MaplePrimes Questions

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Hello everyone,

I am learning how Maple can be applied in academic research, especially for statistical data analysis.

I would like to know:

1. Which Maple packages are recommended for statistical analysis?
2. Can Maple perform descriptive statistics, regression analysis, ANOVA, and hypothesis testing?
3. Is Maple suitable for handling medium to large research datasets?
4. Are there any examples or tutorials demonstrating statistical analysis workflows in Maple?

I would appreciate any guidance, documentation, or example worksheets that could help beginners use Maple effectively for research purposes.

Thank you!

I tried to evaluate the function

convert(BesselJ(nu, x), FormalPowerSeries)

only to obtain the Error message

Error, (in convert/FormalPowerSeries) input contains no or more than one variable.

Seems a rather strange error. I thought it would treat x as a single variable

@aroche 

Is there a Maple Support Update package for Maple 2025 ? If so, how do I download it?

Thanks, Roy

Hi Maple community, and all,

Have a small ask, regarding prime numbers.

see attached

vertical_list_of_prime_numbers.mw

vertical_list_of_prime_numbers.pdf

Thanks in advance.

Regards,

Matt

I cannot find a description of the use of the form of dsolve and the following evaluation of its constants which are found in the downloaded worksheet.

Gnadig_2_problem_177_Rod_moving_on_a_wire_in_B_field.mw

Dear sir how to plots the graphs in three region BC from -1 to 0 and 0 to 1 and 1 to 2 
3_region_work.mw

Dear Maple users

I am testing Maple 2026, which will be used at our school after the summer holiday. I see that AI have now found its way to Maple on a new level. AI can be used internally via the AI Assistant, but as I understand it will also be possible to let ChatGPT use Maple in order to provide an answer, instead of using it's own way to do math. I just cannot figure out how it is done. Having a school license for Maple means we take part in the Maplesoft Elite Maintenance Program (EMP). In Maple 2026 i have found Maple MCP on the "My Maple" båndet. When I click it I am however just referred to the main page of Maplesoft. My question: How can I make Maple MCP work for me?

Kind regards,

Erik V.

Why doesn't a piecewise function plot correctly?

(1)   plot(sin((2*Pi)*100*t), t = 0 .. 1);        Plots correctly

(2)   s1 := t -> sin(200*t*Pi);
       plot(s1(t), t = 0 .. 1);                           Plots correctly

(3)   s2 := t -> piecewise(0 < t, 0, t < 10, sin(200*t*Pi), 0);
       s2 := proc (t) options operator, arrow; piecewise(0 < t, 0, t < 10, sin(200*t*Pi), 0) end proc

       plot(s2(t), t = 0 .. 1);                            Does not plot correctly. Only a blank plot is displayed.

A very stupid question :I want my one-dimensional output from Maple 2016 and also Maple 2026, use as one-dimensional input in my Maple 2026 , which I just bought. I'am very inpatient and think it must be very simple but too complicated for me. Before making a very extensive and timeconsuming study somebody should be able to tell me to perform this  by a few clicks .
Thanking you beforehand and I'll be very gratefull for your help.

Bartele de Jong The netherlands

In 2-D Math input:
In a product of more than two factors space is not allways sufficient to delimite factors when one of the factors is of type numeric.

Just for my interest: Is there a reason or a rule for that?

2-D Math: space interpreted as multiplication

a*b*c

a*b*c

(1)

2*b*c

2*b*c

(2)

With numbers this does not work in these cases

"a 2 c"

Error, missing operation

"a 2 c"

 

"a b 2"

Error, missing operation

"a b 2"

 

Multiplication operators are required

"2 2 c"

 

2*a*c

2*a*c

(3)

2*a*b

2*a*b

(4)

NULL

Download Missing_operation.mw

I am looking to do some gravitational perturbations around a generic background spacetime. But before doing that, I wanted to look at just linearized gravity, and make sure all the standard calculations work with Physics before throwing something a little more complicated at it. I went to ?Physics,Library and found the Linearize command, and I thought this was great! When I was reading through it however, I found that the sign infront of the perturbation in the inverse metric is incorrect. Now, this does not give any invalid results for the Ricci tensor as displayed in the worksheet, since we are only going to linear order, but if we want to go beyond linear order, this will start to cause issues. 

