I am brand new to Maple Cloud and Maple Player.

I have uploaded two worksheets to the cloud, and my wife has just installed Maple Player on her laptop.

In Maple Player, the second worksheet shows the shareable symbol but the first doesn't even thought I uploaded both in the same way by clicking on the upload symbol in the Maple Cloud palette. Why is the first worksheet not shareable?

When my wife displays the second worksheet she is able to move its sliders but they do not change the display as they do when I move the sliders within Maple2016. How can she change the display?

Here is a link to the second worksheet:

Cassinian_oval.mw

Say I have 2 matrices, in which the elements themselves are vectors.

I'm looking for a way to perform matrix multiplication on these so that rather than having the first element as x11y11+x12y21+x13y31

It would be x11.y11+x12.y21+x13.y31 where . is the dot product on the elements of each matrix.

I know I could write a procedure to do this manually but I was wondering if there's any pre-made operations (or modifiers on the Multiply operation) to do this.

How do I plot the optimal control functions in an optimal control problem ?

Hello,

I have been working on Maxima and minima, I am able to extract the eigen values for the expression.
Based on following conditions I am able to find out the critical point is maxima or minima or saddle or inconsistant

If all the eigenvalues are positive, the point is a minimum.
If all the eigenvalues are all negative, it's a maximum.
If some eigenvalues are positive, some are negative, and none are zero, then it's a saddle point.
If any eigenvalues are zero, the test is inconclusive

I want to return all the critical points and their extrema.
just for example : For one perticular function I got a Eigen values as which I can find using sign function.

EigenValues := [[-.381966011250105+0.*I, -2.61803398874989+0.*I], [.414213562373095+0.*I, -2.41421356237309+0.*I]]
signDetails := [seq([seq(sign(EigenValues[i][j]), j = 1 .. nops(EigenValues[i]))], i = 1 .. nops(EigenValues))] #

 signDetails :=[[-1, -1], [1, -1]]

Now if I have a 0 in a list. Sign function returns 1 for 0, which is incorrect. How can I handle such conditions

if I have

EigenValues := [[-.381966011250105+0.*I, -2.61803398874989+0.*I], [.414213562373095+0.*I, -2.41421356237309+0.*I], [0, 2]]

I would like to have output [[-1, -1], [1, -1], [0, 1]],

I would like to know how is it possible return output based on above list

in this case my return shouble something like this [maxima, saddle, inconclusive].

Thank you

Here is a simple procedure that works fine if entered using 1D Maple input
> Q:=proc(x)
sin(x)
end proc;
but if you use 2D math input
> q:=proc(x)
sin(x);

  end proc;

Error, unterminated procedure
    Typesetting:-mambiguous(qAssignTypesetting:-mambiguous(

      procApplyFunction(x) sinApplyFunction(x),

      Typesetting:-merror("unterminated procedure")))
Error, unable to parse
    Typesetting:-mambiguous(  Typesetting:-mambiguous(end,

      Typesetting:-merror("unable to parse")) procsemi)

Ouch! But to confuse things further the following procedures may be entered using 2D math and work fine:
>H := proc (x) x^2*sin(x) end proc;
>K := proc (x) sin(x^2) end proc;
Doesn't make any sense to me. Perhaps 2D math is not ready for prime time?

 

Im trying to plot the graph of g := int(exp(-epsilon(x^4+x^2)), x = -10 .. 10), i want to see how the value of the integrand will change as we let epsilon tend towards infinity, but I am struggling to find a way to do this. Any help would be greatly appreciated.

So I have got the following integral:

around the square with corners 2, 2i, -2, -2i oriented counter-clockwise.

Do I need to tell Maple that z is complex? Do I need to manually parametrize z? Is Maple aware of Cauchy's theorem?

What is the quickest way of evaluating this sort of integrals.

I'm still a maple beginner and looking for a workable solution to the following problem. I have already solved these "by hand" and would now like to control them.

The following equation should be solved:

diff(u(x, t), t, t)-c^2*(diff(u(x, t), x, x))

The approach function would be for example:

u := a(x)*sin(k1*(-c*t+x))

It would be great if maple could give me the solution again in this example:

solution = -c^2*(d^2*a*sin(k1*(-c*t+x))/dx^2+2*d*a*k1*cos(k1*(-c*t+x))/dx)

How exactly do I do this best?
Thanks a lot!
Frank

Hi,

How to simulate 1000 times the Dice1 experiment and put the results in a list to search for the percentils of A=[A1,..,A1000]

( See attachment)

Thanks

QuestionSimulation.mw

how can solve ordinary differenatial equation in maple?

how can use laplce transformation for this equation in maple?

