Given an expression, how to best find the constants _C1, _C2, etc.... in this expression?  Currently I do the following, but I think there should be a better way.

restart;
sol:=dsolve(diff(y(x),x$2)+y(x)=1);
indets(sol);
select(x->`if`(type(x,symbol) and convert(x,string)[1..2]="_C",true,false),indets(sol))

The above works, at least on the few examples I tried it on, but is there a better way to do it?  All constants will have the form _Cn where n is integer. 

Why is this integral so difficult -- so slow to execute?

seq(evalf(Int(1/R*sinh(s*coth(1/2*s))^(2*I/k)/(gam-I*k*cosh(s))^5*
     Int(exp(-(k^2*sinh(s)^2/(gam-I*k*cosh(s))-I*k*(1+cosh(s)))*v)*
     hypergeom([1/k*I],[1],k*v*I)*v^n,v = 0 .. 1/2*R), s=0..s_max)), n=0..1);

Hints to improve the efficiency of execution would be appreciated.

The conversion of a maple formula in latex version 2021 produces some

commands which are not defined , see   primes_latex.mw.

Perhaps more informations should be given  in the news about the command latex in Maple 2021

Hello

I need to find all the variants (I am not sure if this is the correct term to be used but I hope this will be clear in the example) of a specific indeterminate in a given expression.  Here is an example:

alpha[3, 5]*xi[1]*xi[8] + alpha[3, 5]*xi[4]*xi[5] + alpha[3, 3]

For this particular example, xi[1], xi[8], xi[4], and xi[5] are the variables I am looking for. The indexes of xi change depending on the previous calculation.  Also, in some cases, I need the alpha variables instead, that is, alpha[3,5], and alpha[3,3].  

I could not figure out how to use indets in this case.

Many thanks for your help. 

I may be misunderstanding the documentation of implicitplot.  Can someone set me straight?

This is extracted from the implicitplot's help page:

implicitplot(-x^2 + y, x = 0 .. 2, y = 0 .. x);

The plotting range is limited to y ≤ x, as intended.  Let us verify that it does the right thing:

display(
	implicitplot(-x^2 + y, x = 0 .. 2, y = 0 .. x, color=red),
	plot(x, x=0..2, color=blue)
);

Yes, indeed it does.

Now let us try limiting the plotting range to y ≤ 1 − x². Here is what we get:

display(
	implicitplot(-x^2 + y, x = 0 .. 2, y = 0 .. 1-x^2, color=red),
	plot(1-x^2, x=0..2, color=blue)
);

I expected the red curve to lie entirely below the blue curve but it doesn't. Am I misunderstanding implicitplot?

Download worksheet: mw.mw

Sorry to bother you with a basic question again.

How do I get both the indexes and the values of a table?

Table.mw

The following code completes apparently without error on Maple 2020.2, but with Maple 2021.1 I get a deconnection from the kernel:

restart:
with(LinearAlgebra):
A:=Statistics:-Sample(Normal(0,1),[2500,2500]):
U,S,Vt:=SingularValues(A,output=['U','S','Vt']):
Norm(U.DiagonalMatrix(S).Vt-A);

The crash occurs when running the last line.

The code works on both versions with a smaller matrix (256x256).

Is there something obviously wrong with this piece of code, or a change in Maple 2021 that could explain this?

Note: I'm running Maple on Windows 10, the machine has 16 GB RAM, and the memory usage stays low.

Which one of these two versions, is the recommened one to use? For example, to check for type  integer*x, where x is literal x. i.e. identical(x)

restart;
TypeTools:-AddType('type_1', `&*`(integer,identical(x)));
type(3*x,type_1);

restart;
TypeTools:-AddType('type_1', '`*`'({integer,identical(x)}));
type(3*x,type_1)

Both work.  The difference is that `*` needs {} while `&*` does not.

I read the help page  and I do not understand what it says about the difference, and when is one supposed to use `*` vs. `&*`. it says 

              | `*`(type)  a product of factors of the given type
              | `&*`(type*)  a product of factors in which the nth factor is of the nth type specified in type*

Could possibly someone please explain in simple plain english what is the difference? If I use `&*`  vs. `*` with {}, will they work the same all the time or are there cases when to use one vs. the other?  Any rules of thumb to follow?

I need to simplify terms such as (cos(x)^2+sin(x)^2) to 1 if present in input, but I do not Maple to also do any other simplification rewriting polynomials that might be present in the expression.

