I am trying to use Compiler:-Compile with a generated function having a Vector as an argument. My function is (in a simplified form)

Mapt:=proc (pv) options operator, arrow; rtable(1 .. 6, {1 = .998026692438229*pv[1]-.605526643146912*pv[2]+0.154549004943741e-1*pv[6]+0.140818440886566e-5*pv[5], 2 = 0.651134546344795e-2*pv[1]+.998026692438228*pv[2]-0.510903331894809e-1*pv[6]+0.229399145141077e-8*pv[5], 3 = -.500000000000002*pv[3]-5.95856898901206*pv[4], 4 = .125869147673385*pv[3]-.499999999999996*pv[4], 5 = 0.510903331894806e-1*pv[1]-0.154549004943901e-1*pv[2]-40.6778361021594*pv[6]+.999998210424947*pv[5], 6 = .999998210424947*pv[6]-0.229399145141220e-8*pv[1]-0.140818440886565e-5*pv[2]+0.897571581395136e-7*pv[5]}, datatype = anything, subtype = Vector[column], storage = rectangular, order = Fortran_order) end proc;

Since this returns a Vector it cannot be compiled as is. So I split off the first output element (with the idea to later on do this with the others as well):

xpr:=eval(Mapt(<pv[1],pv[2],pv[3],pv[4],pv[5],pv[6]>))[1];
f1:=unapply(xpr,pv);
f1(<0.001,0.001,0,0,0,0>); # test f1: works
          0.000392500049291317

sf1:=Compiler:-Compile(f1); # this works, too

sf1(<0.001,0.001,0,0,0,0>); # but this fails
Error, (in sf1) invalid input: expecting a hardware float[8] rtable, but received 1


The examples in the Helpfile run so I think my installation is ok (my old Maple 15 won't even run the examples, but that is a different story). I have been banging my head against the wall here, trying many different ways, but none of them successfull. Worse, I do not understand the error message at all. Test program attached.

Mac OS X 10.10.5

Thanks for any hint,

M.D.

Compiler.mw

I tried to make a procedure that would find the determinant of any 3x3 matrix but I keep getting unterminated procedure what should I change??

Code:
DeterminantMat:=proc(matA::Array)
    local  i::integer,  j::integer,  x::integer,  y::integer,  A::integer,  B::integer,  indice::integer,  S::integer:
      A:=0:  B:=0:  S:=0: i:= 1:  j:=1:  x:=1:  y := 1:  indice:=1:

for x from 1 to 3 do
   indice := 1:  
   for i  from 1 to  3 do  
       if x = i then
           end do:
        else if x <>i then
           for j from 1 to 3 do
                  if y = j then  end do:  
                  else if indice = 1 then  A := matA(i,j):  indice := 2:  
                     else if indice = 2 then  B:=matA(i,j):  indice :=3:    
                       else if indice = 3 then  B :=B * matA(i,j):  indice :=4:    
                         else if indice = 4 then  A := A * matA(i,j):      
                             if x = 2 then  S := S + (-1 * matA(x,y)) * (A-B):
                               else  S := S + matA(x,y) * (A-B):   end if:   end if:
 end do:
 end if:
 end do:
 end do:
 end proc:

hi.i encounter with another error for solving couple equations.

bcs := {f1(0) = 0, f1(L) = 0, f2(0) = 0, f2(L) = 0, f3(0) = 0, f3(L) = 0, ((D@@1)(f2))(0) = 0, ((D@@1)(f2))(L) = 0, ((D@@1)(f3))(0) = 0, ((D@@1)(f3))(L) = 0}; sys := subs(omega^2 = omega2, {PDE[111], PDE[222], PDE[333]}); sys2 := PDEtools:-dchange({x = L*y, f1(x) = g1(y), f2(x) = g2(y), f3(x) = g3(y)}, sys, [g1, g2, g3, y]); indets(sys2, specfunc(diff)); solve(sys2, {diff(g3(y), `$`(y, 4)), diff(g1(y), y, y, y), diff(g2(y), y, y, y, y)}); sys3 := subs(omega2 = 10^19*omega3, sys2); bcs3 := {g1(0) = 0, g1(1) = 0, g2(0) = 0, g2(1) = 0, g3(0) = 0, g3(1) = 0, ((D@@1)(g2))(0) = 0, ((D@@1)(g2))(1) = 0, ((D@@1)(g3))(0) = 0, ((D@@1)(g3))(1) = 0}; extra_bcs := `minus`({seq(seq(((D@@i)(g1))(a), i = 0 .. 1), a = 0 .. 1), seq(seq(((D@@i)(g2))(a), i = 0 .. 1), a = 0 .. 1), seq(seq(((D@@i)(g3))(a), i = 0 .. 3), a = 0 .. 1)}, `~`[lhs](bcs3)); dsys4 := {bcs3, sys3}; ds := dsys4[1]; indets(dsys4[1], specfunc(diff)); solve(dsys4[1], {diff(g3(y), `$`(y, 4)), diff(g1(y), y, y, y), diff(g2(y), y, y, y, y)}); newsys := {ds[2], ds[3], diff(ds[1], y)}; indets(newsys, specfunc(diff)); S := solve(newsys, {diff(g1(y), `$`(y, 3)), diff(g2(y), `$`(y, 4)), diff(g3(y), `$`(y, 4))}); nops(%); bcs2 := eval[recurse](convert(ds[1], D), `union`({y = 0}, bcs3)); nops(`union`(bcs3, {bcs2}));

Error, (in simpl/eval) numeric exception: division by zero

division_by_zero.mw

please help.thanks..

hi.how i can allocate infinite value for a parameter such as N ,which is attached below ( N := infinite) .i encounter with error.please see it and help

thanks..

infinite.mw

Hi!

