More about e

The exponential function comes up all the time in science. Often the rate of change of some quantity x is proportional to the value x itself:

dx/dt = k * x

where k is a constant. Everyday examples for biology include bacterial growth and radioactive decay. As most of you know, we can solve this problem using a fundamental result from calculus:

d/dx ln(x) = 1/x ln(x) = ∫ 1/x dx

So, rearranging the first equation and then integrating:

dx/x = k * dt ∫ dx/x = ∫ k * dt ln(x) = k * t x = A exp { k * t }

Evaluate the constant A by setting t = 0, so then x = x₀:

x = x0 exp { k * t }

At a special value, t = T (one generation, one half-life):

x = 2 * x0 2 = exp { k * t } ln(2) = k * T

I remember reading something by Francois Jacob, a long time ago, in which he described his and Jacques Monod's excitement at some point that they had done a set of experiments and the value 0.69 appeared from their calculations for all the experiments. What was this magic number? It took a minute before the penny dropped, and they remembered that ln(2) = 0.693:

>>> import math >>> print round(log(2),3) 0.693

I haven't been able to find the reference.

After that long preamble, here is the real subject of this post. Starting from:

(1) d/dx ex = ex

we want to show that:

d/dx ln(x) = 1/x

By definition:

eln(x) = x

Take the derivative on both sides:

d/dx eln(x) = d/dx x

The right side equals 1. That is:

(2) d/dx eln(x) = 1

Now for the left side. It is easier if we make a substitution. Let:

u = ln(x) eln(x) = eu = x

And

d/dx eln(x) = d/dx eu

We use the chain rule:

d/dx eu = d/du eu * du/dx

By the definition of e above in (1):

d/du eu = eu

So:

d/dx eu = eu * du/dx

Here is the trick. Substitute u = ln(x) to go back to x:

d/dx eu = x * du/dx d/dx eln(x) = x * du/dx d/dx eln(x) = x * d/dx ln(x)

But from (2), the left side equals 1:

1 = x * d/dx ln(x) d/dx ln(x) = 1/x

Or, quoting Heath's translation of Euclid: "precisely what was required to be proved."

I would love to credit the source for this, but I've forgotten where I found it.