Is there a way that Maple can handle this? Or do I have to do some sort of double Define for the metric: one with all downstairs indices, and one with all upstairs indices? If so, how do I do that? 

Any help would be greatly appreciated!

restart: with(Physics): with(Library):

Setup(coordinates = cartesian,signature=`-+++`):

`Systems of spacetime coordinates are:`*{X = (t, x, y, z)}

 

_______________________________________________________

(1)

g_[];

g_[mu, nu] = (Matrix(4, 4, {(1, 1) = -1, (1, 2) = 0, (1, 3) = 0, (1, 4) = 0, (2, 1) = 0, (2, 2) = 1, (2, 3) = 0, (2, 4) = 0, (3, 1) = 0, (3, 2) = 0, (3, 3) = 1, (3, 4) = 0, (4, 1) = 0, (4, 2) = 0, (4, 3) = 0, (4, 4) = 1}))

(2)
  

 

Define(h[mu, nu],symmetric)

`Defined objects with tensor properties`

 

{Physics:-Dgamma[mu], Physics:-Psigma[mu], Physics:-d_[mu], Physics:-g_[mu, nu], h[mu, nu], Physics:-LeviCivita[alpha, beta, mu, nu], Physics:-SpaceTimeVector[mu](X)}

(3)
  

 

Define(eta[mu,nu]=rhs((2)))

`Defined objects with tensor properties`

 

{Physics:-Dgamma[mu], Physics:-Psigma[mu], Physics:-d_[mu], eta[mu, nu], Physics:-g_[mu, nu], h[mu, nu], Physics:-LeviCivita[alpha, beta, mu, nu], Physics:-SpaceTimeVector[mu](X)}

(4)

g_[mu,nu]=eta[mu,nu]+epsilon*h[mu,nu]

Physics:-g_[mu, nu] = epsilon*h[mu, nu]+eta[mu, nu]

(5)

Lets "define" the inverse metric as it appears from the Library:-Linearize worksheet.

g_[~mu,~alpha]=eta[~mu,~alpha]+epsilon*h[~mu,~alpha]

Physics:-g_[`~alpha`, `~mu`] = epsilon*h[`~mu`, `~alpha`]+eta[`~alpha`, `~mu`]

(6)

If we multiply the metric and its inverse together, we should expact that we return the KroneckerDelta by definition -- if we consider only to linear order.  

(5)*(6)

Physics:-g_[mu, nu]*Physics:-g_[`~alpha`, `~mu`] = (epsilon*h[mu, nu]+eta[mu, nu])*(epsilon*h[`~mu`, `~alpha`]+eta[`~alpha`, `~mu`])

(7)

expand((7))

Physics:-g_[mu, nu]*Physics:-g_[`~alpha`, `~mu`] = epsilon^2*h[mu, nu]*h[`~mu`, `~alpha`]+epsilon*eta[mu, nu]*h[`~mu`, `~alpha`]+epsilon*eta[`~alpha`, `~mu`]*h[mu, nu]+eta[mu, nu]*eta[`~alpha`, `~mu`]

(8)

Substitute(eta=g_,(8))

Physics:-g_[mu, nu]*Physics:-g_[`~alpha`, `~mu`] = epsilon^2*h[mu, nu]*h[`~mu`, `~alpha`]+epsilon*Physics:-g_[mu, nu]*h[`~mu`, `~alpha`]+epsilon*Physics:-g_[`~alpha`, `~mu`]*h[mu, nu]+Physics:-g_[mu, nu]*Physics:-g_[`~alpha`, `~mu`]

(9)

subs(epsilon^2=0,(9))

Physics:-g_[mu, nu]*Physics:-g_[`~alpha`, `~mu`] = epsilon*Physics:-g_[mu, nu]*h[`~mu`, `~alpha`]+epsilon*Physics:-g_[`~alpha`, `~mu`]*h[mu, nu]+Physics:-g_[mu, nu]*Physics:-g_[`~alpha`, `~mu`]

(10)

Simplify(%)

Physics:-g_[nu, `~alpha`] = 2*epsilon*h[nu, `~alpha`]+Physics:-g_[nu, `~alpha`]

(11)

 

As we can see, we do not get delta alone on the right-hand-side, but instead we still have the perturbation still.

