Unanswered Questions

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The equal sign ":=" in my document sign has changed to a "d".  The equal sign "=" has not changed.  Is there a setting I need to change?

Thank you in advance.

I am trying to find the value of y4 at t=infinity and t=-infinity when lambda1>lambda2 or lambda1<lambda2. But every time I got the same answer. For example, if we do it by hand then the terms which are responsible for making the indeterminate form can be extracted and canceled (see Fig.). 

But in limit.mw y4 is too lengthy-expression and very difficult to do it manually.

The Statistics package contains a function named Specialize (which quite strangely doesn't appear when you expand the sections of this package).
Here is what help(Specialize) says:

The Specialize function takes a random variable or distribution data structure that contains symbolic parameters, and performs a substitution to specialize the given random variable or distribution.

My goal was to work with mixtures of two random variables. There are many ways to do that depending on the what you really want to achieve, but an elegant way is to define such a mixture this way:

  • Let X and Y two random variables representing the two components to be mixed.
    For instance X = Normal(mu, sigma) and Y = Normal(nu, tau).
     
  • Let B a Bernoulli random variable with parameter P.
     
  • Then M = B*X + (1-B)*Y represents a random mixture of the two components in proportions (p, 1-p).
    Note that M is a 5-parameters random variable.

Doing the things this way enables getting a lot of formal informations about M such as its mean, variance, and so on.

In order to illustrate what the mixture is I draw the histogram of a sample of M.
To do this I Specialized the three random variables X, Y, B.

I used parameters

mu=-3, nu=3, sigma=1, tau=1, p=1/2


My first attempt was to draw a sample of the random variable Mspec defined this way

Mspec := Specialize(B, [p=1/2])*Specialize(X, [mu=-3, sigma=1]) + (1-Specialize(B, [p=1/2]))*Specialize(Y, [nu=3, tau=1]);

As you see in the attached file (first plot) the histogram is wrong (so is the variance computed formally).

I changed this into

Mspec := Specialize(B, [p=1/2])*(Specialize(X, [mu=-3, sigma=1])-Specialize(Y, [nu=3, tau=1])) + Specialize(Y, [nu=3, tau=1]);

without more significative success: while the variance is nox corrext the histogram still remains obviously wrong (plot number 2)

My last attempt, which now gives q correct result (plot 3) was:

Bspec := Specialize(B, [p=1/2]);
Mspec := Bspec*Specialize(X, [mu=-3, sigma=1]) + (1-Bspec)*Specialize(Y, [nu=3, tau=1]);

Specialize.mw

I agree that one can easily do this stuff without using  Specialize.
For instance by using the procedure given at the end ofthe attached file.
Or by truly constructing a mixture Distribution (which would be more elegant but more complex).

I also agree that Specialize is in itself of a relative low interest except for educational purposes (you present the theoritical results and next you run a numerical application while giving numeric values to the formal parameters).

But why providing such an anecdotal function if it doesn't do the job correctly?

Note that these results were obtained with Maple 2015, but I doubt they'll be any better for more recent versions, given the confidential nature of Specialize.

I am trying to calculate the line element ds^2, for a de Sitter spacetime in 2+1 dimensions with positive cosmological constant, using the following metric and energy moment tensor:
1- ds² = -A(r) c^2 dt^2 + B(r) dr^2 + r^2d{\theta}^2,
2- T^{\mu \nu} = -(\rho + p) dx^{\mu} dx^{\nu} + p g^{\mu \nu}.

I tried several ways but I can't solve it using Maple 2023, Physics package. Could someone show me step by step how to solve this problem?

Images of the line element we need to find, the metric tensor associated with the problem and the components of the Riemann tensor.
My goal is to calculate the metric tensor, line element and associated components of the Riemann tensor for De Sitter spacetime with positive cosmological constant.

Hello everyone,

I'm trying to learn how to use Shoot Library 9.

Unfortunately, I'm not doing very well. I'm getting an error that I don't understand. I don't know where it comes from.

Please help me solve this odes system using this library.

I attached my Maple worksheet file.

