NetNews Usenet Archive 1992 #26/NN_1992_26.iso/spool/comp/theory/2395

Text File | 1992-11-11 | 1.5 KB | 48 lines

Newsgroups: comp.theory
Path: sparky!uunet!sun-barr!ames!ads.com!marcel
From: [e-mail] (Marcel Schoppers)
Subject: asymptotic value
Message-ID: <[e-mail]>
Sender: Marcel Schoppers
Organization: Advanced Decision Systems, Mtn. View, CA [phone]
Date: Thu, 12 Nov 1992 08:25:50 GMT
Lines: 37

All,
    I need some help figuring out an asymptotically correct value.
Given the equation

        b^s = s + b^(s-2)

set b=1+c and find c in terms of s as s becomes large.  Empirically, c -> 0
from above, but how fast?  I can tell you that b^s >> s, but one has to be
very careful about using this fact.  For example, if you use it in the above
equation you get b=1 and immediately wipe out your chances of finding the
error term.  Rearranging the original equation one can get to

        b^s = s b^2 / (b^2 - 1)
            : s/2c

where ":" means "approximately equal".

At this point if you substitute b=1+c on the left and expand, you get 1+sc
which is nowhere near right: at s = 900, b=1.01, c=.01, b^s=9000 and the
erstwhile approximation to 9000 is 1+sc=10.  So instead, raise both sides
to the power 1/s to get

By defn,    b = 1+c
From above,      : (s/2c)^(1/s)
Taking logs,    c : (1/s) ln (s/2c)
        c*s : ln (s/2c)

Now at least we've got an equation in c and s only, but from here I'm stumped
how to get an asymptotic expression for c.  If we do

        c*s : ln s - ln 2c

we find that at s=900, the terms are all of comparable magnitude, so we can't
just drop one (they are, 9.0 : 6.8 - -3.9).  So now what?  Can anyone help?

Marcel