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Maple 2025.1 on Windows 11: How to resize the wind...

Asked by: C_R 3387
gui resize windows11
Yesterday at 4:41 PM
3  13

I am new to Windows 11 (edit: I was forced to migrate from 10).

I am looking for an option to resize the Window that the interupt button becomes accessible (see red box)

Also: there are no window controls (maximize minimize) visble . (edit: the title bar is clipped off).

Window key + arrow key does not work.

I have also tried the Snap Layout option to resize but cannot find Maple 2025.1 under the selectable tasks.

What else can I try?

This is a test Post. Please ignore.

by: habmanj 0 Maple
openmaple
Yesterday at 9:24 AM
0  0

Testing tesing 123. Testing tesing 123.Testing tesing 123.Testing tesing 123.Testing tesing 123.Testing tesing 123.Testing tesing 123.Testing tesing 123.Testing tesing 123.Testing tesing 123.Testing tesing 123.Testing tesing 123.Testing tesing 123.Testing tesing 123.Testing tesing 123.Testing tesing 123.

Is the ConsistencyTest not work with DEtools?...

Asked by: Ahmed111 135
differential-equations pdetools consistencytest
Yesterday at 5:52 AM
0  2

I checked the ConsistencyTest of the system of equations but no output with 'true' or 'False'. Is it not work in 'DEtools'? Download consistency.mw

Finding the Median for Piecewise Density Function...

Asked by: JoyDivisionMan 35
statistics density median
June 19 2025
0  10

I have a piecewise density function (f).  I am trying to find the median value.

I have tried the Median function and the Percentile function, but neither work for me.  I am not sure why.

I have also tried to integrate the density function on the (0,x) interval such that the area under the curve is 1/2 and then solve for x.  This works for simple problems, but not the one attached.

There is something very simple that I'm not doing, but I am not sure what.

> 

restart
> 

kernelopts(version)

`Maple 2023.2, X86 64 WINDOWS, Nov 24 2023, Build ID 1762575`

 (1)
> 

with(Statistics)

 

 (2)
> 

f := 2*t*piecewise(t <= 0, 0, t <= 1, t^2+Pi-4*t, t <= sqrt(2), -(sqrt(t^2-1)*t^2+2*sqrt(t^2-1)*arcsin((t^2-2)/t^2)-4*t^2+2*sqrt(t^2-1)+4)/sqrt(t^2-1), sqrt(2) < t, 0)

2*t*piecewise(t <= 0, 0, t <= 1, t^2+Pi-4*t, t <= 2^(1/2), -((t^2-1)^(1/2)*t^2+2*(t^2-1)^(1/2)*arcsin((t^2-2)/t^2)-4*t^2+2*(t^2-1)^(1/2)+4)/(t^2-1)^(1/2), 2^(1/2) < t, 0)

 (3)
> 

plot(f, t = 0 .. sqrt(2))

 
> 

Median(f)

2*t*piecewise(t <= 0, 0, t <= 1, t^2+Pi-4*t, t <= 2^(1/2), -((t^2-1)^(1/2)*t^2+2*(t^2-1)^(1/2)*arcsin((t^2-2)/t^2)-4*t^2+2*(t^2-1)^(1/2)+4)/(t^2-1)^(1/2), 2^(1/2) < t, 0)

 (4)
> 

Percentile(f, 50)

2*t*piecewise(t <= 0, 0, t <= 1, t^2+Pi-4*t, t <= 2^(1/2), -((t^2-1)^(1/2)*t^2+2*(t^2-1)^(1/2)*arcsin((t^2-2)/t^2)-4*t^2+2*(t^2-1)^(1/2)+4)/(t^2-1)^(1/2), 2^(1/2) < t, 0)

 (5)
> 

myMedian := solve(int(f, t = 0 .. x) = 1/2, x)

Warning, solutions may have been lost

  

(6)
> 

NULL

Download Median.mw

Beginner question......

Asked by: Alfred_F 305
assumptions assuming assume
June 19 2025
0  7

When calculating limits of real-valued functions, sometimes (especially in competitions) tricky approaches are taken using pen and paper. I repeatedly encountered the simple conclusion that, for example, for natural k, the value sin(k*pi) = 0. Thus, the function value is determined logically without specifying a specific number. There are numerous other examples of this that can easily be constructed.
My question after unsuccessful attempts using "assume" is:
How, for example, does Maple determine the value of sin(k*pi) from the assumption "k is natural" alone? Are such prominent values ​​implemented in tables?

Strange interaction between add and random numbers...

Asked by: mmcdara 7831
add statistics
June 19 2025
2  3

 

As I was comparing visually the first terms of a priori identical sums produced by add , I was surprised to find them different.
So I suspected some error in what I have done, until I realized that add randomly permuted the terms.
Each term is of the form (R + P)2 where R is a random number and P a polynomial.

This behaviour is illustrated in worksheet add_changes_ranks.mw and appears only when random numbers are used (provided the seed is not forced to some constant value)

Does someone ever onserved that or have any idea of what happens here (maybe this behaviour no longer happens in recent versions?) ?

Thanks in advance

Trouble Finding Standard Deviation for Density Fun...

