# 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 $\underset{n\to \infty }{\mathrm{lim}}{\left(1+\frac{1}{n}\right)}^{n}=e$

Consider the following series:  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 ${\left(1+\frac{1}{n}\right)}^{n}$ 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, ${\left(1+\frac{1}{n}\right)}^{n}$ goes to a definite limit.  It can be proved mathematically that ${\left(1+\frac{1}{n}\right)}^{n}$ 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 ${\left(1+\frac{1}{n}\right)}^{n}$ 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:

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

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

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

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, $n\left(n-1\right)/{n}^{2}$ 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:

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

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 ${\left(1+\frac{1}{n}\right)}^{n}$. 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 ex

Taking our definition of $e$ as the infinite $n$ limit of ${\left(1+\frac{1}{n}\right)}^{n},$ it is clear that ${e}^{x}$ is the infinite $n$ limit of ${\left(1+\frac{1}{n}\right)}^{nx}.$.  Let us write this another way: put $y=nx,$  so $1/n=x/y.$   Therefore, ${e}^{x}$ is the infinite $y$ limit of ${\left(1+\frac{x}{y}\right)}^{y}.$  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 ${\left(1+\frac{x}{y}\right)}^{y}$ 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 ${\left(1+\frac{x}{y}\right)}^{y}$ where $y$ is an integer$—$to make this clear, let us write $y=m:$

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

Notice that this has exactly the same form as the binomial expansion of ${\left(1+\frac{1}{n}\right)}^{n}$ 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

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

### Differentiating ex

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

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

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

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

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,

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

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$—$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 $x=0.1,\text{\hspace{0.17em}}\text{\hspace{0.17em}}0.01.$ x = 0.1, 0.01.