Hello guys, i would like to do parallel computation in my code written in the Maple18. The question that can help me is:

Given a procedure that compute an function g, where g = f1+f2+f3+f4+f5+f6+f7+f8, i would like to compute all fi at same time.
Now, i´m using " grid:-seq('f[i]',[i=1,2,3,4,5,6,7,8])" and it works very well. However, i think that for my case an better solution should be;
Calculate the f1 in core 1, f2 in core 2, f3 in core 3 ... f8 in core 8 at same time, and after this, to sum all results(f1+f2+f3+..+f8). How i can do this?

Att,

Griffith.

Greetings to all. The title describes it well, I am writing about testing the limits of the Maple integration engine. A recent discussion at math.stackexchange.com features a family of integrals that involve the product of a power of the natural logarithm and a rational function, more precisely,

int((log(x))^n/(x^3+1), x=0..infinity);

These integrals can be evaluated recursively as described at the MSE link using a technique that generalizes to other types of rational factors. Unfortunately Maple apparently only finds a simple closed form for a few small initial values of n. The following transcript of a Maple session illustrates the problem. Mathematica was successful here. Also observe the memory allocation in the Maple session.

    |\^/|     Maple 18 (X86 64 LINUX)
._|\|   |/|_. Copyright (c) Maplesoft, a division of Waterloo Maple Inc. 2014
 \  MAPLE  /  All rights reserved. Maple is a trademark of
 <____ ____>  Waterloo Maple Inc.
      |       Type ? for help.
> restart; read `cl.maple`;
alpha := (n, k) ->

                                                           n
    -1/3 exp(1/3 I Pi + 2/3 I Pi k) (1/3 I Pi + 2/3 I Pi k)

Q := proc(n)
local res;
option remember;
    if n = 0 then return 2/9*sqrt(3)*Pi end if;
    res := -add(alpha(n + 1, k), k = 0 .. 2)/(n + 1) - add(
        binomial(n + 1, p)*(2*I*Pi)^(n - p)*Q(p),
        p = 0 .. n - 1)/(n + 1);
    simplify(res)
end proc

                               infinity
                              /               n
                             |          log(x)
              VERIF := n ->  |          ------- dx
                             |           3
                            /           x  + 1
                              0

> Q(6);
                                7  1/2
                          910 Pi  3
                          ------------
                              6561

> VERIF(6);
memory used=3.8MB, alloc=40.3MB, time=0.18
       7  1/2
9890 Pi  3       490    5  1/2
------------- + ----- Pi  3    Psi(1, 1/3)
   177147       19683

        490    5  1/2                10    3  1/2            2
     + ----- Pi  3    Psi(1, 2/3) + ---- Pi  3    Psi(1, 1/3)
       19683                        2187

        20   1/2   3
     + ---- 3    Pi  Psi(1, 2/3) Psi(1, 1/3)
       2187

        10    3  1/2            2    40                 4
     + ---- Pi  3    Psi(1, 2/3)  + ----- Psi(2, 2/3) Pi
       2187                         19683

        10   1/2               3
     + ---- 3    Pi Psi(1, 1/3)
       2187

       10               1/2               2
     + --- Psi(1, 2/3) 3    Pi Psi(1, 1/3)
       729

       10   1/2                           2
     + --- 3    Pi Psi(1, 1/3) Psi(1, 2/3)
       729

        10   1/2               3    40     4
     + ---- 3    Pi Psi(1, 2/3)  - ----- Pi  Psi(2, 1/3)
       2187                        19683

        20             2  1/2
     + ---- Psi(2, 2/3)  3    Pi
       6561

        40               1/2
     - ---- Psi(2, 2/3) 3    Psi(2, 1/3) Pi
       6561

        40    2
     + ---- Pi  Psi(2, 2/3) Psi(1, 1/3)
       2187

        40    2
     + ---- Pi  Psi(2, 2/3) Psi(1, 2/3)
       2187

        20   1/2            2
     + ---- 3    Psi(2, 1/3)  Pi
       6561

        40    2
     - ---- Pi  Psi(1, 1/3) Psi(2, 1/3)
       2187

        40    2
     - ---- Pi  Psi(1, 2/3) Psi(2, 1/3)
       2187

> evalf(Q(6));
                          725.5729634

> evalf(VERIF(6));
                          725.5729630

> quit
memory used=22.4MB, alloc=44.3MB, time=0.47
user@host:~/complex-logint$ math
Mathematica 10.0 for Linux x86 (64-bit)
Copyright 1988-2014 Wolfram Research, Inc.