ShootLib_Test.mw

I'm currently addressing a problem related to modified Bessel functions using an older version of Maple (the specific version escapes my memory). In an attempt to resolve issues, I've experimented with the trial version of Maple 2023, but I've encountered an unusual phenomenon. Expressions that were previously simplifiable in Maple now resist simplification. The specific expression provided below, which should equate to 1, fails to be recognized as such by Maple. This poses a concern as it could lead to overly complex expressions in subsequent steps, considering this expression is only an intermediate stage. Is there a recommended approach to overcome this challenge?

f := (BesselI(0, alpha)*alpha-2*BesselI(1, alpha))/(BesselK(0, alpha)*BesselI(1, alpha)*BesselI(0, alpha)*alpha^2+BesselK(1, alpha)*BesselI(0, alpha)^2*alpha^2-2*BesselI(1, alpha))

(BesselI(0, alpha)*alpha-2*BesselI(1, alpha))/(BesselK(0, alpha)*BesselI(1, alpha)*BesselI(0, alpha)*alpha^2+BesselK(1, alpha)*BesselI(0, alpha)^2*alpha^2-2*BesselI(1, alpha))

simplify(f)

(BesselI(0, alpha)*alpha-2*BesselI(1, alpha))/(BesselK(0, alpha)*BesselI(1, alpha)*BesselI(0, alpha)*alpha^2+BesselK(1, alpha)*BesselI(0, alpha)^2*alpha^2-2*BesselI(1, alpha))

eval(f, alpha = .25)

1.000000000

NULL

Download question.mw

this is my model. Please give me how to find the DFE and basic reproduction number from maple.

I've reported this to Maplesoft 6 months ago.

I was wondering if someone with beta version of 2024 could check if these are fixed? (if one is allowed to do so). As these errors keep breaking my program. (not possible to trap).

436

interface(version);

`Standard Worksheet Interface, Maple 2023.2, Windows 10, November 24 2023 Build ID 1762575`

Physics:-Version();

`The "Physics Updates" version in the MapleCloud is 1637 and is the same as the version installed in this computer, created 2023, November 29, 17:28 hours Pacific Time.`

ode:=diff(y(x),x) = (x*y(x)+x^3+x*y(x)^2+y(x)^3)/x^2;
sol:=exp(3*sum(1/(9*_R^2-1)*ln((-_R*x+y(x)-1/3*x)/x),_R = RootOf(27*_Z^3-9*_Z+29)))-c__1*exp(x) = 0;
odetest(sol,ode);

diff(y(x), x) = (x*y(x)+x^3+x*y(x)^2+y(x)^3)/x^2

exp(3*(sum(ln((-_R*x+y(x)-(1/3)*x)/x)/(9*_R^2-1), _R = RootOf(27*_Z^3-9*_Z+29))))-c__1*exp(x) = 0

Error, (in simplify/RootOf) too many levels of recursion

ode:=diff(u(x),x)-1/2*(2*a*u(x)^3+u(x)+2*b)/x = 0;
sol:=2*sum(1/(6*_R^2*a+1)*ln(u(x)-_R),_R = RootOf(2*_Z^3*a+_Z+2*b))-1/2*ln(x)-_C1 = 0;
odetest(sol,ode);

diff(u(x), x)-(1/2)*(2*a*u(x)^3+u(x)+2*b)/x = 0

2*(sum(ln(u(x)-_R)/(6*_R^2*a+1), _R = RootOf(2*_Z^3*a+_Z+2*b)))-(1/2)*ln(x)-_C1 = 0

Error, (in simplify/RootOf) too many levels of recursion

 

Download in_simplify_rootof_too_many_level_of_recursion_jan_6_2024.mw

I was about to update an older discussion with the information that the context pannel now contains an entry "Normalized Expanded".

I only remember that I was participating with another user.

So, I tried C_R AND other_user in the search field. This gives an error.

A space as an implict AND operator does also not work.

