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mmcdara

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Gve more details......

mmcdara 7601
December 14 2024
0 0

@emrantohidikub 

... but I can't guarantee that I'll be capable to answer your questions, or even that I'll have the time to do it if it requires a lot of work.

Rotational kinetic energy, contact, and ...

mmcdara 7601
December 14 2024
0 0

Not accounting for the rotational kinetic energy and assuming the ball has allway a single contact point with the surface is a huge approximation. Even in the absence of gravity (let's say the ball has an internal engine), a single contact requires that the curvature radius of the surface is everywhere larger than the radius of the ball.
More og that the displacement of the ball may exhibt pure rolling, pure sliding, or a a mix of these twos depending of the value of the contat force.

I aggree with Rouben saying it is quite a challenge to account for all these points in the most general way (even if there are probably a lot of papers online about this subject).

For information @C_R  already did something close with MapleSim, see here

old fashion...

mmcdara 7601
December 13 2024
0 0

@acer 

I asked years ago a question about the old fashion way Maple used to represent lengthy expression (using things like %1, %2, ... in the global expression and adding lines to the output of the form %1=something, %2=something-else, ...).
I couldn't find the answer I received but this could lead to something more readable here.

WHY...

mmcdara 7601
December 13 2024
0 0


didn't you just do what I suggested you HERE

MatlabFile := cat(currentdir(), "/STX.txt"):   # change the path and name as you want
ExportMatrix(MatlabFile, STX, target=MATLAB, format=rectangular, mode=ascii, format=entries);

The end of your code is

cat(currentdir()); # Which is a directory NOT a file
                    "C:\Program Files\Maple 2024"

# Below matlabData should be the name of a file: as its not you get an error
# The warning message results from STX being a matrix of complex.

ExportMatrix(matlabData, STX, target = MATLAB, format = rectangular, mode = ascii);

Warning, Matlab may not know how to read complex values stored in 'rectangular' format; try 'entries' format if problems arise
Error, (in ExportMatrix) permission denied

#  matlabData is unassigned
matlabData;
                 matlabData;

Begin to assign matlabData to a file with a legit absolute path.
Next use format=entries as I already told you (if you don't understand why read carefully the ExportMatrix help page): this will suppress the warning.

I advice you to read the help pages about cat and currentdir() too.

PS: Don't be surprised if a moderator deletes this question, which obviously duplicates the one I've already answered.

 



 

A suggestion...

mmcdara 7601
December 13 2024
0 0

@salim-barzani 

Look at he ExportMatrix help page, items MATLAB® Binary Format or MATLAB® ASCII Format.
As you want to export a matrix of complex search (CTRL+F) for the word entries.

Here is an ASCII example

MatlabFile := cat(currentdir(), "/STX.txt"):   # change the path and name as you want
ExportMatrix(MatlabFile, STX, target=MATLAB, format=rectangular, mode=ascii, format=entries);


As I don't have Matlab at home I can't check it can read and plot the exported matrix correctly.


In case you want avoid using the Finance package, here is a Wiener process generator

 

> 

restart
> 

with(Statistics):
> 

G := RandomVariable(Normal(0, 1))

_R

 (1)
> 

N  := 1000:
T  := 10:
dt := T/N:

S  := Sample(G, N):
dW := sqrt(dt) *~ S:
W := Array(<0>):
C := CumulativeSum(dW):
W := ArrayTools:-Extend(W, CumulativeSum(dW), inplace=true):

plot([seq(0..T, dt)], W, style=line, size=[1000, 400]);

 
> 

# Several Wiener processes

rep := 20:
all := NULL:
col := () -> ColorTools:-Color([seq(rand(0. .. 1.)(), i=1..3)]):

for i from 1 to rep do
  S  := Sample(G, N):
  dW := sqrt(dt) *~ S:
  W := Array(<0>):
  C := CumulativeSum(dW):
  W := ArrayTools:-Extend(W, CumulativeSum(dW), inplace=true):

  all := all, plot([seq(0..T, dt)], W, style=line, size=[1000, 400], color=col()):
end do:

plots:-display(all)

 
> 

 


 

Download WienerProcessesGenerator.mw

 

Your initial code with W(t) plugged in...

mmcdara 7601
December 12 2024
0 0

@salim-barzani 

Is this what you want (left = Re, right = Im)
graph-stochastic_mmcdara.mw
 

@dharr Thanks for your explanation....

mmcdara 7601
December 06 2024
0 0

@dharr 

Thanks for your explanation.
See you next time.

