e (idea)

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 de^x/dx = e^x.

Proof of uniqueness and existence:
Suppose that e₁, e₂ both have the desired property. Consider the function f(x) = e₁^xe₂^-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 e^x 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 = lim_n→∞(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) = d^x. 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 (lim_n→∞(1 + 1/n)ⁿ) = lim_n→∞ n*ln(1 + 1/n))

Hence as h → 0, (ln(1+h))/h → 1 and (d^h-1)/h → 1.
Differentiating f from first principles gives

f'(x) = (d^x+h - d^x)/h = d^x(d^h-1)/h = d^x

as h → 0.
So d satisifes the condition that defines e, and hence e = d = lim_n→∞(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)(x^k/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.