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error in dsolve ...

Asked by: KIRAN SAJJAN 40
ode homework dsolve
June 13 2023
0  5

am attaching the worksheet of the problem please help me to solve not able to  compute coupled

error.mw

how to plot this equation?...

Asked by: Aung 65
plotting incomplete-question
June 13 2023
2  7

in this equations...how to plot (a,e)? thank you guys

plot.mw

How do I re-parameterize a triangular distribution...

Asked by: NanaP 40
June 13 2023
2  12

Hi community!

In the mw file attached, I have worked a numerical solution to get the standard parameters of minimum and maximum of the distribution Triangular(a,b,c) given 2 quantiles (and its values) and c.  The code works and was simple.  Now, how do I solve this more elegantly to have the reparameterization directly inputed instead of indirectly by solving a & b first (like the attached file does)? Thanks people,altTriangular2.mw

why LinearAlgebra:-RowDimension works on list?...

Asked by: nm 11218
parameters linearalgebra type
June 13 2023
1  5

From help it says

But in the following it works on list type. Why? Should have not given an error?

158332

> 

restart;

158332

> 

interface(warnlevel=4);
kernelopts('assertlevel'=2):

3

> 

T:=[ [1,2],[3,4],[6,6]];
whattype(T);
type(T,Matrix);
LinearAlgebra:-RowDimension(T);

[[1, 2], [3, 4], [6, 6]]

list

false

3

> 

T:=Matrix([[1,2],[3,4],[6,6]]);
whattype(T);
type(T,Matrix);
LinearAlgebra:-RowDimension(T);

T := Matrix(3, 2, {(1, 1) = 1, (1, 2) = 2, (2, 1) = 3, (2, 2) = 4, (3, 1) = 6, (3, 2) = 6})

Matrix

true

3

> 

 

Download row_dim_june_13_2023.mw

catch "wrong type of arguments"...

Asked by: Anthrazit 885
geometry try-catch
June 13 2023
1  8

Any idea how I catch that error?

(in geometry:-DefinedAs) wrong type of argument |C:/Users/andreas/AppData/Local/Temp/com.maplesoft/NODELibrary-0dc05fe4-0ffc9cfc/lib/NODEFunctions-13751:1976|

Here's the code:

    try
        a := map(coordinates, DefinedAs(segm))[1];
        b := map(coordinates, DefinedAs(segm))[2];
    catch "wrong type of arguments":
        Alert(cat("Error, (in geometry:-DefinedAs) wrong type of arguments, variable segm: ", whattype(segm), segm), table(), 5)
    end try;

better way to write the code for the integration o...

Asked by: Dkunb 70
integration orthogonal
June 12 2023
1  2

how to write the code for integration of the orthogonal cosine function with nu=mu or with nu<>mu

> 

restart;
 
> 

phi:= (mu,Q2)->sqrt(2/l)*sin(mu*Pi*(Q2+l/2)/l);    # E:=mu->(Pi*mu/sqrt(2)/l)^2, mu=1,2...

proc (mu, Q2) options operator, arrow; sqrt(2/l)*sin(mu*Pi*(Q2+(1/2)*l)/l) end proc

 (1)
> 

fh1:=simplify((-1/2/m2*Int(diff(phi(mu,Q2),Q2)*diff(phi(nu,Q2),Q2),Q2=-l/2..l/2)))

-mu*Pi^2*nu*(Int(cos((1/2)*mu*Pi*(2*Q2+l)/l)*cos((1/2)*nu*Pi*(2*Q2+l)/l), Q2 = -(1/2)*l .. (1/2)*l))/(l^3*m2)

 (2)
> 

convert(fh1,int) assuming(mu,integer,nu,integer);

0

 (3)
> 

fh1_subs := simplify(subs(nu = mu, fh1));

-mu^2*Pi^2*(Int(cos((1/2)*mu*Pi*(2*Q2+l)/l)^2, Q2 = -(1/2)*l .. (1/2)*l))/(l^3*m2)

 (4)
> 

convert(fh1_substituted,int)assuming (mu,integer,nu,integer);

-(1/2)*mu^2*Pi^2/(l^2*m2)

 (5)
> 

 

Download test1.mw

How to fix a curve...

Asked by: Saha 80
differential-equations plotting
June 12 2023
0  0

Can you change f(eta) to upflow curve and theta(eta) to downflow curve.

