One hour before noon volume 1 of encyclopedia was placed on the shelf?

Question

1/2 hour before noon volumes 2 and 3 of encyclopedia were added to the shelf, and volume 1 was removed.

1/4 hour before noon volumes 4,5,6,7 of encyclopedia were added to the shelf, and volume 2 was removed.

1/8 hour before noon volumes 8,9,10,11, 12,13,14,15 of encyclopedia were added to the shelf, and volume 3 was removed.

each time when time remainig was halved, twice as many new volumes were added and 1 next volume removed.

How many volumes reamian on the shelf at 12:01 pm?

Answer 1

Each time tn = 12 - 1/2^n the number of volumes is increased, N_n = 2 N_(n-1) - 1. As this sequence tends to infinity, we are tempted to say that the number of volumes at noon will be infinity...

But... for any natural number k, the k_th volume was remove at 12 - 1/2^k... Then, for each volume of the encyclopedia, this particular volume is NOT in the shelf... I think that the answer is zero.

Additional details:

I assumed that the encyclopedia is infinite, as the shell is. And the actions of adding and taking volumes from the shell can be performed instantaneously.

Answer 2

We'll have to assume that all activity has ceased immediately after 12:00 pm, so that there's a minute of "no change". Regardless of what happened in each step 1/2, 1/4, 1/8, 1/16, etc, hour before noon, the number of volumes on the shelf has increased, so therefore after 12:00 pm and at 12:01 pm, there will be an infinity of volumes, even though nobody can find the first infinity of volumes.

Answer 3

I'm going to assume N (the number of volumes on the shelf) a continuous function in time:

N( 1 - 1/2^j ) = 2^(j+1) - j - 1

j = 0, 1, 2, 3, ...

time t is expressed in hours from 11:00 pm, and

t = (2^j - 1 ) / 2^j

So:

j = - ln(1-t) / ln(2)

And:

N(t) = 2 / (1-t) + ln(1-t) / ln(2) - 1

when t = 1+1/60 then:

N = -126.91 + 4.5324i

Answer 4

cumulatively,
at 2^(1-n) hour before noon,

[(2^n) - 1] volume(s) have been added and
(n - 1) volume(s) have been removed.

so up to 2^(1-n) hour before noon,
there's [(2^n) - 1] - (n - 1)
= 2^n - n volumes on shelf.

but 2^(1-n) could go on and on without reaching "absolute" zero, even when n reaching infinity.

so when n becomes so big, 2^n - n will also reach infinity.

Answer 5

Using this mode of reasoning, you would never reach 12:01pm, since you keep halving the "time before noon". There would be, therefore, a theortically infinite number of volumes on the shelf, but all before noon!

Answer 6

0

At some point you will run out of volumes to add and then you will just be removing volumes. Eventually all of the volumes we be removed before noon (since you are halving the time to get to noon, all the add/subtracting will be done before noon).