The Number e and the Exponential Function

Michael Fowler

Disclaimer: these notes are not mathematically rigorous. Instead, they present quick, and, I hope, plausible, derivations of the properties of $e, e^{x}$ and the natural logarithm.

The Limit $\lim_{n \to \infty} {(1 + \frac{1}{n})}^{n} = e$

Consider the following series: $(1 + 1), (1+ \frac{1}{2})^{2}, {(1 + \frac{1}{3})}^{3}, ..., {(1 + \frac{1}{n})}^{n}, ...$ where $n$ runs through the positive integers. What happens as $n$ gets very large?

It’s easy to find out with a calculator using the function x^y. The first three terms are 2, 2.25, 2.37. You can use your calculator to confirm that for $n$ = 10, 100, 1000, 10,000, 100,000, 1,000,000 the values of ${(1 + \frac{1}{n})}^{n}$ are (rounding off) 2.59, 2.70, 2.717, 2.718, 2.71827, 2.718280. These calculations strongly suggest that as $n$ goes up to infinity, ${(1 + \frac{1}{n})}^{n}$ goes to a definite limit. It can be proved mathematically that ${(1 + \frac{1}{n})}^{n}$ does go to a limit, and this limiting value is called $e .$ The value of e is 2.7182818283… .

To try to get a bit more insight into ${(1 + \frac{1}{n})}^{n}$ for large $n,$ let us expand it using the binomial theorem. Recall that the binomial theorem gives all the terms in ${(1 + x)}^{n}$ , as follows:

${(1 + x)}^{n} = 1 + n x + \frac{n (n - 1)}{2!} x^{2} + \frac{n (n - 1) (n - 2)}{3!} x^{3} + ... + x^{n}$

To use this result to find ${(1 + \frac{1}{n})}^{n},$ we obviously need to put $x = 1 / n,$ giving:

${(1 + \frac{1}{n})}^{n} = 1 + n . \frac{1}{n} + \frac{n (n - 1)}{2!} {(\frac{1}{n})}^{2} + \frac{n (n - 1) (n - 2)}{3!} {(\frac{1}{n})}^{3} + ...$

We are particularly interested in what happens to this series when $n$ gets very large, because that’s when we are approaching $e .$ In that limit, $n (n - 1) / n^{2}$ tends to 1, and so does $n (n - 1) (n - 2) / n^{3} .$ . So, for large enough $n,$ we can ignore the $n$ -dependence of these early terms in the series altogether!

When we do that, the series becomes just:

$1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + ...$

And, the larger we take $n,$ the more accurately the terms in the binomial series can be simplified in this way, so as $n$ goes to infinity this simple series represents the limiting value of ${(1 + \frac{1}{n})}^{n}$ . Therefore, $e$ must be just the sum of this infinite series.

(Notice that we can see immediately from this series that $e$ is less than 3, because 1/3! is less than 1/2², and 1/4! is less than 1/2³, and so on, so the whole series adds up to less than 1 + 1 + $½$ + 1/2² + 1/2³ + 1/2⁴ + … = 3.)

The Exponential Function e^x

Taking our definition of $e$ as the infinite $n$ limit of ${(1 + \frac{1}{n})}^{n},$ it is clear that $e^{x}$ is the infinite $n$ limit of ${(1 + \frac{1}{n})}^{n x} .$ Let us write this another way: put $y = n x,$ so $1 / n = x / y .$ Therefore, $e^{x}$ is the infinite $y$ limit of ${(1 + \frac{x}{y})}^{y} .$ The strategy at this point is to expand this using the binomial theorem, as above, and get a power series for $e^{x} .$

(Footnote: there is one tricky technical point. The binomial expansion is only simple if the exponent is a whole number, and for general values of $x, y = n x$ won’t be. But remember we are only interested in the limit of very large $n,$ so if $x$ is a rational number $a / b,$ where $a$ and $b$ are integers, for $n$ ny multiple of $b, y$ will be an integer, and pretty clearly the function ${(1 + \frac{x}{y})}^{y}$ is continuous in $y,$ so we don’t need to worry. If $x$ is an irrational number, we can approximate it arbitrarily well by a sequence of rational numbers to get the same result.)

