Download The Logarithmic Function as a Limit

Applied Mathematical Sciences, Vol. 6, 2012, no. 91, 4511 - 4518
The Logarithmic Function as a Limit
Alvaro H. Salas
Department of Mathematics
Universidad de Caldas, Manizales, Colombia
Universidad Nacional de Colombia, Manizales
FIZMAKO Research Group
[email protected]
Abstract
In this paper we define the logarithmic function of base e and we
establish its basic properties.We also define the exponential function of
base e and we prove the basic properties of these functions.
1
Introduction
There are several ways to define the logarithmic function of base e. Usually,
it is defined as the inverse function of the exponential function of base e. In
this paper we first define the logarithmic function as the limit of a sequence
of functions and then we derive its basic properties without using any calculus
tools.
2
Auxiliary lemmas
Lemma 1. Given a positive integer n and a positive real number x there
exists a positive real number y uniquely defined
such that y n = x. This y is
√
n
called the n−th root of x and it is denoted by x or x1/n .
Proof. We proceed by induction on n. The assertion is valid for n = 1 since
x1 = x and then y = x. Suppose that our lemma is valid for some n. Let us
consider the set
E = {t ≥ 0 | tn+1 < x}.
This set is not empty since 0 ∈ E. On the other hand, if t ∈ E then tn+1 <
x < (x + 1)n+1 . From this, t < x + 1 and x + 1 is an upper bound of E. Let
ρ = sup E ≥ 0. We claim that ρn+1 = x.
A. H. Salas
4512
Indeed, suppose that ρn+1 < x. Choose c with ρn+1 < c < x. Then ρn < c/ρ.
By the inductive hypothesis, we may find b > 0 such that bn = c/ρ. From
ρn < bn we get ρ < b. Choose a so that ρ < a < min(b, xρ/c). Then
an+1 = an a < bn a = ca/ρ < x, so that a ∈ E and then a ≤ ρ. But a > ρ and
we get a contradiction.
Now, suppose that ρn+1 > x. Choose c with ρn+1 > c > x. Then ρn > c/ρ.
By hypothesis, we may find b > 0 such that bn = c/ρ and then ρ > b. Choose
a so that ρ > a > max(b, xρ/c). We have an+1 = an a > bn a = ca/ρ > x, so
that if t ∈ E then tn+1 < x < an+1 and this gives t < a for any t ∈ E, that
is, a is an upper bound of E and ρ = sup E ≤ a. But a < ρ. We again get a
contradiction.
Thus, ρn+1 = x and the lemma is also valid for n + 1. Finally,
√
n
x is unique. This follows from the fact that if y, z > 0 and z n = y n = x then
z = y.
Lemma 2. Let pn+1 (z) = n(z n+1 − 1) − (n + 1)(z n − 1). Then
pn+1 (z) = (z − 1)2 (1 + 2z + 3z 2 + 4z 3 + · · · + nz n−1 ) for n = 1, 2, 3, ... (2.1)
In particular,
n(z n+1 − 1) − (n + 1)(z n − 1) > 0 for any z > 0.
(2.2)
Proof. By induction on n.The identity (2.1) is valid for n = 1 since p2 (z) =
z 2 − 1 − 2(z − 1) = z 2 − 2z + 1 = (z − 1)2 . Suppose that it holds for some n.
Observe that
pn+2 (z) − pn+1 (z) = (n + 1)(z − 1)2 z n .
Then
pn+2 (z) = pn+1 (z) + (n + 1)z n (z − 1)2
= (z − 1)2 (1 + 2z + 3z 2 + 4z 3 + · · · + nz n−1 ) + (n + 1)z n (z − 1)2
= (z − 1)2 (1 + 2z + 3z 2 + 4z 3 + · · · + nz n−1 + (n + 1)z n )
and then equation (2.1) is also valid for n + 1.
Lemma 3. Given any real number x > 0 the sequence fn (x) defined by
√
fn (x) = n( n x − 1), n = 1, 2, 3, ...
(2.3)
has the following properties
1. fn (x) > fn+1 (x) for n = 1, 2, 3, .., that is, this sequence is strictly decreasing.
2. For any n = 1, 2, 3, .. we have
x−1
≤ fn (x) ≤ x − 1
x
(2.4)
The logarithmic function as a limit
4513
and the equality holds only if x = 1.
3. If n ≥ m then
√
fn (x) ≤ m( m x − 1).
(2.5)
1
Proof. 1. Let z = x n(n+1) in (2.2). Then
1
1
n(x n − 1) − (n + 1)(x n+1 − 1) > 0,
which is equivalent to fn (x) > fn+1 (x).
2. Observe that
√
n
x−1= √
n
n−1
x
+
√
n
x−1
x
n−2
+···+
√
n
x+1
< x − 1 for any x > 0 and x = 1.
