I never used Maplets before.

Been learning Explore, which is OK, but the UI does not look good. Too much white spaces between sliders.  So been looking at Maplets to use instead.

Should one just use Maplets to make interactive demos with sliders, buttons, popup menus and so on or use Explore? Which is better?

I do not waste more time learning Explore more if Maplets is a better choice.

If someone here knows Maplet, here is something I just wrote in Explore.  Could this be coded in Maplet to see how it will look like. It took me 30 minutes to make it using Explore and I am no expert in Explore, so hopefully it should not take someone who knows Maplets much time to produce same thing as Maplet. I wanted to see if the UI will look better or same issues with too wasted spaces between sliders.  I assume with Maplets, there is same concept as Explore, with range of variables, and initial values and so on...

I understand one can run Maplet using Maple viewer without needing to have Maple installed on the PC, so this is an advantage.

Download lotka_volterra.mw

Screen shot of the UI

How would the above look using Maplet instead of Explore?

I am not sure whether this should be in "create a post" or "ask a question." Let me know if this is more appropriate in "ask a question"

The standing wave equation is given by:

PDE:=diff(u(x,t),t,t)=c^2*diff(u(x,t),x,x)

IBC := {u(x, 0) = A0*cos(Pi*x/L), D[1](u)(0, t) = 0, D[1](u)(L, t) = 0, D[2](u)(x, 0) = 0}

pdsolve(PDE,IBC,numeric)

In my problem, the material is a magneto-elastic material where c, the speed of the acoustic wave, is a function of a magnetic field H.  The material is nonlinear and saturable.  I define it by a 3 segment piecewise nonlinear function of H.  The material response is a result of a sinusoidal H field.  I am interested in solving u(x,t).  

With that, c in the PDE has to be rewritten as c(H(t)), pdsolve gives an error as PDE has to be expressed as a function of u,t, or x.  So I redefine c(H(t)) as c(t).  I ran into another error in pdsolve as the piecewise has to be based on t or x and not H.  

The problem is that depending on H I can go through all 3 segments and back in one cycle and I have to find the corresponding t's for the piecewise.  Now I am driving the material with 100 cycles, I have to find and list all those piecewise transition points which is hardly practical.

Is there any other ways to approach and solve this problem?

Hello Friends.

I have used Maple to create a Polynomial Regression model.  The model is called "PRModel."  It works fine.  The independent variable is "X" and the dependent variable is "Y."  Both X and Y are of the vector variety.  The model is as follows:

PRModel:=PolynomialFit(10, X, Y, summarize = true);

I would like an estimate of Y for each value of X.  I would like the estimates of Y to be in a variable called "estY."   I have not been successful with this.  I have tried many different variations of the following, but have not been successful.  

estY := eval(PRModel, X);

Any suggestions as to how I can capture the estimates of Y? 

Thank you.

I am very pleased to announce that Volume 6 Number 1 of Maple Transactions has been published.  This is a Special Issue on Matrices and Polynomials in Computer Algebra, and the Guest Editors (our first ever!) were Marc Moreno Maza and Tomas Recio.  There are still some papers that are expected to be added to the issue when they come in, but at this moment there are 8 papers there for you to read (and a description of the issue in the Front Matter section, by the Guest Editors).

A link to this Special Issue

On my journey of discovery through the Maple world, I now want to try out Maple's convenient features in the complex plane, something that used to be laboriously worked out and demonstrated on the blackboard with chalk. I couldn't find a suitable introduction in the help text. I'm interested in whether a package needs to be loaded and how to handle polynomials, series, and line integrals (I have a reasonable understanding only of the theory).

once i founded but  i lost the technique and i can't remember how i can reach the point how to find thus parameter and find the solution of pde

t1.mw

I just found a rather incredible error in versions 2025.1 and 2025.2.

The function works fine in version 2022.

The drag function in solve does not work, even for solving a very simple problem.

For example, in the equation 1 + 5x = 3x - 1, if I try to do 3x from right to left, a box opens with a good suggestion (subtract   1+2x = -1)

# subtract 3*x from both sides
((1)) + ~(3 * x)
we can see that the correct expression should be ((1)) - ~(3 * x)

I finally get:

1 + 8x = 6x -1 (it added and did not subtract 3x from both sides)

Please correct this...

