Gunter Königsmann | 25 May 14:13 2015
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Anoter simplification question

Dear all,


I was convinced that I have asked this question already before or at least read the answer on this list. But in the last weeks I didn't find an answer so I thought I'd better risk asking something obvious:

Desolve sometimes provides me with something like this:

I_G(t)=(%e^(-k)*(%e^k*V_1-V_1-%e^k*V_2+V_2+%e^k*V_3))/R

Is there a way to simplify it to something resembling more the following:

I_G(t)=(1-%e^(-k))*(V_1-V_2)/R+V_3/R;

I am aware of the fact that for every need a different form is the simplest way to express an equation. So if there is no way to do this automatically I continue with doing these things by hand and asking maxima to v erify what I did. But sometimes maxima does magic way beyond anybody could expect and - I thought I'd better ask.

Kind regards,

 Gunter.
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Dimiter Prodanov | 25 May 12:17 2015
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workflow diagrams for Maxima

Dear list,

I am writing a conference paper where I also describe some algorithms in Maxima.
I have short question:

Are there some diagrams/ charts available on how simplification and evaluation work?
Specifically, I would like to discuss the pattern matching and simplification.

best regards,

Dimiter 
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Alessandro Languasco | 25 May 09:33 2015
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Re: subst function

Dear Barton and Pankaj,

thanks for your suggestion. In fact it seems that ratsubst 
is able to do what I need, but I’ll also to check Pankaj’s
packages as soon as possible.

Best regards,
	Alessandro

> On 24/mag/2015, at 23:09, Barton Willis <willisb <at> unk.edu> wrote:
> 
> Maybe ratsubst will do what you want; example:
> 
>    (%i53) qqq : (3*d[0]^2*x[1]^3)/4$
> 
>    (%i54) 2015 + qqq * (18-qqq^2+1)  + 78 * qqq^3;
>    (%o54) (1053*d[0]^6*x[1]^9)/32+(3*d[0]^2*x[1]^3*(19-(9*d[0]^4*x[1]^6)/16))/4+2015
> 
>    (%i55) ratsubst(ppp, qqq,%);
>    (%o55) 77*ppp^3+19*ppp+2015
> 
> 
> —Barton

> On 25/mag/2015, at 04:24, Pankaj Sejwal <pankajsejwal <at> gmail.com> wrote:
> 
> ​​
> Maxima has pattern matching functionality ​same as that of Mathematica which makes it a trivial
problem seeing what it can do.
> 
> eq:(3*d[0]^2*x[1]^3)/4;
> eq/.a_*b_[s_]^c_*d_[k_]^e_->a*diff(d[k]^e,b[s],c);​
> 
> here you can see the transformed expression.
> But diff remains nounified so to evaluate it,
> ev(eq/.a_*b_[s_]^c_*d_[k_]^e_->a*(diff(d[k]^e,b[s],c)),diff); => 0
> 
> 
> It is not yet in core of Maxima so to get this functionality,
> 
> http://www.cs.berkeley.edu/~fateman/lisp/mmasim2015/
> Usage:
> load("infix.mac");
> load("e18a.lisp");
> 
> ​Any problem in loading, please feel free to come back here.
> 
> 
> 
> On Sun, May 24, 2015 at 2:10 PM, <maxima-discuss-request <at> lists.sourceforge.net> wrote:
> Message: 5
> Date: Sun, 24 May 2015 20:47:17 +0200
> From: Alessandro Languasco <languasc <at> math.unipd.it>
> Subject: [Maxima-discuss] subst function
> To: maxima-discuss <at> lists.sourceforge.net
> Message-ID: <69C4682D-2C44-4A5C-B5A3-1FEE36F28C5A <at> math.unipd.it>
> Content-Type: text/plain;       charset=us-ascii
> 
> dear all,
> 
> I have a polynomial whose monomials are of the type
> 
> (3*d[0]^2*x[1]^3)/4
> 
> and I would like to substitute it with
> 
> (3*diff(x[1]^3,d[0],2))/4
> 
> To do so I'm using the subst function which works nicely only if the numerical
> coefficient inside the brackets is 1, i,.e., for monomials like this
> 
> (d[0]^2*x[1]^3)/4
> 
> I guess that this depends on the fact that subst works only on a part of
> an expression and not on a subpart.
> 
> I don't see how to solve this, so I'm asking you for a suggestion.
> 
> thanks in advance. best regards
> 
> Alessandro Languasco
> ​
> 
> 
> -- 
> 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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Dennis P. Weygand | 24 May 16:20 2015
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plotting error in wxmaxima (mac yosemite)