If we instead, use the proper way the inverse should look, which of course comes from the definition of the inverse, it should have minus sign.

g_[~mu,~alpha]=eta[~mu,~alpha]-epsilon*h[~mu,~alpha]

Physics:-g_[`~alpha`, `~mu`] = -epsilon*h[`~mu`, `~alpha`]+eta[`~alpha`, `~mu`]

(12)

subs(epsilon^2=0,expand((5)*(12)))

Physics:-g_[mu, nu]*Physics:-g_[`~alpha`, `~mu`] = -epsilon*eta[mu, nu]*h[`~mu`, `~alpha`]+epsilon*eta[`~alpha`, `~mu`]*h[mu, nu]+eta[mu, nu]*eta[`~alpha`, `~mu`]

(13)

Simplify(Substitute(eta=g_,(13)))

Physics:-g_[nu, `~alpha`] = Physics:-g_[nu, `~alpha`]

(14)

Which is the desired result we want. So, my question: is there a way that Maple can produce the correct inverse metric not only to linear order, but to say quadratic, without explicitly deriving it ourselves?

Here is the Physics:-Library(Linearize) Worksheet/Example with some comments

restart: with(Physics): with(Library):

Setup(coordinates = cartesian);

`Systems of spacetime coordinates are:`*{X = (x, y, z, t)}

 

_______________________________________________________

 

[coordinatesystems = {X}]

(1)
  

The default metric when Physics is loaded is the Minkowski metric, representing a flat (no curvature) spacetime

g_[];

g_[mu, nu] = (Matrix(4, 4, {(1, 1) = -1, (1, 2) = 0, (1, 3) = 0, (1, 4) = 0, (2, 1) = 0, (2, 2) = -1, (2, 3) = 0, (2, 4) = 0, (3, 1) = 0, (3, 2) = 0, (3, 3) = -1, (3, 4) = 0, (4, 1) = 0, (4, 2) = 0, (4, 3) = 0, (4, 4) = 1}))

(2)
  

Suppose you want to define a small perturbation around this metric. For that purpose, define a perturbation tensor h[mu, nu], that in the general case depends on the coordinates and is not diagonal, the only requirement is that it is symmetric (to have it diagonal, change symmetric by diagonal; to have it constant, change delta[i, j](X) by delta[i, j])

h[mu, nu] = Matrix(4, (i, j) -> delta[i, j](X), shape = symmetric);

h[mu, nu] = (Matrix(4, 4, {(1, 1) = delta[1, 1](x, y, z, t), (1, 2) = delta[1, 2](x, y, z, t), (1, 3) = delta[1, 3](x, y, z, t), (1, 4) = delta[1, 4](x, y, z, t), (2, 1) = delta[1, 2](x, y, z, t), (2, 2) = delta[2, 2](x, y, z, t), (2, 3) = delta[2, 3](x, y, z, t), (2, 4) = delta[2, 4](x, y, z, t), (3, 1) = delta[1, 3](x, y, z, t), (3, 2) = delta[2, 3](x, y, z, t), (3, 3) = delta[3, 3](x, y, z, t), (3, 4) = delta[3, 4](x, y, z, t), (4, 1) = delta[1, 4](x, y, z, t), (4, 2) = delta[2, 4](x, y, z, t), (4, 3) = delta[3, 4](x, y, z, t), (4, 4) = delta[4, 4](x, y, z, t)}))

(3)
  

In the above it is understood that delta[i, j] are small quantities, so that quadratic or higher powers of it can be approximated to 0 (i.e., discarded). Define the components of h[mu, nu] accordingly

Define((3));

`Defined objects with tensor properties`

 

{Physics:-Dgamma[mu], Physics:-Psigma[mu], Physics:-d_[mu], Physics:-g_[mu, nu], h[mu, nu], Physics:-LeviCivita[alpha, beta, mu, nu], Physics:-SpaceTimeVector[mu](X)}