In my Problem,Boundary Conditions are

theta(infinity) = 0, (D(f))(infinity) = 1 , (Take, eta =infinity)

Flows will be correct for what value is taken for infinity .

I take  eta = 5. and also tried changing ranges  but could't find it.Please Help to fix the curve.

my code is,

SM.mw

Quiz : How to fix this code ?...

Asked by: jalal 435
complex argument quiz
June 12 2023
1  1

Hi,

I am creating a random question about the trigonometric form of a complex number. My correction code does not accept all possible amplitudes for the argument (for example, it does not accept 13Pi/6 for an argument of Pi/6). Any ideas on how to fix the code?

QComplexCis.mw

Worksheet Viewer for Maple...

Asked by: Roy Hughes 20
worksheet
June 12 2023
1  2

Is there a viewer in Maple to view the .mw files within the Windows or Maple directory [or alternative OS environment], not unlike a pdf or MS Word viewer?

It would make it a lot easier to check through working files without having to load all of them into the system, particulary for developmental work.

regards, RH

Can this equation (x-a)^4 + (x-b)^4 = c has four d...

Asked by: minhthien2016 385
integer
June 11 2023
1  9

I am trying to find three integer a, b, c so that equation (x-a)^4 + (x-b)^4 = c has four different solutions integer x. I tried

restart;
k := 0;
for a from -10 to 10 do
    for b from -10 to 10 do
for c from -10 to 10 do
if a*b*c <> 0 and igcd(a, b) = 1 then X := [solve((x - a)^4 + (x - b)^4 = c, x)]; if type(X[1], integer) and type(X[2], integer) and type(X[3], integer) and type(X[4], integer) and nops({X[1], X[2], X[3], X[4]}) = 4 then k := k + 1; L[k] := [a, b, c, X[]]; end if; end if; end do; end do;
end do;
L := convert(L, list);
k;
L;

I do not get any solution

Can anyone integrate dsolve/numeric with NLPSolve ...

Asked by: Dr. Venkat Subramanian 431
evalhf nlpsolve
June 11 2023
1  4

In a recent question/conversation, I had discussed integrating dsolve/numeric-based codes with NLPsolve at 
https://www.mapleprimes.com/questions/236494-Inconsistent-Behavior-With-Dsolvenumeric-And-NLPSolve-sqp

@acer was able to help resolve the issue by calling NLPSolve with higher optimality tolerance.

I am starting a new question/post to show the need for evalhf as the previous post takes too long to load.
Code is attached for the same.
 

> 

restart:
> 

currentdir();

"C:\Users\Venkat\OneDrive - The University of Texas at Austin\Documents\trial\Learnningsqpoptim"

Test code written by Dr. Venkat Subramanian at UT Austin, 05/31/2023 and has been updated multiple times. This code uses CVP approach (piecwise constant) to perform optimal control. NLPSolve combination with RK2 approach to integrate ODEs (constant step size) works well for this problem. But dsolve numeric based codes work (with optimality tolerance fix from acer), but cannot handle large values of nvar (optimization variables).

> 

Digits:=15;

15

> 

 

> 

eqodes:=[diff(ca(t),t)=-(u+u^2/2)*1.0*ca(t),diff(cb(t),t)=1.0*u*ca(t)-0.1*cb(t)];

[diff(ca(t), t) = -1.0*(u+(1/2)*u^2)*ca(t), diff(cb(t), t) = 1.0*u*ca(t)-.1*cb(t)]

> 

soln:=dsolve({op(eqodes),ca(0)=alpha,cb(0)=beta},type=numeric,'parameters'=[alpha,beta,u],savebinary=true):
> 

 

> 

 

Note that Vector or Array form can be used for RK2h to implement the procedure for any number of variables/ODEs. But the challenge will be when implicit methods are used to integrate ODEs/DAEs and running them in evalhf form (this can be done with Gauss elimination type linear solver, but this will be limited to small number of ODEs/DAEs say 200 or so).