In[1]:= Integrate[Log[z]^6/(1+z^3), {z, 0, Infinity}]

                7
          910 Pi
Out[1]= ------------
        2187 Sqrt[3]

In[2]:= N[Out[1]]

Out[2]= 725.573

In[3]:=
user@host:~/complex-logint$

My question for you all is what the appropriate techniques would be to get Maple to at least simplify the rather involved output from the integration engine to obtain a match of the closed form from the recursive equation.

Best regards, Marko Riedel.

cl-maple.txt

Dear All

It is well known that the package "PDEtools" is helpful in finding infinitesimal transformations for PDEs which I illustrate as follow:

Download Writing_Vector_fields.mw

Regards

Dear All

I have 3D plot

How I can Improve this plot so that it look more aesthetic ?

Secondly, the resulotion of plot is not so good, and when I export this plot as jpeg file, quality of resolution deteriotate even more, I mean pixel tear out on zooming the image. How I fix this issue which I guess might be due to wrong way of exporting plot ?

Download Improving_3D-_plot.mw

Regards

Hi there,
I have a simple task to do with arrays. I have an 8x3000 array, whose columns are sometimes all zeroes (if any element in the column is zero then the entire column is zero). I want to eliminate the zero vectors, so I have this loop (the array is A)

for i from 1 to 3000 do
    while A(1,i)=0 do
        A:=DeleteColumn(A,i);
    end do;
end do:

Setting "A:=DeleteColumn(A,i);" reduces the size of the array, so if all the zero columns have been deleted at "i=200", Maple will keep trying to check the next column, which doesn't exist anymore. This gives an "index out of bounds" error, but also records "A" as the new smaller array, which is fine.

If I run this loop inside a procedure, the error stops the procedure, but it doesn't record the new array, it stays as the older, larger one. Does anyone know of a way around this?

Any help would be much appreciated.

I have the PDE u_{xx}+u_{yy} = 1 with BC: u|_{x^2+y^2=1} =0 ;

how to write down the command of the BC in solving this PDE?, btw can I make maple show me how to solve this PDE analytically?

Thanks in advance.

Here are the lines that I wrote so far:

pde := diff(u(x, y), x, x)+diff(u(x, y), y, y) = 1;

ans := pdsolve(pde)

how to add the BC correctly to pdsolve? I am not sure how to write the condition x^2+y^2=1 and that u will get a value on this boundary.

Dear All

Say I have created a Maple module called MYMODULE. This means that I have a .mpl file called

mymodule.mpl, with inside

   MYMODULE := module()

   export myfunction;

   ****

   ****

   end module:

   savelib(MYMODULE, "MYMODULE.mla"):

Hence I use it in such a way:

> with(MYMODULE);

                                                              [myfunction]

I want to associate a help to the function myfunction. I know that with Maple18 things have changed

but I am not able to find enough information online. Say I have a .txt file called help_myfunction.txt.

How can I enable this help for my function, so that at the command

> ?myfunction;

the content of help_myfunction.txt shows up?

Many thanks,Simone

My query is related to Maple packages "Lie algebra" and "DifferentialGeometry", I wonder can these packages help me to check isomophism between to Lie algebras, for example, suppose I have a 4-dim Lie algebra with commutation relations:

[e2, e3] = e3, [e2, e4] = - e4, [e3, e4] = - e1 

and another 4-dim Lie algebra with commutation relations

[e2, e3] = e1, [e4, e2] = e2, [e3, e4] = - e3

Can I check for isomophism between these Lie algebras if there is any using Lie algebra package ?

Regards

Hello! 

In an assigment I have been asked to use the Eigenvectors command to find the eigenvalues and eigenvectors of a particular matrix. 

As highlighted in the following image, my questions are:

1. What is the meaning of the suffix "+0.I"? Does it mean that there are further decimal digits which are not displayed?

2. How do the first and third eigenvalues, which are equal, result in different eigenvectors? As per my understanding, equal eigenvalues should have equal corresponding eigenvectors. Please help.

Hi Maple community

I'm running an algorithm where a non-linear equation system must be solved, in this case is a 26x26 system.

After 16116 succesful previous computations, fsolve stops giving me results.
I checked why and I was first expecting that, for some reason, the 26x26 system had an error and I ended with something like 25x26 or vice versa. But that was not the case.

So I tried the command solve and it not only worked fine but also gave me two results, but I only need one. I guess I could check for the wrong solution and discard it, but I still wondering why fsolve is failing and if there is anything to help fsolve not to fail.