This post was inspired by the following discussion thread  https://mapleprimes.com/questions/242266-Count-The-Number-Of-Paths , which considered the problem of finding all Hamiltonian paths on an integer lattice in R^2 that connect two distinct vertices. The  AllPaths  procedure solves a more general problem: it finds all self-disjoint paths connecting two distinct vertices not only in the plane  R^2  but also in space  R^3 . Of course, it also finds all Hamiltonian paths or allows one to determine their absence. The procedure does not use commands from GraphTheory package (only direct manipulation of sets and lists).
Required parameters of the procedure: S  is a set or list of lattice vertices specified by their coordinates, Start and Finish are the initial and final vertices. Optional parameter R (defaults it's NULL  if all paths are searched) and any symbol (if only Hamiltonian paths are searched). S can be either a rectangular integer lattice or a union of several such lattices.

Code of the procedure:

restart;
AllPaths:=proc(S::{set(list),listlist},Start::list,Finish::list,R::symbol:=NULL)
local N:=nops(S), S1:=convert(S, set), L, n, m, k, i, j, s, p, q, P, a, b, c;

L:={[Start]};
for n from 2 to N do

if R=NULL then

P:='P';
m:=nops(L);
for k from 1 to m do
if nops(Start)=2 then
i,j:=L[k][-1][];
s:={[i-1,j],[i,j+1],[i+1,j],[i,j-1]} else
a,b,c:=L[k][-1][];
s:={[a-1,b,c],[a,b+1,c],[a+1,b,c],[a,b-1,c],[a,b,c-1],[a,b,c+1]} fi; 
s:=`intersect`(s,S1) minus convert(L[k],set);
if s={} and L[k][-1]=Finish then P[k]:=L[k] else
if s={} and L[k][-1]<>Finish then P[k]:=NULL else
P[k]:=`if`(L[k][-1]=Finish,L[k],seq([L[k][],s[i]],i=1..nops(s)))  fi; fi;
od;
L:=convert(P,set) else

P:='P';
m:=nops(L);
for k from 1 to m do
if nops(Start)=2 then
i,j:=L[k][-1][];
s:={[i-1,j],[i,j+1],[i+1,j],[i,j-1]} else
a,b,c:=L[k][-1][];
s:={[a-1,b,c],[a,b+1,c],[a+1,b,c],[a,b-1,c],[a,b,c-1],[a,b,c+1]} fi;
s:=`intersect`(s,S1) minus convert(L[k],set);
if n<N then 
if s={} or L[k][-1]=Finish then P[k]:=NULL else
P[k]:=seq([L[k][],s[i]],i=1..nops(s)); fi else 
if L[k][-1]=Finish and s<>{} then P[k]:=NULL else P[k]:=seq([L[k][],s[i]],i=1..nops(s));
fi; fi;  
od;
L:=convert(P,set)

fi; od;

L;
end proc:

Examples of use.

In the first example from the post above, we find the number of Hamiltonian paths in 

L:=CodeTools:-Usage(AllPaths({seq(seq([i,j], i=1..11), j=1..3)}, [2,2], [10,2], 'H')):
nops(L);

In this same example, we find the number of all paths from A to B and the possible lengths of these paths.

L:=CodeTools:-Usage(AllPaths({seq(seq([i,j], i=1..11), j=1..3)}, [2,2], [10,2])):
nops(L);
map(t->nops(t), L);
L1:=select(t->nops(t)=%[-1], L):
nops(L1);

In the following example, we find the number of all paths, as well as the number of Hamiltonian paths, and animate these paths (total 24 one's).

S:={seq(seq([i,j],i=1..5),j=1..3)} union {seq(seq([i,j],i=4..7),j=3..5)}: A:=[1,1]: B:=[7,5]:
P:=plots:-display(plots:-pointplot(S, symbol=solidcircle, color=blue, symbolsize=15, view=[0..7.5,0..6.5], size=[600,500], scaling=constrained), plots:-textplot([[A[],"A"],[B[],"B"]], font=[times,bold,22], align=[left,above])):
L:=AllPaths(S,A,B):
nops(L);
map(t->nops(t), L);
L1:=select(t->nops(t)=%[-1], L):
nops(L1);
plots:-animate((plots:-display)@(plottools:-curve),[L1[round(a)], color=red, thickness=4], a=1..%, frames=180, background=P, size=[700,500], paraminfo=false);

In the final example, we search for Hamiltonian paths in a lattice defined on the surface of a cube. Imagine a cube made of wires, and an ant must crawl along these wires from point  A(0,0,0)  to point B(2,2,2) , visiting all nodes of this lattice. Is this possible? We see that it is not. The length of the maximum path is 25, and this lattice has 26 vertices. An animation of one of the maximum paths is provided.

