Bielefeld, May 5, 1997

Exponentials and logarithms: a bit of theory.

0. Generalities. By a (real-valued) exponential we mean a continuous homomorphism E : R^{+} → R^{×}, in other words, a non-zero continuous function E(t) such that

(∗)  E(s + t) = E(s)E(t)

for arbitrary real numbers s and t. This equation immediately entails E(0) = 1 and E(−t) = E(t)^{−1} (see?), as well as the positivity of all function values E(t) = E(t/2)^{2}. Since E(nt) = E(t)^{n} for all natural n, we obtain E(s/n) = E(s)^{1/n} and hence E(qt) = E(t)^{q} for any rational number q = m/n. By continuity, we then have

(∗∗)  E(rt) = lim_{q→r} E(t)^{q} = E(t)^{r} ,

with q running through a sequence of rational numbers converging to r. If r is irrational, this amounts to a definition of the symbol E(t)^{r}. We can now write E(t) = E(1)^{t} for all t.

Suppose that E is non-trivial, i.e., E(1) ≠ 1. By switching from E(t) to E(−t), if necessary, we may assume E(1) > 1. In this case E(t) is monotone increasing, because E(s) = E(1)^{s} > 1 for all positive s, whence E(s + t) > E(t) by (∗). Since E(n) = (1 + a)^{n} > 1 + na for some positive a and all natural n, the function-values E(t) have no upper bound. On the other hand, E(−n) = E(n)^{−1} shows that their greatest lower bound is 0.

Uniqueness. Since R is a connected set, the range of any non-trivial E is an interval, which — by the preceding argument — must be (0, +∞). Therefore, if Ẽ is another exponential, there will be a number r such that E(r) = Ẽ(1). It then follows that Ẽ(t) = Ẽ(1)^{t} = E(rt). Let us summarize.

Given any non-trivial E(t), all exponentials are of the form E(rt) for suitable r.

This gives us the freedom to choose any convenient number z ≠ 1 as our “base” E(1), without loss of generality. Very good — but do non-trivial exponentials exist? We shall take three stabs at this question.

1. Fractional powers. Take a number z > 1 (popular choices are z = 10 or z = 2), and define E(q) = z^{q} for rational q = m/k. Since E(m/k) = E(1/k)^{m}, we get E(m/k + n/k) = E(1/k)^{m+n} = E(m/k)E(n/k), so that (∗) holds for rational s and t.

For computional consistency, we want a continuous function: if s is small, we want E(s + t) to be close to E(t). Because of (∗), this means that s → 0 should imply E(s) → 1. By monotonicity — based on E(q) > 1 for q > 0 as above — it suffices to check E(1/2^{m}) → 1. Since E(1/2^{m}) is the square root of E(1/2^{m−1}), this follows by repeatedly using the fact that (1 + a) = \sqrt{1 + b} implies a < b/2.

We now have a continuous homomorphism E : Q^{+} → R^{×} and can use it for all kinds of practical or playful purposes. We could, for instance, compute logarithms to any degree of accuracy, and nobody would ever know that our domain is not the full real number system. In fact, we do not even need all of Q^{+}: it would suffice to take, for instance, the subgroup B consisting of all finite binary expansions.

To get a full-fledged exponential, we must still define E(r) for irrational r, and we do so by mimicking the second equation in (∗∗). For r > 0, a decreasing rational sequence q_{ν} → r leads to a decreasing sequence E(q_{ν}), which — being bounded below by 1 — has a limit. This should be the desired E(r), but it still seems to depend on the sequence {q_{ν}}. However, any other rational sequence converging to r would have the form {q_{ν} + k_{ν}} with k_{ν} → 0. Hence E(k_{ν}) → 1 by continuity, and the new sequence yields the same value for E(r). For irrational s < 0, we put r = −s and E(s) = E(r)^{−1}.

By standard theorems on limits of sums and products, we now get (∗) for arbitrary real s and t, as well as the monotonicity of the extended function. Its continuity — this time with respect to all of R — follows from its monotonicity on R together with its continuity on Q. Lest you think of this as a pure formality, you should know that there are some mighty strange extensions of E from Q^{+} to R^{+}, for instance by composing it with a (highly discontinuous) additive projection R^{+} → Q^{+}.

