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Re: Re: Two questions (1) Sollve and (2) Precision
*To*: mathgroup at smc.vnet.net
*Subject*: [mg67081] Re: [mg67072] Re: Two questions (1) Sollve and (2) Precision
*From*: Bob Hanlon <hanlonr at cox.net>
*Date*: Fri, 9 Jun 2006 01:06:49 -0400 (EDT)
*Reply-to*: hanlonr at cox.net
*Sender*: owner-wri-mathgroup at wolfram.com
He also needs to force exact methods by rationalizing the coefficients.
eqns={64919121*x-159018721*y==8A1,
41869520.5*x-102558961*y==8A0}//Rationalize;
soln=Solve[eqns,{x,y}]//Flatten
{x -> -16*(159018721*A0 - 102558961*A1),
y -> -8*(129838242*A0 - 83739041*A1)}
soln/.{A0->0,A1->1/8}
{x -> 205117922, y -> 83739041}
Bob Hanlon
---- ben <benjamin.friedrich at gmail.com> wrote:
> Hi Bharat,
>
> (1)
>
> Try == instead of = in your equations.
>
> (2)
>
> Numerical results are usually printed with only a few digits,
> not with full accuracy.
>
> Bye
> Ben
>
> Bharat Bhole schrieb:
>
> > Would appreciate if someone can point out why Mathematica is not giving the
> > expected output in the followng two cases.
> >
> > (1) I was trying to solve the follwing two linear equations using 'Solve'.
> >
> >
> > *In: Solve[{64919121*x-159018721*y=8A1,41869520.5*x-102558961*y=8A0},{x,y}]*
> >
> > *Out: {}*
> >
> > However, the solution exists and is given by x = 205117922, y = 83739041
> >
> > Why is Mathematica unable to solve this simple linear equation? Am I doing
> > something wrong?
> >
> >
> >
> > (2) I suppose that the default precision for numerical calculations is
> > MachinePrecision which is less than 16. If I increase the precision, should
> > I not get more accurate results? The example below seems to contradict that.
> >
> > (i) Exact Calculation
> >
> > *In[1]: 123456789123 * 123456789123*
> >
> > *Out[1]: 15241578780560891109129*
> >
> > (ii) Numerical Calculation with Default Precision
> >
> > *In[2]: 123456789123 * 123456789123.0*
> >
> > *Out[2]: 1.52416 =D7 10^22*
> >
> > (iii) Numerical Calcuation with a higher precision.
> >
> > *In[3]:SetPrecision[ 123456789123 * 123456789123.0 , 50 ]*
> >
> > *Out[3]: 1.5241578780560891838464000000000000000000000000000 x 10^22*
> >
> > Now if I calculate Out[1]-Out[2], I get zero.
> >
> > But if I calculate Out[1]-Out[3], I get -729335.000000000000000000000000000
> > .
> >
> > This seems to suggest that calculation 2 is more accurate even though it has
> > smaller precision. Where am I making a mistake?
> >
> > Thanks very much for your help.
>
--
Bob Hanlon
hanlonr at cox.net
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