# The Natural Base e

## Key Questions

• "e" is a commonly used base for exponential functions.

Any number greater than 0 (except for 1) can be used for an exponential function. "e", called "Euler's Constant" (pronounced oiler) is one of the most commonly used bases, i.e. $y = {e}^{x}$.

It occurs in many applications such as statistics and finance. "e" can be defined as follows. If you take the function $y = {\left(1 + \frac{1}{x}\right)}^{x}$ and then allow x to get larger and larger (approaching infinity), then y will approach the constant e, which is approximately 2.718.

• e = 1+ 1/1 + 1/(2*1) + 1/(3*2*1) + 1/(4*3*2*1) + 1/(5!)+ 1/(6!) + * * *

(For positive integer $n$, we define: n! = n(n-1)(n-2) * * * (3)(2)(1) and 0! = 1

$e$ is the coordinate on the $x$-axis where the area under $y = \frac{1}{x}$ and above the axis, from $1$ to $e$ is $1$

$e = {\lim}_{m \rightarrow \infty} {\left(1 + \frac{1}{m}\right)}^{m}$

$e \approx 2.71828$ it is an irrational number, so its decimal expansion neither terminates nor goes into a cycle.
(It is also transcendental which, among other things, means it cannot be written using finitely many algebraic operations
($\times , \div , + , - , \text{exponents and roots}$) and whole numbers.)

Euler's number is written as e. It is an irrational number and is used as the base of natural logarithms. Its first few digits are 2.718..... More about it is available on web.

#### Explanation:

• Euler's number $e$ makes our lives a little easier since the slope of $f \left(x\right) = {e}^{x}$ at $x = 0$ is exactly one. In other words, $e$ is a base such that

$f ' \left(0\right) = {\lim}_{h \to 0} \frac{{e}^{0 + h} - {e}^{0}}{h} = {\lim}_{h \to 0} \frac{{e}^{h} - 1}{h} = 1$,

which is quite useful in finding $f ' \left(x\right)$. Let us use the definition to find the derivative of $f \left(x\right)$.

$f ' \left(x\right) = {\lim}_{h \to 0} \frac{{e}^{x + h} - {e}^{x}}{h} = {\lim}_{h \to 0} \frac{{e}^{x} \cdot {e}^{h} - {e}^{x}}{h}$

by pulling ${e}^{x}$ out of the limit,

$= {e}^{x} {\lim}_{h \to 0} \frac{{e}^{h} - 1}{h} = {e}^{x} \cdot 1 = {e}^{x}$.

Hence, the derivative of ${e}^{x}$ is itself, which is very convenient.

I hope that this was helpful.