2. A logarithmic integral. If x denotes a positive real variable, the function F (x) = ∫_{1}^{x} du/u is characterized by the properties F′(x) = 1/x and F (1) = 0. For constant w > 0, the modified function F_{w}(x) = F (wx) also satisfies F′_{w}(x) = 1/x, by the Chain Rule — which is very easily checked in this case (see?). Hence F (wx) − F (w) fits the same description as F (x), and we obtain the functional equation

(†)  F (wx) = F (x) + F (w) .

Being an integral, F (x) is continuous, and having a positive derivative, it is monotone increasing. Since (†) implies F (x^{−1}) = −F (x) and F (x^{n}) = nF (x), it is unbounded above and below. Thus its range is the entire real number system R.

Given real numbers s and t, let w and x be such that F (w) = s and F (x) = t. If E denotes the inverse function (i.e. E(t) = x ⟺ F (x) = t), the relation (†) then translates to E(s)E(t) = E(s + t), the functional equation (∗) from Section 0. As the inverse of a continuous and monotone function, E inherits these properties and is therefore an exponential. Reaching a little deeper into analytic know-how or geometric insight, we reckon that E even inherits the differentiability of F.

However, instead of invoking theorems, let us reflect a little on these important properties of E. Monotonicity is obvious, and the functional equation (∗) follows formally from (†). To prove continuity of E, it therefore suffices to show that E(s) falls into an arbitrarily narrow interval I = (1 − ε, 1 + ε) if s is chosen small enough. Now, I is mapped onto the interval J = (F (1 − ε), F (1 + ε)) by the continuous, strictly increasing F. If you choose s inside J, the value E(s) will lie inside I. So much for continuity — what about differentiability?

On account of (∗), the existence of a derivative E′(t) need only be verified for t = 0, as pointed out on the sheet called “Logarithmus naturalis”. Since s → 0 implies E(s) → 1 by continuity, we can write

lim_{s→0} \frac{E(s) − 1}{s} = lim_{E(s)→1} \frac{E(s) − 1}{F (E(s))} = lim_{u→1} \frac{u − 1}{F (u) − F (1)} ,

which is just the reciprocal of F′(1), namely 1. Therefore E′(0) = 1, and since E′(t) = E′(0)E(t) — again as explained loc.cit. — we conclude that E(t) satifies

(‡)  E′(t) = E(t) and E(1) = 0 .

If g(t) is any function such that g′(t) = g(t), differentiating g(t)E(−t) yields g′(t)E(−t) − g(t)E′(t) = 0. Hence g(t)E(−t) is a constant c, and g(t) = cE(t). Moral: E(t) is the only function satisfying (‡).

3. An exponential power series. The polynomials

P_{n}(t) = ∑_{k=0}^{n} t^{k}/k!

form a Cauchy sequence, which converges uniformly on any closed interval [−T, T]. In fact, for n > 2T and ν ≥ 0, the difference P_{n+ν}(T) − P_{n−1}(T) is clearly less than (T^{n}/n!)(1 + T/n + ⋯ + T^{ν}/n^{ν}), which in turn is less than 1/2^{n}(1 + 1/2 + ⋯ + 1/2^{ν}) < 1/2^{n−1}. By the Triangle Inequality, the same bound is valid for |P_{n+ν}(t) − P_{n−1}(t)| and any t ∈ [−T, T].

This uniformity of convergence has two consequences. Firstly, the limit function P_{∞}(t) is continuous, and secondly the integrals (over any finite interval) of the individual terms P_{n}(t) converge to the corresponding integral of P_{∞}(t). Since P′_{n+1}(t) = P_{n}(t), whence P_{n+1}(t) = 1 + ∫_{0}^{t} P_{n}(s)ds (see?), we get

lim_{n→∞} P_{n+1}(t) = 1 + lim_{n→∞} ∫_{0}^{t} P_{n}(s)ds = 1 + ∫_{0}^{t} P_{∞}(s)ds .

The left hand term of this is just P_{∞}(t), and the integral at the other end shows that this function is differentiable. If we now take the derivative, we find P′_{∞}(t) = P_{∞}(t). Since moreover P_{∞}(0) = 1, we conclude that P_{∞}(t) is nothing other than the E(t) as described in Section 2 above — in a more computational guise.

Incidentally, the relation P′_{∞}(t) = P_{∞}(t) does not directly follow from P′_{n+1}(t) = P_{n}(t), in spite of uniform convergence. The function sequence f_{n}(t) = n^{−1} sin nt, for instance, converges to 0 uniformly for all t, but its derivatives go wild!