Hi all,

I've launched a maple program (.mpl) :

nohup maple -F 5_grobner.mpl > 5grobner.out 2>&1 &

on a server to whom I've access through a ssh tunnel.

my code are here :

==========================BEGIN===========================

restart;

with(LinearAlgebra);

F0:=(w*f00+v*f01)/(v+w);

F1:=(u*f10+w*f11)/(w+u);

F2:=(v*f20+u*f21)/(u+v);

T:=(u)^(3)*p0+(v)^(3)*p1+(w)^(3)*p2 + 3*u*v*(u+v)*(u*e01+v*e10) + 3*v*w*(v+w)*(v*e11+w*e20) + 3*w*u*(w+u)*(w*e21+u*e00) + 12*u*v*w*(u*F0+v*F1+w*F2);

coefList:=[p0,p1,p2,e00,e01,e10,e11,e20,e21,f00,f01,f10,f11,f20,f21];

nops(coefList);

for i from 1 to nops(coefList) do

 ind:=coefList[i];

 BF||ind:=subs(w=1-u-v, factor(coeff(T,coefList[i],1)));

end do;

for i from 1 to nops(coefList) do

 ind:=coefList[i];

 DDBF||ind:=factor(diff(BF||ind,u,v));

end do;

QR:=x-> t1*subs(u=u1,v=v1,x) + t2*subs(u=u2,v=v2,x) + t3*subs(u=u3,v=v3,x) + (t4)*subs(u=u4,v=v4,x)+t5*subs(u=u5,v=v5,x);

for i from 1 to nops(coefList) do

 ind:=coefList[i];

 eq[i]:=numer(simplify(int(DDBF||ind, u=0..1-v, v=0..1) - QR(DDBF||ind)));

end do;

sys := [seq(eq[i],i=1..nops(coefList))];

with(Groebner);

res:=Basis(sys,plex(u1,v1,t1,u2,v2,t2,u3,v3,t3,t4,t5,u5,v5));

==========================END===========================

After the program ended, the output file ends with:

......Some outputs of the program....

memory used=1047901.6MB, alloc=32244.7MB, time=147379.36

memory used=1049539.5MB, alloc=32276.7MB, time=147732.14

memory used=1051177.5MB, alloc=32308.7MB, time=148081.18

memory used=1052815.4MB, alloc=32340.7MB, time=148473.81

maple: fatal error, lost connection to kernel

===============================================

I saw some explanations about this message here https://faq.maplesoft.com/hc/en-us/articles/360019491492--Kernel-connection-has-been-lost-error-in-Maple

I guess the reason of my program is " Maple was consuming too much of the RAM or CPU during a calculation;"

But I'm not sure.

Or "Maple sensed that some loop will not terminate,"

So it means there's no solution for my program?

Does someone know what this error means?

Or how can I know the right reason that I got this message?

Thanks!

jun

Hello Everyone

I am working on an itteraative scheme. Durring this i am solving system of algebric equation using fsolve command. In the first itteration all is ok but in second itteration it also produce some extra thing which create problems. Anybode can see this please. I am much worried about it. Thanks i advance.

, where  is a complex parameter,  is a real constant and, , then how to calculate |u|^{2} on maple?

In my worksheet today my intention was to compare the least squares linear regression for three datasets as indicated, but when I right click on the output as seen in the bottom line to select the plot type, all options state there to be independant variables K[0] and K[1], where as the output displays only the variable K as I intended, which part of my code is creating this confusion for maple?

Download ask_maple.mw

Download trc.mw

Hello there, 

Here is a set

thanks

Hello there, 

Here is a set of non-linear equations:

The 'fsolve()' command was able to come up with a solution. 

Then, when 'f9n10 := V[t] * (Ix[c1] - Ix[c2]) + TrainLoad = 0;' becomes 'f9n10 := V[t] * conjugate(Ix[c1] - Ix[c2]) + TrainLoad = 0;', 

the command ('fsolve()') refused to produce a solution. 

Would you tell me how to make the 'fsolve()' command work with the 'conjugate()' operator?

Thank you, 

In Kwon Park

Download no_conjugate.mw

Hello,

I am one of the software coordinators for CUNY and was wondering how would i obtain access to download Maplesoft for the college community?

I have the following program which constructs the multiplication table, CI, for a matrix Lie algebra and evaluates the difference between CI's row dimension and its rank. The code is a little convoluted because "LieTable" formats the entries very strangely and forces incorrect rank values.

The matrix CI is constructed rather quickly (within a few seconds), and everything works well with "small" examples (up to 12 basis elements has evaluated within seconds). However, the example I've included is for a 27-dimensional Lie algebra. As I stated, CI is constructed quickly, even in larger examples, but the rank evaluation (i.e., LinearAlgebra:-Rank(CI)) has never completed for the example I've included. I let it run for about 3 hours before shutting it down.

I have an older Macbook Air which I am using to run these computations. Could this simply be an issue of not enough computing power?

I have attempted to import the matrix CI into Mathematica (to see if it was simply a limitation of Maple), but that's its own headache (reads entries of the matrix incorrectly).

Any recommendations would help. If this is an issue of computing power, I can get access to a more powerful system soon. It doesn't seem that the code itself would cause the issue, since it is not the construction of the matrix which is giving me issues, it is the evaluation of the rank. I am rather naive about Maple (and programming in general) though, so I may be wrong.

