Remembered ther is a command in the LinearAlgebra package,
?LinearAlgebra[GenerateMatrix] may be what you are looking for

For me it seems difficult to see what you intend to achieve (especially from the
C source only), and why.

As suggested in your other thread I first would try to get the Maple script to be
ok, before calling it from something external.

And if I use some external, I carefully care for input and return types, your
function returns an integer, so I would write define_external that way (it is
too long ago, that I remember that wrapper things).

If I would need all the values of the various derivatives I would use (after the
prec-computations) just 1 function with 3 parameters u,v,w and an Array. That
is: why do you need the RTable?

And as already said: I would try to find a way, that the construction already works
in one of the systems (=Maple), before trying to interface with another (=C).


And for finding reasons for crashes I would try to reduce to the most simple case.

There are several things I do not understand in your C source (though it is some
time ago): MATRIX_OFFSET_C_RECT - what is it, have you checked indexing errors?
ALGEB DirectFastPointer - have you check correct type in calling? And so on ...

To locate errors you have to break it down, give the most simple situation with
just one function and most simple calls. Just to be sure, that there are not
multiple problems and that it worked before by lucky incidence.

Just my 2 cents.

But more or less: I do not understand what you want to do.

1. I would prefer to see the code. Not the cryptic thing in your post.

2. Some times ago I did such stuff (replicating Maple commands in C).
It is not so much fun. But it only makes sense (for me), if one has done
the work in Maple first (= have a good code working in Maple) and after
that go that painful route.

3. It may be worth to put all the Maple code into some proc, which
then is being called from C (if ever needed)

4. Maintaining such is a pain.

Why and for what do you you need it?

The appended sheet gives a solution through Maple without more theory:

Re-write as summing +-tan(d°), d = 1,3,5 ...89. Group as tan(d°) + tan(90°-d°),
convert to exp and simplify. Then Maple luckily finds the exact sum, it is 45.

MP_hirnyk_SumTan_ele.mws

The exact answer is 45. And here is a sketch, using (minimal) polynomials for
the sum of those algebraic numbers.

First note that Product( x - a(j), j=1 .. n) has coefficent -Sum(a(j), j=1..n)
in degree n-1 (and the product in degree 0), for example look at

  Product( x - a(j), j=1 .. 4); value(%): expand(%): collect(%, x);
  coeff(%,x, degree(%,x) - 1);

Now there is some luck with that task: the minimal polynomials for cos(...)
are all in x^2, if one takes the same arguments as for the tan(...).

That means, that one can derive the minimal polynomials for the tan(...) in a
simple way (without resolvents and doing the sin case), see Robert Israel's
contribution in http://www.mathkb.com/Uwe/Forum.aspx/math/77031/A-Question-about-Sums-of-Tangents

Addtionally it turns out that for these 45 terms only 6 different polynomials
actually occure they are of degree 1, 2, 4, 6, 8, 24. These degrees add up to
45 and we have 45 different terms as summands.

That means: the summands are exactly the zeros of those 6 polynomials.

Thus adding the coefficients in topdegree - 1 gives the sum (up to sign).

See attached sketchy file

MP_hirnyk_SumTan.mws

You really want 5 = 3*k = 3*1 = 3 ? Why ?

Note that your coefficients cover a wide range, from 1e-16 to 1e+18

Using 15 Digits and 'Isolate' a check for the result shwos, that they are large
and with 75 Digits one gets 1e-28 should be zero.

For 'Bivariate' it works, if one accepts that results are not checked:

BivariatePolynomial(sys , [lam1, lam2], verify=false);

This however returns undefines solution.

Guess, that are the reasons.

May be you want to dig for condition numbers (I hate that, but here
it seems to be needed)

PS: reallly, you should post things in a form ready for copy & paste.

You may look up ?polar (and the link polar coordinates ) and ?evalc

A 2-argument arctan is commonly used and your exp
would have an imaginary I = sqrt(-1)

Where the help pages should be more clear about that

1/myExpr = R*D^2 / ( R*D^2 + s*L + s^2*R*L*C );
isolate(%, myExpr);
expand(%);

                                           2
                                   s L    s  L C
                      myExpr = 1 + ---- + ------
                                      2      2
                                   R D      D

For the adjusted question (powering the Sum) I think it is

Limit( (1+2/n)^n, n=infinity) = exp(1)^2.

For that I used the asymptotic series in the already mentioned Temme paper
in a quite brute way to have hypergeom([1, n],[n+1],2) ~ (-1-2/(n+1)) and
then the Sum reduces to 1 + 2/n.

Ok, some reasoning is needed, why one can do the limit in 2 steps.

Since cos is invertible over that interval one can try:

Int(cos(a*x)*cos(x)^(a-2), x = 0 .. (1/2)*Pi);
Change(%, x=arccos(t),t);

                                  0

It seems that M15 does not find an anti-derivative here?

Numerical tests indicate that -1 is the limit. And an idea to prove it is as follows:

The Sum evaluates in terms of LerchPhi, to be convert to 2F1 and it can be seen,
that one has to show that hypergeom([1, n],[n+1],2) ~ -1.

The usual transformations for 2F1 do not apply due to the parameter constellation.

But having a look into papers of N Temme to get ideas beyond Maple abilities I took
"Large parameter cases of the Gauss hypergeometric function".

Now hypergeom([1, n],[n+1],z) ~ hypergeom([1, n],[n],z)  = (-1+z) for |z| < 1
by taking the summation version of 2F1-functions.

I have not followed his reasoning why this holds for the analytic extensions (and
into the branch cut).

Hopefully somebody finds a more simple way :-(

Edited: this is for the original question (without powering).

I think it is 1/2.

Evaluate the sum, which is a function in sqrt(x), hence use x^2
instead of x.

evalc(Re(%)) assuming 0 < x;

  -1/2 * x*Im(Psi(1,x*I))

Now plotting indicates the limit and it is 'confirmed' by

x*Im(Psi(1,x*I)); 
MultiSeries:-limit(%, x=infinity);

                                  -1

I think you have to be more specific, so that the problem has a unique solution.

alpha^3-alpha*beta^2-beta*alpha^2+beta^3 = 
  0 + (alpha-beta)*(alpha^2-beta^2)

May be you want the automorphism y |-> alpha + beta
x |-> alpha - beta, write in x,y, lambda1, lambda2 and
now the taylor series in x, p1= constant term while for p2
= (fct - constant)/x and finally substitute back. 

Or similar?

Edited: find attached a small example sheet MP_poly_basechang.mws

Edited2: on a 2nd (sigh) thought the following is more 'natural':
  trans:=solve({x=alpha-beta}, {alpha});
  back:= solve(trans, {x});

@joavr

It is still not clear to me what you did and why you can expect a solution
(because more conditions than variables), but they exist, yes.

I played a bit with some specific parameters (after getting rid of denominators).

For m, n, q, r, t, u, v, w, x, y:= 0,0,0,0,0,0,0,0,0,0; the only condition is
r0 = -x0^2/y0,u0 = -x0, all other variables can be choosen arbitrarily,
HilbertDimension of the system equals 2.

For m, n, q, r, t, u, v, w, x, y:= 1,0,1,0,1,0,1,0,1,0; there are many solution
sets and conditions (up to 7), HilbertDimension of the system equals 7.

For me that says, that depending on you parameters the possible solution
sets may jump heavily and Maple may have no chance to find all.

Can not say something towards Java and memory: I did that in the classical
interface, the other one is a pain for me in for that task.

Edited to upload the sheet MP_parametric_polyn_.mws