MaplePrimes Questions

In the attachment is an attempt with undesireable rendering artefacts

I did not use color functions because I intend to export the black fields of the ball to stl format to use them for a better visualisation of spinning spheres in MapleSim.

A related question: The generating function for the chessboard looks complicated to me. Maybe there is a more elegant way to do it.

Chessboard_sphere.mw

i did all the time like that and don't have any issue but i don't know why not take derivative by x and t have a problem when i remove t is ok but when i take duble derivative by x and t not run and did't give me error too what is problem with this?

restart

with(PDEtools)

with(LinearAlgebra)

with(Physics)

with(SolveTools)

undeclare(prime)

`There is no more prime differentiation variable; all derivatives will be displayed as indexed functions`

(1)

tr := {t = tau, x = xi+delta*(tau+1), u(x, t) = U(xi)*exp(I*(-kx+w*(t+1)))}

{t = tau, x = xi+delta*(tau+1), u(x, t) = U(xi)*exp(I*(-kx+w*(t+1)))}

(2)

pde := I*(diff(u(x, t), t))+alpha*(diff(u(x, t), `$`(x, 2)))+beta*(diff(u(x, t), x, t))+gamma*u(x, t)*V(xi) = 0

 

NULL

Download non_sense.mw

As by title, is it possible to check intermediate results found by nodes when using Grid:Map, and stop the execution according to these?

Is there a way to license a Maple package? Protect it from copying?

Thanks

When i am solving for delta...its taking so much time? And the simplyfying command not working showing error
Attaching sheet: Q_simplify_solv.mw

every time i have a small problem 2  why not cancel number 2 in denominator , i don't want see a number with fraction like 1/3  3/4 how i fix this

restart

K := (1/2)*sqrt(2*sqrt(2)*sqrt(lambda*a[5]/a[4])+2*sqrt(-2*a[5]/a[4])*(B[1]*sqrt(-lambda)*sinh(sigma)+B[2]*sqrt(-lambda)*cosh(sigma))/(B[1]*cosh(sigma)+B[2]*sinh(sigma)+mu/lambda)+2*sqrt(-(2*(lambda^2*B[1]^2*a[5]-lambda^2*B[2]^2*a[5]-mu^2*a[5]))/(lambda*a[4]))/(B[1]*cosh(sigma)+B[2]*sinh(sigma)+mu/lambda))*e^(i*psi(x, t))

Warning, if e is meant to be the exponential e, use command/symbol completion or palettes to enter this special symbol, or use the exp function

 

(1/2)*(2*2^(1/2)*(lambda*a[5]/a[4])^(1/2)+2*(-2*a[5]/a[4])^(1/2)*(B[1]*(-lambda)^(1/2)*sinh(sigma)+B[2]*(-lambda)^(1/2)*cosh(sigma))/(B[1]*cosh(sigma)+B[2]*sinh(sigma)+mu/lambda)+2*(-2*(lambda^2*B[1]^2*a[5]-lambda^2*B[2]^2*a[5]-mu^2*a[5])/(lambda*a[4]))^(1/2)/(B[1]*cosh(sigma)+B[2]*sinh(sigma)+mu/lambda))^(1/2)*e^(i*psi(x, t))

(1)

B[1] := 0; mu := 0

0

 

0

(2)

simplify(eval(K))

(1/2)*2^(3/4)*(((coth(sigma)*(-lambda)^(1/2)*(-a[5]/a[4])^(1/2)+(lambda*a[5]/a[4])^(1/2))*B[2]+csch(sigma)*(lambda*a[5]*B[2]^2/a[4])^(1/2))/B[2])^(1/2)*e^(i*psi(x, t))

(3)

simplify((1/2)*2^(3/4)*(((coth(sigma)*(-lambda)^(1/2)*(-a[5]/a[4])^(1/2)+(lambda*a[5]/a[4])^(1/2))*B[2]+csch(sigma)*(lambda*a[5]*B[2]^2/a[4])^(1/2))/B[2])^(1/2)*e^(i*psi(x, t)), 'trig')

(1/2)*2^(3/4)*(((coth(sigma)*(-lambda)^(1/2)*(-a[5]/a[4])^(1/2)+(lambda*a[5]/a[4])^(1/2))*B[2]+csch(sigma)*(lambda*a[5]*B[2]^2/a[4])^(1/2))/B[2])^(1/2)*e^(i*psi(x, t))

