Disclaimer: these notes are
not mathematically rigorous. Instead,
they present quick, and, I hope, plausible, derivations of the properties of e,
ex and the natural logarithm.
Consider the following series:
where n runs
through the positive integers. What happens as n gets very large?
It’s easy to find out if you use a scientific calculator
having the function x^y. The first three
terms are 2, 2.25, 2.37. You can use
your calculator to confirm that for n = 10, 100, 1000, 10,000, 100,000,
1,000,000 the values of
are (rounding off)
2.59, 2.70, 2.717, 2.718, 2.71827, 2.718280.
These calculations strongly suggest that as n goes up to
infinity,
goes to a definite
limit. It can be proved mathematically
that
does go to a
limit, and this limiting value is called e. The value of e is 2.7182818283… .
To try to get a bit more insight into
for large n,
let us expand it using the binomial theorem. Recall that the binomial theorem
gives all the terms in (1 + x)n, as follows:
![]()
To use this result to find
, we obviously need to put x = 1/n, giving:
.
We are particularly interested in what happens to this
series when n gets very large, because that’s when we are approaching e. In that limit,
tends to 1, and so
does
. So, for large enough
n, we can ignore the n-dependence of these early terms in the
series altogether!
When we do that, the series becomes just:
![]()
And, the larger we take n, the more accurately the
terms in the binomial series can be simplified in this way, so as n goes
to infinity this simple series represents the limiting value of
. Therefore, e must be just the sum of this infinite
series.
(Notice that we can see immediately from this series that e is less than 3, because 1/3! is less than 1/22, and 1/4! is less than 1/23, and so on, so the whole series adds up to less than 1 + 1 + ˝ + 1/22 + 1/23 + 1/24 + … = 3.)
Taking our definition of e as the infinite n
limit of
, it is clear that ex is the infinite n
limit of
. Let us write this
another way: put y = nx, so 1/n = x/y. Therefore, ex is the
infinite y limit of
. The strategy at this
point is to expand this using the binomial theorem, as above, and get a power
series for ex.
(Footnote: there is one tricky technical point. The binomial expansion is only simple if the
exponent is a whole number, and for general values of x, y = nx won’t be. But remember we are only interested in the
limit of very large n, so if x is a rational number a/b,
where a and b are integers, for n any multiple of b, y will be an integer, and pretty
clearly the function
is continuous in y,
so we don’t need to worry. If x
is an irrational number, we can approximate it arbitrarily well by a sequence
of rational numbers to get the same result.)
So, we need to do the binomial expansion of
where y is an
integer—to make this clear, let us write y = m.
![]()
Notice that this has exactly the same form as the
binomial expansion of
in the paragraph
above, except that now a power of x appears in each term. Again, we are only interested in the limiting
value as m goes to infinity, and in this limit m(m – 1)/m2
goes to 1, as does m(m-1)(m-2)/m3. Thus, as we take m to infinity, the m
dependence of each term disappears, leaving
![]()
![]()
so when we differentiate ex, we just get ex
back. This means ex is the solution to the equation
, and also the equation
, etc. More generally,
replacing x by ax in the series above gives
![]()
and now differentiating the
series term by term we see
,
, etc., so the function eax is the solution to differential equations of
the form
, or of the form
and so on.
Instead of differentiating term by term, we could have written
![]()
where we have used
in the limit ![]()
We define the natural logarithm function ln x as the inverse of the exponential function, by which we mean
y = ln x, if x = ey
Notice that we’ve switched x
and y from the paragraph above!
Differentiating the exponential function
in this switched
notation,
, so
.
That is to say,
![]()
Therefore, ln x can be written as an integral,
.
You can check that this satisfies the differential equation
by taking the upper limit of the integral to be
then x,
subtracting the second from the first, dividing by
, and taking
very small. But why
have I taken the lower limit of the integral to be 1? In solving the differential equation in this
way, I could have set the lower limit to be any constant and it would still be
a solution—but it would not be the inverse function to ey
unless I take the lower limit 1, since that gives for the value x = 1
that y = ln x = 0. We
need this to be true to be consistent with x = ey,
since e0 = 1.
Exercise: show from the integral form of ln x, that for small x, ln(1 + x) is approximately equal to x. Check with your calculator to see how accurate this is for x = 0.1, 0.01.