@tomleslie Tom, you're right about that syntax error. I assumed that the syntax was correct because plot was happy to plot the real and imaginary parts of DETM2. (I'll have to investigate why that is.) If I correct that syntax error, then fsolve(DETM2 = 0, complex) finds a root in about 10 seconds (in Maple 16 at both Digits = 10 and Digits = 15). That root is

I removed the evalf[4] before doing any of this; however, if I include it, it makes no significant difference in the root that fsolve finds.

@dorna01 

Do you have some reason to believe that a solution exists? Do you have some approximate ranges for the real and imaginary parts? By "numerical techniques", I primarily mean fsolve and DirectSearch combined with some plotting. Unless you're sure that there's a solution and can come up with some ranges, I can't think of anything else in Maple to try.

You can find DirectSearch in the Maple Applications Center. Yes, it runs in Maple 18. It's trivial to install. I need to go to sleep right now. It's very likely that someone else will provide more information very soon.

The error is apparently due to a bug in solve; there's nothing syntactically wrong with your equation. However, there's no hope for obtaining a symbolic solution to such an equation. If there's a solution, only numerical techniques can find it. So, I'm trying with fsolve, and then I'll move to DirectSearch (a third-party add-on package). By plotting, I'm convinced that there's no real solution with magnitude less than 10^8. I will look for complex solutions.

Please post the code and the erroneous use instance.

@kelvin goh But if you include undefined at the end as in the code that I gave, then f(0) <> 0.

Note that "piecewise-continuous" isn't the same as continuous. In case you thought otherwise, there is no need in Maple for a piecewise function to be continuous. Indeed, it can accomodate a step function, as in

piecewise(x < 0, 0, 1)

@sunflower For the third time I'll say that you can't assign to a formal parameter. That means that the line p0:= p produces the error. To correct this, make a local p1 and initialize it to p0. Then never use p0 again.

@sunflower You corrected the first two errors that I pointed out, but not the third: You can't make a direct assignment to a formal parameter. In other words, the line p0:= p is illegal. To correct this, make a local p1 and initialize it to p0. Then never use p0 again.

You'll also need to change diff(f(p0), x) to D(f)(p1).

@sunflower You didn't correctly transcribe what I wrote. The line

det:= LinearAlgebra:-Determinant(A)

needs to be inside the while loop.

@sunflower wrote:

I don't know how to create a matrix which have det=2. I don't understand

Multiplying a row or column of a matrix by c causes the determinant to be multiplied by c. So, multiplying a whole matrix by c^(1/n), c > 0, causes the determinant to be multiplied by c.

@vv Thanks for spotting the problem and suggesting the solution. Correcting the sign with signum(det) doesn't always work. Here's some corrected code:

restart:
macro(LA= LinearAlgebra):
n:= 9:
while not det::positive do
     A:= LA:-RandomMatrix((n,n), datatype= float[8], shape= symmetric);
     det:= LA:-Determinant(A)
end do:
A:= (2/det)^(1/n)*A;

@JohnS Take heart, John S: Your posted solution is far more elegant than my first two attempts, each of which I almost posted. The first of these was to multiply the first row by 2/det. The second was to take the first attempt and multiply by a random permutation matrix and then adjust for the sign of the permutation.

So, I promoted your Reply to an Answer and gave it a vote up.

Note that your code doesn't use RandomTools.

For more-elegant posts, note that you can use Shift-Enter in MaplePrimes to avoid those ugly blank lines between lines of code.

@tomleslie The OP picked up this usage of uses from reading my code examples. 

Stylistically, I object to the form

uses <package 1>, <package 2>, ...;

preferring instead the form

uses <abbrev 1>= <package 1>, <abbrev 2>= <package 2>, ...;

The reason that I object to the former is that it makes it difficult for a reader to know which global names come from which packages and which come from the writer's own definitions.

Likewise, I generally object to using the with command and I use instead, for example,

macro(LA= LinearAlgebra);

I will use a with command to load overloaded operators from a package; using infix operators in prefix form with a package prefix is too much for me.

@Ramakrishnan The midpoint methods avoid evaluating a BVP system at the boundary points, and thus avoid some singularities. You can access them using option method= bvp[midrich] or method= bvp[middefer]. See help page ?dsolve,numeric,bvp.

What makes you (Sunflower) think that there are errors?

@asa12 You are correct that the bracket represents dot product. The symbol that you're transcribing as 1_i_j is almost certainly the Kronecker delta. It's defined as 1 if i = j, and 0 otherwise.

Each entry of the matrix product A.B (for real-valued A and B) comes from taking the dot product of one row of A with one column of B. So yes, the matrix product B.B^* is equivalent to taking the dot product of each vector with each other vector, including with itself.