## How to Calculate the Value of e

ver wonder how the calculator seems to know the numeric value of $e$ ?  How could it possibly know?  $e$ is an Irrational Number which means its decimals don’t repeat! (Specifically $e$ is a Transcendental Number but it falls under the category of an irrational number).  So how can you fit an infinitely large number in a small box?  Or for that matter, how did Leonhard Euler manage to calculate $e$ to 23 decimal places without a calculator?

It was witchcraft! Sorcery!  Just kidding!  🙂  But seriously, what’s the trick?
Skip the explanation, show me the trick!

It all starts with something called a Power Series.  A power series is an infinite series of the form:

$f(x)={\displaystyle \sum_{n=0}^{\infty}c_{n}\cdot(x-a)^{n}=c_{0}\cdot(x-a)^{0}+c_{1}\cdot(x-a)^{1}+c_{2}\cdot(x-a)^{2}+c_{3}\cdot(x-a)^{3}+\cdots}$

where $x$ is a variable, $c_{n}$ are the coefficients (constants), and $a$ is the “center of convergence”.  Fortunately for us, it doesn’t matter where the center of convergence is, for our purposes, since the series converges for all real numbers of $x$.

To save time while investigating the constant coefficients of the power series, we find that each coefficient can be determined as:

$c_{n}= \displaystyle{\frac{f^{(n)}(a)}{n!}}$

where $\displaystyle f^{(n)}(a)$ is the nth derivative. This gives us a bit more direction with a refined formula called a Taylor Series:

$f(x)={\displaystyle \sum_{n=0}^{\infty}\frac{f^{(n)}(a)}{n!}(x-a)^{n}=f(a)+\frac{f^{\prime}(a)}{1!}\cdot(x-a)^{1}+\frac{f^{\prime\prime}(a)}{2!}\cdot(x-a)^{2}+\frac{f^{\prime\prime\prime}(a)}{3!}\cdot(x-a)^{3}+\cdots}$

Since we’re trying to find the value of $e$ we set $f(x)=e^x$. When we take the derivative of $e^x$ it becomes $e^x$ every time. To make things easier, we’ll set $a=0$.   This is a Maclaurin Series of $e^x$.  A Maclaurin Series is a special type of Taylor Series where $a=0$.  This means:

$c_n = \displaystyle{\frac{f^{(n)}(a)}{n!}=\frac{e^0}{n!}=\frac{1}{n!}}$.

We now take this information and apply it to the power series:

$\displaystyle {\displaystyle e^{x}=\sum_{n=0}^{\infty}\frac{1}{n!}\cdot(x-0)^{n}=\sum_{n=0}^{\infty}\frac{x^{n}}{n!}}=\frac{x^{0}}{0!}+\frac{x^{1}}{1!}+\frac{x^{2}}{2!}+\cdots$

which simplifies to:

$\displaystyle {\displaystyle e^{x}=\sum_{n=0}^{\infty}\frac{x^{n}}{n!}=1+x+\frac{x^{2}}{2!}+\cdots}$.

Now that we have the Maclaurin Series for $e^x$, let’s find the value of $\displaystyle e$.

If we let $x=1$ then:

$\displaystyle e={\displaystyle e^{(1)}=\sum_{n=0}^{\infty}\frac{(1)^{n}}{n!}=}1+1+\frac{1}{2}+\frac{1}{6}+\frac{1}{24}+\cdots$.

Of course we can’t evaluate something that goes on forever, but we can find out what the approximate sums are with increasing accuracy.  The Partial Sum or Taylor Polynomial below:

$\displaystyle e\approx S_{k}={\displaystyle \sum_{n=0}^{k}\frac{(1)^{n}}{n!}=}1+1+\frac{1}{2}+\cdots+\frac{(1)^{k}}{k!}$

shows us how to calculate the value of $\displaystyle e$.  As a matter of fact, depending on a calculator’s limitations, this partial sum is how your calculator calculates the  value of $e$ to display onscreen.

Let’s watch!

$latex {\displaystyle \begin{array}{rllcc} S_{0}= & 1 & =1 & \text{when} & k=0\\ S_{1}= & 1+\frac{1}{1!} & =2 & \text{when} & k=1\\ S_{2}= & 1+\frac{1}{1!}+\frac{1}{2!} & =2.5 & \text{when} & k=2\\ S_{3}= & 1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!} & =2.66666666667 & \text{when} & k=3\\ S_{4}= & 1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!} & =2.70833333333 & \text{when} & k=4\\ S_{5}= & 1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\frac{1}{5!} & =2.71666666667 & \text{when} & k=5\\ S_{6}= & 1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\frac{1}{5!}+\frac{1}{6!} & =2.71805555556 & \text{when} & k=6\\ S_{7}= & 1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\frac{1}{5!}+\frac{1}{6!}+\frac{1}{7!} & =2.71825396825 & \text{when} & k=7\\ \vdots & \vdots & \vdots & \vdots & \vdots\\ e= & 1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\frac{1}{5!}+\frac{1}{6!}+\frac{1}{7!}+\cdots & =2.71828182846\ldots & \text{as} & k\rightarrow\infty \end{array}}$

It’s Mathemagic!