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

**References**:**Re: Solve's Strange Output***From:*Szabolcs <szhorvat@gmail.com>