Kitonum

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These are questions asked by Kitonum

In Maple 2015.1 we have

restart;

solve([sin(2*x)/cos(x+3*Pi/2)=1,  x>-4*Pi, x<-5*Pi/2], x, allsolutions, explicit);

solve([sin(2*x)/cos(x+3*Pi/2)=1, x>0, x<2*Pi], x, allsolutions, explicit);

 

 

In the first example, the error message is not clear (actually there exists a unique root  x=-11*Pi/3), in the second example, one root  (x=5*Pi/3) is lost.

 

Is there in Maple a command to verify that one segment on the real axis is a subset of another segment?  For example  RealRange(2, 4)  is subset of  RealRange(2, 5)

Of course easy to write a procedure that makes it:

Subset:=proc(r1::RealRange, r2::RealRange)

local a, b, c, d;

a, c:= op~(1,[r1, r2])[];

b, d:= op~(2,[r1, r2])[];

if a>=c and b<=d then true else false fi;

end proc:

 

Example of use:

Subset(RealRange(2,4), RealRange(2,5));

Subset(RealRange(2,5), RealRange(2,4));

                            true

                            false

 

Maple does not cope with the following simple example:

with(geom3d):

point(A,0,0,0), point(B,1,0,0), point(C,2,0,0), point(E,2,1,0):

AreCoplanar(A,B,C,E);

           Error, (in geom3d:-plane) the points may not be AreCollinear

 

Should we interpret this behavior as a bug? I think I met with this yet 10-12 years ago, but unfortunately since then nothing has changed.

When solving a simple assignment problem in Maple 2015.1 the bug occurs:

 

In Maple 2012 there are no problems:

A := Matrix([[1, 7, 1, 3], [1, 6, 4, 6], [17, 1, 5, 1], [1, 6, 10, 4]]):

n:=4:

z:=add(add(A[i,j]*x[i,j], j=1..n), i=1..n):

restr:={seq(add(x[i,j], i=1..n)=1, j=1..n),seq(add(x[i,j], j=1..n)=1, i=1..n)};

sol:=Optimization[LPSolve](z, restr, assume=binary); 

 

 

Can Maple look for the limit of a sequence  f(n)  for n=1,2,3, ... ? Of course, if there is a limit of the function of continuous argument, the limit of the sequence is the same number. But it is easy to give examples of when there is the limit of the sequence, and the limit of the function does not exist:

assume(n, posint):

 limit(sin((n^2+1)/n*Pi), n=infinity);  # Obviously the correct answer is  0

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