Maple 2026 Questions and Posts

These are Posts and Questions associated with the product, Maple 2026

I was trying to use the debugger into a proc that has this call

P:=plots:-contourplot(RHS,':-colorbar' = false,':-contours' = L):

Even though the proc has : at its end, and the above call to plots also ends with :, the debugger insists in printing to the debugger window the contour cuves lines. i.e the value of P

Is there a way to tell the debugger not to do this? i.e. not show the value of P. It seems it does that automatically.

Here is the worksheet. Simply evaluate the call foo(); this will open a debugger windows. Then click on next button and now debugger will print  the output of plots:-contourplot(RHS,':-colorbar' = false,':-contours' = L): 

restart;

kernelopts('assertlevel'=2):
kernelopts(numcpus=1);

32

interface(version);

`Standard Worksheet Interface, Maple 2026.1, Windows 10, April 28 2026 Build ID 2011354`

foo:=proc()
local L := [$ -4 .. 4]:
local RHS:=y/tan(x):
local P,T:

DEBUG();
P:=plots:-contourplot(RHS,':-colorbar' = false,':-contours' = L):
T:= timelimit(60,plottools:-getdata(P,'rangesonly')):
end proc:

 

foo();  #this will open a debugger window

 

Download hang_maple_2026_1_on_timelimit.mw

Here is screen shot

In my main actual code, the output was so large that it hanged Maple UI when stepping into the debugger and hitting that line.  I had to kill Maple from task manager as Java UI got stuck due to large output.

Why is it showing value of P when I have : at the end? Is there an option to turn automatic display of variables in debugger as one steps in?

The integral shown in the attached file "test" was posted on another forum for calculation. I unsuccessfully attempted to apply Green's theorem in Maple and—as befits a Maple beginner—failed. Does Maple offer a sequence of commands to carry this out? I would appreciate some advice. If this is possible, I would then tackle the line integral using the residue theorem.

restart

NULL

NULL

 

``

Download test.mw

I am implementing the Fokas (unified transform) method for the heat equation on a finite interval [0,1]. The solution is expressed as a contour integral in the complex k-plane and I evaluate it numerically in Maple.

When I call plot3d I get the following error, even though approx_u(0.5, 0.1)

Is there a way to make plot3d work? Any help appreciated.

restart; with(plots)

V := exp(-k^2*t)*((-4*k*(sin(k*x)/exp(1)+sin(k*(1-x)))*(k^2-1))*(1/((k^2+1)*(k^4+1))))/((2*Pi)*(exp(I*k)-exp(-I*k)))

-2*exp(-k^2*t)*k*(sin(k*x)/exp(1)+sin(k*(1-x)))*(k^2-1)/(Pi*(exp(I*k)-exp(-I*k))*(k^2+1)*(k^4+1))

(1)

L := 3/4; k1 := proc (r) options operator, arrow; L+(3/4)*I+r*exp(((1/6)*I)*Pi) end proc; k2 := proc (r) options operator, arrow; -L+(3/4)*I+r*exp(((5/6)*I)*Pi) end proc; k3 := proc (s) options operator, arrow; s+(3/4)*I end proc; dk1 := D(k1); dk2 := D(k2); dk3 := D(k3)

3/4

 

proc (r) options operator, arrow; L+(3/4)*I+r*exp(((1/6)*I)*Pi) end proc

 

proc (r) options operator, arrow; -L+(3/4)*I+r*exp(((5/6)*I)*Pi) end proc

 

proc (s) options operator, arrow; s+(3/4)*I end proc

 

proc (r) options operator, arrow; exp(((1/6)*I)*Pi) end proc

 

proc (r) options operator, arrow; exp(((5/6)*I)*Pi) end proc

 

1

(2)

integrand1 := Re((eval(V, k = k1(r)))*dk1(r)-(eval(V, k = k2(r)))*dk2(r)); integrand3 := Re((eval(V, k = k3(s)))*dk3(s)); integrand2 := simplify(evalc(integrand1)); integrand4 := simplify(evalc(integrand3))

approx_u := proc (x, t) local temp1, temp2; temp1 := Int(eval(integrand2, [:-x = x, :-t = t]), r = 0 .. infinity, method = _d01amc); temp2 := Int(eval(integrand4, [:-x = x, :-t = t]), s = -L .. L, method = _d01ajc); evalf(temp1+temp2) end proc

approx_u(.5, .1)