(4)

simplify((1/2)*2^(3/4)*(((coth(sigma)*(-lambda)^(1/2)*(-a[5]/a[4])^(1/2)+(lambda*a[5]/a[4])^(1/2))*B[2]+csch(sigma)*(lambda*a[5]*B[2]^2/a[4])^(1/2))/B[2])^(1/2)*e^(i*psi(x, t)))

(1/2)*2^(3/4)*(((coth(sigma)*(-lambda)^(1/2)*(-a[5]/a[4])^(1/2)+(lambda*a[5]/a[4])^(1/2))*B[2]+csch(sigma)*(lambda*a[5]*B[2]^2/a[4])^(1/2))/B[2])^(1/2)*e^(i*psi(x, t))

(5)

simplify((1/2)*2^(3/4)*(((coth(sigma)*(-lambda)^(1/2)*(-a[5]/a[4])^(1/2)+(lambda*a[5]/a[4])^(1/2))*B[2]+csch(sigma)*(lambda*a[5]*B[2]^2/a[4])^(1/2))/B[2])^(1/2)*e^(i*psi(x, t)), 'constant')

(1/2)*2^(3/4)*(((coth(sigma)*(-lambda)^(1/2)*(-a[5]/a[4])^(1/2)+(lambda*a[5]/a[4])^(1/2))*B[2]+csch(sigma)*(lambda*a[5]*B[2]^2/a[4])^(1/2))/B[2])^(1/2)*e^(i*psi(x, t))

(6)
 

 

 

NULL

Download cancelation.mw

Recently a new configuration of a computer, with a better cpu - AMD Ryzen 7 8700G w/ Radeon 780M Graphics, but maple computation is particularly slow, cpu call special less computation is particularly slow, is there any way to improve?

How we can change identity like 1/sin(x)=csc(x) or 1/cos(x)=sec(x) sometime our function is beger than this and radical come in how i can do thus simplification?

restart

M := sin(x)/cos(x)

sin(x)/cos(x)

(1)

convert(M, trig)

sin(x)/cos(x)

(2)

tan(x)

tan(x)

(3)

simplify(M)

sin(x)/cos(x)

(4)

K := 1/sinh(x)

1/sinh(x)

(5)

simplify(convert(K, trig))

1/sinh(x)

(6)

csch(x)

csch(x)

(7)

Q := sqrt(beta[0]/(B[1]*cosh(xi*sqrt(-lambda))))

(beta[0]/(B[1]*cosh(xi*(-lambda)^(1/2))))^(1/2)

(8)

(beta[0]/(B[1]*cosh(xi*(-lambda)^(1/2))))^(1/2)

(9)

simplify((beta[0]/(B[1]*cosh(xi*(-lambda)^(1/2))))^(1/2), 'trig')

(beta[0]/(B[1]*cosh(xi*(-lambda)^(1/2))))^(1/2)

(10)
 

NULL

Download identity_change.mw

Since my problem involving a 2d heat equation was answered so well,

(hats off to nm !!) I thought I would try another PDE question.

 

Question: can pdsolve handle a combination of the function and its derivative for

the boundary conditions, i.e., Robin boundary conditions?

 

I looked through the document example, pdsolve_boundaryconditions, which was

beautifully written. However, I saw no example with Robin BCs.

 

"restart; interface(imaginaryunit = i):      StartTemp(x):= `T__0`*(e)^(-`alpha__0` )*cos(Pi*(x)/(`L__0`));   `L__0` := 10: #` Length of rod` `T__0` := 100: #` max temperature` `alpha__0` := 6/(100): #`  Dampening factor` plot(StartTemp(x) , x = 0 .. `L__0`,  thickness = 5,             view = [0..`L__0`, -100..100],  labels = ["Position on rod", "Temperature"], labelfont = [Times, 14],             labeldirections = [horizontal, vertical], title = "Temperature in rod at t = 0" , titlefont = [Times, 16],             size = [500, 225]);"

proc (x) options operator, arrow, function_assign; T__0*exp(-alpha__0)*cos(Pi*x/L__0) end proc

 

 

The equation and the initial condition:

heat_equation := diff(T(x, t), t) = k*(diff(T(x, t), x, x)); initial_condition := T(x, 0) = T__i(x)

Using a linear combination of the function and its derivative, i.e., Robin BCs.

