Re: a big limit of mathematica?

Dana wrote:
Hi. Looks like as the size of your matrix gets bigger, the Determinant tends towards zero.
However, I can't explain why with a size 100*100 (the first output), it returns "0." and with the largest sizes, numbers that are very small.

In[1]:=
Simplify[(i-j)+3/j^4+50*i]

Out[1]=
51*i+3/j^4-j

In[2]:=
Table[Det[Table[51.*i+3/j^4-j,{i,n},{j,n}]],{n,100,400,100}]


Out[2]=
{0.,
-1.1058552925176574276594077*^-2291,
 7.9451068970561274753784538*^-3316,
-2.0828859268407618803830820*^-4315
}




In Mathematica 5.0.0 on an Intel Pentium 4,

{
 -2.3072190778544011494885203792388`15.954589770191005\
*^-1432,
 1.6138130601828444080943999966955787`15.9545897701910\
05*^-2817,
 -2.17670639763548416057774043151`15.954589770191005*^\
-4180,
 1.283924316663441664176128821`15.954589770191005*^-55\
02}

So the result "0.0" is a recent innovation.  On my
system, the Precision of the first number is 15.9546
(decimal digits),  not MachinePrecision, which makes
one wonder what is going to happen in version 5.3 or 5.4.
Mathematica's numerical model has been criticized by
several people (including me), since Mathematica 1.0.


If you do the calculation EXACTLY  by using 51 instead of "51."
the determinant of the 100X100 matrix comes out exactly 0.
In fact, the determinant of the 10x10, 20x20 etc all come out
exactly 0.

The advantage of doing exact arithmetic  (and using a CAS
like Mathematica, Maple,Macsyma etc) over Matlab,  is that the results
can often be exactly right. In Matlab the result will
be exactly right only by coincidence.

For example, is there anything special about 51.0,  51.000000000000000 or
exactly 51?   Well, why not just put in a symbol, like A instead
of 51. and compute the determinant. the 10X10 determinant
is exactly zero. I assume that knowing the answer you can prove
the determinant is zero always.

Try that in Matlab.  Actually, matlab can be coerced to doing
it, but only by calling Maple.  A much more versatile system
then would actually be sometime like Mathematica, which could,
if necessary for efficiency, call Matlab.  The Mathematica
people claim to be doing that, or maybe they claim to have
already done that.

So, for the original poster:  your complaint about mathematica
running out of memory has not been solved, but your reason for
trying that computation is probably no longer compelling.

RJF


.

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