Unanswered Questions

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Hi everyone.

I am trying to build a package in Maple out of a bunch of procedures that I have. To create the procedures, I write them on a Vim editor, than copy and paste into the Maple worksheet to test. In this way, it works (I end the procedure with ; and then I see all the text in blue and I afterwords test it).

I then gathered these procedures to a module, option package, and I do the same: edit on Vim, copy paste it to Maple worksheet and press enter to see a blue message "module() ... end module" in blue, and then I start playing.

However, I decided to add some more procedures to the module that was working. I have a new procedure that works (meaning when I copy to the worksheet it gives the blue text and I can call normally), but when I copy the contents of this new procedure inside the package, with all the others, and I copy everything to the worksheet, the package now fails to work. It gives a message "Error," in pink, no more text, and that is it. Does this sort of error sound familiar to anyone? Thanks in Advance,

Marcelo

i have two functions , first naive function has error after used.
 
if run FromMatlab, does it mean that do not need to copy the result and run again because it had already run?
as i do not understand the output code after translated.
 
there is syntax error when translate second function
 
with(Matlab):
FromMatlab("function X = sylv_naive(A,B,Q)
% X=SYLV_NAIVE(A,B,Q) solves the Sylvester equation AX + XB = Q
%    A, B, Q: matrix coefficients
%    X : solution of AX + XB = Q
[m,n] = size(Q);
H = kron(eye(n), A) + kron(B.', eye(m));
Qvec = reshape(Q,m*n,1);
Xvec = H\Qvec;
X = reshape(Xvec,m,n);
");
sylv_naive(Jesus7,Jesus7,Matrix([[0,0],[0,0]]));
sylv_naive(Jesus7,Jesus7,Matrix([[0],[0]]));
Error, (in ArrayTools:-Reshape) the desired output contains a different number of elements than the input
 
 
 
with(Matlab):
FromMatlab("function X = sylvester(A,B,Q)
% X=SYLVESTER(A,B,Q) solves the Sylvester equation AX + XB = Q
% by using the Bartels and Stewart algorithm based on the complex
% Schur decomposition
%    A, B, Q: matrix coefficients
%    X : solution of AX + XB = Q
[m,n] = size(Q);
[U,A1] = schur(A,'complex');
[V,B1] = schur(B.','complex');
Q1 = U'*Q*conj(V);
X = zeros(m,n);
X(:,n) = (A1 + B1(n,n)*eye(m))\Q1(:,n);
for i = n-1:-1:1
    v = Q1(:,i) - X(:,i+1:n)*B1(i,i+1:n).';
    X(:,i) = (A1 + B1(i,i)*eye(m))\v;
end
X = U*X*V.';");

Error, (in Matlab:-FromMatlab) on line 15, syntax error
    X(:,i) = (A1 + B1(i,i)*eye(m))
 
 

Good day. Please can someone kindly help to reduce the result of this code. Thank you and kind regards

restart:
s:=(sum(a[j]*x^j,j=0..3)+sum(a[j]*exp(-(j-3)*x),j=4..7)):
F:=diff(s,x):
p1:=simplify(eval(s,x=q))=y[n]:
p2:=simplify(eval(F,x=q))=f[n]:
p3:=simplify(eval(F,x=q+h/3))=f[n+1/3]:
p4:=simplify(eval(F,x=q+h))=f[n+1]:
p5:=simplify(eval(F,x=q+5*h/3))=f[n+5/3]:
p6:=simplify(eval(F,x=q+2*h))=f[n+2]:
p7:=simplify(eval(F,x=q+7*h/3))=f[n+7/3]:
p8:=simplify(eval(F,x=q+3*h))=f[n+3]:


vars:= seq(a[i],i=0..7):
Cc:=eval(<vars>, solve({p||(1..8)}, {vars})):
for i from 1 to 8 do
	a[i-1]:=Cc[i]:
end do:
Cf:=s:
L:=collect(simplify(simplify(expand(eval(Cf,x=q+3*h)),size)), [y[n],f[n],f[n+1/3],f[n+1],f[n+5/3],f[n+2],f[n+7/3],f[n+3]], factor):
length(L);
H := ee -> collect(numer(ee),[exp],h->simplify(simplify(h),size))/collect(denom(ee),[exp],h->simplify(simplify(h),size)):
M:=y[n+3]=(H@expand)(L);
length(M);

 

Error, (in is/internal) too many levels of recursion

I get the error above when attempting to solve for the roots of partial derivative.  My results are below:


 

S3 := -(1/2*I)*(-(2*I)*exp(I*Pi*k*tau/T)*Pi*k-exp(I*Pi*k*tau/T)*T+I*exp(I*Pi*k*tau/T)*Pi*k*tau+(4*I)*Pi*k-(2*I)*exp(-I*Pi*k*tau/T)*Pi*k+exp(-I*Pi*k*tau/T)*T+I*exp(-I*Pi*k*tau/T)*Pi*k*tau)*sin(2*Pi*k*x/T)/(Pi^2*k^2)