Asked by: JoyDivisionMan 35
statistics integration evalf
June 18 2025
0  2

I am trying to find the standard deviation for a piecewise density function (f).  The interval is (0, sqrt(3))  I am convinced the density function is what I want.  I calculate the expected value of the density function (evE) and the answer is correct -- about 0.66145

When I try to calculate the standard deviation (stdE), I get an answer that is "off" by a large degree.  Via simulating values, I should get a value of about 0.24936.  Each time I "re-run" the calculation, I get varying results, all of which are "off" by a large degree.

I am only guessing, but my integration function might be missing some sort of assumption and/or option.

My work is attached.  Does anyone know what I am doing wrong?

> 

restart
> 

with(Statistics)
> 

f := -t*piecewise(t <= 0, 0, t <= 1, t*(t^3+6*Pi*t-8*t^2-4*Pi), t <= sqrt(2), -(2*(4*t^2*sqrt(t^2-1)*arcsin((t^2-2)/t^2)+4*t^2*arctan(sqrt(t^2-1))*sqrt(t^2-1)+(t^4+(2*Pi+3)*t^2-4*t*Pi+3*Pi-1/2)*sqrt(t^2-1)-8*t^4+4*t^2+4))/sqrt(t^2-1), t <= sqrt(3), (8*arctan((t^3+t^2-3*t-1)/(sqrt(t^2-2)*(t^3-t^2-t-1)))*t*sqrt(t^2-2)-4*arccot((t^2-t-1)/sqrt(t^2-2))*t^2*sqrt(t^2-2)+4*t*sqrt(t^2-2)*(t+2)*arctan((t^2+t-1)/sqrt(t^2-2))+(8*t^2-4)*sqrt(t^2-2)*arcsin((t^2-3)/(t^2-1))+16*arcsin(sqrt(t^2-2)/sqrt(t^2-1))*sqrt(t^2-2)-16*arcsin(1/sqrt(t^2-1))*sqrt(t^2-2)+((t^2+5)*sqrt(t^2-2)-8*t^2+16)*(t^2+1))/sqrt(t^2-2), sqrt(3) < t, 0)

-t*piecewise(t <= 0, 0, t <= 1, t*(t^3+6*Pi*t-8*t^2-4*Pi), t <= 2^(1/2), -2*(4*t^2*(t^2-1)^(1/2)*arcsin((t^2-2)/t^2)+4*t^2*arctan((t^2-1)^(1/2))*(t^2-1)^(1/2)+(t^4+(2*Pi+3)*t^2-4*t*Pi+3*Pi-1/2)*(t^2-1)^(1/2)-8*t^4+4*t^2+4)/(t^2-1)^(1/2), t <= 3^(1/2), (8*arctan((t^3+t^2-3*t-1)/((t^2-2)^(1/2)*(t^3-t^2-t-1)))*t*(t^2-2)^(1/2)-4*arccot((t^2-t-1)/(t^2-2)^(1/2))*t^2*(t^2-2)^(1/2)+4*t*(t^2-2)^(1/2)*(t+2)*arctan((t^2+t-1)/(t^2-2)^(1/2))+(8*t^2-4)*(t^2-2)^(1/2)*arcsin((t^2-3)/(t^2-1))+16*arcsin((t^2-2)^(1/2)/(t^2-1)^(1/2))*(t^2-2)^(1/2)-16*arcsin(1/(t^2-1)^(1/2))*(t^2-2)^(1/2)+((t^2+5)*(t^2-2)^(1/2)-8*t^2+16)*(t^2+1))/(t^2-2)^(1/2), 3^(1/2) < t, 0)

 (1)
> 

plot(f, t = 0 .. sqrt(3))

 
> 

evE := simplify(int(t*f, t = 0 .. sqrt(3)))

(1/105)*(294*(2^(1/2)-3/2)*(-2+3^(1/2))*ln(1+2^(1/2))+168*(2^(1/2)-3/2)*(-2+3^(1/2))*ln(1+3^(1/2))-252*(2^(1/2)-3/2)*(-2+3^(1/2))*arccoth(2^(1/2))+((-14*Pi-84*ln(2)-19)*3^(1/2)+28*Pi+168*ln(2)+50)*2^(1/2)+(21*Pi+126*ln(2)+20)*3^(1/2)-42*Pi-252*ln(2)-58)/((2*2^(1/2)-3)*(-2+3^(1/2)))

 (2)
> 

evalf(evE)

.6617071838

 (3)
> 

stdE := simplify(sqrt(int(f*(t-evE)^2, t = 0 .. sqrt(3))))