> 

RK2h:=proc(NN,u,dt,y0::Vector(datatype=float[8]))#this procedure can be made efficient in vector form
local j,c1mid,c2mid,c10,c20,c1,c2;
option hfloat;
c10:=y0[1];c20:=y0[2];
for j from 1 to NN do
  c1mid:=c10+dt/NN*(-(u+u^2/2)*c10):
  c2mid:=c20+dt/NN*(u*c10)-0.1*dt/NN*c20:
  c1:=c10/2+c1mid/2+dt/NN/2*(-(u+u^2/2)*c1mid):
  c2:=c20/2+c2mid/2+dt/NN/2*(u*c1mid)-0.1*dt/NN/2*c2mid:
  c10:=c1:c20:=c2:
  od:
y0[1]:=c10:y0[2]:=c20:
end proc:
> 

 

> 

soln('parameters'=[1,0,0.1]);soln(0.1);

[alpha = 1., beta = 0., u = .1]

[t = .1, ca(t) = HFloat(0.9895549324188543), cb(t) = HFloat(0.009898024276129616)]

> 

 

> 

 

> 

ssdsolve:=proc(x)
interface(warnlevel=0):

#if  type(x,Vector)#if type is not needed for this problem, might be needed for other problems
#then


local z1,n1,i,c10,c20,dt,u;
global soln,nvar;
dt:=evalf(1.0/nvar):
c10:=1.0:c20:=0.0:
for i from 1 to nvar do
u:=x[i]:
soln('parameters'=[c10,c20,u]):
z1:=soln(dt):
c10:=subs(z1,ca(t)):c20:=subs(z1,cb(t)):
od:
-c20;
 #else 'procname'(args):

#end if:

end proc:
> 

 

> 

 

> 

ssRK2:=proc(x)#based on RK2
#interface(warnlevel=0):
option hfloat;
#if  type(x,Vector)
#then

local z1,n1,i,c10,c20,dt,u,NN,y0;
global nvar,RK2h;
y0:=Array(1..2,[1.0,0.0],datatype=float[8]):
#y0[1]:=1.0:y0[2]:=0.0:
dt:=evalf(1.0/nvar):
NN:=256*2/nvar;#NN is hardcode based on the assumption that nvar will be a multiple of 2 <=256


for i from 1 to nvar do
u:=x[i]:
evalhf(RK2h(NN,u,dt,y0)):
od:
-y0[2];
 #else 'procname'(args):

#end if:

end proc:
> 

nvar:=2;

2

> 

ic0:=Vector(nvar,[seq(0.1,i=1..nvar)],datatype=float):
> 

bl := Vector(nvar,[seq(0.,i=1..nvar)],datatype=float):bu := Vector(nvar,[seq(5.,i=1..nvar)],datatype=float):
> 

infolevel[Optimization]:=15;

15

> 

CodeTools:-Usage(Optimization:-NLPSolve(nvar,evalhf(ssRK2),[],initialpoint=ic0,[bl,bu],optimalitytolerance=1e-6)):

NLPSolve: calling NLP solver

NLPSolve: using method=sqp

NLPSolve: number of problem variables 2

NLPSolve: number of nonlinear inequality constraints 0

NLPSolve: number of nonlinear equality constraints 0

NLPSolve: number of general linear constraints 0

NLPSolve: feasibility tolerance set to 0.1053671213e-7

NLPSolve: optimality tolerance set to 0.1e-5

NLPSolve: iteration limit set to 50

NLPSolve: infinite bound set to 0.10e21

NLPSolve: trying evalhf mode

NLPSolve: trying evalf mode

attemptsolution: number of major iterations taken 11

memory used=0.96MiB, alloc change=0 bytes, cpu time=16.00ms, real time=121.00ms, gc time=0ns

> 

 

Calling the procedures based on RK2 with NLPSolve uses evalf for numerical gradient (fdiff). Providing a procedure for gradient solves this issue.

> 

gradRK2 := proc (x,g)
option hfloat;
local base, i, xnew, del;
global nvar,ssRK2;
xnew:=Array(1..nvar,datatype=float[8]):
base := ssRK2(x);
for i to nvar do
xnew[i] := x[i]; end do;
for i to nvar do
del := max(1e-5,.1e-6*x[i]);
xnew[i] := xnew[i]+del;
g[i] := (ssRK2(xnew)-base)/del;
xnew[i] := xnew[i]-del;
end do;
end proc:
> 

CodeTools:-Usage(Optimization:-NLPSolve(nvar,evalhf(ssRK2),[],initialpoint=ic0,[bl,bu],objectivegradient=gradRK2,optimalitytolerance=1e-6)):

NLPSolve: calling NLP solver

NLPSolve: using method=sqp

NLPSolve: number of problem variables 2

NLPSolve: number of nonlinear inequality constraints 0

NLPSolve: number of nonlinear equality constraints 0

NLPSolve: number of general linear constraints 0

NLPSolve: feasibility tolerance set to 0.1053671213e-7

NLPSolve: optimality tolerance set to 0.1e-5

NLPSolve: iteration limit set to 50

NLPSolve: infinite bound set to 0.10e21

NLPSolve: trying evalhf mode

attemptsolution: number of major iterations taken 11

memory used=184.77KiB, alloc change=0 bytes, cpu time=31.00ms, real time=110.00ms, gc time=0ns

Significant saving in memory used is seen. CPU time is also less, which is more apparent at larger valeus of nvar.

