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From: Brent Meeker <meekerdb.domain.name.hidden>

Date: Mon, 25 Jun 2001 21:14:55 -0700

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.
*

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
*

*> R.Standish.domain.name.hidden Room 2075, Red Centre
*

*> http://parallel.hpc.unsw.edu.au/rks
*

*>
*

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Regards

Received on Mon Jun 25 2001 - 22:21:26 PDT

Date: Mon, 25 Jun 2001 21:14:55 -0700

On 25-Jun-01, Russell Standish wrote:

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

I don't think that's a counterexample, as Euler proved the sequence to

diverge (although *very* slowly).

Brent Meeker

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Regards

Received on Mon Jun 25 2001 - 22:21:26 PDT

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