Your expressions have multiple (at least two) different assumed names that print as p~.

indets(Q[6][2,3],name);

                           {p~, p~, x}

Those do not combine or cancel under `simplify` (which is usual).

I have not yet looked to see where they arise. It may be that you could use `additionally` rather than repeated calls to `assume`. Or you might be able to proceed with something crude like,

simplify(convert(Q[6][2,3],`global`));

[update]

You just have to move the line which calls `assume(p,real)` out of the P_chain procedure. It can be called just once at the top-level (since your P_chain already declares p as global), at before the loop. Here is an edited version,

June_18_modif.mw

acer

With Maple 18 one could try and use the new InertForm package for this.

The dot appearing in the imaginary term looks a little heavy to me. (It shows in the Std GUI, but not the TTY).

One nice thing is that the extra brackets don't appear, when done in the GUI.

For example,

  restart:

  c:=a+b*I:
  expr:=c^2:

  U := `%+`(evalc(Re(expr)),evalc(I*Im(expr))):

  with(InertForm):

  Display(U);

  Display(U, inert=false);

  Value(U);

acer

Are you using deprecated matrix rather than Matrix, and if so why?

Are you using Maple 18 and have you declared D as a top level local? If you are using name D at the top level (ie. this code is not appearing inside any procedure of your own) then perhaps it would simplify your issue if you switched that to say DD. 

How about something simpler such as this, if the Q[i] are all Matrices,

DD := add( evalf( (1/3)*Q[i]*(z(i+1)^3-z(i)^3) ), i=1..4 );

Note that even if you kept the loop, rather than use add, then you would not need evalm if you used Matrix instead of matrix.

If you still get problems with the diagonal elements, then perhaps you code upload the code so that we could investigate directly.

acer

I don't yet see a way to get an exact chatcterization of all the roots.

You may be able to obtain float approximations to roots. It may help to scale the expression (according to its magnitude on the requested range? Unclear to me.)

restart:

expr:=-9999990000000000000000*cos(166*beta)*sinh(166*beta)*cosh(88*beta)^2
 -9999990000000000000000*cos(88*beta)^2*sin(166*beta)*cosh(166*beta)
 +9999990000000000000000*sinh(166*beta)*cos(88*beta)^2*cos(166*beta)
 +9999990000000000000000*cosh(88*beta)^2*cosh(166*beta)*sin(166*beta)
 +10000010000000000000000*cos(166*beta)*sinh(166*beta)
 +10000010000000000000000*sin(166*beta)*cosh(166*beta)
 +9999990000000000000000*sinh(88*beta)*cos(166*beta)*cosh(166*beta)*cosh(88*beta)
 -9999990000000000000000*sinh(88*beta)*sin(166*beta)*sinh(166*beta)*cosh(88*beta)
 +9999990000000000000000*sin(88*beta)*cos(88*beta)*sinh(166*beta)*sin(166*beta)
 +9999990000000000000000*cos(88*beta)*cos(166*beta)*sin(88*beta)*cosh(166*beta)
 -9980010000000000000000*cosh(88*beta)^2*sinh(166*beta)*cos(88*beta)^2*cos(166*beta)
 -9980010000000000000000*cosh(88*beta)^2*cosh(166*beta)*sin(166*beta)
 *cos(88*beta)^2
 +9980010000000000000000*sinh(88*beta)*cos(88*beta)^2*sin(166*beta)
 *sinh(166*beta)*cosh(88*beta)
 -9980010000000000000000*cos(88*beta)*cosh(88*beta)^2*sin(88*beta)
 *sin(166*beta)*sinh(166*beta)
 +9980010000000000000000*sinh(88*beta)*cosh(88*beta)*cosh(166*beta)
 *cos(88*beta)^2*cos(166*beta)
 +9980010000000000000000*cosh(88*beta)^2*cosh(166*beta)*cos(88*beta)
 *sin(88*beta)*cos(166*beta)-9980010000000000000000*cos(88*beta)
 *sinh(88*beta)*cos(166*beta)*sin(88*beta)*sinh(166*beta)*cosh(88*beta)
 +9980010000000000000000*cos(88*beta)*cosh(88*beta)*sin(88*beta)
 *sin(166*beta)*cosh(166*beta)*sinh(88*beta):

