Logarithms were first considered by
Napier^{5}
and were based on an assumption that an exponential function exists. However, by the
latter half of the seventeenth century, it was realized that the best way to consider
exponential functions and logarithms is through the general approach discussed in this
section where the natural logarithm ln is defined first and used to obtain the exponential
function.
Let A
(a,b)
denote the area of the region R
(a,b)
which is under the graph of y = 1∕x,
above the x axis, and between the graphs of the lines x = a and x = b. A picture of this
region is shown below along with approximating rectangles whose sum is too large and
rectangles whose sum is too small.
PICT
The area is defined to be the number which is between all such upper sums and all
such lower sums. Denote by C_{n}
(a,b)
the sum of the areas of the rectangles which enclose
R
(a,b)
in case there are n of them and B_{n}
(a,b)
the sum of the areas of the rectangles
which are contained in R
(a,b)
.
In the picture n = 10. What is C_{n}
(a,b)
−B_{n}
(a,b)
, the discrepancy between the two
sums of areas of rectangles? The width of each rectangle is
(b− a)
∕n. Thus C_{n}
(a,b)
=
−1b−-a- ( b−-a)− 1 b−-a ( b-− a-)−1 b−-a
a n + a+ n n + ⋅⋅⋅+ a+ (n− 1) n n
n−1( ) −1
= ∑ a + kb−-a- b−-a-
k=0 n n
a similar formula holding for B_{n}
(a,b)
. Hence,
Cn (a,b)− Bn (a,b) =
n∑−1( b− a) −1 b − a ∑n ( b− a )−1 b− a
a + k-n--- --n--− a+ k--n-- -n---
k(=0 ) k=1
= a−1 − (a+ b − a )−1 b−-a-=-1(b− a)2 1-
n ab n
This shows that if n is large, the sum of the areas of the small rectangles inside the region
and the sum of the areas of the large rectangles enclosing the region, are both
approximately equal to what should be defined as the area of the region just
described.
Now notice that if r > 0,
Cn (ar,br) = Cn (a,b), Bn(ar,br) = Bn (a,b).
The reason for this is that the width of the rectangles in the sum for C_{n}
(ar,br)
is
multiplied by r and the height is multiplied by 1∕r, leaving the area unchanged for
each rectangle in the sum. Similar considerations apply to B_{n}
(a,b)
. Therefore,
A(a,b)− A (ra,rb) ≤ Cn (a,b)− Bn (ra,rb)
= Cn (a,b)− Bn (a,b) ≤ 1-(b− a)2 1
ab n
Similarly,
A(ra,rb)− A (a,b) ≤ Cn (ra,rb)− Bn (a,b)
1 1
= Cn (a,b)− Bn (a,b) ≤ --(b− a)2--
ab n
Thus
1- 2 1
|A (ra,rb) − A (a,b)| ≤ ab (b− a) n
and since n is arbitrary,
A(ra,rb) = A (a,b).
This fundamental result is summarized in the following lemma.
Lemma 2.5.1Let a,b > 0 and denote by A
(a,b)
the area of the region which isbounded by the lines x = a,x = b, the graph of y = 1∕x, and the x axis. Then ifr > 0,
A (ra,rb) = A (a,b)
Now with this preparation, the following function is defined.
Definition 2.5.2For x > 0 define L
(x)
as follows. Letting A
(a,b)
be as in theabove lemma,
{
A (1,x) for x ≥ 1
L(x) = − A (x,1) for x < 1
PICT
Theorem 2.5.3The function of Definition 2.5.2for x > 0 has thefollowing properties.
L
(xy)
= L
(x)
+ L
(y)
, L
(1)
= 0,
If lim_{n→∞}x_{n} = x, then lim_{n→∞}L
(xn)
= L
(x)
L : (0,∞) → ℝ is one to one and onto.