(4)
  

Define also a tensor eta[mu, nu] representing the unperturbed Minkowski metric

eta[mu, nu] = rhs((2));

eta[mu, nu] = (Matrix(4, 4, {(1, 1) = -1, (1, 2) = 0, (1, 3) = 0, (1, 4) = 0, (2, 1) = 0, (2, 2) = -1, (2, 3) = 0, (2, 4) = 0, (3, 1) = 0, (3, 2) = 0, (3, 3) = -1, (3, 4) = 0, (4, 1) = 0, (4, 2) = 0, (4, 3) = 0, (4, 4) = 1}))

(5)

Define((5));

`Defined objects with tensor properties`

 

{Physics:-Dgamma[mu], Physics:-Psigma[mu], Physics:-d_[mu], eta[mu, nu], Physics:-g_[mu, nu], h[mu, nu], Physics:-LeviCivita[alpha, beta, mu, nu], Physics:-SpaceTimeVector[mu](X)}

(6)
  

The weakly perturbed metric is given by

g_[mu, nu] = eta[mu, nu] + h[mu, nu];

Physics:-g_[mu, nu] = eta[mu, nu]+h[mu, nu]

(7)
  

Make this be the definition of the metric

Define((7));

_______________________________________________________

 

`Coordinates: `[x, y, z, t]*`. Signature: `(`- - - +`)

 

_______________________________________________________

 

Physics:-g_[mu, nu] = Matrix(%id = 36893488152142178892)

 

_______________________________________________________

 

`Setting `*lowercaselatin_is*` letters to represent `*space*` indices`

 

`Defined objects with tensor properties`

 

{Physics:-D_[mu], Physics:-Dgamma[mu], Physics:-Psigma[mu], Physics:-Ricci[mu, nu], Physics:-Riemann[mu, nu, alpha, beta], Physics:-Weyl[mu, nu, alpha, beta], Physics:-d_[mu], eta[mu, nu], Physics:-g_[mu, nu], Physics:-gamma_[i, j], h[mu, nu], Physics:-Christoffel[mu, nu, alpha], Physics:-Einstein[mu, nu], Physics:-LeviCivita[alpha, beta, mu, nu], Physics:-SpaceTimeVector[mu](X)}

(8)
  

The linearized form of the Ricci tensor is computed by introducing this weakly perturbed metric in the expression of the Ricci  tensor as a function of the metric. This can be accomplished in different ways, the simpler being to use the conversion network between tensors, but for illustration purposes, showing steps one at time, a substitution of definitions one into the other one is used

Ricci[definition];

Physics:-Ricci[mu, nu] = Physics:-d_[alpha](Physics:-Christoffel[`~alpha`, mu, nu], [X])-Physics:-d_[nu](Physics:-Christoffel[`~alpha`, mu, alpha], [X])+Physics:-Christoffel[`~beta`, mu, nu]*Physics:-Christoffel[`~alpha`, beta, alpha]-Physics:-Christoffel[`~beta`, mu, alpha]*Physics:-Christoffel[`~alpha`, nu, beta]

(9)

Christoffel[~alpha, mu, nu, definition];

Physics:-Christoffel[`~alpha`, mu, nu] = (1/2)*Physics:-g_[`~alpha`, `~beta`]*(Physics:-d_[nu](Physics:-g_[beta, mu], [X])+Physics:-d_[mu](Physics:-g_[beta, nu], [X])-Physics:-d_[beta](Physics:-g_[mu, nu], [X]))

(10)

Substitute((10), (9));

