Maple Questions and Posts

These are Posts and Questions associated with the product, Maple

Dear Maplesoft,

I inquired about this problem 4 years ago, but never really was able to fix my problem based on the response at the time. This has to do with plotting a parametrized curve where the parametrization involves the numerical solution of a condition. 

Consider the family of cardioids  
           "r = 1 + c*sin(theta), theta = 0 .. 2*Pi"
 in polar coordinates for 
                         "c = 0 .. 2.5"

In this example we find the polar angle 
                           "theta(c)"
 on the evolving family of cardioids where the slope is 
                              "-1"
 as a function of the shape parameter 
                              "c"
 of this family by a procedure involving fsolve, but then try to plot the parametrized curve 
                  "r(c) = 1 + c*sin(theta(c))"
. No direct plotting method works because of evaluation order problems that I do not understand. The first plot is my desired plot but I used an ugly workaround to get the gray curve. Can you fix the direct method with delayed evaluation or something? 

Maplesoft Response. We don't help with this kind of problem. Ask MaplePrimes. 

This evaluation order problem pops up every time you want to plot a curve determined by numerically solving a condition, yet Maplesoft seems to think this is too sophisticated a problem to respond to. Naively trying to animate such curves always derails, so it reveals a weakness of Maple for users who do not belong to the elite class of Maple experts. I have been using Maple for a quarter century, and have made some pretty intricate animations and plots over the years, but always run up against this problem with animating numerically determined curves. Is there a Maple pro out there who can help?
Since I can't find a way to attach my Maple worksheet, here is the URL:
http://www34.homepage.villanova.edu/robert.jantzen/maple/misc/cardioidfamily.mw

 

Hi, I have the following problem:

I want to plot the cone given by 1/16*(3x^2+10xz+3z^2-16y^2) and x>=0, z>=0. I tried it with

implicitplot3d([1/16*(3*x^2+10*x*z-16*y^2+3*z^2), x >= 0, z >= 0], x = -5 .. 15, y = -15 .. 15, z = -5 .. 15, grid = [30, 30, 30], style = surface);

But the result is one surface for each inequaility and not the cone.

If i restrict the range of x and z to be 0..15 and dismiss the additonal inequalities, a big part of the cone surface is missing somehow:

Does anybody know how to fix this? Do I have to use another plotcommand?

Thanks for your help!

Hi, 

This thread is more or less related to a previous one about the Statistics:-Sample procedure.
(see https://www.mapleprimes.com/questions/228421-A-Serious-Problem-With-StatisticsSample )
I've just implemented two variant of the Box-Muller procedure to sample normal rvs.
The source is "The art of computer programming", Donald E. Knuth, 2nd edition, p117 (aka "algorithm P").

The first implementation (BoxMuller_1) is basically what D.E. Knuth writes, except that I "vectorize" some operations in order to avoid using an if.. then..else structure (as a minor consequence I generate a little bit more numbers than required).
This procedure uses the build-in Maple's procedure select (please see the link above and acer's and carl love's replies).
It appears to be relatively slow.
More of this CodeTools:-Usage(BoxMuller_1(10^6)) generates a "conenction to kernel lost" for some unknown reason.

The variant named BoxMuller_2 uses sort and ListTools:-BinaryPlace instead of select.
Ir appears to be around 3 times faster than BoxMuller_1, but remains 10 time slower than Statistics:-Sample(..., method=envelope) (here again, see the link above to understand why method=envelope is needed).

I wonder how it could be possible to speed up this procedure. In particular acer showed in one of his reply (again see the link above) how using hfloats can improve the efficiency of a procedure, but I'm very incompetent on this point.
Does anyone have any idea?

Thanks in advance.

