# Re: The worm problem (was Looking for a new soapbox)

And one last word from me on the rubber worm...
Aha! Uri's result for the discrete case points me to the proof of
the result that I had noticed numerically. As he found, the
fraction of the rope covered after step n in the discrete case is
f = (sum from k=1 to n of (1/k)) / C
where C is the single parameter for the problem which I defined as
(rope speed/worm speed). Combining this with my result that the
time (analogous to n) required to cover a fraction of the rope in
the continuous case is
(t = ) n_cont = e^(fC) - 1
we get the ratio of time (or number of steps) required for a given
f in the continuous case to that in the discrete case is
r = n_cont/n
= (e^(fC) - 1)/n
= [e^(sum from k=1 to n of (1/k)) - 1]/n
A ratio that is independent of C as I had noticed numerically.
Now I was interested in the limit of this ratio
r_inf = limit as n -> infinity of r
= lim as n -> inf of e^(sum from k=1 to n of (1/k))/n
Taking natural logs of both sides
ln(r_inf) = lim as n -> inf of [(sum from k=1 to n of (1/k)) - ln(n)]
which is one of the known representations of Euler's constant, gamma.
Thus,
r_inf = exp(gamma)
QED
Finally, I can get back to work!
John
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A. John Mallinckrodt email: mallinckrodt@csupomona.edu
Professor of Physics voice: 909-869-4054
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