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Bug corrected in inttrans[hilbert]...

vv 11977 Maple
inttrans hilbert
April 10 2018
2 2

There is a bug in inttrans:-hilbert:

> 

restart;
> 

inttrans:-hilbert(sin(a)*sin(t+b), t, s);
# should be:
sin(a)*cos(s+b);   expand(%);

sin(a)*cos(s)

  

sin(a)*cos(s+b)

  

sin(a)*cos(s)*cos(b)-sin(a)*sin(s)*sin(b)

 (1)
> 

########## correction ##############
> 

`inttrans/expandc` := proc(expr, t)
local xpr, j, econst, op1, op2;
      xpr := expr;      
      for j in indets(xpr,specfunc(`+`,exp)) do
          econst := select(type,op(j),('freeof')(t));
          if 0 < nops(econst) and econst <> 0 then
              xpr := subs(j = ('exp')(econst)*combine(j/('exp')(econst),exp),xpr)
          end if
      end do;
      for j in indets(xpr,{('cos')(linear(t)), ('sin')(linear(t))}) do
          if type(op(j),`+`) then
              op1:=select(has, op(j),t); ##
              op2:=op(j)-op1;            ##
              #op1 := op(1,op(j));
              #op2 := op(2,op(j));
              if op(0,j) = sin then
                  xpr := subs(j = cos(op2)*sin(op1)+sin(op2)*cos(op1),xpr)
              else
                  xpr := subs(j = cos(op1)*cos(op2)-sin(op1)*sin(op2),xpr)
              end if
          end if
      end do;
      return xpr
end proc:
> 

#######################################
> 

inttrans:-hilbert(sin(a)*sin(t+b), t, s); expand(%);

-(1/2)*cos(a-b)*sin(s)+(1/2)*sin(a-b)*cos(s)+(1/2)*cos(a+b)*sin(s)+(1/2)*sin(a+b)*cos(s)

  

sin(a)*cos(s)*cos(b)-sin(a)*sin(s)*sin(b)

 (2)
> 

 


Download hilbert.mw

 

Multiple integrals over sets defined by ...

vv 11977 Maple 2017
October 28 2017
6 11

I have just posted an article with this title at Maplesoft Application Center here.
It was motivated by a question posed by  Markiyan Hirnyk  here and a test problem proposed there by Kitonum.

Now I just want to give the promissed complete solution to Kitonum's test:

Compute the plane area of the region defined by the inequalities:

R := [ (x-4)^2+y^2 <= 25, x^2+(y-3)^2 >= 9, (x+sqrt(7))^2+y^2 >= 16 ];

plots:-inequal(R, x=-7..10, y=-6..6, scaling=constrained);

The used procedures (for details see the mentioned article):

ranges:=proc(simpledom::list(relation), X::list(name))
local rez:=NULL, r,z,k,r1,r2;
if nops(simpledom)<>2*nops(X) then error "Domain not simple!" fi;
for k to nops(X) do    r1,r2:=simpledom[2*k-1..2*k][]; z:=X[k];
  if   rhs(r1)=z and lhs(r2)=z then rez:=z=lhs(r1)..rhs(r2),rez; #a<z,z<b
  elif lhs(r1)=z and rhs(r2)=z then rez:=z=lhs(r2)..rhs(r1),rez  #z<b,a<z
  else error "Strange order in a simple domain" fi
od;
rez
end proc:

MultiIntPoly:=proc(f, rels::list(relation(ratpoly)), X::list(name))
local r,rez,sol,irr,wirr, rels1, w;
irr:=[indets(rels,{function,realcons^realcons})[]];
wirr:=[seq(w[i],i=1..nops(irr))];
rels1:=eval(rels, irr=~wirr);
sol:=SolveTools:-SemiAlgebraic(rels1,X,parameters=wirr):
sol:=remove(hastype, eval(sol,wirr=~irr), `=`); 
add(Int(f,ranges(r,X)),r=sol)
end proc:

MeasApp:=proc(rel::{set,list}(relation), Q::list(name='range'(realcons)), N::posint)
local r, n:=0, X, t, frel:=evalf(rel)[];
if indets(rel,name) <> indets(Q,name)  then error "Non matching variables" fi;
r:=[seq(rand(evalf(rhs(t))), t=Q)];
X:=[seq(lhs(t),t=Q)];
to N do
  if evalb(eval(`and`(frel), X=~r())) then n:=n+1 fi;
od;
evalf( n/N*mul((rhs-lhs)(rhs(t)),t=Q) );
end proc:

Problem's solution:

MultiIntPoly(1, R, [x,y]):  # Unfortunately it's slow; patience needed!
radnormal(simplify(value(%)));

evalf(%) = MeasApp(R, [x=-7..10,y=-6..6], 10000); # A rough numerical check
           61.16217534 = 59.91480000

 

Riemann hypothesis is false! (simple pro...

vv 11977 Maple
October 01 2017
2 2

# Riemann hypothesis is false! (simple proof)
 

restart;
assume( s>0, s<1/2, t>0 );
coulditbe(abs(Zeta(s+I*t))=0);
                              true

# Q.E.D.

Unfortunately coulditbe(Zeta(s+I*t)=0) returns FAIL, but our assertion is already demonstrated!

The moral: the assume facility deserves a much more careful implementation.

A geometric construction for the Summer ...

vv 11977 Maple
geometry fsolve
August 02 2017
1 2


 

A geometric construction for the Summer Holiday

 
Does every plane simple closed curve contain all four vertices of some square?
 This is an old classical conjecture. See:
https://en.wikipedia.org/wiki/Inscribed_square_problem

Maybe someone finds a counterexample (for non-analytic curves) using the next procedure and becomes famous!

 

> 

SQ:=proc(X::procedure, Y::procedure, rng::range(realcons), r:=0.49)
local t1:=lhs(rng), t2:=rhs(rng), a,b,c,d,s;
s:=fsolve({ X(a)+X(c) = X(b)+X(d),
            Y(a)+Y(c) = Y(b)+Y(d),
            (X(a)-X(c))^2+(Y(a)-Y(c))^2 = (X(b)-X(d))^2+(Y(b)-Y(d))^2,
            (X(a)-X(c))*(X(b)-X(d)) + (Y(a)-Y(c))*(Y(b)-Y(d)) = 0},
          {a=t1..t1+r*(t2-t1),b=rng,c=rng,d=t2-r*(t2-t1)..t2});  #lprint(s);
if type(s,set) then s:=rhs~(s)[];[s,s[1]] else WARNING("No solution found"); {} fi;
end:

 

Example

 

> 

X := t->(10-sin(7*t)*exp(-t))*cos(t);
Y := t->(10+sin(6*t))*sin(t);
rng := 0..2*Pi;

proc (t) options operator, arrow; (10-sin(7*t)*exp(-t))*cos(t) end proc

  

proc (t) options operator, arrow; (10+sin(6*t))*sin(t) end proc

  

0 .. 2*Pi

 (1)
> 

s:=SQ(X, Y, rng):
plots:-display(
   plot([X,Y,rng], scaling=constrained),
   plot([seq( eval([X(t),Y(t)],t=u),u=s)], color=blue, thickness=2));

 

A bug/typo with correct result!...

vv 11977
July 05 2017
1 2

The help page for  ?procedure has the following example.

> 

addList := proc(a::list,b::integer)::integer;
    local x,i,s;
    description "add a list of numbers and multiply by a constant";
    x:=b;
    s:=0;
    for i in a do
        s:=s+a[i];
    end do;
    s:=s*x:
end proc:
> 

sumList:=addList([1,2,3,4,5],2);

30

 (1)

Of course it's a bug (actually a typo) here:  s:=s+a[i];   should be   s:=s+i;

A strange/funny fact is that the result is correct but, for almost any other list the call will produce an error or a wrong answer.
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