PS: this file has been written with Maple 2015.2


 

restart:

BoxMuller_1 := proc(N)
  local V, S, T, L, X1, X2:
  V  := Statistics:-Sample(Uniform(-1, 1), [ceil(2*N/3), 2]):
  S  := V[..,1]^~2 +~ V[..,2]^~2;
  T  := < subs(NULL=infinity, select(`<`, S, 1)) | V >;
  T[.., 1]  := (-2 *~ log~(T[.., 1]) /~ T[.., 1])^~(0.5);
  X1 := select[flatten](type, T[.., 1] *~  T[..,2], 'float');
  X2 := select[flatten](type, T[.., 1] *~  T[..,3], 'float');
  return <X1 , X2>:
end proc:


BoxMuller_2 := proc(N)
  local V1, V2, S, W, u, r, T, X1, X2:
  V1 := Statistics:-Sample(Uniform(-1, 1), ceil(2*N/3)):
  V2 := Statistics:-Sample(Uniform(-1, 1), ceil(2*N/3)):
  S  := V1^~2 +~ V2^~2;
  W  := sort(S, output = [sorted, permutation]):
  u  := ListTools:-BinaryPlace(W[1], 1);
  r  := [$1..u]:
  T  := (-2 *~ log~(W[1][r]) /~ W[1][r])^~(0.5);
  X1 := T *~ V1[W[2][r]];
  X2 := T *~ V2[W[2][r]];
  return <X1 | X2>^+:
end proc:  

CodeTools:-Usage(Statistics:-Sample(Uniform(-1, 1), 10^5)):
CodeTools:-Usage(Statistics:-Sample(Uniform(-1, 1), 10^5, method=envelope)):
# CodeTools:-Usage(BoxMuller_1(10^6)): # generates a "connection to kernel lost" error msg
CodeTools:-Usage(BoxMuller_2(10^6)):

print():
CodeTools:-Usage(BoxMuller_1(10^5)):
CodeTools:-Usage(BoxMuller_2(10^5)):
print():

S1 := BoxMuller_1(10^5):
S2 := BoxMuller_2(10^5):

memory used=1.17MiB, alloc change=0 bytes, cpu time=12.00ms, real time=12.00ms, gc time=0ns
memory used=3.32MiB, alloc change=32.00MiB, cpu time=64.00ms, real time=64.00ms, gc time=0ns

memory used=148.17MiB, alloc change=119.93MiB, cpu time=706.00ms, real time=637.00ms, gc time=14.63ms

 

 

memory used=67.43MiB, alloc change=186.80MiB, cpu time=946.00ms, real time=762.00ms, gc time=285.14ms
memory used=14.81MiB, alloc change=0 bytes, cpu time=67.00ms, real time=62.00ms, gc time=0ns

 

(1)

if false then
DocumentTools:-Tabulate(
  [
    plots:-display(
      Statistics:-Histogram(S1),
      plot(Statistics:-PDF(Normal(0, 1), x), x=-4..4, color=red,thickness=3)
    ),
    plots:-display(
      Statistics:-Histogram(S2),
      plot(Statistics:-PDF(Normal(0, 1), x), x=-4..4, color=red,thickness=3)
    )
  ], width=60
)
end if:
  

# carl love's procedure with slight modifications t

SampleCheck := proc(X, f, N::posint)
uses St= Statistics;
  local S, M, O, E:
  S:= f(N):
  M:= numelems(S):
  O:= <St:-TallyInto(S, <[$(floor@min-1..ceil@max)(S)]>)>:
  E:= St:-Probability~(X <~ (rhs@lhs)~(O), 'numeric') * M:
  <rhs~(O)[2..] | E[2..] - E[..-2]>
end proc:
 

interface(rtablesize=20):
SampleCheck(Statistics:-RandomVariable(Normal(0,1)), 'BoxMuller_1', 10^5),
SampleCheck(Statistics:-RandomVariable(Normal(0,1)), 'BoxMuller_2', 10^5);