(2.6)
On the other hand,
√
n
x−1= √
n
x
n−1
+
√
n
x−1
n−2
x
+···+
√
n
x+1
>
x−1
for any x > 1.
nx
(2.7)
From (2.6) and (2.7) it is clear that
√
x−1
< n( n x − 1) < x − 1 for any x > 1.
x
Let 0 < x < 1. Then
1
x
(2.8)
> 1 and (2.8) gives
n(
n
1
1
− 1) < − 1
x
x
from where
n(1 −
√
n
x) <
√
x−1
1−x√
1−x
n
and then
< n( n x − 1).
x<
x
x
x
(2.9)
Inequalities (2.6) and (2.9) give
√
x−1
< n( n x − 1) < x − 1 for 0 < x < 1.
x
Finally, for x = 1 we obtain
√
x−1
= n( n x − 1) = x − 1 = 0.
x
√
3. This follows from 1 since fn (x) ≤ fm (x) = m( m x − 1) for any n ≥ m.
A. H. Salas
4514
3
The logarithmic function.
√
Since n( n x − 1)√
= fn (x) > fn+1 (x) ≥ x−1
for n = 1, 2, 3, .. we see that there
x
n
exists limn→∞ n( x − 1). We will call it the natural logarithm of x and we will
denote it by log x. Thus,
√
(3.1)
log x = lim n( n x − 1), x > 0.
n→∞
Equation (2.9) defines a function on (0, ∞) as the limit of the sequence of
functions fn (x) given by (2.3) . This function is called logarithmic function.
It is also denoted by ln x.
3.1
Properties of logarithmic function.
√
Property 1. log 1 = 0.This is obvious since log 1 = limn→∞ n( n 1 − 1) = 0.
Property 2. For any t > 0 and x > 0,
log(tx) = log t + log x.
(3.2)
Indeed, from (2.4) we obtain
x − 1
} for any x > 0.
|fn (x)| ≤ cx , where cx = max{|x − 1| , x (3.3)
On the other hand,
√
√
n
|fn (tx) − fn (t) − fn (x)| = n t − 1 n x − 1
= n·
|fn (t)| |fn (x)|
ct cx
·
≤
.
n
n
n
Thus,
0 ≤ |fn (tx) − fn (t) − fn (x)| ≤
ct cx
for any t, x > 0 (n = 1, 2, 3, ...)
n
(3.4)
Letting n → ∞ in (3.4) and taking into account (3.1) we obtain
|log(tx) − log(t) − log(x)| = 0,
which proves (3.2).
Property 3. For any positive integer n, log(xn ) = n log x. This may be
proved by induction by using Property 2.
Property 4. For any x > 0, log x1 = − log x. Indeed, 0 = log 1 = log(x · x1 ) =
log x + log x1 .
The logarithmic function as a limit
4515
Property 5. For any t > 0 and x > 0,
t
log( ) = log t − log x.
x
(3.5)
Indeed, log( xt ) = log(t · x1 ) = log t + log x1 = log t − log x.
Property 5.The logarithmic function is strictly increasing on (0, ∞).
Indeed, letting n → ∞ in (2.4) gives
x−1
≤ log x ≤ x − 1 for all x > 0.
x
Let 0 < x1 < x2 . Then x =
0<
x2
x1
(3.6)
> 1 and from (3.6) we obtain
x−1
x2
= log x2 − log x1 , from where log x1 < log x2 .
< log
x
x1
Property 6. The logarithmic function is continuous on (0, ∞).
ε
Indeed, ε > 0 and consider any t > 0 subject to |t − 1| < ε+1
. Then
1 − ε/(ε + 1) < t < 1 + ε/(ε + 1).
From (3.6) we see that
log t ≤ t − 1 <
ε
< ε.
ε+1
On the other hand, t > 1 − ε/(ε + 1) = 1 −
have
log t ≥
ε
ε+1
=
(3.7)
1
ε+1
and then by (3.6) we
t−1
1
= 1 − > 1 − (ε + 1) = −ε.
t
t
(3.8)
Conditions (3.7) and (3.8) give
|log t| < ε for any t > 0 satisfying |t − 1| <
ε
.
ε+1
(3.9)
aε
.
ε+1
Consider any
Now, let et a > 0 be a fixed number and ε > 0. Define δ =
x such that |x − a| < δ. Let t = xa .Then
x
|x − a|
δ
ε
< =
− 1 = |t − 1| =
a
a
a
ε+1
(3.10)
and then from (3.9) and (3.10) it follows that |log x − log a| = log xa =
ε
|log t| < ε if |x − a| < ε+1
.This proves that limx→a log x = log a.
Property 7. Given any real number ξ there exists a positive real number x
such that log x < ξ.
