@student_md 

So, you want to define a sequence of matrices.
It would be better to use as index the position of the matrix in the sequence. I.e. C(k) is the k-th matrix (having the order 2^(k-1) in your case.

You need to define clearly H(k), c(k). With your notation for c_a = c_(m/2) it is not clear what is e.g.  c_3  in the sequence c_0, c_1,...,c_(m/2-1). That's because with your notation, c(m) is defined only for m = "a power of 2".

In other words, use consistent math notations. It will be easy then to obtain the maple code.

@Kitonum 

Of course, but the workaround often works when the locus is not known.
And after all your example is a (degenerate) ellipse.

@Kitonum 

But once we are aware of the (inherent!) problem, workarounds are possible. E.g.
plots:-implicitplot(f(x,y)=d+1e-5, x=0..6, y=0..6, gridrefine=5);

@student_md 

The main advantage is that you can edit it (or even create it with a text editor).

An .m file (internal format) loads faster, but your file is small and does not matter.

@peter2108 
You already have writebytes for this.

@ecterrab 

I just want to note that in the first integral (J), the change of variables was not done at all. J was simply rewritten using intat.

@Rouben Rostamian  

The EPS export for 3D is (and was) totally wrong (at least in Windows 64 bit).
plot3d(x^2+y^2, x=-1..1,y=-1..1);
(I have posted twice the result in the past.)

@tomleslie 
The same version and OS.

The worksheet executes correctly for me.

@Christian Wolinski

collect(evala(Norm((lhs-rhs)(A))),Y,`convert/parfrac`);

@Carl Love 
Yes, I have corrected.

@Carl Love 

No, I din not think much about it. But it seems that a more general fact is true.
If p is a fixed prime and 2 <= k <= p-2 is also fixed then the events
A = {a in {1,...,p-1} :  a is prime} and
B = {a in {1,...,p-1} : a^k mod p  is prime}
are "asymptotically independent". For k=p-2 one obtains the inverses mod p.

@Carl Love 

Actually the two events are not independent (for a fixed p). They seem to have some kind of "asymptotic independence" (p --> oo).
It's interesting your intuition about this!

@acer 

It could be that the integral is exactly 1/8 but I have not time now to check.

@digerdiga 

The numerical integration is easier for rectangular domains and it is always possible to reduce the integral to such domains via a change of variables.
Note that we may include the domain D in a larger rectangle R and define the function 0 in R \ D but this is not the best solution because the new function is (generally) not continuous.