.6536368264

(3)

``

forget(`evalf/int`); forget(evalf); CodeTools:-Usage(plot3d(approx_u, 0 .. 1, 0 .. 2*Pi, grid = [10, 10], axes = boxed, labels = ["x", "t", "u(x,t)"], title = "Fokas Method of solution", shading = zhue))

Warning, unable to evaluate the function to numeric values in the region; see the plotting command's help page to ensure the calling sequence is correct

 

memory used=14.27GiB, alloc change=-16.00MiB, cpu time=92.96s, real time=88.85s, gc time=8.99s

 

 

 

Download heat_equation_on_finte_interval.mw

Any one knows a trick to help Maple obtain this much simpler solution to this ode obtained using AI?

ode := 4*(-1 + sqrt(1 - 1/x^2)*x^2)*sec(4 + 4*x + 4*arccsc(x))^2 - sqrt(1 - 1/x^2)*x^2*diff(f(x), x) = 0

ode:= 4*(-1 + sqrt(1 - 1/x^2)*x^2)*sec(4 + 4*x + 4*arccsc(x))^2 - sqrt(1 - 1/x^2)*x^2*diff(f(x), x) = 0;

4*(-1+(1-1/x^2)^(1/2)*x^2)*sec(4+4*x+4*arccsc(x))^2-(1-1/x^2)^(1/2)*x^2*(diff(f(x), x)) = 0

maple_sol:=dsolve(ode);

f(x) = c__1+Int(8*(-1+(1-1/x^2)^(1/2)*x^2)*x^6/((1-1/x^2)^(1/2)*(-8*((x^2-1)/x^2)^(1/2)*x^7*sin(8+8*x)+x^8*cos(8+8*x)+x^8+80*((x^2-1)/x^2)^(1/2)*x^5*sin(8+8*x)-32*x^6*cos(8+8*x)-192*((x^2-1)/x^2)^(1/2)*x^3*sin(8+8*x)+160*x^4*cos(8+8*x)+128*((x^2-1)/x^2)^(1/2)*x*sin(8+8*x)-256*cos(8+8*x)*x^2+128*cos(8+8*x))), x)

odetest(maple_sol,ode);

0

AI_sol:=f(x)=_C1+tan(4*(1+x+arccsc(x)));

f(x) = c__1+tan(4+4*x+4*arccsc(x))

odetest(AI_sol,ode)

0

 

 

Download AI_sol.mw

given

ode:=2*x^(1/2)*diff(y(x),x)-y(x) = -sin(x^(1/2))-cos(x^(1/2)); 
ic:=y(infinity) = y__0; 
sol:=dsolve([ode,ic]);

It gives  

This solution satisfies the ode itself. Now cos(sqrt(x)) when x=infinity is  -1..+1

But IC says y(infinity)=y0  so odetest do not verify the IC and gives this

odetest(sol,[ode,ic]);

I think dsolve should not have returned a solution at all. 

What do the experts here think of this result?

Maple 2026.1 on windows 10

I asked an LLM to provide an expansion of the MacDonald function of arbitrary order (a modified Bessel function of the second kind with purely imaginary order and positive argument), K(I*y,r), as a weighted sum of MacDonald functions of integer order. It came back with

         K(I*y,z)=2*sinh(Pi*y)/Pi* [K(0,r)/2*y+sum( (-1)^n*y*BesselK(0,r)/(y^2+n^2),n=1..infinity)]

(see below for more readable text)

I evaluated the LHS and RHS using Maple 2026 for various choices of y and r and found numerical agreement using both "sum" and "Sum".  I was very pleased until I realized that the RHS isn't a convergent series!

Can anyone explain to me how Maple pulls this off! 

(I asked Maplesoft Tech Support but they said it is above their pay grade... I suspect that Maple is using Borel summability to evaluate the RHS but I haven't been able to verify that)

I apologize, but I can't see how to attach a .mw file, so I've cut and pasted the code below

WHen I open many worksheets at same time, say 10. The new UI do not stack them all (i.e. the tab at the top), forcing one to use the small arrow to navigate to each worksheet.

Is there a way to tell the UI to show all tabs (may be double rows and 3 rows as needed) to make it easier to jump from one worksheet to the other?

I do not know if this is new feature in the new ribbon UI or not. 