(1/210)*(-4808607312140100*(2^(1/2)-3880899/2744210)*(3^(1/2)-70226/40545)*ln(1+2^(1/2))^2+(-5495551213874400*(2^(1/2)-3880899/2744210)*(3^(1/2)-70226/40545)*ln(1+3^(1/2))+8243326820811600*(2^(1/2)-3880899/2744210)*(3^(1/2)-70226/40545)*arccoth(2^(1/2))+((457962601156200*Pi+2747775606937200*ln(2)+631290043332900)*3^(1/2)-793214493249360*Pi-4759286959496160*ln(2)-1093426429714584)*2^(1/2)+(-647656921614780*Pi-3885941529688680*ln(2)-892778941072368)*3^(1/2)+1121774694224184*Pi+6730648165345104*ln(2)+1546338486359352)*ln(1+2^(1/2))-1570157489678400*(2^(1/2)-3880899/2744210)*(3^(1/2)-70226/40545)*ln(1+3^(1/2))^2+(4710472469035200*(2^(1/2)-3880899/2744210)*(3^(1/2)-70226/40545)*arccoth(2^(1/2))+((261692914946400*Pi+1570157489678400*ln(2)+360737167618800)*3^(1/2)-453265424713920*Pi-2719592548283520*ln(2)-624815102694048)*2^(1/2)+(-370089669494160*Pi-2220538016964960*ln(2)-510159394898496)*3^(1/2)+641014110985248*Pi+3846084665911488*ln(2)+883621992205344)*ln(1+3^(1/2))-3532854351776400*(2^(1/2)-3880899/2744210)*(3^(1/2)-70226/40545)*arccoth(2^(1/2))^2+(((-392539372419600*Pi-2355236234517600*ln(2)-541105751428200)*3^(1/2)+679898137070880*Pi+4079388822425280*ln(2)+937222654041072)*2^(1/2)+(555134504241240*Pi+3330807025447440*ln(2)+765239092347744)*3^(1/2)-961521166477872*Pi-5769126998867232*ln(2)-1325432988308016)*arccoth(2^(1/2))+((-392539372419600*ln(2)^2+(-130846457473200*Pi-180368583809400)*ln(2)-10903871456100*Pi^2-30061430634900*Pi+1205966071986042)*3^(1/2)+679898137070880*ln(2)^2+(226632712356960*Pi+312407551347024)*ln(2)+18886059363080*Pi^2+52067925224504*Pi-2088794509063308)*2^(1/2)+(555134504241240*ln(2)^2+(185044834747080*Pi+255079697449248)*ln(2)+15420402895590*Pi^2+42513282908208*Pi-1705493574764511)*3^(1/2)-961521166477872*ln(2)^2+(-320507055492624*Pi-441810996102672)*ln(2)-26708921291052*Pi^2-73635166017112*Pi+2954001523727854)^(1/2)*(5*2^(1/2)-7)*(-5+3*3^(1/2))/((70*2^(1/2)-99)*(15*3^(1/2)-26)*(-2+3^(1/2))*(4*3^(1/2)-7)*(12*2^(1/2)-17)*(2*2^(1/2)-3))

 (4)
> 

evalf(stdE)

7031348.295

 (5)
 

NULL

Download StdE.mw

How do i best work with Polar vectors in Maple?...

Asked by: Mads758 10
vector polar vectorcalculus
June 18 2025
2  3

Hello,

I’m currently a student working with vectors as part of my electrical AC calculations. Up until now, I’ve been using GeoGebra to add and plot vectors, but I’m trying to transition fully to Maple, since I already use it for everything else in my studies.

All of my assignments are either given in — or require answers in — polar form (magnitude and angle), so I would really like to work directly in polar coordinates without converting everything to Cartesian and back.

I’ve already tried to figure this out on my own, but so far, I’ve only been able to make it work using Cartesian notation. I’ve attached a PDF with a typical example of the kind of tasks I work on, in case that helps clarify my needs.

Could you guide me on how to best set up and work with polar vectors in Maple — including how to define, add, and plot them directly in polar notation?

Vector_Help.pdf

Thanks in advance for your help!

Best regards,
Mads Bach Nielsen

- Yes this was written with the help of Chat GPT

ODE error in plots ...

Asked by: KIRAN SAJJAN 40
ode dsolve numeric
June 17 2025
0  13

why this error is getting. but in the published paper for the same parameter it is converging. what is the mistake in this worksheet please rectify it in sachin_base_paper.mw

Laplace Adomian Decomposition method for solving...

by: dharr 8060 Maple
symbolic pde differential-equations
June 17 2025
5  1

Thanks to @salim-barzani, I became interested in the Laplace Adomian Decompositon method to solve partial differential equations. It produces a sum of analytical components that converge to the exact solution. Probably it is not that efficient for numerical solutions, but the possibility of finding an exact solution to nonlinear PDEs by finding a formula for the components and therefore the infinite sum is interesting.

The version here covers many of the common forms of PDEs, but is still a work in progress. The method for finding the Adomian polynomials could be improved by using one of the reccurence formulas in the literature. Extension to more PDE forms, ODEs etc are possible and I will attempt some of these before uploading an improved version to the application centre. In the meantime, feedback about bugs, usability etc. are appreciated.

Laplace Adomian Decomposition method to solve partial differential equations.
 D.A. Harrington, June 2025, v 1.16.
 Procedure LAD is in the startup code region.

Arguments:

1. 

Partial differential equation in one "time" variable and several other variables (called spatial variables here). Specified as an equation, or an expression implicitly equal to zero.
Comprising: (1) one time derivative term of order m (denoted L), (2) linear terms that are derivatives in the spatial variables (R), (3) nonlinear terms with derivatives in the spatial variables (N). Together, (1)-(3) is a multivariate polynomial in the dependent variable, its derivatives, and any parameters. (4) inhomogenous terms in the time and spatial variables that do not contain the dependent variable or its derivatives (G). The general form is L = R+N-G.
Any symbolic parameters are assumed to be real. Any numeric coefficients or parameters should be of type algebraic. Floating point values will be converted to rationals with convert(value, rational).
2. 

