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Samuel L. Marateck[^1]\
[*Courant Institute of Mathematical Sciences, New York University, New York, N.Y. 10012*]{}

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[**1.  Introduction.**]{}\
Traditionally the value of $e$ has been obtained, for instance, by taking the limit of ever-decreasing interest intervals in the compound interest formula (see Greenleaf \[1\]) or linear interpolation (see Flanders and Price \[2\]).  We describe an alternate technique of obtaining $e$ that should have pedagogic value.  In this section we give an approximation of $e$ using this technique and generalize it in the next section. 

If $f'(x)$, the derivative of $f(x),$ exists at point $x$, and you start at point $x$ and move a distance $\Delta x$, the value at the point $x + \Delta$ is given by

$$f(x + \Delta x) \cong f(x) + f'(x)\cdot \Delta x$$

We want to find a constant, let’s call it $e$, such that when it’s raised to the power $x$ obtaining the function $e^x$, the function’s derivative is also $e^x$[^2]. 

Since $f'(x)$ equals $f(x),$ we rewrite equation (1) as

$$f(x + \Delta x) \cong f(x)(1 + \Delta x)$$

We will analyse this in the interval \[1,2\].  Let’s take x = 1 and $\Delta x$ = 0.1.  So $x + \Delta x$ is 1.1.  Equation (2) gives

$$f(1.1) \cong f(1)(1 + 0.1)$$

or

$$e^{1.1} \cong 1.1e$$

. 

Now take $x = 1.1$ and use the same value of $\Delta x$, i.e., 0.1.  We will be using the same increment in $x$ in this and all subsequent steps since eventually we will let $\Delta x$ approach zero.  Continuing in this way

$$f(1.1 + 0.1) \cong 1.1f(1.1)$$

So $f(1.2) \cong 1.1e^{1.1}$. Or

$$e^{1.2} \cong (1.1)^2e$$

Eventually we will get $e^2$ on the left side of the equation, so we can solve for e. So let’s compute $e^{1.3}$. We get $e^{1.3} \cong 1.1e^{1.2}$ But this equals $(1.1)^3e$. If we extrapolate to $x = 1.8$, we see that

$$e^{1.9} \cong (1.1)^9e$$

and finally that

$$e^2 \cong (1.1)^{10}e$$

Solving for $e$ we get $e \cong (1.1)^{10}$ or $e$ equals 2.59 to three digits, where the 10 corresponds to dividing 1 by 0.1.  Equation (1) presupposes that $\Delta x$ approaches zero.  If we let $\Delta x= .00000001$, or $10^{-8}$, we raise $(1 + .00000001)$ to $10^{8}$. The answer for $e$ is 2.71828 to five significan figures. 

[**2.  Generalization.**]{}\
We now sketch the steps that describe the preceding method in general.  Using equation (2), and setting $x = 1$, we write

$$e^{1 + \Delta x} \cong e(1 + \Delta x)$$

We continue, letting $x = x+ \Delta x$ and keeping $\Delta x$ the same, and write

$$e^{1 + \Delta x + \Delta x} \cong e^{1 + \Delta x}(1 + \Delta
  x)$$

or

$$e^{1 + 2\Delta x} \cong e(1 + \Delta x)^2$$

We have to add $\Delta x$ to $x$ $1/\Delta x$ times to get $e^2$ on the left side of these equations.  So we get

$$e^{1 + (1/ \Delta x) \cdot \Delta x} \cong e(1 + \Delta x)^{1/
\Delta x}$$

or

$$e^2 = e(1 + \Delta x)^{1/\Delta x}$$

Solve for e and since the definition of the derivative in equation (1) lets $\Delta x \to 0$, take the same limit here.  We get

$$e = \displaystyle \lim_{\Delta x \to 0}(1 + \Delta x)^{1/
\Delta x}$$

which is one of the definitions of $e$.

[**3.  References.**]{}\
\[1\] Greenleaf, Fredrick E.. [*Quantatative Reasoning, 3/e*]{}, McGraw Hill (2006). 

\[2\] Flanders, Harley and Price, Justin J., [*Calculus with Analytic Geometry*]{}, Academic Press (1978). 

[^1]: [*email: marateck@courant.nyu.edu*]{}

[^2]: Our analysis also holds if $f(x) = Ce^x$ where $C$ is a constant.