Maple Questions and Posts

These are Posts and Questions associated with the product, Maple

Hi, I am using ArchLinux and used Maple's official installer to install it. Whenever I export my Maple files as PDF all my "-" symbols are converted to "K" for some reason. Has anyone else had this issue or have an idea on how to fix it?

Below is a screenshot of what I mean. "-" has been replaced with "K"

Hi! Do you know how can I plot revolution of this curve:

plot(1 + 1/4*sin(8*(1 + 1/8*sin(16*(1 + 1/16*sin(32*t))))), t = 0 .. 2*Pi, coords = polar)

to get a wrinkled torus like this:

Abelprijs 2015 voor John Nash en Louis Nirenberg - NEMO Kennislink

Do u have any idea? (I've tried to convert this polar curve to cartesian coords but I don't get it)

Hello

I have a list with n positive integer numbers and a positive integer k.

I would like to write a Malple programm, that calculates, in how many ways k can be written as a sum of numbers in the list.

Easy examles: let the list be [2,3,6,9].

k=9 can be written in 4 ways: 9=9, 9=6+3, 9=3+3+3, 9=3+2+2+2. The result therefore shoud be 4.

k=8 can be written in 3 ways: 8=6+2, 8=3+3+2, 8=2+2+2+2. The result therefore shoud be 3.

Thank you very much your help!

Hi everyone,

I am using Maple for quite a while, now I would like to do something new.

I am a teacher at a german highschool and I would like to generate class tests with maple. Say for an example, I need 30 tests,  where the first task is calculating the first derivative of a function. This function should be different for every person, for example f(x)=a*x^3+3x^2, where the value of a is different in every test (probably random).

So what I need is a way to generate nice looking sheets in Maple (including page breaks, titles and so on). Is there a way to generate such a thing in maple or do I have to export to TeX and adjust there quite a bit?

Maybe someone has done a similar thing. Actually I am looking for something like serial letters in Maple.

Any idea, how I could do such a thing?

Thanx for your help,

Axel

GraphTheory:-GraphEqual says that G1 and G2 are equal, but GraphTheory:-AllPairsDistance gives different results instead: 

restart;

with(GraphTheory)

M := `<|>`(`<,>`(0, 0, 0), `<,>`(1, 0, 0), `<,>`(1, 1, 0))

G__1 := Graph(convert(-M, Matrix, datatype = integer[8]))

G__2 := Graph(convert(-M, Matrix, datatype = integer))

GraphEqual(G__1, G__2)

true

(1)

AllPairsDistance(G__1)

AllPairsDistance(G__2)

Matrix(%id = 36893491227039185244)

 

Error, (in GraphTheory:-AllPairsDistanceExt) negative cycle detected

 

 

Download allpairs.mw

So, which one is incorrect? Any reasons?

restart

with(PolynomialIdeals):

``

``

randomzero := proc (ListVar) local A, G, i, lm, B, f, g; option trace; A := ListVar; G := NULL; for i to nops(A) do f := A[i]^(i+1)+randpoly([op(`minus`({op(A)}, {A[i]}))], terms = 1, coeffs = rand(-4 .. -1), degree = i); G := G, f end do; print(IsZeroDimensional(`<,>`(G))); RETURN([G]) end proc:

randomzero([x, y])

{--> enter randomzero, args = [x, y]

 

[x, y]

 

"G:="

 

x^2-y-4

 

x^2-y-4

 

y^3-3*x

 

x^2-y-4, y^3-3*x

 

true

 

<-- exit randomzero (now at top level) = [x^2-y-4, y^3-3*x]}

 

[x^2-y-4, y^3-3*x]

(1)

``

Download bug.mw

Hi

I think there is a bug in the "randpoly" command. please see the attached file line 7 of my procedure "randomzero". Why x^2-y-4 is created while terms=1 is considered and the outputs must contain binomial?

Hi
It's been years since expressions like A %* B %+ C involving inert arithmetic operators used in infix form are correctly understood (parsed) when written on a 1D-Math input line. The idea is simple: have the operators %. %*, %+, %-, %^, %/ work on input as infix operators the same way their active forms: ., *, +, -, ^, / do. This useful functionality, however, remained elusive when using 2D-Math input notation, so one would have to resort to using the functional form of the operators. E.g., input the above as `%+`(`%*`(A, B) ,C), which for me is really ugly. Besides being a bit demoralizing: we do all this fuzz about how great computer algebra and 2D-Math input notation is, and then input things in that way …

So this is to mention that this elusive functionality of inert arithmetic operators used in infix form within a 2D-Math input line now works. The novelty is present in the latest Maplesoft Physics Updates for Maple 2023, which is version 1490. As usual, to install the Updates open a Maple worksheet and input Physics:-Version(latest).