Here is screen show where I have 10 worksheets open

There is also a pull down menu, but it only shows 8 worksheets and one can have more open but they do not show. So have to scroll down looking for the rest. Even that does not work well. many times when I try to scroll down, the window closes. It will not give me time to move the mouse to the scroll bar to move it before it closes.

Both of these solutions are not good. Having to use the arrow key to look and navigate for a different worksheet is bad UI design.

How to see all tabs for all open worksheet in same UI?  If the tabs do not fit on one row, why not make second row? If two rows do not fit, make 3rd row. This should be an option for the user. But I did not see one so far. But will keep looking.

I find tabs where all worksheet show much better design that this UI design.   

I only use worksheet and not document mode. Windows 10.

To give you idea what I mean, These are examples found on the net of stacked tabs

 

 

Where in Maple, each tab above will have the name of the worksheet open. Font can be small, is OK.

Is it possible to have this in the new UI for open worksheets?

One option I might try to make my worksheets names much shorter. May be then they will fit all in same window.

Dear all,

I'm reporting what seems to me as a bug in the SMTLIB library in maple. 

    |\^/|     Maple 2026 (X86 64 LINUX)
._|\|   |/|_. Copyright (c) Maplesoft, a division of Waterloo Maple Inc. 2026
 \  MAPLE  /  All rights reserved. Maple is a trademark of
 <____ ____>  Waterloo Maple Inc.
      |       Type ? for help.
> SMTLIB:-Satisfiable({x^2=2,y^2=2,x<y});
                                     true

> SMTLIB:-Satisfiable({x^2=2,y^2=2,y<x});
                                     false

> SMTLIB:-Satisfiable({x^2=2,a^2=2,a<x});
                                     true

The Satisfiable command do not output the correct decision on two formulas of equivalent realization by switching x<y (output SAT) to y<x (output UNSAT). I suspect this is because some alphabetical order depandance in the variables as for a<y we get SAT again.

I tried to feed Z3 with the code given by ToString on the problematic formula and I get two different outputs :

  • on the Z3 version 4.8.12 from the ubuntu repository (apt install) I also get the wrong UNSAT output;
  • one the Z3 version 4.17.0 build from the official github repository I finally get the correct SAT output.

Thus, I suspect a version problem in SMTLIB that do not take in account the last updates made in SMT solvers (Z3?).

Many thanks for considering my problem!

Recently @salim-barzani asked a question about a paper that involved analysing the different types of roots of polynomials. The appendix in that paper gave the example of the roots of x^4+x^2*e[2]+x*e[1]+e[0]using the analysis in Lu et al, "A complete discrimination system for polynomials", Science in China (Ser. E), 39 (1996) 628-646. The analysis uses the discriminant sequence and extensions. Maple provides this through RegularChains:-ParametricSystemTools:-DiscrminantSequence. For example for this polynomial we find there is a real root of multiplicity 2 and a complex conjugate pair when D__2*D__3 < 0 and D__4 = 0 where the D__i are the ith entries in the discriminant sequence [1, -e[2], -2*e[2]^3+8*e[0]*e[2]-9*e[1]^2, 16*e[0]*e[2]^4-4*e[1]^2*e[2]^3-128*e[0]^2*e[2]^2+144*e[0]*e[1]^2*e[2]-27*e[1]^4+256*e[0]^3].

 

The problem with these conditions is that they are in terms of the D__i and not directly in terms of the e__i parameters. One can derive these conditions and then solve them to find the conditions on the parameters, but Maple has various routines in the RegularChains, RootFinding:-Parametric and SolveTools packages that directly find conditions on parameters to find when there are specified numbers of real or complex roots for polynomial systems. So this post is my attempt to use these tools to find the conditions on the parameters of the above polynomial that give various types of roots. One immediate difficulty is that generally these routines count distinct roots irrespective of multiplicity, and so some indirect analysis is required. There are several different types of commands and analyses that could be used, and my choices here are more to do with my learning experience than an optimum analysis.

 

The first conclusion is that it is possible, although RealComprehensiveTriangularize did not work as I expected when asking for zero real roots (see cases (a) and (b)) (bug?). Assuming it had worked, RealComprehensiveTriangularize could cover all the cases here, though that will not be true for higher-degree polynomials with more parameters. There doesn't seem to be an obvious systematic way of doing this analysis, which is a downside. Another downside is the large number of subcase conditions, which look as if they could be combined into fewer subcases. CellDecomposition works well for cases without multiplicity.


