Maple 16 Questions and Posts

These are Posts and Questions associated with the product, Maple 16

Maple show "

Error, (in simplify/hypergeom/baseHG) Maple was unable to allocate enough memory to complete this computation. Please see ?alloc"

after I finish the equations set and run it.

what's the reason?

Hello, my problem is that I've changed in tools maple default mode to "Maple input" from "2-d" I've aplied it globbaly and it doesnt'work. Still when type Enter it switch automatically to 2-d Math, and I have to change it by Ctrl-M to 1-d. Any ideas how to solve it? I want 2-d just to output.


this is the second time I'm writing.

I posted this question in June

This time I have  a similar problem because I'm trying to find a solution for a parametric system of equations but the number of equations and parameters is much bigger and using the tips you gave me last time I couldn't reach any result.

Here is the system:

1) alpha[1]=v*a*u*b ;
2) alpha[2]=v*a*u*(1-b);
3) alpha[3]= v*z*c*(1-a) ;
4) alpha[4]=v*z*(1-a)*(1-c) ;
5) alpha[11]=1/2*v*a* u* b* (-p*u*b+p*u*b*a+b*g-g);
6) alpha[22]=1/2*v*a*u*(1-b)* (p u b-p u b a-b g-p u+p u a);
7) alpha[33] =1/2*v*c*z*(1-a)* (c* (-z*p*a+q)-q);
8) alpha[44]=1/2*v*z*((1-a)*(1-c)* (c*z*p*a-z*p*a-q*c);
9) alpha[12]=v*a*u*b*(1- b)*(-p*u+p*u*a+g) ;
10) alpha[13]=v*a*u*b*z*c*p*(1-a) ;
11) alpha[14]=a*u*b*z*(1-a)*(1-c) ;
12) alpha[23]=a*u*z*c*(1-a)*(1-b);
13) alpha[24]=v*a*u*z*p*(1-a)*(1-b)*(1-c);
14) alpha[34]= v*c*z*(1-a)*(1-c)*(-z*p*a+q);


I have 14 equations/unknowns and 8 parameters (a, b, c, u, v, z, p, q).

I would like to write this system only in terms of alphas. In order to do so, I usually try to find the value for the parameters and the substitute them into the equations (and I have already found b,c,g,q using this technique) but I couldn't manage to find all of them. 

Howveer, as you suggested me, with Maple there is the command "eliminate" that implement exactly what I'm looking for but I can't make it work.

This is my code:

> sys := {alpha[1] = v*a*u*(1-b), alpha[2] = v*a*u*b, alpha[3] = v*z*c*(1-a), alpha[4] = v*z*(1-a)*(1-c), alpha[11] = (1/2)*v*a*u*(1-b)*(p*u*b-p*u*b*a-b*g-p*u+p*u*a), alpha[12] = v*a*u*b*(1-b)*(-p*u+p*u*a+g), alpha[13] =      z*c*a*u*(1-a)*(1-b), alpha[14] = v*z*a*u*p*(1-a)*(1-b)*(1-c), alpha[22] = (1/2)*v*a*u*b*(-p*u*b+p*u*b*a+b*g-g), alpha[23] = v*z*c*a*u*b*p*(1-a), alpha[24] = z*a*u*b*(1-a)*(1-c), alpha[33] = (1/2)*v*c*z*(1-a)*(c*(-z*p*a+q)-q), alpha[34] = v*c*z*(1-a)*(1-c)*(-z*p*a+q), alpha[44] = (1/2)*v*z*(1-a)*(1-c)*(c*z*p*a-z*p*a-q*c)};

> eliminate(sys, {a,b,c, p, q, u, v, z});

> simplify(%, size);


I also tries to substitute in the system the four parameters I already found but still I can't find a solution.

What am I doing wrong? Or the problem is that it is too complicated?


Thank you for your attention,


I wish to  plot 2D animate for the soltion of this equation here is the code


for i from 0 to N/h do
for j from 0 to M/h do
for i from 1 to N/h do
for j from 1 to M/h do
psi[i,j]:=-psi[i-1, j-1]+(1-(1/8)*h^2*V((1/2)*h*(j-i-1)))*psi[i, j-1]+(1-(1/8)*h^2*V((1/2)*h*(j- i+1)))*psi[i-1,j]:


I am having trouble printing out a limit cylce on maple 16.  I have the attached file and if anybody could look at it and perhaps help me out it would be greatly appreciated.  The first limit cycle is supposed to look somewhat like the second one.  I'v tried many different things but nothing seems to be working.  an explenation would also be nice too.  if the file does not open correctly also let me know. thank you very much.

I've been poking around with convolutions on Maple, and some weird behavior came up---if I let it compute the convolution of a piecewise function, then take the convolution of that, it comes out differently than if I enter a function from scratch as the middle step---file attached (  I'm not really a Maple pro, so am I'm doing something crazy here?


