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WHETSTONE 45
Geometric Progression, Applications: Part 1
The area under a curve – Fermat’s technique
Finding areas of shapes with straight sides, such as rectangles, triangles, and other polygons involves
practising techniques that have been known for thousands of years. Finding areas of curved shapes
proves more challenging and it was largely due to developments in the calculus of the late 17th
Century that such problems were largely solved. In the lead up to the invention of the calculus a
number of ingenious methods for finding areas under curves had been discovered. (When we talk
about areas under curves we are really referring to closed regions of the Cartesian plane, bounded
by the curve of some function, the 𝑥-axis, and vertical lines 𝑥 = 𝑎 and 𝑥 = 𝑏 with 𝑏 > 𝑎).
One such discovery concerns the area under the curve 𝑦 = 𝑥 2 between 𝑥 = 0 and 𝑥 = 𝑎. Pierre de
Fermat, sometime after 1629, used a geometric sequence and a sum to arrive at an answer. Here is
how he did it.
Fermat broke up the area into circumscribed rectangles of diminishing width, by putting the first
rectangle with its right hand vertical edge at 𝑥 = 𝑎. (The idea is shown in the diagram below.) The
second rectangle, to the left of the first, has its right hand edge at 𝑥 = 𝑎𝑟 where 𝑟 is a positive
geometric ratio less than 1. This means that the second rectangle will be slightly narrower than the
first. The third rectangle’s right hand edge is then drawn at 𝑥 = 𝑎𝑟 2 and the fourth rectangle’s edge
is at 𝑥 = 𝑎𝑟 3 and this continues indefinitely with each new rectangle narrower than the one that
preceded it. Note that no finite number of rectangles will ever reach the origin if we continue in this
manner.
𝑦 = 𝑥2
𝑎
𝑎𝑟 4 𝑎𝑟 3 𝑎𝑟 2 𝑎𝑟
The diminishing width rectangles on 𝑦 = 𝑥 2
The sum of the areas of these rectangles is given by 𝑆𝐴 where
𝑆𝐴 = (𝑎 − 𝑎𝑟) ∙ 𝑎2 + (𝑎𝑟 − 𝑎𝑟 2 ) ∙ (𝑎𝑟)2 + (𝑎𝑟 2 − 𝑎𝑟 3 ) ∙ (𝑎𝑟 2 )2 + ⋯
When factorised, this is 𝑆𝐴 = 𝑎3 (1 − 𝑟){1 + 𝑟 3 + 𝑟 6 + ⋯ }. Then, using the limiting sum formula for
𝑎
a geometric sequence, 𝑆∞ = 1−𝑟 , we have
1
𝑆𝐴 = 𝑎3 (1 − 𝑟) (
)
1 − 𝑟3
= 𝑎3 (1 − 𝑟)
1
(1 − 𝑟)(1 + 𝑟 + 𝑟 2 )
177
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GEOMETRIC PROGRESSION, APPLICATIONS: Part 1
=
𝑎3
(1 + 𝑟 + 𝑟 2 )
Fermat now made the ratio 𝑟 approach 1. This caused the values {𝑎, 𝑎𝑟, 𝑎𝑟 2 , … } to draw closer
together. Since 𝑟 is to be only slightly less than 1, the rectangles become very narrow, but the width
of each successive rectangle continues to reduce indefinitely. Fermat saw that as 𝑟 → 1, the total
area of all the rectangles would approach the area under the curve as the limiting value.
From the derived equation for 𝑆𝐴 it is clear that (1 + 𝑟 + 𝑟 2 ) is approaching 3 and so putting
everything together, Fermat could say that the area under the curve 𝑦 = 𝑥 2 between 𝑥 = 0 and
𝑥 = 𝑎 and above the 𝑥 axis, is 𝑆𝐴 =
𝑎3
.
3
This result agrees with the one we now find routinely in
integral calculus.
The Rule of 72
An investment of 𝑎 = $100 made at the beginning of the year in a Bank offering 3% pa for a year
will have a future value (𝐹𝑉) at the beginning of the 2nd year of $103. A way to write this
calculation is 𝐹𝑉2 = $100(1.03) = $103. Under the same conditions, left alone till the beginning of
the third year, the future value would become 𝐹𝑉3 = (𝐹𝑉2 ) × 1.03 and so 𝐹𝑉3 = $100(1.03)2 =
$106.09. We realise then that 𝐹𝑉𝑛 = $100(1.03)𝑛−1
In general, the future value (in dollars) at the beginning of the 𝑛th year of an investment of $𝑎 at 𝑟%
will be 𝐹𝑉𝑛 = 𝑎(1 + 𝑟)𝑛−1 . This is the compound interest formula but it is also the general term of a
geometric sequence with first term 𝑎 and common ratio (1 + 𝑟).
We might wonder how long it will take to double the investment. We want to know for what value
of 𝑛 would make 𝑎(1 + 𝑟)𝑛−1 = 2𝑎. Thus, we must solve for 𝑛 the equation
(1 + 𝑟)𝑛−1 = 2
1
Raising each side of this equation to the power 𝑛−1 reveals the relationship between the bank rate
and the term to be
𝑟=
𝑛−1
√2 − 1
For example, suppose it was required to double an investment in 12 years. By the beginning of the
12
13th year, 𝑛 = 13 years and thus 𝑟 = √2 − 1 = 0.0594630943593, or about 6% pa. For doubling
7
in 7 years, the rate would have to be 𝑟 = √2 − 1 ≈ 10.41%
Bankers constructed a ready reckoner of this doubling relationship similar to the following table, to
inform customers about investments.
Doubling time in
years
3
4
5
6
7
8
9
10
11
12
13
Bank Rate
25.99
18.92
14.87
12.25
10.41
9.05
8.01
7.18
6.5
5.95
5.08
Product
77.97
75.68
74.35
73.5
72.87
72.4
72.09
71.8
71.5
71.4
71.12
178
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WHETSTONE 45
The numbers in the third row in the table, the product of the bank rate and the time required to
double the investment, did not go unnoticed. The rule of 72 became a rule of thumb in the banking
world. The rule, stated as
72
𝑛=
𝑟
was used to estimate loosely the number of years 𝑛 it took for an investment at 𝑟% to double in
size.
From the table it is clear that the rule is fairly accurate across the terms shown. The idea can be used
in other contexts as well. If a house doubles in value every ten years, a phenomenon that seemed to
ring true in the late 1900s in Australia, then from the rule of 72 the average annual price inflation for
houses during that time was around 7.2% pa.
This graph shows the difference between the actual values and those shown by the rule of 72 for
various terms between 3 and 10 years. The rule gives a useful approximation.
30
28
26
24
22
20
Rate %
18
16
14
Actual
12
Rule of 72
10
8
6
4
2
0
0
2
4
6
8
10
Term in years
The equation (1 + 𝑟)𝑛−1 = 2 introduced earlier, can be rearranged by taking logarithms and
simplifying, so that
log 2
𝑛−1=
log(𝑟 + 1)
remembering that 1 < (𝑟 + 1) < 2 and that (𝑛 − 1) is the actual number of years required to
double the investment.
log 2
So, for example, at 8% pa interest we have, 𝑛 − 1 = log 1.08 ≈ 9.0065 years.
179
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GEOMETRIC PROGRESSION, APPLICATIONS: Part 1
Questions
1. Use Fermat’s technique to determine the area under the curve 𝑦 = 𝑥 between 𝑥 = 0 and 𝑥 = 𝑎.
Draw a diagram and carefully establish the sum of the geometric progression.
2. Using the rule of 72, an investment will double in size approximately every
72
𝑟
years where 𝑟 is
the interest rate per annum. If 𝑟 = 6% what is the future value of $25,000 predicted by the rule
𝑟
𝑛
after 48 years? Using the formula 𝐹𝑉 = 𝐴 (1 + 100) what is the exact future value, and
determine the percentage error on the rule of 72 in this case.
Answers
1
1.
Area = 𝑎2
2.
$400,000 by rule. $409,846.79 with a percentage error of about 2.4%
2
180