And example will make it clear. Given this

expr:= (cos(x)^2+sin(x)^2)+5+(1+x+x^2+x^3)*(cos(x)^2+sin(x)^2)*exp(x);

I want expr to become  6+(1+x+x^2+x^3)*exp(x).   i.e. only simplify trig terms

But simplify(expr,trig); gives

                 6 + (x + 1)*(x^2 + 1)*exp(x)

Which is not what I want. Then I tried the trick of thaw and freeze to tell Maple to freeze polynomial type, like this

restart;
expr:= (cos(x)^2+sin(x)^2)+5+(1+x+x^2+x^3)*(cos(x)^2+sin(x)^2)*exp(x);
thaw(simplify(subsindets[flat](expr,satisfies(Z->type(Z,polynom(integer, x))),(freeze))));

And it actually worked, giving

           6 + (x^3 + x^2 + x + 1)*exp(x)

Question is: Why did simplify(expr,trig) not do what expected, which is to only simplify trig terms in expression and not mess around with the polynomial there? 

Is the above method of thaw/freeze to control which parts of expression gets simplify a recommended way to work around this, or is there a better way?

I can't figure how to make my own type, which is rational and greater than one.

There are buildin types for postive and posint, and so on. But what if I want to make for rational and greater than some value, say 1?

This is easy to do using patmatch, using the conditional. But do not know how to do it to make my own type.

Here is a simple example. I want to check for sin(x)^n, where n is rational and must be >1.  Using pathmatch

restart;
patmatch(sin(x)^(1/2),conditional(sin(x)^a::rational,a>1),'la');la;

does it. Using structured type, the best I could do is this

restart;

TypeTools:-AddType('my_sin',specfunc(sin)^And(rational,positive));
type(sin(x)^(1/2),'my_sin');

But this does not check for >1, only positive.

Any suggestions? I know I could do this using other means, by direct parsing, using op, and so on. But I'd like to learn how to do it using structured type, just to learn the syntax if there is one.

is it possible to use conditional with structured types? But need  name to do that, like with patmatch, but there is no such syntax in structured types. 

ps. I think conditional does not make much sense with structured type. But I need to figure how to make my own type, which is rational and say >1, or integer and say >2 and so on. I just do not know how to do that yet. But I am sure there is a way. Will try to figure it out.

I can't get pdsolve to solve this pde. Here are my tries below. One of them works, but the analytical solution Maple gives is wrong. So I am not sure if it needs some additional hints or some other help to make it give the correct solution.

This is Laplace pde but with non-zero on RHS. On disk centered at origin and with radius 1, with given BC on edge of disk.

restart;
the_rhs := r^2*cos(theta)*sin(theta);
pde := VectorCalculus:-Laplacian(u(r,theta),'polar'[r,theta])=the_rhs;
bc := u(1, theta) = cos(theta)*sin(theta);
sol:=pdsolve([pde, bc], u(r, theta));

Maple gives

But this does not look right. There is a complex exponential on its own there. The solution should be real. 

I tried to solve it using pdsolve numerically, but could not make it work. It kepts complaining about missing BC. I am not good with numerical solvers in Maple. May be someone could try to do it.

evalf(eval(rhs(sol),[r= sqrt(1/4 + 1/4),theta=Pi/4]))

gives 0.5951891582 which is wrong,. It should be  0.23958

It tried to give it the symmetry conditions on theta, but now it did not solve it

restart;
the_rhs := r^2*cos(theta)*sin(theta);
pde := VectorCalculus:-Laplacian(u(r,theta),'polar'[r,theta])=the_rhs;
bc := u(1, theta) = cos(theta)*sin(theta),u(r,0) = u(r,2*Pi),(D[2](u))(r, 0) = (D[2](u))(r, 2*Pi);
sol:=pdsolve([pde, bc], u(r, theta));

No solution. I tried adding HINT = boundedseries(r=0) but that also did not help. No solution.

Could someone solve this PDE numerically in Maple and check the solution at the above location? It should be 0.23958 which shows the analytical solution given is not correct. Can Maple solve this numerically?

Maple 2021.1

ps. Analytical solution was obtained by Mathematica. I did not solve this by hand.

I was expanding an expressions, and found that Maple expands trig also. Then found there is a way to turn that off. Which is great. I tested it in my worksheet and it works. But when I put the same code inside my module, I get an error.  It seems like some scoping issue which I do not understand.

Here is a MWE

restart;
expr:=(1+x)*sin(3*x+k);
forget(expand);
expand(expandoff());
expandoff(cos,sin,exp);
F:=expand(expr);
forget(expand);

which gives 

Which does what I want. But inside a proc, the code fails

foo:=proc(expr)
local F;

forget(expand);
expand(expandoff());
expandoff(cos,sin,exp);
F:=expand(expr);
forget(expand);

end proc;

And now when calling it as 

expr:=x*sin(3*x+k);
foo(expr)

What Am I doing wrong? And how to make it work? I need to prevent expand() from expanding these functions. Help says that expand has memory table, and I want to clear that before and after doing this, so it does not affect later code.

