David and Michele Holmgren | 1 Mar 02:45 2003
Picon

Re: New user question

Hi - Even if you try eliminate([f1,f2],[B]) or eliminate([f1,f2],[F]), you
get some very ugly expressions which seem to defy simplfication or solution.
Also eliminate([f1,f2],[B,F]) produces [0] as a result.  So, this system may
defy symbolic solution (unfortunately), and a numerical method seems the
only recourse.

 Dave Holmgren

> > I'm new to Maxima
>
> Thanks for your interest in Maxima!
>
> > f1: (Rb/Bmin) * B * (B - Bmin) * (1 - B/Bmax) - A * B * F = 0;
> > f2: (Rf/Fmin) * F * (F - Fmin) * (1 - F/Fmax) - A * B * F = 0;
> > res: solve ([f1, f2], [B, F]);
> >
> > Now if I input just that, and try to solve it symbolically,
> > Maxima hangs.
>
> I think you have found a bug.
>
> Your system reduces to something of the form:
>
>   q1: b2 * B^2 + b1 * B + b0 + bf * F = 0
>
>   q2: f2 * F^2 + f1 * F + f0 + fb * B = 0
>
> and Maxima seems to run forever on that, too.
>
> I tried solving the system step by step.
(Continue reading)

Dan Stanger | 1 Mar 04:06 2003
Picon

Re: New user question

The commercial macsyma can solve this, however it leaves the
answer in terms of rootof.  However, it looks like the rootof expression
is a polynomial.
Following is the result:
[[b = 0, f = fmin], [b = 0, f = fmax], [b = bmin,
f = (( - bmin^2 * rb + bmin * (bmin + bmax) * rb - bmax * bmin * rb)/(a * bmax *
bmin))], [b = bmax,
f = ((bmax * (bmin + bmax) * rb - bmax * bmin * rb - bmax^2 * rb)/(a * bmax *
bmin))], [f = 0, b = 0],
[b = root_of(rf * rb^2 * b^4 + ( - 2 * rf * bmin - 2 * rf * bmax) * rb^2 * b^3
 + ((rf * bmin^2 + 4 * rf * bmax * bmin + rf * bmax^2) * rb^2 + (fmin + fmax) *
rf * a * bmax * bmin * rb) * b^2
 + (( - 2 * rf * bmax * bmin^2 - 2 * rf * bmax^2 * bmin) * rb^2
 + (( - fmin - fmax) * rf * a * bmax * bmin^2 + ( - fmin - fmax) * rf * a *
bmax^2 * bmin)
 * rb + fmax * fmin * a^3 * bmax^2 * bmin^2) * b + rf * bmax^2 * bmin^2
 * rb^2 + (fmin + fmax) * rf * a * bmax^2 * bmin^2 * rb + fmax * fmin * rf * a^2
* bmax^2 * bmin^2, b),
f = ((b * (bmin + bmax) * rb - bmax * bmin * rb - b^2 * rb)/(a * bmax * bmin))]]

David and Michele Holmgren wrote:

> Hi - Even if you try eliminate([f1,f2],[B]) or eliminate([f1,f2],[F]), you
> get some very ugly expressions which seem to defy simplfication or solution.
> Also eliminate([f1,f2],[B,F]) produces [0] as a result.  So, this system may
> defy symbolic solution (unfortunately), and a numerical method seems the
> only recourse.
>
>  Dave Holmgren
>
(Continue reading)

wang yin | 1 Mar 06:53 2003
Picon

how to do this?

Hi,

I want to deduce n[v]=2*n-n[c]-2 from

n[e]=1/2*(3*n[v]+n[c])
	and
n-n[e]+n[v]=1

Another words: I want to express n[v] in terms of other variables.

But solve does't work this for me.
What function should I use?
	
--

-- 
Wang Yin
DA Lab, Tsinghua University,
100084
Beijing China
Stavros Macrakis | 1 Mar 07:01 2003
Picon
Picon

RE: New user question

Oops.  I said:

   ...substituting into f2; Solve solves the resulting equation
easily...

Which is true.  Solve *did* solve the equation easily.  Unfortunately,
it did not solve it completely -- there is still an F on the right-hand
side.  This brings us back to the messy quartic, which did not magically
cancel away....  And of course the general symbolic solution of a
quartic is a huge mess.  Maxima can calculate it (trust me, I did it),
but the result is enormous and presumably useless.  One way to approach
it is to just take the resulting quartic and run radcan over it for a
few days, in the hopes that something will simplify.  I wouldn't be too
optimistic about that.

Another approach is to approximate.  For example, for A << 1, I get the
following solution (among others):

           A Bmax Bmin (Fmin + Fmax)
B = Bmax - -------------------------
               (Bmax - Bmin) Rb

I applied the approximation after the substitution, and before the
second Solve.

But I have been quick and dirty and sloppy in deriving this, so it is
perfectly possible I've blundered again.  It does come out to within
0.1% of one of the numeric solutions -- is that good enough?

       -s
(Continue reading)

Stavros Macrakis | 1 Mar 07:05 2003
Picon
Picon

RE: how to do this?

