# 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 **$e,\text{\hspace{0.17em}}{e}^{x}$** and the natural logarithm.*

### The Limit $\underset{n\to \infty}{\mathrm{lim}}{(1+{\scriptscriptstyle \frac{1}{n}})}^{n}=e$

Consider the following series: $(1+1),\text{(1+}{\scriptscriptstyle \frac{\text{1}}{\text{2}}}{)}^{2},\text{}{(1+{\scriptscriptstyle \frac{1}{3}})}^{3},\text{}\mathrm{...}\text{,}{(1+{\scriptscriptstyle \frac{1}{n}})}^{n},\mathrm{...}$ 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 ${(1+{\scriptscriptstyle \frac{1}{n}})}^{n}$ 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, ${(1+{\scriptscriptstyle \frac{1}{n}})}^{n}$ goes to a definite limit. It can be proved mathematically that ${(1+{\scriptscriptstyle \frac{1}{n}})}^{n}$ *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 ${(1+{\scriptscriptstyle \frac{1}{n}})}^{n}$ for large $n,$ let us expand it using the binomial theorem. Recall that the binomial theorem gives all the terms in ${\left(1+x\right)}^{n}$, as follows:

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

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

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

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, $n\left(n-1\right)/{n}^{2}$ tends to 1, and so does $n\left(n-1\right)\left(n-2\right)/{n}^{3}.$. 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:

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

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 ${(1+{\scriptscriptstyle \frac{1}{n}})}^{n}$.
*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 1/2^{2},
and 1/4! is less than 1/2^{3}, and so on, so the whole series adds up
to less than 1 + 1 + $\mathrm{\xbd}$ + 1/2^{2} + 1/2^{3} + 1/2^{4} + … = 3.)

### The Exponential Function *e*^{x}

^{x}

Taking our definition of $e$ as the infinite $n$ limit of ${(1+{\scriptscriptstyle \frac{1}{n}})}^{n},$ it is clear that ${e}^{x}$ is the infinite $n$ limit of ${(1+{\scriptscriptstyle \frac{1}{n}})}^{nx}.$. Let us write this another way: put $y=nx,$ so $1/n=x/y.$ Therefore, ${e}^{x}$ is the infinite $y$ limit of ${(1+{\scriptscriptstyle \frac{x}{y}})}^{y}.$ 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 $x,\text{\hspace{0.17em}}y=nx$ 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 $b,\text{\hspace{0.17em}}y$ *will* be an integer, and pretty clearly
the function ${(1+{\scriptscriptstyle \frac{x}{y}})}^{y}$ 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 ${(1+{\scriptscriptstyle \frac{x}{y}})}^{y}$ where $y$ is an integer$\u2014$to make this clear, let us write $y=m:$

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

Notice that this has exactly the same form as the binomial expansion of ${(1+{\scriptscriptstyle \frac{1}{n}})}^{n}$ 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 $m\left(m-1\right)/{m}^{2}$ goes to 1, as does $m\left(m-1\right)\left(m-2\right)/{m}^{3}.$ Thus, as we take $m$ to infinity, the $m$ dependence of each term disappears, leaving

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

### Differentiating *e*^{x}

^{x}

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

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

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

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

$$\frac{d}{dx}{e}^{ax}=\underset{\Delta x\to 0}{\mathrm{lim}}\frac{{e}^{a\left(x+\Delta x\right)}-{e}^{ax}}{\Delta x}=\underset{\Delta x\to 0}{\mathrm{lim}}\frac{{e}^{ax}\left({e}^{a\Delta x}-1\right)}{\Delta x}=a{e}^{ax}$$

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

### The Natural Logarithm

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

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

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

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

That is to say,

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

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

$\mathrm{ln}x={\displaystyle \underset{1}{\overset{x}{\int}}\frac{dz}{z}}$.

You can check that this satisfies the differential equation by taking the upper limit of the integral to be $x+\Delta x,$ then $x,$ subtracting the second from the first, dividing by $\Delta x,$ and taking $\Delta x$ 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$\u2014$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 $y=\mathrm{ln}x=0.$ We need this to be true to be consistent with $x={e}^{y},$ since ${e}^{0}=1.$

*Exercise*: show from
the integral form of $\mathrm{ln}x$ that for small $x,\text{\hspace{0.17em}}\mathrm{ln}\left(1+x\right)$ is approximately equal to $x.$ Check with your calculator to see how accurate
this is for $x=0.1,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathrm{0.01.}$