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. :) ))

Is it even possible to use a replacement list as attempted below to convert a z-transform to a finite difference expression?

Download replace_Y_of_z_by_Y_of_q.mw

Dear all

I have an equation, I would like to introduce a variable R[0], how can I use subs in my equaiton 

subs_equation.mw

Thank you

This code fails in Maple 2023.  It used to work in earlier versions.  Did I miss something in the release notes?

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

Greetings,

I am trying to display an array of plots. I did a search and found this: "https://www.maplesoft.com/support/help/view.aspx?path=plot%2Farrayplot".

However, when I try it, I get an error message: "Error, (in plots:-display) element 1 of the rtable is not a valid plot structure".

Here's the code identical to the sample page. If I display(A__1) or display(A__2) individually, it works.

with(plots);
A := Array(1 .. 2);
A__1 := plot(sin(x), x = -Pi .. Pi, axes = framed, labels = [x, sin(x)]);
A__2 := plot(cos(x), x = -Pi .. Pi, axes = boxed, style = point);
display(A);
 

Running Maple 2023 on MacBook Pro Monterey V12.6.5 (latest as far as I know).

Thanks for any assistance.

Hello 

How to remove this error 

3d_plots.mw

The following program:

with(RealDomain):
sol1 := solve(x-1 > 0, x);
sol2 := solve(1/x^2 > 0, x);
sol3 := solve(1/(x^2-1) > 0, x);
sol4 := solve(x^2+1 > 0, x);
sol5 := solve(x^2+1 < 0, x);

Gives the following output:

sol1 := RealRange(Open(1),infinity)
sol2 := {x <> 0}
sol3 := RealRange(-infinity,Open(-1)), RealRange(Open(1),infinity)
sol4 := x
sol5 := NULL

I would like to force some consistency in the output of solve because I need to feed the output to a third-party program. For example, what's the deal with sol2 and sol3? Why are they formatted differently?

Ideally, I would like the previous output to be:

sol1 := RealRange(Open(1),infinity)
sol2 := RealRange(-infinity,Open(0)), RealRange(Open(0),infinity)
sol3 := RealRange(-infinity,Open(-1)), RealRange(Open(1),infinity)
sol4 := RealRange(-infinity,infinity)
sol5 := NULL

How do I achieve this?

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.

Hello, 

Ive followed the examples, and yet not the result the book gave. The question before this one, same type, and i did get the right answer there. 

Would you have a clue why it is not going as planned? 

Translation: "Through a resistancewire with the measured resistance R= (1.5+-0.5)Ohm goes an electric current with the measured strength I= (2.5+-0.05)A, During a timespan of t=(10+-0.5)s develops an amount of heat Q=RI^2*t. Give the estimate of the maximum relative deviation in Q."

Here i got the right answer of .42, and yet when doing the exact same thing in assignment 4, i got the wrong answers. 

Translation: "Two bodies with mass M and m have a massmidpointdistance r. For the gravitational force between these bodies counts: F=y*M*m/r^2, at which y is the gravitational constant. Give an estimation of the maximum realative deviation in F on ground of the following measurementdata: M=(7.0+-0.05)kg, m=(2.0+-0.02)kg and r=(5.0)+-0.05)m."

I had the answer -0.0029. The books says its wrong, but i cant see what i did wrong. 

Translation of the example: "According to the law of Ohm the currentstrength in the stream(/current)circle(circuit) with tension(voltage) U and resistance R is equal to I=U/R. For a given currentcircle((part of a)circuit) is given: U=(30.0+-0.05)Volt and R=(2.5+-0.03)Ohm. Determine the maximum absolute and relative deviation in I. 

Solution

The maximum absolute deviation in I is

This gives the currentstrength:

The maximum realtive deviation in I is: .... or about 1.4%"

Could somebody tell me what i am doing wrong?

Thank you!

Greetings,

The Function

Download Mapleprimes_Question_Book_2_Paragraph_5.13_Question_4.mw

Why does this not return a quick solution?

sol := solve(.2894606222 = .4090662520/((1+0.1481289280e-2*x)^5.034901366*(1/(1+0.1481289280e-2*x)^5.034901366)^.1986136226), x)

mw.mw

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.