MaplePrimes Questions

Dear team,

Here below you can see a simple program containing 3 loops. The problem is  that even for low iterations the execution time seems endless. I kept this program running for over 10 minutes to no avail.   Could you help me? 

Here you are the technical details of my HP laptopper

 

Hi everyone,

I am trying to visualize the integral of x3 over the interval x=−1 to x=1. I tried using:

with(plots):
display(
   plot(x^3, x = -1 .. 1, color = black, thickness = 2),
   shadebetween(x^3, 0, x = -1 .. 1, color = cyan)
);

This works, but I wondered if there’s a better or more elegant way to visualize definite integrals. For example:

  • Can I add transparency to the shaded region?
  • Is there a built-in function that directly plots definite integrals with shading?
  • Any tips for improving the aesthetics of such plots?

Thanks in advance for any help!

Download εμβαδόν_χωρίου.mw

Windows 10. From normal command line window:

"C:\Program Files\Maple 2024\bin.X86_64_WINDOWS\mint.exe" foo.mpl

Gives this

The file foo.mpl is

export module ODE() 
    option object; 
    export ode::`=`;
end module;

Notice the funny looking characters in the output.

Why does it happen?

restart

with(PDEtools)

undeclare(prime, quiet); declare(u(x, y, z, t), quiet); declare(f(x, y, z, t), quiet)

``

(1)

thetai := k[i]*(t*w[i]+y*l[i]+z*r[i]+x)

eqw := w[i] = (-1+sqrt(-4*beta*mu*l[i]-4*delta*mu*r[i]-4*mu*k[i]^2-4*alpha*mu+1))/(2*mu)

Bij := proc (i, j) options operator, arrow; -24*mu/(sqrt(1+(-4*beta*l[j]-4*delta*r[j]-4*alpha)*mu)*sqrt(1+(-4*beta*l[i]-4*delta*r[i]-4*alpha)*mu)-1+((2*r[i]+2*r[j])*delta+(2*l[i]+2*l[j])*beta+4*alpha)*mu) end proc

NULL

theta1 := normal(eval(eval(thetai, eqw), i = 1)); theta2 := normal(eval(eval(thetai, eqw), i = 2))

eqf := f(x, y, z, t) = theta1*theta2+Bij(1, 2)

eqfcomplex := collect(evalc(eval(eval(eqf, l[2] = conjugate(l[1])), l[1] = a+I*b)), t)

eq17 := u(x, y, z, t) = 2*(diff(diff(f(x, y, z, t), x), x))/f(x, y, z, t)-2*(diff(f(x, y, z, t), x))^2/f(x, y, z, t)^2; equ := simplify(eval(eq17, eqfcomplex))

So we want to find a substitution that removes the time dependence from u. One way is to find the maximum and see how it moves. Here, the first solution gives what we want.

ans := solve({diff(rhs(equ), x), diff(rhs(equ), y), diff(rhs(equ), z)}, {x, y, z}, explicit)

 

NULL

Download hfz.mw

Hi! An basic and interesting issue.

First, why I run worksheet again and again, I will get different output (even seems understandable)? I guess maybe the memory is not renewed?

Second, how to obtain the periodic solution without adding ```k*Pi``` manually?

`assuming`([solve(sqrt(2)*sin(2*x-(1/6)*Pi) = 1)], [`in`(x, real)])

5/24

(1)

x := solve(sqrt(2)*sin(2*x-(1/6)*Pi) = 1)+k*Pi

5/24+k*Pi

(2)
 

NULL

Download periodic_solution_of_a_simply_Sine_function.mw

i need the result for (eqt33) but i can reach the result there is any  other way for finding? i need to plot 3D of that function but without have the function how i can do explore on it

w1.mw

Is this a bug in mint?

Given this foo.mpl

foo:=proc()
    local x:=2,y:=3;       
        
    if MmaTranslator[Mma][LeafCount](x)<MmaTranslator[Mma][LeafCount](y)  then        
        0;
    else
       1;
    fi;  

end proc;   

It says

Procedure foo() on lines 2 to 11
  These names were used as global names but were not declared:  LeafCount, Mma

But if I rewrite the above using :- instead:

foo:=proc()
    local x:=2,y:=3;       
        
    if MmaTranslator:-Mma:-LeafCount(x)<MmaTranslator:-Mma:-LeafCount(y)  then        
        0;
    else
       1;
    fi;  

end proc;  

Now mint is happy and no message are given.

