Re: infinity
- From: David C. Ullrich <ullrich@xxxxxxxxxxxxxxxx>
- Date: Thu, 04 Aug 2005 09:28:31 -0500
On 4 Aug 2005 01:36:00 -0700, snapdragon31@xxxxxxxxx wrote:
>Hi Theo. Before answering your question, I would like to do some math
>on infinite series first and then explain to you why the vase is not
>empty at the noon time.
>
>Let S = 1 + 2 + 3 + 4 + 5 + 6 + ... + oo oo = infinity
>Then 2*S = 2 + 4 + 6 + 8 + .... + oo
>Subtract S from 2*S, we have
> 2*S - S = (2 + 4 + 6 + ... + oo) - (1 + 2 + 3 + 4 + 5 + 6 + ... +
>oo)
>==> S = (2 + 4 + 6 + ... + oo) - (1 + 3 + 5 + ... + oo + 2 + 4 + 6 +
>... + oo)
>==> S = (2 + 4 + 6 + ... + oo) - (1 + 3 + 5 + ... + oo) - (2 + 4 + 6 +
>... + oo)
>==> S = -(1 + 3 + 5 + 7 .... + oo)
>What? The summation of all integers is a negative number? Obviously
>there must be something wrong. The error is that we cannot rearrange
>numbers like that for an infinite series.
>
>Now go back to your question
>Let S1 be the balls being put in the vase
>Let S2 be the balls being taken out from the vase
>
>At time = noon
>S1 = {1, 2, 3, ... , oo} n(S1) = oo
>S2 = {1, 2, 3, ... , oo} n(S2) = oo
># of balls in the vase would be n(S1) - n(S2) = oo - oo which is
>undefined.
>As seen from the previous example, we cannot just cancel out same terms
>from each set and come up with result zero. S1 and S2 are infinite
>series with different properties. We cannot simply say every number in
>the set S1 also exists in S2 then S1 - S2 = 0. If you do that then you
>are rearrangeing the numbers in the infinite series.
Reading the start of this post I was all set to reply that it
was all nonsense. But the fact that the calculations above
are wrong was exactly your point, great.
>In mathematics, usually if at 'noon' the question is undefined then we
>would take the limit of the time --> noon to see the tendency of the
>result
>As time --> noon
>S1 = {1, 2, 3, ... , 10n) Number of S1 = 10n
>S2 = {1, 2, 3, ... , n) Number of S2 = n
>The remaining balls are {n+1, n+2, ... , 10n}
># of balls remained in the vase would be n(S1) - n(S2) = 10n - n = 9n >
>0
>Therefore, the vase is not empty and actually it is growing rapidly.
Alas, this is wrong. The number of balls in the vase at noon is
simply not equal to the limit as n -> infinity of the number of
balls at the n-th stage. It's exactly right that the number of
balls tends to infinity as we approach noon - it does not follow
that the vase is not empty at noon. It _is_ empty at noon.
(Which one of the balls does remain at noon, in your opinion?)
>>>From a physical point of view, the number of balls in the vase are
>growing rapidly as the time approaches noon. Even though, right at
>the noon time, the problem is undefined but we would expect the vase is
>not empty.
************************
David C. Ullrich
.
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