## Friday, July 19, 2013

### The Logarithmic Function From Scratch

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)

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)

Thus

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.