S3 = ((-I)*(1/2))*((I*Pi*k*tau-(2*I)*Pi*k)*(exp(I*Pi*k*tau/T)+exp(-I*Pi*k*tau/T))-T*(exp(I*Pi*k*tau/T)-exp(-I*Pi*k*tau/T))+(4*I)*Pi*k)*sin(2*Pi*k*x/T)/(Pi^2*k^2)
"(->)"true"(->)"true

Despite the the simplifications MAPLE generated above MAPLE does not recognize the expressions as being equivalent.  So set the following for the Fourier coeffs:NULL

Ck := (-I*((2*(I*Pi*k*tau-(2*I)*Pi*k))*cos(Pi*k*tau/T)-(2*I)*T*sin(Pi*k*tau/T)+(4*I)*Pi*k))*(1/(2*Pi^2*k^2))

-((1/2)*I)*(2*(I*Pi*k*tau-(2*I)*Pi*k)*cos(Pi*k*tau/T)-(2*I)*T*sin(Pi*k*tau/T)+(4*I)*Pi*k)/(Pi^2*k^2)

(1)

The Fourier series can now be expressed as a 1 sided series employing both the Sum vs sum vs add commands for comparison:``

T0 := kernelopts(cputime), time(), time[real]();

204175839, 9435456, 8.998352053

(2)

S7 := a[0]+Sum(Ck*sin(2*Pi*k*x/T), k = 1 .. m):

1600991, 0, 0.

(3)

solve([diff(S5, x) = 0, 0 < x, x < 10*T/(2*Pi*m)], x)

Error, (in is/internal) too many levels of recursion

 

solve([diff(S5, x) = 0, 0 < x, x < 1/8], x)

Error, (in is/internal) too many levels of recursion

 

``


 

Download 2_many_levels_of_recursion.mw

Hi, 

I'm currently studying the Rossler Attractor, which is the following system:

diff(x(t), t) = -y(t)-z(t)

 diff(y(t), t) = x(t)+a*y(t)

diff(z(t), t) = b+x(t)*z(t)-c*z(t).

I have found the points of equilibrium of the system.

But now I need to draw bifurcation diagrams: one for varying a, one for varying b, and one for varying c.

 

Can someone please help me with a procedure of how to do this?

Thanks in advance.

 

I have a relatively complicated ODE that i am plotting. One of the variables in particular (B[2211], purple line in the graph) should under go an exponential-like decay to zero, but instead flies off into negative territory  (see graph below) despite having a very simple equation:

problem variable rate of change= -problem variable*constant +linear combination of variables that  are always posotive.

 

My intuition is that this is because i have somehow used the odeplot tool wrong - possibly due to a problem with stepsizes.

Here is a MWE i've made:

aa_problem_MWE.mw

 

Maple 2018.

I am surprised Maple pdsolve can't solve this basic heat PDE. it is heat PDE on bar, with left end boundary condition being time dependent is only difference from basic heat PDE's on a bar.

May be a Maple expert can find a work around? I tried all the HINTS I know about.

restart;
#infolevel[pdsolve] := 3:
pde:=diff(u(x,t),t)=diff(u(x,t),x$2);
bc:=u(0,t)=t,u(Pi,t)=0:
ic:=u(x,0)=0:
sol:=pdsolve([pde,bc,ic],u(x,t)) assuming t>0 and x>0;

 

I also hope this question of mine do not get deleted as well, like the question I posted last night asking why pdsolve ignores assumptions that showed number of examples, was deleted few hrs after I posted it. 

If this question gets deleted, I will get the message that posts showing any problem in Maple software are not welcome here by Maplesoft and I will stop coming here.

 

Hi, I've been doing a few small explore plots where the ranges for the parameters are something like [-1,1], but 0 is a special value which will often need to be set.  The slider, gauge components, ... are a bit fiddly to set with exactly zero using a mouse to drag the value.  Thought about a few different ways to do this

* Right click and set value from the component properties option - a bit fiddly for the user and the graph doesn't immediately update when OK'ed.

* Extra control with a tick to set a zero value (fiddly programming and possible confusing interpretation)

* Slider component and snaptoticks=true

Last option seems to be the easiest all round, but I was wondering why it is only the slider component that has a snaptoticks property - none of the other components seem to support it and a gauge would probably be a more natural component to use.

Thanks in advance

Hi!

