How to Calculate the Value of e

ever 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!

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