Here is an image (worksheet at the end) showing the new thing


The implementation is pretty new; reports of anything related to these inert operators not displayed/working as you'd expect are much appreciated. 


Download Inert_arithmetic_operators_in_2D_Math.mw

Edgardo S. Cheb-Terrab
Physics, Differential Equations and Mathematical Functions, Maplesoft

Some work, while others don't work. 
 

restart;

`assuming`([is(1/(sqrt(n*(n+1))*(sqrt(n)+sqrt(n+1))) = 1/sqrt(n)-1/sqrt(n+1))], [n::posint])

true

(1)

sum(1/sqrt(n)-1/sqrt(n+1), n = 1 .. infinity)

1

(2)

sum(1/(sqrt(n*(n+1))*(sqrt(n)+sqrt(n+1))), n = 1 .. infinity)

sum(1/((n*(n+1))^(1/2)*(n^(1/2)+(n+1)^(1/2))), n = 1 .. infinity)

(3)

`assuming`([sum(binomial(3*n, n)*x^n/(2*n+1), n = 0 .. infinity)], [abs(x) <= 4/27])

(2/3)*3^(1/2)*sin((1/3)*arcsin((3/2)*3^(1/2)*x^(1/2)))/x^(1/2)

(4)

`assuming`([eval(binomial(3*n, n)*x^n/(2*n+1), n = 0)+sum(binomial(3*n, n)*x^n/(2*n+1), n = 1 .. infinity, parametric)], [abs(x) <= 4/27])

1+x*hypergeom([1, 4/3, 5/3], [2, 5/2], (27/4)*x)

(5)

simplify(convert(1+x*hypergeom([1, 4/3, 5/3], [2, 5/2], (27/4)*x)-hypergeom([1/3, 2/3], [3/2], (27/4)*x), elementary), symbolic)

(1/3)*(3*x^(1/2)+3*x^(3/2)*hypergeom([1, 4/3, 5/3], [2, 5/2], (27/4)*x)-2*3^(1/2)*sin((1/3)*arcsin((3/2)*3^(1/2)*x^(1/2))))/x^(1/2)

(6)

plot((1/3)*(3*x^(1/2)+3*x^(3/2)*hypergeom([1, 4/3, 5/3], [2, 5/2], (27/4)*x)-2*3^(1/2)*sin((1/3)*arcsin((3/2)*3^(1/2)*x^(1/2))))/x^(1/2), x = -4/27 .. 4/27)

 

verify(0, (1/3)*(3*x^(1/2)+3*x^(3/2)*hypergeom([1, 4/3, 5/3], [2, 5/2], (27/4)*x)-2*3^(1/2)*sin((1/3)*arcsin((3/2)*3^(1/2)*x^(1/2))))/x^(1/2), 'equal')

FAIL

(7)

NULL


 

Download unable_to_sum.mw

How to explain this behavior?
I have read the help page, but I can not get the point.

Please, is there anyway I can solve this problem with a nice output? The solution I got is complicated. Please, helper is need

f := exp(-beta*x)*x^(5-alpha)

int(f, x)

beta^(alpha-6)*(-x^(-alpha)*beta^(-alpha)*(alpha^4-14*alpha^3+71*alpha^2-154*alpha+120)*(alpha-6)*(beta*x)^((1/2)*alpha)*exp(-(1/2)*beta*x)*WhittakerM(-(1/2)*alpha, -(1/2)*alpha+1/2, beta*x)/(-alpha+6)+x^(-alpha)*beta^(-alpha)*(beta^4*x^4-alpha*beta^3*x^3+alpha^2*beta^2*x^2+5*beta^3*x^3-alpha^3*beta*x-9*alpha*beta^2*x^2+alpha^4+12*alpha^2*beta*x+20*beta^2*x^2-14*alpha^3-47*alpha*beta*x+71*alpha^2+60*beta*x-154*alpha+120)*(alpha-6)*(beta*x)^((1/2)*alpha)*exp(-(1/2)*beta*x)*WhittakerM(-(1/2)*alpha+1, -(1/2)*alpha+1/2, beta*x)/(-alpha+6))

How to write a program that obtains the value of the function by giving different values
Like the example below:

 

I rephrased my previous question in a more synthetic form
(there was probably a lot in it that I thought was important for understanding the problem, but I realized afterwards that it only added confusion).