Is not  MmaTranslator:-Mma:-LeafCount(x)  the same as MmaTranslator[Mma][LeafCount](x) ?

Maple itself is happy with both. So why is mint complaining?

I am finding many problems with mint. Will post more problems found when I have more time.

Is mint still actively  maintained by Maplesoft? 

Hi Guys,

I encountered this rather unexpected behavior of Maple 18 when using the "read" function as shown below, where Maple fails to reduce the answer to 0, especially for (5). I'm curious if there's any deeper reason for this, and if there's a better or more standard solution than mine, that is, to use "parse( convert( expr, string ) )" as in (6)?

Thanks a lot!

``

restart;

 

assume(x>0);

 

f := x^2;

x^2

(1)

 

save f, "read_test.m";

 

restart;

 

f;

f

(2)

 

read "read_test.m";

 

f;

x^2

(3)

 

f - x^2;

-x^2+x^2

(4)

 

assume(x>0);

 

f - x^2;

x^2-x^2

(5)

 

f1 := parse( convert( f, string ) );

x^2

(6)

 

f1;

x^2

(7)

 

f1 - x^2;

0

(8)

 

NULL

Download test1.mw

Hi,

I am creating a multiple-choice questionnaire (MCQ) on functions, but the display of logarithmic functions is automatically converted to natural logarithms. How can I keep the logarithms displayed in base a without conversion? Thank you for your help.

S5_QCM_Expo_Log.mw

I found system remember tables mentioned under Details in ?CacheOption.

What exactly are they? Anything special about them?

The ProgrammingGuide does not given more details. It only mentions that there are "procedures that cannot use option system remember tables". For what reason a procedure cannot use them?

For the following limit determination, I do not get the known result 2, but only a term with LerchPhi. Please give me some advice.
lim(n-->oo)[(n/(2^n))*sum(k=1 to n) (2^k/k)]

I wish to plot two line segments over the intervals [0,1] and [2,3].  Maple 2024 draws a single line segment over [0,3].  Oddly enough, if I specify a color option, then it produces the expected result!

This used to work correctly in Maple 2023.  Perhaps this can be corrected in the forthcoming release?

restart;

kernelopts(version);

`Maple 2024.2, X86 64 LINUX, Oct 29 2024, Build ID 1872373`

plots:-display(
        plot(1, x=0..1),
         plot(1, x=2..3), thickness=5);

plots:-display(
        plot(1, x=0..1),
         plot(1, x=2..3),thickness=5, color=red);


Download mw.mw

Does this happen to you (ChatGPT session):

I am taking a course in quantum mechanics and trying to do the calculations in Maple.

In the worksheet below I try to solve a system of equations to find some constants. 

It isn't very relevant that the domain is physics. 

The issues I have are with manipulations of expressions (specifically, simplifications).

restart

Levine - Ch. 2.4 - Particle in a Rectangular Well

NULL

Define

s__1 := sqrt(2*m*(V__0-E))/`&hbar;` = 2^(1/2)*(m*(V__0-E))^(1/2)/`&hbar;`NULL

s__2 := -s__1 = -2^(1/2)*(m*(V__0-E))^(1/2)/`&hbar;`NULL

NULLs := sqrt(2*m*E)/`&hbar;` 

2^(1/2)*(m*E)^(1/2)/`&hbar;`

(1)

`&psi;__1`, `&psi;__2` and `&psi;__3` are wavefunctions.

`&psi;__1` := proc (x) options operator, arrow; C*exp('s__1'*x) end proc = proc (x) options operator, arrow; C*exp('s__1'*x) end procNULL

`&psi;__2` := proc (x) options operator, arrow; A*cos('s'*x)+B*sin('s'*x) end proc = proc (x) options operator, arrow; A*cos('s'*x)+B*sin('s'*x) end procNULL

`&psi;__3` := proc (x) options operator, arrow; G*exp('s__2'*x) end proc = proc (x) options operator, arrow; G*exp('s__2'*x) end procNULL

NULL

assume(G <> 0, C <> 0)

NULL

I would also like to assume that A and B cannot both be zero simultaneously.