#Digits:=500:
plot(expr,beta=0.0..0.2, view=0..1e-100);
#Digits:=10:

findroots:=proc(expr,a,b,{guard::posint:=5,maxtries::posint:=50})
local F,x,sols,i,res,start,t;
   x:=indets(expr,name) minus {constants};
   if nops(x)>1 then error "too many indeterminates"; end if;
   F:=subs(__F=unapply(expr,x[1]),__G=guard,proc(t)
      Digits:=Digits+__G;
      __F(t);
   end proc);
   sols,i,start:=table([]),0,a;
   to maxtries do
      i:=i+1;
      res:=RootFinding:-NextZero(F,start,
                                 'maxdistance'=b-start);
      if type(res,numeric) then
         sols[i]:=fnormal(res);
         if sols[i]=sols[i-1] then
            start:=sols[i]+1.0*10^(-Digits);
            i:=i-1;
         else
            start:=sols[i];
         end if;
      else
         break;
      end if;
   end do;
   op({entries(sols,'nolist')});
end proc:

findroots(expr, 0, 0.2); # missing some hinted at by the plot

  0.01641624893, 0.03039949207, 0.05572565081, 0.06691672633, 0.1327837617


findroots(expr*1e-90, 0, 0.2); # agreeing with the plot

 0.01641624893, 0.03039949207, 0.05572565081, 0.06691672633, 0.09465903392, 

   0.1039052938, 0.1327837505, 0.1327837510, 0.1417059480, 0.1700140953, 

   0.1804062466

acer

expr1 := p+p^(-1/(theta-1))*sum(q[i]^(theta/(theta-1)),i=(1..n));

                                         /  n                  \
                           /      1    \ |-----     /  theta  \|
                           |- ---------| | \        |---------||
                           \  theta - 1/ |  )       \theta - 1/|
             expr1 := p + p              | /    q[i]           |
                                         |-----                |
                                         \i = 1                /

expr2 := p^(-1/(theta-1))*(p^(theta/(theta-1))
         +sum(q[i]^(theta/(theta-1)),i=1..n));

                              /               /  n                  \\
                /      1    \ | /  theta  \   |-----     /  theta  \||
                |- ---------| | |---------|   | \        |---------|||
                \  theta - 1/ | \theta - 1/   |  )       \theta - 1/||
      expr2 := p              |p            + | /    q[i]           ||
                              |               |-----                ||
                              \               \i = 1                //

H:=(ee,a) -> a*simplify(1/a*ee):

H(expr1, p^(-1/(theta-1)));

                          /               /  n                  \\
            /      1    \ | /  theta  \   |-----     /  theta  \||
            |- ---------| | |---------|   | \        |---------|||
            \  theta - 1/ | \theta - 1/   |  )       \theta - 1/||
           p              |p            + | /    q[i]           ||
                          |               |-----                ||
                          \               \i = 1                //

Did you want to also programmaticaly "discover" that second argument to H above, the special term, perhaps as the coefficient of the Sum subexpression?

acer

Is this what you're after?

FirstList:=[A,B,C,D,E,F]:

combinat:-choose(FirstList,4);
   [[A, B, C, D], [A, B, C, E], [A, B, C, F], [A, B, D, E], 

     [A, B, D, F], [A, B, E, F], [A, C, D, E], [A, C, D, F], 

     [A, C, E, F], [A, D, E, F], [B, C, D, E], [B, C, D, F], 

     [B, C, E, F], [B, D, E, F], [C, D, E, F]]

nops(%);
                               15

acer

 Another possibility might be to use Maple's CodeGeneration[Matlab] command to produce code for a Matlab .m function from the Maple procedure `symb_sol` (then written out to a file), so that all the subsequent pieces are done right in the running Matlab session. 

