1 Mar 02:45 2003

### 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.
```

1 Mar 04:06 2003

### 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
>
```

1 Mar 06:53 2003

### 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
```
1 Mar 07:01 2003

### 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

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
```

1 Mar 07:05 2003

### 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
```
1 Mar 07:35 2003

### 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
```
1 Mar 07:42 2003

### 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
```

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
```

1 Mar 18:14 2003

### 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
```
1 Mar 19:48 2003

### 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.
>
```