As seen in the other write-ups there are many ways to describe the
number e. But how do they fit together? Let's begin by using a
definition which shows why e is such an important
constant. The
proofs explain a fair deal about
exponential functions.
Definition:
e is the unique number such that dex/dx = ex.
Proof of uniqueness and existence:
Suppose that e1, e2 both have the desired property. Consider the function f(x) = e1xe2-x. f'(x) = 0, so e1/e2 = f(1) = f(0) = 1, and thus e1 = e2. So e is unique.
The existence is ascertained by the propositions below. QED
How do we find the value of e? Well, we know that ex describes exponential growth where the growth rate per unit time is equal to the current value. We can approximate this exponential growth by dividing a unit time interval into n equal subintervals. In each subinterval the increase is approximately by a factor 1 + 1/n. Thus we would expect the expression (1 + 1/n)n to give a good approximation of e for large n. Hence we are led to
Proposition:
e = limn→∞(1 + 1/n)n
Proof:
First we need to show that the limit exists.
(1 - 1/(n+1)2)n+1 > 1 - 1/(n+1) ⇒
(1 + 1/(n+1))n+1 > (1 + 1/n)n
so the sequence is increasing. Using the binomial expansion we find that
(1 + 1/n)n < SUM(k = 0, n)(1/k!) < SUM(k = 0, ∞)(1/k!)
for all n, so the sequence is bounded above. Thus the limit exists, and we may call it d.
Consider f : R -> R+, f(x) = dx. This is a continuous strictly increasing function, and therefore has a continuous inverse which we call ln.
Since ln is the inverse we have
1 = ln d = ln (limn→∞(1 + 1/n)n) = limn→∞ n*ln(1 + 1/n))
Hence as h → 0, (ln(1+h))/h → 1 and (dh-1)/h → 1.
Differentiating f from first principles gives
f'(x) = (dx+h - dx)/h = dx(dh-1)/h = dx
as h → 0.
So d satisifes the condition that defines e, and hence e = d = limn→∞(1 + 1/n)n. QED.
The other approach to take when exploring exponentials is to use power series.
Proposition:
e = SUM(k = 0, ∞)(1/k!)
Proof:
Define exp: R -> R by exp x = SUM(k = 0, oo)(xk/k!). Termwise differentiation gives that exp' x = exp x.
Consider f(x) = (exp x)(exp a+b-x). f'(x) = 0, so (exp a)(exp b) = f(a) = f(0) = exp a+b. We can consider this as a functional equation, and using that exp is continuous we find that exp x = (exp 1)x.
Thus exp 1 satisfies the condition that defines e and e = exp 1 = SUM(k = 0, ∞)(1/k!). QED.