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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.
>
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