How to make ChatGPT use Maple via MCP

erik10 — Fri, 03 Jul 2026 23:00:04 Z

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.

plotting piecewise function

anthonymlinar — Fri, 03 Jul 2026 17:52:49 Z

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.

Upgrading from Maple 2016.2 to Maple 2026

Bart_de_Jong — Wed, 01 Jul 2026 13:09:05 Z

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

Missing operation (multiplication) in 2D-Math

C_R — Sat, 27 Jun 2026 11:34:01 Z

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?

Download Missing_operation.mw

How can I get the correct inverse metric with Physics?

Hullzie16 — Tue, 23 Jun 2026 21:44:57 Z

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 h 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=`-+++`):

(1)

g_[];

(2)

>	Define(h[mu, nu],symmetric)

(3)

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

(4)

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

(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)

(7)

>	expand((7))

(8)

>	Substitute(eta=g_,(8))

(9)

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

(10)

>	Simplify(%)

(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]

(12)

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

(13)

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

(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);

(1)

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

g_[];

(2)

Suppose you want to define a small perturbation around this metric. For that purpose, define a perturbation tensor , 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 by )

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

(3)

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

>	Define((3));

(4)

Define also a tensor representing the unperturbed Minkowski metric

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

(5)

>	Define((5));

(6)

The weakly perturbed metric is given by

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

(7)

Make this be the definition of the metric

>	Define((7));

(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];

(9)

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

(10)

>	Substitute((10), (9));

(11)

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

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

(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 . 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]

(13)

>	Linearize((13),h)

(14)

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

(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]

(16)

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

(17)

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

>	Linearize((12), h);

(18)

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

g_[min];

(19)

Replace in the expression for the Ricci tensor the intermediate Minkowski by

>	subs(eta = g_, (18));

(20)

Simplifying, results in the linearized form of the Ricci tensor

>	Simplify((20));

(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

What if my permutation has a fixed point?

wkehowski — Tue, 23 Jun 2026 06:13:03 Z

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]]