@Joe Riel The following doesn't split mixed assumptions and temporary assignments, but does handle both classes.

restart;
p:=subsop(3=overload,eval(`assuming`)):
`assuming`:=overload(
     [
        proc(x::uneval,a::list({equation,set(equation),list(equation)}),$) option overload;
            op(eval(x,ListTools:-Flatten(evalindets([a], set, [op]))));
        end proc,
        eval(p)
      ]
):
plot(a*sin(x),x=0..b) assuming {a=-2,b=7};
param := {a=2,b=7}:
plot(a*sin(x),x=0..b) assuming param;
plot(a*sin(x),x=0..b) assuming a=-2,b=7;
plot(a*sin(x),x=0..b) assuming [a=-2],{b=Pi};
simplify(sqrt((x*y)^2)) assuming x>0,y>0;
int(exp(a*t),t=0..infinity) assuming a<0;

#Interestingly, this one works (with or without overloading), but in the result b is not replaced by 2:

int(exp(b*a*t),t=0..infinity) assuming b=2,a<0;

#This one doesn't work, because the main procedure doesn't accept sets or lists of properties.

int(exp(b*a*t),t=0..infinity) assuming {a<0,b=2};

@acer You are right. 'Using' sounds better.

And I have to get used to using the 'use' statement. I had almost forgotten its existence.

Unfortunately, with

param:={a=-2,b=7};

neither one of
use param in plot(a*sin(x),x=0..b) end use;
use op(param) in plot(a*sin(x),x=0..b) end use;

works. It is convenient to have the parameters defined as a set or list since 'eval' doesn't accept a sequence, as does subs.

@pchin Thanks to Joe Riel and pchin. The wikipedia link about reentrancy is difficult reading, by I think I got the gist of it.

@pchin Thanks to Joe Riel and pchin. The wikipedia link about reentrancy is difficult reading, by I think I got the gist of it.

@pvrbik I always use 1d-input (also known as Maple Input), so that is why.

I tried doing the whole thing with 2d-input and the result was that the input in the tex-file looked like this:

\begin{mapleinput}
\mapleinline{active}{2d}{for i to 10 do f := proc (x) options operator, arrow; x^2 end proc; f(x) end do; 1}{\[\]}
\end{mapleinput}

but nothing of that turned up in the dvi-file, only the 10 results.

@pvrbik I always use 1d-input (also known as Maple Input), so that is why.

I tried doing the whole thing with 2d-input and the result was that the input in the tex-file looked like this:

\begin{mapleinput}
\mapleinline{active}{2d}{for i to 10 do f := proc (x) options operator, arrow; x^2 end proc; f(x) end do; 1}{\[\]}
\end{mapleinput}

but nothing of that turned up in the dvi-file, only the 10 results.

test.pdf

@pvrbik 

No images. See the file test.pdf which I produced from your Maple code. It should be attached.

test.pdf

@pvrbik 

No images. See the file test.pdf which I produced from your Maple code. It should be attached.

@pvrbik Well, I'm using Windows.

My latex is handled by Scientific Workplace. It produced a dvi-file, which I printed to file (postscript).

The postscript file I converted to pdf using gsview/ghostscript.

The result looked nice.

Of course you can export to pdf directly from Maple, but that is not what you want, I guess.

@pvrbik Well, I'm using Windows.

My latex is handled by Scientific Workplace. It produced a dvi-file, which I printed to file (postscript).

The postscript file I converted to pdf using gsview/ghostscript.

The result looked nice.

Of course you can export to pdf directly from Maple, but that is not what you want, I guess.

Same thing happens with plot3d.

Maybe animate is finding a coordinate system in which all frames can be displayed?

It would be quite annoying if the following animation changed the view between frames.

animate(plot,[sin(x),x=-a..a],a=1..10);

which it doesn't.

In range = a..b the length of the real interval [a, b] is b-a. The code could be written from that point of view (not unreasonable).

In your example the matrix has integral values (1,2,3,4), so b-a would result in 3.

In range = a..b the length of the real interval [a, b] is b-a. The code could be written from that point of view (not unreasonable).

In your example the matrix has integral values (1,2,3,4), so b-a would result in 3.

@hirnyk 

Yes, the two sides have simple poles at 2 with the same residues. But so what?

series(lhs(eq),x=2);
                       1        (-1)   1
                       - (x - 2)     + -
                       2               2
series(rhs(eq),x=2,2);
           1        (-1)   3   1           /       2\
           - (x - 2)     + - + -- x - 2 + O\(x - 2) /
           2               8   32                    

@hirnyk 

Yes, the two sides have simple poles at 2 with the same residues. But so what?

series(lhs(eq),x=2);
                       1        (-1)   1
                       - (x - 2)     + -
                       2               2
series(rhs(eq),x=2,2);
           1        (-1)   3   1           /       2\
           - (x - 2)     + - + -- x - 2 + O\(x - 2) /
           2               8   32