I solve a set of equations in this way and I have three set of answers ,but I don`t know wich one is true.

and I have another question ,how can I assume v[0] like a constant?

alpha[2]:= 2.727272728*10^5: alpha[4]:= 3738.685337: alpha[6]:= -30.18675539: alpha[7] := -4.116375735*10^6: alpha[8] := 1.859504132*10^10: alpha[9]:= 2.489142857*10^(-12):

l10:=(alpha[7]*v[0]^2+1)*gamma[i*n]^4+(-alpha[4]*beta[n]^2+alpha[8]*v[0]^2-alpha[9])*gamma[i*n]^2+(2*I)*gamma[i*n]*alpha[2]*beta[n]*v[0]+(2*I)*gamma[i*n]^3*alpha[6]*beta[n]*v[0]-beta[n]^2 = 0:

l11 := subs(i = 1, l10);

l12 := subs(i = 2, l10);

l13 := subs(i = 3, l10);

l14 := subs(i = 4, l10);

l15 := (exp(I*(gamma[n]+gamma[2*n]))+exp(I*(gamma[3*n]+gamma[4*n])))*(gamma[3*n]^2-gamma[4*n]^2)*(gamma[n]^2-gamma[2*n]^2)+(exp(I*(gamma[n]+gamma[4*n]))+exp(I*(gamma[2*n]+gamma[3*n])))*(gamma[2*n]^2-gamma[3*n]^2)*(gamma[n]^2-gamma[4*n]^2)+(exp(I*(gamma[2*n]+gamma[4*n]))+exp(I*(gamma[n]+gamma[3*n])))*(gamma[2*n]^2-gamma[4*n]^2)*(-gamma[n]^2+gamma[3*n]^2) = 0;

l1 := combine(expand(evalc(l15)), trig):

l2 := combine(expand(evalc(Re(l15))), trig):

l3 := combine(expand(evalc(Im(l15))), trig): v[0] := 1; 1

fsolve({l1, l11, l12, l13, l14}, {beta[n], gamma[n], gamma[2*n], gamma[3*n], gamma[4*n]}):

fsolve({l11, l12, l13, l14, l2}):

solve({l11, l12, l13, l14, l3}):

thanks

Reyhaneh

Friends in Maple

I have a Vehicle Routing Problem I wish to cast as a integer model. VRP is a kind of TSP and knapsack problem hybrid  I would be grateful if someone can finish it. answers included. This isn't homework BTW.

VRP_IP.mw

with(plots):with(plottools):
p1:=plot( x^3,x=-1..1,thickness=20,color=red):
p2:=plot(-x^3,x=-1..1,thickness=20,color=blue):
p3:=display(rectangle([0.5, 1],[0.75,-1],color=green)):
p4:=display(rectangle([-1,-0.1],[1,0.1],color=yellow)):

display(p1,p2,p3,p4); # order = 4312

#  display(p3,p4,p1,p2); # order = 4312 (the same)

It seems that the rectangles are plotted first, in reversed order, and then the curves, in direct order.
Has someone an explanation?

Cmaple always print the source code first and then print output , how to print output only without printing heading logo and source code by default when use cmaple

When print a formula , if variable has power, the power will print in another row

how can it print like a^2 in one row 

Dear all

I would like to minimuze the following function 4 x ^2 + 4 x y  under constraint  16=x^2 y  and both x , y  nonnegative real number 

this is my code

with(Optimization)

Minimize(4*x^2+4*x*y, {x^2*y = 16}, assume = nonnegative)

The result obtained from this code is strange

because when i do directly the computation without maple by substituting y =16/x^2 and simple derivation i get the minimum is at x=2 and so y=4 and therefore the minimum is 48

But as i say using my code I obtained a different solution

whats is the problem occurs in this situation

Many thanks

I am using MAPLE 2016.1.

I have created an animation of points on a polarplot (2D), but would like it to be projected into 3D, with time t being the 3rd axis.  Here is my code for the polarplot animation:

with(plots);
Repltlist := proc (t) options operator, arrow; [[sin(t), 0], [cos(t), (1/3)*Pi], [cos(2*t), (2/3)*Pi]] end proc;
Impltlist := proc (t) options operator, arrow; [[-sin(3*t), 0], [-2*cos(t), (1/3)*Pi], [-cos(4*t), (2/3)*Pi]] end proc;
titleName := "Polar representation of time-dependence of degrees of freedom in configuration space"; captionName := "Re - red, Im - blue";
P := animate(polarplot, [Repltlist(t), color = "Red"], t = -Pi .. Pi, symbol = solidcircle, style = point, color = red, symbolsize = 12, frames = 100, gridlines = true);
Q := animate(polarplot, [Impltlist(t), color = "Blue"], t = -Pi .. Pi, symbol = solidcircle, style = point, color = blue, symbolsize = 12, frames = 100, gridlines = true);
R := display([P, Q], title = titleName, caption = captionName); R;

I have been trying use the plots[transform] to 'lift' this polarplot into 3D: 

with(plottools); with(plots);

plrPt(2); f := proc (tt) options operator, arrow; plottools:-transform(proc (r, theta) options operator, arrow; [r, theta, tt] end proc) end proc; display((f(2))(plrPt(2)));

but the last command gives no output.

Can anyone help?

MRB
PS:  I am now attemptiing to do the lift to 3D by using a cylinderplot but would like know why the transform function is not able able to lift the polarplot into 3D.

