Hi! Can anyone show me a quick example of a procedure with local and global variables?!? It seems that i'm not getting the hang of it, because i keep receiving "unable to parse" messages. 

During evaluating psi0 (see eq, (7)), why we need two values of f (i.e., f,0 and f,3)? I asked one of my seniors and according to him these values are arbitrary. Can anyone explain why we need two values and why f,0 and f,3?

rwo.mw  

Hi, Maple users 

I hope you are doing well.
Here I solved one of the ode problems by dsolve.
But I am getting  "Error, invalid subscript selector" error.
Kindly do the needful to find a solution.
Thank you.

JVB.mw

 I want to plot phase portrait but i get some errors. I didn't understand errors? Can anyone help me? zuhal_faz_portresi.mw

I have been studying the solutions of the equation below using Maple and Mathematica. 

Here, I want to express the solutions for k. In Maple, the solution is imaginary. As one may see below.

On the other hand, if one solves that  exact equation with Mathematica, then one has real solutions. That is,

Based on the above, I have a few questions:

1. What is the correct solution for the first equation above? That is, are the solutions real or complex?

2. How may I obtain the real solutions, k_{1,2}, above using Maple?

Attached below you may find my mwfile with my doubts.

Mydoubts.mw

Thanks in advance.

Hi! I'm having doubts about the "automatic spacing" in maple. I would like to know two things:

1-can incorrect spacing ruin my code? (Ex by typing "spacebar" too many times )

2- i know that to type spacebar is obligatory after: proc()" "...end proc;    Is there any other situation in wich to use spacebar isn't optional? Thanks for your answers

I want to reverse a catenated expression (eg. xxtt ->x,x,t,t). I don't know the command to achieve this. Anyone with an idea please help

Thanks

Hi! I've got a set of subsets S with A_i being the i-subset. I need to sum up all the x's in all the A_i's. How can this be achieved?

(ex):

sum( { {a, b, c}, {a, b, d} } )= 2a+2b+c+d

Hello,

I am writing an arrow procedure and will like to know if there is a way to implement the following 
 

bj := (G, y) ->  (`@`(seq((t -> u -> v -> G(u, t) - v)(args[1 + nargs - i]), i = 1 .. nargs - 2)))(y)
bj(F, y, a, b, c, d);
v:=[p, q, r, s]
the output

My question is, how can I replace the v with each element of the list to get the following as output
F(F(F(F(y,a)-p),b)-q,c)-r,d)-s

Aany suggestion will be highly appreciated

Hi! I'm new to Maplesoft. I would like to use it for my research. However, in order to do so, i have to construct an n-dimensional vector space, with vectors of the form A_1*B, A_2*B, ..., A_k*B, where A_i is a set of  n components from a Field (or ring) and B is the basis for the vector space.

It would REALLY help if you could give some ideas for me to bring this up, or else i will be forced to typset every one of the vectors i'll use in a matrix-multiplication form and will make considerably harder my work.

I am trying to develop a recursive code to the above equations.  Note, U,X&Y are multivariate functions (in this case bivariate functions of (x,y)) i.e. U=U(x,y), X=X(x,y) etc.

Given a list L=[x,y,z], what is the best way to generate the following sequential expression?
L₁=[[x],[y],[z],[x,x],[x,y],[[x,z],[y,y],[y,z],[z,z],[x,x,x],[x,x,y],[x,x,z],[x,y,y],[x,y,z],[y,y,y],[y,y,z],[y,z,z],[z,z,z]].

With L₁ in mind (a list of list of at most three variables), how can one generate a list of lists of any number of variable from L.

I am currently out of options, a response will be highly appreciated.

Thanks

Dear all,

I do not understand the output of the "Dual" command in the Logic package.

According to the help:

"The Dual command returns the dual of the Boolean expression b, that is, the expression generated by replacing &and with &or, &or with &and, leaving &not fixed, and extending to the remaining Boolean operators by their formulas in terms of &and, &or, and &not."

Applying Dual to "a &implies b" gives:
  `Logic:-&implies`(`&not`(a), `&not`(b))

But "a &implies b" is equivalent to "&not(a) &or b" and its Dual would be "&not(a) &and b", which is not equivalent to the result Maple returns.

Best regards,
 Anthei

restart;
u := (H(x, t, z)+sqrt(R))*exp(I*R*x);
                /              (1/2)\           
                \H(x, t, z) + R     / exp(I R x)

I*(Diff(u, z))+Diff(u, `$`(x, 2))+Diff(u, `$`(t, 2))+(abs(u)*abs(u))*u-((abs(u)*abs(u))*abs(u)*abs(u))*u;
  / d  //              (1/2)\           \\
I |--- \\H(x, t, z) + R     / exp(I R x)/|
  \ dz                                   /

     / 2                                   \
     |d  //              (1/2)\           \|
   + |-- \\H(x, t, z) + R     / exp(I R x)/|
     \                                     /

     / 2                                   \                    
     |d  //              (1/2)\           \|                  2 
   + |-- \\H(x, t, z) + R     / exp(I R x)/| + (exp(-Im(R x)))  
     \                                     /                    

                       2                                    
  |              (1/2)|  /              (1/2)\              
  |H(x, t, z) + R     |  \H(x, t, z) + R     / exp(I R x) - 

                                        4                       
                 4 |              (1/2)|  /              (1/2)\ 
  (exp(-Im(R x)))  |H(x, t, z) + R     |  \H(x, t, z) + R     / 

  exp(I R x)
value(%);
  / d            \              / d  / d            \\           
I |--- H(x, t, z)| exp(I R x) + |--- |--- H(x, t, z)|| exp(I R x)
  \ dz           /              \ dx \ dx           //           

         / d            \             
   + 2 I |--- H(x, t, z)| R exp(I R x)
         \ dx           /             

     /              (1/2)\  2           
   - \H(x, t, z) + R     / R  exp(I R x)

     / d  / d            \\                             2 
   + |--- |--- H(x, t, z)|| exp(I R x) + (exp(-Im(R x)))  
     \ dt \ dt           //                               

                       2                                    
  |              (1/2)|  /              (1/2)\              
  |H(x, t, z) + R     |  \H(x, t, z) + R     / exp(I R x) - 

                                        4                       
                 4 |              (1/2)|  /              (1/2)\ 
  (exp(-Im(R x)))  |H(x, t, z) + R     |  \H(x, t, z) + R     / 

  exp(I R x)
simplify(%);
  / d            \              / d  / d            \\           
I |--- H(x, t, z)| exp(I R x) + |--- |--- H(x, t, z)|| exp(I R x)
  \ dz           /              \ dx \ dx           //           

         / d            \                 2                      
   + 2 I |--- H(x, t, z)| R exp(I R x) - R  exp(I R x) H(x, t, z)
         \ dx           /                                        

      (5/2)              / d  / d            \\           
   - R      exp(I R x) + |--- |--- H(x, t, z)|| exp(I R x)
                         \ dt \ dt           //           

                                                  2           
                             |              (1/2)|            
   + exp(-2 Im(R x) + I R x) |H(x, t, z) + R     |  H(x, t, z)

                                                  2       
                             |              (1/2)|   (1/2)
   + exp(-2 Im(R x) + I R x) |H(x, t, z) + R     |  R     

                                                  4           
                             |              (1/2)|            
   - exp(-4 Im(R x) + I R x) |H(x, t, z) + R     |  H(x, t, z)

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

        / d            \   / d  / d            \\
    + I |--- H(x, t, z)| + |--- |--- H(x, t, z)||
        \ dz           /   \ dx \ dx           //

      / d  / d            \\\           
    + |--- |--- H(x, t, z)||| exp(I R x)
      \ dt \ dt           ///           

                                                   2           
                              |              (1/2)|            
    + exp(-2 Im(R x) + I R x) |H(x, t, z) + R     |  H(x, t, z)

                                                   2       
                              |              (1/2)|   (1/2)
    + exp(-2 Im(R x) + I R x) |H(x, t, z) + R     |  R     

                                                   4           
                              |              (1/2)|            
    - exp(-4 Im(R x) + I R x) |H(x, t, z) + R     |  H(x, t, z)

                                                   4       
                              |              (1/2)|   (1/2)
    - exp(-4 Im(R x) + I R x) |H(x, t, z) + R     |  R     
 

restart;
H := a__1*exp(I*(-Omega*t+k*x+z*k[1]))+a__2*exp(-I*(-Omega*t+k*x+z*k[1]));
I*(diff(H, z))+diff(H, x, x)+diff(H, t, t)+R*(H+conjugate(H))+R^2*(H+conjugate(H))*H;
value(%);
simplify(%);

Download m18bs.mw