U0 := proc (x) options operator, arrow; cosh(sqrt(2)*x)-1 end proc;
sys_ode := diff(Uc(x), x, x)-(2*(1+U0(x)))*Uc(x) = 0;

Download a.mw

Hi,

Persay I have a sinusodal function or dampened sinusodial; it is understand that the domain is from negative infinity to infinity.  Say I wish to find the zeros or a certain y value of a sinusodial on a restricted domain - say from A to B. How would I tell maple -syntax wise- to solve on that restricted domain? Using the solve feature on maple, it only yields 1 value that satisfies the condition I give it .

Thank you all for the support; you al l are allowing to me develop a better understanding for maple.

say I have the functions:

y=e^x*sin(x)

y=e

i wish to find all the x values that satisfy that system on the x domain of 0 to 100

ode_problem.mw Can someone please look at this? I am having a problem with the ordinary differential equations. 

Hello there,

I'm pretty new to MAPLE so this may probably an "easy" mistake, but I don't know what's the problem anyway...

I'd like to find a solution to a system of four equations with four unknowns:

fsolve({(T[1]-T[0])/(10000-T[1]+T[0]) = -2.000000000, (T[2]-T[0])/(20000-T[2]+T[0]) = 0, (T[3]-T[0])/(50000-T[3]+T[0]) = 50, .1*T[0]+.3*T[1]+.55*T[2]+0.5e-1*T[3]-5000 = 0}, {T[0], T[1], T[2], T[3]})

Problem: Instead of providing the single solutions, MAPLE is simply just rewriting the fsolve-statement and does not solve. There is no error-message.

Does anybody know, what the problem is here? Is there no solution after all?

An n*n matrix A is called an MDS matrix over an arbitrary field if all determinant of square sub-matrices of A are non-zero over the field. It is not difficult to prove that the number of all square sub-matrices of A is binomial(2*n, n)-1. The code that I use to check whether A is an MDS matrix is in the following form 

 u := 1;
 for k to n do
 P := choose(n, k);
   for i to nops(P) do
    for j to nops(P) do
         F := A(P[i], P[j]);
         r := Determinant(F);
        if r = 0 then

           u := 0; k:=n+1;

            i := nops(P)+1; j := nops(P)+1;

        end if;

    end do;

      end do; 
   if u = 1 then

      print(A is an MDS Matrix) 

   end if; 
  end do:

When I run the mentioned code for n=16, it takes long time since we need to check binomial(32, 16)-1=601080389 cases to verify that A is an MDS matrix or not. 

My Question: Is there a modified procedure which can be used to check that an n*n matrix is  whether an MDS matrix for n>=16. 

i'm doing an optimaztion for my final project. when i execute the coding, it says: 

Warning, initial point [m1 = .900000000000000, m2 = .900000000000000, m3 = .900000000000000, sigma1 = .900000000000000, sigma2 = .900000000000000, sigma3 = .900000000000000] does not satisfy the inequality constraints; trying to find a feasible initial point
Error, (in DirectSearch:-Search) cannot find feasible initial point; last infeasible trial point is [m1 = HFloat(0.9), m2 = HFloat(1.9), m3 = HFloat(1.9), sigma1 = HFloat(1.9), sigma2 = HFloat(0.9), sigma3 = HFloat(0.9)].

here my full coding:

restart;
with(linalg);
with(Optimization);
with(DirectSearch);
[BoundedObjective, CompromiseProgramming, DataFit, 