Parametric equation of second-order curve in 3d. Draghilev method.
PLAN_CURVE_3d_1.mw
Examples:
x1^2+x1*x3+13*x2^2+x3-1=0;
x1+x2+x3=0;

 x1^2+0.1*x2^2+x3^2-9=0;
 x1+3*x3+1=0;

 x1^2-0.1*x2^2+x3^2-9=0;
 x1+3*x3+1=0;

Hello; I found another ODE which Maple gives division by zero on.  Is this also a bug? 

dsolve(x*(a^2*x+(x^2-y(x)^2)*y(x))*diff(y(x),x)^2-(2*a^2*x*y(x)-(x^2-y(x)^2)^2)*diff(y(x),x)+a^2*y(x)^2-x*y(x)*(x^2-y(x)^2) = 0, y(x));

Error, (in dsolve) numeric exception: division by zero

This is from a book. Using Maple 2016.1 on windows.

Please Sir/Ma, I'm trying to generate a recurrent relations of this series and I try to use "if" "else" condition but I didn't get it right. Any one with useful suggestions. Appreciate 

restart;
Y[0] := A; Y[1] := B;
if k = a then delta(k-a) := 1 else 0 end if;
                               0
for k from 0 to 10 do Y[k+2] := solve(add(delta(i-1)*(k-i+1)*(k-i+2)*Y[k-i+2], i = 0 .. k)+add((delta(i)-delta(i-1))*(k-i+1)*Y[k-i+1], i = 0 .. k)+lambda*Y[k] = 0, Y[k+2]) end do;
y := sum(Y[j]*x^j, j = 0 .. 10);
 

I use maple 12 on my dual core laptop and i plan to buy maple 2016 and an 8 core system to get advandage of multiple processors. 

Does maple 2016 functions get advandage of multiprocessor systems or it will be the same as having one processor?

Maple 2016.1 on windows. This ODE from a book, and Maple gives division by zero. Is this a bug or expected?

ode:=(a^2*x+(x^2-y(x)^2)*y(x))*diff(y(x),x)+x*(x^2-y(x)^2) = a^2*y(x);
dsolve(ode, y(x));

Error, (in dsolve) numeric exception: division by zero

Mathematica gives this to same ODE, but no division by zero.

DSolve[x*(x^2 - y[x]^2) + (a^2*x + y[x]*(x^2 - y[x]^2))*y'[x] == a^2*y[x], y[x], x]

Solve[x^2/2 - (1/2)*a^2*Log[x - y[x]] + (1/2)*a^2*Log[x + y[x]] +y[x]^2/2 == C[1], y[x]]

Where is the division by zero coming from in Maple?

2016.1 on windows.

Please, anybody with a good answer. Thanks in advance

restart;
U[0] := 1; U[1] := a; U[2] := 3/2; U[3] := (1/6)*b;
for k from 0 to 20 do U[k+4] := (4/factorial(k)+U[k])/((k+1)*(k+2)*(k+3)*(k+4)) end do;
u := sum(U[j]*y^j, j = 0 .. 20);
           3  2   1    3   5   4   /1     1   \  5    7   6
 1 + a y + - y  + - b y  + -- y  + |-- + --- a| y  + --- y 
           2      6        24      \30   120  /      720   

      / 1      1    \  7    1    8   /  1       1     \  9
    + |---- + ---- b| y  + ---- y  + |----- + ------ a| y 
      \1260   5040  /      4480      \45360   362880  /   

        11     10   /   1         1      \  11      13      12
    + ------- y   + |------- + -------- b| y   + --------- y  
      3628800       \4989600   39916800  /       479001600    

      /    1           1       \  13       1       14
    + |--------- + ---------- a| y   + ---------- y  
      \518918400   6227020800  /       5811886080    

      /     1               1        \  15         17        16
    + |------------ + ------------- b| y   + -------------- y  
      \108972864000   1307674368000  /       20922789888000    

      /      1                 1         \  17
    + |-------------- + --------------- a| y  
      \22230464256000   355687428096000  /    

             19         18
    + ---------------- y  
      6402373705728000    

      /       1                   1           \  19
    + |---------------- + ------------------ b| y  
      \7602818775552000   121645100408832000  /    

              1           20
    + ------------------ y  
      115852476579840000    
solve({(sum(U[j]*y^j, j = 0 .. 20), y = 1) = 2*exp(1), (sum(j*(j-1)*U[j]*y^j, j = 0 .. 20), y = 1) = 4*exp(1)}, [a, b]);
 

Here is one that the students generated which caused confusion. 

a := 0.76;
eq1 := 2*cot(a*sqrt(2*E)) = (2*E-5.4)/(sqrt(E*(5.4-E));
solve(eq1, E)

And the results are: 0., 4.411954070, 2.423743792

The problem is with the second answer because it does not exist. If we plot the LHS and the RHS of eq1 vs E where E=0..5.4

And it gets more interesting, if we calculate:

solve(evalf(eq1), E)

The answers are: 0., 2.423743793, 14.33807304+27.39159712*I

where the 3rd answer is again incorrect.

Finally, if a = 0.8 or larger, the incorrect answers disappear. 

Note - fsolve does handle this problem correctly. And despite my attempts to remind them to use fsolve, they see the solve command as the universal truth. Apparently this will be another teaching moment for next year.

So any thoughts about why this happens and why there is a difference in the outcomes between 0.76 and 0.8 for the value of a?

hi .how i can fsolve 8 equations, in which obtained from

for k from 2 to m do eq1[k_] end do

AND

for k from 2 to m do eq2[k_] end do  and so on........

And gain unknown functions as f3[2], f4[4],....

thanks?

fdm-maple.mw
 

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