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


 

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...

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.


 

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!! 

both_of_them_ac.mw

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)+`ρA`*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)+`ρA`*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

D[1,2](1/x);

what does this mean?

How do I find where 𝑓(𝑥) = (2*𝑥 ^2) tan(2*𝑥), is discontinuous in the interval [0,2𝜋], and find the discontinuities. I know you need ot use the commands: discount(f(x),x), iscont(f(x),x=a..b,’open’), iscont(f(x),x=a..b,’close’) and fdiscont(f(x),x=a..b, resolution) which will help me find the list of ranges, each of width resolution, in which there appears to be a discontinuity in the function or its first derivative.However, what is next ?