I use  Maple 2015 and I try to understand how the simplification rules apply in the case of the expression 

f := n -> (ln(x)^n)^(1/n)

Here n is assumed to be a strictly positive and I consider only the cases "n is an integer" or "1/n is an integer".

All the questions are orange written in the attached file and resumed below:

1. Why simplify(f(2)) simplifies f(2) whereas simplify(f(n)) doesn't simplifies f(n) for any integer n > 2?
    
2. Why simplify(f(1/n)) simplifies f(1/n)?
    
3. Why simplify(f(3)) with adhoc assumptions returns a simplified expression of some form whereas, for any integer n > 3,  simplify(f(n)) with (the same corresponding) adhoc assumptions returns a simplified expression of a complete different form than with n=3?

Can you please have a look to it and give me some clarifications?
Simplification_rules.mw

Thanks in advance

As I was numerically investigating this recent question I incidentally discovered a strange behaviour of Maple 2015 (which maybe exists in more recent versions?)

The attached worksheet presents an erratic behaviour (plus a remanance isssue because saving it, and opening it again changes the displays).
Note that this strange behaviour seems to occur only when tickmarks use the atomic name `#mo("2")`.

display_issue.mw

Here is a pdf print of this same worksheet: as I hope you will see (because I don't know what you are about to see when opening the attached worksheet) its content differs from the worksheet's. 

display_issue.pdf

Here are 3 screen captures which show what MY worksheet looks like

PAGE 1


PAGE 2
There is a typo in the comment below: read "void" instead of "coid", sorry for the mistake.


PAGE 3


Is this a Maple 2015 issue which has been fixed in earlier versions?
Is there a way to fix these issues?

( squircle is the humoristic name for the 2D open ball of center 0 and radius 1 in L_n norm ).
The equation of the squircle in L_n norm writes  |x|ⁿ+|y|ⁿ = 1

The attached file gives the exact values of the areas of squircles in norms L₂, L₄, L₁₀₀, L₁. 
Unless for n=2 the results are dramatically poor (evalf/Int gives the same wrong results).

The function a(n) gives the exact expression of the squircle area in  L_n norm.

squircle.mw

I have some cubic and quartic equations with complex cofficients. Maple 2015 is able to solve these and returns the roots as labelled sets, so I can do things like "plot S[1]". I want to vary some parameters in the coefficients, and see what happens to the roots.
My problem is when I log out and then rerun the code, the labels 1,2,3,(4) are frequently attached to different roots than they were the first time. This is both unexpected and inconvenient. Is there any way to ensure that the same roots are always given the same labels?

[moderator: see also this Question from 2023]

Can anyone explain me the reason of the last result?
Thanks in advance

Download What-does-happen-here.mw

Hi!

I am using a proceure to conpute de integral of a function by he Simpson's rule. My function is defined from a function and a procedure, but I am getting the error  "Error, (in w) invalid input: hfun2 expects its 1st argument, t, to be of type numeric, but received (1/10)*i+1/20"

As you can see in the attaxhed file, I have tried several ways to compute the integral but always returns the above error. Please, can yo help me?

Thanks

forum.mw

Hi Dear,

I hope everyone is fine here. In the attached file, I have generated a square matrix "Q" using two-dimensional polynomials. The dimension of the square matrix "Q" depends on M1 and M2 parameters. In my simulation, sometimes I need this matrix of 1000 by 1000 dimensions. Using the attached method, it took a lot of time to compute two-dimensional polynomials and then to compute the general square matrix "Q." I wanted to write this matrix using proc (procedures). Maybe by using this way, I don't need to compute the polynomials, and it took less time to compute the square matrix "Q." I know how to generate a matrix using proc when its dimension depends on one parameter. However, here, the dimension of matrix "Q" depends on two parameters, M1 and M2. So, I am a little bit confused about how to adjust them in proc. Please see the attached file and share your useful ideas. 

help.mw

Thanks in advance

Hi
I hope you are doing well. I have plotted (in the attached file) the contour plot of the function and its density plot; both have the same behavior but different appearances (error in direction may be rotation needs to apply). I don't know why it happens because this code works well for other solutions. Kindly have a look and fix the issue. I shall be waiting for your positive response. Please take care.
Help.mw

It seems that Maple needs more help than necessary:

Download limits.mw

Why don't the commands labelled "Shouldn't this return.." do the job?

