hi,

 Does anyone know why my code is not running?

 eqs.mw

I am not able to make a MWE for this error, as it only shows in the debugger. So it seems Maple internal memory changes or some other library is loaded to cause this. Inside the debugger, I get to a function which does this

DBG> simplify(JacobiDN(x,k)^2*n)

Error, invalid input: simplify/Jacobi/JacobiDN expects its 1st argument, k, to be of type posint, but received 0

Version 2023 on windows 10

In a worksheet, the above works just fine

restart;
simplify(JacobiDN(x,k)^2*n)

Back to the debugger, if I write (2*n) instead of 2*n, then the error goes away

DBG> simplify(JacobiDN(x,k)^(2*n))
JacobiDN(x,k)^(2*n)

The values of x,k,n are all symbols and have no values in the code running:

I have no idea why this happend when I run the code only. It think x is zero in the above for some reason.

Sorry can't make MWE, I wish I can. Something strange happens when I run the code that does not show otherwise. 

Any suggestions how to invetigate this more? Stepping into the simplify code it fails in

DBG> next
`simplify/check_constant`:
   3   return type(r,'And(constant,Or(Not(function),satisfies(f -> evalb(op(f)
         <> NULL))))')

DBG> r
JacobiDN(x,k)^2*n^2

DBG> type(r,'And(constant,Or(Not(function),satisfies(f -> evalb(op(f)          <> NULL))))')
false

DBG> step
`simplify/Jacobi`
`simplify/do`:
  84               userinfo(1,simplify,'applying',new_simp,
                     `function to expression`);

 85               new_r := new_simp(r,symb_mode);  

Here it generate the error.

It has nothing to do with simplify. Here is a call to integrate which gives same error

DBG> lhs(ode)
diff(diff(xi(x),x),x)-k^2*JacobiSN(x,k)*JacobiCN(x,k)/JacobiDN(x,k)*diff(xi(x),x)+(-k^2*JacobiCN(x,k)^2+k^2*JacobiSN(x,k)^2-k^4*JacobiSN(x,k)^2*JacobiCN(x,k)^2/JacobiDN(x,k)^2-JacobiSN(x,k)^2*k^2*n^2+n^2)*xi(x)

DBG> int(lhs(ode),x)
Error, invalid input: simplify/Jacobi/JacobiDN expects its 1st argument, k, to be of type posint, but received 0

DBG> x
x

DBG> k
k

DBG> xi(x)
xi(x)

Hei

Vet noen om Windows 11 støtter Maple 2020? Eller støttes det bare av Windows 10.

If I understand right, the form  is equivalent to  (where the optional index variable is omitted), which produces a sequence of n occurrences of y. But how to explain the following output (of )? 

Download singular_behaviour_of_`seq`.mw

Main code: 

p1 := proc()
    local a := 1, b := 2;
    seq('assign(('a', 'b') = (a + 1, 2*b))', 3);
    print(a, b)
end:

p1():

 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!!!!

 sy01.mw

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.

uestionRevolution.mw

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

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. :) ))

Hi

How can I obtain the following equation

from

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

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. 

G:=Graph(6,{{1,2},{1,4},{4,5},{2,5},{2,3},{3,6},{5,6}});
DrawGraph(G);
vp:=[[0,0],[0.5,0],[1,0],[0,0.5],[0.5,0.5],[1,0.5]]:
SetVertexPositions(G,vp):
DrawGraph(G);

Latex(G, terminal, 100, 100)

\documentclass{amsart}
\begin{document}
\begin{picture}(100,100)
\qbezier(12.40,87.60)(12.40,50.00)(12.40,12.40)
\qbezier(12.40,87.60)(31.20,50.00)(50.00,12.40)
\qbezier(12.40,12.40)(31.20,50.00)(50.00,87.60)
\qbezier(12.40,12.40)(50.00,50.00)(87.60,87.60)
\qbezier(50.00,87.60)(68.80,50.00)(87.60,12.40)
\qbezier(50.00,12.40)(68.80,50.00)(87.60,87.60)
\qbezier(87.60,87.60)(87.60,50.00)(87.60,12.40)
\put(12.400000,87.600000){\circle*{6}}
\put(10.194,96.943){\makebox(0,0){1}}
\put(12.400000,12.400000){\circle*{6}}
\put(8.726,3.531){\makebox(0,0){2}}
\put(50.000000,87.600000){\circle*{6}}
\put(50.000,97.200){\makebox(0,0){3}}
\put(50.000000,12.400000){\circle*{6}}
\put(50.000,2.800){\makebox(0,0){4}}
\put(87.600000,87.600000){\circle*{6}}
\put(91.274,96.469){\makebox(0,0){5}}
\put(87.600000,12.400000){\circle*{6}}
\put(89.806,3.057){\makebox(0,0){6}}
\end{picture}
\end{document}
 

DrawGraph(G,
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.

120523_problem_parallel.mw

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).

g:=GraphTheory[SpecialGraphs]:-FriendshipGraph(2);
DrawGraph(g)

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. 

• https://cs.stackexchange.com/questions/43731/travelling-salesman-which-can-repeat-cities

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?

Imposing_IC_v.msim