Consider, fixed an integer m>1, the mapping given by the following procedure:

 

G := proc (t) local k, C; C := NULL; C := t; for k from 2 to d do C := C, 1/2-(1/2)*cos(Pi*m^(k-1)*t) end do; return [C] end proc

Then, it can be proved that given x in the cube [0,1]^{d} there is t in [0,1] such that the norm of x-G(t) is less, or equal, than sqrt(d-1)/m. Indeed, dividing the cube [0,1]^{d} into m^{d-1} subcubes of side-length 1/m x ... x 1/m x 1, the point x belongs to some of these subcubes, say J. As, by the properties of the cosines function, the curve G(t) lies in J whenever t in certain subinterval of [0,1], the result follows.

In other words, computing all the solutions of the equation

1/2*(1-cos(Pi*m^(d-1)*t)) = x[d], (j-1)/m <= t and t <= j/m

for some of these solutions the desired t is obtained, where j is such that x1 in [(j-1)/m,j/m] (x1 is the first coordinate of the point x). However, for large values of m and d, the above equation have many solutions, I have tried find all of them and the process is extremely slow....Other way to find such a t can be the following: find a t satisfying the following system of inequalities

EQ := abs(t-x[1]) <= 1/m; for k from 2 to d do EQ := EQ, abs(1/2*(1-cos(Pi*m^(k-1)*t))-x[k]) <= 1/m end do

 

and then, a solution of this system is a such t. I do not know how to find, efficiently, a t such that of x-G(t) is less, or equal, than sqrt(d-1)/m   :(

Some idea?

Many thanks for your comments in advance.

 

 

I do not know how I can solve this equation, you can help me 

I am having trouble getting Maple 2017.3 with latest Physics update to give solution to Burger's PDE for viscous fluid flow with the following initial condition. May be I am not doing something right. I tried different HINTS, but no luck.

Maple can solve the PDE without the initial conditions.

May be a Maple expert can find work around or show what I might be doing wrong.

restart;
pde := diff(u(x, t), t) + u(x, t)*diff(u(x, t), x) = mu*diff(u(x,t),x$2);
ic  := u(x,0) = PIECEWISE([0,x>=0],[1,x<0]);
sol := pdsolve({pde,ic}, u(x, t)) assuming mu>0;

Maple returns () as solution.

This PDE can be solved analytically. Here is Mathematica' solution

ClearAll[u,x,y,mu]
pde = D[u[x,t],{t}]+u[x,t]*D[u[x,t],{x}]==mu*D[u[x,t],{x,2}];
ic  = u[x,0]==Piecewise[{{1,x<0},{0,x>=1}}];
sol = DSolve[{pde,ic},u[x,t],{x,t},Assumptions->mu>0]

 

 

Is it possible to use Regex to generate a complete list of patterns?

for search pattern in a list of characters

Strange (inconsistent) behavior of eval in piecewise

 

restart;#####################

a:=2:

p:=piecewise(x<a, 1/(x-a), x=a, 22, 33);

p := piecewise(x < 2, 1/(-2+x), x = 2, 22, 33)

(1)

eval(p, x=a);

22

(2)

restart;#####################

a:=Pi:

p:=piecewise(x<a, 1/(x-a), x=a, 22, 33);

p := piecewise(x < Pi, 1/(-Pi+x), x = Pi, 22, 33)

(3)

eval(p, x=a);

Error, (in eval/piecewise) numeric exception: division by zero

 

restart;#####################

a:=Pi+ln(2):

p:=piecewise(x<a, 1/(x-a), x=a, 22, 33);

p := piecewise(x < Pi+ln(2), 1/(-Pi-ln(2)+x), x = Pi+ln(2), 22, 33)

(4)

eval(p, x=a);

22

(5)

restart;#####################

a:=Pi+1:

p:=piecewise(x<a, 1/(x-a), x=a, 22, 33);

p := piecewise(x < Pi+1, 1/(-Pi-1+x), x = Pi+1, 22, 33)

(6)

eval(p, x=a);

Error, (in eval/piecewise) numeric exception: division by zero

 

restart;#####################

a:=gamma:

p:=piecewise(x<a, 1/(x-a), x=a, 22, 33);

p := piecewise(x < gamma, 1/(-gamma+x), x = gamma, 22, 33)

(7)

eval(p, x=a);

Error, (in eval/piecewise) numeric exception: division by zero

 

#######################

f:=x->piecewise(x<Pi, 1/(x-Pi), x=Pi, 22, 33);

f := proc (x) options operator, arrow; piecewise(x < Pi, 1/(x-Pi), x = Pi, 22, 33) end proc

(8)

f(Pi);

22

(9)

 


Download piecewise-strange.mw

how do i solve the following expression in maple and how do i plot sigma wrt a?

sigma = 4*sigma[s](1/4-(int((BesselJ(1, y))^2/y(1+y(400*10^(-10)/a)^2+sqrt(1+y(400*10^(-10)/a)^2)), y = 0 .. 10^4)))

 

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