As can be seen above, there are four unknown constants, "A,B,C,"and G.

 

We can obtain values for these unknowns by imposing boundary conditions.

 

Some of the boundary conditions involve the first derivatives of the wavefunctions.

`&psi;__1,d` := D(`&psi;__1`) = proc (z) options operator, arrow; C*s__1*exp(s__1*z) end procNULL

`&psi;__2,d` := D(`&psi;__2`) = proc (z) options operator, arrow; -A*s*sin(s*z)+B*s*cos(s*z) end procNULL

`&psi;__3,d` := D(`&psi;__3`) = proc (z) options operator, arrow; G*s__2*exp(s__2*z) end procNULL

NULL

The boundary conditions are

NULL

`&psi;__1`(0) = `&psi;__2`(0)*`&psi;__2`(l) and `&psi;__2`(0)*`&psi;__2`(l) = `&psi;__3`(l)*(D(`&psi;__1`))(0) and `&psi;__3`(l)*(D(`&psi;__1`))(0) = (D(`&psi;__2`))(0)*(D(`&psi;__2`))(l) and (D(`&psi;__2`))(0)*(D(`&psi;__2`))(l) = (D(`&psi;__3`))(l)

NULL

My question is how to find the constants in Maple.

NULLNULL

In this problem, E < V__0 and m > 0.

 

assume(E < V__0, m > 0)

 

Solving the first boundary condition is easy.

expr1 := solve(`&psi;__1`(0) = `&psi;__2`(0), {C}) = {C = A}

NULL

Next, I solve the third boundary condition and sub in the result from solving the first boundary condition. We get B in terms of A.

expr3 := eval(solve(`&psi;__1,d`(0) = `&psi;__2,d`(0), {B}), expr1) = {B = A*(m*(V__0-E))^(1/2)/(m*E)^(1/2)}NULL

NULL

Already here it is not clear to me why Maple does not cancel the m's.

 

Next, consider the second boundary condition

expr2 := solve(`&psi;__2`(l) = `&psi;__3`(l), {G}) = {G = (A*cos(s*l)+B*sin(s*l))/exp(s__2*l)}

 

We can obtain G in terms of just A by subbing in previously found relationships, as follows

expr2 := simplify(subs(expr3, expr2))

{G = exp(2^(1/2)*m^(1/2)*(V__0-E)^(1/2)*l/`&hbar;`)*A*(sin(2^(1/2)*m^(1/2)*E^(1/2)*l/`&hbar;`)*(V__0-E)^(1/2)+cos(2^(1/2)*m^(1/2)*E^(1/2)*l/`&hbar;`)*E^(1/2))/E^(1/2)}

(4)

NULL

Finally, consider the fourth boundary condition.

solve(`&psi;__2,d`(l) = `&psi;__3,d`(l), {G})

{G = (m*E)^(1/2)*(A*sin(2^(1/2)*(m*E)^(1/2)*l/`&hbar;`)-B*cos(2^(1/2)*(m*E)^(1/2)*l/`&hbar;`))/((m*(V__0-E))^(1/2)*exp(-2^(1/2)*(m*(V__0-E))^(1/2)*l/`&hbar;`))}

(5)

expr4 := simplify(subs(expr3, solve(`&psi;__2,d`(l) = `&psi;__3,d`(l), {G})))

{G = (sin(2^(1/2)*m^(1/2)*E^(1/2)*l/`&hbar;`)*E^(1/2)-cos(2^(1/2)*m^(1/2)*E^(1/2)*l/`&hbar;`)*(V__0-E)^(1/2))*exp(2^(1/2)*m^(1/2)*(V__0-E)^(1/2)*l/`&hbar;`)*A/(V__0-E)^(1/2)}

(6)

result := simplify(subs(expr2, expr4))

{exp(2^(1/2)*m^(1/2)*(V__0-E)^(1/2)*l/`&hbar;`)*A*(sin(2^(1/2)*m^(1/2)*E^(1/2)*l/`&hbar;`)*(V__0-E)^(1/2)+cos(2^(1/2)*m^(1/2)*E^(1/2)*l/`&hbar;`)*E^(1/2))/E^(1/2) = (sin(2^(1/2)*m^(1/2)*E^(1/2)*l/`&hbar;`)*E^(1/2)-cos(2^(1/2)*m^(1/2)*E^(1/2)*l/`&hbar;`)*(V__0-E)^(1/2))*exp(2^(1/2)*m^(1/2)*(V__0-E)^(1/2)*l/`&hbar;`)*A/(V__0-E)^(1/2)}

(7)

At this point, we have this complicated expression.