L
(x)
= −L
( )
1x
Proof: Consider the first claim. Suppose that x,y ≥ 1. Then 1 ≤ x ≤ xy and so,
using the definition of L and Lemma 2.5.1,
L (xy ) ≡ A(1,xy) = A (1,x)+ A(x,xy)
= A(1,x)+ A (1,y) ≡ L(x)+ L (y )
Next suppose that x, the smaller of the two numbers x,y is less than 1. Then consider
several cases. In the first case, 0 < xy < y < 1. Then
A (xy,1) = A (xy,y)+ A(y,1)
and from the definition,
− L (xy) = A (x,1) + A(y,1) = − L(x)− L (y)
and so L
(xy)
= L
(x)
+ L
(y)
. Next suppose 0 < xy < 1 < y. Then from Lemma 2.5.1
and the definition,
A (x,1) = A (xy,y) = A (xy,1)+ A (1,y)
so from the definition,
− L(x) = − L (xy )+ L(y)
which implies again that L
(xy)
= L
(x )
+ L
(y)
. Finally, say 0 < 1 < xy < y. Then
L(y) = A (1,y) = A(1,xy)+ A (xy,y)
= L (xy) +A (x,1) = L (xy )− L(x)
thus, again it follows that L
(xy)
= L
(x)
+ L
(y)
. It is obvious from the definition that
L
(1)
= 0. Note that this implies that L
(1∕x)
+ L
(x)
= L
(1)
= 0.
Observe that from this claim, if a < b, then
L(b∕a)+ L(a) = L (b)
and so
0 ≤ L (b) − L (a) = L (b∕a) = A(1,b∕a)
From the definition,
b− a
L(b)− L (a) = L(b∕a) ≤ b∕a− 1 ≤----.
a
PICT
Now consider the second claim. If x_{n}< x. Then
x − x |x − x |
|L (x )− L(xn)| = L (x) − L (xn) ≤---n = -----n-
xn xn
If x_{n}> x, then
|L (x )− L(x )| = L (x ) − L (x) ≤ xn-−-x = |x-−-xn|
n n x x
For all n large enough, x_{n}> x∕2 and so, for such n,
2|x−-xn|
|L (x) − L (xn)| ≤ |x|
which converges to 0.
Finally consider the last claim. From the definition, it follows that L
(2)
> 1∕2. (Draw
a picture.) Therefore, from the first claim,
Now let y ∈ ℝ. Choose n such that n∕2 > y. Then it follows that for x > 2^{n},L
(x)
> y.
Also, there exists m such that −m∕2 < y. Then if x < 1∕2^{m}, it must be the case that
L
(x)
< y.
S ≡ {x : L (x) > y}
Thus S has a greatest lower bound because it is bounded below. Let l denote this
greatest lower bound. Then there exists x_{n}∈ [l,l + n^{−1}) ∩S since otherwise l is not the
greatest lower bound. l + n^{−1} is a larger lower bound. Also let y_{n}∈
(l− 1∕n,l)
. Then
y_{n}
∕∈
S because l is a lower bound for S and so L
(yn)
≤ y while L
(xn)
> y. Then by the
second claim and the limit theorems,
y ≤ lim L(xn) = L(l) , y ≥ lim L(yn) = L (l)
n→ ∞ n→ ∞
and so L
(l)
= y. Therefore, L is onto ℝ. If x < y, then L
(y)
− L
(x)
= L
(y∕x)
by the
first claim, and L
(y∕x)
> 0 so it follows that L is strictly increasing. Hence the function
L is also one to one.
The last claim is obvious from part 1. Indeed, L
(x)
+ L
(1)
x
= L
(1)
= 0.
■
This function L will be denoted as ln. It is the natural logarithm.
Definition 2.5.4Since ln is one to one and onto, there exists a real numbere such that ln
(e)
= 1.This number e is called Euler’s number.
It can be computed directly from the above definition of ln. You would get a good
table of values of ln using the above definition and then go backwards in the table to
obtain e. There are of course much more sophisticated ways to find it, and here it is to
several decimal places. e = 2.7183
From knowledge of ln and its properties, it is easy to get the existence of an
exponential function. Let exp : ℝ
↦→
(0,∞) be the inverse of ln.
exp is defined on all of ℝ (2.17)
(2.17)
exp (x + y) = exp(x)exp(y) (2.18)
(2.18)
exp : ℝ ↦→ (0,∞ ) is one to one and onto (2.19)
(2.19)
exp (0) = 1 (2.20)
(2.20)
Each is satisfied. Since ln maps (0,∞) onto ℝ, its inverse function is defined on ℝ
and has values which are positive numbers. This inverse function exp is one to
one and onto. Indeed, if exp
(y1)
= exp
(y2)
, then since each is positive, you
can take ln of both sides and conclude that y_{1} = y_{2}. Thus exp is one to one.
If y ∈ (0,∞), then ln
(y)
∈ ℝ and so y = exp
(ln (y))
. Thus exp is also onto.
ln(exp(x+ y)) = x+ y
ln(exp(x)exp(y)) = ln(exp (x)) +ln(exp(y))
= x+ y