I've been tasked with generating "phase plots" which are visualizations of complex functions. 

A 2D phase plot is easy to create: Given a complex function F : C -> C colour points (x, y) in R^2 by [ arg(F(x+I*y)), 1, 1, colortype=HSV].   Something like the following seems to do the trick
    
    p := plot3d( 
        1,
        x=-5..5,
        y=-5..5,
        scaling=constrained,
        color=[ argument(f(x+I*y))/2/Pi, 1, 1, colortype=HSV],
        axes=none,
        style=patchnogrid,
    ):
    
    g := plottools[transform]((x,y,z)->[x,y],p);
    plots[display]( g(p) ):

Now, "colouring each point of R^2" is only possible using some type of bijection onto the Riemann sphere or Pseudosphere.

The pseudosphere is:

    x := (u,v) -> sech(u)*cos(v);
    y := sech(u)*sin(v);
    z := u - tanh(u);
    
    return  plot3d(
        [x(u,v),y(u,v),z(u,v)],
        u=0..3,
        v=0..2*Pi,
        numpoints=2^10,
        lightmodel=none,
        color=ColorFunc(u,v) );

In this case I need  ColorFunc to be:

ColorFunc := proc(u,v)
    x, y := v, exp(u);
    ans  := Re( f(x+I*y) );
    if ans > 2*Pi then
       return [0,0,0,colortype=HSV];
    end if;
    return [ans/2/Pi,1,1,colortype=HSV];
end proc;

But it seems that "ColorFunc" cannot be very sophisticated.  Namely, it cannot contain "frems" or even "if" statements because (as far as I can tell) of the order of evaluations.  

It seems possible that I can generate a psuedosphere then change colours AFTER by swapping out the COLOR information in a more sophisticated way.  How can I do this?  I really just need to know how to identify and swap out points from a MESH.

h1_y_h2.mw
 

im trying to build a matrix starting from a function, so i can later use this matrix to get a more simple function using linearfit from the statistics package, like a kind of regression.

i want to get a matrix starting from h__1(T1,T__1s) so it has to be a 3 columns matrix (T__1,T__1s,h__1). so as you can see, i have got the functions h__1 and h__2, but i need to evaluate it at differents values for T__1 and T__1s and building a kind of value-table in matrix form.

for h__1, T__1 must be higher than T__1s, or you could get imaginary values, don't know if that important for building the matrix.

thank you very much for your help.

Download h1_y_h2.mw

I want to plot some color points in CIE 1976 color space with SpatterPlot command exactly as showed on the help page. When I try:

restart:with(ColorTools):with(ImageTools):
barvy:=Color("Lab",[80.38,13.50,7.96]); 	
SpatterPlot(barvy,symbol="box");

I got the error:

Error, invalid input: ColorTools:-SpatterPlot expects its 1st argument, colors, to be of type list({ColorTools:-Color, name, string, list({float, nonnegint}), specfunc({COLOR, COLOUR})}), but received _m2194815429568

Where is the problem?
 

with(IterativeMaps);
with(ImageTools);
Logistic := Bifurcation([x], [r*x*(1-x)], [.5], 2.5, 4);
ArrayTools:-Dimensions(Logistic);
ColouringProcedures:-HueToRGB(Logistic);
Embed(Logistic);

This is the code for Bifurcation program of the Logistic map. How can I change the black background color of this figure, and How can I save this figure.

Download Sum_Sum.mw
 

Download Sum_Sum.mw

Please anyone, I have been battling with this problem for a while yet the error message keeps coming. Would be happy if responded to.

Thanks 
 

Download kuikennnnnn.mw
 

Let us consider the improper integral

int((abs(sin(2*x))-abs(sin(x)))/x, x = 0 .. infinity);

Si(Pi)-Si((1/2)*Pi)+sum(-(-1)^_k*Si(Pi*_k)+signum(sin((1/2)*Pi*_k))*Si((1/2)*Pi*_k)+Si(Pi*_k+Pi)*(-1)^_k-signum(cos((1/2)*Pi*_k))*Si((1/2)*Pi*_k+(1/2)*Pi), _k = 1 .. infinity)
                    

Mathematica 11 produces a similar expression and a warning

Integrate::isub: Warning: infinite subdivision of the integration domain has been used in computation of the definite integral \!\(\*SubsuperscriptBox[\(\[Integral]\), \(0\), \(\[Infinity]\)]\(\*FractionBox[\(\(-Abs[Sin[x]]\) + Abs[Sin[2\ x]]\), \(x\)] \[DifferentialD]x\)\). If the integral is not absolutely convergent, the result may be incorrect.

Up to Pedro Tamaroff http://math.stackexchange.com/questions/61828/proof-of-frullanis-theorem , the answer is 2/Pi*ln(2) because of 

J := int(abs(sin(2*x))-abs(sin(x)), x = 0 .. T) assuming T>2;
-1/2-signum(sin(T))*signum(cos(T))*cos(T)^2+(1/2)*signum(sin(T))*signum(cos(T))+cos(T)*signum(sin(T))+floor(2*T/Pi)

B := limit(J/T, T = infinity);
                               2 /Pi

K := x*(int((abs(sin(2*t))-abs(sin(t)))/t^2, t = x .. 1)) assuming x>0,x<1;

     2*sin(x)*cos(x)-2*Ci(2*x)*x+Ci(x)*x+sin(1)*x-sin(2)*x+2*Ci(2)*x-Ci(1)*x-sin(x)

                         
A := limit(K, x = 0, right);
                               0

Its numeric calculation results 

evalf(Int((abs(sin(2*x))-abs(sin(x)))/x, x = 0 .. infinity));
                        Float(undefined)

which seems not to be true.