I suppose you would have to programme this, because I couldn't find anything relating to vector spaces in my Maplesoft Linear Algebra tutor or package.

My question really is, you have a problem like, the set of all pairs of real numbers of the form (1,x) with the operations (1,y)+(1,y')=(1,y+y') and k(1,y)=(1,ky) 

( ^ taken straight out of my homework haha)

and you run through the 10 vecctor space axioms to determine whether it fails and it not a vector space, or it passes all and it is a vector space.


Is there a way to do this in Maplesoft?


In ode solve command i generated a large array data. The output shows a large order matrix of this form


[110001x6 Matrix





I want to export this matrix into a notepad. Which can then be used for plotting in TecPlot. 


Looking for good response



I want to begin by saying hello! im new to the forums i hope some one can give me a push in the right direction with some of my maple homework. im sort of stuck on a few of these questions and would be greatfull for some help. 


Let .

a) Let g be the tangent line to f when x = c. Use Maple to find g as a function of c.

b) Use Maple to plot f and g(3) using view = [0..5, -25..100].

c) Define a function called plot_tan that plots both f and g(c) where f is blue and g(c) is red. Also use the same view as in part (b). Note that plot_tan is also a function of c.

d) Using your function plot_tan, the following all in one graph using the display command:

plot_tan(1), plot_tan(1.5), plot_tan(2), plot_tan(2.5), plot_tan(3), plot_tan(3.5), plot_tan(4), plot_tan(4.5), plot_tan(5).

e) Try using the option insequence = true in the display command. What does this option do? (You will need to click on the graph and play around with some buttons).

now it seems to me i have to use the point slope formula to get to a fuction g of c. thanks in advance! i hope you can help





I'm using the Global Optimization Toolbox to solve some examples and fit equations to a given data, finding "unknown" parameters. I generated the data on Excel, and I already know the values of these parameters.

The XY case is (there is no problem here, I just put as a example I follow):

> with(GlobalOptimization);
> with(plots);

> X := ExcelTools:-Import("F:\\Data.xlsx", "Plan1", "I5:I25");
> Y := ExcelTools:-Import("F:\\Data.xlsx", "Plan1", "J5:J25");

> XY := zip( (X, Y) -> [X, Y] , X, Y);
> fig1 := plot(XY, style = point, view = [.9 .. 3.1, 6 .. 40]);

> Model := A+B*x+C*x^2+D*cos(x)+E*exp(x):
> VarInterv := [A = 0 .. 10, B = 0 .. 10, C = -10 .. 10, D = 0 .. 10, E = 0 .. 10];

> ModelSubs := proc (x, val)

    subs({x = val}, Model)

    end proc;

> SqEr := expand(add((ModelSubs(x, X(i))-Y(i))^2, i = 1 .. 21));
> CoefList := GlobalSolve(SqEr, op(VarInterv), timelimit = 5000);

> Model := subs(CoefList[2], Model):


I could find the right values of A, B, C, D and E. 


My problem is in the XYZ case, where I don't know how to "write" the right instruction. My last attempt was:

> with(GlobalOptimization);
> with(plots);

> X := ExcelTools:-Import("F:\\Data.xlsx", "Plan1", "Q5:Q25"); X2 := convert(X, list);
> Y := ExcelTools:-Import("F:\\Data.xlsx", "Plan1", "R5:R25"); Y2 := convert(Y, list);
> Z := ExcelTools:-Import("F:\\Data.xlsx", "Plan1", "S5:S25"); Z2 := convert(Z, list);
> NElem := numelems(X);

> pointplot3d(X2, Y2, Z2, axes = normal, labels = ["X", "Y", "Z"], symbol = box, color = red);


> Model := A*x+B*y+C*sin(x*y)+D*exp(x/y);

> VarInterv := [A = 0 .. 10, B = 0 .. 10, C = 0 .. 10, D = 0 .. 10];

> ModelSubs:=proc({x,y},val)


end proc:
Error, missing default value for option(s)

> SqEr := expand(add((ModelSubs(x, y, X(i), Y(i))-Z(i))^2, i = 1 .. NElem));
> CoefList := GlobalSolve(SqEr, op(Range), timelimit = 5000);
Error, (in GlobalOptimization:-GlobalSolve) finite bounds must be provided for all variables


My actual problem involves six equations, six parameters and four or five independent variables on each equation, but I alread developed a way to solve two or more equations simultaneously.




A:=Matrix([[ a , b ], [ c , d ]]);

a:=1; b:=0; c:=0; d:=1;


produces differents results under MAPLE 16  linux i386 and under MAPLE 16  linux amd64

in the first case the last evalution has the following printed output:


on the second machine the printed output is


Does anybody has an explication; I thought that the "coorect behaviour was the first one since tables use last-name evalutation. But now I am puzzled.


hello. before I used Mapple 15. But then I`ve run Mapple 16 and now I`ve a problem. I can`t use this program. I open the program, everthing is in the rule, but if I want to write any mathemathical function, or a letter, such as- x or x+2, the program does`t give any reaction. program only gives reaction the numbers.