Main worksheet [not all is displayed below]:

Download RootAnalysis4.mw

restart

with(RegularChains); with(ParametricSystemTools); with(RootFinding:-Parametric)

Consider the following polynomial in x with the three real parameters e[0], e[1], e[2]. We would like to know the conditions on these parameters that lead to different numbers of real and complex-conjugate pairs of roots of different multiplicities.

p := x^4+x^2*e[2]+x*e[1]+e[0]

x^4+x^2*e[2]+x*e[1]+e[0]

Consider first how many cases there are. We can set this up as a combinatorial problem in the combstruct package.

sys := {C = Atom, R = Atom, realrts = Set(multiplereal), rts = Prod(realrts, complexrts), complexpr = Prod(C, C), complexrts = Set(multiplecomplex), multiplecomplex = Sequence(complexpr, card > 0), multiplereal = Sequence(R, card > 0)}

Draw := proc (q) options operator, arrow; eval(q, {Epsilon = NULL, Prod = `[]`, Set = (proc () options operator, arrow; args end proc), Sequence = `*`}) end proc

For a degree 4 polynomial there are 9 different cases to consider. Here [C, C] means a (non-real) complex-conjugate pair of roots and R means a real root; the exponents indicate the multiplicities.

all := combstruct:-allstructs([rts, sys], size = degree(p, x)); nops(%); `~`[Draw](all)

9

[[[C, C]^2], [[C, C], [C, C]], [R^2, [C, C]], [R^4], [R, R, [C, C]], [R^2, R^2], [R^3, R], [R, R, R^2], [R, R, R, R]]

These are (in order)
(a) A duplicate pair of complex-conjugate roots

(b) Two distinct pairs of complex-conjugate roots

(c) A real root of multiplicity 2 and a pair of complex-conjugate roots

(d) A real root of multiplicity 4

(e) Two distinct real roots of multiplicity 1 and a complex-conjugate pair

(f) Two distinct real roots each of multiplicity 2

(g) A real root of multiplicity 3 and a real root of multiplicity 1

(h) Three distinct real roots, of multiplicities 2, 1, and 1

(i) Four distinct real roots of multiplicity 1

Declare the variables first and parameters last. np is the number of parameters. Use the suggested order.

vp := SuggestVariableOrder([p = 0], [x]); R := PolynomialRing(vp); np := nops(`minus`({vp[]}, {x}))

[x, e[0], e[1], e[2]]

polynomial_ring

3

Define derivative polynomials. p2 = 0 when there is a root of multiplicity 2 or more; p3 = 0 when there is a root of multiplicity 3 or more and p4 = 0when there is a root of multiplicity 4.

p2 := diff(p, x); p3 := diff(p2, x); p4 := diff(p3, x)

4*x^3+2*x*e[2]+e[1]

12*x^2+2*e[2]

24*x

Discriminant is zero if and only if there are repeated roots.

Delta := discrim(p, x)

16*e[0]*e[2]^4-4*e[1]^2*e[2]^3-128*e[0]^2*e[2]^2+144*e[0]*e[1]^2*e[2]-27*e[1]^4+256*e[0]^3

(d) A real root of multiplicity 4

 

This is perhaps the simplest case and can be done using RealComprehensiveTriangularize. Specifying np = 3 means use the last 3 variables in the PolynomialRing as parameters. The argument 1 means we want the cases where there is one distinct real root. We specify that all of p, p2, p3, p4 are zero so that the 1 real root is the common root of these polynomials, i.e., is a root on multiplicity 4. (I find the cadcell output a little easier to use, but it is not critical.)

rct := RealComprehensiveTriangularize({p = 0, p2 = 0, p3 = 0, p4 = 0}, np, R, 1, output = cadcell); Display(rct, R)

PiecewiseTools:-Is, "Wrong kind of parameters in piecewise"

This means that the conditions on the parameters to get a single real root of multiplicity 4 are

conds := Info(rct[2][1][1], R)

[e[2] = 0, e[1] = 0, e[0] = 0]

and that the polynomial to solve to find this root under these conditions is

poly := Info(rct[1][][2], R)