As I assumed 'n' and 'm' are real, eta is complex. But still, there is a bar on these discrete independent variables. Secondly, the substitution of (8) applies in some terms of 'r2', and the remaining terms remain as is it.

restart

with(LinearAlgebra); with(PDEtools); with(plots); with(LREtools)

setup(mathematicalnotation = true)

setup(mathematicalnotation = true)

(1)

assume(n::real); assume(m::real)

A := proc (n, m) options operator, arrow; Matrix([[eta*phi(n, m), conjugate(eta)*conjugate(psi(n, m))], [phi(n, m), conjugate(psi(n, m))]]) end proc; Adet := Determinant(A(n, m))

eta*phi(n, m)*conjugate(psi(n, m))-conjugate(eta)*conjugate(psi(n, m))*phi(n, m)

(2)

B := proc (n, m) options operator, arrow; Matrix([[phi(n, m), conjugate(psi(n, m))], [-psi(n, m), conjugate(phi(n, m))]]) end proc; Bdet := Determinant(B(n, m))

phi(n, m)*conjugate(phi(n, m))+conjugate(psi(n, m))*psi(n, m)

(3)

r := Adet/Bdet

(eta*phi(n, m)*conjugate(psi(n, m))-conjugate(eta)*conjugate(psi(n, m))*phi(n, m))/(phi(n, m)*conjugate(phi(n, m))+conjugate(psi(n, m))*psi(n, m))

(4)

p := {eta = 1+I, phi(n, m) = (1+I*a*eta)^n*(1+I*b*eta^2)^m, psi(n, m) = (1-I*a*eta)^n*(1-I*b*eta^2)^m, conjugate(eta) = 1-I, conjugate(phi(n, m)) = (1-I*a*conjugate(eta))^n*(1-I*b*conjugate(eta)^2)^m, conjugate(phi(n, m)) = (1+I*a*conjugate(eta))^n*(1+I*b*conjugate(eta)^2)^m}

{eta = 1+I, phi(n, m) = (1+I*a*eta)^n*(1+I*b*eta^2)^m, psi(n, m) = (1-I*a*eta)^n*(1-I*b*eta^2)^m, conjugate(eta) = 1-I, conjugate(phi(n, m)) = (1-I*a*conjugate(eta))^n*(1-I*b*conjugate(eta)^2)^m, conjugate(phi(n, m)) = (1+I*a*conjugate(eta))^n*(1+I*b*conjugate(eta)^2)^m}

(5)

r1 := simplify(subs(p, r))

(2*I)*(1+I*a*eta)^n*(1+I*b*eta^2)^m*conjugate((1-I*a*eta)^n*(1-I*b*eta^2)^m)/((1+I*a*eta)^n*(1+I*b*eta^2)^m*(1-I*a*conjugate(eta))^n*(1-I*b*conjugate(eta)^2)^m+abs(-1+I*a*eta)^(2*n)*abs(I*b*eta^2-1)^(2*m))

(6)

r2 := 1-I*delta(r1, n)

1-I*((2*I)*(1+I*a*eta)^(n+1)*(1+I*b*eta^2)^m*conjugate((1-I*a*eta)^(n+1)*(1-I*b*eta^2)^m)/((1+I*a*eta)^(n+1)*(1+I*b*eta^2)^m*(1-I*a*conjugate(eta))^(n+1)*(1-I*b*conjugate(eta)^2)^m+abs(-1+I*a*eta)^(2*n+2)*abs(I*b*eta^2-1)^(2*m))-(2*I)*(1+I*a*eta)^n*(1+I*b*eta^2)^m*conjugate((1-I*a*eta)^n*(1-I*b*eta^2)^m)/((1+I*a*eta)^n*(1+I*b*eta^2)^m*(1-I*a*conjugate(eta))^n*(1-I*b*conjugate(eta)^2)^m+abs(-1+I*a*eta)^(2*n)*abs(I*b*eta^2-1)^(2*m)))

(7)

exp_expr := subs({(1+I*b*eta^2)^m = exp(I*eta^2*t)}, r2)

