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Re: Re: Solve's Strange Output
*To*: mathgroup at smc.vnet.net
*Subject*: [mg89124] Re: [mg89096] Re: Solve's Strange Output
*From*: Andrzej Kozlowski <akoz at mimuw.edu.pl>
*Date*: Mon, 26 May 2008 01:31:47 -0400 (EDT)
*References*: <g1avr4$fev$1@smc.vnet.net> <200805251027.GAA22888@smc.vnet.net>
On 25 May 2008, at 19:27, Szabolcs wrote:
> On May 25, 9:09 am, Bruce Colletti <bwcolle... at verizon.net> wrote:
>> Re 6.0.2 under WinXP.
>>
>> This code's output is strange: what does 0.-7.9424 g mean? Ditto
>> for=
> all values returned by Solve.
>>
>> Thankx.
>>
>> Bruce
>>
>> {x[0],y[0]}={10.5,6.08};
>> {x[1],y[1]}={3.23,14.4};
>> {x[2],y[2]}={18,12.7};
>> m=16.1;
>>
>> Solve[{a+c==0,b+d==m*g,d(x[2]-x[0])==c(y[2]-y[0]),a(y[1]-y[0])=
> ==b(x[1]-x[0])},{a,b,c,d}]
>>
>> Out[11]= {{a->0.-7.9424 g,b->0.+9.08951 g,c->0.+7.9424 g,d-
>> >0.+7.01049 g=
> }}
>
> It has been mentioned many times that using Solve with inexact numbers
> invites trouble. Though in this specific case nothing bad happens, it
> is better to Rationalize the numbers before solving:
>
> Solve[Rationalize[{a + c == 0, b + d == m*g,
> d (x[2] - x[0]) == c (y[2] - y[0]),
> a (y[1] - y[0]) == b (x[1] - x[0])}], {a, b, c, d}]
>
> 0 is not the same as 0.0. The latter is an inexact zero (we only know
> that it is closer to 0 than $MinMachineNumber), so Mathematica does
> not simplify 0. + g.
>
The advice not to mix symbolic algebraic methods with approximate
numbers is a sound one in general, but in this particular case no
serious symbolic algebra is involved, so one can simply apply Chop to
the answer returned by Solve:
Chop[Solve[{a + c == 0, b + d == m*g,
d*(x[2] - x[0]) == c*(y[2] - y[0]),
a*(y[1] - y[0]) == b*(x[1] - x[0])},
{a, b, c, d}]]
{{a -> -7.942397088866652*g, b -> 9.089510836227038*g,
c -> 7.942397088866652*g, d -> 7.010489163772964*g}}
In more complicated situations the alternative to rationalizing is to
use NSolve, which can be very much faster and is intended for dealing
with algebraic-numeric issues:
Chop[NSolve[{a + c == 0, b + d == m*g, d*(x[2] - x[0]) ==
c*(y[2] - y[0]), a*(y[1] - y[0]) == b*(x[1] - x[0])}, {a, b,
c, d}]]
{{a -> -7.942397088866652*g, b -> 9.089510836227038*g,
c -> 7.942397088866652*g, d -> 7.010489163772964*g}}
but of course in this very simple case it makes not difference which
method we use.
Andrzej Kozlowski
Andrzej Kozlowski
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