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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.  

  r_inf = exp(gamma)


Finally, I can get back to work!

  A. John Mallinckrodt      email:
  Professor of Physics      voice:  909-869-4054
  Cal Poly Pomona             fax:  909-869-5090
  Pomona, CA 91768         office:  Building 8, Room 223