So, we need to do the binomial expansion of ${(1 + \frac{x}{y})}^{y}$ where $y$ is an integer $—$ to make this clear, let us write $y = m :$

${(1 + \frac{x}{m})}^{m} = 1 + m . \frac{x}{m} + \frac{m (m - 1)}{2!} {(\frac{x}{m})}^{2} + \frac{m (m - 1) (m - 2)}{3!} {(\frac{x}{m})}^{3} + ...$

Notice that this has exactly the same form as the binomial expansion of ${(1 + \frac{1}{n})}^{n}$ in the paragraph above, except that now a power of $x$ appears in each term. Again, we are only interested in the limiting value as $m$ goes to infinity, and in this limit $m (m - 1) / m^{2}$ goes to 1, as does $m (m - 1) (m - 2) / m^{3} .$ Thus, as we take $m$ to infinity, the $m$ dependence of each term disappears, leaving

$e^{x} = \lim_{m \to \infty} {(1 + \frac{x}{m})}^{m} = 1 + x + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + ...$

Differentiating e^x

$\frac{d}{d x} e^{x} = \frac{d}{d x} (1 + x + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + ...) = 1 + x + \frac{x^{2}}{2!} + ...$

so when we differentiate $e^{x},$ we just get $e^{x}$ back. This means $e^{x}$ is the solution to the equation $\frac{d y}{d x} = y,$ and also the equation $\frac{d^{2} y}{d x^{2}} = y,$ etc. More generally, replacing $x$ by $a x$ in the series above gives

$e^{a x} = 1 + a x + \frac{a^{2} x^{2}}{2!} + \frac{a^{3} x^{3}}{3!} + ...$

and now differentiating the series term by term we see $\frac{d}{d x} e^{a x} = a e^{a x},$ $\frac{d^{2}}{d x^{2}} e^{a x} = a^{2} e^{a x},$ etc., so the function $e^{a x}$ is the solution to differential equations of the form $\frac{d y}{d x} = a y,$ or of the form $\frac{d^{2} y}{d x^{2}} = a^{2} y$ and so on.

Instead of differentiating term by term, we could have written

$\frac{d}{d x} e^{a x} = \lim_{Δ x \to 0} \frac{e^{a (x + Δ x)} - e^{a x}}{Δ x} = \lim_{Δ x \to 0} \frac{e^{a x} (e^{a Δ x} - 1)}{Δ x} = a e^{a x}$

where we have used $(e^{a Δ x} - 1) \to a Δ x$ in the limit $Δ x \to 0.$

The Natural Logarithm

We define the natural logarithm function $\ln x$ as the inverse of the exponential function, by which we mean

$y = \ln x if x = e^{y} .$

Notice that we’ve switched $x$ and $y$ from the paragraph above! Differentiating the exponential function $x = e^{y}$ in this switched notation,

$\frac{d x}{d y} = e^{y} = x$ so $\frac{d y}{d x} = \frac{1}{x} .$

That is to say,

$\frac{d}{d x} \ln x = \frac{1}{x} .$

Therefore, $\ln x$ can be written as an integral,

$\ln x = \int_{1}^{x} \frac{d z}{z}$ .

You can check that this satisfies the differential equation by taking the upper limit of the integral to be $x + Δ x,$ then $x,$ subtracting the second from the first, dividing by $Δ x,$ and taking $Δ x$ very small. But why have I taken the lower limit of the integral to be 1? In solving the differential equation in this way, I could have set the lower limit to be any constant and it would still be a solution $—$ but it would not be the inverse function to $e^{y}$ unless the integral has lower limit 1, since that gives for the value $x = 1$ that $y = \ln x = 0.$ We need this to be true to be consistent with $x = e^{y},$ since $e^{0} = 1.$

Exercise: show from the integral form of $\ln x$ that for small $x, \ln (1 + x)$ is approximately equal to $x .$ Check with your calculator to see how accurate this is for $x = 0.1, 0.01.$

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The Number e and the Exponential Function

The Limit lim n→∞ (1+ 1 n ) n =e

The Exponential Function ex

Differentiating ex

The Natural Logarithm

The Limit $\lim_{n \to \infty} {(1 + \frac{1}{n})}^{n} = e$

The Exponential Function e^x

Differentiating e^x