[x = 0]

which we can check:

eval(p, conds); solve(%, x)

x^4

0, 0, 0, 0

(g) A real root of multiplicity 3 and a real root of multiplicity 1

   

(f) Two distinct real roots each of multiplicity 2

   

(i) Four distinct real roots of multiplicity 1

   

(h) Three distinct real roots, of multiplicities 2, 1, and 1

   

(e) Two distinct real roots of multiplicity 1 and a complex-conjugate pair

   

(c) A real root of multiplicity 2 and a pair of complex-conjugate roots

   

(b) Two distinct pairs of complex-conjugate roots

   

(a) A duplicate pair of complex-conjugate roots

 

Here we want multiplicity 2 but no real roots. The discriminant is expected to be zero, but in most cases is not.

rct := RealComprehensiveTriangularize({p = 0, p2 = 0, p3 <> 0, p4 <> 0}, np, R, 0, output = cadcell); nops(rct[2]); Display(rct[2][1 .. 4], R)

40

[[PIECEWISE([e[0] < RootOf(256*_Z^3-128*e[2]^2*_Z^2+(16*e[2]^4+144*e[1]^2*e[2])*_Z-4*e[1]^2*e[2]^3-27*e[1]^4, index = real[1]), ``], [e[1] < -(2/9)*(-6*e[2]^3)^(1/2), ``], [e[2] < 0, ``]), []], [PIECEWISE([RootOf(256*_Z^3-128*e[2]^2*_Z^2+(16*e[2]^4+144*e[1]^2*e[2])*_Z-4*e[1]^2*e[2]^3-27*e[1]^4, index = real[1]) < e[0], ``], [e[1] < -(2/9)*(-6*e[2]^3)^(1/2), ``], [e[2] < 0, ``]), []], [PIECEWISE([e[0] < -(1/12)*e[2]^2, ``], [e[1] = -(2/9)*(-6*e[2]^3)^(1/2), ``], [e[2] < 0, ``]), []], [PIECEWISE([e[0] = -(1/12)*e[2]^2, ``], [e[1] = -(2/9)*(-6*e[2]^3)^(1/2), ``], [e[2] < 0, ``]), []]]

Consider one of the cases with an equality for e[0], suggesting a zero discriminant.
Find a sample point satisfying the conditions, and see what the roots are like

j := 37; cell := rct[2][j][1]; conds := Info(cell, R); pts := Info(SamplePoints(cell, R), R)[1]; `~`[is](eval(conds, pts))

37

cad_cell

[0 < e[2], e[1] = 0, e[0] = (1/4)*e[2]^2]

[e[0] = 1/16, e[1] = 0, e[2] = 1/2]

[true, true, true]

We find two complex roots of multiplicity 2, which are complex conjugates, as expected

eval(p, pts); solve(%, x)

x^4+(1/2)*x^2+1/16

(1/2)*I, -(1/2)*I, (1/2)*I, -(1/2)*I

Check that discriminant is zero.

eval(Delta, pts)

0

But, for example, the first cell does not have the discriminant zero, and does not give a correct result.

j := 1; cell := rct[2][j][1]; conds := Info(cell, R); pts := Info(SamplePoints(cell, R), R)[1]; `~`[is](eval(conds, pts))

1

cad_cell

[e[2] < 0, e[1] < -(2/9)*(-6*e[2]^3)^(1/2), e[0] < RootOf(256*_Z^3-128*e[2]^2*_Z^2+(16*e[2]^4+144*e[1]^2*e[2])*_Z-4*e[1]^2*e[2]^3-27*e[1]^4, index = real[1])]

[e[0] = 1/2, e[1] = -3/2, e[2] = -1/2]

[true, true, true]

We find a complex conjugate pair and two distinct real roots, which is not expected for this case.

eval(p, pts); fsolve(%, x, complex)

x^4-(1/2)*x^2-(3/2)*x+1/2

-.7473459056-.9001675303*I, -.7473459056+.9001675303*I, .3077440660, 1.186947745

The discriminant is indeed nonzero.

eval(Delta, pts)

-3073/16

We did not yet find the conditions for this case. We can ask for the conditions for different numbers of complex roots (complex in this context includes real).

cmplx := ComplexRootClassification([p], np, R)

[[constructible_set, 1], [constructible_set, 2], [constructible_set, 3], [constructible_set, 4]]

We are interested in the conditions for exactly two distinct roots, which is found from the second constructible_set. There are two subcases.