For some reason my three equation with three unknowns won't be solved. Anyone who can help? I tried the following:

Really appreciate anyone who can help me out!

I an experimenting with Groebner basis. Have a set 17 equations. There are 12 unknowns. Using 14 of the equations and tdeg, the system produced 65 equations. How do I get a list of the monomials in this new list?

how to do optimization for two equations in terms of two variables

LPSolve({eq1}, {eq2}, assume = {nonnegative});

eq1 is a rational function and eq2 is a very large rational function

after run , it return error, objective function must be specified as a linear polynomial or vector

da := [LengthSplit(Flatten([[1,m],[2,m2],[seq([i+1,close3[i][1]], i=2..4)]]),2)];
f := PolynomialInterpolation(da, z):
solution := solve(f=-z, z, explicit);
zz := [x1,x2,x3,x4];
sigma := symMonomial(zz); #sigma := [x1+x2+x3+x4, x1*x2+x1*x3+x1*x4+x2*x3+x2*x4+x3*x4, x1*x2*x3+x1*x2*x4+x1*x3*x4+x2*x3*x4, x1*x2*x3*x4]
sys1 := subs([x1=solution[1],x2=solution[2],x3=solution[3],x4=solution[4]], sigma[1]):
sys2 := subs([x1=solution[1],x2=solution[2],x3=solution[3],x4=solution[4]], sigma[2]):
sys3 := subs([x1=solution[1],x2=solution[2],x3=solution[3],x4=solution[4]], sigma[3]):
sys4 := subs([x1=solution[1],x2=solution[2],x3=solution[3],x4=solution[4]], sigma[4]):
da := [seq([i,close3[i][1]], i=1..5)];

with(Optimization):
LPSolve(sys1, {sys2}, assume = {nonnegative});

Hi all,

I would like to plot a function U(x,t) in an interval [-2,0] such that

U(x,t)=u0(z)+t*u1(z) where  z=x/t,

u0(z)=z^2/3+sqrt(-z) and

diff(u1(z), z)=9/4*(84*(-z)^(11/2)-8*z^7+540*(-z)^(5/2)+324*z^4-324*z)*u1(z)/((z^2+3*sqrt(-z))*(2*(-z)^(3/2)+3)*(8*z^6+66*(-z)^(9/2)-189*z^3+216*(-z)^(3/2)+81))

I used numeric method rkf45 to solve and plot u1(z) with initial condition u1(-2)=1.

I can also use dchange to change variable z to x as variable and t as parameter.

But, I have difficullty to go back to U(x,t).

If someone have any idea.

Thanks in advance.

how to further remove _Z with explicit in two variables case?

da := [[1,m],[2,m2],[3,1],[4,2],[5,3],[6,4]];
f := PolynomialInterpolation(da, z):
solution := solve(f=-z, z, explicit);

Assume we have a sreach with 10 for, for example

answer=0:

for i[1] from 1 by 1 to 5 do

    for i[2] from 1 by 1 to 5 do 

        ...

        if .... then answer=1: print(i[1],...,i[10]); "quitting the search"

        ...

     end do

end do

If I write break at qutting part, it will only exit from one for, one idea is putting if answer=1 break before end do of the rest for, that is why I used the local extra variable "answer" but is this the best idea? Any better idea which contains checking less if is apprecied.

Tengo una figura obtenida con ScatterPlot3D a partir de una Matriz de 3 dimensiones ( X,Y,Z ). ¿ Cómo puedo obtener la función polinómica z(x,y) que reprenta la superficie que se obtiene en el ScatterPlot3D ?

Is there a library to draw state machine diagram from logic table in Maple?

Hello,

I am trying to plot 3d scattered data as a 2d scatterplot with the third dimension being displayed as colour or a line that separates the value of the two functions (for example, the higher value and lower value at that point). I attached the file. I have two functions and several parameters that are constant. Here, I want to use P[A] on the x-axis, and P[B] on the y-axis. The z dimension is the profit functions, either Profit[1] or Profit [2]. I want to show them in colors like the highest one is red and the lower one is green, or something similar.  P[A] and P[B] are probability values and P[B] is between 0 to 0.50 with 0.01 increments, while P[A] is always less than P[B]. The file is here: example_file.mw

Example;

P[A] := 0 ..P[B]-0.01

P[B] := 0 ..0.50

Anyone please help me to code this problem.Thanks,

YC

Hey

I have a simple question. 

In maple - when working with Ohm's Law.

Maple know how to calculate with e, but how can I show the result

I have calculated the following calculation:

(2e-4)/2

Maple returns the result:0.0001000000000

How do I maple show the result as 1e-4?