> 

 

dsolvenumeric based codes work for optimization, but evalf is invoked possibly for both the objective and gradient

> 

CodeTools:-Usage(Optimization:-NLPSolve(nvar,(ssdsolve),[],initialpoint=ic0,[bl,bu],optimalitytolerance=1e-6)):

NLPSolve: calling NLP solver

NLPSolve: using method=sqp

NLPSolve: number of problem variables 2

NLPSolve: number of nonlinear inequality constraints 0

NLPSolve: number of nonlinear equality constraints 0

NLPSolve: number of general linear constraints 0

NLPSolve: feasibility tolerance set to 0.1053671213e-7

NLPSolve: optimality tolerance set to 0.1e-5

NLPSolve: iteration limit set to 50

NLPSolve: infinite bound set to 0.10e21

NLPSolve: trying evalhf mode

NLPSolve: trying evalf mode

attemptsolution: number of major iterations taken 11

memory used=14.61MiB, alloc change=37.00MiB, cpu time=94.00ms, real time=213.00ms, gc time=46.88ms

> 

 

Providing gradient procedure doesn't help with evalhf computaiton for dsolve/numeric based procedures.

> 

graddsolve := proc (x,g)
local base, i, xnew, del;
global nvar,ssdsolve;
#xnew:=Vector(nvar,datatype=float[8]):
xnew:=Array(1..nvar,datatype=float[8]):
base := ssdsolve(x);
for i to nvar do
xnew[i] := x[i]; end do;
for i to nvar do
del := max(1e-5,.1e-6*x[i]);
xnew[i] := xnew[i]+del;
g[i] := (ssdsolve(xnew)-base)/del;
xnew[i] := xnew[i]-del;
end do;
end proc:
> 

CodeTools:-Usage(Optimization:-NLPSolve(nvar,(ssdsolve),[],initialpoint=ic0,[bl,bu],objectivegradient=graddsolve,optimalitytolerance=1e-6)):

NLPSolve: calling NLP solver

NLPSolve: using method=sqp

NLPSolve: number of problem variables 2

NLPSolve: number of nonlinear inequality constraints 0

NLPSolve: number of nonlinear equality constraints 0

NLPSolve: number of general linear constraints 0

NLPSolve: feasibility tolerance set to 0.1053671213e-7

NLPSolve: optimality tolerance set to 0.1e-5

NLPSolve: iteration limit set to 50

NLPSolve: infinite bound set to 0.10e21

NLPSolve: trying evalhf mode

NLPSolve: trying evalf mode

attemptsolution: number of major iterations taken 11

memory used=3.82MiB, alloc change=0 bytes, cpu time=31.00ms, real time=129.00ms, gc time=0ns

> 

 

Calling both RK2 and dsolve based procedures again to check the values and computation meterics

> 

s1RK2:=CodeTools:-Usage(Optimization:-NLPSolve(nvar,evalhf(ssRK2),[],initialpoint=ic0,[bl,bu],objectivegradient=gradRK2,optimalitytolerance=1e-6)):

NLPSolve: calling NLP solver

NLPSolve: using method=sqp

NLPSolve: number of problem variables 2

NLPSolve: number of nonlinear inequality constraints 0

NLPSolve: number of nonlinear equality constraints 0

NLPSolve: number of general linear constraints 0

NLPSolve: feasibility tolerance set to 0.1053671213e-7

NLPSolve: optimality tolerance set to 0.1e-5

NLPSolve: iteration limit set to 50

NLPSolve: infinite bound set to 0.10e21

NLPSolve: trying evalhf mode

attemptsolution: number of major iterations taken 11

memory used=185.09KiB, alloc change=0 bytes, cpu time=16.00ms, real time=95.00ms, gc time=0ns

> 

s1dsolve:=CodeTools:-Usage(Optimization:-NLPSolve(nvar,ssdsolve,[],initialpoint=ic0,[bl,bu],objectivegradient=graddsolve,optimalitytolerance=1e-6)):

NLPSolve: calling NLP solver

NLPSolve: using method=sqp

NLPSolve: number of problem variables 2

NLPSolve: number of nonlinear inequality constraints 0

NLPSolve: number of nonlinear equality constraints 0

NLPSolve: number of general linear constraints 0

NLPSolve: feasibility tolerance set to 0.1053671213e-7

NLPSolve: optimality tolerance set to 0.1e-5

NLPSolve: iteration limit set to 50

NLPSolve: infinite bound set to 0.10e21

NLPSolve: trying evalhf mode

NLPSolve: trying evalf mode

attemptsolution: number of major iterations taken 11

memory used=3.82MiB, alloc change=0 bytes, cpu time=31.00ms, real time=122.00ms, gc time=0ns

> 

s1RK2[1];s1dsolve[1];

-.523900304316377463

-.523901163953022553

> 

 

Next, a for loop is written to optimize for increasing values of nvar. One can see that evalhf is important for larger values of nvar. While dsolve/numeric is a superior code, not being able to use it in evalhf format is a significant weakness and should be addressed. Note that dsolve numeric evalutes procedures that are evaluated in evalhf or compiled form, so hopefully this is an easy fix.

> 

infolevel[Optimization]:=0:
> 

for j from 1 to 9 do
nvar:=2^(j-1):
ic0:=Vector(nvar,[seq(0.1,i=1..nvar)],datatype=float):
bl := Vector(nvar,[seq(0.,i=1..nvar)],datatype=float):bu := Vector(nvar,[seq(5.,i=1..nvar)],datatype=float):
soptRK[j]:=CodeTools:-Usage(Optimization:-NLPSolve(nvar,evalhf(ssRK2),[],initialpoint=ic0,[bl,bu],objectivegradient=gradRK2,optimalitytolerance=1e-6)):
print(2^(j-1),soptRK[j][1]);
> 

od:

memory used=88.40KiB, alloc change=0 bytes, cpu time=0ns, real time=10.00ms, gc time=0ns

1, HFloat(-0.5008626988793192)

memory used=155.66KiB, alloc change=0 bytes, cpu time=16.00ms, real time=30.00ms, gc time=0ns

2, -.523900304316377463

memory used=175.36KiB, alloc change=0 bytes, cpu time=62.00ms, real time=80.00ms, gc time=0ns

4, -.535152497919956782

memory used=243.69KiB, alloc change=0 bytes, cpu time=156.00ms, real time=239.00ms, gc time=0ns

8, -.540546896131004706

memory used=260.09KiB, alloc change=0 bytes, cpu time=141.00ms, real time=260.00ms, gc time=0ns

16, -.542695734426874465

memory used=385.84KiB, alloc change=0 bytes, cpu time=313.00ms, real time=545.00ms, gc time=0ns

32, -.542932877726400531

memory used=0.65MiB, alloc change=0 bytes, cpu time=734.00ms, real time=1.18s, gc time=0ns

64, -.543011976841572652

memory used=1.24MiB, alloc change=0 bytes, cpu time=1.55s, real time=2.40s, gc time=0ns

128, -.543035276649319276

memory used=2.99MiB, alloc change=0 bytes, cpu time=3.45s, real time=5.92s, gc time=0ns

256, -.543041496228812814

> 

for j from 1 to 6 do
nvar:=2^(j-1):
ic0:=Vector(nvar,[seq(0.1,i=1..nvar)],datatype=float):
bl := Vector(nvar,[seq(0.,i=1..nvar)],datatype=float):bu := Vector(nvar,[seq(5.,i=1..nvar)],datatype=float):
soptdsolve[j]:=CodeTools:-Usage(Optimization:-NLPSolve(nvar,evalf(ssdsolve),[],initialpoint=ic0,[bl,bu],objectivegradient=graddsolve,optimalitytolerance=1e-6)):
print(2^(j-1),soptdsolve[j][1]);
> 

od:

memory used=0.66MiB, alloc change=0 bytes, cpu time=16.00ms, real time=11.00ms, gc time=0ns

1, HFloat(-0.5008631947224957)

memory used=3.79MiB, alloc change=0 bytes, cpu time=15.00ms, real time=52.00ms, gc time=0ns

2, -.523901163953022553

memory used=21.28MiB, alloc change=0 bytes, cpu time=78.00ms, real time=236.00ms, gc time=0ns