By the way: congratulations, you've passed me :-)

@Ronan  The denomintor then is&nbs...

mmcdara 7601
December 06 2024
0 0

@Ronan 

The denomintor then is 100*n^(33/20).

For smart display only...

mmcdara 7601
November 25 2024
0 0
> 

restart:
> 

edp := diff(u(x, t), t)+lambda*diff(u(x, t), x$2) = s

diff(u(x, t), t)+lambda*(diff(diff(u(x, t), x), x)) = s

 (1)
> 

eDp := convert(edp, D):
inD := select(has, indets(eDp), D):
rw  := map(i -> i =  eval(op([0, 0], i), {D=u, 1=x, 2=t}), inD):

rwedp := eval(eDp, rw);

lambda*u[x, x]+u[t] = s

 (2)
> 

original_edp := convert(eval(rwedp, (rhs=lhs)~(rw)), diff)

diff(u(x, t), t)+lambda*(diff(diff(u(x, t), x), x)) = s

 (3)
> 

# Or, if you want to get rid of the commas:

rw  := map(i -> i = cat('u__', eval(op(op([0, 0], i)), {1=x, 2=t})), inD):
rwedp := eval(eDp, rw);

lambda*u__xx+u__t = s

 (4)
 

 

Download smart_display.mw

Data file...

mmcdara 7601
November 25 2024
0 0

Is located in your own directories.
Please upload the file or provide us some link we can acccess.

Alread: writting this in your procedure SSE is illegal

.
.
.
local solution := dsolve({ode1,ode2,S(0) = S0,T(0) = i0},parameters=[beta,Gamma,S0,i0],[S(t),T(t)],numeric);
solution(parameters = [b,g,x,y]);
local difference := add((rhs(solution(k)[3])-sum_total[k])^2,k=1..N);
return difference;
end proc:

A correct syntaxt is

.
.
.
local solution := dsolve({ode1,ode2,S(0) = S0,T(0) = i0},parameters=[beta,Gamma,S0,i0],[S(t),T(t)],numeric);
solution(parameters = [b,g,x,y]);
add((rhs(solution(k)[3])-sum_total[k])^2,k=1..N);
end proc:


Last but not least: if you expect finding 4 parameters such that the brown curve is close to the empirical data, this is a dead end: don't need to be a great scientist to understand why.

@Carl Love  @dharr @acerT...

mmcdara 7601
November 23 2024
0 0

@Carl Love  @dharr @acer

Thanks you all for your involvement.
I particularly appreciated Carl's explanations and embedded comments by Dharr (I missed the usedpts=1044 for method=_MonteCarlo).

There are stlll a few messages that bother me.

For instance method=_CubaSuave returns 
cuba: chi-square probability that the error is not reliable: 1... which means that there are 100% of chances that the error (cuba: estimated (absolute) error: 6.73167e-05) ! Are you sure the first message is correct?

All the more that method=_CubaVegas gives
cuba: estimated (absolute) error: 6.58857e-05
cuba: chi-square probability that the error is not reliable: 0.0023283
and that CubaSuave is aimed to be an improvement of CubaVegas (see  Hahn for instance)

By the way, as CubaVegas uses a Quasi Random Numbers Generator [QRNG] (the Sobol' one, see previous ref) don't you think it would have been a good idea for the core development team to also offer this quasi-random number generator (see NAG library QMC) ?

