The natural logarithm

1. Introduction

A previous web page defined two particular sequences converging to a number - which by experiment can be shown to be approximately 2.718281828. The very reasonable question arises asking if this number really is the Euler number e familiar from exponentials and natural logarithms. One approach is to define for each real number x the value E ( x ) = lim n 1 + x n n and prove that this function has the expected properties of the expontential function exp ( x ) , such as

  • E ( 0 ) = 1
  • E ( x + y ) = E ( x ) · E ( y )
  • lim h 0 E ( x + h ) - E ( x ) h = E ( x )

for all x , y .

An alternative approach is to define the limit of the infinite series

1 + x + x 2 2 + x 3 3 +

and prove that this limit is precisely E ( x ) . Both these methods can be made precise, and will (with some effort) work to make rigorous the real exponential function. Much of this work goes beyond the scope of the present sequence of web pages, however. In particular the notion of a limit of the form lim h 0 has not been discussed here.

For any doubting students of analysis, or indeed just for interested readers, this web page explores these ideas further. Just for fun, I have chosen to take a very different approach to either of the more normal ones just outlined and will define and use the natural logarithm function, and apply that to the function E ( x ) defined above. Because logarithms tend to use additions where exponentials use multiplication, many of the algebraic manipulations turn out to be easier. However, the material in this web page is sketched in less detail than you will normally find on these web pages and is not part of an official sequences and series module, so is for interested readers only. You may safely skip to the next page if you wish.

2. The logarithm

We start by defining two functions using a limiting process similar to things you have already seen.

Suppose x is a positive real number. Then we define

l ( x ) = lim n n 1 - x -1 n

and

L ( x ) = lim n n x 1 n - 1

We will spend a bit of time proving that both these limits exist and in fact they are equal.

To this end, we fix some positive x , and let a n = n 1 - x -1 n and b n = n x 1 n - 1 .

Lemma.

For all n , we have b n a n .

Proof.

Indeed, we have x 2 n - 2 x 1 n + 1 = ( x 1 n - 1 ) 2 0 whence x 1 n - 1 1 - x -1 n which gives b n a n .

Lemma.

For all sufficiently large k and all n k we have b n b k .

Proof.

Let k be sufficiently large so that | x 1 k - 1 | < 1 . Then

x 1 n = ( 1 + ( x 1 k - 1 ) ) k n 1 + k n ( x 1 k - 1 )

by the exponential inequality, hence

n ( x 1 n - 1 ) k ( x 1 k - 1 )

as required.

Lemma.

For all sufficiently large k and all n k we have a k a n .

Proof.

Let k be sufficiently large so that | x 1 k - 1 | < 1 . Then

x -1 n = ( 1 - ( 1 - x 1 k ) ) k n 1 - k n ( 1 - x -1 k )

by the exponential inequality, hence

n ( x -1 n - 1 ) - k ( 1 - x -1 k )

and so

n ( 1 - x -1 n ) k ( 1 - x -1 k )

as required.

Proposition.

For all real x > 0 the functions l ( x ) and L ( x ) are defined and equal.

Proof.

The monotone convergence theorem together with the three preceding lemmas already show that the limits l ( x ) and L ( x ) are defined with l ( x ) L ( x ) , since they show that the sequences ( a n ) and ( b n ) are monotonic and bounded. If x > 1 then 0 < a 1 l ( x ) so l ( x ) a 1 > 0 , and if x < 1 then L ( x ) b 1 < 0 so L ( x ) b 1 < 0 . Moreover, it is easy to see that l ( 1 ) = L ( 1 ) = 0 . Suppose x 1 and a = lim a n and b = lim b n . Then

b n a n = x 1 n - 1 1 - x -1 n = x 1 n 1 - x -1 n 1 - x -1 n = x 1 n 1

Therefore by continuty of division at ( b , a ) and uniqueness of limits b a = 1 and hence a = b .

Definition.

For x > 0 we define log ( x ) , the natural logarithm of x , to be the value of l ( x ) or L ( x ) .

3. Properties of the logarithm function

Having defined this function, we want to explore some of its properties. In particular, we would like to know that it behaves as a logarithm function and is the inverse to the E ( x ) function. Some properties emerge remarkably quickly. For example, if x > 0 then log ( x 2 ) = 2 log ( x ) . More generally,

Lemma.

For each x > 0 and k we have log ( x k ) = k log ( x ) .

Proof.

log ( x k ) = lim n n ( x k n - 1 ) = lim n ( k n k ( x k n - 1 ) ) = k lim n n k ( x k n - 1 ) = k log ( x )

using the continuity of scalar multiplication.

Lemma.

For each x > 0 we have log ( x -1 ) = - log ( x ) .

Proof.

We have

L ( x -1 ) = lim n n x -1 1 n - 1 = lim n n x -1 n - 1 = lim n - n 1 - x -1 n = - l ( x )

and this together with l ( x ) = L ( x ) proves the result.

Proposition.

The function log : ( 0 , ) is everywhere nondecreasing and unbounded.

Proof.

Suppose x y 1 . Then n ( x 1 n - 1 ) n ( y 1 n - 1 ) hence taking limits log ( x ) log ( y ) . A similar argument applies to x , y 1 . To see that log is unbounded let x = 2 and note that log ( x ) > 0 . So log ( x k ) = k log ( x ) can be taken as large or as small as we like by suitable choice of large positive k or negative k .

The argument just given does not show that log is increasing as < is not preserved in limits. However we will prove that log is in fact increasing in a moment.

Proposition.

For each x we have log ( E ( x ) ) = x .

Proof.

By definition E ( x ) = lim n ( 1 + x n ) n = lim n ( 1 + x n ) n + 1 with the first limit being eventually monotonic nondecreasing and the second limit eventually monotonic nonincreasing. Furthermore log ( x ) = lim n n ( 1 - x -1 n ) = lim n n ( x 1 n - 1 ) with the first eventually monotonic nondecreasing and the second eventually monotonic nonincreasing. Putting these together we have

n 1 + x n n +1 n - 1 log ( E ( x ) ) n 1 - 1 + x n n -1 n = n 1 - 1 1 + x n

for eventually all n . The right hand side here equals n ( 1 - n n + x ) = n x n + x which converges to x as n . The left hand side is

n 1 + x n 1 + x n 1 n - 1 = n 1 + x n 1 n - 1 + x 1 + x n 1 n

and x 1 + x n 1 n x as n as 1 + x n 1 and hence 1 + x n 1 n 1 . Also n 1 + x n 1 n - 1 0 . This can be proved by taking an arbitrary ε > 0 and choosing N such that x N < ε . Then for all n N we have 1 + x n 1 n < 1 + 1 n x n by the exponential inequality and so n 1 + x n 1 n - 1 < x n < ε . This shows that

r n log ( E ( x ) ) s n

where both sequences r n and s n converge to x . So log ( E ( x ) ) = x , as required.

Proposition.

The function log : ( 0 , ) is everywhere increasing, bijective, and continuous.

Proof.

We already know log is nondecreasing. To see that it is increasing suppose to get a contradiction 1 < x < y with log ( x ) = log ( y ) . From x 1 n 1 we know that there is n such that x 1 n < y x . In particular letting u be x 1 n we have u > 1 and x = u n < u n + 1 y . By the fact that log is nondecreasing we deduce that log ( x ) = n log ( u ) ( n + 1 ) log ( u ) log ( y ) . As by supposition log ( x ) = log ( y ) we deduce that log ( u ) = 0 . But as u > 1 the sequence ( u n ) is unbounded so for each z > 1 in reals there is k such that z u k and hence 0 log ( z ) log ( u k ) = k log ( u ) = 0 . It follows that log ( z ) = 0 for all z > 1 . This is a contradiction to the previous proposition.

The case when x < y < 1 is similar.

Continuity follows from a general result that says a nondecreasing function log : ( 0 , ) which takes all values in is automatically continuous.

The notion of derivative is outside the scope of these notes. However, you may be familiar enough with it to understand the following argument.

Proposition.

The function log : ( 0 , ) has derivative log ( x ) = 1 x .

Proof.

By definition,

log ( x ) = lim h 0 log ( x + h ) - log ( x ) h

or, as lim n x 1 n = 1 , we can use x n +1 n - x as a possible value for a positive h . (This is justified by the facts that x n +1 n - x 0 and the fact that log is continuous and increasing. Strictly speaking since we are only looking at positive h we are computing a one-sided derivative. The derivative on the left can be computed and found equal to the right derivative by a similar argument.) This gives

log ( x ) = lim n log ( x n +1 n ) - log ( x ) x n +1 n - x

which equals

lim n 1 n log ( x ) x ( x 1 n - 1 ) = log ( x ) x · lim n n ( x 1 n - 1 ) = 1 x

as required.

Proposition.

The function log : ( 0 , ) satisfies log ( x y ) = log ( x ) + log ( y ) for all x , y > 0 .

Proof.

(Sketch.)

Given x and y which for simplicity we assume are greater than 1 , let a n = x 1 n so a n n = x and let k n be the unique integer with a n k n y < a n 1 + k n . Then y = lim n a n k n and

log ( x y ) = log ( a n n · lim n a n k n ) = lim n log ( a n n · a n k n ) = lim n n log ( a n ) + k n log ( a n ) = lim n log ( x ) + log ( a n k n ) = log ( x ) + log ( lim n a n k n ) = log ( x ) + log ( y )

by continuity of various functions including log .