I just installed wxmaxima, maxima, gnuplot on my iMac
when trying to generate a simple 2d plot in wxmaxima, I get the following error:


How do I set the terminal? And to what?
Thanks in advance,
D.P. Weygand
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Alessandro Languasco | 24 May 20:47 2015
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subst function

dear all,

I have a polynomial whose monomials are of the type

(3*d[0]^2*x[1]^3)/4

and I would like to substitute it with 

(3*diff(x[1]^3,d[0],2))/4

To do so I'm using the subst function which works nicely only if the numerical 
coefficient inside the brackets is 1, i,.e., for monomials like this 

(d[0]^2*x[1]^3)/4

I guess that this depends on the fact that subst works only on a part of 
an expression and not on a subpart.  

I don't see how to solve this, so I'm asking you for a suggestion.

thanks in advance. best regards

Alessandro Languasco
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Alessandro Languasco | 24 May 08:01 2015
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monomials printed in a specified order

dear all,

I'm a beginner user of maxima so I'm sorry if my question is naive. 

Here is the problem:

I have a multivariate polynomial p(x0,...xn,y0,...,yn) and I would
like that maxima prints it having every monomial

 constant*xi^a * yj^b

 in this particular order; so with the x-variables printed before of the y-variables. 

That would be useful for further manipulations I have to 
perform on such a polynomial.

Some experimentation let me guess
that in fact maxima prints first the indeterminate with maximal 
exponent in the specified monomial (this is just a guess; maybe I'm wrong).

So I'm wondering if there's a way to force maxima to print
monomials this way.

Thanks in advance.
Best regards,
Alessandro Languasco
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(1-x)^m problem

Valenzuela's problem brings up a very annoying bug in Maxima. With m declared as integer, all these are find:

factor(1-x) => -(x-1)
factor((1-x)^5) => -(x-1)^5
factor((2-2*x)^m) => (-2)^m*(x-1)^m

but when the exponent is symbolic:

factor((1-x)^m) => (1-x)^m

and not

... => (-1)^m*(x-1)^m

More
​annoying (presumably related) behavior:


factor((1-x)^(m+2)) => (1-x)^m*(x-1)^2
factor((1-x)^(m+3)) => -(1-x)^m*(x-1)^3

In fact, there seems to be no easy way to pass between the two forms in either direction!

This seems like a bug that should be pretty easy to fix.

             -s

On Sat, May 23, 2015 at 3:55 AM, Mike Valenzuela <mickle.mouse <at> gmail.com> wrote:
Hello all,

I keep running into problem where

(i1) desired: (1-x)^(m-2) / (1 - (1-x)^(m-1));
(i2) starting_form: fullratsimp(desired);

(o1) (1-x)^(m-2)/(1-(1-x)^(m-1))
(o2) (1-x)^m/(x^2+((1-x)^m-2)*x-(1-x)^m+1)

I know the desired form I want the expression in, but almost anything ruins that form. So once the cat's out of the bag, how do I put it back in? I've tried factor, factorsum, gcfac, scanmap( simplification_function, starting_form), etc.


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Nijso Beishuizen | 23 May 15:32 2015
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working with list or array

Dear all,

Recently I have been working on some large systems and computational
efficiency becomes now an issue. For my understanding, I am wondering if
the following is correct: 

maxima lists are linked lists and maxima arrays are 'random access'
arrays.

If L is a list, is this:
(%)for i in L do (stuff on i)

different from this:
(%)for i: 1 thru length(L) do (stuff on L[i]) 

I expect that the first statement will go through the list L only once,
but I'm not sure about the second statement. Because L is a linked list,
I expect that if I need L[i], it needs to traverse through the list from
start to the ith element every time, so the second for loop above would
be less efficient.

