Good Morning,

This is the script that I have used to to an integration, as the command ''int'' didn't work because the coputation was really heavy.

p_old := 0;

g := 0; #index

E_tot_net_Point := 0;

for g to 2 do

 h := g+1;

 loopstart := SP_time[g];

 loopend := SP_time[h];

  for tt from loopstart by 0.01 to loopend do

   t := tt;

   P_tot_net_Fkt := t -> C_motore[g]*omega_motore[g]

if t≠SP_time[g] and t ≠ SP_time[h] then

 P_tot_net_Point := P_tot_net_Fkt(t);

 P_tot_net_Point_next := P_tot_net_Fkt(t+0.001);

 plt_P_tot_net_plt[t] := pointplot([t, P_tot_net_Point]);

 E_tot_net_Point := E_tot_net_Point+(P_tot_net_Point*0.01);

 plt_E_tot_net_plt[t] := pointplot([t, E_tot_net_Point]);

 p_old := P_tot_net_Point;

end if;      

  end do:

end do:

plots:-display(entries(plt_E_tot_net_plt, 'nolist'), title = "Energy", color = "Red", symbol = point, symbolsize = 20, gridlines = true, size = [1000, 500], labels = ['[s]', '[Watt*s]'], font = [axes, 12])


I have plotted the function as a series of points over the orizontal axis (t).

Now I would like to calculate the last value of the curve (which is itself a point) as a function of the orizontal axes t. How can I do it?

Something like E_tot_net_Point = f(t)

Thank you for the help

This post is my attempt to answer the question from   here : how to find all integer points (all points with integer coordinates) in the intersection of two cubes. The following procedure  IntegerPoints  solves a more general problem: it finds all the integer points of a bounded polyhedral region of arbitrary dimension, defined by a system of linear inequalities and / or equations.

Required parameters of the procedure: SN is a set or a list of linear inequalities and/or equations with any number of variables, the Var is the list of variables. The procedure returns the set of all integer points, satisfying the conditions  SN .

Code of the procedure:

restart;

IntegerPoints := proc (SN::{list, set}, Var::list)

local SN1, sn, n, Sol, k, i, s, S, R;

uses PolyhedralSets, SolveTools[Inequality];

SN1 := convert(evalf(SN), fraction);

for sn in SN1 do

if type(sn, `<`) then SN1 := subs(sn = (`<=`(op(sn))), SN1)

end if; end do;

if IsBounded(PolyhedralSet(SN1)) = false then error "The region should be bounded" end if;

n := nops(Var);

Sol := LinearMultivariateSystem(SN, Var);

if Sol = {} then return {} else

k := 0;

for s in Sol do if nops(indets(s[1])) = 1 then

S[0] := [[]];

for i to n do

S[i] := [seq(seq([op(j1), op(j2)], j2 = [isolve(eval(s[i], j1))]), j1 = S[i-1])] end do;

k := k+1; R[k] := op(S[n]);

end if; end do;

convert(R, set);

map(t->rhs~(t), %);

end if;

end proc:

Examples of use:

IntegerPoints({x > 0, y > 0, z > 0, 2*x+3*y+z < 12}, [x, y, z]);

  {[1, 1, 1], [1, 1, 2], [1, 1, 3], [1, 1, 4], [1, 1, 5], [1, 1, 6], [1, 2, 1], [1, 2, 2], [1, 2, 3], [2, 1, 1], [2, 1, 2],

                                   [2, 1, 3], [2, 1, 4], [2, 2, 1], [3, 1, 1], [3, 1, 2]}

IntegerPoints({x > 0, y > 0, z > 0, 2*x+3*y+z = 12}, [x, y, z]);

                                    {[1, 1, 7], [1, 2, 4], [1, 3, 1], [2, 1, 5], [2, 2, 2], [3, 1, 3], [4, 1, 1]}

IntegerPoints([x > 0, y > 0, z > 0, 2*x+3*y+z = 12, x+y+z <= 6], [x, y, z]);

