Exponential functions, \(e\), and slope

Introduction

Wikipedia defines \(e\) in terms of a limit: \[ e = \lim _{n\to \infty }\left(1+{\frac {1}{n}}\right)^{n}. \] As it turns out, this definition makes \(e^x\) as simple to differentiate as possible!

Exponential functions

An exponential function is a function of the form \(f(x) = b^x\). The number \(b\) is called the base and can be anything larger than zero.

It’s pretty easy to compute the value of an exponential function at an integer, since it’s just \[b^n=\overbrace{b\times b\times\cdots\times b}^{n\text{ times}}.\]

As a result, it’s pretty easy to graph an exponential function: just plot some values at a few integers and connect the dots!

viewof fs = Inputs.checkbox(
    ["2^x", "e^x", "3^x"], 
    {format: d => tex`${d}`, label: "Plot:"}
);

plots_with_dots(fs, state);

state = [];

Note that the larger the base, the steeper the function:

viewof b = Inputs.range([0.25,4], {value: 2, step: 0.001, label: "b:"})

simple_plot(b)

The derivative

We measure steepness, of course, with the derivative. When trying to really understand the derivative of a new function, we generally apply the difference quotient. Here’s how:

\[\begin{aligned} \frac{d}{dx} b^x &= \lim_{h\to0} \frac{b^{x+h}-b^x}{h} = \lim_{h\to0} \frac{b^xb^h-b^x}{h} \\ &= b^x \lim_{h\to0}\frac{b^h-1}{h} = c_b \, b^x. \end{aligned}\]

The value of that limit \(\lim_{h\to0}(b^h-1)/h\) is presumably some number, which I’ve chosen to denote with a \(c_b\). Thus, this computation suggests that the derivative of any exponential function is simply a constant times that exponential function you started with.

The limit that this leads to \[ \lim_{h\to0}\frac{b^h-1}{h} \] is not quite as elementary as those limits that arise when dealing with algebraic functions like polynomials. There’s no evident way to cancel the \(h\) in the denominator.

We’re going to tackle this from a couple different directions:

• Graphically and
• Numerically

The geometric look

The value of \(c_b\) has a simple geometric interpretation. If \(f(x) = b^x\), then \[ f'(x) = c_bb^x \text{ so that } f'(0) = c_b. \] Thus, \(c_b\) represents the slope of \(y=b^x\) as the graph crosses the \(y\)-axis.

Let’s take a closer look at the graphs of \(2^x\), \(3^x\), and \(e^x\):

viewof parameters = Inputs.form([
  Inputs.toggle({label:"Base two:", value: true}),
  Inputs.toggle({label:"Base three:", value: false}),
  Inputs.toggle({label:"Base e:", value: false})
]);
viewof h = Inputs.range([-2,2], {value: 1, step: 0.01, label: 'h:'});

html`
<div style="position: relative; height: 400px">
  <div style="position: absolute; top:110px; left: 90px">
    ${parameters[0]
      ? tex`m_2 \approx ${d3.format("0.3f")((2 ** h - 1) / h)}`
      : ""}
  </div>
  <div style="position: absolute; top:135px; left: 90px">
    ${parameters[1]
      ? tex`m_3 \approx ${d3.format("0.3f")((3 ** h - 1) / h)}`
      : ""}
  </div>
  <div style="position: absolute; top:160px; left: 90px">
    ${parameters[2]
      ? tex`m_e \approx ${d3.format("0.3f")((Math.exp(h) - 1) / h)}`
      : ""}
  </div>
  <div style="position: absolute; top:0; left:0">
    ${plots_with_slopes(parameters, h)}
  </div>
</div>`

Note that \(2^x\) crosses the \(y\)-axis with slope less than one; \(3^x\) crosses the \(y\)-axis with slope just a little more than one. It stands to reason that there should be a number between the two where the slope is exactly one. The number \(e\) is, by definition, exactly that number!

The numeric look

Let’s take a closer look at this limit numerically. To this end, Table 1 lists values of \[(2^h-1)/h \text{ and }(3^h-1)/h\] near \(h=0\).

Table 1: Values of \((2^h-1)/h\) and \((3^h-1)/h\) near \(h=0\)

\(h\)	0.100000	0.010000	0.001000	0.000100	0.000010	0.000001
\((2^h-1)/h\)	0.717735	0.695555	0.693387	0.693171	0.693150	0.693147
\((3^h-1)/h\)	1.161232	1.104669	1.099216	1.098673	1.098618	1.098613

The table indicates that \[ \lim_{h\to0} \frac{2^h-1}{h} \approx 0.6931 \text{ and } \lim_{h\to0} \frac{3^h-1}{h} \approx 1.0986. \] This suggests that there must be some value of \(b\) between \(2\) and \(3\) so that the limit \[ \lim_{h\to0}\frac{b^h-1}{h} \text{ is exactly } 1. \] We denote this value of \(b\) with the letter \(e\). That is, the number \(e\) is chosen, by definition, to be the unique number such that \[ \lim_{h\to0}\frac{e^h-1}{h} = 1. \] The great thing about this from the perspective of calculus is that it explains the simpler formula for the derivative of the exponential function, namely \[ \frac{d}{dx} e^x = e^x. \] Thus \(f(x) = e^x\) is the exponential function whose derivative is as simple as possible.

The limit definition

So, how does this approach to \(e\) as the natural exponential base relate to Wikipedia’s claim that \[ e = \lim _{n\to \infty }\left(1+{\frac {1}{n}}\right)^{n}? \] Well, if \[ \lim_{h\to0}\frac{e^h-1}{h} = 1, \] then for small values of \(h\) we must have \[ \frac{e^h-1}{h} \approx 1. \] Solving for \(e\), we get \[ e \approx (1+h)^{1/h} \] or, maybe \[ e = \lim_{h\to0} (1+h)^{1/h}. \] The advantage here is that it makes it much easier to compute estimates for the numerical value of \(e\); simply plug values of \(h\) that are close to zero into \((1+h)^{1/h}\). The common definition stated by Wikipedia is simply a sequential version of this obtained with the identification \(n=1/h\). The limit then becomes \[ e = \lim_{n\to\infty} \left(1+\frac{1}{n}\right)^{n}. \] We can then approximate \(e\) by plugging in large values of \(n\) into \((1+1/n)^n\).

Table 2: Values of \((1+1/n)^n\) for large \(n\)

\(n\)	10	100	1000	10000	100000	1000000
\((1+1/n)^n\)	2.593742	2.704814	2.716924	2.718146	2.718268	2.718280

It looks like \(e\approx 2.718\). A more precise approximation is \[ e \approx 2.7182818284590452353602874714. \]

Building a limit table

Here’s a little code (in a language called SageMath) that allows you to define your base b and then prints a table of values of \(h\) vs \((b^h - 1)/h\) for various values of \(h\):

That’s live code so that you can execute it or even edit it and then execute it. Could prove handy for a homework problem!

import {plots_with_dots, plots_with_slopes, simple_plot} from './plots.js';