Chronology Current Month Current Thread Current Date
[Year List] [Month List (current year)] [Date Index] [Thread Index] [Thread Prev] [Thread Next] [Date Prev] [Date Next]

Re: worm problem, A PARADOX?



On Fri, 7 Mar 1997 kowalskil@alpha.montclair.edu wrote:

At time t=n the tractor's location (along the rigid x axis) is

L{n}=10^5+10^5*n=10^5*(n+1) (A)

The location of the bug, at t=n, is given by iterative formula

x{n}=[x{n-1}+1]*(1+1/n) (B)

That formula yields exactly the same values as the non-iterative formula o
which contains a sum of 1/k terms (Uri's notation). As everybody else, I
say that the bug will catch the tractor when L{n}=x{n}. At that moment

x{n}-x{n-1}=LST=10^5 (C)

where LST is the length of the last step.

No. I assume that you mean for LST to be *defined* to be x{n} - x{n-1}.
If so, then LST is approximately equal to but slightly greater than 10^5.
In fact LST is exactly equal to 100001. There is a world of difference
between approximately equal and equal: You needn't exercise any
particular care in applying an equality; you must exercise extreme care in
applying an approximation.

(the last speed of the bug
is 10^5 cm/s, the same as the speed of the tractor).

I don't know how to interpret this in any precise way, but it doesn't
appear to be critical to your argument so I'll simply let it slide.

I know, from the
sollutions of Chip, Uri or John, that x{n} and x{n-1} are gigantic in
comparison with 10^5. What is one kilometer (10^5 cm) in comparison with
zilions of trip around the universe? Therefore I can assume, while writing
L{n}=x{n}, that x{n}=x{n-1}.

No; x{n} is approximately equal to but slightly larger than x{n-1}. It is
in fact larger by 100001. Again, there is a world of difference between
approximately equal and equal.

Once I do this I conclude that 1/n --> 0 and
n --> infinity, as shown below. What is wrong?

L{n}=[x{n-1}+1]*(1+1/n) or (D)

Yes

L{n}=[L{n}-LST]*(1+1/n) or (E)

No; L{n} is approximately equal to but slightly larger than
[L{n}-LST]*(1+1/n). In fact, L{n}=[L{n}-LST]*(1+1/n) + (1+1/n)

L{n}=[L{n}-0]*(1+1/n) (F)

No; L{n} is approximately equal to but slightly less than
[L{n}-0]*(1+1/n). In fact, L{n}=[L{n}-0]*(1+1/n) - 10^5*(1+1/n). If you
subtract the enormous value L{n} from both sides and then simplify this
equation you get L{n} = 10^5*(n+1) as we already knew. No paradox here.

I know that n=10^43429.443 and n--> infinity are practically identical.

(This might qualify as the most incorrect thing you've said.)

But conceptually they are very different. How can a neglection of an
unsignificant term, LST, change a finite solution of a problem into
infinity? Where did I err?

The "significance" of a term is always judged only in relation to
something else. No term is ever "insignificant" in absolute terms.
Fundamentally, this is where you erred.

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/