@GeorgesL In any case, it is not a bug. It is a decision that was made. What do you think 0^k should evaluate to, when k is unassigned?

Good question.

I think the solution f = constant and g = 0 is unstable, so if your solution approaches such a solution the numerical solution may run into severe problems.

So you got work to do :-)

Good question.

I think the solution f = constant and g = 0 is unstable, so if your solution approaches such a solution the numerical solution may run into severe problems.

So you got work to do :-)

@soechristian Here is a simple example.

sys:={y=x^2-1,y=x};
res:=solve(sys,{x,y},explicit);

#This is just an example of the use of evaluating some expression at res[1] or res[2]:

expand(eval(x^2+y^2-1,res[1]));
expand(eval(x^2+y^2-1,res[2]));

#Now actually assigning
assign(res[1]);
x,y;

res[1] refers to the first result in the sequence res.

@soechristian Here is a simple example.

sys:={y=x^2-1,y=x};
res:=solve(sys,{x,y},explicit);

#This is just an example of the use of evaluating some expression at res[1] or res[2]:

expand(eval(x^2+y^2-1,res[1]));
expand(eval(x^2+y^2-1,res[2]));

#Now actually assigning
assign(res[1]);
x,y;

res[1] refers to the first result in the sequence res.

@hirnyk For some functions coords = polar makes the pictures look nice:

plots:-conformal(z,z=0..1+Pi*I,numxy=[20,100],grid=[10,25],coords=polar,scaling=constrained,caption="The z-plane"):

plots:-conformal(sec(z),z=0..1+Pi*I,numxy=[20,100],grid=[10,25],coords=polar,scaling=constrained,caption="The w-plane"):

plots:-display(Array([%%,%]));

@hirnyk For some functions coords = polar makes the pictures look nice:

plots:-conformal(z,z=0..1+Pi*I,numxy=[20,100],grid=[10,25],coords=polar,scaling=constrained,caption="The z-plane"):

plots:-conformal(sec(z),z=0..1+Pi*I,numxy=[20,100],grid=[10,25],coords=polar,scaling=constrained,caption="The w-plane"):

plots:-display(Array([%%,%]));

@Orion Is it going to help any to replace ``  with f (assuming that f doesn't evaluate to anything but its own name)? As in

eval(p2(u),``=f);

At least you only get a warning when doing

CodeGeneration:-Fortran(eval(p2(u),``=f));

You would have to remove f all over the place though. But that could be done in a text editor with Edit/replace.

@Orion Is it going to help any to replace ``  with f (assuming that f doesn't evaluate to anything but its own name)? As in

eval(p2(u),``=f);

At least you only get a warning when doing

CodeGeneration:-Fortran(eval(p2(u),``=f));

You would have to remove f all over the place though. But that could be done in a text editor with Edit/replace.

Extending the above procedure to arbitrary integral powers:

CombineLikePowers:=proc(u::algebraic,L::{posint,set(posint),list(posint)}:=infinity) local S0,S,k,u0,u1,u2,u3,u4;
    if type(u,`+`) then
       map(procname,u,L)
    elif type(u,`*`) then
       u0:=evalindets(u,function,freeze);
       if not member(true,map(type,{op(u0)},{name^posint,name^negint})) then return u end if;
       S0:=map(abs,map2(op,2,indets(u0,`^`(anything,integer)))) minus {1};
       if L=infinity then
          S:=S0
       elif type(L,posint) then
          S:={L} intersect S0
       else
          S:=convert(L,set) intersect S0
       end if;
       for k in S do
       u1:=evalindets(u0,{algebraic^k,algebraic^(-k)},x->``(root[k](x,symbolic)));
       u2,u3:=selectremove(type,u1,specfunc(algebraic,``));
       u4:=``(eval(u2,``=(()->args)))^k*u3;
       u0:=evalindets(u4,specfunc({name,`^`},``),expand);
       end do;
       thaw(u0)      
    else
       u
    end if
end proc;

You may try it on something like

w:=expand(randpoly([x,y,sin(x^2)],degree=4,dense)/a^4/b^3);

You could do

CombineLikePowers(w);

or if you only want powers 2 and 3:

