Unanswered Questions

This page lists MaplePrimes questions that have not yet received an answer

Any semantic differences for building a set on the fly between

A:="123"; B:="456"; C:="789"; F:="10";


L:= L1 union L2


L:= { op(L1) , op(L2) }

They both do the same thing. Is there a reason to prefer one over the other?


A:="123"; B:="456"; C:="789"; F:="10";





{"123", "456", "789"}

{"10", "789"}

L:= L1 union L2

{"10", "123", "456", "789"}

L:= { op(L1) , op(L2) }

{"10", "123", "456", "789"}


Download union.mw


I want to animate a ball rolling on the surface cos(abs(x)+abs(y)).  The ball mass m is 1kg, radius r is 0.1meters starts at (0.5,0.5,cos(abs(5)+abs(5))) meters using g=9.8. 

If we say the initial velocity of the ball is pushed in some random direction. 

How do I show the path of the ball and animate?



 I run the attached code, but after waiting for a few minutes, there is no response, does anyone know the reason?

 It only shows Evaluating!!!!


I'm trying to do a scatter diagram together with a line graph in one. But the function I putting in is not taking it. Is there anyone that can help me with this problem. And thank you.

How to solve an initial value or boundary value fractional differential equation problem by Maple or, a Maple code is required for solving fractional differential initial value or boundary value problem.


Hi, I am creating an activity to illustrate various rotations around different axes. My goal is to enhance the display by showing the axes (one portion as a dashed line and the other as a vector) and to insert the texts (Ox, Oy, Oz) in an optimal manner. Any ideas to optimize the rendering? Thank you.


In this paper (see the footnote on the page), the authors pointed out:

We use -v16 as it is faster than the -v2019 for real root isolation on our benchmarks.

Since I don't have Maple 16 at present, I cannot confirm such an assertion personally. However, I'd like to ascertain if old Maple 16 (released in 2012) is still faster in that specialized realm now (i.e., in 2023). Is there any benchmark test that supports this claim (or tells an opposite story instead?)?

Note. Here I just try to justify the authors' statement (which has caused confusion). Incidentally, will this library be a part of Maple in some future version? (I find that some recent papers provide new "powerful"—in accordance with their "Experiment" sections—Maple packages to tackle difficult problems efficaciously, but such functionalities are not built into the Maple kernel as yet. (In fact, if Maple is a single integrated all-in-one technical platform, then I will no longer need to change tools and formats at each stage. :) ))


How can I obtain the following equation


under  the condition  and  at  by Maple and what is the form of ? (All symbols are constant with respect to  except -exp(-  .

I find that no matter how I modify the style in the Maple command, the LaTeX output seems to be always fixed. This is a bit frustrating. 


Latex(G, terminal, 100, 100)



stylesheet=[vertexborder=false,vertexpadding=20,edgecolor = "Blue",
vertexcolor="Gold",edgethickness=3], font=["Courier",10],size=[180,180])

Even when I change their colors (any other thing), they always go their own way. I don't know if there is a setting that can make the LaTeX output more flexible and in line with our expectations after adjustment.


Last execution block is not producing any output. Why?

The 3x3 nonlinear system I am trying to solve is already a stylized version of my problem, as I already:

  1. Calibrated my equations before attempting to solve for them (search for "Calib_1" in my script)
  2. Split the original 6x6 system into two 3x3 sub-systems (since 3 out of 6 variables only appear in 3 out of 6 equations) and solved for one sub-system

What else can you think of? Should I instead use the parallel solver on the whole 6x6 system rather than just the unsolved 3x3 sub-system?

How to find, if exist, singular solutions? That is, some valuation of some parameters that will yield a solution that cannot be obtained by applying the same valuation to a general solution. Carl Love (who I cannot tag) once mentioned: "The parameter valuations that lead to singular solutions can often be guessed by using valuations that would produce zeros in denominators in the general solution. A singular solution can't be expressed as any instantiation of a generic symbolic solution. By instantiation I mean an assigment of numeric values to some parameters. Here's an example:"

#2x2 matrix and 2x1 vector. 5 parameters (a, b, d, x, y). The 2 decision variables are
#unseen and unnamed in this pure matrix-vector form. Their values are the two entries 
#in the solution vectors S0 and S1.

A:= <a, b; 0, d>;  B:= <x, y>;

#Get a generic solution:
S0:= LinearAlgebra:-LinearSolve(A, B);

#Instantiate 3 parameters (a, d, y) to 0 and solve again:
S1:= LinearAlgebra:-LinearSolve(eval([A, B], [a, d, y]=~ 0)[]);

#Note that no possible instantiation of S0 can produce S1.

Thank you!

A Hamiltonian walk on a connected graph is a closed walk of minimal length which visits every vertex of a graph (and may visit vertices and edges multiple times). (See https://mathworld.wolfram.com/HamiltonianWalk.html)


Please note that there is a distinction between a Hamiltonian walk and a Hamiltonian cycle. Any graph can have a Hamiltonian walk, but not necessarily a Hamiltonian cycle.  The Hamiltonian number h(n) of a connected G is the length of a Hamiltonian walk. 

For example, let's consider the graph FriendshipGraph(2).


Then its Hamiltonian number is 6.  A Hamiltonian walk is as shown in the following picture.




PS: After being reminded by others, I learned that a method in the following post seems to work, but I don't understand why it works for now. 

it's possible to formulate this as a Travelling Salesman Problem by augmenting the sparse graph into a complete graph by adding edges. These new edges represent shortest paths between any two vertices in the original graph and have weight equal to the path length.


It seems that the correctness of this transformation requires rigorous proof.


I was not able to apply ICs (initial conditions) using the advanced parameter settings of the flexible beam component in the attached file. My intention is to have a beam moving at t=0s (velocity of end frame <> 0). The computed initial values for the elastic coordinates (see output console for the disabled component below encircled in red) indicate that the flexible beam has no kinetic energy at t=0s.

A workaround is to attach a rigid body to the end frame and apply the ICs to the body. (Inertia and mass of the rigid body can be set to zero.)

How to apply initial conditions to the disabled component so that it moves as the workaround?


Given Graph G and Graph H how to store the list of all the various Subgraphs of Shape H in G with their vertex labels.

I have write a code but it may not be the optimal way to do it and it may be time consuming for even medium graphs too

Any way to optimize

Let From this list of subgraphs H of say [G0,G1,G2,.....] we pick NumberOfVertices(G) Graphs
say H= [H0,H1,H2,H3,....,Hn]

i) Now any two graph Hi,Hj have excatly one edge interection if (i,j) is an edge in G

 ii) An edge in  the list H will occur exactly twice only in the entire list of graphs

If these two conditions satifies all such H I need to save


Toycode until picking graphs

From that how to pick lists which satify conditions i) and ii) is another need help


How to solve this RLC Electric Circuit  with this initial conditions,I couldn't plot this equation.Help me to do this problem.


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