And how sad, that the docu is not more clear on Powmod and powmod in its statements (i.e. telling also in Math terms, not only by Maple commands).

Thank you, but unfortunately I do not read Russian.

If you do not mind: What does his code give for the example in char p = 3?

As Robert Israel pointed our, the multiplicities 'forbid' a diagonalization
(the algebraic multiplicities for the charateristic polynomial change modulo 3)

Thank you, but unfortunately I do not read Russian.

If you do not mind: What does his code give for the example in char p = 3?

As Robert Israel pointed our, the multiplicities 'forbid' a diagonalization
(the algebraic multiplicities for the charateristic polynomial change modulo 3)

I have not tried.

If you only want the discriminant then I think you can use:

IntegerCharacteristicPolynomial(M,x);
discrim(%,x);
% mod p;

                                  4

I have not tried.

If you only want the discriminant then I think you can use:

IntegerCharacteristicPolynomial(M,x);
discrim(%,x);
% mod p;

                                  4

You mean symmetric, I suppose. Then you must have the eigenvalues in the scalars, since
they are the same as for D (do not use that symbol ...) and here it is obvious.

So you must allow a finite extension.

In your specific example the roots of the char polynomial are 1,2,3,4 and it splits in the field.

IIRC that diagonalization can be done by Maple, I have posted that for the Reals/Complex,
http://www.mapleprimes.com/blog/axelvogt/normalformsquadraticfunctionsvectorspaces

I think you can immitate it (but there is something special in char = 2 for quadratic forms,
can not remember, but it has to do with the difference between functions (=evaluations)
and polynomials, the polynomial fct may be zero while the polynomial is not, like x^2 + 1).

Correction: not x^2 + 1, but (x-0)*(x+1) = x^2 + x

You mean symmetric, I suppose. Then you must have the eigenvalues in the scalars, since
they are the same as for D (do not use that symbol ...) and here it is obvious.

So you must allow a finite extension.

In your specific example the roots of the char polynomial are 1,2,3,4 and it splits in the field.

IIRC that diagonalization can be done by Maple, I have posted that for the Reals/Complex,
http://www.mapleprimes.com/blog/axelvogt/normalformsquadraticfunctionsvectorspaces

I think you can immitate it (but there is something special in char = 2 for quadratic forms,
can not remember, but it has to do with the difference between functions (=evaluations)
and polynomials, the polynomial fct may be zero while the polynomial is not, like x^2 + 1).

Correction: not x^2 + 1, but (x-0)*(x+1) = x^2 + x

Killed? Please do not say that:

For me it has advantages, that I do not want to miss.

And if I can not have it with 64 bits? Well, then I will not use 64 bits, certainly.

Killed? Please do not say that:

For me it has advantages, that I do not want to miss.

And if I can not have it with 64 bits? Well, then I will not use 64 bits, certainly.

Yes, it is intended for helping and sharing informations. But not solving homework. You should try to do a little bit of work by yourself and a least look up the handbook or lecture notes. Already the first question of the teaching assistant may debunk you - a bad idea facing oral exams.

BTW: the answer may be "at most 2, if you mean the Reals" (I think).

Yes, it is intended for helping and sharing informations. But not solving homework. You should try to do a little bit of work by yourself and a least look up the handbook or lecture notes. Already the first question of the teaching assistant may debunk you - a bad idea facing oral exams.

BTW: the answer may be "at most 2, if you mean the Reals" (I think).

An inverse search for 49-241-980919-30 finally gives me nedeco GmbH, with same adress as Scientific Computers.

?

Ok, that seems to be not enough. It involves roots in the model and one
either formulates it in a way, that it is safe or has to boundaries for
the parameters.

And you made me to look closer:


At least if a brute fit does not work as expected one can ask whether it
is a model which has a chance to approximate data. One way is to look at
simple geometric as the data reflect.

  M*(  (1 -  ( 1 - q*(0.0012/M)^(1-q))*exp(-k*q*x)) /q)^(1/(1-q)) :
  convert(%, rational):
  #subs(q = r+1, %);
  h:=unapply(%, x);
                                                          /  1  \
                                                          |-----|
                                                          \1 - q/
                 /    /      /  3   \(1 - q)\            \
                 |1 - |1 - q |------|       | exp(-k q x)|
                 |    \      \2500 M/       /            |
     h := x -> M |---------------------------------------|
                 \                   q                   /


The 2nd factor is a root (and q must be different from 1), hence positive
(and its input has to be positive as well (!) ), thus 0 < M.

Moreover

  h(0); simplify(%) assuming ( 0 < M, q::real, r::real); evalf(%);

                          0.0012000000000000

But that is the y-value of the 1st data point, which has x = 13.2621 and
that never can work.

Thus either the data or the x must be shifted by that.

For large x the data indicate h ---> infinity (the user must know, if that
can be correct, fitting may go that way):

If both k and q are positive then the limit is finite (I think), 
and for k and q both negative one can get complex values (k=-2, q=-1/2),
if k < 0 and 0 < q then ... etc.

Also it looks as if the data have 'derivative = 0' at the left (and which
should have x=0, not x=13.2), they reach it almost flat:

  D(h)(0); simplify(%) assuming ( 0 < M, q::real, r::real);
  subs(q = r+1, %);
  `M` = solve(%, M);
  simplify(%) assuming r::real;

                                      (- 1/r)
                             3 (r + 1)
                         M = ----------------
                                   2500

This shows, that M and r = q - 1 are closely related and one would think
about the need to use some M at all (they are not independent).

After all that it would be the right moment to retry fitting again.

And I do not think it is primarily the weakness of Maple's standard
routines that it does not immediatly work for the original form (so it
is inviting enough to try it first).

Ok, that seems to be not enough. It involves roots in the model and one
either formulates it in a way, that it is safe or has to boundaries for
the parameters.

And you made me to look closer:


At least if a brute fit does not work as expected one can ask whether it
is a model which has a chance to approximate data. One way is to look at
simple geometric as the data reflect.

  M*(  (1 -  ( 1 - q*(0.0012/M)^(1-q))*exp(-k*q*x)) /q)^(1/(1-q)) :
  convert(%, rational):
  #subs(q = r+1, %);
  h:=unapply(%, x);
                                                          /  1  \
                                                          |-----|
                                                          \1 - q/
                 /    /      /  3   \(1 - q)\            \
                 |1 - |1 - q |------|       | exp(-k q x)|
                 |    \      \2500 M/       /            |
     h := x -> M |---------------------------------------|
                 \                   q                   /


The 2nd factor is a root (and q must be different from 1), hence positive
(and its input has to be positive as well (!) ), thus 0 < M.

Moreover

  h(0); simplify(%) assuming ( 0 < M, q::real, r::real); evalf(%);

                          0.0012000000000000

But that is the y-value of the 1st data point, which has x = 13.2621 and
that never can work.

Thus either the data or the x must be shifted by that.

For large x the data indicate h ---> infinity (the user must know, if that
can be correct, fitting may go that way):

If both k and q are positive then the limit is finite (I think), 
and for k and q both negative one can get complex values (k=-2, q=-1/2),
if k < 0 and 0 < q then ... etc.

Also it looks as if the data have 'derivative = 0' at the left (and which
should have x=0, not x=13.2), they reach it almost flat:

  D(h)(0); simplify(%) assuming ( 0 < M, q::real, r::real);
  subs(q = r+1, %);
  `M` = solve(%, M);
  simplify(%) assuming r::real;

                                      (- 1/r)
                             3 (r + 1)
                         M = ----------------
                                   2500

This shows, that M and r = q - 1 are closely related and one would think
about the need to use some M at all (they are not independent).

After all that it would be the right moment to retry fitting again.

And I do not think it is primarily the weakness of Maple's standard
routines that it does not immediatly work for the original form (so it
is inviting enough to try it first).

That 'error' typically can occure, if functions are used which are not defined for all Reals.

A brute and easy way is log( abs(t) ) instead of log(t) or similar for (square) roots etc.