Display(cmplx[2][1], R)

PIECEWISE([12*e[0]+e[2]^2 = 0, ``], [27*e[1]^2+8*e[2]^3 = 0, ``], [e[2] <> 0, ``]), PIECEWISE([4*e[0]-e[2]^2 = 0, ``], [e[1] = 0, ``], [e[2] <> 0, ``])

Consider the second case

case2 := Info(cmplx[2][1], R)[2]

[[4*e[0]-e[2]^2, e[1]], [e[2]]]

solve(case2[1], {e[0], e[1]}); q1 := eval(p, %); rts2 := [solve(%, x)]

{e[0] = (1/4)*e[2]^2, e[1] = 0}

x^4+x^2*e[2]+(1/4)*e[2]^2

[(1/2)*(-2*e[2])^(1/2), -(1/2)*(-2*e[2])^(1/2), (1/2)*(-2*e[2])^(1/2), -(1/2)*(-2*e[2])^(1/2)]

We saw this before when e[2]<0 as case (f) with real roots of multiplicity 2. Now, for e[2]>0 we indeed have two duplicate pairs of complex conjugate roots

`assuming`([simplify(rts2)], [e[2] > 0])

[((1/2)*I)*2^(1/2)*e[2]^(1/2), -((1/2)*I)*2^(1/2)*e[2]^(1/2), ((1/2)*I)*2^(1/2)*e[2]^(1/2), -((1/2)*I)*2^(1/2)*e[2]^(1/2)]

Consider the first case

case1 := Info(cmplx[2][1], R)[1]

[[12*e[0]+e[2]^2, 27*e[1]^2+8*e[2]^3], [e[2]]]

ans1 := {solve(case1[1], {e[0], e[1]}, explicit)}; q11 := eval(p, ans1[1]); rts11 := [solve(%, x, explicit)]

{{e[0] = -(1/12)*e[2]^2, e[1] = -(2/9)*(-6*e[2])^(1/2)*e[2]}, {e[0] = -(1/12)*e[2]^2, e[1] = (2/9)*(-6*e[2])^(1/2)*e[2]}}

x^4+x^2*e[2]-(2/9)*x*(-6*e[2])^(1/2)*e[2]-(1/12)*e[2]^2

[(1/6)*(-6*e[2])^(1/2), (1/6)*(-6*e[2])^(1/2), (1/6)*(-6*e[2])^(1/2), -(1/2)*(-6*e[2])^(1/2)]

The rts11 subcase polynomial q11 must have real coefficients and therefore only applies for e[2]<0. The roots are real with multiplicity 3 and multiplicity 1 and this case is just case (g) above. The rts12 subcase polynomial q12 (below) also requires e[2]<0 and corresponds to case (g), but with signs reversed.

q12 := eval(p, ans1[2]); rts12 := [solve(%, x, explicit)]

x^4+x^2*e[2]+(2/9)*x*(-6*e[2])^(1/2)*e[2]-(1/12)*e[2]^2

[(1/2)*(-6*e[2])^(1/2), -(1/6)*(-6*e[2])^(1/2), -(1/6)*(-6*e[2])^(1/2), -(1/6)*(-6*e[2])^(1/2)]

Therefore the rts2 case is the solution for case (a); the cell37 example was a special case of this. In fact if we have a duplicate pair of complex-conjugate roots, the polynomial must be a pefect square, as we see it is

factor(q1)

(1/4)*(2*x^2+e[2])^2

NULL

Download RootAnalysis5.mw

 

Any explanation why this happens? notice, I did not supply the x and y ranges, let Maple decide.

restart;

interface(version);

`Standard Worksheet Interface, Maple 2026.1, Windows 10, April 28 2026 Build ID 2011354`

plots:-contourplot(y+sin(x),'colorbar'=false,':-contours' = 2,size=[100,100]);

plots:-contourplot(y+sin(x),'colorbar'=false,':-contours' = 1,size=[100,100]);

Error, (in plot/iplot2d) numeric exception: division by zero

 

 

Download bug_in_contourplot.mw

I asked Maple AI what a glyph is. Then I prompted this

A kernel lost message was returned and the AI pannel became irresponsive.

Maple is still running well in exsisting and new tabs. 

Can the AI service be restarted from the user interface?

(Is that crash reproducible?)