We haven't been shown the equations which are solved ( via eliminate) so its not possible to tell yet whether all of `symb_sol` can be so handled.

acer

It's difficult to figure out the whole situation, as we do not seem to know what SystemEquations_NoDriver are.

But I don't under stand why you are forming all the symbolic input equations and holding BB_Driverlist_set(i) for all `i` in Maple memory at the same time.

You run (in Matlab), a first loop to store every symbolic equation set, in the BB_Driverlist_set(i). This is the step you've described as loading the control points into Maple.

Then you run another loop (in Matlab), to apply the Maple symb_sol function to every data set previously stored. Each time through the loop this produces a set of results as equations. And then you take rhs's and stuff the numeric results into the next part of a single Matrix. It's a bit dodgy about ordering of equations in a set, according to the name on the lhs, but it works.

Using a single Matrix is probably a good idea. How about a single loop which takes the input set (for a given time point), applies symb_sol, and then puts the result into the appropriate slot in the single result Matrix. And do not store or assign BB_Driverlist_set(i) for all i or for all SimNum.

And especially do not ever create any equation of the form name=expression, in a loop.

I also do not understand why you have to form the symbolic equations in sets. Perhaps I misunderstand the inputs. (Recall that we don't know what your SystemEquations_NoDriver is.)

As I understand it, you are calling from a running Matlab session into Maple to execute various steps of solving and manipulation. By "solving" you seem to mean that you generate an explicit general symbolic solution, and then evaluate this at each timestep's input data. (We'll have to take your word on this being more desriable than repeated numeric solving.)

It seems to me that you are doing something a bit like this,

pars:={ICE_T, BAT_P, EM2_T, EM2_W}:

eqs:={ICE_T=var1+cos(var2),BAT_P=sin(var1),EM2_T=var1^2,EM2_W=var3^2};

                 /                               2              2  
         eqs := { BAT_P = sin(var1), EM2_T = var1 , EM2_W = var3 , 
                 \                                                 

                                   \ 
           ICE_T = var1 + cos(var2) }
                                   / 

symb_sol:=unapply(eqs,[var1,var2,var3]):

ansMat := Matrix(nops(pars), 50000): #Matrix(m, numtimesteps):

# Loop over these next steps...

for i from 1 to 1 do

  ans := symb_sol(1.1,2.3,3.7);  print(ans);

  ansMat[..,i] := Vector([seq( rhs(ans[i]), i=1..nops(pars) )]):

end do:

  {BAT_P = 0.8912073601, EM2_T = 1.21, EM2_W = 13.69, ICE_T = 0.4337239787}

ansMat[..,1];

                               [0.8912073601]
                               [            ]
                               [        1.21]
                               [            ]
                               [       13.69]
                               [            ]
                               [0.4337239787]

How about making it a bit more like this,

pars:={ICE_T, BAT_P, EM2_T, EM2_W}:

eqs:={ICE_T=var1+cos(var2),BAT_P=sin(var1),EM2_T=var1^2,EM2_W=var3^2}:

exprs:=eval([BAT_P, EM2_T, EM2_W, ICE_T], eqs);

                     [               2      2                  ]
            exprs := [sin(var1), var1 , var3 , var1 + cos(var2)]

symb_sol:=unapply(exprs,[var1,var2,var3]):

ansMat := Matrix(nops(pars), 50000): #Matrix(m, numtimesteps):

# Loop over these next steps...

for i from 1 to 1 do

  ans := symb_sol(1.1,2.3,3.7);  print(ans);

  ansMat[..,i] := Vector(ans,datatype=float[8]):

end do:

                  [0.8912073601, 1.21, 13.69, 0.4337239787]

ansMat[..,1];

                             [0.891207360100000]
                             [                 ]
                             [ 1.21000000000000]
                             [                 ]
                             [ 13.6900000000000]
                             [                 ]
                             [0.433723978700000]

Those are just ideas for improvement. Better still might be to have no named equations, anywhere. Pass float[8] Vectors or Matrices of timestep data into Maple. Pump these directly into a procedure. (You may need help in altering your simple `unapply` into another kind of procedure...) And put the results directly into the slots of a float[8] answer Matrix. No equations, no calls to `rhs`, no sets, no lists.