(For the record there exist several QRNG and I implemented Faure's in Maple a few years ago. But I believe fast built-in functions for QRNG would be welcome.)

 

@segfault Good, I'm glad I coul...

mmcdara 7601
November 19 2024
0 0

@segfault 

Good, I'm glad I could help.

Can you pay attention to what you write?...

mmcdara 7601
November 19 2024
0 0

@Aung 

YourModel :

epsilon[0] + epsilon[0] * alpha[theta] * B[1] * (-1 + exp(-beta[1] * t)) 
           + epsilon[0] * alpha[theta] * B[2] * (-1 + exp(-beta[2] * t))
           + epsilon[0] * alpha[theta] * B[3] * (-1 + exp(-beta[3] * t))

Where  epsilon[0] > 0 and alpha[theta] > 0
As you impose B[1] >= 0, B[2] >= 0, B[3] >= 0, beta[1] >= 0, beta[2] >= 0, beta[3] >= 0, 
the 3 last terms are always negative (t >=0) or null in the "best" case.
Given that epsilon[0] = 0.5581381911e-2 the maximum value of your model cannot exceed this value:
So how can you even think this model could represent your data?




My advice, and do with it what you will, is summarized in the attached file:
(the model I use is similar to yours after sign correction)
 

> 

restart;
> 

with(Statistics):with(plots):with(Optimization):with(LinearAlgebra):
> 


# given data from strain rate curve
E_0[theta] := 7.883352314*10^9;
alpha[theta]:= 0.982

7883352314.

  

.982

 (1)
> 


# experimental creep data under 44 at 100 degree celcius
c_strain := Vector ([<<0>,<0.0284698>,<0.0533808>,<0.0782918>,<0.0996441>,<0.124555>,<0.142349>,<0.156584>,<0.16726>,<0.177936>,<0.181495>,<0.188612>,<0.192171>,<0.19573>,<0.19573>,<0.202847>,<0.206406>,<0.206406>,<0.209964>,<0.209964>,<0.209964>,<0.206406>,<0.209964>>]):

c_time := Vector ([<<0>,<0>,<0.048>,<0.192>,<0.352>,<0.544>,<0.704>,<0.896>,<1.088>,<1.312>,<1.52>,<1.76>,<1.984>,<2.208>,<2.464>,<2.736>,<3.088>,<3.392>,<3.664>,<4.016>,<4.352>,<4.592>,<4.832>>]):
sigma[0] := 44*10^6;
epsilon[0] := sigma[0]/E_0[theta];

44000000

  

0.5581381911e-2

 (2)
> 


# change vector to list
c_strain := convert(c_strain,list):
c_time := convert(c_time,list):

# extract zero from list
c_strain := c_strain [2..-1];
c_time := c_time [2..-1];

[0.284698e-1, 0.533808e-1, 0.782918e-1, 0.996441e-1, .124555, .142349, .156584, .16726, .177936, .181495, .188612, .192171, .19573, .19573, .202847, .206406, .206406, .209964, .209964, .209964, .206406, .209964]

  

[0, 0.48e-1, .192, .352, .544, .704, .896, 1.088, 1.312, 1.52, 1.76, 1.984, 2.208, 2.464, 2.736, 3.088, 3.392, 3.664, 4.016, 4.352, 4.592, 4.832]

 (3)
> 


# for further calculation need to know how many elements are in the list
M := nops(c_strain);
N := nops(c_time);

22

  

22

 (4)


WHAT I SUGGESTED YOU

> 

CreepCompliance := (t, b) ->  (1 -exp(-b*t)) / b;

NComponentModel := n -> unapply( shift + add((s||i)*CreepCompliance(t, b||i) , i=1..n), t):

NComponentModel(3);

eps := 1e-6:

Phi := 0;
obj := add( ( NComponentModel(3)~(c_time) - c_strain )^~2 )
           +
           Phi / (1e-4+add( (s||i - add(s||j, j=1..3)/3)^2, i=1..3)):

 opt_3 := Optimization:-NLPSolve( obj, {seq(s||i >= eps, i=1..3), seq(b||i >= eps, i=1..3)}, iterationlimit=1000);

model_3 := eval(NComponentModel(3)(t), (opt_3)[2]):

[ seq(isolate(epsilon[0]*alpha[theta]*B[i] = eval(s||i / b||i, opt_3[2]), B[i]), i=1..3) ];

plots:-display(
   Statistics:-ScatterPlot(c_time, c_strain, symbol=circle)
   , plot(model_3, t=0. .. max(c_time), color=red,  legend="3 components")
);

proc (t, b) options operator, arrow; (1-exp(-b*t))/b end proc

  

proc (t) options operator, arrow; shift+s1*(1-exp(-b1*t))/b1+s2*(1-exp(-b2*t))/b2+s3*(1-exp(-b3*t))/b3 end proc

  

0

  

[0.225314772163480432e-3, [b1 = HFloat(1.3084255135648692), b2 = HFloat(1.3084255135598823), b3 = HFloat(1.308425513559216), s1 = HFloat(0.07458889456019333), s2 = HFloat(0.07458889455231471), s3 = HFloat(0.07458889455084408), shift = HFloat(0.03700542692440384)]]

  

[B[1] = HFloat(10.400924369210232), B[2] = HFloat(10.400924368151253), B[3] = HFloat(10.40092436795148)]

  

 
> 

Phi := 1;
obj := add( ( NComponentModel(3)~(c_time) - c_strain )^~2 )
           +
           Phi / (1e-4+add( (s||i - add(s||j, j=1..3)/3)^2, i=1..3)):

 opt_3 := Optimization:-NLPSolve( obj, {seq(s||i >= eps, i=1..3), seq(b||i >= eps, i=1..3)}, iterationlimit=1000);

model_3 := eval(NComponentModel(3)(t), (opt_3)[2]):

[ seq(isolate(epsilon[0]*alpha[theta]*B[i] = eval(s||i / b||i, opt_3[2]), B[i]), i=1..3) ];

plots:-display(
   Statistics:-ScatterPlot(c_time, c_strain, symbol=circle)
   , plot(model_3, t=0. .. max(c_time), color=red,  legend="3 components")
);

1

  

[0.934443784351978465e-4, [b1 = HFloat(370724.4386808285), b2 = HFloat(1.2325972978366215), b3 = HFloat(817606.2552664457), s1 = HFloat(4721.29039619488), s2 = HFloat(0.2034032484701095), s3 = HFloat(2179.004303789217), shift = HFloat(0.028462609697833815)]]

  

[B[1] = HFloat(2.3235727208121157), B[2] = HFloat(30.108106594130508), B[3] = HFloat(0.48625116430224813)]

  

 
> 

 

 


 

Download Accept_it_or_not_but_I_am_done.mw

A suggestion...

mmcdara 7601
November 14 2024
0 0

@C_R 

Let F =~ 0.
Convert F into exponential and apply the substitution

X2Y := { exp(I*x4) = y4, exp((I*x5) = y5}

to the result.
This will avoid the square roots.
Then F is a rational fraction whose denominator, equal to y4*y5, is not null (y4 and Y( are complex numbers). So consider the numerator N (N =~ 0)
Maybe this could this could make Isolate's task easier ???

