seq(Y[i-1],i=1..N+1)];
> 

  maxmin(A,maxv,minv); # USE OF SECOND PROCEDURE --------------
> 

  xmax:={};xmin:={};
> 

  for i from 0 to N do
> 

    if Y[i]=maxv then xmax:=xmax union {X[i]} end if;
> 

    if Y[i]=minv then xmin:=xmin union {X[i]} end if;
> 

  end do; #----------------------------------------------------
> 

  pic:=display({seq(L[i],i=1..N)}):
> 

  print(`maximum y value is`,maxv,`and occurs at these x values`,xmax);
> 

  print(`minimum y value is`,minv,`and occurs at these x values`,xmin);
> 

  end proc:
> 

 

Note that you must execute the code that defines maxmin before the procedure optimize will work. This only make sense.

 

We test the optimize procedure with the following function

 

> 

f:=x->3+10*(-x^2+x^4)*exp(-x^2);

proc (x) options operator, arrow; 3+10*(-x^2+x^4)*exp(-x^2) end proc

 (1)

on the interval [-1, 4].

 

> 

optimize(f,-1,4,150,pic);

`maximum y value is`, 6.088066652, `and occurs at these x values`, {1.633333333}

  

`minimum y value is`, 1.391538737, `and occurs at these x values`, {-.6333333333, .6333333333}

 (2)
> 

pic;

 

 

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