CombineLikePowers(w,{2,3});

Extending the above procedure to arbitrary integral powers:

CombineLikePowers:=proc(u::algebraic,L::{posint,set(posint),list(posint)}:=infinity) local S0,S,k,u0,u1,u2,u3,u4;
    if type(u,`+`) then
       map(procname,u,L)
    elif type(u,`*`) then
       u0:=evalindets(u,function,freeze);
       if not member(true,map(type,{op(u0)},{name^posint,name^negint})) then return u end if;
       S0:=map(abs,map2(op,2,indets(u0,`^`(anything,integer)))) minus {1};
       if L=infinity then
          S:=S0
       elif type(L,posint) then
          S:={L} intersect S0
       else
          S:=convert(L,set) intersect S0
       end if;
       for k in S do
       u1:=evalindets(u0,{algebraic^k,algebraic^(-k)},x->``(root[k](x,symbolic)));
       u2,u3:=selectremove(type,u1,specfunc(algebraic,``));
       u4:=``(eval(u2,``=(()->args)))^k*u3;
       u0:=evalindets(u4,specfunc({name,`^`},``),expand);
       end do;
       thaw(u0)      
    else
       u
    end if
end proc;

You may try it on something like

w:=expand(randpoly([x,y,sin(x^2)],degree=4,dense)/a^4/b^3);

You could do

CombineLikePowers(w);

or if you only want powers 2 and 3:

CombineLikePowers(w,{2,3});

@Alex Smith You are missing the point, which (as I understod it) was to integrate term by term, which is NOT done when you do

value(Int((sum(x^(2*k+1), k = 1 .. infinity))*sqrt(1-x), x = 0 .. 1));

because

Int((sum(x^(2*k+1), k = 1 .. infinity))*sqrt(1-x), x = 0 .. 1);

is not the integral of an infinite sum as I pointed out in my answer above (In the remark: "This is not really an infinite summation example since Maple first computes the sum").

@Alex Smith You are missing the point, which (as I understod it) was to integrate term by term, which is NOT done when you do

value(Int((sum(x^(2*k+1), k = 1 .. infinity))*sqrt(1-x), x = 0 .. 1));

because

Int((sum(x^(2*k+1), k = 1 .. infinity))*sqrt(1-x), x = 0 .. 1);

is not the integral of an infinite sum as I pointed out in my answer above (In the remark: "This is not really an infinite summation example since Maple first computes the sum").

@hirnyk You are quite right. Something seems to suggest that the exact answer is correct, but the numerical evaluation of it is rather sensitive to the value of Digits. According to Maple the sum

sum(int(x^(2*k+1)*sqrt(1-x), x = 0 .. 1), k = 1 .. infinity);

has the value

(1/8)*sqrt(Pi)*sqrt(2)*MeijerG([[1], [11/4, 13/4]], [[5/2, 2, 1], []], -1)

If you do

evalf(sum(int(x^(2*k+1)*sqrt(1-x), x = 0 .. 1), k = 1 .. infinity),6);

you are using Digits = 6 (regardless of the value of Digits).

Interestingly, you get a better (and roughly correct) result than if you Digits = 10 or 20 (in the latter case the result is 4.0448194072012987198*10^11-3.4793277886925613872*10^10*I, which obviously is far off.

@hirnyk You are quite right. Something seems to suggest that the exact answer is correct, but the numerical evaluation of it is rather sensitive to the value of Digits. According to Maple the sum

sum(int(x^(2*k+1)*sqrt(1-x), x = 0 .. 1), k = 1 .. infinity);

has the value

(1/8)*sqrt(Pi)*sqrt(2)*MeijerG([[1], [11/4, 13/4]], [[5/2, 2, 1], []], -1)

If you do

evalf(sum(int(x^(2*k+1)*sqrt(1-x), x = 0 .. 1), k = 1 .. infinity),6);

you are using Digits = 6 (regardless of the value of Digits).

Interestingly, you get a better (and roughly correct) result than if you Digits = 10 or 20 (in the latter case the result is 4.0448194072012987198*10^11-3.4793277886925613872*10^10*I, which obviously is far off.