Another idea: instead of solving {F1, F2, F3, F4, F5} try solving {F1, F2-F1, F3-F1, F4-F1, F5-F1}

Does this suit you?...

mmcdara 7601
November 09 2024
0 0

@AHSAN 

> 

restart
> 

``
> 

P0 := -6*k/((k+1)*(k-1)*(k*x-k-x)^2)-6/((k-1)*(k*x-k-x))-6/(k^2-1):
> 

lambda[0] := k/(k+1):
> 

P1 := -6*x*((x-1)^4*(A+(-1/3-4*Br*(1/21))*m)*k^8-(1/5)*(4*(((A*beta+15*Br*m*(1/56))*lambda[0]^3+(-9*A*beta*(1/5)-15*Br*m*(1/28))*lambda[0]^2+4*A*beta*lambda[0]*(1/3)+(-7*beta*(1/18)+5)*A-(1/21)*(20*(Br+7/4))*m)*x+(-2*A*beta-15*Br*m*(1/28))*lambda[0]^3+(18*A*beta*(1/5)+15*Br*m*(1/14))*lambda[0]^2-8*A*beta*lambda[0]*(1/3)+7*A*beta*(1/9)))*(x-1)^3*k^7+12*(x-1)^2*(((A*beta+15*Br*m*(1/56))*lambda[0]^3+(-9*A*beta*(1/5)-5*Br*m*(1/7))*lambda[0]^2+4*A*beta*lambda[0]*(1/3)+(-14*beta*(1/27)+5/2)*A-(1/21)*(10*(Br+7/4))*m)*x^2+((-A*beta-15*Br*m*(1/56))*lambda[0]^3+(9*A*beta*(1/5)+15*Br*m*(1/14))*lambda[0]^2-4*A*beta*lambda[0]*(1/3)+7*A*beta*(1/9))*x+(-4*A*beta*(1/3)-5*Br*m*(1/14))*lambda[0]^3+(12*A*beta*(1/5)+5*Br*m*(1/28))*lambda[0]^2-16*A*beta*lambda[0]*(1/9)+7*A*beta*(1/54))*k^6*(1/5)-(1/5)*(12*(((A*beta+5*Br*m*(1/28))*lambda[0]^3+(-9*A*beta*(1/5)-15*Br*m*(1/14))*lambda[0]^2+8*A*beta*lambda[0]*(1/9)+(-7*beta*(1/9)+5/3)*A-(1/63)*(20*(Br+7/4))*m)*x^3+((5/14)*Br*m*lambda[0]^3+(15/14)*Br*m*lambda[0]^2+(16/9)*A*beta*lambda[0]+(7/9)*A*beta)*x^2+((-A*beta-45*Br*m*(1/56))*lambda[0]^3+(9*A*beta*(1/5)+15*Br*m*(1/28))*lambda[0]^2-4*A*beta*lambda[0]+7*A*beta*(1/18))*x-2*lambda[0]*((A*beta+5*Br*m*(1/56))*lambda[0]^2-9*A*beta*lambda[0]*(1/5)+4*A*beta*(1/9))))*(x-1)*k^5+(((4*A*beta*(1/5)-3*Br*m*(1/7))*lambda[0]^3+(-72*A*beta*(1/25)-12*Br*m*(1/7))*lambda[0]^2-32*A*beta*lambda[0]*(1/15)+(1-56*beta*(1/45))*A-(1/21)*(4*(Br+7/4))*m)*x^4+((15*Br*m*(1/7)+4*A*beta*(1/5))*lambda[0]^3+(6*Br*m*(1/7)+144*A*beta*(1/25))*lambda[0]^2+32*A*beta*lambda[0]*(1/3)+28*A*beta*(1/45))*x^3+((-15*Br*m*(1/14)+4*A*beta*(1/5))*lambda[0]^3+(-396*A*beta*(1/25)+9*Br*m*(1/7))*lambda[0]^2-16*A*beta*lambda[0]*(1/3)+14*A*beta*(1/15))*x^2+(1/5)*(4*((-75*Br*m*(1/56)+A*beta)*lambda[0]^2+81*A*beta*lambda[0]*(1/5)-20*A*beta*(1/3)))*lambda[0]*x-32*A*lambda[0]^2*(-27/40+lambda[0])*beta*(1/5))*k^4+(((4*A*beta*(1/5)+9*Br*m*(1/14))*lambda[0]^3+(3*Br*m*(1/7)+108*A*beta*(1/25))*lambda[0]^2+16*A*beta*lambda[0]*(1/5)+14*A*beta*(1/45))*x^4+((-4*A*beta-9*Br*m*(1/14))*lambda[0]^3+(-324*A*beta*(1/25)+3*Br*m*(1/7))*lambda[0]^2-16*A*beta*lambda[0]*(1/5)+14*A*beta*(1/45))*x^3+8*lambda[0]*((A*beta-3*Br*m*(1/28))*lambda[0]^2+27*A*beta*lambda[0]*(1/25)-8*A*beta*(1/15))*x^2-8*A*lambda[0]^2*(lambda[0]-27/25)*beta*x-16*A*beta*lambda[0]^3*(1/5))*k^3-12*x*lambda[0]*(((A*beta+5*Br*m*(1/56))*lambda[0]^2+9*A*beta*lambda[0]*(1/5)+4*A*beta*(1/9))*x^3+((-3*A*beta+5*Br*m*(1/56))*lambda[0]^2-9*A*beta*lambda[0]*(1/5)+4*A*beta*(1/9))*x^2+2*A*lambda[0]*(lambda[0]-6/5)*beta*x+2*A*beta*lambda[0]^2)*k^2*(1/5)+12*x^2*A*((lambda[0]+3/5)*x^2+(-lambda[0]+3/5)*x-4*lambda[0]*(1/3))*lambda[0]^2*beta*k*(1/5)-4*A*x^3*beta*lambda[0]^3*(x+1)*(1/5))*(x-1)*(k-1)/(k^4*(k+1)*((x-1)*k-x)^6):
> 

