K M | 21 Oct 16:33 2014
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diagonalize matrix

Hello Everyone!

I started to use wxMaxima today :) and now trying to diagonalize a 3x3 matrix looks like:
matrix( [0,0,a], [0,b,-a], [a,-a,c]).
(a,b,c are all symbolic).

with an example:
assume(a>0,b>0); M : matrix([a,a+b,a+b],[a+b,a,a+b],[a+b,a+b,a]); load("eigen"); [myeigval,myeigvec]:similaritytransform(ev(M,hermitianmatrix));
; however end up with errors
 (%o3) "C:/Maxima-5.31.2/share/maxima/5.31.2/share/matrix/eigen.mac"
length: argument cannot be a symbol; found M
#0: eigenvectors(mat=M)(eigen.mac line 107)
#1: uniteigenvectors(m=M)(eigen.mac line 154)
#2: similaritytransform(mat=M)(eigen.mac line 169)
 -- an error. To debug this try: debugmode(true);

Could you help me with this?

Thanks!
/Erik
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Klaus Rohe | 21 Oct 15:44 2014
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Get an explanation from Maxima which integration rules have been applied to solve an indefinite integral

Hi all,

 

is it possible to get an explanation (or list) from Maxima which integration rules have been applied to solve an indefinite integral?

 

Thanks in advance

 

Klaus

 

Dipl.-Phys. Klaus Rohe

Adolf-Kolping-Strasse 10a

85625 Glonn

E-Mail: klaus-rohe <at> t-online.de

Tel.:      +49 (0) 8093 5402

Mobil:  +49 (0) 170 8133634

 

 

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mm82999-oh@yahoo.de | 20 Oct 19:25 2014
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LISP Compilation

Dear Sirs,

I'm working with wxMaxima (Version 13.04.02, Maxima 5.31.2).
Regarding Compilation of Maxima-Code (Calculation in Maxima) is there
a question. I have no problem to compile the code into LISP. I get the
o-File. But how can I create an executable? I tried already SBCL, GCL
of Maxima and extra installed one.
How is it possible to get an exe-File from Maxima's language?

Of course Maxima is remarkable fast. Much faster than I expected. Thus I thought do some simulations. For this reason I need much more speed.
Are there any further suggestions how the speed could be increased?

Thank you for your attention.

Regards,
Markus



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Zaskia Anzalota | 20 Oct 10:40 2014

Problem installing Maxima in a windows surfer

Hi, I have been trying to install and run maxima at a windows surfer without any success.  I download the
program but when trying to run it a message appears indicating: This app can't run on your PC
To find apps for this PC, open the Windows store

Can you help me with this.

Thanks, 

Zaskia Anzalota

Sent from my iPod
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Isaac Haïk Dunn | 20 Oct 03:42 2014
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Unable to solve an integral? Related to the polygarithm function.

Hi people,
I am using Maxima 5.34.0 and I am unsure whether I'm not doing things
the right way or if Maxima is unable to solve an integral.
Here are the 2 commands:

(%i1) f(z):=-4/%pi*integrate(x^2*log(1-z*exp(-x^2)),x,0,inf);
                   - 4            2                  2
(%o1)      f(z) := --- integrate(x  log(1 - z exp(- x )), x, 0, inf)
                   %pi
(%i2) f(1);
                             inf
                            /                    2
                            [     2           - x
                          4 I    x  log(1 - %e    ) dx
                            ]
                            /
                             0
(%o2)                   - ----------------------------
                                      %pi
(%i3) float(f(1));
                              inf
                             /
                             [     2                   1.0
(%o3)    - 1.273239544735163 I    x  log(1.0 - -------------------) dx
                             ]                                   2
                             /                                  x
                              0.0              2.718281828459045

Where f(z) is supposed to be the polygarithm function of order 5/2. I
wanted to plot it for real values of z between 0 and 1, but I am
unable to do it properly (Maxima would think for too long and I had to
kill the process).
Maybe I should specify that I want z to be real valued?

