arbitrary precision linear algebra

[Robin Becker]

On 02/03/2011 16:39, Ben123 wrote:
...........
>> Languages can't support infinitely large or small numbers, so try to
>> multiply the inner variables by 10^n to increase their values if this
>> will not involve on the method. For example, I did this when was
>> calculating geometric means of computer benchmarks.
>
> Currently I have values between 1 and 1E-300 (not infinitely small). I
> don't see how scaling by powers of 10 will increase precision.
>
>> In such way you will be storing the number of zeros as n.
>
> Are you saying python cares whether I express a number as 0.001 or
> scaled by 10^5 to read 100? If this is the case, I'm still stuck. I
> need the full range of eigenvalues from 1 to 1E-300, so the entire
> range could be scaled by 1E300 but I would still need better precision
> than 1E19
>
.......

If you enter a number as 1e-19 then python will treat as a float; by default I 
think that float is IEEE double precision so you're getting a 48 bit mantissa 
(others may have better details). That means you've already lost any idea of 
arbitrary precision.

When you say you have numbers like 1E-300 are those actually numerically zero or 
have you some valid inputs that vary over a huge range. It should be possible to 
compute determinants/inverses etc to arbitrary precision as those are known to 
be polynomial functions of the input elements. However, eigenvalues are not.

eg

[0 2]
[1 0]

has eigenvalues +/- sqrt(2) so even though you can represent the matrix in 
finite precision the eigenvalues require infinite precision.

Eigenvalues are roots of a polynomial in the elements and root solving may 
require an infinite number of steps so it will be difficult with arbitrary 
matrices to keep arbitrary precision.
-- 
Robin Becker