Physics:-Ricci[mu, nu] = Physics:-d_[alpha]((1/2)*Physics:-g_[`~alpha`, `~kappa`]*(Physics:-d_[nu](Physics:-g_[kappa, mu], [X])+Physics:-d_[mu](Physics:-g_[kappa, nu], [X])-Physics:-d_[kappa](Physics:-g_[mu, nu], [X])), [X])-Physics:-d_[nu]((1/2)*Physics:-g_[`~alpha`, `~tau`]*(Physics:-d_[mu](Physics:-g_[tau, alpha], [X])+Physics:-d_[alpha](Physics:-g_[tau, mu], [X])-Physics:-d_[tau](Physics:-g_[alpha, mu], [X])), [X])+(1/4)*Physics:-g_[`~beta`, `~iota`]*(Physics:-d_[nu](Physics:-g_[iota, mu], [X])+Physics:-d_[mu](Physics:-g_[iota, nu], [X])-Physics:-d_[iota](Physics:-g_[mu, nu], [X]))*Physics:-g_[`~alpha`, `~lambda`]*(Physics:-d_[beta](Physics:-g_[lambda, alpha], [X])+Physics:-d_[alpha](Physics:-g_[lambda, beta], [X])-Physics:-d_[lambda](Physics:-g_[alpha, beta], [X]))-(1/4)*Physics:-g_[`~beta`, `~omega`]*(Physics:-d_[mu](Physics:-g_[omega, alpha], [X])+Physics:-d_[alpha](Physics:-g_[omega, mu], [X])-Physics:-d_[omega](Physics:-g_[alpha, mu], [X]))*Physics:-g_[`~alpha`, `~chi`]*(Physics:-d_[nu](Physics:-g_[chi, beta], [X])+Physics:-d_[beta](Physics:-g_[chi, nu], [X])-Physics:-d_[chi](Physics:-g_[beta, nu], [X]))

(11)
  

Introducing the perturbed metric, and the inert form of Ricci for simplification purposes

Substitute((7), Ricci = %Ricci, (11));

%Ricci[mu, nu] = (1/2)*Physics:-d_[alpha](eta[`~alpha`, `~kappa`]+h[`~alpha`, `~kappa`], [X])*(Physics:-d_[nu](eta[kappa, mu]+h[kappa, mu], [X])+Physics:-d_[mu](eta[kappa, nu]+h[kappa, nu], [X])-Physics:-d_[kappa](eta[mu, nu]+h[mu, nu], [X]))+(1/2)*(eta[`~alpha`, `~kappa`]+h[`~alpha`, `~kappa`])*(Physics:-d_[alpha](Physics:-d_[nu](eta[kappa, mu]+h[kappa, mu], [X]), [X])+Physics:-d_[alpha](Physics:-d_[mu](eta[kappa, nu]+h[kappa, nu], [X]), [X])-Physics:-d_[alpha](Physics:-d_[kappa](eta[mu, nu]+h[mu, nu], [X]), [X]))-(1/2)*Physics:-d_[nu](eta[`~alpha`, `~tau`]+h[`~alpha`, `~tau`], [X])*(Physics:-d_[mu](eta[alpha, tau]+h[alpha, tau], [X])+Physics:-d_[alpha](eta[mu, tau]+h[mu, tau], [X])-Physics:-d_[tau](eta[alpha, mu]+h[alpha, mu], [X]))-(1/2)*(eta[`~alpha`, `~tau`]+h[`~alpha`, `~tau`])*(Physics:-d_[mu](Physics:-d_[nu](eta[alpha, tau]+h[alpha, tau], [X]), [X])+Physics:-d_[alpha](Physics:-d_[nu](eta[mu, tau]+h[mu, tau], [X]), [X])-Physics:-d_[nu](Physics:-d_[tau](eta[alpha, mu]+h[alpha, mu], [X]), [X]))+(1/4)*(eta[`~beta`, `~iota`]+h[`~beta`, `~iota`])*(Physics:-d_[nu](eta[iota, mu]+h[iota, mu], [X])+Physics:-d_[mu](eta[iota, nu]+h[iota, nu], [X])-Physics:-d_[iota](eta[mu, nu]+h[mu, nu], [X]))*(eta[`~alpha`, `~lambda`]+h[`~alpha`, `~lambda`])*(Physics:-d_[beta](eta[alpha, lambda]+h[alpha, lambda], [X])+Physics:-d_[alpha](eta[beta, lambda]+h[beta, lambda], [X])-Physics:-d_[lambda](eta[alpha, beta]+h[alpha, beta], [X]))-(1/4)*(eta[`~beta`, `~omega`]+h[`~beta`, `~omega`])*(Physics:-d_[mu](eta[alpha, omega]+h[alpha, omega], [X])+Physics:-d_[alpha](eta[mu, omega]+h[mu, omega], [X])-Physics:-d_[omega](eta[alpha, mu]+h[alpha, mu], [X]))*(eta[`~alpha`, `~chi`]+h[`~alpha`, `~chi`])*(Physics:-d_[nu](eta[beta, chi]+h[beta, chi], [X])+Physics:-d_[beta](eta[chi, nu]+h[chi, nu], [X])-Physics:-d_[chi](eta[beta, nu]+h[beta, nu], [X]))

