Thank you :-)

Using sequences L:=NULL, then L:=L,tmp and finally calling {L} should do it
for older Maple versions as well, you may try it for x^11 + y^11 + z^11 = 11.

I would not mind to get fooled (grrrrrrrrrrrrrrrr ...): do you want to say, that there is no solution?

I would not mind to get fooled (grrrrrrrrrrrrrrrr ...): do you want to say, that there is no solution?

At least a sketch for modulo 2013 = 3*11*61 (my script is quite ugly, I would
have to clean it up for uploading).

The idea is to find x,y,z it for every factor and then use 'fundamental Thm for
abelian groups' (where I later have seen, that the command 'chrem' will do it)
and finally add the periods with nx,ny,nz (binomial theorem).

Do it for p=61, for example. 

First note that Maple is not able to reduce x^2013 mod p using little Fermat,
but doing it by hand: x^2013 = (x^p)^(3*11) = x^33 or one does it as code by
subs(2013 = 2013/p, equation).

[Later I saw for p = 3 or 11 one should have done better ...].

Then modulo p the equation writes as 4*x^33+2*y^33+41*z^33+2 +61*k, where some
integer variable k is introduced (and vanishes mod p), since isolve returns NO
solution without k.

Now isolve returns a solution. It is not quite in the integers, but staring at
it one sees, that the parameters for should by at least multiplied by p^2 in
order to have x,y,z as integers: this is, because there is division by 3721 =
61*61 in the result of isolve.

Crosschecking (with according k) shows, that this is not enough, so try 3rd power.
That does and allows proper calculation modulo p.

For p=61 one gets x = 0,y = 4,z = 8 modulo p.

The route is the same for p=3 and 11.

At least a sketch for modulo 2013 = 3*11*61 (my script is quite ugly, I would
have to clean it up for uploading).

The idea is to find x,y,z it for every factor and then use 'fundamental Thm for
abelian groups' (where I later have seen, that the command 'chrem' will do it)
and finally add the periods with nx,ny,nz (binomial theorem).

Do it for p=61, for example. 

First note that Maple is not able to reduce x^2013 mod p using little Fermat,
but doing it by hand: x^2013 = (x^p)^(3*11) = x^33 or one does it as code by
subs(2013 = 2013/p, equation).

[Later I saw for p = 3 or 11 one should have done better ...].

Then modulo p the equation writes as 4*x^33+2*y^33+41*z^33+2 +61*k, where some
integer variable k is introduced (and vanishes mod p), since isolve returns NO
solution without k.

Now isolve returns a solution. It is not quite in the integers, but staring at
it one sees, that the parameters for should by at least multiplied by p^2 in
order to have x,y,z as integers: this is, because there is division by 3721 =
61*61 in the result of isolve.

Crosschecking (with according k) shows, that this is not enough, so try 3rd power.
That does and allows proper calculation modulo p.

For p=61 one gets x = 0,y = 4,z = 8 modulo p.

The route is the same for p=3 and 11.

I do not even get to that error, since I start with(LinearAlgebra); with(Finance);

Then I get "Error, (in Matrix) unable to store 'd^(1/2)' when datatype=float[8]"
regarding the definition of alpha.

Which seems to tell: you need numerical values (the help is not clear on that).

If I choose d as 3.0 (do not forget to write as decimal !), then it runs until
the error message you reported.

Checking the help: "ExpectedValue(expression, opts) ... All stochastic variables
involved in expression should be of the form X(t) ... "

Yours is of the form alpha1(0)[1]*alpha1(0)[1] = _W3(0)[1]^2

That's a purely formal observation: I am not used to the language of processes,
but may be it helps you, at least a bit.

And as far as I understood the help the expectation comes through a numerical
simulation. That would need a numerical value for sqrt(d), of course.

So may be you have to give a formula for the expected value in (i,j) for your
matrix. Isn't there something in literature for that?

Roughly the recursion must have integer steps and an initial value.

So one would need to 'rescale' time by 1000, i.e. the old unit time now is 1000
steps of 1 (and not 1000 steps of 0.001), and providing some value in time = 0.

But as already said: I do not think this leads to a solution: even if you have
done that you first want to check what happens in simple cases, say for just
integrating in dimension = 1 (instead of dim=12) and see if it makes sense.

For that my feeling (looking at p1) is you have a density in dim=4 and you want
to integrate symbolically.

If it would do with success in dim=1 (for the pdf, choose mean and std deviation
in a proper way as being Reals) then you can try dim=2.

Which would be a bi-variate density. Maple does not "know" it (even if involved
correlations are non-degenerated), it is not a "known function" for Maple and
has to be done numerically, in general.


But all that is the very task: your noise is numerical, no? So it is very likely
that it makes good sense to filter it with a numerical routine. And not with a
symbolic approach.


Where one should be aware, that the symbolic solution for a 12-dim integral may
give a combinatorical explosion (too much time needed and solution too lengthy).


So it would be better to look for tested (!) code, which already implemented the
filter in a numerical way.

Roughly the recursion must have integer steps and an initial value.

So one would need to 'rescale' time by 1000, i.e. the old unit time now is 1000
steps of 1 (and not 1000 steps of 0.001), and providing some value in time = 0.

But as already said: I do not think this leads to a solution: even if you have
done that you first want to check what happens in simple cases, say for just
integrating in dimension = 1 (instead of dim=12) and see if it makes sense.

For that my feeling (looking at p1) is you have a density in dim=4 and you want
to integrate symbolically.

If it would do with success in dim=1 (for the pdf, choose mean and std deviation
in a proper way as being Reals) then you can try dim=2.

Which would be a bi-variate density. Maple does not "know" it (even if involved
correlations are non-degenerated), it is not a "known function" for Maple and
has to be done numerically, in general.


But all that is the very task: your noise is numerical, no? So it is very likely
that it makes good sense to filter it with a numerical routine. And not with a
symbolic approach.


Where one should be aware, that the symbolic solution for a 12-dim integral may
give a combinatorical explosion (too much time needed and solution too lengthy).


So it would be better to look for tested (!) code, which already implemented the
filter in a numerical way.

why not entering it in the second form?

which also gives the chance to state more clearly where some and or some or is meant
and will give Maple the chance to simplify your input?

No, Maple will do the recursion using integers - as far as I know.

Even if you switch to dNew and some scaling by 1000 (in your example)
my feeling is: that is the wrong approach.

Looking into the given Wikipedia link at keywords / references I would
go along that route, it's a numerical problem.

May be you can find something more at engineering sources.

No, Maple will do the recursion using integers - as far as I know.

Even if you switch to dNew and some scaling by 1000 (in your example)
my feeling is: that is the wrong approach.

Looking into the given Wikipedia link at keywords / references I would
go along that route, it's a numerical problem.

May be you can find something more at engineering sources.

d should be an integer, else I am afraid one can not write it that way

Upper Case: please check manual, it is writing an integral down,
without evaluating it (just to 'see' it).

Quotation marks more or less do the same, if you have any other
(composed) function.

May be you want to explain in words, what this intends.
Or what is the task given in your project / assignment.

PS: If you use fixed fonts please be aware, that you have to
break lines manually in such a way, that they are displayed
(=care for smaller screens, there will be no automatic through
HTML, that's intrinsic to fixed size)

d should be an integer, else I am afraid one can not write it that way

Upper Case: please check manual, it is writing an integral down,
without evaluating it (just to 'see' it).

Quotation marks more or less do the same, if you have any other
(composed) function.

May be you want to explain in words, what this intends.
Or what is the task given in your project / assignment.

PS: If you use fixed fonts please be aware, that you have to
break lines manually in such a way, that they are displayed
(=care for smaller screens, there will be no automatic through
HTML, that's intrinsic to fixed size)

If you write(note quotation marks and UperCase Int))

p3:=Int(subs(seq('U'(k-d)[i]=W[i],i=1..12),'p1'(k)*p0(k-1)),
        seq(W[i]=-infinity..infinity,i=1..12));

then you see, that Maple understands so far: integrate p1(k)*p0(k-1) 
over IR^12, where the variables are w1 ... w12.

Now p1 uses U (and some non-defined mu1, sigma1 to locate the Gaussian).

But looking at that and into your procedure U: what is d ?
I think that is a reason, why Maple can not resolve the
recursive call. Hope I got you right.

Edited to add: where I do not see any of the integration variables w,
so at least p1(k) is constant w.r.t. to each of the w, no ?

If you write(note quotation marks and UperCase Int))

p3:=Int(subs(seq('U'(k-d)[i]=W[i],i=1..12),'p1'(k)*p0(k-1)),
        seq(W[i]=-infinity..infinity,i=1..12));

then you see, that Maple understands so far: integrate p1(k)*p0(k-1) 
over IR^12, where the variables are w1 ... w12.

Now p1 uses U (and some non-defined mu1, sigma1 to locate the Gaussian).

But looking at that and into your procedure U: what is d ?
I think that is a reason, why Maple can not resolve the
recursive call. Hope I got you right.

Edited to add: where I do not see any of the integration variables w,
so at least p1(k) is constant w.r.t. to each of the w, no ?