Function to solve for, e.g., u(x, y, t).
3. 

m initial condition(s) (at time zero). For m > 1, these must be in a list, and are the values of the function and its successive derivatives with respect to time, evaluated at time zero.
4. 

name of time variable.
5. 

number of iterations.
6. 

(optional) order for a fractional derivative, specified as fracorder = name or fracorder = numeric*value (floating point values are converted to type rational). The permissible values alpha are related to the value of m: m-1 < alpha and alpha <= m, i.e., an mth order time derivative is specified in the pde and m initial conditions are provided, as well as the value of alpha in the correct range. A symbolic alpha is assumed to be in this range.

infolevel[LAD] may be set to different values to print out additional information as follows. Greater numbers include the information from smaller values.
1. 

The nonlinear expansion variables (those in the Adomian polynomials).
2. 

The different parts of the the pde (L, R, N, G) in jet notation.
3. 

Progress is indicated, as the time at each iteration.
4. 

Values of the components of the solution as they are produced (one per iteration).
> 

restart

It may be useful to load the Physics package if derivatives contain abs. LAD converts these to a form without abs but with the conjugate of the dependent variable. The Physics package affects the internal simplifications and may affect the form of the output or the running time.

> 

PDEtools:-declare(U(x, t), quiet)

Example from A-M. Wazwaz, Applied Mathematics and Computation 111 (2000) 53. Sec. 4.

> 

pde := diff(U(x, t), t)+U(x, t)^2*(diff(U(x, t), x)); inx := 3*x

diff(U(x, t), t)+U(x, t)^2*(diff(U(x, t), x))

3*x

> 

infolevel[LAD] := 2
> 

approx := LAD(pde, U(x, t), inx, t, 10)

LAD: L = U[t]; R = 0; N = -U^2*U[x]; G = 0.

LAD: nonlinear expansion variables (conjugated functions denoted by C) are: [U, U[x]]

175692092397588*t^10*x^11-5650915252554*t^9*x^10+184670433090*t^8*x^9-6155681103*t^7*x^8+210450636*t^6*x^7-7440174*t^5*x^6+275562*t^4*x^5-10935*t^3*x^4+486*t^2*x^3-27*t*x^2+3*x

By extrapolating to an infinite number of terms, we can find an exact solution. (Wazwaz calculates the coefficient of x^4*t^3 incorrectly and deduces an incorrect exact solution.) Multiplying by t gives a series in y = x*t for which guessgf can use the coefficients to find the exact solution.

> 

algsubs(x*t = y, expand(approx*t)); [seq(coeff(%, y, i), i = 0 .. degree(%, y))]; gfun:-guessgf(%, y)[1]; exact := (eval(%, y = x*t))/t

175692092397588*y^11-5650915252554*y^10+184670433090*y^9-6155681103*y^8+210450636*y^7-7440174*y^6+275562*y^5-10935*y^4+486*y^3-27*y^2+3*y

[0, 3, -27, 486, -10935, 275562, -7440174, 210450636, -6155681103, 184670433090, -5650915252554, 175692092397588]

6*y/(1+(1+36*y)^(1/2))

6*x/(1+(36*t*x+1)^(1/2))

Check

> 

pdetest(U(x, t) = exact, [pde, U(x, 0) = inx])

[0, 0]

An example with inhomogeneous terms (but no nonlinear part). Ex. 3 from R. Shah et al, Entropy 21 (2019) 335.

> 

pde := diff(U(x, t), t)+diff(U(x, t), `$`(x, 3)) = -sin(Pi*x)*sin(t)-Pi^3*cos(Pi*x)*cos(t); inx := sin(Pi*x); exact := sin(Pi*x)*cos(t); pdetest(U(x, t) = exact, [pde, U(x, 0) = inx])

diff(U(x, t), t)+diff(diff(diff(U(x, t), x), x), x) = -sin(Pi*x)*sin(t)-Pi^3*cos(Pi*x)*cos(t)

sin(Pi*x)

sin(Pi*x)*cos(t)

[0, 0]

> 

approx := LAD(pde, U(x, t), inx, t, 10)

LAD: L = U[t]; R = -U[x,x,x]; N = 0; G = -sin(Pi*x)*sin(t)-Pi^3*cos(Pi*x)*cos(t).

LAD: nonlinear expansion variables (conjugated functions denoted by C) are: []