A. H. Salas
4516
Indeed, choose a positive integer m for which ξ > −m. Then 1+ξ/m > 0.√ We
m
x−
define x = 12 (1 + ξ/m)m . Taking into account (2.5) we obtain fn (x) ≤ m(
√
m
1) for any n ≥ m. Letting n → ∞ in this inequality gives log x ≤ m( x − 1).
Then
√
log x ≤ m( m x − 1) < m( m (1 + ξ/m)m − 1) = ξ.
(3.11)
Property 8. The range of logarithmic function is the set of real numbers.
Indeed, let y be any real number. By Property 7, we may find a number a > 0
and a number b such that log a < y and log b < −y. Then
1
.
b
It is clear that a < b. Let H = {t > 0 such that log t < y }. This set is
not empty since a ∈ H . If t ∈ H then log t < y < log b and this gives t < b.
Thus, H is bounded from above by b. Let x = sup H. We claim that log x = y.
Indeed, suppose that log x < y. Let ε = y − log x. There exists δ > 0 such that
|log x − log t| < ε if |x − t| < δ. Let t = x+δ/2 > x. Then log t < log x+ε = y
and t ∈ H so that t ≤ x. But t > x. Contradiction.
On the other hand, suppose that log x > y. Let ε = log x − y. There exists
δ > 0 such that |log x − log s| < ε if |x − s| < δ. Choose some s such that
0 < s < x and |x − s| < δ. Then log s > log x − ε = y > log t so that t < s for
any t ∈ H. We conclude that s is an upper bound of H and then x ≤ s. But
x > s. Contradiction.
We have proved that log x = y. Since y was arbitrarily chosen, we conclude
that the range of log is the set of real numbers.
log a < y < − log b = log b, where b =
4
The exponential function.
By Properties 5 and 6 of previous section, the logarithmic function log defined
by (3.8) is strictly increasing and continuous. On the other hand, by virtue of
Property 8, its range is the set of real numbers. Thus, log : (0, ∞) → (−∞, ∞)
is one to one and onto. The inverse function of log is denoted by exp so that
exp : (−∞, ∞) → (0, ∞) is strictly increasing, one to one and onto. Moreover,
exp(log(x)) = x for any x > 0 and log(exp(x)) = x for any real x.
(4.1)
Observe that
exp(x1 + x2 ) = exp(x1 ) exp(x2 ) for any real numbers x1 and x2 .
(4.2)
Indeed, let y1 = exp(x1 ) and y2 = exp(x2 ) so that x1 = log y1 and x2 = log y2 .
We have
exp(x1 + x2 ) = exp(log y1 + log y2 ) = exp(log(y1 y2 )) = y1 y2 = exp(x1 ) exp(x2 ).
The logarithmic function as a limit
4517
In particular, exp(x) exp(−x) = exp 0 = 1. This gives exp(−x) = 1/ exp(x).
Then
exp(x1 − x2 ) = exp (x1 ) exp(−x2 ) =
exp(x1 )
.
exp(x2 )
(4.3)
The function exp is continuous. To prove this, let ε > 0. Let us fix a real
number a and define δ = log(1 + ε exp(−a)). Consider any real x for which
|x − a| < δ. Then −δ < x − a < δ and exp(−δ) < exp(x − a) < exp(δ). This
inequality gives
1
exp(x)
< exp(x − a) =
< 1 + ε exp(−a)
1 + ε/ exp(a)
exp(a)
(4.4)
From this inequality it is easy to see that
−ε < −
ε
< exp(x) − exp(a) < ε.
1 + ε exp(−a)
We have proved that for any ε > 0 and any real number a there exists δ =
log(1 + ε exp(−a)) such that if |x − a| < δ then |exp(x) − exp(a)| < ε, that is,
limx→a exp(x) = exp(a).
In view of Property 8, the equation log x = y has unique solution on (0, ∞)
for any real number y. The solution of the equation log x = 1 is denoted by e.
This is the base of natural logarithms.
5
Conclusions.
We have defined the logarithmic function as a limit of sequence of functions and
we established its basic properties. We also defined the exponential function as
the inverse of the logarithmic function. These facts allow to establish the law
of exponents starting from the equation ax = exp(x log a) for any real number
x and a > 0. In particular, ex = exp(x log e) = exp x. Finally, inequality (3.6)
allows us to prove following important limits :
x
log(1 + t)
1
lim
= 1, lim 1 +
=e
t→0
x→∞
t
x
and
x n
for any real number x.
ex = lim 1 +
n→∞
n
References
[1] Hardy G. H., A Course of Pure mathematics, Ninth Edition, University of
Cambridge (1945).
4518
A. H. Salas
[2]Mitrinovic D.S., Elementary Inequalities, Belgrade-Yugoslavia, P. Noordhoff Ltd - Groningen - The Netherlands (1964)
[4] Rudin W., Principles of Mathematical Analysis, Third Edition, McGraw Hill
(1976).
Received: April, 2012