PP := P1*epsilon+P0:
> 

ind := [indets(PP, name)[]];

[A, Br, beta, epsilon, k, m, x]

 (1)
> 

ind := convert(indets(PP) minus {x}, list);

epsilon_values := [.1, .3, .5, .7, .9];
k_values       := [seq(i, i = 2.5 .. 20, .5)];

data := (eval, kval) -> ind =~ [0.5, 0.2, 0.5, eval, kval, 1.5]:

# example

data(.2, 2.5)

[A, Br, beta, epsilon, k, m]

  

[.1, .3, .5, .7, .9]

  

[2.5, 3.0, 3.5, 4.0, 4.5, 5.0, 5.5, 6.0, 6.5, 7.0, 7.5, 8.0, 8.5, 9.0, 9.5, 10.0, 10.5, 11.0, 11.5, 12.0, 12.5, 13.0, 13.5, 14.0, 14.5, 15.0, 15.5, 16.0, 16.5, 17.0, 17.5, 18.0, 18.5, 19.0, 19.5, 20.0]

  

[A = .5, Br = .2, beta = .5, epsilon = .2, k = 2.5, m = 1.5]

 (2)
> 

KE     := numelems(k_values)*numelems(epsilon_values);
AllMax := Matrix(KE, 3):

counter := 1:
for varepsilon in epsilon_values do
  for kappa in k_values do
    __PP             := eval(PP, data(varepsilon, kappa));
    Pre_prime        := diff(__PP, x);
    Pre_double_prime := diff(Pre_prime, x);
    critical_points  := solve(Pre_prime = 0, x, real, maxsols = 4);

    local_max := -infinity:
    for cp in [critical_points] do
      if eval(Pre_double_prime, x=cp) < 0 then
        local_max := max(local_max, eval(__PP, x=cp))
      end if:
    end do:
    AllMax[counter, 1] := varepsilon:
    AllMax[counter, 2] := kappa:
    AllMax[counter, 3] := local_max:
    counter := counter+1:
  end do:
end do;

AllMax;

Statistics:-ScatterPlot3D(AllMax, labels=[``(epsilon), ``(k), ``])

KE := 180

  

Matrix(%id = 18446744078491512094)

  

 
> 

col := 1;
map(ke -> if AllMax[ke, col]=0.3 then AllMax[ke, [3-col, 3]] end if, [$1..KE]);

col := 1

  