How about passing a single float[8] Matrix of size numtimesteps-by-numinputs in one call from Matlab to Maple. Then make another call from Matlab to Maple which runs a Maple procedure (which you write) that evaluates the "solver" on each row of input data, and then places the results into a row of a results float[8] Matrix. Then pass the whole float[8] answer Matrix back to Maple. Ie, about one or three calls from Matlab to Maple, not 50000+.

acer

Note:

restart:             

'10.2^20';

                                              21
                               0.1485947396 10

Even special evaluation rules (to prevent premature evaluation of arguments in a procedure call, say) will not prevent automatic simplification. Having said that,

restart:

p := proc() 10.2^20; end proc;

                         p := proc() 10.2^20 end proc

evalf[50](p());
                                                             21
               0.14859473959783543420355740092833203224576 10


restart:       

evalf[50](`^`(10.2,20));

                                                             21
               0.14859473959783543420355740092833203224576 10

acer

Using the parameters and compile option of dsolve/numeric allows the same two last plots to be produced, with the overall computation time reduced from about 75 sec to 38 sec.

I also changed the 2D Math of the DE system to 1D Maple Notation. This seems to make the GUI use less resources (and respond better while editing). Your mileage may vary.

Benzene_mod.mw

acer

Note that one can generally ony display colors from what is displayable within the RGB colorspace, which is a smaller triangular region within the visible gamut of the CIE xyY colorspace (which you seem to be constructing).

But it's possible to get an approximation of the visible xyY gamut by shading the non-displayable points with colors obtained by regressing back to the edge of the displayable RGB gamut. Below, this is attempted for the given xy points (presumed as some approximation of the edge of the xyY gamut -- let me know it that's not right).

The final plot does not look fuzzy when run in my Maple 18 Standard GUI. I have used at least one command new to Maple 18 (ColorTools:-ToDisplayable), so this code wouldn't work in an earlier version.

In the followup interpolation I also added the line of purples to what I presume you intend as the spectral color edge.

Download cie_xyY_gamut.mw

A nice task would be to use Maple to produce an ImageTools image or a densityplot, with the whole gamut shaded.

acer

restart:

FF := Q-1+(1/5)*K*dp^3*h^5+(1/3)*dp*h^3+h+h1*h:
DDP:=[solve(FF,dp)]:
h:=1+phi*cos(2*Pi*x):
h1:=2*Pi*alpha*beta*phi*cos(2*Pi*x):
beta:=1:alpha:=0:
phi:=0.5:
dpdx:=evalf(DDP[1]):
dpp:=Int(Re(dpdx),x=0..1):

plot([subs(K=-0.1,dpp)],Q=-1..1,axes=box,color=[blue]);

acer

The answer is not the one you expected, but it is not completely invalid. Degrees can represent an absolute or a relative scale.

A temperature of 32.0 degC is the same as a temperture of 89.6 degF.

But if the temperature is increased by 32.0 degC then it is increased by 57.6 degF.

The context menu does the latter conversion. Perhaps it could be augmented with a way to do either.

convert( 32.0*Unit(degC), units, degF);

                57.60000000 Units:-Unit('degF')

convert( 32.0*Unit(degC), temperature, degF);

                 89.6000000 Units:-Unit('degF')

acer

Could you not transform the plots with cartesian coordinates?

Download polarde.mw

acer

eval(irem(m,4),m=21);

                               1

eval('`mod`'(m, 4), m=21);

                               1

Even if you were to obtain an unevaluated return from the call `mod`(m,4), then you'd still need an evaluation after substituting with `subs`.

acer