A look at the Euler’s number: e

I’ve never particularly liked the way e, Euler’s number, is introduced in textbooks. Most approaches give me a very limited intuitive feel for what e actually is. Modern textbooks appear to use one of four common ways to introduce e, and only one of them gives me a semblance of an intuitive feel for e. These approaches include computing compound interest (I think this is one of the worst), integrating 1/x and noting that the area from 1 to the value of e is one (amazing but so what?), looking at the slope of a^x at x = 0, the slope is one when a = e (yes ok, and…) and finally the one I like is starting with dy/dt = y and finding the power series solution and then using this to define e.

Before continuing let’s state what e is numerically equal to:

e = 2.71828….

For normal algebra, all we need are addition, and multiplication (subtraction and division are just alternative forms of these). For convenience, we also introduce a power notation such as x^2 and x^n, but these are just a short-hand notation for doing lots of multiplications at once. With the introduction of trigonometry which involves relationships between sides and angles of triangles, the basic algebra becomes cumbersome because it involves the use of infinite series. Rather than writing the series down all the time, we define short-cut names such as sine, cosine etc. Calculus brings us another special type of series which involves the solution to differential equations. The short-cut notation for this series is e^x.

Note that in all these cases we are still only doing addition and multiplication. The functions sine and e^x are just short-hand for particular combinations of addition and multiplication that happen to be in the form of an infinite series.

Given that e pops up in the form of e^x when solving differential equations, I think this is the place to start. Let’s consider the simplest possible non-trivial differential equation:

    \[ \frac{dy}{dx} = y \]

This equation is saying something quite interesting, that the derivative of y is the same as y. What this means is that if we were to find a solution to this differential equation, the solution would have to also equal dy/dt. We can express the above equation in the form:

    \[ f'(x) = f(x) \]

this perhaps makes it more obvious that the derivative is the same as the function f(x). Can we find an actual function that is like this? The way to find this answer is to define f(x) as a general power series:

    \[ f(x) = a_o + a_1 x + a_2 x^2 + a_3 x^3 + \ldots + a_n x^n \]

We now differentiate this with respect to x to give:

    \[ f'(x) = a_1 + 2 a_2 x + 3 a_3 x^2 + \ldots + (n + 1) a_n x^n \]

To find the function f(x) that also equals f'(x) we need to discover the values for the coefficients, a_o, a_1, etc. To make it easier, let’s decide that when x=0, the value of f(0) is one, i.e: f(0) = 1. This will mean that a_o = 1. What we’ll now do is match up the pairs of terms in f(x) and f'(x), that is set: a_o = a_1, a_1 x = 2 a_2 x, a_2 x^2 = 3 a_3 x^2 and so on.

This allows us to state that: a_1 = 1, a_2 = 1/2, a_3 = a_2/3, a_4 = a_3/4 and so on. Working back by subsituting a_3 into the a_4 equation and a_2 into the a_3 equation, and so on leads to the result that:

    \[ a_n = \frac{1}{n!} \]

We therefore conclude that the equation, f(x) that satisfies f(x) = f'(x) is:

    \[ f(x) = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} \hdots \]

You can easily check by differentiating f(x), you’ll get f(x) again. The solution to dy/dt = y must therefore also be:

    \[ y = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} \hdots \qquad\qquad (1) \]

Not to labor the point too much, but differentiate y and you’ll get the equivalence dy/dx = y.

Let us define the value of the series when x = 1 to be:

    \[ f(1) = 1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \hdots + \frac{1}{n!} \]

For convenience we will call this value e:

    \[ e = f(1) = 1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \hdots + \frac{1}{n!} \]

The next question is, if f(1) = e then what does f(x) equal?

I’m going to make a jump here and propose that f(x) = e^x. From first principles, we can work out the derivative of e^x and remarkably it is e^x! You’ll find a proof at the excellent Paul’s Online Math Notes

Let’s now use the Maclaurin series to find an approximation to e^x. The Maclaurin series is a Taylor series centered on zero, that is:

    \[ f(x)=f(0)+xf'(0)+\frac{x^2}{2 !}f''(0)+...+\frac{x^n}{n !}f^{(n)}(0)+...\]

Since the derivative of e^x is e^x, as well as the third, fourth fifth derivatives etc and noting that e^x at x = 0 is one then we can write:

    \[ f(x) = 1 + 1 \cdot x + \frac{1 \cdot x^2}{2} + \frac{1 \cdot x^3}{6} + \ldots = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \ldots = \sum_{n=0}^{\infty} \frac{x^n}{n~} \]

But this is just the solution to dy/dx= y, — see equation (1) — therefore we conclude that:

    \[ y = e^x \]

It is also worth noting that the derivative for the general exponential a^x is given by:

    \[ \frac{d a^x}{dx} = a^x \log (a) \]

In other words the derivative of e^x is the special case where \log (a) = 1. One reason why e^x is special is that its the purest exponential in the sense that the derivative has no scaling factor. It’s the canonical exponential, the simplest possible. It’s the only exponential function where the function and its derivative are identical.

What else can we say about e^x? What about its rate of increase? We know that the rate of increase is itself, e^x, but how fast is that? One way to look at this is to derive the relative increase in e^x. The relative growth rate is defined by:

    \[ (dy/y)/(dx/y) = \frac{dy}{dx} \frac{x}{y} \]

Given this definition, let’s look first at how fast a^x is increasing by computing

    \[ \frac{d a^x}{dx} \frac{x}{a^x} = a^x \log (x) \frac{x}{a^x} = \log (a) x \]

If we now do the same for e^x we find:

    \[ \frac{d e^x}{dx} \frac{x}{e^x} = e^x \log (e) \frac{x}{e^x} = x \]

In other words e^x is the only exponential function that increases at a rate equal to x. For example if x = 1, it means that a 1% increase in x leads to a 1% increase in e^x while if x = 10 then a 1% increase in x leads to a 10% increase in e^x. This is the characteristic of exponential functions, they increase at an ever increasing rate. Contrast this with a power term such as x^2. If we compute the relative increase for x^2 we find it has a fixed increase of 2:

    \[ \frac{d x^2}{dx} \frac{x}{x^2} = 2 x\frac{x}{x^2} = 2 \]

The growth of a bacterial colony follows this pattern, where the growth in the colony is a fixed percentage over time.

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