[Vector[row](%id = 18446744078473882910), Vector[row](%id = 18446744078473883030), Vector[row](%id = 18446744078473883150), Vector[row](%id = 18446744078473883270), Vector[row](%id = 18446744078473883390), Vector[row](%id = 18446744078473883510), Vector[row](%id = 18446744078473871358), Vector[row](%id = 18446744078473871478), Vector[row](%id = 18446744078473871598), Vector[row](%id = 18446744078473871718), Vector[row](%id = 18446744078473871838), Vector[row](%id = 18446744078473871958), Vector[row](%id = 18446744078473872078), Vector[row](%id = 18446744078473872198), Vector[row](%id = 18446744078473872318), Vector[row](%id = 18446744078473872438), Vector[row](%id = 18446744078473872558), Vector[row](%id = 18446744078473872678), Vector[row](%id = 18446744078473872798), Vector[row](%id = 18446744078473872918), Vector[row](%id = 18446744078473873038), Vector[row](%id = 18446744078473873158), Vector[row](%id = 18446744078473873278), Vector[row](%id = 18446744078473873398), Vector[row](%id = 18446744078473873518), Vector[row](%id = 18446744078473873638), Vector[row](%id = 18446744078473874478), Vector[row](%id = 18446744078473874598), Vector[row](%id = 18446744078473874718), Vector[row](%id = 18446744078473874838), Vector[row](%id = 18446744078473874958), Vector[row](%id = 18446744078473875078), Vector[row](%id = 18446744078473875198), Vector[row](%id = 18446744078473875318), Vector[row](%id = 18446744078473867262), Vector[row](%id = 18446744078473867382)]

 (3)
> 

CurveBundle := proc()
  description "arg 1: fixed parameter\narg 2: fixed value\narg 3: color\narg 4: legend location";

  local minames, col, val, pts:

  minames := [`#mi("&epsilon;")`, `#mi("k")`];

  if _passed[1] = epsilon then
    col := 1:
  elif _passed[1] = k then
    col := 2:
  else
    error "first parameter passed should be epsilon or k"
  end if:

  if _passed[1] = epsilon
     and
     `not`(member(_passed[2], convert(epsilon_values, set))) then
     error cat(_passed[2], " is not a value in epsilon_values")
  end if:

  if _passed[1] = k
     and
     `not`(member(_passed[2], convert(k_values, set))) then
     error cat(_passed[2], " is not a value in k_values")
  end if:
  
  # same type of code could be written to ask for the value of epsilon or k
  # (I'm too lazzy to write it)

  val := _passed[2]:
  pts := map(ke -> if AllMax[ke, col]=val then convert(AllMax[ke, [3-col, 3]], list) end if, [$1..KE]);

  return plot(
           pts
           , color=_passed[3]
           , labels=[minames[3-col], "Max"]
           , legend=typeset(minames[col]=val)
           , legendstyle=[location=_passed[4]]
         )

end proc:
  
> 

Describe(CurveBundle):


# arg 1: fixed parameter
# arg 2: fixed value
# arg 3: color
# arg 4: legend location
CurveBundle( )

  
> 

CurveBundle(epsilon, 3.14, red, right);
PlotOneCurve(k, 3.14, red, right);

Error, (in CurveBundle) 3.14 is not a value in epsilon_values

  

Error, (in PlotOneCurve) 3.14 is not a value in k_values

  
> 

RandColor := () -> ColorTools:-Color([seq(rand(0. .. 1.)(), i=1..3)]):

plots:-display(
  seq(CurveBundle(epsilon, val, RandColor(), right), val in epsilon_values)
  , size=[500, 400]
)

 
> 

RandColor := () -> ColorTools:-Color([seq(rand(0. .. 1.)(), i=1..3)]):

plots:-display(
  seq(CurveBundle(k, val, RandColor(), right), val in k_values[[seq](1..numelems(k_values), 2)])
  , size=[600, 400]
)

 
 

 

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