Since people startedposting their solutions to the "worm problem" we were
given by Leigh, I became curious: Chip Sample had one solution, John
Mallincrodt a different one, so I thought I would post mine - Leigh, you
will have to give us grades now...
Anyway: doing the discrete version as posed by Leigh, I find that after n
steps the worm will be at a distance d(n), such that:
d(n)= (sum from k=1 to k=n of (1/k))*(n+1), in centimeters, while the rope
will be simply
(n+1) kilometers long.This is nice, since it means that we simply have to
find n such that
(sum from k=1 to k=n of (1/k))=10^5
The sum can easily be found in tables: it is the harmonic series, which
diverges to infinity as I remember from my freshman maths, so we are
certain the worm will get there. A _good approximation_ is to replace the
sum by an integral and obtain that
log(n)=10^5, (log=natural logarithm)which leads to n=10^43429.448...
This was done with paper and pencil (no computer, although I am using a
PowerMac9500). I am curious about the (+/-) in John's answer.
Uri
Prof. Uri Ganiel
Head, Department of Science Teaching
The Weizmann Institute of Science
Rehovot 76100, ISRAEL