Please, help me. (my english isn`t very good, and I don`t know I`ve explained my opinion).


I get the error in the following code





eq2:=diff(T(eta),eta,eta)+1/(k(eta)/k1[w])*(-2/(1-zet^2)*rho(eta)*c(eta)*u(eta)/(p2*10000)+( (a[k1]+2*b[k1]*phi(eta))/(1+a[k1]*phi1[w]+b[k1]*phi1[w]^2)*diff(phi(eta),eta)+k(eta)/k1[w]/(eta)*diff(T(eta),eta) ));
      /  d   /  d         \\   mu1[w] (1 - u(eta))
      |----- |----- u(eta)|| + -------------------
      \ deta \ deta       //         mu(eta)      

           /             /  d           \\               
           |      mu_phi |----- phi(eta)||               
           | 1           \ deta         /| /  d         \
         + |--- + -----------------------| |----- u(eta)|
           \eta           mu(eta)        / \ deta       /
                                /      /                        
                                |      |                        
/  d   /  d         \\     1    |      |  rho(eta) c(eta) u(eta)
|----- |----- T(eta)|| + ------ |k1[w] |- ----------------------
\ deta \ deta       //   k(eta) |      |         5000 p2        
                                \      \                        

                                /  d           \
     (a[k1] + 2 b[k1] phi(eta)) |----- phi(eta)|
                                \ deta         /
   + -------------------------------------------
         1 + a[k1] phi1[w] + b[k1] phi1[w]      

            /  d         \\\
     k(eta) |----- T(eta)|||
            \ deta       /||
   + ---------------------||
           k1[w] eta      ||
                                      /  d         \
                             phi(eta) |----- T(eta)|
          /  d           \            \ deta       /
          |----- phi(eta)| + ------------------------
          \ deta         /                          2
                             N[bt] (1 + 0.01 T(eta))
rho:=unapply(  phi(eta)*rhop+(1-phi(eta))*rhobf ,eta):
c:=unapply(  (phi(eta)*rhop*cp+(1-phi(eta))*rhobf*cbf )/rho(eta) ,eta):



#A somewhat speedier version uses the fact that you really need only compute 2 integrals not 3, since one of the integrals can be written as a linear combination of the other 2:
Q:=proc(pp2,fi0) local res,F0,F1,F2,a,INT0,INT10,B;
global Q1,Q2;
if not type([pp2,fi0],list(numeric)) then return 'procname(_passed)' end if:
res := dsolve(subs(p2=pp2,phi0=fi0,{eq1=0,eq2=0,eq3=0,u(1)=lambda/(phi(1)*rhop/rhobf+(1-phi(1)))*D(u)(1),D(u)(0)=0,phi(1)=phi0,T(1)=0,D(T)(1)=1}), numeric,output=listprocedure):
B:=(-cbf*rhobf+cp*rhop)*INT10+ rhobf*cbf*INT0;
if type(procname,indexed) then a[op(procname)] else a[1],a[2] end if
end proc;
#The result agrees very well with the fsolve result.
#Now I did use a better initial point. But if I start with the same as in fsolve I get the same result in just about 2 minutes, i.e. more than 20 times as fast as fsolve:

Q1:=proc(pp2,fi0) Q[1](_passed) end proc;
Q2:=proc(pp2,fi0) Q[2](_passed) end proc;

proc(pp2, fi0)  ...  end;
proc(pp2, fi0)  ...  end;
proc(pp2, fi0)  ...  end;
              HFloat(6.5), HFloat(0.006737946999)



the error is :

Error, (in Optimization:-LSSolve) system is singular at left endpoint, use midpoint method instead

how can I fix it.




When you use the slider without Do(%MathContainer1 = StandardError(Variance, R)):
everything works ok but when you add Do(%MathContainer1 = StandardError(Variance, R)):
Maple Crashes.....



Here is the question:Prior to this question I was given f(z)=z^2+1, N(z)=(z^2-1)/(2z), T(z)=z-I/z+I such that T(N^k (z))=(T(z))^2^k. And L is a set of number on the real axis. Now the question is that given we have two regions of the complext plane as follow:

R+ = {z : Nk{z) -> i as k -> ∞}; R- = {z : Nk(z) -> -i as k -> ∞}.

Draw a diagram to illustrate these regions, the line L and the roots i and -i. We call R+ the basin of attractionfor the root +i, and similarly R-is the basin of attraction for the root -i.

 Show that if z is on the set L (the common boundary of the two regions R+ and R_, then Nk(z) stays on L for all values of k. (This is easy once you identify what L is.) So in this case iteration does not produce a root at all.

So basically my problem is that the fact I'm not very familar with the commands to draw such diagram, and I don't know much about Newton's method to compute complex roots. It would be appreciated if anyone can help me how to get start with the question. Thanks.


1 2 3 4 5 6 7 Last Page 2 of 40