boundary_conditions := `α__r`*T(0, t)+`β__r`*(D[1](T))(0, t) = f(t), `α__r`*T(L__0, t)+`β__r`*(D[1](T))(L__0, t) = g(t)

 

Setting the constants

"  `alpha__r` := 1:  #` Constant for boundary condition`  `beta__r` := 2: #` Constant for boundary condition`     `T__s`:= 10:  #` Time interval`  k:=9/(10):  #` Heat-conductivity of material`    f(t) :=-1/(10) t:                  g(t):=1/(10) t:  "

Solving using nm's technique of not defining the function before calling pdsolve

sols := pdsolve({heat_equation, boundary_conditions, initial_condition}, T(x, t))

 

Now include the initial temperature function

"`T__i`(x) := StartTemp(x):  "

Extracting the solution, and using 20 terms

sol := eval(rhs(sols), infinity = 20)

-134/135+(1/270)*x^3-(7/90)*x^2+(67/135)*x+Sum(exp(-(9/1000)*Pi^2*n^2*t)*(Pi*n*cos((1/10)*n*Pi*x)-5*sin((1/10)*n*Pi*x))*(Int(-(x^3-21*x^2+134*x-27000*exp(-3/50)*cos((1/10)*Pi*x)-268)*(Pi*n*cos((1/10)*n*Pi*x)-5*sin((1/10)*n*Pi*x)), x = 0 .. 10))/(1350*Pi^2*n^2+33750), n = 1 .. 20)+(1/50)*t*x-(7/50)*t

(1)

Evaluate the integrals and build the solution to the temperature as a function of position and time

T__sol := MakeFunction(value(sol), x, t)

 

Here is the problem:

in plotting the solution at t = 0, we do NOT reproduce the initial temperature function.

 

plot(T__sol(x, 0), x = 0 .. L__0, -120 .. 120, thickness = 5, size = [500, 225])

 

If I use boundary conditions for either Dirichlet or Neumann, i.e., T(0, t) & T(L__0, t) 

or "`T__x`(0,t) & `T__x`(`L__0`,t)", the initial function is reproduced at t = 0.

 

Is this example one of those situations that was addressed in the notes?

Given a problem with PDEs and some BCs, note the following:

• 

Depending on how you write a general PDE solution, it becomes
possible or nearly impossible to adjust it to match the BCs.

 

NULL

Download Heat_Equation_1D_Robin_Conditions.mw

Using Maple to prove one of the axioms in mathematics.

The idea is to prove that (-1)*(-1)= 1.  The mere fact that 0's are involved in the proof makes it harder.  So it's easy enough on paper to solve, as I will show in the steps below.

It is known 0*x=0 for all values of x,
and 1-1=0,
If we let x=-1, then 0*(-1)=0,
sub in for zero   (1-1)(-1)=0,
expand (1)*(-1) + (-1)*(-1)=0, also we agree that 1*y=y for all values of y,
so ......  (-1) + (-1)*(-1)=0,
and therefore (-1)*(-1) = 1

Does anyone have an elegant way of showing that in Maple?

OneFrame := proc(k)
local Courbe, T, a, b, c, t, P, Q, NormM, F, Ell, sol, N1, N2, dr, tx;
a := 11; b := 7; c := sqrt(a^2 - b^2); t := 1/3*Pi;
Ell := x^2/a^2 + y^2/b^2 = 1;
geometry:-point(T, (a^2 - b^2)*cos(t)^3/a, -(a^2 - b^2)*sin(t)^3/b);
Courbe := plots:-implicitplot(Ell, x = -a - 10 .. a + 10, y = -b - 10 .. b + 10, scaling = constrained, color = blue);
NormM := plots:-implicitplot(y - b*sin(t) = a*sin(t)*(x - a*cos(t))/(b*cos(t)), x = -a - 5 .. a + 10, y = -b - 10 .. b + 10, color = orange); geometry:-line(Per, y - b*sin(t) = a*sin(t)*(x - a*cos(t))/(b*cos(t)), [x, y]);
geometry:-point(P, subs(y = 0
, geometry:-Equation(Per), 0));
geometry:-point(Q, 0, subs(x = 0, geometry:-Equation(Per)));
geometry:-point(M, a*cos(t), b*sin(t));
geometry:-point(N1, a*cos(k), b*sin(k));
geometry:-point(F, 2.329411765, -2.567510609);
geometry:-line(L, N1, F);
sol := solve({geometry:-Equation(L), Ell}, {x, y},explicit);
geometry:-point(N2, subs(sol[2], x), subs(sol[2], y));
geometry:-segment(sg, N1, N2);
tx := plots:-textplot([[geometry:-coordinates(M)[], "M"],
[geometry:-coordinates(N1)[], "N1"], [geometry:-coordinates(N2)[], "N2"],
[geometry:-coordinates(P)[], "P"],
[geometry:-coordinates(Q)[], "Q"],
[geometry:-coordinates(F)[], "F point de Frégier"],
[geometry:-coordinates(T)[], "T"]], font = [times, bold, 16], align = [above, left]);
dr := geometry:-draw([sg(color = magenta, linestyle = dash),
Per(color = black), P(color = red, symbol = solidcircle, symbolsize = 12),
Q(color = red, symbol = solidcircle, symbolsize = 12),
M(color = black, symbol = solidcircle, symbolsize = 12),
F(color = red, symbol = solidcircle, symbolsize = 12),
N1(color = black, symbol = solidcircle, symbolsize = 8),
N2(color = black, symbol = solidcircle, symbolsize = 8),
T(color = black, symbol = solidcircle, symbolsize = 8)]);
plots:-display(Courbe, tx, dr, scaling = constrained, axes = none); end proc;