TIA

I have an expression equal to the sum of N terms of the form Int(f_n=1..N(x), x) and I want to replace each f_n(x) by its Taylor (or series) expansion.

When the integrals are definite, like J1 below, I can easily obtain a new expression (K1) where the integrand has been replaced by some expansion.
But when the integral is indefinite, like J2, I get an evaluated expression for K2.

It seems I have to do some gymnastic (J3 --> K3) to get what I want

Download Integration.mw

Why do I get this unevaluatedform for K2?
Do I have to use Intat to get K3?

Thanks in advance

For years I observe that package orthopoly is not considered as a package within a procedure.
Finally I have decided to ask for clarifications: Can someone explain me why procedure f generates an error?
 

Download meaning.mw

TIA

This post is inspired by minhthien2016's question.

The problem, denoted 2/N/1, for reasons that will appear clearly further on, is to pack N disks into the unit square in such a way that the sum of their radii is maximum.

I replied this problem using Optimization-NLPSolve for N from 1 (obvious solution) to 16, which motivated a few questions, in particular:

• @Carl Love: "Can we confirm that the maxima are global (NLPSolve tends to return local optima)?" 
  Using NLPSolve indeed does not guarantee that the solution found is the (a?) global maximum. In fact packing problems are generaly tackled by using evolutionnary algorithms, greedy algorithms, or specific heuristic strategies.
  Nevertheless, running NLPSolve a large number of times from different initial points may provide several different optima whose the largest one can be reasonably considered as the global maximum you are looking for.
  Of course this may come to a large price in term of computational time.
   
• @acer: "How tight are [some constraints], at different N? Are they always equality?"
  The fact some inequality constraints type always end to equality constraints (which means that in an optimal packing each disk touches at least one other annd, possibly the boundary of the box) seems hard to prove mathematically, but I gave here a sketch of informal proof.

I found 2/N/1 funny enough to spend some time digging into it and looking to some generalizations I will refer to as D/N/M:  How to pack N D-hypersheres into the unit D-hypercube such that the sum of the M-th power of their radii is maximum?
For the sake of simplicity I will say ball instead of disk/sphere/hypersphere and box instead of square/cube/hypercube.

The first point is that problems D/N/1 do not have a unique solution as soon as N > 1 , indeed any solution can be transformed into another one using symmetries with respect to medians and diagonals of the box. Hereafter I use this convention:

Two solutions s and s' are fundamental solutions if:

1. the ordered lists of radii s and s'  contain are identical but there is no composition of symmetries from s to s',
2. or, the ordered lists of radii s and s'  contain are not identical.
    

It is easy to prove that 2/2/1 and 3/2/1, and likely D/2/1, have an infinity of fundamental solutions: see directory FOCUS__2|2|1_and_3|2|1 in the attached zip file..
At the same time 2/N/2, 3/N/3, and likely D/N/D, have only one fundamental solution (see directory FOCUS__2|N|2 for more details and a simple way to characterize these solutions

 (Indeed the strategy ito find the solution of D/N/D  in placing the biggest possible ball in the largest void D/N-1/D contains. Unfortunately this characterization is not algorithmically constructive in the sense that findind this biggest void is a very complex geometrical and combinatorial problem.
 it requires finding the largest void  in a pack of balls)

Let M_{d, 1}(N)  the maximum value of the sum of balls radii for problem d/N/1.
The first question I asked myself is: How does M_{d, 1}(N) grows with N?
 

(M_{d, 1}(N) is obviously a strictly increasing function of N: indeed the solution of problem d/N/1 contains several voids where a ball of strictly positive radius can be placed, then  M_{d+1, 1}(N) > M_{d, 1}(N) )

The answer seems amazing as intensive numerical computations suggest that
                                      

See D|N|M__Growth_law.mw in the attached sip file.
This formula fits very well the set of points  { [n, S_{d, 1}(n) , n=1..48) } for d=2..6.
I have the feeling that this conjecture might be proven (rejected?) by rigourous mathematical arguments.