Matrix(11, 2, {(1, 1) = 1, (1, 2) = HFloat(0.029943974977392658), (2, 1) = 5, (2, 2) = HFloat(3.28979552036377), (3, 1) = 121, (3, 2) = HFloat(138.1791685600983), (4, 1) = 2276, (4, 2) = HFloat(2243.2153196005097), (5, 1) = 14488, (5, 2) = HFloat(14245.846696531142), (6, 1) = 35630, (6, 2) = HFloat(35780.43897239682), (7, 1) = 35886, (7, 2) = HFloat(35780.43897239682), (8, 1) = 14071, (8, 2) = HFloat(14245.846696531138), (9, 1) = 2212, (9, 2) = HFloat(2243.2153196005092), (10, 1) = 130, (10, 2) = HFloat(138.1791685601056), (11, 1) = 2, (11, 2) = HFloat(3.2897955203516176)}), Matrix(10, 2, {(1, 1) = 1, (1, 2) = HFloat(3.293310594473029), (2, 1) = 127, (2, 2) = HFloat(138.32680996055558), (3, 1) = 2241, (3, 2) = HFloat(2245.6121457991635), (4, 1) = 14226, (4, 2) = HFloat(14261.068070193269), (5, 1) = 35632, (5, 2) = HFloat(35818.66958395649), (6, 1) = 35864, (6, 2) = HFloat(35818.66958395649), (7, 1) = 14451, (7, 2) = HFloat(14261.068070193272), (8, 1) = 2266, (8, 2) = HFloat(2245.6121457991685), (9, 1) = 125, (9, 2) = HFloat(138.32680996054842), (10, 1) = 1, (10, 2) = HFloat(3.293310594468494)})

(2)

 


 

Download Box-Muller.mw

How to solve this DE by using  Differential transformation method?

diff(f(x),x$3)+1/2 * f(x) *diff(f(x),x$2)=0;

with boundary conditions

  f(0)=1 ; f '(0)= lamda * f ''(0)   and   f ' (x) -> 1  as x -> infinity
where lamda is some constant...

This update fixes the problems inadvertently introduced in Maple 2019.2, namely:

  • Maple failed to run the code in the maple.ini/.mapleinit initialization files when loading existing worksheets containing a restart() command
  • Installing some packages from the MapleCloud was unsuccessful

For anyone who installed the 2019.2 update, installing 2019.2.1 will fix these problems.

If you are at Maple 2019.1 or earlier, installing this update will bring you straight to Maple 2019.2.1.

This update is available through Tools>Check for Updates in Maple, and is also available from our website on the Maple 2019.2.1 download page.

If you are a MapleSim user, please note that these problems do not affect your use of MapleSim. If you use Maple on its own, and if you use Maple command initialization files and/or you need to install a package from the MapleCloud that does not work, please contact Maplesoft Technical Support for assistance.

We sincerely apologize for the inconvenience and thank you for your patience as we worked through this issue.

Explore the values of km_digit(n,m) using km_ for all m,0 < or equal to m and less than or equal to 8. 

Look at the output until you can make a conjecture that concerns the pattern obtained.

Hint use km_list(m,6,20) when m is not a multiple of 4 and km_list(m,6,50) when m is a multiple of 4

Hi, 
The choice method=default (the implicit choice) gives dramatically false results as demonstrated here for the sampling of a normal r.v.
It's likely to be the case for any continuous distribution.
In a few words the tails of the distribution are not correctly sampled (far from the mean by 4 standard deviations for a normal r.v., not an extremely rare event in practical applications).

I think the attached file is sufficiently documented for you to understand the problem.