sin(Pi*x)-cos(Pi*x)*Pi^3*sin(t)+sin(Pi*x)*(-1+cos(t))-Pi^3*(Pi^3*(-1+cos(t))*sin(Pi*x)-cos(Pi*x)*sin(t))+(Pi^3*(-sin(t)+t)*cos(Pi*x)+sin(Pi*x)*(-1+cos(t)))*Pi^6-(1/2)*Pi^9*(Pi^3*(t^2+2*cos(t)-2)*sin(Pi*x)+2*cos(Pi*x)*(-sin(t)+t))-(1/6)*Pi^12*(Pi^3*(t^3+6*sin(t)-6*t)*cos(Pi*x)-3*sin(Pi*x)*(t^2+2*cos(t)-2))+(1/24)*(Pi^3*(t^4-12*t^2-24*cos(t)+24)*sin(Pi*x)+4*cos(Pi*x)*(t^3+6*sin(t)-6*t))*Pi^15+(1/120)*Pi^18*(Pi^3*(t^5-20*t^3-120*sin(t)+120*t)*cos(Pi*x)-5*sin(Pi*x)*(t^4-12*t^2-24*cos(t)+24))-(1/720)*Pi^21*(Pi^3*(t^6-30*t^4+360*t^2+720*cos(t)-720)*sin(Pi*x)+6*cos(Pi*x)*(t^5-20*t^3-120*sin(t)+120*t))-(1/5040)*(Pi^3*(t^7-42*t^5+840*t^3+5040*sin(t)-5040*t)*cos(Pi*x)-7*sin(Pi*x)*(t^6-30*t^4+360*t^2+720*cos(t)-720))*Pi^24+(1/40320)*Pi^27*(Pi^3*(t^8-56*t^6+1680*t^4-20160*t^2-40320*cos(t)+40320)*sin(Pi*x)+8*cos(Pi*x)*(t^7-42*t^5+840*t^3+5040*sin(t)-5040*t))+(1/362880)*Pi^30*(Pi^3*(t^9-72*t^7+3024*t^5-60480*t^3-362880*sin(t)+362880*t)*cos(Pi*x)-9*sin(Pi*x)*(t^8-56*t^6+1680*t^4-20160*t^2-40320*cos(t)+40320))

Expand the exact solution as a series in t to the same order and verify that it is the same as above.

> 

series(exact-approx, t, 11, oterm = false)

0

An example with a complex solution

> 

pde := I*(diff(U(x, t), t))+diff(U(x, t), `$`(x, 2))+2*abs(U(x, t))^2*U(x, t); inx := sech(x); exact := sech(x)*exp(I*t); `assuming`([simplify(pdetest(U(x, t) = exact, [pde, U(x, 0) = inx]))], [real])

I*(diff(U(x, t), t))+diff(diff(U(x, t), x), x)+2*abs(U(x, t))^2*U(x, t)

sech(x)

sech(x)*exp(I*t)

[0, 0]

> 

approx := LAD(pde, U(x, t), inx, t, 10)

LAD: L = I*U[t]; R = -U[x,x]; N = -2*U^2*C; G = 0.

LAD: nonlinear expansion variables (conjugated functions denoted by C) are: [C, U]

sech(x)+I*sech(x)*t-(1/2)*sech(x)*t^2-((1/6)*I)*sech(x)*t^3+(1/24)*sech(x)*t^4+((1/120)*I)*sech(x)*t^5-(1/720)*sech(x)*t^6-((1/5040)*I)*sech(x)*t^7+(1/40320)*sech(x)*t^8+((1/362880)*I)*sech(x)*t^9-(1/3628800)*sech(x)*t^10

The series for exp(I*t) is apparent.

> 

collect(approx, sech); series(approx-exact, t, 11, oterm = false)

(1+I*t-(1/2)*t^2-((1/6)*I)*t^3+(1/24)*t^4+((1/120)*I)*t^5-(1/720)*t^6-((1/5040)*I)*t^7+(1/40320)*t^8+((1/362880)*I)*t^9-(1/3628800)*t^10)*sech(x)

0

An example with fractional order differentiation wrt t of order alpha. For m-1 < alpha and alpha <= m, the derivative must be entered with order m and there must be a list of m initial conditions:f(x, 0), (D[2](f))(x, 0), (D[2, 2](f))(x, 0), () .. ()

 Ex. 1 from R. Shah et al, Entropy 21 (2019) 335

> 

pde := diff(U(x, t), t) = -2*(diff(U(x, t), x))-(diff(U(x, t), `$`(x, 3))); inx := sin(x)

diff(U(x, t), t) = -2*(diff(U(x, t), x))-(diff(diff(diff(U(x, t), x), x), x))

sin(x)

Symbolic order is accepted.

> 

approx := LAD(pde, U(x, t), inx, t, 6, fracorder = alpha)

LAD: L = U[t]; R = -2*U[x]-U[x,x,x]; N = 0; G = 0.

LAD: nonlinear expansion variables (conjugated functions denoted by C) are: []

sin(x)-cos(x)*t^alpha/GAMMA(1+alpha)-sin(x)*t^(2*alpha)/GAMMA(1+2*alpha)+cos(x)*t^(3*alpha)/GAMMA(1+3*alpha)+sin(x)*t^(4*alpha)/GAMMA(1+4*alpha)-cos(x)*t^(5*alpha)/GAMMA(1+5*alpha)-sin(x)*t^(6*alpha)/GAMMA(1+6*alpha)

Giving a numerical value for the order will generally be more efficient. For alpha = 1/2, we can find an exact solution.

> 

approx := collect(LAD(pde, U(x, t), inx, t, 20, fracorder = 1/2), [sin, cos])

LAD: L = U[t]; R = -2*U[x]-U[x,x,x]; N = 0; G = 0.

