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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 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 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));

 

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.

Here is a problem from SEEMOUS 2017 (South Eastern European Mathematical Olympiad for University Students)
which Maple can solve (with a little help).

For k a fixed nonnegative integer, compute:

Sum( binomial(i,k) * ( exp(1) - Sum(1/j!, j=0..i) ), i=k..infinity );

(It is the last one, theoretically the most difficult.)

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