 

restart:


The problem presented here also exists with Maple 2018 and Maple 2019

interface(version)

`Standard Worksheet Interface, Maple 2015.2, Mac OS X, December 21 2015 Build ID 1097895`

(1)

with(Statistics):

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


Generate a sample of X of size 10^6

Here is the R code for those who would like to verify the results and the performances

 

N <- 10^6

R <- 20

Q <- c(0:5)

 

M <- matrix(nrow=R, ncol=length(Q))

 

for (r in c(1:20)) {

  S <- rnorm(N, mean=0, sd=1)

  for (q in Q) {

    M[r, q+1] <-length(S[S>q])

  }

}

 

Expected.Number.of.Outliers <- floor(N - pnorm(Q, mean=0, sd=1) * N)

Observed.Number.of.Outliers <-  round(colMeans(M))

 

Expected.Number.of.Outliers

Observed.Number.of.Outliers

 

Absolute.Differences <- M - kronecker(t(Expected.Number.of.Outliers), rep(1, R))

boxplot(differences)

Remark the loop below takes about 30 seconds to run (to be compared to the 20 minutes
it would take am I used the build-in procedure Select, and to  the 3 seconds R demands).

N := 10^6:
R := 20:

Q  := [$0..5]:
nQ := numelems(Q):
M  := Matrix(R, nQ, 0):

for r from 1 to R do
  S := Sample(X, N):
  for q in Q do
    M[r, q+1] := add(Heaviside~(S -~ q)):
  end do:
end do:

Expected_Number_Of_Outliers := Vector[row](nQ, i -> Probability(X > Q[i], numeric)*N);

Expected_Number_Of_Outliers := Vector[row](6, {(1) = HFloat(500000.0), (2) = HFloat(158655.25393145697), (3) = HFloat(22750.13194817921), (4) = HFloat(1349.8980316301036), (5) = HFloat(31.67124183311998), (6) = HFloat(0.2866515719235352)})

(2)

Observed_Number_Of_Outliers := Mean(M);

Observed_Number_Of_Outliers := Vector[row](6, {(1) = 500291.4, (2) = 158782.3, (3) = 22761.35, (4) = 1349.5, (5) = 51.6, (6) = 1.9}, datatype = float[8])

(3)


While the values of  the Observed_Number_of_Outliers seem to be reasonably correct for q =0, 1, 2 and 3
there are highly suspicious for q = 4 and likely for q = 5 (but N is too small to be categorical)

So let's examine more closely rhe results for q = 4

`q=4`:= convert(M[..,5], list);
 min(`q=4`)

[HFloat(49.0), HFloat(63.0), HFloat(42.0), HFloat(47.0), HFloat(44.0), HFloat(68.0), HFloat(46.0), HFloat(51.0), HFloat(58.0), HFloat(55.0), HFloat(50.0), HFloat(51.0), HFloat(68.0), HFloat(45.0), HFloat(58.0), HFloat(44.0), HFloat(59.0), HFloat(49.0), HFloat(47.0), HFloat(38.0)]

 

HFloat(38.0)

(4)


This means we obtained 20 estimations out of 20 of the probability that X exceeds q=4.
There are only about on chance out of 1 million to get such a result (roughly (1/2)^20 ).

Here is the result R returned for the same quantile q=4

 M[,5]

 [1] 24 37 23 38 30 43 23 30 25 39 25 29 29 36 39 41 35 37 36 34

(9 values out of 20 are less than the expected value of 31.67)

A very simple test based on the Binomial distribution will prove, without any doubt, that the
distribution of the 20 numbers of outliers for q = 4 implies the sampling algorithm is of
is of very poor quality.



Is it possible to obtain correct results with Maple ?


(simulations done below for the case q=4 only)

N := 10^6:
R := 20:

Menv := Vector[column](R, 0):

for r from 1 to R do
  S := Sample(X, N, method=envelope):
  Menv[r] := add(Heaviside~(S -~ 4)):
end do:

`q=4`:= convert(Menv, list);
Mean(Menv);
Variance(Menv);

[HFloat(26.0), HFloat(26.0), HFloat(35.0), HFloat(34.0), HFloat(30.0), HFloat(30.0), HFloat(28.0), HFloat(24.0), HFloat(27.0), HFloat(31.0), HFloat(28.0), HFloat(21.0), HFloat(34.0), HFloat(33.0), HFloat(29.0), HFloat(38.0), HFloat(38.0), HFloat(33.0), HFloat(39.0), HFloat(25.0)]

 

HFloat(30.45)

 

HFloat(24.892105263157895)

(5)



Conclusion, the "default" sampling strategy suffers strong flaw.

Remark: it's well known that one of the best method to sample normal random variables
is Box-Mueller's: why it has not been programmed by default ?

 

 


 

Download Bug_Normal_Sampler.mw


 

I have to do a homework on the euler explicite and when I am trying to test it I get an erreor can someone help me please :)

restart;
eulerexp := proc (fin, condin, h, tmax)

local i, n, j, tab, N; N := tmax/h;

for j to 5 do

tab[1, j] := condin[1, j]

end do;

for n from 2 to N do

tab[n, 1] := tab[n-1, 1]+h;

tab[n, 2] = tab[n-1, 2]+h*fin[1](tab[n-1, 1], tab[n-1, 2], tab[n-1, 3], tab[n-1, 4], tab[n-1, 5]);

tab[n, 3] = tab[n-1, 3]+h*fin[2](tab[n-1, 1], tab[n-1, 2], tab[n-1, 3], tab[n-1, 4], tab[n-1, 5]);

tab[n, 4] = tab[n-1, 4]+h*fin[3](tab[n-1, 1], tab[n-1, 2], tab[n-1, 3], tab[n-1, 4], tab[n-1, 5]);

tab[n, 5] = tab[n-1, 5]+h*fin[4](tab[n-1, 1], tab[n-1, 2], tab[n-1, 3], tab[n-1, 4], tab[n-1, 5]);

end do;

return tab end proc;

condin := [25, 1, 2, 3, 4];
                        [25, 1, 2, 3, 4]


fin := proc (t, w, x, y, z) options operator, arrow; [2*t-4*w+5*x-6*y-z, x, z, t] end proc;
(t, w, x, y, z) -> [2 t - 4 w + 5 x - 6 y - z, x, z, t]

h := .1;
                              0.1
tmax := 20;
                               20


eulerexp(fin, condin, h, tmax);
Error, (in eulerexp) invalid subscript selector


 

I've always had this issue. 1. Double click to highlight term to get help on. 2. Hit F1 3. Help pops up and I have to type in that term and find the info.

 

Why can't maple just combine all these steps? I've tried F2, various alt's, ctrl's, and shift combos and nothing... Surely something as basic as this has a hot key? And ideally I'd like to assign F1 to it!

 

The  plots help page contains an entry for ternaryplot. But it is empty (Maple 2018&2019) and there is no such command.
Does somebody know about it?

restart;
Digits:=4:
n:=11:
M := 2:
Le := 5:
Lb := 2:
L:= 1:
l := 0.5:
Pr := 1:
Pe := 2:
Nt := 0.5:
Nb := 0.8:
F[0]:=0:
F[1]:=l*F[2]:
F[2]:=a:
T[0]:=1:
T[1]:=b:
d:=k->piecewise(k<>0,0,k=0,1):
for k from 0 to n do F[k+3]:=-1/(d(k)+(k+1)*(k+2)*F(k+2))*((sum(F[k-m]*(m+1)*(m+2)*F[m+2],m=0..k))-(sum((k-m+1)*(m+1)*F[k-m+1]*F[m+1],m=0..k))-M*(k+1)*F[k+1])*(factorial(k)/factorial(k+3));
T[k+2]:=-1*(Pr/(k+1)*(k+2))*((sum(F[k-m]*(m+1)*T[m+1],m=0..k))+Nt*(sum((k-m+1)*(m+1)*T[k-m+1]*T[m+1],m=0..k))+Nb*(sum((k-m+1)*(m+1)*T[k+1]*F[k-m+1],m=0..k))) end do: 
f:=sum(F[k]*y^k,k=0..n); 
t:=sum(T[k]*y^k,k=0..n); 
with(numapprox):
pade(diff(f,y),y,[4,4]):
pade(t,y,[4,4]):
solve({limit(pade(diff(f,y),y,[4,4]),y=infinity)=0.,limit(pade(t,y,[4,4]),y=infinity)=0.,[a,b]);

I'm new to using Maple and am trying to create the procedure above. My code so far:

with(combinat):
g:=proc(n)
 local x; local setG; local setG2;
    for x from 1 to n do
        setG:={seq(x+1,x=1..n-1)}; setG2:=choose(setG,3); nops(setG2);
    end do;
end proc; 

Although this doesn't seem to work even though

setG:={seq(x+1,x=1..5)}; setG2:=choose(setG,3); nops(setG2);

Any help would be greatly appreciated!! 

 

I am studying the motion of a beam coupled to piezoelectric strips. This continuous system is modelled by two DE:

YI*diff(w(x,t), x$4)-N[0]*cos(2*omega*t)*diff(w(x,t), x$2)+c*diff(w(x,t), t)+`&rho;A`*diff(w(x,t), t$2)+C[em,m]*v(t) = 0;

And:

C[p]*diff(v(t), t)+1/R[l]*v(t) = C[em,e]*(D[1,2](w)(0,t)-D[1,2](w)(ell,t));
 

where "w(x,t)" stands for the beam's vibration and "v(t)" means the electric voltage, which is constant throught the beam. I would like to numerically solve both DE simultaneosly, but maple will not let me do it. I would like to know why. I am getting the following error:

Error, (in pdsolve/numeric/process_PDEs) number of dependent variables and number of PDE must be the same
 

I suppose it is because "w(x,t)" depends on "x" and "t", while "v(t)" depends solely on time, but I am not sure. Could someone help me out? Here is my current code:

restart:
with(PDEtools):
declare(w(x,t), v(t)):

YI*diff(w(x,t), x$4)-N[0]*cos(2*omega*t)*diff(w(x,t), x$2)+c*diff(w(x,t), t)+`&rho;A`*diff(w(x,t), t$2)+C[em,m]*v(t) = 0;
pde1:= subs([YI = 1e4, N[0] = 5e3, c = 300, omega = 3.2233993, C[em,m] = 1], %):
ibc1:= w(0,t) = 0, D[1,1](w)(0,t) = 0, w(ell,t) = 0, D[1,1](w)(ell,t) = 0, D[2](w)(x,0) = 0, w(x,0) = sin(Pi*x/ell):

C[p]*diff(v(t), t)+1/R[l]*v(t) = C[em,e]*(D[1,2](w)(0,t)-D[1,2](w)(ell,t));
pde2:= subs([C[p] = 10, R[l] = 1000, C[em,e] = 1, ell = 5], %):
ibc2:= v(0) = 0:

pdsolve({pde1, pde2}, {ibc1, ibc2}, numeric);

Thanks.

Hello everyone!

I'm interesting in "zcoloring" funciton in colorscheme option.

I wrote simple programm which compares two results: spectrogram of signal drawn with "colormap" list and spectrogram which was plotted with zcoloring function. I use red, green, blue functions to construct JET-colormap: list and expressions in "zcoloring".

My result:

As I understand, when I use:

colorscheme = ["zcoloring", [z-> Red color function(z), z-> Green color function(z), z-> Blue color function(z)], colorspace = "RGB"]

Maple plots z-value with color RGB color coordinates defined from "zcoloring". For example, if "zcoloring" is

colorscheme = ["zcoloring", [z-> 5*z, z-> 3*z, z-> 2*z], colorspace = "RGB"]

and z value is 10, then 10 value will correspond [50,30,20]-RGB color.

My test program:

Spectrogram_zcoloring.mw

Spectrogram of my test signal:

list_test.txt

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