1-I*((2*I)*(1+I*a*eta)^(n+1)*exp(I*eta^2*t)*conjugate((1-I*a*eta)^(n+1)*(1-I*b*eta^2)^m)/((1+I*a*eta)^(n+1)*exp(I*eta^2*t)*(1-I*a*conjugate(eta))^(n+1)*(1-I*b*conjugate(eta)^2)^m+abs(-1+I*a*eta)^(2*n+2)*abs(I*b*eta^2-1)^(2*m))-(2*I)*(1+I*a*eta)^n*exp(I*eta^2*t)*conjugate((1-I*a*eta)^n*(1-I*b*eta^2)^m)/((1+I*a*eta)^n*exp(I*eta^2*t)*(1-I*a*conjugate(eta))^n*(1-I*b*conjugate(eta)^2)^m+abs(-1+I*a*eta)^(2*n)*abs(I*b*eta^2-1)^(2*m)))

(8)

``

NULL

NULL

NULL

plot3d(abs(exp_expr), n = -5 .. 5, t = -5 .. 5, eta = 1+I)

Error, (in plot3d) unexpected option: eta = 1+I

 
 

Download soldis.mw

Although I still prefer applyrule (as evalindets/subsindets is not as intuitive as applyrule),  I have heard that it is regarded as being more or less antiquated in modern Maple. I notice that a lot of (yet not all) examples given in the help pages of evalindets/subsindets can be reformulated by applyrule, but does any use of applyrule also correspond to using evalindets/subsindets? And if so, how to equivalently rewrite those transformation rules (especially complicated ones like nested function applications) in the syntax of evalindets/subsindets?

There are things that seem simple but rapidly turn into a nightmare.

Here is an example: what I want is to the expression given at equation (4) in the attached file.

Using Int gives a wrong result.
Using int gives a right one but not of the desired form (some double integrals are nested while others are not).

I've been stuck on this problem for hours, can you please help me to fix it?

TIA

restart

use Statistics in
  # For more generality defina an abstract probability distribution.
  AbstractDistribution := proc(N)
    Distribution(
      PDF = (x -> varphi(seq(x[n], n=1..N)))
    )
  end proc:

  # Define two random variables pf AbstractDistribution type.
  X__1 := RandomVariable(AbstractDistribution(2)):
  X__2 := RandomVariable(AbstractDistribution(2)):

end use;

proc (N) Statistics:-Distribution(Statistics:-PDF = (proc (x) options operator, arrow; varphi(seq(x[n], n = 1 .. N)) end proc)) end proc

 

_R

 

_R0

(1)

F := (U1, U2) -> U1/(U1+U2);
T := mtaylor(F(X__1, X__2), [X__1=1, X__2=1], 2):

proc (U1, U2) options operator, arrow; U1/(U1+U2) end proc

(2)


Error: x[2] is droped out of the double integral in the rightmost term

use IntegrationTools in

J := eval([op(expand(T))], [seq(X__||i=x[i], i=1..2)]);
L := add(
       map(
         j ->  
         if j::numeric then
           j
         else
           (Expand@CollapseNested)(
             Int(
               j * Statistics:-PDF(X__1, x)
               , seq(x[i]=-infinity..+infinity, i=1..2)
             )
           )
         end if
         , J
       )  
     ):
ET := %
end use;

[1/2, (1/4)*x[1], -(1/4)*x[2]]

 

1/2+(1/4)*(Int(x[1]*varphi(x[1], x[2]), [x[1] = -infinity .. infinity, x[2] = -infinity .. infinity]))-(1/4)*x[2]*(Int(varphi(x[1], x[2]), [x[1] = -infinity .. infinity, x[2] = -infinity .. infinity]))

 

1/2+(1/4)*(Int(x[1]*varphi(x[1], x[2]), [x[1] = -infinity .. infinity, x[2] = -infinity .. infinity]))-(1/4)*x[2]*(Int(varphi(x[1], x[2]), [x[1] = -infinity .. infinity, x[2] = -infinity .. infinity]))

(3)


I want this

'ET' = 1/2
       +
       (1/4)*(Int(Int(x[1]*varphi(x[1], x[2]), x[1] = -infinity .. infinity), x[2] = -infinity .. infinity))
       -(1/4)*(Int(Int(x[2]*varphi(x[1], x[2]), x[1] = -infinity .. infinity), x[2] = -infinity .. infinity))

ET = 1/2+(1/4)*(Int(Int(x[1]*varphi(x[1], x[2]), x[1] = -infinity .. infinity), x[2] = -infinity .. infinity))-(1/4)*(Int(Int(x[2]*varphi(x[1], x[2]), x[1] = -infinity .. infinity), x[2] = -infinity .. infinity))