LAD: nonlinear expansion variables (conjugated functions denoted by C) are: []

(1-t+(1/2)*t^2-(1/6)*t^3+(1/24)*t^4-(1/120)*t^5+(1/720)*t^6-(1/5040)*t^7+(1/40320)*t^8-(1/362880)*t^9+(1/3628800)*t^10)*sin(x)+(-2*t^(1/2)/Pi^(1/2)+(4/3)*t^(3/2)/Pi^(1/2)-(8/15)*t^(5/2)/Pi^(1/2)+(16/105)*t^(7/2)/Pi^(1/2)-(32/945)*t^(9/2)/Pi^(1/2)+(64/10395)*t^(11/2)/Pi^(1/2)-(128/135135)*t^(13/2)/Pi^(1/2)+(256/2027025)*t^(15/2)/Pi^(1/2)-(512/34459425)*t^(17/2)/Pi^(1/2)+(1024/654729075)*t^(19/2)/Pi^(1/2))*cos(x)

The first series is exp(-t), the second series is not so obvious, After some scaling, the second series may be passed to guessgf

> 

ser2 := `assuming`([simplify(expand(sqrt(Pi)*op(2, approx)/(sqrt(t)*cos(x))))], [t > 0]); [seq(coeff(ser2, t, i), i = 0 .. 9)]; ser2ex := gfun:-guessgf(%, t)[1]

-2+(4/3)*t-(8/15)*t^2+(16/105)*t^3-(32/945)*t^4+(64/10395)*t^5-(128/135135)*t^6+(256/2027025)*t^7-(512/34459425)*t^8+(1024/654729075)*t^9

-exp(-t)*Pi^(1/2)*erfi(t^(1/2))/t^(1/2)

So the full solution is

> 

exact := exp(-t)*sin(x)+ser2ex*sqrt(t)*cos(x)/sqrt(Pi)

exp(-t)*sin(x)-exp(-t)*erfi(t^(1/2))*cos(x)

Check. fracdiff's default method returns an integral, method = laplace fails, but method = series works with cancellation of all terms but the "left over" last half-integral-order one.

> 

Ord := 20; fracdiff(exact, t, 1/2, method = series, methodoptions = [order = Ord])-series(eval(rhs(pde), U(x, t) = exact), t, Ord)

20

-(1048576/319830986772877770815625)*sin(x)*t^(39/2)/Pi^(1/2)+O(t^(39/2))-O(t^20)

A second-order example (linear)

> 

pde := diff(U(x, t), t, t) = -c^2*(diff(U(x, t), `$`(x, 2))); inx := [exp(I*x/c), exp(I*x/c)]; exact := exp(t+I*x/c)

diff(diff(U(x, t), t), t) = -c^2*(diff(diff(U(x, t), x), x))

[exp(I*x/c), exp(I*x/c)]

exp(t+I*x/c)

> 

pdetest(U(x, t) = exact, [pde, U(x, 0) = inx[1], (D[2](U))(x, 0) = inx[2]])

[0, 0, 0]

> 

approx := LAD(pde, U(x, t), inx, t, 6)

LAD: L = U[t,t]; R = -c^2*U[x,x]; N = 0; G = 0.

LAD: nonlinear expansion variables (conjugated functions denoted by C) are: []

exp(I*x/c)*(1+t)+(1/6)*exp(I*x/c)*t^2*(3+t)+(1/120)*exp(I*x/c)*t^4*(t+5)+(1/5040)*exp(I*x/c)*t^6*(7+t)+(1/362880)*exp(I*x/c)*t^8*(9+t)+(1/39916800)*exp(I*x/c)*t^10*(t+11)+(1/6227020800)*exp(I*x/c)*t^12*(13+t)

> 

series(exact-approx, t, 13, oterm = false)

0

Fractional order example

> 

approx := collect(LAD(pde, U(x, t), inx, t, 10, fracorder = 3/2), [exp, t])

LAD: L = U[t,t]; R = -c^2*U[x,x]; N = 0; G = 0.

LAD: nonlinear expansion variables (conjugated functions denoted by C) are: []

(1+(1/20922789888000)*t^16+(1/1307674368000)*t^15+(32768/6190283353629375)*t^(29/2)/Pi^(1/2)+(16384/213458046676875)*t^(27/2)/Pi^(1/2)+(1/6227020800)*t^13+(1/479001600)*t^12+(4096/316234143225)*t^(23/2)/Pi^(1/2)+(2048/13749310575)*t^(21/2)/Pi^(1/2)+(1/3628800)*t^10+(1/362880)*t^9+(512/34459425)*t^(17/2)/Pi^(1/2)+(256/2027025)*t^(15/2)/Pi^(1/2)+(1/5040)*t^7+(1/720)*t^6+(64/10395)*t^(11/2)/Pi^(1/2)+(32/945)*t^(9/2)/Pi^(1/2)+(1/24)*t^4+(1/6)*t^3+(8/15)*t^(5/2)/Pi^(1/2)+(4/3)*t^(3/2)/Pi^(1/2)+t)*exp(I*x/c)

There appear to be two separate series in t, with integer and half-integer powers.