(12)
  

The sign infront of the perturbation in the inverse metric is wrong, it should be minus.

  

This expression contains several terms quadratic in the small perturbation h[mu, nu]. The routine to filter out those terms is Linearize, that takes as second argument the symbol representing the small quantities (perturbation)

Lets look at the metric times inverse in this setup

g_[mu,nu,definition]*g_[~mu,~alpha,definition]

Physics:-g_[mu, nu]*Physics:-g_[`~mu`, `~alpha`] = (eta[mu, nu]+h[mu, nu])*(eta[`~mu`, `~alpha`]+h[`~mu`, `~alpha`])

(13)

Linearize((13),h)

Physics:-g_[mu, nu]*Physics:-g_[`~alpha`, `~mu`] = eta[mu, nu]*eta[`~alpha`, `~mu`]+eta[mu, nu]*h[`~alpha`, `~mu`]+eta[`~alpha`, `~mu`]*h[mu, nu]

(14)

Simplify(subs(eta=g_,(14)))

Physics:-g_[nu, `~alpha`] = Physics:-g_[nu, `~alpha`]+2*h[nu, `~alpha`]

(15)

The result is not correct, left-hand-side does not match right-hand-side, this is because the inverse metric has the wrong. If it were a minus, we would get:

g_[mu, nu]*g_[~alpha, ~mu] = eta[mu, nu]*eta[~alpha, ~mu] - eta[mu, nu]*h[~alpha, ~mu] + eta[~alpha, ~mu]*h[mu, nu]

Physics:-g_[mu, nu]*Physics:-g_[`~alpha`, `~mu`] = eta[mu, nu]*eta[`~alpha`, `~mu`]-eta[mu, nu]*h[`~alpha`, `~mu`]+eta[`~alpha`, `~mu`]*h[mu, nu]

(16)

Simplify(subs(eta=g_,(16)))

Physics:-g_[nu, `~alpha`] = Physics:-g_[nu, `~alpha`]

(17)

Which is correct. The continued calculation from the Help page is below.

 

Linearize((12), h);

%Ricci[mu, nu] = (1/2)*eta[`~alpha`, `~tau`]*Physics:-d_[nu](Physics:-d_[tau](h[alpha, mu], [X]), [X])-(1/2)*eta[`~alpha`, `~tau`]*Physics:-d_[mu](Physics:-d_[nu](h[alpha, tau], [X]), [X])-(1/2)*eta[`~alpha`, `~kappa`]*Physics:-d_[alpha](Physics:-d_[kappa](h[mu, nu], [X]), [X])+(1/2)*eta[`~alpha`, `~kappa`]*Physics:-d_[alpha](Physics:-d_[nu](h[kappa, mu], [X]), [X])+(1/2)*eta[`~alpha`, `~kappa`]*Physics:-d_[alpha](Physics:-d_[mu](h[kappa, nu], [X]), [X])-(1/2)*eta[`~alpha`, `~tau`]*Physics:-d_[alpha](Physics:-d_[nu](h[mu, tau], [X]), [X])

(18)
 
  

In this result, eta[mu, nu] is the flat Minkowski metric. To further simplify this expression using the internal algorithms for a flat metric it is practical to reintroduce g[mu, nu] representing that Minkowski metric

g_[min];

_______________________________________________________

 

`The Minkowski metric in coordinates `*[x, y, z, t]

 

`Signature: `(`- - - +`)

 

_______________________________________________________

 