Also notice that float(f(1));  does not return a float. So it seems
that Maxima is not even able to get a numerical approximation of the
integral for z=1, or I am missing something and I can ask for a
numerical value?

I can't even get the Taylor expansion really:
(%i4) taylor(f(z),z,0,3);
Is z positive, negative or zero?

pos;
                              !                                   !
              inf             !                 inf               !
             /            2   !                /              2   !
             [     2   - x    !                [     2   - 2 x    !         2
         (4 (I    x  %e     dx!     )) z   (2 (I    x  %e       dx!     )) z
             ]                !                ]                  !
             /                !                /                  !
              0               !                 0                 !
                              !z = 0                              !z = 0
(%o4)/T/ ------------------------------- + ----------------------------------
                       %pi                                %pi
                                                            !
                                          inf               !
                                         /              2   !
                                         [     2   - 3 x    !         3
                                     (4 (I    x  %e       dx!     )) z
                                         ]                  !
                                         /                  !
                                          0                 !
                                                            !z = 0
                                   + ---------------------------------- + . . .
                                                   3 %pi

Thanks guys.

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Dimiter Prodanov | 19 Oct 23:57 2014
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inf vs infinity for integration

Hello,
my piece of mind ...
I think that inf , minf and infinity should be treated in a special way e.g. as Richard points out with this enumeration of infinite variables.

best regards,

Dimiter

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Pankaj Sejwal | 18 Oct 05:15 2014
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inf vs infinity for integration

I found out that in using "inf" and "minf" the integration gives 0(it's wrong as its divergent) but in case of using complex infinity, "infinity" or "-infinity" it returns message that lower bound(ll) and upper bound(ul) to integration must be a real which is still acceptable.

But in my opinion it must give similar error message for "inf"/"minf" or it should say its divergent.
​Or considering the iota/complex part gone​
​ one should always refer to complex infinity(hence reduced to real infinity) as standard infinity in maxima ?​

--
Regards,
Pankaj Sejwal
____________________________________________________
"The more I read, the more I acquire, the more certain I am that I know nothing.” -
Voltaire
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pranav kr | 17 Oct 13:38 2014
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incorrect integration

sir,
i got -(2)^1.5*cos(t/2) on integrating "m" manually
while software gives us upon simplification -(2)^1.5*(cos(t/2))^2

where is the mistake...

regards pranav


Attachment (maxima.pdf): application/pdf, 6855 KiB
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Tamas Papp | 17 Oct 10:31 2014
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simplify fractional powers

Consider

declare(alpha, noninteger)$
assume(alpha > 0)$
assume(alpha < 1)$
x: (alpha^(2*alpha/(alpha-1))-alpha^((alpha+1)/(alpha-1)))/((alpha-1)*alpha^(2/(alpha-1)));

I know that x=alpha, but could only verify it with

factor(map(radcan,distrib(x)));

radcan, ratsimp, factor don't simplify the expression. I was wondering
if I am missing a flag or something else that would help, or if the
solution above is the idiomatic one in Maxima.

Best,

Tamas

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Nijso Beishuizen | 16 Oct 23:46 2014
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using partial derivative symbol in wxmaxima

Dear all,

I would like wxmaxima to show the latex symbol \partial in a partial
differential equation. Is it possible to: 
1) show the \partial symbol (del) in wxmaxima?
2) tell maxima that diff(u(x,y),x) is a partial derivative and that the
\partial symbol should be used?

Best,
Nijso

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pranav kr | 16 Oct 10:52 2014
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draw3d animation

hi,
 i tried to plot the following function for animation of spring but alas i get each frame instead which needs to be closed to get another frame  for thru loop closes after each loop in draw3d graph when i close it anotherr frame appears ...please see to it why its happening
 
for k :0 thru 20 step .01 do
draw3d(
xrange=[-2,2],yrange=[-2,2],zrange=[0,50],
nticks=100,
parametric(cos(t),sin(t),(sin(k)+1.2)*t,t,0,8*%pi));
regards
pranav
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Gmane