Axel Vogt

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20 years, 228 days
Munich, Bavaria, Germany

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These are replies submitted by Axel Vogt

I got the 'same' 10 solutions (for the example) considering that as a
system of polynomial equations (in degree 2 and 3), sorting out the
problem caused by the original denominators.

I am not sure that it is a bug, it may be caused by numerical issues as
well and suggest to use convert/rational for the data (and evalf[100]
to look at the result).

I got the following 'structure' for the solution (fE2=x, fT=y, fDHT=z)

data := [KsT = 10000000000, KaT = 460000, KsE2 = 3140000000,
         KaE2 = 421000, KsDHT = 300000, KaDHT = 350000];
params:=[E2=1, T=2, DHT=3, Ca=4, Cshbg=5];

y = root of a polynomial of degree 10 (8 real and 2 complex roots,
abs <= 12, some are quite close to zero),

and x, z = polynomial of degree 9 in y with rational coefficients,
those may be large ( ~ 10 ^10 )

I liked that. And have a minor question: why Maple uses 1 (integer) instead of 1.0 (float)? Though I guess (!) that a compiler would optimize that away to avoid type conversion ...

1) Because of symmetry, just plot it to get the idea (choosing some upper). And there you can solve sin(phi) = x for phi, in case Maple makes an error that will help to correct it. I forgot to post that, sorry:

'eval(A1, phi=Pi/2 - alpha) = eval(A1, phi=Pi/2 + alpha)'; is(%);

                                 true

2) Just fill in (by hand = paper + pencil) and you see, what cancels out. You have to make sure that you do not divide by zero and check those cases separately

yes, and I think that carries over to any n, it is of form exp(x) = const <> 0, if y is fixed

hm ... Int(%, x = 0 .. l); is what you write - why you expect an x in the answer?

I use Maple 18.

Edited: likewise

K; # = the integrand
[op(%)]; # the summands
map('q -> Int(q, x = 0 .. l)', %); # for each summand
value(%); # find the values
convert(%, `+`); # add them
evalf(%);

I know. But it solves the problem, no?

To have it more simple I would solve the 1st linear eq for x and feed
it to the other. Then solve the 2nd for z (in terms of y, conjugate(y)),
to be feed into the 3rd.

This gives a a compact solution for x and z and it remains to solve the last.

That is a linear system over R^2 (as Markiyan 'says').

For specific values that might be very short (see above, acer)

@Markiyan Hirnyk I can believe that (i.e. it uses double precision).

But if I read "But in practice, I don't have the time to do that. And I dont think I have the knowledge either." by the poster then my interest becomes exhausted.

There are some issues with the task, even in its re-formulated version.

There is no 'rank' per se, as we talk about a matrix over a ring and not
over a field. In the very case one can do it (for example looking at the
quotient field ~ rational functions), like rank = rank Image(linear).
But over rings the cokernel is not free (like in linear algebra) though
it can be healed here (rank over integral domains).

However one can prove that the cokernel is free except an algebraic set,
hence an analytic meager set, measure = 0.

Outside it is constant and maximal. Shooting (exact) numbers means to hit
that open set ("generic") and shooting again for modulo is a second step.

So from Math I think it is not quite clear what the poster wants to be
done by Maple in a justified manner. Actually it is not a Maple task to
make the it a precise question, I would say.

Note that for this case 'it does not matter' whether using analytic or
algebraic approaches.

I think that Carl Love meant: a generic (closed) point should do (using
exact, symbolic values) - an integer - and then to proceed from there by
modulo p, using enough primes.
Sorry, if I am too rusty.
Delta[epsilon]:= (T,z) -> sin(T*z);
'Delta[epsilon](T,z)': '%'=%;

One can divide out 1 factor of r in the DE. Then one needs that BesselY does not appear (this is what makes problems in r=0), so I would set its _C=0.  Now using that for diff(res1, r) in r=0 leads to -C1/G3 = 0 and setting C1 in the initial task works (I think). So for r=0 that C1 = zero or not is the 'obstruction'. Hm ?

@radzys 

If you use view = 0 .. 30 (or omit it at all) in the above then you may guess what I mean, the zeros are 'spurious' [and there is the ugly scaling by -1/500000*EI*2^(1/4)*k_r*k_h ~ 1E28]. I have no reasonable idea how to cut off the high values off omega ~ zero.

And no, it is unlikely that an analytic solution exists: though omega enters M in a 'polynomial way' it also enters through beta into the trigonometric functions (which are periodic. I have not tried to look deeper for symmetries, only looked for x = -x, but that would not help much.

Thanks for the correction.

First: please do not modify initial posts, just add modifications
or explanations.


I think the behaviour for the rank is correct (have not checked which
algorithm is used): it is a morphism between free modules.

Here is a basic sketch for you desire to use Groebner, starting with
your f: C^n -> C^p. You need to write down the according morphism of
algebras F: C_p -> C_n, C_p, C_n = polynomial ring in p or n variables.
This is q |-> q°f, if you 'read' a polynomial as function.

Define J:= kernel(F). This is a prime ideal: given a,b with a*b in J
you have 0  = F(a*b) = F(a)*F(b), so one of it must be zero, i.e. one
of the elements is in J. Especially it is irreducible and the dimension
is no problem of components.

Now C_p/J as factor algebra defines an irreducible subspace in C^p and
that should be the image of f, I would say (at least set-theoretic it
should not be too hard).

Then you want the dimension of J, using Groebner. May be the package
finds it. However the main task before will be to find the ideal.
Enough said.

You can, but I do not know an easy way (becoming rusty on such), it is an exercise in Hartshorne, Chap II 3 11

Hartshorne, image of a morphism

 

PS: you may look for related stuff through http://math.stackexchange.com/questions/283397/the-image-of-a-morphism-between-affine-algebraic-varieties and links from there (which may need to describe yor morphism at the level of algebras, as already said)

PPS: for dimension I would try Ex. 3.22, the morphism is dominant w.r.t. to the image (as it is defined as the closure of the set theoretic image). Now that excercise reduces it to study the dimension of fibres in generic (=open) cases.

PPPS: there should be some algebraic stuff for rank through exterior algebras, I forgot it (like det = scalar on induced exterior product in degree = dimension)

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