Maple 2021.1

Update

I just found out, if I add a restart before defining my function, then the error goes away

restart;
expr:=(1+x)*sin(3*x+k);
forget(expand);
expand(expandoff());
expandoff(cos,sin,exp);
F:=expand(expr);
forget(expand);

restart;  # had to add this. But why??
foo:=proc(expr)
local F;

forget(expand);
expand(expandoff());
expandoff(cos,sin,exp);
F:=expand(expr);
forget(expand);
end proc;

#
expr:=(1+x)*sin(3*x+k);
foo(expr)

#no error. now it works

I need this to work repeatedly inside a module, so I can't do restart each time ofcourse like the above.

Even if I do restart before defining the function, next time I call it, it will fail:

restart; 
foo:=proc(expr)
local F;

forget(expand);
expand(expandoff());
expandoff(cos,sin,exp);
F:=expand(expr);
forget(expand);
end proc;

#
expr:=(1+x)*sin(3*x+k);
foo(expr); #OK
foo(expr); #FAILED

Very strange. 

I can't find how to use Maple structured type, to check for say sin(x) and say sin(x)^2 without having to duplicate the code and make a structured type for each of the two cases. An example will explain.

I need to check if expression is of this form   m*sin(anything)^n*cos(anything)^r  Where in this example below, m,n,r can be integer or rational.     

The problem is that 'specfunc'(sin)^Or(integer,rational) will not match sin(x) but will only match if sin(x) is raised to an actual power different from default 1.

So I have to duplicate the check below, by writing Or('specfunc'(sin),'specfunc'(sin)^Or(integer,rational) and the same for cos(x). I am trying to avoid all this duplication, because as the structured type becomes more involved, it will hard to read and work with like this.

Is there a way to eliminate this? In Mathematica for example, this is done using the pattern   x_. where the little dot at the end there, allows for zero of more. So for the above, it will be  Sin[x_]^n_. and this will match Sin[anything] and also match Sin[anything]^2 

I know it is possible to use patmatch in Maple to do this, but is it possible without using pathmatch and just using structured types? as I am trying to do everything without having to use patmatch now.

restart;
my_type:=''`*`'( { Or('specfunc'(sin),'specfunc'(sin)^Or(integer,rational)),
                   Or('specfunc'(cos),'specfunc'(cos)^Or(integer,rational)),
                   Or(integer,rational)})';
type(3*sin(x)^2*cos(x)^3,my_type);
type(sin(x)^2*99*cos(x),my_type);
type(sin(x)*cos(x),my_type);
type(cos(x)*sin(x)^(1/2),my_type);

gives

btw, the help page for structured type mentiones zero or more occurances of but do not know how to do this for the above example

$-\Delta u=1$ in $D$

$\frac{\partial u}{\partial\nu}+(x_1^2+1)u=0$ on $\partial D$

where $D=\{(x_1,x_2) : x_1^2+x_2^2<1\}$.

Ok, I think I am starting to get the hang of this. So in Maple, structured types is like a pattern in Mathematica. To find some subexpression one needs to first define a structured type which match that subexpression, and then use  select(type,.....).  

This works well so far (but it is not as easy as setting up a pattern).

But one small problem is that select() starts looking at the operands of the expression to look for a match.

So if the whole expression itself matches the structured type, it will not detect the whole expression, since select goes over each operand, missing that the whole thing actually matches the type.

May be an example helps shows what I mean. I want to find if an expression has cos(anything)*sin(anything) so I made a structured type for this 

mytype_3 :=  ''`*`'( {'specfunc'(cos),'specfunc'(sin)})';

btw, I do not know if I should use `&*` or '`*`' but that is a side point, I'll try to find this out.   

Now I want to use the above to check if expression has the same "structured type", or "pattern". The expression is expr:=cos(x)*sin(x); clearly it matches it. But since select looks at the operands, it will only see cos(x) by itself, and then sin(x) by itself and hence not find the structured type. 

restart;
mytype_3 :=  ''`*`'( {'specfunc'(cos),'specfunc'(sin)})';
expr:=cos(x)*sin(x);
type(expr,mytype_3);  # true
select(type, expr, mytype_3); # does not work, does not find it.

Since I am doing this in code, and I do not know what the expression is, I think I have to now do the following 

restart;
mytype_3 :=  ''`*`'( {'specfunc'(cos),'specfunc'(sin)})';
expr:=cos(x)*sin(x);
if type(expr,mytype_3) then
   print("The whole expression is a match! nothing to do. no need to use select");
else
   select(type, expr, mytype_3);
fi;

Which is OK. I can do this, But it will be nice if there was a way to have select (or another function like it) starts at the top level before looking at the operands?  

How do others handle such cases? does the above sound like an OK solution to this?