The following is a long post regarding the solution to the "rubber
worm problem." Only a few of you (Leigh, Chip, Uri, ??) are likely
to be interested in it.
"Uri's value is correct for the continuous case but is high by a
factor of about 10^.25 for the discrete case. Here's what I did:
One can solve the continuous problem directly and find that
df(t)/dt = B/L(t)
where f(t) is the fraction of the rope covered at time t, B is the
worm speed, and L(t) is the length of the rope at time t. Now
L(t) = Lo + At, where Lo is the original length and A is the "rope
stretch speed." So we find
f(t) = C*ln(1+t)
where C = B/A and I have expressed t in units of Lo/A which is 1
second for the given problem. This expression can be inverted to
obtain
t(f) = e^(fC) - 1
With C = 10^5 and f = 1 this gives the solution that Uri
mentioned, i.e., 10^43429.448... units of time. BTW, this number
is *so* big that, as Leigh hinted in his original post, one can
reasonably say that the answer is 10^(4.34x10^4) times *any*
imagineable unit of time from typical strong interaction decay
times all the way up to the age of the universe.
No computer is capable of directly performing the discrete analog
of this case, but for parameters C of less than 8 or 9 a
spreadsheet works fine. What I found when I did this was
surprising. The discrete answer was always smaller by about a
factor of 1.78 as long as the parameter was greater than 3. I
started looking for simple analytical expressions involving
exponentials and ln's that would give me such a number without
success at first.
So I went back to the spreadsheet and programmed it to add another
couple of columns. At the end of each step I was already
calculating f_d (the discrete value of f) so I used it and the
last equation above to calculate--at each step--the value of t for
which the continuous problem would give the same value of f. Then
I divided the required continuous value of t by the discrete value
of t and obtained a ratio that quickly converged, as I had already
noticed, to a value of about 1.78. Interestingly, this ratio is a
function *only* of step number and *not* of the parameter C.
Unfortunately, the value still increases slowly even after 4000
iterations.
So I decided to look at the *rate* of convergence by subtracting
the value at step 500 from that at step 1000, doing the same for
steps 1500 and 1000, 2000 and 1500, 2500 and 2000, etc. What I
found was a series of differences that decreased in proportion to
the series 1, 1/3, 1/6, 1/10, 1/15, 1/21, 1/28, ... This is a
well known infinite series and it sums to exactly 2. Thus,
assuming that I was not being mislead by the numerical
implications, all I needed to do to get an accurate limiting value
of the ratio was to find the difference between its value at step
4000 (for instance) and that at step 2000, multiply that
difference by 2 and add it to the value at step 2000. Doing this,
I got the value 1.781072409. I then went to the library and
looked the number up in a wonderful book by the Borweins (whom
Leigh tells me live just down the Hall from him!) called "The
Dictionary of Real Numbers." The book reminded me that exp(gamma)
= 1.78107241 (where gamma is Euler's constant, .5772156649...)
This is *not* likely a coincidence. In fact, I note that Euler's
constant can be defined in terms of sums that bear striking
similarity to the kinds of things going on in the discrete case.
So, now I "know" that to get the discrete value of t from the
continuous value of t in the case of large enough parameters C
(where large enough means anything greater than 5 for several
digit accuracy), one need only divide the continuous result by
1.78107241... In the case C = 10^5, this gives the discrete
answer as 10^43429.1975...
Finally, a few words about the uncertainty I quoted: I posted my
response last night before consulting the Borweins so I was trying
to be conservative about my knowledge of the ratio. It is worth
noting that the exact value of the ratio is always something less
than exp(gamma) but that that value is good to around six places
after *only* 4000 steps.
John
----------------------------------------------------------------
A. John Mallinckrodt email: mallinckrodt@csupomona.edu
Professor of Physics voice: 909-869-4054
Cal Poly Pomona fax: 909-869-5090
Pomona, CA 91768 office: Building 8, Room 223
web: http://www.sci.csupomona.edu/~mallinckrodt/