Are the following statements correct for maxima?

- if I always go through a list from start to end, I should use a list.
- if I need random access, I should switch to 'array'.

Best regards,
Nijso

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sys | 23 May 13:45 2015

about use 'algsys' to solve big equations system

HI:
   Can Maximal solve out the quadratic polynomial equations system,equations is 151, vars is 75. I at least hope to know whether there 
is a solution. After input the equations, the maximal's state is unknown.
The equation is follow(because it is rejected with too long, I have truncated it):
eq1 :  a_021 * a_122+ a_022 * a_121+  a_031 * a_132+ a_032 * a_131+  a_041 * a_142+ a_042 * a_141;
eq2 :  a_021 * a_123+ a_023 * a_121+  a_031 * a_133+ a_033 * a_131+  a_041 * a_143+ a_043 * a_141;
eq3 :  a_021 * a_124+ a_024 * a_121+  a_031 * a_134+ a_034 * a_131+  a_041 * a_144+ a_044 * a_141;
eq4 :  a_021 * a_125+ a_025 * a_121+  a_031 * a_135+ a_035 * a_131+  a_041 * a_145+ a_045 * a_141;
eq5 :  a_022 * a_123+ a_023 * a_122+  a_032 * a_133+ a_033 * a_132+  a_042 * a_143+ a_043 * a_142;
eq6 :  a_022 * a_124+ a_024 * a_122+  a_032 * a_134+ a_034 * a_132+  a_042 * a_144+ a_044 * a_142;
eq7 :  a_022 * a_125+ a_025 * a_122+  a_032 * a_135+ a_035 * a_132+  a_042 * a_145+ a_045 * a_142;
eq8 :  a_023 * a_124+ a_024 * a_123+  a_033 * a_134+ a_034 * a_133+  a_043 * a_144+ a_044 * a_143;
eq9 :  a_023 * a_125+ a_025 * a_123+  a_033 * a_135+ a_035 * a_133+  a_043 * a_145+ a_045 * a_143;
eq10 :  a_024 * a_125+ a_025 * a_124+  a_034 * a_135+ a_035 * a_134+  a_044 * a_145+ a_045 * a_144;
eq11 :  a_021 * a_121+ a_031 * a_131+ a_041 * a_141;
eq12 :  a_022 * a_122+ a_032 * a_132+ a_042 * a_142;
eq13 :  a_023 * a_123+ a_033 * a_133+ a_043 * a_143;
eq14 :  a_024 * a_124+ a_034 * a_134+ a_044 * a_144;
eq15 :  a_025 * a_125+ a_035 * a_135+ a_045 * a_145;
eq16 : - a_011 * a_122- a_012 * a_121+  a_031 * a_232+ a_032 * a_231+  a_041 * a_242+ a_042 * a_241;
eq17 : - a_011 * a_123- a_013 * a_121+  a_031 * a_233+ a_033 * a_231+  a_041 * a_243+ a_043 * a_241;
eq18 : - a_011 * a_124- a_014 * a_121+  a_031 * a_234+ a_034 * a_231+  a_041 * a_244+ a_044 * a_241;
eq19 : - a_011 * a_125- a_015 * a_121+  a_031 * a_235+ a_035 * a_231+  a_041 * a_245+ a_045 * a_241;
eq20 : - a_012 * a_123- a_013 * a_122+  a_032 * a_233+ a_033 * a_232+  a_042 * a_243+ a_043 * a_242;
eq21 : - a_012 * a_124- a_014 * a_122+  a_032 * a_234+ a_034 * a_232+  a_042 * a_244+ a_044 * a_242;
eq22 : - a_012 * a_125- a_015 * a_122+  a_032 * a_235+ a_035 * a_232+  a_042 * a_245+ a_045 * a_242;
eq23 : - a_013 * a_124- a_014 * a_123+  a_033 * a_234+ a_034 * a_233+  a_043 * a_244+ a_044 * a_243;
eq24 : - a_013 * a_125- a_015 * a_123+  a_033 * a_235+ a_035 * a_233+  a_043 * a_245+ a_045 * a_243;
eq25 : - a_014 * a_125- a_015 * a_124+  a_034 * a_235+ a_035 * a_234+  a_044 * a_245+ a_045 * a_244;