4, -.535153942647626391

memory used=144.82MiB, alloc change=-4.00MiB, cpu time=469.00ms, real time=1.60s, gc time=46.88ms

8, -.540549407239521607

memory used=347.30MiB, alloc change=16.00MiB, cpu time=1.27s, real time=3.45s, gc time=140.62ms

16, -.542699055038344258

memory used=1.33GiB, alloc change=-8.00MiB, cpu time=7.66s, real time=15.92s, gc time=750.00ms

32, -.542936165630524603

> 

 

> 

SS:=[seq(soptRK[j][1],j=1..9)];

[HFloat(-0.5008626988793192), -.523900304316377463, -.535152497919956782, -.540546896131004706, -.542695734426874465, -.542932877726400531, -.543011976841572652, -.543035276649319276, -.543041496228812814]

> 

 

> 

E1:=[seq(SS[i]-SS[i+1],i=1..nops(SS)-1)];

[HFloat(0.02303760543705824), 0.11252193603580e-1, 0.5394398211048e-2, 0.2148838295869e-2, 0.237143299527e-3, 0.79099115172e-4, 0.23299807746e-4, 0.6219579494e-5]

To get 6 Digits accuracy we need nvar = 256 which may not be attainable with dsolve/numeric approach unless we are able to call it evalhf format.

> 

 

> 

 


 

Download ImportanceofevalhfNLPSolve.mw

Why a separate input line is required to convert i...

Asked by: C_R 3332
convert
June 11 2023
1  6

Eval(`&varphi;`(t), t = 0) = `&varphi;__0`, Eval(diff(`&varphi;`(t), t), t = 0) = 0

Eval(varphi(t), t = 0) = varphi__0, Eval(diff(varphi(t), t), t = 0) = 0

 (1)

`~`[value](convert({Eval(diff(varphi(t), t), t = 0) = 0, Eval(varphi(t), t = 0) = varphi__0}, D))

{eval((D(varphi))(t), t = 0) = 0, varphi(0) = varphi__0}

 (2)

map(value, convert({Eval(diff(varphi(t), t), t = 0) = 0, Eval(varphi(t), t = 0) = varphi__0}, D))

{eval((D(varphi))(t), t = 0) = 0, varphi(0) = varphi__0}

 (3)

convert({Eval(diff(varphi(t), t), t = 0) = 0, Eval(varphi(t), t = 0) = varphi__0}, D); `~`[value](%)

{varphi(0) = varphi__0, (D(varphi))(0) = 0}

 (4)

NULL

I have expected elementwise and map to be effective. Why aren't they for this example?

(Evaluation at a point (see help(D)): In output (3) it is the first time that I see a composition of the D operator with the Eval function (vertical bar). Is that an undocumented feature or part of an answer that I could not figure out myself?)

Download Convert_inert_IC_to_D_notation.mw

System requirements / older versions...

Asked by: Axel Vogt 5936
version
June 11 2023
1  2

I am aware of https://www.maplesoft.com/products/system_requirements.aspx which shows requirements for the concurrent version (the old page https://www.maplesoft.com/products/roadmap.aspx does no longer exist)

Does anybody know a list or URL telling which Maple version (supposedly) needs what Windows version and which combinations will not work?

I am asking because I switch to a new NB and consider to setup a virtual machine for old programs.

How do you color a parametric curve as a function ...

Asked by: Jaqr 90
spacecurve plotting colorscheme
June 11 2023
0  2

Hi,

Here is the code I would like to complete:

with(plots):
theta:= s -> -2*Pi*n*s/ell;
v:=s->piecewise(And(0<=s,s<=ell/2), c*s, ell/2<s, c*(ell-s));
ell:=1;
c:=1;
n:=2;
spacecurve([cos(theta(s)),v(s),sin(theta(s))],s=0..ell,colorscheme = ["valuesplit",???????]);

What to put in place of the question marks so that the color of the curve is, for example, red if s is between 0 and ell/2, and green if s is between ell/2 and ell?

Thanks.

Error when I change discrete set to be an interval...

Asked by: Zeineb 540
interval domain
June 10 2023
0  9

Dear all

I have a code that compute the elements of the sigma algebra generated by C in the domain X 
If X is a discrete set and C is also a discrete set 

I get a good results 

But in change X to be an interval and C will be set of three elements

No result obtained 

sigma_algebra.mw

Thank you

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