On 25-Jun-01, Russell Standish wrote: > Obviously, what you're looking for is some kind of counter example. I > think the problem lies in not being able to determine at any point of > the calculation just how many digits of the limit you have found.

## Advertising

OK, I can understand that. In order for a number to be considered computable not only must there be an index m beyond which the the first n places don't change, but we must be able to put a bound on m as a function of n. For > the counterexample what we need is a computable series, which we know > converges, yet we cannot compute the limit. > > Perhaps the more mathematically nimble might try the following example > x_i = \sum_{j=0}^i 1/p_j, where p_j is the jth prime number. I suspect > this is a convergent sequence, yet converges too slowly to compute the > limit. I don't think that's a counterexample, as Euler proved the sequence to diverge (although *very* slowly). Brent Meeker > > Cheers > > Brent Meeker wrote: >> >> On 25-Jun-01, Russell Standish wrote: >> No - the set of computable numbers does not form a continuum. >> Continuity is related to the concept of limits: {x_i} is a >> convergent sequence if >> \forall \epsilon>0, \exist N: |x_i-x_N|<\epsilon. >> A continuous space is one for which every convergent sequence >> converges to a limit, ie >> >> \exists x: \forall\epsilon>0\exists N: |x_i-x|<\epsilon \forall i>N. >> >> we commonly denote x by lim_{i->\infty}x_i. >> >> There are many convergent sequences x_i whose limits cannot be >> computed (uncountably many, in fact). >> >> Thanks for the education vis a vis definition of a continuum, but now >> I don't see why the computable numbers don't form a continuum. >> Suppose a0,a1,a2,...ai,... is a convergent sequence of computable >> numbers with limit A. Then it seems that A must be a computable >> number since given any number of decimal places (or bits) n there is >> a value of i=m such that am is equal to A for the first n places and >>> m. Isn't this >> the definition of a computable number - one whose representation can >> be computed to a given accuracy in a finite number of steps? So the >> computability of the sequence ai entails computability of the >> sequences limit. >> >> thnx, Brent Meeker >> I am very interested in the Universe - I am specializing in the >> Universe and all that surrounds it. >> --- Peter Cook >> >> > > > > ---------------------------------------------------------------------------- > Dr. Russell Standish Director High Performance Computing Support Unit, > Phone 9385 6967 UNSW SYDNEY 2052 Fax 9385 6965 Australia > [EMAIL PROTECTED] Room 2075, Red Centre > http://parallel.hpc.unsw.edu.au/rks > ---------------------------------------------------------------------------- Regards