(4)


With int instead of Int one integral is double the other is double-nested

L := add(
       map(
         j ->  
         if j::numeric then
           j
         else
             int(
               j * Statistics:-PDF(X__1, x)
               , seq(x[i]=-infinity..+infinity, i=1..2)
             )
         end if
         , J
       )  
     ):
ET := %

1/2+int(int((1/4)*x[1]*varphi(x[1], x[2]), x[1] = -infinity .. infinity), x[2] = -infinity .. infinity)+int(-(1/4)*x[2]*(int(varphi(x[1], x[2]), x[1] = -infinity .. infinity)), x[2] = -infinity .. infinity)

(5)


As the expression of ET is now correct, I tried to use IntegrationTools to get the
form I want (equation (4)).

But as soon as I replace int by Int x[2] is again droped out.

So it's not even worth thinking about using CollapseNested!

 

use IntegrationTools in
  eval(ET, int=Int);  
end use;

1/2+Int(Int((1/4)*x[1]*varphi(x[1], x[2]), x[1] = -infinity .. infinity), x[2] = -infinity .. infinity)+Int(-(1/4)*x[2]*(Int(varphi(x[1], x[2]), x[1] = -infinity .. infinity)), x[2] = -infinity .. infinity)

(6)

 

Download Int_int.mw

Hi,

is there fix for the following quirk in the Maple 2023 editor:

randomly (hours, days) some characters change their appearance like p.e. the = sign becomes d-bold or sigma becomes s-bold. I have never experienced this in previous versions.

Thanks in advance.

I would like to take advantage from the powerful command "SSTransformation" of the DynamicSystems package to reuse the corresponding output.

For example, if we use the following shape:

         > SSTransformation( Amat, Bmat, Cmat, Dmat, form = ModalCanon, output=['A','B','C','D','T'] );

How to do to assign names to the outputs A,B,C,D and T to subsequently reuse them?

is it possible to find why Maple fails to solve these two equations in two unknowns? Has this always been the case? I do not have older versions of Maple to check. The trace shows that it found solution but then itg says no solution was found. This is very strange.

17020

interface(version)

`Standard Worksheet Interface, Maple 2023.2, Windows 10, November 24 2023 Build ID 1762575`

Physics:-Version()

`The "Physics Updates" version in the MapleCloud is 1622. The version installed in this computer is 1618 created 2023, November 29, 17:28 hours Pacific Time, found in the directory C:\Users\Owner\maple\toolbox\2023\Physics Updates\lib\`

restart;

17020

sol:=1/4*exp(-t) * (c2*(-1+exp(4*t)) + c1*(3+exp(4*t))):
expand(simplify(sol));

-(1/4)*c2/exp(t)+(1/4)*(exp(t))^3*c2+(3/4)*c1/exp(t)+(1/4)*(exp(t))^3*c1

eq1:=-3=eval(sol,t=4):
expand(simplify(eq1));

-3 = (1/4)*c1*exp(12)+(1/4)*c2*exp(12)+(3/4)*exp(-4)*c1-(1/4)*exp(-4)*c2

eq1:=-17=eval(diff(sol,t),t=4);
expand(simplify(eq1));

-17 = -(1/4)*exp(-4)*(c2*(-1+exp(16))+c1*(3+exp(16)))+(1/4)*exp(-4)*(4*c2*exp(16)+4*c1*exp(16))

-17 = (1/4)*exp(-4)*c2+(3/4)*c2*exp(12)-(3/4)*exp(-4)*c1+(3/4)*c1*exp(12)

infolevel[solve]:=5;
solve([eq1,eq2],[c1,c2])

5

Main: Entering solver with 2 equations in 2 variables

Main: attempting to solve as a linear system

Linear: solving 2 linear equations

Algebraic: # equations is: 2

Main: Linear solver successful. Exiting solver returning 1 solution

solve: Warning: no solutions found

[]

Download unable_to_solve_2_equations_dec_26_2023.mw

For reference this is the solution given by Mathematica

 

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