> I want to deduce n[v]=2*n-n[c]-2 from
> 
> n[e]=1/2*(3*n[v]+n[c])
> 	and
> n-n[e]+n[v]=1
> 
> Another words: I want to express n[v] in terms of other variables.
> 
> But solve does't work this for me.

You need to solve for the *two* variables:

(C1) n[e]=1/2*(3*n[v]+n[c]);

                                 3 n  + n
                                    v    C
(D1)                        n  = ---------
                             e       2
(C2) n-n[e]+n[v]=1;

(D2)                       n  + n - n  = 1
                            v        e

(C3) solve([d1,d2],[n[v],n[e]]);

(D3)           [[n  = 2 n - n  - 2, n  = 3 n - n  - 3]]
                  v          C       e          C
Wolfgang Jenkner | 1 Mar 07:35 2003
Picon

Re: how to do this?

wang yin <wang-y01 <at> mails.tsinghua.edu.cn> writes:

> I want to deduce n[v]=2*n-n[c]-2 from
> 
> n[e]=1/2*(3*n[v]+n[c])
> 	and
> n-n[e]+n[v]=1
> 
> Another words: I want to express n[v] in terms of other variables.
> 
> But solve does't work this for me.
> What function should I use?

Maybe something like

(C1) eliminate([n[e]=1/2*(3*n[v]+n[c]),n-n[e]+n[v]=1],[n[e]]);
(D1) 			     [- n  + 2 n - n  - 2]
				 v	    c
(C2) solve(%,n[v]);
(D2) 			      [n  = 2 n - n  - 2]
				v	   c
(C3) 

Wolfgang
wang yin | 1 Mar 07:42 2003
Picon

another simple question

Thank you all! :)
One more simple question:

How can I do something like the "apart" operation in Mathematica?

(C1) a/c+b/d;

                                     b   a
(D1)                                 - + -
                                     d   C
(C2) rat(%);

                                   a d + b C
(D2)/R/                            ---------
                                      C d

Now given the form like [D2], how can I get something like D1?

I'm not good at Mathematics but I'm drawing a lot of people in my unversity to MAXIMA.  Formerly they don't
even know this wonderful tool. They use Mathematica a lot.

They'll ask me a lot of questions like "How to do ... like that in Mathematica in MAXIMA?" I sometimes don't
know. So I post the questions here. I wish you could help.

Thanks.

--

-- 
Wang Yin
DA Lab, Tsinghua University,
100084
(Continue reading)

Mike Thomas | 1 Mar 13:09 2003

RE: Windows - maxima.bat

Hi Michel.

It works here on XP, and does not seem to have changed, but I believe that
it could depend on whether you use command or cmd as a shell.  Frankly, I
don't know which I use - I'm not a DOS guru - but either way if yours is
better for W98 that's great!

Cheers

Mike Thomas

| -----Original Message-----
| From: michel.lavaud <at> univ-orleans.fr
| [mailto:michel.lavaud <at> univ-orleans.fr]
| Sent: Thursday, February 27, 2003 9:13 PM
| To: Mike Thomas
| Cc: gcl-devel <at> gnu.org; maxima <at> www.ma.utexas.edu
| Subject: RE: [Maxima] Windows - maxima.bat
|
|
| Hello Mike,
|
| > Very difficult and fragile.
| >
| > Here is an example based on one I found on the web:
| >
| >  <at> SET cd=
| >  <at> SET promp%prompt%
| >  <at> PROMPT SET cd
| >  <at> CALL>%temp%.\setdir.bat
(Continue reading)

Martin RUBEY | 1 Mar 18:14 2003
Picon

Re: another simple question

(C1) a/c+b/d;

                                     b   a
(D1)                                 - + -
                                     d   C
(C2) rat(%);

                                   a d + b C
(D2)/R/                            ---------
                                      C d
(C3)  PARTFRAC(%,c,d);

                                     b   a
(D3)                                 - + -
                                     d   C
 - Function: PARTFRAC (exp, var)
     expands the expression exp in partial fractions with respect to
     the main variable, var.  PARTFRAC does a complete partial fraction
     decomposition.  The algorithm employed is based on the fact that
     the denominators of the partial fraction expansion (the factors of
     the original denominator) are relatively prime.  The numerators
     can be written as linear combinations of denominators, and the
     expansion falls out.  See EXAMPLE(PARTFRAC); for examples.

Martin
Richard Fateman | 1 Mar 19:48 2003
Picon

Re: another simple question

try partial fraction expansion, partfrac.

wang yin wrote:
> Thank you all! :)
> One more simple question:
> 
> How can I do something like the "apart" operation in Mathematica?
> 
> (C1) a/c+b/d;
> 
>                                      b   a
> (D1)                                 - + -
>                                      d   C
> (C2) rat(%);
> 
>                                    a d + b C
> (D2)/R/                            ---------
>                                       C d
> 
> Now given the form like [D2], how can I get something like D1?
> 
> I'm not good at Mathematics but I'm drawing a lot of people in my unversity to MAXIMA.  Formerly they don't
even know this wonderful tool. They use Mathematica a lot.
> 
> They'll ask me a lot of questions like "How to do ... like that in Mathematica in MAXIMA?" I sometimes don't
know. So I post the questions here. I wish you could help.
> 
> Thanks.
> 
(Continue reading)


Gmane