                                                           {[1, 3, 1], [2, 2, 2], [4, 1, 1]}

isolve({x > 0, y > 0, z > 0, 2*x+3*y+z < 12});  #  isolve fails with these examples

              Warning, solutions may have been lost

isolve({x > 0, y > 0, z > 0, 2*x+3*y+z = 12});

              Warning, solutions may have been lost

In the following example (with a visualization) we find all integer point in the intersection of a square and a triangle:

S1 := {x > 0, y > 0, x < 13/2, y < 13/2}:

S2 := {y > (1/4)*x+1, y < 2*x, y+x < 12}:

S := IntegerPoints(`union`(S1, S2), [x, y]):

Region := plots[inequal](`union`(S1, S2), x = 0 .. 7, y = 0 .. 7, color = "LightGreen", nolines):

Points := plot([op(S)], style = point, color = red, symbol = solidcircle):

Square := plottools[curve]([[0, 0], [13/2, 0], [13/2, 13/2], [0, 13/2], [0, 0]], color = blue, thickness = 3):

Triangle := plottools[curve]([[4/7, 8/7], [4, 8], [44/5, 16/5], [4/7, 8/7]], color = blue, thickness = 3):

plots[display](Square, Triangle, Points, Region, scaling = constrained);

In the following example (with a visualization) we find all integer point in the intersection of two cubes. The second cube is obtained from the first cube by rotation with orthogonal matrix  A  and by a translation:

A := <1/3, 2/3, 2/3; -2/3, 2/3, -1/3; -2/3, -1/3, 2/3>:

f := unapply(A^(-1).<x+5, y-4, z-7>, x, y, z):

S1 := {x > 0, y > 0, z > 0, x < 6, y < 6, z < 6}:

S2 := eval(S1, {x = f(x, y, z)[1], y = f(x, y, z)[2], z = f(x, y, z)[3]}):

S := IntegerPoints(`union`(S1, S2), [x, y, z]);

Points := plots[pointplot3d](S, color = red, symbol = box):

Cube := plottools[cuboid]([0, 0, 0], [6, 6, 6], color = blue, linestyle = solid):

F := plottools[transform]((x, y, z)->convert(A.<x, y, z>+<-5, 4, 7>, list)):

plots[display](Cube,  F(Cube), Points, scaling = constrained, linestyle = solid, transparency = 0.7, orientation = [25, 75], axes = normal);

In the example below, all the ways to exchange $ 1 coins of 1, 5, 10, 25 and 50 cents, if the number of coins no more than 8, there is no pennies and there is at least one 50-cent coin:

IntegerPoints({x1 = 0, x2 >= 0, x3 >= 0, x4 >= 0, x5 >= 1,  x1+5*x2+10*x3+25*x4+50*x5 = 100, x1+x2+x3+x4+x5 <= 8}, [x1, x2, x3, x4, x5]);

nops(%);

                              {[0, 0, 0, 0, 2], [0, 0, 0, 2, 1], [0, 0, 5, 0, 1], [0, 1, 2, 1, 1], [0, 2, 4, 0, 1],

                                                 [0, 3, 1, 1, 1], [0, 4, 3, 0, 1], [0, 5, 0, 1, 1]}

                                                                                    8

Integer_points.mw

Addition: Below in my comments another procedure  IntegerPoints1  is presented that solves the same problem.

I assume that I'm not providing the correct input to the simplify command to get the simplification that I want.  In particular, for the following code:

assume(n, positive);
simplify(3^(-(1/2)*n)*2^((1/6)*n)-2^((2/3)*n)*6^(-(1/2)*n));
simplify(log(3^(-(1/2)*n)*2^((1/6)*n))-log(2^((2/3)*n)*6^(-(1/2)*n))); 

The expression should evaluate to 0.  However, the first expression does not simplify to 0 (it does not simplify at all in Maple) while the second expression simplifies to 0.