> 

t_series := approx/exp(I*x/c); t1, t2 := selectremove(proc (w) options operator, arrow; (degree(w, t))::integer end proc, t_series)

1+(1/20922789888000)*t^16+(1/1307674368000)*t^15+(1/6227020800)*t^13+(1/479001600)*t^12+(1/3628800)*t^10+(1/362880)*t^9+(1/5040)*t^7+(1/720)*t^6+(1/24)*t^4+(1/6)*t^3+t, (32768/6190283353629375)*t^(29/2)/Pi^(1/2)+(16384/213458046676875)*t^(27/2)/Pi^(1/2)+(4096/316234143225)*t^(23/2)/Pi^(1/2)+(2048/13749310575)*t^(21/2)/Pi^(1/2)+(512/34459425)*t^(17/2)/Pi^(1/2)+(256/2027025)*t^(15/2)/Pi^(1/2)+(64/10395)*t^(11/2)/Pi^(1/2)+(32/945)*t^(9/2)/Pi^(1/2)+(8/15)*t^(5/2)/Pi^(1/2)+(4/3)*t^(3/2)/Pi^(1/2)

The integer-powers series is the series for exp(t) with every third term missing: missing powers 2, 5, 8, 11, ...

> 

t1sum := simplify(exp(t)-(sum(t^(3*i+2)/factorial(3*i+2), i = 0 .. infinity))); series(%, t, 17)

(2/3)*exp(t)+(1/3)*(cos((1/2)*3^(1/2)*t)+3^(1/2)*sin((1/2)*3^(1/2)*t))*exp(-(1/2)*t)

series(1+t+(1/6)*t^3+(1/24)*t^4+(1/720)*t^6+(1/5040)*t^7+(1/362880)*t^9+(1/3628800)*t^10+(1/479001600)*t^12+(1/6227020800)*t^13+(1/1307674368000)*t^15+(1/20922789888000)*t^16+O(t^17),t,17)

And for the other one, the following scaled series shows coefficients with powers of two divided by double factorials, every third term missing: missing powers 2,5,8,11,...

> 

expand(sqrt(Pi)*t2/(4*t^(3/2))); all := add(2^i*t^i/doublefactorial(2*(i+1)+1), i = 0 .. 9); missing := add(eval(2^i*t^i/doublefactorial(2*(i+1)+1), i = 3*j+2), j = 0 .. 3)

(8192/6190283353629375)*t^13+(4096/213458046676875)*t^12+(1024/316234143225)*t^10+(512/13749310575)*t^9+(128/34459425)*t^7+(64/2027025)*t^6+(16/10395)*t^4+(8/945)*t^3+(2/15)*t+1/3

1/3+(2/15)*t+(4/105)*t^2+(8/945)*t^3+(16/10395)*t^4+(32/135135)*t^5+(64/2027025)*t^6+(128/34459425)*t^7+(256/654729075)*t^8+(512/13749310575)*t^9

(4/105)*t^2+(32/135135)*t^5+(256/654729075)*t^8+(2048/7905853580625)*t^11

> 

t2sum := 4*t^(3/2)*(sum(2^i*t^i/doublefactorial(2*(i+1)+1), i = 0 .. infinity)-(sum(eval(2^i*t^i/doublefactorial(2*(i+1)+1), i = 3*j+2), j = 0 .. infinity)))/sqrt(Pi)

4*t^(3/2)*((1/4)*(Pi^(1/2)*exp(t)-2*t^(1/2)-Pi^(1/2)*erfc(t^(1/2))*exp(t))/t^(3/2)-(4/105)*t^2*hypergeom([1], [3/2, 11/6, 13/6], (1/27)*t^3))/Pi^(1/2)

Putting it together suggests

> 

exact := exp(I*x/c)*simplify(t1sum+t2sum)

(1/3)*exp(I*x/c)*(Pi^(1/2)*exp(-(1/2)*t)*sin((1/2)*3^(1/2)*t)*3^(1/2)+Pi^(1/2)*exp(-(1/2)*t)*cos((1/2)*3^(1/2)*t)+3*Pi^(1/2)*exp(t)*erf(t^(1/2))-(16/35)*t^(7/2)*hypergeom([1], [3/2, 11/6, 13/6], (1/27)*t^3)+2*Pi^(1/2)*exp(t)-6*t^(1/2))/Pi^(1/2)

Check.

> 

Ord := 20; fracdiff(exact, t, 3/2, method = series, methodoptions = [order = Ord+1])-series(eval(rhs(pde), U(x, t) = exact), t, Ord)

20

O(t^(39/2))-(1048576/319830986772877770815625)*exp(I*x/c)*t^(39/2)/Pi^(1/2)-O(t^(41/2))

Numerical calculation of a soliton

> 

pde := I*(diff(U(x, t), t))+diff(U(x, t), `$`(x, 2))+U(x, t)^2*conjugate(U(x, t)); exact := sqrt(2*B)*sech(sqrt(B)*(2*k*t-x-x__0))*exp(I*(k*x+(-k^2+B)*t)); `assuming`([simplify(pdetest(U(x, t) = exact, pde))], [real, B > 0]); inx := eval(exact, t = 0)

I*(diff(U(x, t), t))+diff(diff(U(x, t), x), x)+U(x, t)^2*conjugate(U(x, t))

2^(1/2)*B^(1/2)*sech(B^(1/2)*(2*k*t-x-x__0))*exp(I*(k*x+(-k^2+B)*t))

0

2^(1/2)*B^(1/2)*sech(B^(1/2)*(-x-x__0))*exp(I*k*x)

With numerical parameters it runs faster. Float parameters will be converted to rationals.

> 

params := {B = 2, k = -1/2, x__0 = -2}; inxnum := eval(inx, params); exactnum := eval(exact, params)

2*sech(2^(1/2)*(-x+2))*exp(-((1/2)*I)*x)

2*sech(2^(1/2)*(-t-x+2))*exp(I*(-(1/2)*x+(7/4)*t))

> 

infolevel[LAD] := 3; approx := LAD(pde, U(x, t), inxnum, t, 28)

LAD: L = I*U[t]; R = -U[x,x]; N = -U^2*C; G = 0.

LAD: nonlinear expansion variables (conjugated functions denoted by C) are: [C, U]

LAD: calculating component 1 at time 0.

LAD: calculating component 2 at time .221

[Edited for brevity]

LAD: calculating component 27 at time 90.414

LAD: calculating component 28 at time 103.870

LAD: completed at time 105.309

For larger x and t, the approximation diverges at a point that depends on the number of iterations.

> 

plot3d([abs(exactnum), abs(approx)], t = 0 .. 2, x = -3 .. 3, color = [red, blue], view = [default, default, 0 .. 5], orientation = [-30, 60])

> 

NULL

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Where is the "Stop Execution" icon?...

Asked by: WD0HHU 40
interrupt gui duplicate-question
June 17 2025
0  1

Maple 2024 had a very well defined "Stop Execution" symbol on the desktop.  

Maple 2025 doesn't have an obvious symbol somewhere.  Also, the two little "dots" that I once had in the lower left bottom in Maple 2025 have disappeared! There is only a bar at the bottom which seems to do nothing.

there is anyway for making fast the loop?...

Asked by: salim-barzani 1555
performance simplification conjugate
June 17 2025
2  4

i need a way for making my loop working fast, i need something for calculating such term fast without showing result but thus term make a lot of time for coming out ?

fasting.mw

substitute equation to equation...

Asked by: ashy 35
substitution incomplete-question
June 17 2025
0  4

I want to substitute equation to equation, how?

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Computation of left and right eigenvectors...

Asked by: Elisha 50
bifurcation eigenvector
June 17 2025
0  0

Someone should please help me compute the left and right eigenvectors of the system below. The purpose is to compute values for 'a' and 'b' in the bifurcation formula.

Thank you

``

> 

with(VectorCalculus)

 

 (1)
> 

interface(imaginaryunit = I)

I

 (2)
> 

I

I

 (3)
 
> 

diff(S(t), t) := `&Lambda;__p`-(`#mrow(mi("&varphi;",fontstyle = "normal"),mo("&sdot;"),msub(mi("&theta;",fontstyle = "normal"),mi("B")),mo("&sdot;"),msub(mi("I"),mi("B")))`/N[p]+µ__C)*S+`&omega;__B`*I__B

Lambda__p-(`#mrow(mi("&varphi;",fontstyle = "normal"),mo("&sdot;"),msub(mi("&theta;",fontstyle = "normal"),mi("B")),mo("&sdot;"),msub(mi("I"),mi("B")))`/N[p]+µ__C)*S+omega__B*I__B

 (4)
> 

diff(I__B(t), t) := `#mrow(mi("&varphi;",fontstyle = "normal"),mo("&sdot;"),msub(mi("&theta;",fontstyle = "normal"),mi("B")),mo("&sdot;"),msub(mi("I"),mi("B")))`*S/N[p]-`&omega;__B`*I__B-(`&sigma;__B`+µ__C)*I__B

`#mrow(mi("&varphi;",fontstyle = "normal"),mo("&sdot;"),msub(mi("&theta;",fontstyle = "normal"),mi("B")),mo("&sdot;"),msub(mi("I"),mi("B")))`*S/N[p]-omega__B*I__B-(sigma__B+µ__C)*I__B

 (5)
> 

NULL

``

 (6)
> 

diff(S__A(t), t) := `&Lambda;__A`-(µ__A+`#mrow(mi("&varphi;",fontstyle = "normal"),mo("&sdot;"),msub(mi("&alpha;",fontstyle = "normal"),mi("B")),mo("&sdot;"),msub(mi("I"),mi("B")))`/N[p])*S__A+`&delta;__A`*I__A

Lambda__A-(µ__A+`#mrow(mi("&varphi;",fontstyle = "normal"),mo("&sdot;"),msub(mi("&alpha;",fontstyle = "normal"),mi("B")),mo("&sdot;"),msub(mi("I"),mi("B")))`/N[p])*S__A+delta__A*I__A

 (7)
> 

diff(I__A(t), t) := `#mrow(mi("&varphi;",fontstyle = "normal"),mo("&sdot;"),msub(mi("&alpha;",fontstyle = "normal"),mi("B")),mo("&sdot;"),msub(mi("I"),mi("B")))`*S__A/N[p]-(µ__A+`&delta;__A`)*I__A

`#mrow(mi("&varphi;",fontstyle = "normal"),mo("&sdot;"),msub(mi("&alpha;",fontstyle = "normal"),mi("B")),mo("&sdot;"),msub(mi("I"),mi("B")))`*S__A/N[p]-(µ__A+delta__A)*I__A

 (8)
> 

NULL

``

 (9)

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remove the parentheses (for the dᵢ values)...

Asked by: jalal 435
table typesetting task
June 17 2025
0  6

Hi,

I’m trying to remove the parentheses (for the dᵢ values) to achieve optimal display. Ideas ? Thanks

S4_Droites_Implicite.mw

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