If you don't care whether or not your matrix is symmetric or Hermitian, then you can use any nonsingular matrix, R to transform your diagonal matrix of eigenvalues.

R.D.R^(-1);

So you can generate R using RandomMatrix, but you should check that it is nonsingular. This is still not the most general way, but is useful for most purposes (the matrices generated are diagonalizable, with diagonal Jordon forms).

I don't have v11, so I can't see everything on your worksheet, but the following idea should work. If your global "constant" is r and the functions defining your parametric plot are something like

alpha:=x->r*sin(x);
beta:=x->r*cos(x);

then you can plot curves with several r values by

plots[display](seq(plot([alpha,beta,0..2*Pi]),r=[1,3,5]));

If you want more individual control over the plots, e.g. different colors, then you can store the plots in variables and then use display to put them together .

David.

 

Eval(y,x=0);

gives

Eval(y,x=0);

(though the Maple tag seems not to display it well in the post). This is the mathematical equivalent of your statement in words.

 

interface(displayprecision=3);

should do the trick.

 

 

Try phaseportrait or DEplot in the DEtools package.

There is a missing line in your output - a list of integers which can be converted to known mersenne primes. My output has the following line after the first "end if":

w := [2, 3, 5, 7, 13, 17, 19, 31, 61, 89, 107, 127, 521, 607, 1279, 2203, 2281, 3217, 4253, 4423, 9689, 9941, 11213, 19937, 21701, 23209, 44497, 86243, 110503, 132049, 216091, 756839, 859433, 1257787, 1398269, 2976221, 3021377, 6972593, 13466917, 20996011, 24036583, 25964951, NULL];

So the routine has no smarts... Try showstat if you can't see it on another try.

 

If you think about eval([x,y],{x=0,y=-1}) giving [0,-1], then you can work this up with map2 to give what you want:

q:={x = 0, y = -1}, {x = 3/4, y = -7/16}:

map2(eval,[x,y],[q])[];

gives

[0,1],[3/4,-7/16]

as required. This works for any number of solutions returned by solve.

 

Sorry, bad copying job; it is op(0,q) which gives y. But note that op(q) is the same as op(1,q) in this context.

q:=y[2]; Then op(q); gives 2 and op(1,q) gives y

As Robert Israel points out, in a Maple plot, you need to specify all parameters. If you want to see how the plot changes with parameter k, then you can plot a 3d plot, with k as one of the axes plot3d(k*x^2 - k*x^4,x=-10..10,k=-10..10); If a plot has a universal shape, then it can be non-dimensionalized, and the non-dimensionalized plot then encapsulates all possible behaviors. For your case of x=k, one defines the (nondimensional variable) X=x/k, and then plots X=1. For the case of y=k*x^2 - k*x^4, let Y=y/k, then Y=x^2 - x^4 has no parameters and can be plotted.

Maybe the @ notation is a nice alternative

(f@x)(t);
           f(x(t))

D(f@x)(t);
           (D(f))(x(t))*(D(x))(t)

convert(%,diff);

           (eval(diff(f(t1), t1), t1 = x(t)))*(diff(x(t), t))

X:=(u,T)->u*cos(T);
Y:=(u,T)->u*sin(T);
plot3d([X(u,T),Y(u,T),u],T=0..2*Pi,u=0..1,axes=boxed);

For these sorts of problems, I usually rename the file to *.txt; often the virus checker is more worried about the end user double-clicking on *.exe or other dangerous filetypes than actually looking closely at the content (unless of course there is a known virus in there!).

I have certainly zipped (using PKZIP under Windows) directories containing *.lib, *.ind, *.hdb and other types of files without any difficulty. Haven't tried *.mla yet. Note that if you create with march, you get to specify the size, so this gives you some control - see help under repositories or march.

I think the original question was about the diagonalizable nature of matrices, i.e., whether a similarity transform converts them to a diagonal matrix. Here again there is no correlation,

[1 0]
[0 0]

is trivially diagonalizable (because it is already diagonal) and has determinant zero, but

[0 1]
[0 0]

is not diagonalizable and still has determinant zero. Diagonalizability has more to do with whether there are missing eigenvectors. When the eigenvalues are all distinct, the matrix will be diagonalizable, but if some eigenvalues are equal then the matrix may be non-diagonalizable (or may be diagonalizable). Whether or not the degenerate (equal) eigenvalues are zero (determinant will then be zero) or not is not relevant to the question of diagonalizability. To check diagonalizability, one looks at the Jordan form to see if it is diagonal - see Maple's JordanForm.