The simplification is fairly easy for the first expression by factoring 6 and combining terms; it seems like I'm not entering the command to simplify in this way.

After manually working out answer for problem 4-4 in Mathews & Walker's Mathematical Methods of Physics , I tried to check my solution with maple2015. Briefly the problem involves inputs periodic with period T, being transformed into outputs, through a kernal G.  The net result is that all input frequencies omega periodic in T are multiplied by (omega_0/omega)^2, except for constant frequency which transforms to zero.  The problem asks to evaluate the kernal G.

Maple2015 correctly evaluated the integral for a constant input, a cosine input, and a sine input, but gave undefined when I tried an exponential(i*x) input which is just a linear combination of the two previous inputs.  I found this interesting because the integral is finite, well defined, and only has an absolute function (in the kernal), which may cause Maple problems, as it correctly evaluated integral when I split it into two regions.  Interestingly if instead of working with a period of T, I used 2*pi, and redfined my G function accordingly, Maple evaluated the exp input integral without any problems.  So the problem appears to be with the T variable, but I correctly used assumptions of T>0, and 0<t<T, so I am not sure why it would work correctly when I use T=2*pi, but failed when using a general period T.  Any help would be welcome.

Download MathewsWalkerProblem4-4.mwMathewsWalkerProblem4-4.mw

I hoped that Maple would return the value of 1 in all commands (see below). However, introducing a scaling parameter, sigma, yields the unevaluated expression. Why? I still think it should evaluate to the value of 1.

kind regards,

Harry (not a mathematician, but a psychologist)

integral.mw

How to increase the size of toolbar icons in Maple 2015 32-bit Classic on Windows?

Hello everyone! I got some trouble in process a list. Hope you can help:

Assume i got a list like this:

{{k = k, l = RootOf(_Z^2+_Z*k+k^2-1), o = -k-RootOf(_Z^2+_Z*k+k^2-1)}, {k = k, l = RootOf(_Z^2+_Z*k+k^2+1), o = -k-RootOf(_Z^2+_Z*k+k^2+1)}, {k = 0, l = 1, o = -1}, {k = 0, l = -1, o = 1}, {k = 1, l = 0, o = -1}, {k = 1, l = -1, o = 0}, {k = -1, l = 0, o = 1}, {k = -1, l = 1, o = 0}, {k = RootOf(_Z^2+1), l = 0, o = -RootOf(_Z^2+1)}, {k = RootOf(_Z^2+1), l = -RootOf(_Z^2+1), o = 0}}

Now all i want is remove Complex and RootOf from this list, how can i do that?

Thank for your reading adn your help!

I have the following expression

f=u/(sqrt(u*(u-1)))

and I want to simplify it. Eventhough that I tell Maple that u is real and greater than 1 but it does not simplify the expression. What is wrong? Please see the attached file.

Radical.mw

How do you put labels on individual columns in ColumnGraph?

Hello! Hope every is fine. I want to expand the following expression

exp(2*c*t+2*d*n-d)*alpha*c*a[0]*b[1]^2-exp(2*c*t+2*d*n-d)*alpha*c*a[1]*b[0]*b[1]-exp(2*c*t+2*d*n-d)*alpha*a[0]*a[1]*b[1]+exp(2*c*t+2*d*n-d)*alpha*a[1]^2*b[0]+exp(2*c*t+2*d*n)*alpha*a[0]*a[1]*b[1]-exp(2*c*t+2*d*n)*alpha*a[1]^2*b[0]+exp(c*t+d*n)*alpha*c*a[0]*b[0]*b[1]-exp(c*t+d*n)*alpha*c*a[1]*b[0]^2-exp(2*c*t+2*d*n-d)*a[0]*b[1]^2+exp(2*c*t+2*d*n-d)*a[1]*b[0]*b[1]-exp(c*t+d*n-d)*alpha*a[0]^2*b[1]+exp(c*t+d*n-d)*alpha*a[0]*a[1]*b[0]+exp(2*c*t+2*d*n)*a[0]*b[1]^2-exp(2*c*t+2*d*n)*a[1]*b[0]*b[1]+exp(c*t+d*n)*alpha*a[0]^2*b[1]-exp(c*t+d*n)*alpha*a[0]*a[1]*b[0]-exp(c*t+d*n-d)*a[0]*b[0]*b[1]+exp(c*t+d*n-d)*a[1]*b[0]^2+exp(c*t+d*n)*a[0]*b[0]*b[1]-exp(c*t+d*n)*a[1]*b[0]^2

