One hour before noon volume 1 of encyclopedia was placed on the shelf? - 9 1/2 shelf and closet rod bracket
1 / 2 hours before noon, 2 a.m. to 3 p.m. volumes of the encyclopedia has been added to the platform, and volume 1 has been deleted.
1 / 4 hour before noon 4,5,6,7 volumes of the encyclopedia has been added on the platform, and Volume 2 has been withdrawn.
Added 1 / 8 hour before noon volumes 8,9,10,11, 12,13,14,15 was free, the platform, and Volume 3 has been withdrawn.
remains when the time is divided by two times, many new volumes and 1 volume next eliminated.
Reamian many books on the platform of 12.01 clock?
Tuesday, January 19, 2010
9 1/2 Shelf And Closet Rod Bracket One Hour Before Noon Volume 1 Of Encyclopedia Was Placed On The Shelf?
6 comments:
Every time when t = 12 to 1 / 2 ^ n is the number of volumes increased n_n N_ n = 2 (-1) - 1 In this sequence tends to infinity, we are tempted to say that the number of volumes will be at noon infinitely ...
But ... k for every natural number, volume k_th retire at 12 - 1 / 2 h. .. Then, for each volume of the encyclopedia, this work is not in the drawer ... I think the answer is zero.
More information:
I think free is infinite, and deposition. And the actions that can be done with the scope and the dish immediately.
We must assume that all activities to immediately discontinue after 12:00 clock, making it one minutes "no change". Regardless of what happens at each step, 1 / 2 1 / 4, 1 / 8, 1 / 16, etc. hours before noon, the number of volumes increased on the platform, so that after 12:00 Clock-12: 01 hours, a large number of volumes, even if nobody can find the first infinite quantities.
0
At some point you have to add more volume, then you will only be available volumes. Finally, to eliminate all volumes in the morning (as they are to halve the time to arrive about noon, all add, subtract / instead) before noon.
I will assume N (the number of volumes on the shelves) is a continuous function in time:
N (1 1 / 2 ^ j) = 2 ^ (j +1) - J - 1
j = 0, 1, 2, 3, ...
t is the time in hours between 11:00 and Clock
t = (2 ^ j - 1) / 2 ^ j
So:
J = - ln (1-t) / ln (2)
Y:
N (t) = 2 / (1-t) + ln (1-t) / ln (2) - 1
if t = 1 1 / 60 at:
N = -126.91 + 4.5324i
By reducing this argument that would never come, at 12.01 clock, in time to "before noon, at the half. It would therefore theortically countless volumes on the shelf, but all before noon!
Cumulative
2 ^ (n-1) hour before noon,
[(2 ^ n) - 1], (volume s) were added and
(n - 1) amount (s) are eliminated.
well, n to 2 ^ (-1) hour before noon,
is [(2 ^ n) - 1] - (n - 1)
= 2 ^ n - volumes on the shelf.
but 2 ^ (1-N) could go on and on, without absolute zero, although the scope of infinite n.
n, so do that if n is very large, 2 ^ - and the absolute perfection.
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