# 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  $lim n→∞ (1+ 1 n ) n =e$

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 $(1+ 1 n ) n$ 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, $(1+ 1 n ) n$ goes to a definite limit.  It can be proved mathematically that $(1+ 1 n ) n$ 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 $(1+ 1 n ) n$ 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:

$(1+x) n =1+nx+ n(n−1) 2! x 2 + n(n−1)(n−2) 3! x 3 +...+ x n$

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

$(1+ 1 n ) n =1+n. 1 n + n(n−1) 2! ( 1 n ) 2 + n(n−1)(n−2) 3! ( 1 n ) 3 +...$

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:

$1+1+ 1 2! + 1 3! + 1 4! +...$

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 $(1+ 1 n ) n$. 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 ex

Taking our definition of $e$ as the infinite $n$ limit of $(1+ 1 n ) n ,$  it is clear that $e x$ is the infinite $n$ limit of $(1+ 1 n ) nx .$  Let us write this another way: put $y=nx,$  so $1/n=x/y.$   Therefore, $e x$ is the infinite $y$ limit of $(1+ x y ) y .$  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 $x, y=nx$ 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 $b, y$ will be an integer, and pretty clearly the function $(1+ x y ) y$  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 $(1+ x y ) y$ where $y$ is an integer$—$to make this clear, let us write $y=m:$

$(1+ x m ) m =1+m. x m + m(m−1) 2! ( x m ) 2 + m(m−1)(m−2) 3! ( x m ) 3 +...$

Notice that this has exactly the same form as the binomial expansion of $(1+ 1 n ) n$ 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

$e x = lim m→∞ (1+ x m ) m =1+x+ x 2 2! + x 3 3! +...$

### Differentiating ex

$d dx e x = d dx (1+x+ x 2 2! + x 3 3! +...)=1+x+ x 2 2! +...$

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

$e ax =1+ax+ a 2 x 2 2! + a 3 x 3 3! +...$

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

Instead of differentiating term by term, we could have written

$d dx e ax = lim Δx→0 e a( x+Δx ) − e ax Δx = lim Δx→0 e ax ( e aΔx −1 ) Δx =a e ax$

where we have used $( e aΔx −1 )→aΔx$ in the limit $Δx→0.$

### The Natural Logarithm

We define the natural logarithm function $lnx$ 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,

$dx dy = e y =x$  so $dy dx = 1 x .$

That is to say,

$d dx lnx= 1 x .$

Therefore, $lnx$ can be written as an integral,

$lnx= ∫ 1 x dz z$.

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

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