The true question is yellow-highlighted in the code below

restart

# The result below seems natural: we were taught in school that exp 
# being a bijective function we can get rid of it in the equality to
# solve and write simply x=Pi.

x = solve(exp(x)=exp(Pi), x)

x = Pi

(1)

# But the solution method solve uses is not that natural (and I
# don't really understand it).
# infolevel[solve] := 10:
# x = solve(exp(x)=exp(Pi), x);

# Replacing now exp by some undefined function f produces a
# kind of "no-solution" answer: this seems quite normal because
# not knowing the properties of f one cannot simply get rid of it.

infolevel[solve] := 0:
x = solve(f(x)=f(Pi), x)

x = RootOf(f(_Z)-f(Pi))

(2)

# Finally replace f by a bijective function with no analytic expression.

s = solve(erf(x)=erf(Pi), x) assuming x::real

s = RootOf(erf(_Z)-erf(Pi))

(3)

# It would have seem reasonable for Maple to answer x=Pi, or
# at least it is what I would have done given the properties
# of the erf function.
#
# How can I "force" Maple to "simplify" it's RootOf result to get
# x=Pi?


Download solve_erf.mw

For those interested in the motivations of this quastion, see here Where_does_the_question_come_from.mw

The original question is here Original_question.pdf

How is the expansion of trigonometric functions in Maple like the following function?

that in Kip Thorne's book Maple is mentioned through out ?

Kip Thorne and Roger Blandford : "Modern Classical Physics: Optics, Fluids, Plasmas, Elasticity, Relativity, and Statistical Physics"

also, courtesy of Caltech in Chapter 24

restart;
alias(u = u(x, z, t), f = f(x, z, t));
                              u, f
u := (f+sqrt(R))*exp(I*R*x);
                    /     (1/2)\           
                    \f + R     / exp(I R x)
pde1 := I*(diff(u, z))+diff(u, x, x)+diff(u, t, t)+u*abs(u)*abs(u)-(u*abs(u)*abs(u))*abs(u)*abs(u);
    / d   \              / d  / d   \\           
  I |--- f| exp(I R x) + |--- |--- f|| exp(I R x)
    \ dz  /              \ dx \ dx  //           

           / d   \                /     (1/2)\  2           
     + 2 I |--- f| R exp(I R x) - \f + R     / R  exp(I R x)
           \ dx  /                                          

       / d  / d   \\           
     + |--- |--- f|| exp(I R x)
       \ dt \ dt  //           

                                                            2
       /     (1/2)\                           2 |     (1/2)| 
     + \f + R     / exp(I R x) (exp(-Im(R x)))  |f + R     | 

                                                            4
       /     (1/2)\                           4 |     (1/2)| 
     - \f + R     / exp(I R x) (exp(-Im(R x)))  |f + R     | 

simplify(%);
         / d   \              / d  / d   \\           
       I |--- f| exp(I R x) + |--- |--- f|| exp(I R x)
         \ dz  /              \ dx \ dx  //           

                / d   \                 2             
          + 2 I |--- f| R exp(I R x) - R  exp(I R x) f
                \ dx  /                               

             (5/2)              / d  / d   \\           
          - R      exp(I R x) + |--- |--- f|| exp(I R x)
                                \ dt \ dt  //           

                                               2  
                                   |     (1/2)|   
          + exp(I R x - 2 Im(R x)) |f + R     |  f

                                               2       
                                   |     (1/2)|   (1/2)
          + exp(I R x - 2 Im(R x)) |f + R     |  R     

                                               4  
                                   |     (1/2)|   
          - exp(I R x - 4 Im(R x)) |f + R     |  f

                                               4       
                                   |     (1/2)|   (1/2)
          - exp(I R x - 4 Im(R x)) |f + R     |  R     
collect(%, exp(I*R*x));
  /  (5/2)       / d   \      2       / d   \   / d  / d   \\
  |-R      + 2 I |--- f| R - R  f + I |--- f| + |--- |--- f||
  \              \ dx  /              \ dz  /   \ dx \ dx  //

       / d  / d   \\\           
     + |--- |--- f||| exp(I R x)
       \ dt \ dt  ///           

                                          2  
                              |     (1/2)|   
     + exp(I R x - 2 Im(R x)) |f + R     |  f

                                          2       
                              |     (1/2)|   (1/2)
     + exp(I R x - 2 Im(R x)) |f + R     |  R     

                                          4  
                              |     (1/2)|   
     - exp(I R x - 4 Im(R x)) |f + R     |  f

                                          4       
                              |     (1/2)|   (1/2)
     - exp(I R x - 4 Im(R x)) |f + R     |  R     
 

Hi Dears

I need some random zero-dimensional binomial ideals (20 ideals or more) with two, three, or four ... generators with 4 variables atmost. Then I want to regenerate each of them such that some of their generators are not binomial and the obtained ideals are equal to the first corresponding original binomial ideals. How can do I this automatically?

As a simple example let I be an ideal generated by {x-1, y-1, z-1} which is zero-dim. We can obtain J=<x-z, x+z-2, y+z-2> that is equal to I.

Thank you in advance.

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