These are the set of equations if somebody wants to check them:

EQ[16117][1] := W[1, 16117]*(-0.3860115660e-1*HRa[1, 16117]-0.1876793978e-1*ga[1, 16117]+0.7836678184e-1) = 2.040147478*10^6*SR[1, 16118], W[1, 16117]*(-0.3915554290e-1*HRa[1, 16117]-0.1903748329e-1*ga[1, 16117]+0.8260795999e-1) = 3.876387504, W[1, 16117]*(-0.1876794098e-1*HRa[1, 16117]-0.9892449327e-2*ga[1, 16117]+0.3810204607e-1) = 2.040147478*10^6*v[1, 16118], HLa[1, 16117] = .9724029753*ga[1, 16117]+HRa[1, 16117], NRa[1, 16117] = 0.7006679273e-1*HRa[1, 16117]-.1803623678*ga[1, 16117]+1.002451672, NLa[1, 16117] = 0.7006679273e-1*HRa[1, 16117]+.2484955248*ga[1, 16117]+1.002451672, SL[2, 16118] = SR[1, 16118], fra[1, 16117] = HRa[1, 16117]-HLa[2, 16117], fra[1, 16117] = .25*NRa[1, 16117]+.25*NLa[2, 16117], ga[1, 16117] = 0.;

EQ[16117][2] := W[2, 16117]*(-0.3860115660e-1*HRa[2, 16117]-0.1876793978e-1*ga[2, 16117]+0.7836678184e-1) = -2.040147478*10^6*SL[2, 16118]+7.152482840, W[2, 16117]*(-0.3915554290e-1*HRa[2, 16117]-0.1903748329e-1*ga[2, 16117]+0.8260795999e-1) = 3.876387504, W[2, 16117]*(-0.1876794098e-1*HRa[2, 16117]-0.9892449327e-2*ga[2, 16117]+0.3810204607e-1) = -1.983845478*10^6*SL[2, 16118]+5.221405977, HLa[2, 16117] = .9724029753*ga[2, 16117]+HRa[2, 16117], NRa[2, 16117] = 0.7006679273e-1*HRa[2, 16117]-.1803623678*ga[2, 16117]+1.002451672, NLa[2, 16117] = 0.7006679273e-1*HRa[2, 16117]+.2484955248*ga[2, 16117]+1.002451672, SL[3, 16118] = 0.3505865589e-5, fra[2, 16117] = HRa[2, 16117]-HLa[3, 16117];

EQ[16117][3] := W[3, 16117]*(-0.3860115660e-1*HRa[3, 16117]-0.1876793978e-1*ga[3, 16117]+0.7836678184e-1) = -2.040147478*10^6*SL[3, 16118]+10.82168541, W[3, 16117]*(-0.3915554290e-1*HRa[3, 16117]-0.1903748329e-1*ga[3, 16117]+0.8260795999e-1) = 3.876387504, W[3, 16117]*(-0.1876794098e-1*HRa[3, 16117]-0.9892449327e-2*ga[3, 16117]+0.3810204607e-1) = -1.983845478*10^6*SL[3, 16118]+8.751240594, HLa[3, 16117] = .9724029753*ga[3, 16117]+HRa[3, 16117], NRa[3, 16117] = 0.7006679273e-1*HRa[3, 16117]-.1803623678*ga[3, 16117]+1.002451672, NLa[3, 16117] = 0.7006679273e-1*HRa[3, 16117]+.2484955248*ga[3, 16117]+1.002451672, SL[4, 16118] = 0.5304364281e-5, fra[3, 16117] = HRa[3, 16117];

And after these the solving command that I used was:

SOL[j]:=fsolve({seq(EQ[j][n],n=1..N)},indets({entries(EQ[j],nolist)},assignable(name)));

Which returns

SOL[j]:=

As I said, then I tried the solve command:

SOL[j]:=solve({seq(EQ[j][n],n=1..N)},indets({entries(EQ[j],nolist)},assignable(name)));

which returns:

SOL[16117] :=

{HLa[1, 16117] = 1.011251860, HLa[2, 16117] = .5007913055, HLa[3, 16117] = -0.4240068535e-1, HRa[1, 16117] = 1.011251860, HRa[2, 16117] = .8728245835, HRa[3, 16117] = .2686716410, NLa[1, 16117] = 1.073306847, NLa[2, 16117] = .9685353734, NLa[3, 16117] = .9417827567, NRa[1, 16117] = 1.073306847, NRa[2, 16117] = 1.132612831, NRa[3, 16117] = 1.078974668, SL[2, 16118] = 0.1737463747e-5, SL[3, 16118] = 0.3505865589e-5, SL[4, 16118] = 0.5304364281e-5, SR[1, 16118] = 0.1737463747e-5, W[1, 16117] = 90.12372195, W[2, 16117] = 69.57451714, W[3, 16117] = 49.58407210, fra[1, 16117] = .5104605550, fra[2, 16117] = .9152252689, fra[3, 16117] = .2686716410, ga[1, 16117] = 0., ga[2, 16117] = -.3825916698, ga[3, 16117] = -.3199006320, v[1, 16118] = 8.447574110*10^(-7)},