Index_and_Contact.mw

I want to do the substitution f(t) - ff(t) = epsilon for any variable t in Maple:

expand(myerror);
    2 f(x - 2 h)   f(x)   3 f(x + 3 h)   2 ff(x - 2 h)   ff(x)
  - ------------ - ---- + ------------ + ------------- + -----
        15 h       6 h        10 h           15 h         6 h

       3 ff(x + 3 h)
     - -------------
           10 h     

NULL;
myfunc := t -> f(t) - ff(t) = epsilon;
 myfunc := proc (t) options operator, arrow, function_assign;

    f(t)-ff(t) = epsilon end proc

algsubs(myfunc(t), myerror);
          2               1        3            
        - -- f(x - 2 h) - - f(x) + -- f(x + 3 h)
          15              6        10           
        ----------------------------------------
                           h                    

               2                1         3             
             - -- ff(x - 2 h) - - ff(x) + -- ff(x + 3 h)
               15               6         10            
           - -------------------------------------------
                                  h                     

NULL;
subs(f(-h*n + x) = 1, ff(-h*n + x) = 0, f(x) = 1, ff(x) = 0, f(h*m + x) = 1, ff(h*m + x) = 0, myerror)*epsilon;
                           4 epsilon
                           ---------
                             15 h   

How do I change equations from the 2D-output into definitions?

#! Change the order (this case second):
solve({-alpha[-1] + 2*alpha[2] = 1, 1/2*alpha[-1]*h + 2*alpha[2]*h = 0, alpha[-1]/h + alpha[0]/h + alpha[2]/h = 0}, {alpha[a], alpha[b], alpha[c]});
          /            -2             1             1\
         { alpha[-1] = --, alpha[0] = -, alpha[2] = - }
          \            3              2             6/

NULL;
lhs({alpha[-1] = -2/3, alpha[0] = 1/2, alpha[2] = 1/6}[1]) := rhs({alpha[-1] = -2/3, alpha[0] = 1/2, alpha[2] = 1/6}[1]);
     /            -2\       / /            -2             1  
  lhs|alpha[-1] = --| := rhs|{ alpha[-1] = --, alpha[0] = -,
     \            3 /       \ \            3              2  

               1\    \
    alpha[2] = - }[1]|
               6/    /

Why am I not able to replace O(h^3), and in what other way can I achieve the result (I want to get the coefficients of the f(x), D(f)(x), and D^(2)(f)(x) variables).

Q(h);
1 /          /                                       2    / 3\\
- |alpha[-2] \f(x) - 2 D(f)(x) h + 2 @@(D, 2)(f)(x) h  + O\h //
h \                                                            

   + alpha[0] f(x)

              /                     9                 2    / 3\\\
   + alpha[3] |f(x) + 3 D(f)(x) h + - @@(D, 2)(f)(x) h  + O\h /||
              \                     2                          //

NULL;

subs(f(x) = 1, D(f)(x) = 0, (D@@2)(f)(x) = 0, O(h^3) = 0, collect(Q(h), f(x)));
                         /     / 3\\            /     / 3\\
    alpha[0]   alpha[-2] \1 + O\h // + alpha[3] \1 + O\h //
    -------- + --------------------------------------------
       h                            h                      

How do I combine greek letters and latin letters in variable names? e.g. here I want to combine the greek letter Delta and the latin letter x to be used as one variable name:

Delta*x;
                            Delta x

Deltax;
                             Deltax

The original question (crossed through below) was too vague, so I tried to clarified it:

I have the following expression
Q(h);
 1 /  2                    1                 2    / 3\   1     
 - |- - f(x) - D(f)(x) h + - @@(D, 2)(f)(x) h  + O\h / + - f(x)
 h \  3                    2                             2     

      1                                        2    / 3\\
    + - f(x) + 2 D(f)(x) h + 2 @@(D, 2)(f)(x) h  + O\h /|
      6                                                 /

It doesn't expand this following way...
expand(Q(h));
      /                   1                 2    / 3\\       
    2 |f(x) - D(f)(x) h + - @@(D, 2)(f)(x) h  + O\h /|       
      \                   2                          /   f(x)
  - -------------------------------------------------- + ----
                           3 h                           2 h

                                              2    / 3\
       f(x) + 2 D(f)(x) h + 2 @@(D, 2)(f)(x) h  + O\h /
     + ------------------------------------------------
                             6 h                       

But it does expand when I add the multiplication symbols manually:
expand(2*(f(x) - D(f)(x)*h + 1/2*(D@@2)(f)(x)*h^2 + O(h^3))/(3*h) + f(x)/(2*h) + (f(x) + 2*D(f)(x)*h + 2*(D@@2)(f)(x)*h^2 + O(h^3))/(6*h));
                                                    / 3\
       4 f(x)   1           2                    5 O\h /
       ------ - - D(f)(x) + - h @@(D, 2)(f)(x) + -------
        3 h     3           3                      6 h  

How can I expand my output without having to add the multiplication symbols?

------------------------------------old question below, please disregard----------------------

I found there is assume=[...] syntax. So I changed my code to use it, instead of .... assuming 

Now I find that some tests fail.  Should assuming and assume not give same result?

restart;
sol :=y(x)^(1/2) = -1+2*exp(x):
ode:=diff(y(x),x)-2*y(x) = 2*y(x)^(1/2):

odetest(sol,ode) assuming x::real, x>0

gives 0

but

odetest(sol,ode,assume=[x::real, x >0])

does not give zero. Please see screen shot below.

Reason I changed, since I started saving assumptions to use in a list. So it was easier to use assume=[....] than assuming ...., since assuming wants expression sequence in front of it and I did not know how to convert the list of assumptions I have collected to an expression sequence on the fly to use assuming, otherwise I would not have changed the code.

But my question is: Should both give same result? Why is the result different?

Help says

The assume routine sets variable properties and relationships between variables. Similar functionality is provided by the assuming command.

I am probably not using assume= correctly.   I thought I could use the same thing I had on assuming, just put it in a list.