 

Edit:

For quite some time, I have wanted to solve the system attached in "test" using Maple. The smallest solution in natural numbers x, y, and z test.mw

restart

kernelopts(version)

`Maple 2026.0, X86 64 WINDOWS, Apr 28 2026, Build ID 2011354`

(1)

interface(version)

`Standard Worksheet Interface, Maple 2026.1, Windows 11, April 28 2026 Build ID 2011354`

(2)

with(NumberTheory)

isolve({x*y*z = w^2, x+y+z = u^2, x*y+x*z+y*z = v^2})

{u = _Z1, v = 0, w = 0, x = _Z1^2, y = 0, z = 0}

(3)

"(->)"

{u = _Z1, v = 0, w = 0, x = _Z1^2, y = 0, z = 0}

(4)

``

Download test.mw

is known, and all these numbers are less than 4 × 10¹². Is this possible in Maple?

(x=1633780814400; y=252782198228; z=3474741058973)

Hi MaplePrimes, and all,

Here is a new, to me, set of numbers
defined by ,
the first three numbers are {1,2,3}
and then, 
the next number is the sum of the three 
previous numbers,
so,
{1,2,3,6,11,20, ... }
but can only calculate a finite number of numbers
the, so called, Tribonacci numbers
could start with {0,1,0}
see online
https://oeis.org/A001590

and
triple_recursive_sequence_simple_first.mw

triple_recursive_sequence_simple_first.pdf

regards,
Matt
 

I am trying to select from a default list and a modified list to get a third combined list of the unmodified default elements and the mdfified elements plus extra elements. This is for plot data. I have managed to extract the colour data using something I did a couple of years ago. Though at this stage I don't relly know how that works either but it works.

restart

with(ListTools)

[BinaryPlace, BinarySearch, Categorize, Classify, Collect, Deal, DotProduct, Enumerate, FindMaximalElement, FindMinimalElement, FindRepetitions, Flatten, FlattenOnce, Group, Interleave, InversePermutation, Join, JoinSequence, LengthSplit, MakeUnique, Occurrences, Pad, PartialSums, Reverse, Rotate, Search, SearchAll, SelectFirst, SelectLast, Slice, Sorted, Split, Transpose, Unpermute]

(1)

PlotDefaults:=['colour' = ':-blue', symbol = ':-solidcircle', ':-symbolsize' = 8,thickness=2]; #for points and lines

[colour = blue, symbol = solidcircle, symbolsize = 8, thickness = 2]

(2)

Inputs:=['color'=[red, black, blue],symbol=square,':-linestyle'=dash]; #'color'=[red, black, blue],

 

[color = [red, black, blue], symbol = square, linestyle = dash]

(3)

if has(Inputs,{colour,color}) then
Colourlist:=remove(has,remove(has,Flatten(eval([':-color',':-colour'],Inputs)),':-colour'),':-color')
else
Colourlist:=remove(has,remove(has,Flatten(eval([':-color',':-colour'],PlotDefaults)),':-colour'),':-color');
end if

[red, black, blue]

(4)


Make this list

plotdata:=[symbol=square,':-linestyle'=dash,':-symbolsize' = 8,thickness=2]

[symbol = square, linestyle = dash, symbolsize = 8, thickness = 2]

(5)

mx:=max(nops(PlotDefaults),nops(Inputs))

4

(6)

Plotdata:=[];
for i to mx do

end do

 

 

[]

(7)
 

 

Download 2026-05-18_Q_Select_from_Two_Lists_to_get_New_List.mw

Currently I have Maple versions 2023,2025, and 2026 installed on Windows 11. Today I installed a workbook package containing a module that I just completed using the PackageTools installer in Maple 2026.. To my surprise, I found that a package installed from Maple 2026 was also available in Maple 2023 and, conversely, a package installed in Maple 2023 was automatically available in Maple 2026. i noticed that, with the exception of the Maple Customer Support Updates, the toolbox directory is no longer broken down by versions. I also noticed that the directory containing the module installed by Maple 2026 was named by the workbook instead of the module name (ie. hopfwords.maple). As I recall, the toolboxes used to be version dependent. 

The question is to what extent can one assume that a package created in Maple 2026 will be compatible with at least the more recent versions of Maple, I am also wondering why the directory name is now the workbook name instead of the module name. 

1 2 3 4 Page 1 of 4