Physics:-g_[mu, nu] = Matrix(%id = 36893488152069364060)

(19)
  

Replace in the expression for the Ricci tensor the intermediate Minkowski eta[mu, nu]by g[mu, nu]

subs(eta = g_, (18));

%Ricci[mu, nu] = (1/2)*Physics:-g_[`~alpha`, `~tau`]*Physics:-d_[nu](Physics:-d_[tau](h[alpha, mu], [X]), [X])-(1/2)*Physics:-g_[`~alpha`, `~tau`]*Physics:-d_[mu](Physics:-d_[nu](h[alpha, tau], [X]), [X])-(1/2)*Physics:-g_[`~alpha`, `~kappa`]*Physics:-d_[alpha](Physics:-d_[kappa](h[mu, nu], [X]), [X])+(1/2)*Physics:-g_[`~alpha`, `~kappa`]*Physics:-d_[alpha](Physics:-d_[nu](h[kappa, mu], [X]), [X])+(1/2)*Physics:-g_[`~alpha`, `~kappa`]*Physics:-d_[alpha](Physics:-d_[mu](h[kappa, nu], [X]), [X])-(1/2)*Physics:-g_[`~alpha`, `~tau`]*Physics:-d_[alpha](Physics:-d_[nu](h[mu, tau], [X]), [X])

(20)
  

Simplifying, results in the linearized form of the Ricci tensor

Simplify((20));

%Ricci[mu, nu] = -(1/2)*Physics:-d_[mu](Physics:-d_[nu](h[tau, `~tau`], [X]), [X])-(1/2)*Physics:-dAlembertian(h[mu, nu], [X])+(1/2)*Physics:-d_[nu](Physics:-d_[tau](h[mu, `~tau`], [X]), [X])+(1/2)*Physics:-d_[mu](Physics:-d_[tau](h[nu, `~tau`], [X]), [X])

(21)

This is correct result, because we are going to linear order only the +/- does not have an effect on the end result.

Download LinearizedWorksheet-Comments.mw

Download LinearQuestion.mw

I am trying to use the Perm command in the GroupTheory package to create permutations. The problem is when the permutation has fixed points. For example, neither of the forms

[[1,4,7],[2,8,5],[3],[6]]

[[1,4,7],[2,8,5],[3,3],[6,6]]

will work. Any suggestions?

Has anybody seen something like that? I do not use Maple 2026 very often.

Does this vanish when the document is executed on another machine?

 

repeated_equation_labels.mw

Update:

  • expanding the document block by "show command" makes the equation labels disappear.
  • copying the input to another document block seems to fix the problem

Consider these two output, both for solving system of 2 first order different equations.

Why is the first result is put in a list, then each solution is in a set inside the list, while the second one is just a set of the two solutions?

My guess is that because the first system is non-linear.  Is this why?

This makes it little harder to parse the result later on, as it can change each time. 

Is there a way to get same output for the first example as in the second example?

Mapkle 2026.1

ode:=diff(x(t),t) = x(t)^2, diff(y(t),t) = exp(t);
sol:=dsolve([ode],[x(t),y(t)])

ode:=diff(x(t),t) = x(t), diff(y(t),t) = t;
sol:=dsolve([ode],[x(t),y(t)])

ps. the ode's are not even coupled in these example. So each can be solved on its own if needed.

And if there is one ode with multiple solutions, now dsolve returns expression sequence. No set, no list.

ode:=2*x*diff(y(x),x)*diff(diff(y(x),x),x) = -1+diff(y(x),x)^2; 
dsolve(ode,y(x));

This whole thing is a mess. 

There should be one consistent way to return solutions for all cases. 

Regadless if it is one ode with one solution, or one ode with mutliple solutions, or coupled systems of odes, linear or not and so on.

The output should be the same form in all cases. A list of lists or list of sets or whatever it is decided on.

But it should not change.

i do same trasnformation but i don't know what is issue  the result is near to same but parameter (t) appear in my which that make my ode  not be correct so  i can't see the problem in here  and i am intrested in finding this, regarding to this i will put here my result and the papers result rregarding to resolve the issue 

pde1

pde2

 

T1.mw

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