 

The calculations above come from the book "Quantum Chemistry" by Levine. The variable "result" above should, after a few more steps of manipulating the expression, give us

 

(2*E-V__0)*sin(sqrt(2*mE)*l/`&hbar;`) = 2*sqrt(-E^2+E*V__0)*cos(sqrt(2*mE)*l/`&hbar;`)

(2*E-V__0)*sin(2^(1/2)*mE^(1/2)*l/`&hbar;`) = 2*(-E^2+E*V__0)^(1/2)*cos(2^(1/2)*mE^(1/2)*l/`&hbar;`)

(8)

 

I've done the calculation by hand. It is not difficult. The exponentials and the A cancel on each side and we are left with an equation involving sin(sl) and cos(sl) which easily comes out to the desired equation above.

I would like to be able to do it here in Maple.

The first thing I would do at this point is to start eliminating some of the clutter. For example, I would like to replace sqrt(2*mE)/`&hbar;` with s.

 

This is a first task (that I have asked about in a separate question).

 

Then, I would want Maple to do the cancellations for me, but I have already used "simplify".

 

Finally, I think I would like to collect terms involving cos and sin.

 

Here is what happens when I try to get Maple to solve the equations for me in one go

 

solve({`&psi;__1`(0) = `&psi;__2`(0), `&psi;__1,d`(0) = `&psi;__2,d`(0), `&psi;__2`(l) = `&psi;__3`(l), `&psi;__2,d`(l) = `&psi;__3,d`(l)}, {A, B, C, G})

{A = 0, B = 0, C = 0, G = 0}

(9)

Download solve_constants.mw

In a calculation, I obtained the following expression from Maple.

expr1:=A*sqrt(-m*(-V__0 + E))/sqrt(m*E)

A*(-m*(-V__0+E))^(1/2)/(m*E)^(1/2)

(1)

The first question I have is why the "m's don't cancel."

Next, consider the expression

expr2:=exp(sqrt(2)*sqrt(m)*sqrt(-V__0 + E)*l*I/`&hbar;`)

exp(I*2^(1/2)*m^(1/2)*(-V__0+E)^(1/2)*l/`&hbar;`)

(2)

simplify(expr2,[E-V__0=y])

exp(I*2^(1/2)*m^(1/2)*y^(1/2)*l/`&hbar;`)

(3)

simplify(expr2,[2*m=y])

exp(I*y^(1/2)*(-V__0+E)^(1/2)*l/`&hbar;`)

(4)

Why doesn't the simplification below work?

simplify(expr2,[2*m*(E-V__0)=y])

exp(I*2^(1/2)*m^(1/2)*(-V__0+E)^(1/2)*l/`&hbar;`)

(5)

I also tried with other commands

eval(expr2,2*m=y)

exp(I*2^(1/2)*m^(1/2)*(-V__0+E)^(1/2)*l/`&hbar;`)

(6)

eval(expr2,E-V__0=y)

exp(I*2^(1/2)*m^(1/2)*y^(1/2)*l/`&hbar;`)

(7)

eval(expr2,2*m*(E-V__0)=y)

exp(I*2^(1/2)*m^(1/2)*(-V__0+E)^(1/2)*l/`&hbar;`)

(8)

algsubs(2*m=y,expr2)

exp(I*y^(1/2)*(-V__0+E)^(1/2)*l/`&hbar;`)

(9)

algsubs(E-V__0=y,expr2)

exp(I*2^(1/2)*m^(1/2)*y^(1/2)*l/`&hbar;`)

(10)

algsubs(2*m*(E-V__0)=y,expr2)

exp(I*2^(1/2)*m^(1/2)*(-V__0+E)^(1/2)*l/`&hbar;`)

(11)
 

NULL

In reality, I would like to the entirety of sqrt(2m(E-V__0)/hbar) a variable by the name of s.

Download simplify_roots.mw

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