  ExponentialWeightedSum, GlobalOptima, GlobalSearch, Minimax, 

  ModifiedTchebycheff, Search, SolveEquations, WeightedProduct, 

  WeightedSum]
q := Array(0 .. .111, [0.586e-2, 0.67475e-3, 0.52476e-3, 0.419785e-3, 0.354814e-3, 0.324859e-3, 0.319948e-3, 0.31e-3, 0.295013e-3, 0.28e-3, 0.259974e-3, 0.25496e-3, 0.26e-3, 0.270027e-3, 0.280026e-3, 0.284987e-3, 0.274934e-3, 0.279893e-3, 0.294824e-3, 0.324765e-3, 0.374672e-3, 0.439555e-3, 0.509439e-3, 0.56934e-3, 0.60923e-3, 0.634201e-3, 0.634233e-3, 0.624319e-3, 0.61442e-3, 0.624491e-3, 0.6495e-3, 0.684465e-3, 0.714452e-3, 0.729463e-3, 0.749451e-3, 0.789387e-3, 0.864347e-3, 0.964331e-3, 0.1064261e-2, 0.119413e-2, 0.1333876e-2, 0.150354e-2, 0.1683293e-2, 0.1883023e-2, 0.2102591e-2, 0.2356929e-2, 0.2656062e-2, 0.3004981e-2, 0.3403637e-2, 0.3841874e-2, 0.4349482e-2, 0.4931339e-2, 0.5587622e-2, 0.6293742e-2, 0.7044467e-2, 0.780967e-2, 0.8563865e-2, 0.9291539e-2, 0.9987678e-2, 0.10677379e-1, 0.1139506e-1, 0.12195833e-1, 0.13108819e-1, 0.14262384e-1, 0.15557439e-1, 0.17002799e-1, 0.18574592e-1, 0.20263971e-1, 0.22103523e-1, 0.24043337e-1, 0.26165465e-1, 0.28548531e-1, 0.31489568e-1, 0.3458584e-1, 0.37891201e-1, 0.41511365e-1, 0.45444649e-1, 0.49867283e-1, 0.54877269e-1, 0.60478547e-1, 0.66532745e-1, 0.7397857e-1, 0.81900977e-1, 0.90034141e-1, 0.98921462e-1, .108746024, .11839162, .130752255, .14368791, .157201822, .169770195, .177233464, .188823516, .204733363, .224639756, .249617627, .273409998, .298674064, .304378882, .319592743, .342821691, .367512872, .395257421, .426270115, .460795318, .498484563, .537417055, .577542345, .618093683, .660441764, .703856817, 1]);
variables = [m1, m2, m3, sigma1, sigma2, sigma3];
        variables = [m1, m2, m3, sigma1, sigma2, sigma3]
psi1 := 0.11480601e-1; psi2 := 0.8890123e-2; psi3 := .979629276;
                          0.011480601
                          0.008890123
                          0.979629276
s1(x):=exp(-(x/(m1))^((m1/(sigma1)))):  s1(x+1):=exp(-((x+1)/(m1))^((m1/(sigma1)))):  s2(x):=1-exp(-(x/(m2))^((-m2/(sigma2)))):  s2(x+1):=1-exp(-((x+1)/(m2))^((-m2/(sigma2)))):  s3(x):=exp(exp(-m3/(sigma3))-exp((x-m3)/(sigma3))): s3(x+1):=exp(exp(-m3/(sigma3))-exp((x+1-m3)/(sigma3)))  :  s(x):=(psi1*s1(x)+psi2*s2(x)+psi3*s3(x)): s(x+1):=(psi1*s1(x+1)+psi2*s2(x+1)+psi3*s3(x+1)):   qtopi(x):=1-((s(x+1))/(s(x))):
fungsikerugian := add((1-qtopi(x)/q[x])^2, x = 0 .. .110);
for x from 0 to 111 do c1[x] := 0 <= s1(x+1); c2[x] := 1 >= s1(x+1); c3[x] := 0 <= s2(x+1); c4[x] := 1 >= s2(x+1); c5[x] := 0 <= s3(x+1); c6[x] := 1 >= s3(x+1); c7[x] := 0 <= s1(x); c8[x] := 1 >= s1(x); c9[x] := 0 <= s2(x); c10[x] := 1 >= s2(x); c11[x] := 0 <= s3(x); c12[x] := 1 >= s3(x); c13[x] := 0 <= s(x); c14[x] := 1 >= s(x); c15[x] := 0 <= s(x+1); c16[x] := 1 >= s(x+1); c17[x] := 0 <= qtopi(x); c18[x] := 1 >= qtopi(x) end do;

constr := {m2 > sigma2, m3 > sigma3, seq(c1[x], x = 0 .. .110), seq(c10[x], x = 0 .. .111), seq(c11[x], x = 0 .. .111), seq(c12[x], x = 0 .. .111), seq(c13[x], x = 0 .. .111), seq(c14[x], x = 0 .. .111), seq(c15[x], x = 0 .. .110), seq(c16[x], x = 0 .. .110), seq(c17[x], x = 0 .. .110), seq(c18[x], x = 0 .. .110), seq(c2[x], x = 0 .. .110), seq(c3[x], x = 0 .. .110), seq(c4[x], x = 0 .. .110), seq(c5[x], x = 0 .. .110), seq(c6[x], x = 0 .. .110), seq(c7[x], x = 0 .. .111), seq(c8[x], x = 0 .. .111), seq(c9[x], x = 0 .. .111), m1 < sigma1, 0 <= m1 and m1 < 17, 17 <= m2 and m2 < 33, 33 <= m3 and m3 < 111};
solusi := DirectSearch:-Search(fungsikerugian, constr, assume = positive, evaluationlimit = 5555);
Warning, initial point [m1 = .900000000000000, m2 = .900000000000000, m3 = .900000000000000, sigma1 = .900000000000000, sigma2 = .900000000000000, sigma3 = .900000000000000] does not satisfy the inequality constraints; trying to find a feasible initial point
Error, (in DirectSearch:-Search) cannot find feasible initial point; last infeasible trial point is [m1 = HFloat(0.9), m2 = HFloat(1.9), m3 = HFloat(1.9), sigma1 = HFloat(1.9), sigma2 = HFloat(0.9), sigma3 = HFloat(0.9)]

thanks before