That counterpart to last week's development of the exponential function is the logarithm. As anyone who ever played with the antique known as a slide rule, the logarithmic function converts multiplication into addition:
Log(a b) = Log(a) + Log(b) , (1)
but that's getting ahead of ourselves.
Assume that (1) is all we know: Consider a real function G defined on at least some part of the real numbers, and having the property
G(a b) = G(a) + G(b) , (2)
for any real numbers a and b. What can we say about this function?
First, take a = 0. Then we have
G(0 b) = G(0) + G(b) , or
G(0) = G(0) + G(b) . (3)
So either G(b) = 0 for all b, or G(0) is undefined. (Yes, we can say that G(0) = ∞, but I'd rather not go there right now.) Since G = 0 is uninteresting, we'll assume that G(0) is undefined. For now, at least, we'll restrict ourselves to x > 0 and so avoid the problem.
If we set a = 1 in (2) we get
G(b) = G(1) + G(b) , which means that
G(1) = 0 , (4)
which implies G[a (1/a)] = G(a) + G(1/a) = G(1) = 0 , or
G(1/a) = - G(a) . (5)
For any integer M we can repeatedly apply (2) and get
G(aM) = M G(a) . (6)
If we let b = aM, or a = b1/M, then by turning around (6) we get
G(b1/M) = (1/M) G(b) , (7)
and combining the two we get
G(xM/N) = (M/N) G(x) (8) for any integers M and N.
Furthermore, we can play the same trick as we did last week and extend this to all positive real numbers y:
G(xy) = y G(x) . (9)
What about continuity? I'm going to be a little more physicist-like this week (though last week was probably not properly mathematical anyway), and go through it rather quickly. Let's take two numbers, x > 0 and y > -x. Then we can always write
x + y = x Q , Q = 1+y/x > 0 ,
and we want to study the behavior of
G(x + y) = G(x Q) = G(x) + G(Q) as y → 0 or Q → 1.
When we put it like that, the answer is obvious. Write Q = h1/N. Then as N → ∞, Q → ∞ 1, and
limN → ∞ G(Q) = limN → ∞ G(h)/N → 0 . (10)
limy→0 |G(x+y) - G(x)| = 0 , (11)
and G(x) is a continuous function for x > 0.
Using the same tricks, let's look at the derivative, if it exists:
G'(x) = limy→∞0 [G(x+y)-G(x)]/y , (12)
With the above substitutions we can write this as
G'(x) = limN→∞ [G(x h1/N) - G(x)]/[x (h1/N-1) , (13)
which reduces to G'(x) = [G(h)/x]/ [limN→∞ N (h1/N - 1)] . (14)
Last week we showed that the limit in the denominator of (14) is finite, so it follows that G'(x) is well defined for x > 0.
To evaluate the derivative, we do the standard tricks:
d/dy G(x y) = x G'(x y) = d/dy [G(x) + G(y)] = G'(y)
and set y = 1. Then
G'(x) = G'(1)/x , x > 0 . (15)
Thus each and every log function (2) is uniquely defined once we define G'(1). Since G(1) = 0, we can integrate (15) to find
G(x) = G'(1) ∫1x dt / t . (16)
The logical prototype logarithm function (which we could call the, ahem, natural logarithm) is then the function with G'(1) = 1:
ln x = ∫1x dt / t . (17)
Which is exactly how my college calculus course started out.
One more thing: we want to show that ln x and exp x are inverse functions of one another. Write
H(x) = ln(exp x) . (18)
Obviously H(0) = ln(exp 0) = ln(1) = 0. Furthermore,
H'(x) = [d/dx exp x]/exp x = exp x/exp x = 1 , (19)
so H(x) = x. We could go the opposite way as well and show that
exp(ln x) = x . (20)
That concludes this little exploration of the log and exponential functions. Next week we'll take a look at sines and cosines.