Damian has two
identical cups, cup 1 and cup 2. Cup 1 is initially full of water, and cup 2 is
initially empty. He begins a procedure in which he exchanges the water among
the cups. For the first turn of the process, he pours half of the water in cup
1 into cup 2. Next, he pours half of the water in cup 2 back into cup 1, and
then half of the total water in cup 1 back into cup 2 (see illustration below).
He continues this process for some time.
If Damian were to
continue this process indefinitely (for an infinite amount of turns), how much
water would remain in each cup?
Set up
Cup 1 is initially
full, so the total volume of water it holds is considered to be 1. A “turn” is
each time Damian pours half of the water in one cup into the other. Let’s
create a table to see how the volume of water changes after a certain number of
turns.
It seems that the
volume of water in cup 1 is approximately 1/3 for odd numbered turns and 2/3
for even numbered turns, and the volume in cup 2 is 2/3 for odd numbered turns
and 1/3 for even numbered turns. This gives us a pretty good idea that if this
process were continued indefinitely, the volume of water in each cup should
oscillate from 1/3 and 2/3, but how can we actually prove this?
What to do in a nutshell
We’ll have to figure
out a formula for the volume of water in each cup after some number of turns
and take its limit as the number of turns approaches infinity.
Getting started
We’ll have to express
the volume of water in each cup in a different manner. It looks more
complicated, but a very interesting pattern becomes evident. Closely examine
the way the volume in each cup is expressed on the chart below:
Because the total
volume for both cups must always be 1, we couldn't end up with more or less than what we started out with. After we've expressed the volume of water in one
cup, we can simply subtract it from 1 in order to have an expression for the
volume of water in the other cup. In
addition, the exponent in the denominator represents the number of times the
volume of water in each cup has been halved.
Let us from now on
focus on even numbered turns of cup 1. Consider the pattern seen in turn 6, an
even turn. The first term that appears in the numerator (2n-1 ) is
raised to a power that is one less than the number of the turn (turn n), and
this is true for the denominator as well. The exponent also lessens by 1 as we progress
through the series. Upon analyzing this pattern, we can write volume of water
in cup 1 for the nth even term as the following series Vn:
If we distribute the
negatives and remove the parentheses, we get the following alternating series:
*First of all, let me
point out why the quantity (1-1/2) is negative. Look at the chart above. If we
examine the expression of the volume in cup 1 for even turns, we’ll notice that
there are an odd number of negatives that get distributed, ultimately leaving
the quantity (1-1/2) as negative. We can conclude that for even turns, this
portion of the series will always be negative, and because Vn
applies specifically to even turns of cup 1, its final term will also be
negative; however, we will soon see that this portion of the series can be
ignored regardless of its sign.
Since that’s out of
the way, let’s go ahead and break up Vn into two parts:
Let’s go ahead and
divide the terms of the series by 2n-1:
This is where it gets
exciting. Notice that the terms of the series share a fixed ratio of 1/4; it’s
a geometric series! It was pretty evident earlier that this is a geometric
series, but doesn't it look so much cleaner now? To make Vn even
more compact, we will rewrite it using sigma notation. Below I've provided the
formula for the nth term of a geometric sequence because we will use it to rewrite
the sum. The term a1 is the first term of the sequence and r is the common ratio
This looks
significantly better. This takes us back to the quantity (1-1/2) that became 2-n
after dividing by 2n-1. Remember that we are working with an
infinite series. As n becomes very large, 2-n approaches 0. That’s
why this term of the series is ignored, and it doesn’t matter to us whether
this quantity is positive or negative; it’s zero either way.
We are interested in
calculating the sum of this geometric series, so we can apply the following
formula:
Notice that the
geometric series we are dealing with has a common ratio that has an absolute
value less than 1, and when such a number is raised to the power of some
positive number greater than 1, the output decreases. For example,
0.52=0.25
and 0.53 is even smaller. If we take the limit of Sn as n
approaches infinity, we will get the following formula:
This is actually the
standard formula for an infinite geometric series. Note that it is only valid
if the absolute value of the common ratio is less than 1.
We are now ready to
apply this formula to Vn. We know that a1 is equal to 1
and r is equal to 1/4:
This is our final
solution, and it seems that our conjecture from the very beginning is correct!
The volume of water held by cup 1 after many even turns is 2/3. From this, we
can conclude that the volume held by cup 2 must be 1/3 for even turns. Well,
what about odd turns? Look at the chart below:
The volume of water
held by each cup will oscillate from 1/3 and 2/3 over and over and over and
over…
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