I'm using Firefox 8.0 under Windows 7, and I don't see any problem on that page.

Here's a screen-capture.

It appears from the Maple output that you did not assign values to ABC and Werte, so LSSolve is just taking these all to be 0.

It appears from the Maple output that you did not assign values to ABC and Werte, so LSSolve is just taking these all to be 0.

Your commands have unassigned n's in them.

For convenience, you might define

M1:= n -> Matrix(n,n,shape=Circulant[[0,1/2,0$n-3,-1/2]]);
M2:= n -> Matrix(n,n, (i,j) -> piecewise(i-j mod n = 1, -1/2, i-j mod n = n -1, 1/2, 0));

and then try

M1(8);

M2(8);

Your commands have unassigned n's in them.

For convenience, you might define

M1:= n -> Matrix(n,n,shape=Circulant[[0,1/2,0$n-3,-1/2]]);
M2:= n -> Matrix(n,n, (i,j) -> piecewise(i-j mod n = 1, -1/2, i-j mod n = n -1, 1/2, 0));

and then try

M1(8);

M2(8);

@LijiH : the "parameters" are considered as variables, on the same footing as x.  Maple is indeed quite reliable at finding whatever factorization is possible (into irreducible multivariate polynomials over the field generated by the coefficients).  For example:

> factor(x^4 - 2*a*x^2 + a^2);
 

If it is really long and has lots of variables, that may slow Maple down somewhat (although the algorithms are pretty efficient), but shouldn't affect the reliability. 

@LijiH : the "parameters" are considered as variables, on the same footing as x.  Maple is indeed quite reliable at finding whatever factorization is possible (into irreducible multivariate polynomials over the field generated by the coefficients).  For example:

> factor(x^4 - 2*a*x^2 + a^2);
 

If it is really long and has lots of variables, that may slow Maple down somewhat (although the algorithms are pretty efficient), but shouldn't affect the reliability. 

The PDE is second order in r, so Maple says you need 2 boundary conditions.   For any problem where "time-based" numerical methods have a chance of working, I think that will in fact be true.

The PDE is second order in r, so Maple says you need 2 boundary conditions.   For any problem where "time-based" numerical methods have a chance of working, I think that will in fact be true.

@acer : Ah, I see.  I somehow got the threads tangled.

@acer : Ah, I see.  I somehow got the threads tangled.

Of course, if you're going to actually print out each Matrix in full, efficiency is not going to be much of a concern: the time required to print it in the worksheet and scan it with your eyes will be much larger than the time to compute it. 

Of course, if you're going to actually print out each Matrix in full, efficiency is not going to be much of a concern: the time required to print it in the worksheet and scan it with your eyes will be much larger than the time to compute it. 

I'm sure my response was not just

for n to 20 do 'A'^n = A^n end do;

which was the same code as the original.  I think it was something like

for k from 1 to 10 do   
    n:= 2*k-1;
    print('A'^n = A^n, 'A'^(n+1)=A^(n+1))
 end do:

Did somebody edit it?

I'm sure my response was not just

for n to 20 do 'A'^n = A^n end do;

which was the same code as the original.  I think it was something like

for k from 1 to 10 do   
    n:= 2*k-1;
    print('A'^n = A^n, 'A'^(n+1)=A^(n+1))
 end do:

Did somebody edit it?