plots:-animate(OneFrame, [k], k = Pi/3 .. Pi, frames = 50);
Error, (in plots/animate) wrong type of arguments
Why this animation does't work ? Thank you very much.
 

Dear Maple users

I am defining various geometrical items in the plottools package and finally plot them using the display command. Let's say I have defined:

Circle1:=circle(...):

Circle2:=circle(...):

Line1:=line(...):

etc.

and finally want to plot them with the display command:

display(Circle1,Circle2,Line1,...)

It works, however sometimes I only want some of these items to be plotted. Certain variables at the beginning of the document do control if those items should be plotted or not. Let's say a variable CircleOrNot can be set to have the values "no" or "yes", and if I set it to "yes", it should plot it and if the value is "no", it should not plot it. I know that I can make a conditional statement in the end when dealing with the display command, but it can be quite a mess (condition after condition ...).

What I would like is making a conditional assignment like the following:

If CircleOrNot="yes" then Circle1:=circle(...) else Circle1:=null end if

or something like that, and afterwards use the full display command including all objects:

display(Circle1,Circle2,Line1,...)

But it does not work. I receive an error. There doesn't seem to be an empty statement, which will just be ignored by the display command. How can I do it? Probably someone has a good way to handle this situation.

Best regards,

Erik

What is the correct way to find all terms of the form  _Cn^m where n is non negative integer and m is exponent which can be 1?

The problem is finding _Cn with no exponent. Maple's   _Cn^anything does not match _Cn but only when there is an actual exponent present.

It is easy to find _Cn^m with m present and easy to find _Cn with no exponent, but how to combine the two is the problem. Here is an example.

Given e:=_C1^2+_C1+_C2^3+a+b; I want to obtain the set {_C1^2,_C1,_C2^3}  but I get instead {_C1, _C2, _C1^2, _C2^3} using ther following code

e:=_C1^2+_C1+_C2^3+a+b;
indets(e,   Or(  And(symbol, suffixed(_C, nonnegint)),
                 And(symbol, suffixed(_C, nonnegint))^anything  ));

If instead I do this

e:=_C1^2+_C1+_C2^3+a+b;
indets(e, And(symbol, suffixed(_C, nonnegint))^anything);

This gives {_C1^2, _C2^3} , so it did not pick the _C1 term since exponent is missing. If I do this instead

e:=_C1^2+_C1+_C2^3+a+b;
indets(e, And(symbol, suffixed(_C, nonnegint)));

This it gives {_C1, _C2} and if I do 

e:=_C1^2+_C1+_C2^3+a+b;
indets(e,   And(symbol, suffixed(_C, nonnegint))^Or(anything,identical(1)))

it gives {_C1^2, _C2^3}

What to do to obtain {_C1^2,_C1,_C2^3} 

I am trying to find all _Cn^m in an expression where m can be anything including not being there (or 1 basically) but in Maple, exponent 1 is not counted as exponent.

Update

Another input example is e:=_C1^2+_C1+ _C1*_C2^3+a+b;  This should be produce the set {_C1^2,_C1,_C2^3} 

Any trick to use for this?

Maple 2024.1

 

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