B:= Array((0..n-1)$2, [ListTools:-LengthSplit(L,n)]);

Edit: Corrected error pointed out by Kitonum.

Your symbolic dsolve solution contains a complicated RootOf expression whose principal value has nonzero imaginary part for the values of t that you're plotting. You can avoid this by using a numeric dsolve with parameters, like this:

sol:= dsolve(
   {eq1b, eq2b, r(0) = r__0, theta(0) = theta__0}, 
   parameters= [r__0, theta__0], numeric
):
sol(parameters= [theta__0= Pi/4, r__0= 0.1]):
plot1:= plots:-odeplot(
   sol, r(t)*~[cos,sin](theta(t)), t= 0..10,
   axiscoordinates= polar, color= red, labels= [``$2], tickmarks= [4,3]
);

Note the explicit conversion from polar coordinates: r(t)*~[cos,sin](theta(t)).

If the matrix's entries can be expressed as a function of the indices, then you can use an initializer function like this:

U:= Matrix((5,5), (i,j)-> `if`(i=j, 1, 0));

The command solve doesn't understand domain restrictions; however, it does understand inequalities if they're put with the equations in the first argument:

solve({eq4, eq5, theta > 0}, {Fh1, theta});

The numerical solver fsolve does respect domain restrictions entered as ranges:

fsolve({eq4, eq5}, {Fh1, theta}, {theta= 0..infinity});

The following code shows all the equivalence classes of adjacency matrices.This output is much neater looking in a Maple worksheet. With a small modification, it could show a drawing of a representative of each class.

n:= 3: N:= n*(n-1):
<ListTools:-Categorize(
   (M1,M2)-> GraphTheory:-IsIsomorphic(GraphTheory:-Digraph~([M1,M2], n)[]),
   map(
      L-> Matrix((n$2), [0, seq([L[n*k+1..n*(k+1)][], 0][], k= 0..n-2)]), 
      combinat:-permute([0$N, 1$N], N)
   )
)>; 

Your model function is not differentiable at t=0, yet it appears that t=0 is one of the data points. Try removing this one data point. Of course, I can't test this without having your actual data.

If all of the parameters being estimated are coefficients, then you can and should use linear regression. The model function doesn't need to be a linear function of the data variables in order to use linear regression; it merely needs to be a linear function of the parameters being estimated.

If A is the matrix, then you can use convert(A^+, listlist). The first argument, A^+, means the transpose of A. Using this doesn't require foreknowledge of the number of columns of A.

k1:= seq(add(X[h,t], t= 1..23) >= 9, h= 1..6);

foldl(gcd, 0, P[]);

The title says it all. The command that you want is assign~(A =~ B):.

But I can't imagine any situation where I would want to or where it would be beneficial to make 1 million assignment statements. I suspect that Maple will crash or will become extremely slow.

What you're asking for---to reverse the effect of KroeneckerDelta---is mathematically impossible. It would be like multiplying ME by 0 and then trying to reverse the effect of that.

Use "Export as"  (rather than "Save" or "Save as") from the File menu to save the inner file as type Maple Input. Simply changing the file extension to .mpl is not enough.

The answer Dirac(X - 0.5) is what you get if X hasn't been previously assigned. So it seems that the X in you PDF command is not the same as the X in your RandomVariable command. One of the Xs may contain a nonprinting character.

This may be a bug that has been corrected in Maple 2016; I can't duplicate it. Try using restart. If that doesn't work, here's a workaround that should work in any Maple with the Statistics package:

eval(PDF(X, _t), _t= 0.5);

Let me know how it goes.