Maple 18 Questions and Posts

These are Posts and Questions associated with the product, Maple 18

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 Abeing the i-subset. I need to sum up all the x's in all the Ai'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 Lin 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(%);

restart;

H := a__1*exp(I*(-Omega*t+k*x+z*k[1]))+a__2*exp(-I*(-Omega*t+k*x+z*k[1]));

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;

I*(I*a__1*k[1]*exp(I*(-Omega*t+k*x+z*k[1]))-I*a__2*k[1]*exp(-I*(-Omega*t+k*x+z*k[1])))-a__1*k^2*exp(I*(-Omega*t+k*x+z*k[1]))-a__2*k^2*exp(-I*(-Omega*t+k*x+z*k[1]))-a__1*Omega^2*exp(I*(-Omega*t+k*x+z*k[1]))-a__2*Omega^2*exp(-I*(-Omega*t+k*x+z*k[1]))+R*(a__1*exp(I*(-Omega*t+k*x+z*k[1]))+a__2*exp(-I*(-Omega*t+k*x+z*k[1]))+conjugate(a__1*exp(I*(-Omega*t+k*x+z*k[1]))+a__2*exp(-I*(-Omega*t+k*x+z*k[1]))))+R^2*(a__1*exp(I*(-Omega*t+k*x+z*k[1]))+a__2*exp(-I*(-Omega*t+k*x+z*k[1]))+conjugate(a__1*exp(I*(-Omega*t+k*x+z*k[1]))+a__2*exp(-I*(-Omega*t+k*x+z*k[1]))))*(a__1*exp(I*(-Omega*t+k*x+z*k[1]))+a__2*exp(-I*(-Omega*t+k*x+z*k[1])))

value(%);

I*(I*a__1*k[1]*exp(I*(-Omega*t+k*x+z*k[1]))-I*a__2*k[1]*exp(-I*(-Omega*t+k*x+z*k[1])))-a__1*k^2*exp(I*(-Omega*t+k*x+z*k[1]))-a__2*k^2*exp(-I*(-Omega*t+k*x+z*k[1]))-a__1*Omega^2*exp(I*(-Omega*t+k*x+z*k[1]))-a__2*Omega^2*exp(-I*(-Omega*t+k*x+z*k[1]))+R*(a__1*exp(I*(-Omega*t+k*x+z*k[1]))+a__2*exp(-I*(-Omega*t+k*x+z*k[1]))+conjugate(a__1*exp(I*(-Omega*t+k*x+z*k[1]))+a__2*exp(-I*(-Omega*t+k*x+z*k[1]))))+R^2*(a__1*exp(I*(-Omega*t+k*x+z*k[1]))+a__2*exp(-I*(-Omega*t+k*x+z*k[1]))+conjugate(a__1*exp(I*(-Omega*t+k*x+z*k[1]))+a__2*exp(-I*(-Omega*t+k*x+z*k[1]))))*(a__1*exp(I*(-Omega*t+k*x+z*k[1]))+a__2*exp(-I*(-Omega*t+k*x+z*k[1])))

simplify(%);

a__1^2*exp(-(2*I)*(Omega*t-k*x-z*k[1]))*R^2+2*a__1*a__2*R^2+a__2^2*exp((2*I)*(Omega*t-k*x-z*k[1]))*R^2-exp(-I*(Omega*t-k*x-z*k[1]))*a__1*k[1]+a__2*k[1]*exp(I*(Omega*t-k*x-z*k[1]))+a__1*exp(-I*(Omega*t-k*x-z*k[1]))*conjugate(a__1*exp(-I*(Omega*t-k*x-z*k[1]))+a__2*exp(I*(Omega*t-k*x-z*k[1])))*R^2+a__2*exp(I*(Omega*t-k*x-z*k[1]))*conjugate(a__1*exp(-I*(Omega*t-k*x-z*k[1]))+a__2*exp(I*(Omega*t-k*x-z*k[1])))*R^2-a__1*Omega^2*exp(-I*(Omega*t-k*x-z*k[1]))-a__1*k^2*exp(-I*(Omega*t-k*x-z*k[1]))-a__2*Omega^2*exp(I*(Omega*t-k*x-z*k[1]))-a__2*k^2*exp(I*(Omega*t-k*x-z*k[1]))+R*a__1*exp(-I*(Omega*t-k*x-z*k[1]))+R*a__2*exp(I*(Omega*t-k*x-z*k[1]))+R*conjugate(a__1*exp(-I*(Omega*t-k*x-z*k[1]))+a__2*exp(I*(Omega*t-k*x-z*k[1])))

 

Download m18bs.mw

I try to simplify the expression by putting exponential function into equation but the maple shows error which I can't fix it.simplification.mw

I was able to fix the problem. But I wanna use if statements to differentiate between complex and real roots in the attached file below:SOLVING_SYS.mw.

I am trying to find the solution (\Psi) as approaches zero. However, after applying the limit the solution becomes zero. See the attached .mw file.limit.mw

I have tried a code in python for Francis QR algorithm but didn't desire the result. I don't know if it is possible to code in maple.

Given that A^0 = [[3.0, 1.0, 4.0], [1.0, 2.0, 2.0], [0., 13.0, 2]].

1. Write a little program that computes 1 step of Francis QR and test your program by starting from

A^0 = [[3.0, 1.0, 4.0], [1.0, 2.0, 2.0], [0., 13.0, 2]]  and compute A^1, A^2 ...A^6.

I expected to get:

A^0 = [[3.0, 1.0, 4.0], [1.0, 2.0, 2.0], [0., 13.0, 2]], 

A^1 = [[3.5,  -4.264, 0.2688], [-9.206, 1.577, 9.197], [0., -1.41, 1.923]], 

... A^6 = [[8.056,  1.596, 8.584], [0.3596, -2.01, -7.86], [0., 2.576*10^(-16), 0.9542]]. 

But didn't get the same results.

Here is my python code:

# Import packages
import numpy as np
from numpy.linalg import qr # QR from Linear Algebra Library
import scipy.linalg   # SciPy Linear Algebra Library
 

A = np.array([[3.0, 1.0, 4.0], [1.0, 2.0, 2.0], [0., 13.0, 2]])
p = [1, 2, 3, 4, 5, 6]
for i in range(30):
    q, r = qr(A)
    a = np.dot(r, q)
    if i+1 in p:
        print("Iteration {i+1}")
        print(A)

I would really appreciate your help.

Thank you.

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