like this 

exp(2*c*t+2*d*n)*exp(-d)*alpha*c*a[0]*b[1]^2-exp(2*c*t+2*d*n)*exp(-d)*alpha*c*a[1]*b[0]*b[1]-exp(2*c*t+2*d*n)*exp(-d)*alpha*a[0]*a[1]*b[1]+exp(2*c*t+2*d*n)*exp(-d)*alpha*a[1]^2*b[0]+exp(2*c*t+2*d*n)*alpha*a[0]*a[1]*b[1]-exp(2*c*t+2*d*n)*alpha*a[1]^2*b[0]+exp(c*t+d*n)*alpha*c*a[0]*b[0]*b[1]-exp(c*t+d*n)*alpha*c*a[1]*b[0]^2-exp(2*c*t+2*d*n)*exp(-d)*a[0]*b[1]^2+exp(2*c*t+2*d*n)*exp(-d)*a[1]*b[0]*b[1]-exp(c*t+d*n)*exp(-d)*alpha*a[0]^2*b[1]+exp(c*t+d*n)*exp(-d)*alpha*a[0]*a[1]*b[0]+exp(2*c*t+2*d*n)*a[0]*b[1]^2-exp(2*c*t+2*d*n)*a[1]*b[0]*b[1]+exp(c*t+d*n)*alpha*a[0]^2*b[1]-exp(c*t+d*n)*alpha*a[0]*a[1]*b[0]-exp(c*t+d*n)*exp(-d)*a[0]*b[0]*b[1]+exp(c*t+d*n)*exp(-d)*a[1]*b[0]^2+exp(c*t+d*n)*a[0]*b[0]*b[1]-exp(c*t+d*n)*a[1]*b[0]^2

i.e., expand exp(2*c*t+2*d*n-d) into exp(2*c*t+2*d*n)*exp(-d) 

waiting your kind response 

hi.please help me for solve this equations

thanks...

Tur.mw

Non dimensionalisation is a vary common task, and I was suprised that I couldn't find a maple tool to automate it . Has anyone developed their own package for it?

I want to automatically do it to the system equations for some Dynamical systems to make some of the other processing I do with them easier.

I was hoping to start with somehting in the form of 

Diff(x[1],t)=f[1](p[1]....p[n],x[1]...x[m])

...

Diff(x[m],t)=f[m](p[1]....p[n],x[1]...x[m])

where each f[i] is some kind of quotient of multivariate polynomials in the variables and parameters:
and end up with something like

Diff(y[1],s)=f[1](q[1]....q[p],y[1]...y[m])

...

Diff(y[m],s)=f[m](q[1]....q[p],y[1]...y[m])

where p<n

I want to cancel some expressions in numerator and denominator of a quotient. But Maple deos not cancel it!

Please see the attached file.

Simplifying_Radicals.mw

in LinearAlgebra Eigenvectors calculation.

So the above output startled me.  I have used the Maple Linear Algebra Eigenvalues, Eigenvectors commands many times with no problem.  Can any one explain to me what is going on.  The program correctly calculates the eigenvalues for the matrix which are all distinct for a real symmetric matrix, and thus should have three distinct non-zero eigenvectors, yet the eigenvectore command returns only zeros for the eigenvectors.  I calculated an eigenvector by hand corresponding to the eigenvalue of 1 and obtained (1, -sqrt(2)/sqrt(3), -1/sqrt(3).

So this is either a serious bug or I am going completely insane.