# [rc5] Re: OS/2 ramdisk and spindown

Eric Gindrup gindrup at okway.okstate.edu
Wed Oct 15 12:51:39 EDT 1997

```     Actually, this isn't a terribly good way to do this.  The result of
your method depends on the indexing of the set.  A better way would be
in analogy to Riemann sums:
Let S be the set of values whose median we desire.
Let n be a non-negative integer.
Let C_n be the collection of n-element subsets of S (rather a lot).
Let M(x) compute the median of the finite set x.
Let K be the collection of sequences of elements taken successively
from the C_n where each sequence member is a subset of the following
sequence member.
Let K_n be the collection of n-element sequence heads of elements of
K.
NOW...
Consider lim sup_{k_n in K_n, n->inf} M(k_n) and the same lim inf.  So
we're ,looking at the upper and lower limit medians as sequences from
S are evaluated for successively longer lengths.  If the two lims
converge to the same value, then we might say that the median of the
set converges.  For finite sets the medians converge to the same value
because all of the above sets are finite.
It's not entirely clear that these limits are identical very often
for infinite sets.
-- Eric Gindrup ! gindrup at Okway.okstate.edu

Oh, the analogy:
Riemann sums are (fully rigorously) the coincident value of the lim
sup and lim inf of the finite sums constructed by taking the interval
of integration and breaking it into partitions and successively
refining the partitions by adding more division lines.  So S goes with
the region of integration, n is the number of regions we are breaking
it into at the moment, C_n is the set of all n-part partitions, K is
the set of all refinement sequences of partitions, and K_n goes with
the finite intermediate partitions used to compute the lims sup anf
inf.

Subject: Re: [rc5] Re: OS/2 ramdisk and spindown
Author:  <rc5 at llamas.net > at SMTP
Date:    1997/10/15 12:09

In order to try to resolve the objection raised by James Mastros

The median of an infinite set is m if and only if the limit as n goes
to infinity of the ratio of (number of set elements with indices less
than n and values less than m)/(number of set elements with indices
less than n and values greater than n) is 1.

or

lim    (# elements <m)/(# elements >m)  = 1
n->inf

This uses the pre-existing definition of the limit to take care of
problems with the infinitude of the set.  Of course, the set has to be
indexed...

(The problem of a set which doesn't contain elements on both sides of
its median can be solved by first adjoining (m-1) and (m+1) -- they
allow the ratio to be well-defined, but they won't affect the value of
the median itself.)

Sorry for the rather non-topical discussion; I'm so far behind that I
don't even really know how the thread got started.

--
Seth David Schoen L&S '01 (undeclared) / schoen at uclink4.berkeley.edu
Magna dis immortalibus habenda est atque huic ipsi Iovi Statori, antiquissimo
custodi huius urbis, gratia, quod hanc tam taetram, tam horribilem tamque
infestam rei publicae pestem totiens iam effugimus.  -- Cicero, in Catilinam I
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```