Re: Two questions (1) Sollve and (2) Precision

*To*: mathgroup at smc.vnet.net*Subject*: [mg67043] Re: Two questions (1) Sollve and (2) Precision*From*: "Bharat Bhole" <bbhole at gmail.com>*Date*: Thu, 8 Jun 2006 04:52:56 -0400 (EDT)*References*: <a3224ef10606061329v71bacd73q75928f8db772e99c@mail.gmail.com>*Sender*: owner-wri-mathgroup at wolfram.com

Hello All: Thank you for those who replied. I will try your suggestions. However, there seems to be one problem in the Solve problem I posted. I originally wrote the equation as 64919121*x-159018721*y ==1 and 41869520.5*x-102558961*y == 0 but in your responses I see that it has changed to "=8A1" and "=8A0" respectively (without the quotes). I wonder why? So as far as entering the equation is concerned, it had been entered correctly. I am surprised that it is working out well for Jean-Marc Gulliet. Does this mean there is something wrong with my computer. I am also using Mathematica version 5.2. I will try again. It is still unclear why its not working on my computer. On 6/6/06, Bharat Bhole <bbhole at gmail.com> wrote: > > 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. >