I am having hard time understanding why Maple does this.

I have an integral, which maple could not integrate. So it returns int(....,x). Which is all fine. Then I used subsindets to force all terms in form e^(ln()+ln()+...) to expand in order to simplify the integrand, just for display purposes. 

subsindets returns back the integral unevaluated (as expected) but with integrand a little simpler, again as expected.

All is well so far.  

Then I find to my surprise, if I type the result back one more type, now the the integral actually evaluates.

Why? 

I have 2 questions on this.

1) If the original integral did not evaluate, why simplifying e^(ln()+ln()+...) makes it now evaluate? Did not Maple know this allready?

2) Why result back from subsindets remained unevaluated integral, and I had to type it again to see it now evaluates?

I found this after long time debugging, since this was done in code, not looking at screen. 

What happened is this: I store the result  of subsindets in a variable, and when I look at it in the debugger, I see int() still there, as expected.

I call a function to return back this result. I now see int() is gone!   So the act of just returning the result back, caused it to evaluate. Even though the orginal variable still had int() in it as. So returning the expression back, was equivalent to typing it again on the screen in the example below, which caused it to evaluate. 

Please see worksheet below.

Download why_it_now_evaluates.mw

Another simpler example, is just doing assignment to new variable. This causes evaluation.

restart;
expr:=int(-1/2*exp((ln(a+cos(1/2*x)^2)+(-2*a-2)*ln(cos(1/2*x)))/a)*(-1+cos(x))/sin(1/2*x)/cos(1/2*x)/(a+cos(1/2*x)^2),x);
expr:=subsindets(expr,'specfunc( anything, exp )',f->(`if`(has(op(1,f),'ln'),expand(f),f)));
#the above still has int() in it.

#this assignment, also causes evaluation
A:=expr;

What is the logic behind this. 

restart;
expr:=int(f(x),x);
lprint(algsubs(int=Int,expr));
lprint(subs(int=Int,expr));

gives

   int(f(x),x)

   Int(f(x),x)

So algsubs failed to replace int by Int

Looked at help. and see nothing. But I might have overlooked something. It says

It is a generalization of the subs command, which only handles syntactic substitution.

Generalization? If so, I expected it to work here. But may be there is a subtle reason why it did not? May be with algsubs, the replacement has to be algebraic expression and "int" is not, it is just a name.

Maple 2020.2
 

Hello Maple experts.

According to our teacher class notes, the ODE   y'=2*sqrt(y) with IC  y(0)=0 has 2 solutions. y(0)=0 and sqrt(y)=x

I am not able to get Maple to give the second solution,. It only gives y(0)=0.

Is there an option I am overlooking to make it give the other solution sqrt(y)=x ?

ode := diff(y(x),x) = 2*sqrt(y(x));
ic:=y(0)=0;
sol:=dsolve([ode,ic],y(x));

One can see the other solution by doing this

ode := diff(y(x),x) = 2*sqrt(y(x));
ic:=y(0)=A;
sol:=dsolve([ode,ic],y(x));
subs(A=0,sol)

I tried this in Mathematica. Mathematica does not give y=0 but it gives the second solution

I tried the singsol=all also, but it had no effect. Maple only shows the y(0)=0 solution.

Any suggestions?

Maple 2020.2

inside a local proc, when calling a  function such as map using the syntax map(x->x^2, target) does one need to declare x as local inside the proc?

Same for other Maple calls, which uses something similar. For example 

subsindets(expr,'specfunc( anything, csgn )', f->simplify(f));

does one need to declare f local to the proc where the above call is made?

Which one of these two example is more correct?

restart;
foo:=proc(expr)
  local x;
  map(x->x^2,expr);
end proc;

foo([x,y])

vs.

restart;
foo:=proc(expr)
    map(x->x^2,expr);
end proc;
foo([x,y])

In Mathematica for example, such symbols used by similar functions of the system (for example, Plot command, and others) are automatically localized to the system call itself avoiding any conflict with user own symbols.

I am not sure how Maple handles this. Should one always declare these symbols?

Maple 2020.2

I have an expression with number of csgn(arg) in it.

I'd like to scan this expression, telling Maple to assume arg is positive, in order to replace csgn(arg) by 1.

I am trying to do this using subsindets. But I do not know why it is not working. I am doing this in code, without looking at the expression. So I do not know what the arg's are and how many such cases could be.

Here is an example

expr:=(1+csgn(a)*a)/(3*csgn(b)*b);

To get to each csgn(arg), I am using the type 'specfunc( anything, csgn )'. I checked this is correct type by doing

type(csgn(a),'specfunc( anything, csgn )');

           true

Then

subsindets(expr,'specfunc( anything, csgn )', f->simplify(f) assuming positive);

But this does not change anything. 

I also tried

subsindets(expr,'specfunc( anything, csgn )',f->simplify(f) assuming op(1,f)::positive);

No change, But if I do 

simplify(csgn(a)) assuming positive;

it works. And Maple returns 1. Also this works

simplify(expr) assuming a>0,b>0;

But since I am do not before hand what the arguments to csgn() in the expression are, I can't do the above. I do not know even if expression has csgn() in it even. I am trying to simplify a result I obtain inside the program by doing the above.

What is wrong with my use of  subsindets above?

I think the problem is with using assumptions inside subsindents. As this below works

subsindets(expr,'specfunc( anything, csgn )',f->1);

So the call to subsidents was OK, it is just that the assumptions do not seem to be somehow effective inside.  May be name scoping issue?

Maple 2020.2

edit:

For now and as workaround, I am doing this

restart;
expr:=(1+csgn(a)*a)/(3*csgn(b)*b):
fun:=selectfun(expr,'csgn'); #find csgn if any

if numelems(fun)>0 then
    the_args:= op~(1,fun);
    simplify(expr) assuming map(x->x::positive,the_args)[];
fi;