Fundamental solutions raise several open problems:

• Are D/2/1 problems the only one with more than one fundamental solutions?
  
  The truth is that I have not been capable to find any other example (which does not mean they do not exist).
  A quite strange thing is the behaviour of NLPSolve: as all the solutions of D/2/1 are equally likely, the histogram of the solutions provided by a large number of NLPSolve runs from different initial points is far from being uniform.
  For more detail refer ro directory FOCUS__2|2|1_and_3|2|1 in the attached zip file. 
  I do not understand where this bias comes from: is it due to the implementation of SQP in NLPSolve, or to SQP itself?
   
• For some couples (D, N) the solution of D/N/1 is made of balls of same radius.
  For N from 1 to 48 this is (numerically) the case for 2/1/1 and2/2/1, but the three dimensional case is reacher as it contains  3/1/1, 3/2/1,  3/3/1,  3/4/1 and 3/14/1 (this latter being quite surprising).
  Is there only a finite number of values N such that D/N/1 is made of balls with identical radii?
  If it is so, is this number increasing with D?
  It is worth noting that those values of N mean that the solution of problems D/N/1 are identical to those of a more classic packing problem: "What is the largest radius N identical balls packed in a unit bow may have?".
  For an exhaustive survey of this latter problem see Packomania.
   
• A related question is "How does the number of different radii evolves as N increases dor given values of D and M?
  Displays of 2D and 3D packings show that the set of radii has significantly less elements than N... at least for values of N not too large. So might we expect that solution of, let us say, 2/100/1 can contain 100 balls of 10 different radii, or it is more reasonable to expect it contains 100 balls of 100 different radii?
   
• At the opposite numerical investigations of  2/N/1 and  3/N/1 suggest that the number of different radii a fundamental solution contains increases with N (more a trend than a continuous growth).
  So, is it true that very large values of N correspond to solutions where the number of different radii is also very large?
  Or could it be that the growth of the number of different radii I observed is simply the consequence of partially converged results?
   
• Numerical investigations show that for a given dimension d and a given number of balls n,  solutions of d/n/1 and d/n/M (1 < M < d) problems are rather often the same. Is this a rule with a few exceptions or a false impression due to the fact that I did not pushed the simulations to values of n large enough to draw a more solid picture)?

It is likely that some of the questions above could be adressed by using a more powerful machine than mine.


All the codes and results are gathered in  a zip file you can download from OneDrive Google  (link at the end of this post, 262 Mb, 570 Mb when unzipped, 1119 files).
Install this zip file in the directory of your choice and unzip it to get a directory named PACKING. 
Within it:

• README.mw contains a description of the different codes and directories
• Repository.rtf must contain a string repesenting the absolute path of directory PACKING. 

Follow this link OneDrive Google

Hi,

I hope everyone is fine here. I want to write the following nested loop in my code

restart; TOL := 10^(-3); v[i] := 10^(-4); u[i] := 10^(-4);
if abs(v[i]-1)<=TOL then    if abs(u[i])<=TOL then break    else t[i+1]:=t[i]+alpha; else; fi; fi;

it is not working. If the first condition (v[i]-1)<=TOL is fulfilled, then we have to check the second condition, u[i]<=TOL. That’s why I don’t use “and” here. If both conditions are fulfilled, then stop the loop and if second the condition not fulfilled (but first fulfill) then t[j+1]=t[j]+alpha. 

I shall be waiting for your positive response.

Why sorting these 4 vectors wrt L(+oo) norm returns a correct result bur sorting them wrt L2 norm doesn't (unless if I evaluate the norms as floats)?
 

Download SortingVectors.mw

TIA

I have a surface defined by C(x, y, x) = 0 that I visualize with implicitplot3d.
Using shading=shue does not suits me and I would like to define my own coloring function F(x, y, z).

The first error I got made me think that a coloring function cannot depend on 3 parameters.
But a simpler (and not visually satysfying function) F(x, y) already leads to an error, which makes me wonder if it is possible to use a colorig function with  implicitplot3d?

implicitplot3d_coloring.mw

Thanks in advance