........................................
eq26 : - a_011 * a_121+ a_031 * a_231+ a_041 * a_241;
eq27 : - a_012 * a_122+ a_032 * a_232+ a_042 * a_242;
eq150 :  a_035 * a_045+ a_135 * a_145+ a_235 * a_245;
algsys ([eq1,eq2,eq3,eq4,eq5,eq6,eq7,eq8,eq9,eq10,eq11,eq12,eq13,eq14,eq15,eq16,eq17,eq18,eq19,eq20,eq21,eq22,eq23,eq24,eq25,eq2
6,eq27,eq28,eq29,eq30,eq31,eq32,eq33,eq34,eq35,eq36,eq37,eq38,eq39,eq40,eq41,eq42,eq43,eq44,eq45,eq46,eq47,eq48,eq49,eq50,eq51,e
q52,eq53,eq54,eq55,eq56,eq57,eq58,eq59,eq60,eq61,eq62,eq63,eq64,eq65,eq66,eq67,eq68,eq69,eq70,eq71,eq72,eq73,eq74,eq75,eq76,eq77
,eq78,eq79,eq80,eq81,eq82,eq83,eq84,eq85,eq86,eq87,eq88,eq89,eq90,eq91,eq92,eq93,eq94,eq95,eq96,eq97,eq98,eq99,eq100,eq101,eq102
,eq103,eq104,eq105,eq106,eq107,eq108,eq109,eq110,eq111,eq112,eq113,eq114,eq115,eq116,eq117,eq118,eq119,eq120,eq121,eq122,eq123,e
q124,eq125,eq126,eq127,eq128,eq129,eq130,eq131,eq132,eq133,eq134,eq135,eq136,eq137,eq138,eq139,eq140,eq141,eq142,eq143,eq144,eq1
45,eq146,eq147,eq148,eq149,eq150],[a_011,a_012,a_013,a_014,a_015,a_021,a_022,a_023,a_024,a_025,a_031,a_032,a_033,a_034,a_035,a_0
41,a_042,a_043,a_044,a_045,a_051,a_052,a_053,a_054,a_055,a_121,a_122,a_123,a_124,a_125,a_131,a_132,a_133,a_134,a_135,a_141,a_142
,a_143,a_144,a_145,a_151,a_152,a_153,a_154,a_155,a_231,a_232,a_233,a_234,a_235,a_241,a_242,a_243,a_244,a_245,a_251,a_252,a_253,a
_254,a_255,a_341,a_342,a_343,a_344,a_345,a_351,a_352,a_353,a_354,a_355,a_451,a_452,a_453,a_454,a_455]);




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Nicholas Jankowski | 23 May 12:28 2015

maxima with Windows DEP compliant LISP?

I know there's been some recent discussion about compiling different LISP versions with Maxima. Was wondering if anyone knows if there is a LISP version that is able to run without setting the DEP exception.  Been running Maxima for years, but a recent group policy update at work clears all DEP exceptions, sees it as a security vulnerability. I'm currently going through the process of appealing a standing exception for Maxima, but they may disallow it. I didn't see much discussion on the matter since back around 2006 or so, and I assume it's just been 'the way it is' since then. 

nickj
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Mike Valenzuela | 23 May 09:55 2015
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Factoring and exponents

Hello all,

I keep running into problem where

(i1) desired: (1-x)^(m-2) / (1 - (1-x)^(m-1));
(i2) starting_form: fullratsimp(desired);

(o1) (1-x)^(m-2)/(1-(1-x)^(m-1))
(o2) (1-x)^m/(x^2+((1-x)^m-2)*x-(1-x)^m+1)

I know the desired form I want the expression in, but almost anything ruins that form. So once the cat's out of the bag, how do I put it back in? I've tried factor, factorsum, gcfac, scanmap( simplification_function, starting_form), etc.

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Gmane