{HLa[1, 16117] = 3.043461992, HLa[2, 16117] = 2.386862361, HLa[3, 16117] = -0.4240068535e-1, HRa[1, 16117] = 3.043461992, HRa[2, 16117] = 1.087485894, HRa[3, 16117] = .2686716410, NLa[1, 16117] = 1.215697293, NLa[2, 16117] = 1.410701230, NLa[3, 16117] = .9417827567, NRa[1, 16117] = 1.215697293, NRa[2, 16117] = .8376385519, NRa[3, 16117] = 1.078974668, SL[2, 16118] = 0.2032780481e-5, SL[3, 16118] = 0.3505865589e-5, SL[4, 16118] = 0.5304364281e-5, SR[1, 16118] = 0.2032780481e-5, W[1, 16117] = -106.0268094, W[2, 16117] = 265.7250566, W[3, 16117] = 49.58407210, fra[1, 16117] = .6565996307, fra[2, 16117] = 1.129886580, fra[3, 16117] = .2686716410, ga[1, 16117] = 0., ga[2, 16117] = 1.336253076, ga[3, 16117] = -.3199006320, v[1, 16118] = 9.883410782*10^(-7)}

Thanks in advance for any recommendations and suggestions.
 

Here, I attached my maple code. I need to find root. I am using fsolve. But I am not geting the root. Please any one help me... to find the root.

Download root.mw

Hi everyone, 

I am pretty new in MAPLE and I am having trouble doing a really easy thing. 

I would like to do a plot and add a legend. 

I know how to do a plot and I succeeded but I don't know how to write the legend. I want a legend that looks like:

I have stored in my K__V the value 0.2. How can I do that? 

I tried with legend = 'K__V' 

This is ok to write the first part, but then how do I write the second one: ' = 0.2 ' ?

Of course I don't want to write 0.2 directly but I want to write K__V in this way the legend will change according to the value of K__V. 

In MATLAB what I want to write is really easy to do: 
legend('K__V=%d',K__V)

I hope I was clear.

Thanks

with(plots);

implicitplot3d((x+y+z)^2=0, x=-100..100,y=-100..100,z=-100..100)

can be plot, and only show me an empty box.

and i get the same empty box when i reduce the interval.

for example :

implicitplot3d((x+y+z)^2=0, x=-5..5,y=-5..5,z=-5..5)

and

implicitplot3d((x+y+z)^2=0, x=-0.1..0.1,y=-0.1..0.1,z=-0.1..0.1)

please help me.

thanks for reading this question, and i hope i get an answer.

Hello guys,

I was just playing around with the Shanks transformation of a power series, when I noticed that polynomials aren't evaluated as I would expect.
I created this minimal working example; the function s should evaluate for z=0 to a[0], however it return simply 0.
Is there something I messed up?

Download evaluate_sum.mw

Thanks for your help,

Sören

Greetings to all.

With the following matter I am betting on there being a simple mistake on my part due to fatigue owing to a challenging session of intense computing. The following link at Math.Stackexchange.Com points to a computation involving complex residues. Consult the link for additional details.

I usually verify my computations with Maple, I did the same this time. Thereby I happened on a curious phenomenon which I have documented below. Please study the session data provided, I believe it speaks for itself.

user@host:~$ math
Mathematica 10.0 for Linux x86 (64-bit)
Copyright 1988-2014 Wolfram Research, Inc.

In[1]:= Residue[z^2/(z^4 + 2*z^2 + 2)^2, {z, 2^(1/4)*Exp[3*Pi*I/8]}]

            1/8      1/4
        (-1)    ((-1)    + Sqrt[2])
Out[1]= ----------------------------
            1/4      1/4           3
        16 2    ((-1)    - Sqrt[2])

In[2]:= N[%]

Out[2]= 0.117223 - 0.0083308 I

In[3]:=
user@host:~$ maple
    |\^/|     Maple 18 (X86 64 LINUX)
._|\|   |/|_. Copyright (c) Maplesoft, a division of Waterloo Maple Inc. 2014
 \  MAPLE  /  All rights reserved. Maple is a trademark of
 <____ ____>  Waterloo Maple Inc.
      |       Type ? for help.
> residue(z^2/(z^4 + 2*z^2 + 2)^2, z=2^(1/4)*exp(3*Pi*I/8));
                                       0

> quit
memory used=0.9MB, alloc=8.3MB, time=0.07
user@host:~$

I am looking forward to learning what the correct syntax is to get the residue in this case and I hope I can assist other users who might have run into the same problem. I will cancel the question should it turn out to be trivial and of little potential use to the community.

Best regards,

Marko Riedel

Post Scriptum. Being a programmer myself I would be curious to learn more about the algorithms that are deployed here and how and why they did not succeed.