S:={seq(seq(seq([i,j,k],i=0..2),j=0..2),k=0..2)} minus {[1,1,1]}: A:=[0,0,0]: B:=[2,2,2]:
P:=plots:-display(plots:-pointplot3d(S, symbol=solidcircle, color=blue, symbolsize=20, scaling=constrained), plots:-textplot3d([[A[],"A"],[B[],"B"]], font=[times,bold,22], align=[left,above]), plottools:-curve([[2,0,1],[2,2,1],[0,2,1],[0,0,1],[2,0,1]], color=black,thickness=0),plottools:-curve([[1,0,2],[1,0,0],[1,2,0],[1,2,2],[1,0,2]], color=black, thickness=0),plottools:-curve([[2,1,0],[0,1,0],[0,1,2],[2,1,2],[2,1,0]], color=black,thickness=0), tickmarks=[3,3,3]):
L:=AllPaths(S,A,B):
nops(L);
map(t->nops(t), L);
L1:=select(t->nops(t)=%[-1], L):
nops(L1);
plots:-animate((plots:-display)@(plottools:-curve),[L1[-1][1..round(a)], color=red, thickness=4], a=1..25, frames=240, background=P, paraminfo=false, axes=box, labels=[x,y,z]);

We can see from this animation that the path does not pass through the vertex (0, 2, 2) .

Edit. A code error that could cause incorrect operation when using the  R  option has been fixed. Everything now works correctly. If there is no Hamiltonian path passing through all vertices, the procedure returns the empty set { } .

Paths1.mw

I would like to plot multiple curves in a single graph with proper legends for different values of θ, similar to the sample figure. However, I am encountering an error while generating the legend. What modifications should I make to correct this? Additionally, how can I display labels such as θ=0.4, θ=0.5 etc., directly inside the plot as shown in the sample?

Graph_legend_error.mw

Although the worksheets provided here have been developed under Maple 2015, they should work correctly with newer versions, except perhaps for commands that use the 'op' function ('piecewise' mainly).

In the sequel the acronym 'pdf' stands for 'probability density function'.


CONTEXT

This post originates from a recent question by @JoyDivisionMan and the ensuing discussion. 

In a few words, the OP noticed Maple 2025 failed to return a result and asked why. In his reply, @acer identified a code regression somewhere in between Maple 2023 and Maple 2025.
Like Maple 2023 "my"  Maple 2015 does not fail but provide... a wrong answer: Do versions in between 2015 and 2024 return wrong results too?
 

In this post I explain how we can calculate the result by hand (only elementary maths required), why Maple 2015 (and likely newer versions) returns an incorrect result, why Maple generally fails in returning a result, and finally provide several examples to illustrate that even mathematically simple.
The various test cases I present are all equally simple for a skilled human agent, but conversely all beyond the reach of Statistics:-PDF.
This raises the question: Can a robust algorithm for calculating this PDF be developed without resorting to an expert system (I don't like the term AI) that mimics human reasoning?


THE MAIN OBSERVATION

Let X some continuous univariate random variable (CURV) and 𝟇 a real valued function from defined over the support of  . Let Y the random variable defined by Y =  𝟇(X).

The main observation about Statistics:-PDF is that unless very specific situations, this procedure does not build the correct expression of pdf(Y) as soon as 𝟇 is not a strictly monotone function

Here are a fex exceptions to this claim:

X any CURV , 𝟇 : x ⟼ xⁿ  , n positive integer (correct solution even if 𝟇 is not monotone)

X ~ Uniform(-1, 1) , 𝟇 : x ⟼ arctanh(x)  (no result returned even if 𝟇 is strictly monotone).

It is worth saying that, only by chance, Statistics:-PDF may sometimes return the correct result even if 𝟇 is a non monotone function.
For instance, in @JoyDivisionMan's question, the procedure was indeed capable to provide a correct result for the case X ~ Uniform(0, 2𝜋) , 𝟇 : x ⟼ arctanh(x) but this was only because two errors balanced each other out. Replace X ~ Uniform(0, 2𝜋)  by X ~ Uniform(a, a+2𝜋) and Maple is wrong (notice that the pdf of 𝟇(X) remains unchanged whatever the value of a).


A GOOD DRAWING WORTH A THOUSAND WORDS

Here is a picture to help understand how to get the pdf of Y =  𝟇(X) for a non monotone function 𝟇 (the case of a monotone function directly comes from this latter).

In this illustration X ~ Uniform(0, 2𝜋) , 𝟇 : x ⟼ sine(x).
To ease the explanation, I write X as a mixture of three uniform random variables X₁, X₂, X₃, whose supports are the intervals of the three branches of 𝟇. More formally, X  = (1/4)∙X₁ + (1/2)∙X₂ + (1/4)∙X₃.
The restrictions  of 𝟇 to these three branches are denoted 𝟇₁, 𝟇₂, 𝟇₃.

The large rectangles below the horizontal axis represent the pdf of X₁, X₂, X₃  and the blue curve the 𝟇 function.
The image of the interval [y-dy, y+dy]  by the inverse functions 𝟇₁^(-1) and 𝟇₂^(-1) of 𝟇₁ and 𝟇₂  are represented by the vertical rectangles [x₁-dy, x₁+dy] and [x₂-dy, x₂+dy] .
These two intervals bring a contribution to the pdf of Y (light gray blue on the right) represented by the horizontal violet rectangle on the right side of the picture.

The probability Prob(Y ∊ [y-dy, y+dy]) that Y belongs to the interval [y-dy, y+dy]) is simply the sum 
                 Prob(Y ∊ [y-dy, y+dy])  = Prob(X₁ ∊ 𝟇₁^(-1)([y-dy, y+dy]))  + Prob(X₂ ∊ 𝟇₂^(-1)([y-dy, y+dy]))

Let 𝟇'_b^(-1)(y) denote the derivative of 𝟇_b^(-1)(y).
Making dy tends to 0 gives
                pdf(Y = y)  =  pdf(X₁ = 𝟇₁^(-1)(y)) ×  | 𝟇'₁^(-1)(y) | + pdf(X₂ = 𝟇₂^(-1)(y)) ×  | 𝟇_'2^(-1)(y) |

As I said before there is truly no big math behind this, Except maybe those absolute values?
To understand where they come from zoom in on the rectangle 𝜔 = [x₁-dx, x₁+dx] ╳ [y-dy, y+dy] and denote X_𝜔 and Y_𝜔 the restrictions of X₁ and Y to 𝜔.
Locally Y_𝜔 is proportional to A+B∙X_𝜔 where constant B = 𝟇'(x₁) and the value of constant A does not matter here.
So the pdf of Y_𝜔 is (a classical result)  pdf(Y_𝜔 = y) = pdf(X_𝜔 = (y-A)/C) / |C|.
Few details can be found Here.


MAPLE FAILURES AND WEAKNESSES

So why did Maple, at last some versions, produce a wrong result and why some versions are not even capable to return one?
The reason is that there is no big math only at first sight...because determining the inverse function of 𝟇 can be quite tricky as soon as 𝟇 is not one-to-one map, for instance when 𝟇 is not strictly monotone.
When it is so 𝟇^(-1) must be defined for all the branches whose definition intervals intersect the support of X.

I spent a lot of time debugging the procedure Statistics:-PDF to understand why it either fails or produces incorrec results.
The sine_debug_nodebugoutput.mw  worksheet presents the "X ~ Uniform(0, 2𝜋) , 𝟇 : x ⟼ sine(x)" case (as Mapleprimes stubbornly refuses to upload the worksheet containing the debugger trace, I convert it to sine_debug.pdf to help you see this trace). 
To orient the core development team correcting this procedure (assumming they care), the critical procedures are

Statistics:-RandomVariables:-PDF:-Univariate:-GetValueTab[anything]
and  Statistics:-RandomVariables:-GetInverse
 

At the very end it is this second procedure which is truly responsible of Maple failing to provide a result or returning an incorrect on, because it does not correctly build the inverse functions 𝟇_b^(-1)  for all the branches b which matter.
 

I wrote above that "it is only by chance that Maple provided the correct result for the non monotone function 𝟇 = cosine". Indeed Statistics:-PDF returns a wrong result when X ~ Uniform(z, z+2𝜋) and z is not a multiple of 𝜋/2 (see cosine.mw).

Other important situations where Maple fails returning a result are those where 𝟇 is a polynomial function with different zeros located in the support of X.
I did not trace them but it seems that Statistics:-PDF does not know how to build the 𝟇_b^(-1) in this case (even though it is quite simple, see "Polynomial" examples below).


A SELECTION OF EXAMPLES

Here is a selection of examples to demonstrate that even in rather complex cases the pdfs of 𝟇(X) can be constructed quite easily (note that Maple either fails to compute them or to provide a correct result):

• Three examples where X is a uniform random variable and the restriction of 𝟇 to the support of X is a non monotone polynomial function of increasing degree (the number of branches to consider is equal to the degree of 𝟇):
  Polynomial_function_of_a_random_variable_1.mw,
  Polynomial_function_of_a_random_variable_2.mw,
  Polynomial_function_of_a_random_variable_3.mw
   
• Two examples where the support o  X is unbounded examples and 𝟇 = sine. Here an infinity number of branches must be accounted for, one of these two examples (X ~ Exponential) can be treated in a comple analytic way while the other (X ~ Normal) cannot thus leading to a truncated approximation of 𝟇(X) pdf.
  Infinte_support_sine.mw

OPEN QUESTION

Tracing Statistics:-PDF reveals an already complex algorithm designed to handle a broad variety of "canonical" situations. Even this the algorithm fails in almost all non-toy-problem such as this compilation proves Maple_failures.mw.
At first sight, there seems to be a contradiction between the (apparent?) simplicity with which one can obtain, by hand in sometimes in an ad hoc way, the expression of pdf(𝟇(X)), whether it is exact or truncated, and the complexity of the Statistics:-PDF algorithm, which results in failure in all non trivial cases.

This observation leads to the important question "Is it possible to rewrite Statistics:-PDF in order to enlarge its domain of success?".
I have the feeling that this means designing an algorithm which focuses more on mimicing the human reasoning than identifying "canonical" situations (as it is done today). 
An AI-driven algorithm maybe?

I repeat here that the maths are very simple, and all the more simple if you represent the random variable X as a mixture of components X₁, ... X_B, both having the same (truncated) distribution than X, and whose supports identify to the intervals of definition of the B branches of 𝟇 over the whole support of X.
The only difficulty lies in the identification of these branches and in the construction of the functions 𝟇_b^(-1) over the supports of each X_b.

I have no answer to this question.

These are the original set up to solve for second order partial differetial equations:

E := 1.456*10^11;
rho := 7900;
a := sqrt(E/rho);
L := 0.03639;

f0 := 1/(2*L)*a;
f := f0 - 50;
A0 := 11.75*L/2*10^(-6);

d := 0.8*1000;
T0 := 5/f;
NULL;
PDE := diff(u(x, t), t, t) = a^2*diff(u(x, t), x, x) - d*diff(u(x, t), t);
IBC := {u(x, 0) = A0*cos(Pi*x/L), D[1](u)(0, t) = 0, D[1](u)(L, t) = 0, D[2](u)(x, 0) = 0};
pds := pdsolve(PDE, IBC, numeric, timestep = 2*10^(-7));

 I know it solves properly because I can plot it using pds:-plot without problem.  The question is how to extract the value at specific x and t.  I started to try some of the more intuitive way such as u(0,0) without getting an answer.  I tried a number of eval expression such as eval(pds,[x=0,t=0]), eval(u(x,t),[x=0,t=0]) and so on without success.  

Then I tried using :-value as mentioned in an example from the internet, also resulting in an error.

val:=pds:-value(x=0,t=0)

It seems to be simple.  I guess I am too new with maple so be gentle.  Thank you.

Hi,

My student licence  expired yesterday mar1. According to the description of my student licence it renews automatically. So I did not get active. But apparently it did not occur automatically. When starting Maple I get the message that my licence expired yesterday. To check whether the automatic renewal applies but did not work or whether I have to purchase a new licence for the next 12 months I tried to contact the customer support through online form. 

I do not see any other channel to contact customer support by mail or phone, what can I do ?

Manfred

When I solve for Pr and q simultaneously using Maple, I obtain a solution that differs from my manual derivation. I suspect there may be an issue with the solve function or the way it is specified. Could anyone please suggest how this can be corrected?
Manually solving q* = (Cv-Cd+Ce*tau-Ci)/(4Cr-t) which is different from maple results.

Sheet:Question_1.mw

Hello Friends,

I have a function f(t) which I would like to convolve with itself.  I have no problem doing that typically, but my function contains two elliptic integrals (EllipticE and EllipticK), which clearly complicates matters.

My work in included.  As you can see, the plot on the (0,2) interval is continuous, and the area under the curve sums to unity via numerical integration.

Download CircleDerivation2.mw

Are there any sort of transformations that I can employ which will permit me to convole this function with itself?  I realize a closed-form solution is probably not likely, but something approaching closed-form would be nice.  Via simulation, I have learned that the self-convolved function will have a continuous boundary on the (0, 2*sqrt(2)) interval.

nans:=dsolve({(-y^4+y^2)*diff(U(y),y,y)+(-2*y^2+14.2)*diff(U(y),y)*y-15.2*U(y)*(U(y)-1)*(U(y)+1) = 0, DU(.99999) = -.3401375821,U(.99999)= 0.3000034013},numeric);
Error, (in dsolve/numeric/process_input) invalid specification of initial conditions, got DU(.99999) = -.3401375821