The question is: how to obtain the reliable results for it with Maple, both symbolic and numeric? 

Hello. Two teams A and B (consisting of 2, 3 or even 4 players) compete and the outcome of each game is either a win or a loss. I have a to process the new (gaussian) laws of players given whom beats whom.

Given the initial means and standard deviations of the players, I have a algorithm (RC) which computes the new laws of the players. [the actual algorithm i use is different to the one i am showing here]. 

By way of example consider two teams with 2 players per side. Each person plays 2 games.

The initial laws of A1,A2,B1,B2 are given.

(A1,B1) --->(A1',B1').......[iteration 1. A1 beats B1, resulting in new laws A1' and B1'(computed by RC)]

(B2, A1') --->(B2', A1").......[ iteration 2. B2 beats A1. using A1' from iteration 1. A1 has played twice and is denoted by A1"]

(B2',A2) --->(B2",A2').......[ iteration 3. B2 beats A2. using B2' from iteration 2.  B2 has finished and is denoted by B2"]

(A2',B1') --->( A2",B1")......[ iteration 4. A2 beats B1. using B1' from iteration 1 and A2' from iteration 3. now all players have played their matches]

new laws A1" ,A2", B1" & B2" should be outputted.

my first code gets the result, but it is tedious to enter the iterations in the right place .especially for teams of 3 or more.

my second with parameters gets errors.

So what i want is to enter who beat who:

eg  [[B1, A1], [B2,A1], [B2, A2], [A2, B1]];
and the final laws are computed automatically.

bb_processing_edit.mw

Good morning Maple Expert,

I would like to explain the concept of piecewise curves in 2D & 3D to my students with plots.In this regard I request the associated expetrs in Maple to provide the appropriate Maple commands to get piecewise curves in 3D space.

With warm regards.

Mr.M.Anand

Associate Professor in Mathematics.

Hyderabad Institute of Technology and Management

What shall I do to install Maple 15 and MapleSym 5 in a Windos 10 environment.

I find the procedure in Maplesoft site up to Windows 8.

May I follow it for Windows 10.

Thank you for any help. Kind regards

Hi everyone,

I created a procedure "SIM" which depends on two formal parameters x and y. I write Threads:-Map(SIM, x, y) in order to execute it with the Threads package.

I would like to create .mpl files automatically, each one made of the code Threads:-Map(SIM, x, y) with specific values of x and y. For instance:

Threads:-Map(SIM, 5000, 1), then Threads:-Map(SIM, 5000, 2) ... and so on until Threads:-Map(SIM, 5000, 5000).

The fact is, I tried writing the following:

for y from 1 to 5000 do
a[y] := Threads:-Map(SIM, 5000, y);
save a[y], sprintf("SIM_%d.mpl", y)
end do

However, it does not work. The error message says "Error, save can only save names". I also tried, but without success:

for y from 1 to 5000 do
a[y] := Threads:-Map(SIM, 5000, y);
save convert(a[y], name), sprintf("SIM_%d.mpl", y)
end do

Any idea? Thanks a lot.

I want to plot the argument for a complex function. The input (x,y) represented in polar coordinates (r,phi) by default puts the cut at -I*Pi. Likewise the argument function:

argument(f(x)) plots the range -Pi..Pi.

However the function f(x)=x^2 could typically be plotted with 2 riemann surfaces on top of each other. When phi becomes 2Pi f(x) becomes 4Pi and only then I want to identify the 0 with 4Pi again since the points are equivalent in the preimage.

On the other hand the function f(x)=sqrt(x) never surpasses its own domain. The values always stay within the argument range of (0,2Pi) (in fact it only goes till Pi, or -Pi/2..Pi/2 in maple) when the preimage is taken to be (0,2Pi). Thus when plotting a preimage value of (x,y) with argument phi and 2Pi+phi they will have the same value since phi=2Pi+phi and I see a step in the plot. This step is actually there since the function has a cut at this point.

This step in the plotting image is also shown for f(x)=x^2 (e.g. at phi=+-Pi/2) but it is not of importance since it just comes from the argument function being constrained to -Pi..Pi.

So is it possible to change this behaviour?

Can i make Explore with number of parameters differs from parameter in explore?
For excemple i whant to make sum with 'l' values wich will declorate in explore, and 'l' - count of them will declorate in same explore, but it dont work, so is it posible?
 

Download Explore_Problem_Exemple.mw

I need yours hepl.  I work with the physics paсkage and I set:

with(Physics)

Setup(mathematicalnotation = true)

 Coordinates(X)

Setup(Dgammarepresentation = standard)

Setup(spaceindices = uppercaselatin)

Define(M, aa, mu, mu5, Pi, eta)

M_[mu, mu5] := Dgamma[mu]*d_[mu]+M+Psigma[A]*aa[A]-mu*Dgamma[0]-mu5*Dgamma[0]*Dgamma[5]+i*Dgamma[5]*Psigma[B]*Pi[B]+i*Dgamma[5]*eta

And next:

Dagger(M_[mu, mu5])

How is Maple explained that  

Dagger(d_[mu])=d_[mu]

conjugate(M)=M

conjugate(aa[A])=aa[A]

conjugate(i)=i

and so on?

Dear Friends, I work with physics paсkage. I have a quation. I don't understend how one works with metrics. For example, let: