Oulun Lyseon lukio / Galois club 2010-2011 / On the extraordinary constant e / TL
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1. Where does the constant e come from?
Euler’s number, denoted by e, also known as Napier’s number, is one of the most important constants in all
mathematics. Below we sketch a few natural contexts in which the constant e emerges.
Example 1.1. (Continuous compound interest)
Assume a hypothetical bank account with an annual interest rate 100 %. Assume that at the beginning of
the year the account holds exactly 1 euro (or any monetary unit). Assume that the interest is credited to
the account n times a year at regular intervals. We calculate the bank account’s final saldo 𝑓𝑛 after one year
for n = 1, 2, 3.
1
n = 1:
𝑓1 = (1 + 1) ∙ 1 = 2
n = 2:
𝑓2 = (1 + 2) ∙ (1 + 2) ∙ 1 = (1 + 2) = 2.25
n = 3:
𝑓3 = (1 + 3) ∙ (1 + 3) ∙ (1 + 3) ∙ 1 = (1 + 3) ≈ 2.37
1
1
1
1
1 2
1 3
1
You notice the pattern that arises. We have in general
1 𝑛
𝑓𝑛 = (1 + 𝑛)
[equation 1.1]
It seems that the final amount might actually increase towards infinity as we reduce the crediting period
towards zero (in which case we may speak of continuous compounding of interest ), or equivalently let 𝑛 →
∞? However, armed with the formula [1] and calculator we may convince ourselves that this is not the
case. We have, for example:
𝑓12 = (1 +
1 12
)
12
1 52
𝑓52 = (1 + 52)
≈ 2.61
<monthly crediting>
≈ 2.69
<weekly crediting>
365
1
𝑓365 = (1 + 365)
1
≈ 2.71
8760
𝑓8760 = (1 + 8760)
1
≈ 2.72
525600
𝑓525600 = (1 + 525600)
<daily crediting>
≈ 2.72
<hourly crediting>
<minutely crediting>
END OF EXAMPLE 1.1
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Oulun Lyseon lukio / Galois club 2010-2011 / On the extraordinary constant e / TL
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Motivated by the above example we define
1 𝑛
𝑒 = lim (1 + 𝑛)
𝑛→∞
1 𝑥
or equivalently 𝑒 = lim (1 + 𝑥)
𝑥→∞
It will be shown later that this is a valid definition i.e. that the limit really exists.
The number 𝑒 is irrational (to be proved later) and its approximation to two decimal places is 2.72 as
indicated already in the example above.
The symbol 𝑒 refers to Euler, a Swiss mathematician (1707-1783), who was the first to see its enormous
significance in many fields of mathematics although the Scottish mathematician John Napier, aka Neper
(1550-1617), had introduced 𝑒 earlier as a natural basis for logarithms (which he actually invented).
Example 1.2. (Continuous compound interest in general)
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Assume a bank account with an annual interest rate 𝑝 %. Let 𝛼 = 𝑝 be the corresponding growth factor
(e.g. if 𝑝 = 5 then 𝛼 = 0.05). Assume again that the account holds 1 euro at the beginning of the year and
the interest is credited to the account n times a year at regular intervals. Below we calculate the final
amount f after one year for values of n = 1, 2, 3.
n = 1:
𝑓1 = (1 + 𝛼) ∙ 1 = 1 + 𝛼
n = 2:
𝑓2 = (1 + 2 𝛼) ∙ (1 + 2 𝛼) ∙ 1 = (1 + 2 𝛼)
n = 3:
𝑓3 = (1 + 3 𝛼) ∙ (1 + 3 𝛼) ∙ (1 + 3 𝛼) ∙ 1 = (1 + 3 𝛼)
1
1
1
1
1
2
1
1
3
In general we have
1
𝑓𝑛 = (1 + 𝑛 𝛼)
𝑛
When 𝑛 → ∞ the interest is compounded continuously and the final amount will be
1
𝑛
1
𝑛
𝛼
𝑓∞ = lim (1 + 𝛼) = lim ((1 + 𝑛 ) )𝛼 = 𝑒 𝛼 .
𝑛
𝑛→∞
𝑛→∞
𝛼
For example, if 𝑝 = 5 and 𝛼 = 0.05, we have
𝑓∞ = 𝑒 𝛼 = 𝑒 0.05 ≈ 1.05127
while
𝑓1 = 1 + 𝛼 = 1.05 and 𝑓2 = (1 + 0.025)2 = 1.050625.
END OF EXAMPLE 1.2
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Oulun Lyseon lukio / Galois club 2010-2011 / On the extraordinary constant e / TL
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Example 1.3. (Bernoulli trials)
Jacob Bernoulli (1654-1705), another Swiss mathematician and Euler’s predecessor, considered the
following problem:
In a bowl there are n balls of which all except one are white (the exceptional one being black). A
gambler picks one ball at random, checks its colour and puts it back into the bowl. This is repeated n
times. What is the probability 𝑃𝑛 that the gambler never gets the black ball? What limit does this
probability approach when n increases towards infinity?
At each draw the probability of not getting the black ball is
𝑛−1
𝑛
1
= 1 − 𝑛 . Because successive draws are
independent of each other we may apply the product rule to obtain the required probability (that the same
results is repeated at each of the n draws)
1
𝑃𝑛 = (1 − 𝑛)𝑛 .
For example
𝑃100 = (1 −
1 100
)
100
≈ 0.366 .
It may be proved that
1
lim 𝑃𝑛 = lim (1 − 𝑛)𝑛 = 1/𝑒 ≈ 0.36787944.
𝑛→∞
𝑛→∞
2. Infinite series representation for e
The number e has many remarkable properties. Besides the limits of sequences (or functions) considered
above, the number e can be expressed as infinite sums, of which the most useful is
𝑒 =1+
1
1
1
1
+ + + +
1!
2!
3!
4!
⋯
[equation 2.1]
This follows from a more general result that the exponential function 𝑒 𝑥 can be represented as a power
series
𝑥
𝑒 𝑥 = 1 + 1! +
𝑥2
2!
+
𝑥3
3!
+
𝑥4
4!
+⋯
[equation 2.2]
It can be proved that this representation holds true for all real values of x.
Note that from [equation 2.2] we obtain, by setting 𝑥 = −1 ,
1
𝑒
1
1
1
1
= 1 − 1! + 2! − 3! + 4! − + ⋯ .
The equation 2.1 can also be justified (if not rigorously proved) directly by applying the binomial theorem
1 𝑛
to the definition 𝑒 = lim (1 + 𝑛) .
𝑛→∞
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Oulun Lyseon lukio / Galois club 2010-2011 / On the extraordinary constant e / TL
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By the binomial theorem we have
𝑛
𝑛
1 𝑛
1
1
(1 + 𝑛) = 1𝑛 + ( ) ∙ 1𝑛−1 ∙ (𝑛)1 + ( ) ∙ 1𝑛−2 ∙ (𝑛)2 + ⋯
1
2
=1+1+
𝑛(𝑛 − 1) 𝑛(𝑛 − 1)(𝑛 − 2)
+
+⋯
1 ∙ 2 ∙ 𝑛2
1 ∙ 2 ∙ 3 ∙ 𝑛3
and consequently
1 𝑛
𝑛
𝑒 = lim (1 + ) = 1 +
𝑛→∞
1
1
1
+ + +
1!
2!
3!
…
Analogous justification works for the more general equation 2.2 when we notice (as we already did in the
example 1.2 where we had 𝛼 for 𝑥) that
𝑥 𝑛
𝑒 𝑥 = lim (1 + 𝑛)
𝑛→∞
3. Irrationality of e
Theorem 3.1. The Euler-Napier number e is irrational
Proof: Assume to the contrary that e is rational, that is 𝑒 = 𝑝/𝑞 for some positive integers 𝑝, 𝑞. Using the
power series representation of e we have then
∞
𝑞
∞
𝑘=0
𝑘=0
𝑘=𝑞+1
𝑝
1
1
1
=∑ =∑ + ∑
𝑞
𝑘!
𝑘!
𝑘!
and consequently
𝑞
∞
𝑘=0
𝑘=𝑞+1
𝑝
1
1
−∑ = ∑
𝑞
𝑘!
𝑘!
Now multiply both sides of this equation by 𝑞! to get
𝑞
∞
𝑘=0
𝑘=𝑞+1
𝑞!
𝑞!
𝑝(𝑞 − 1)! − ∑ = ∑
𝑘!
𝑘!
where the right hand side is positive and can also be evaluated from the above as follows
∞
∞
𝑘=𝑞+1
𝑘=1
𝑞!
1
1
1
1
∑
=
+
+⋯ < ∑
=
< 1
𝑘
𝑘! 𝑞 + 1 (𝑞 + 1)(𝑞 + 2)
(𝑞 + 1)
𝑞
So we have
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Oulun Lyseon lukio / Galois club 2010-2011 / On the extraordinary constant e / TL
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∞
0 < ∑
𝑘=𝑞+1
𝑞!
< 1
𝑘!
But then the equation
𝑞
∞
𝑘=0
𝑘=𝑞+1
𝑞!
𝑞!
𝑝(𝑞 − 1)! − ∑ = ∑
𝑘!
𝑘!
is impossible because the left hand side is an integer while the right hand side lies between 0 and 1.
But this equation followed from our assumption that e is rational. Hence the assumption must be false.
QED
Note: Charles Hermite, a French mathematician proved in 1873 that the number e is not only irrational but
even transcendental (non-algebraic). Nine years later, in 1882, Ferdinand von Lindemann, a German
mathematician, proved that another important mathematical constant 𝜋 is also transcendental. Georg
Cantor, creator of the modern set theory, had proved in 1874 that actually almost all real numbers are
transcendental.
4. Number e represented as a simple continued fraction
Remember that simple1 continued fractions are of the form
2+
1
3
, 2+
1
1
3+7
,
2+
1
3+
1
,
1
7+
2
2+
1
3+
1
1
7+ 1
2+
4
which are usually denoted by a special square-bracketed coordinate notation as follows
[2; 2] , [2; 2, 3] , [2; 2, 3, 7] , [2; 2, 3, 7, 2] , [2; 2, 3, 7, 2, 4]
All positive rational numbers can be expressed as finite (or terminating) continued fractions (like those
above).
Irrational numbers, in contrast, can be expressed as non-terminating (infinite) continued fractions like
√2 = [1; 2, 2, 2, . . . ]
It can be shown that the Euler number e also has a quite regular representation
𝑒 = [2; 1, 2, 1, 1, 4, 1, 1, 6, 1, 1, 8, 1, 1, . . . ]
1
A continued fraction is simple if all of its numerators are equal to 1.
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