Euler Number

I. Definition

It can be shown that sum of the infinite series:

$1 + \dfrac{1}{1} + \dfrac{1}{1\cdot 2} + \dfrac{1}{1\cdot 2 \cdot 3}... \infty$

$= \dfrac{1}{0!} + \dfrac{1}{1!} + \dfrac{1}{2!} + \dfrac{1}{3!}... \infty$

$= \sum\limits_{k \rightarrow 0}^{\infty} \dfrac{1}{k!}$

is bounded by an upper limit which is defined as the Euler number, denoted as $e$ which is an irrational number $\approx 2.718...$.

The following widget shows how the value of the above expression changes between $k = (0, 20)$.

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II. Limit Of $\left(1 + \dfrac{1}{n}\right)^n$:

Using the above definition of Euler number we will see that the value:

$\left(1 + \dfrac{1}{n}\right)^n$ also approaches $e$ as $n$ approaches $\infty$, that is:

$\lim\limits_{n \rightarrow \infty} \left(1 + \dfrac{1}{n}\right)^n = e$

Using binomial expansion, the $k\xasuper{th}$ term of $\left(1 + \dfrac{1}{n}\right)^n$ can be written as:

$\xacomb{n}{k} \dfrac{1}{n^{k}}$

$= \dfrac{n!}{k!(n - k)!} \times \dfrac{1}{n^{k}}$

$= \dfrac{n \cdot (n - 1) \cdot (n-2)...(n - k + 1)}{k!} \times \dfrac{1}{n^{k}}$

$= \dfrac{n}{n} \times \dfrac{n-1}{n} \cdot \dfrac{n-2}{n}...\dfrac{n - k + 1}{n} \times \dfrac{1}{k!}$

$= \left(\dfrac{n}{n} \right) \times \left(1 - \dfrac{1}{n} \right) \times \left(1 - \dfrac{2}{n} \right) ... \times \left(1 - \dfrac{k-1}{n} \right) \times \dfrac{1}{k!}$

$= \left(1 \right) \times \left(1 - \dfrac{1}{n} \right) \times \left(1 - \dfrac{2}{n} \right) ... \times \left(1 - \dfrac{k-1}{n} \right) \times \dfrac{1}{k!}$

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We can substitute $m = \dfrac{1}{n}$ and get the $k\xasuper{th}$ term as:

$\left(1 \right) \times \left(1 - 1 \cdot m \right) \times \left(1 - 2 \cdot m \right) ... \times \left(1 - (k - 1) \cdot m \right) \times \dfrac{1}{k!}$

As $n \rightarrow \infty$, $m \rightarrow 0$, therefore, we get the limit of the $k\xasuper{th}$ term as:

$\lim\limits_{n \rightarrow \infty} \xacomb{n}{k} \left(\dfrac{1}{n^k}\right)$

$= \lim\limits_{n \rightarrow \infty} \dfrac{n!}{k!(n - k)!} \times \dfrac{1}{n^{k}}$

$= \lim\limits_{m \rightarrow 0} \left(1 \right) \times \left(1 - 1 \cdot m \right) \times \left(1 - 2 \cdot m \right) ... \times \left(1 - (k - 1) \cdot m \right) \times \dfrac{1}{k!}$

$= 1 \times 1 \times 1... \times 1 \times \dfrac{1}{k!}$

$= \dfrac{1}{k!}$

Now, we can rewrite the complete binomial expansion as:

$\lim\limits_{n \rightarrow \infty} \left(1 + \dfrac{1}{n}\right)^n$

$= \lim\limits_{n \rightarrow \infty} \displaystyle{\sum\limits_{k = 0}^{n}}\ \xacomb{n}{k} \left(\dfrac{1}{n^k}\right)$

$= \displaystyle{\sum\limits_{k = 0}^{\infty}}\ \lim\limits_{n \rightarrow \infty} \xacomb{n}{k} \left(\dfrac{1}{n^k}\right)$

$= \displaystyle{\sum\limits_{k = 0}^{\infty}} \dfrac{1}{k!}$

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The sum of this infinite series, by definition, is the Euler number $e$.

Therefore,

$\lim\limits_{n \rightarrow \infty} \left(1 + \dfrac{1}{n}\right)^n = \displaystyle{\sum\limits_{k = 0}^{\infty}} \dfrac{1}{k!} = e$

A more general form of this relationship is:

$e^x = \lim \limits_{n \rightarrow \infty} \left(1 + \dfrac{x}{n} \right)^n = \dfrac{x^0}{0!} + \dfrac{x^1}{1!} + \dfrac{x^2}{2!} + .... \infty$

$\Rightarrow e^x = \sum \limits_{k = 0}^{\infty} \dfrac{x^k}{k!}$

The proof of this requires knowledge of Taylor series expansion, and will be covered in a future topic.

III. Application Of Euler Number In Compound Interest:

The Euler number finds a wide range of applications in mathematical and scientific analysis.

One of the common application, that we will see here is the application of this for calculating compound interest.

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Let,

$P$ be the principal amount,

$r$ be the annual interest rate in fraction,

$n$ be the number of periods per year for which interest is calculated,

$T$ be the total time period in years,

$A$ be the total amount after $T$ years

The amount after the first year is $P\left(1 + \dfrac{r}{n}\right)^n$

The final amount $A$ after $T$ years is $P\left(1 + \dfrac{r}{n}\right)^{nT}$

As $n$ grows larger, the interest calculation is done more frequently, with a smaller per period interest.

What happens if interest is calculated every instant, that is continuously?

This can be found by taking the value of $n$ as close to infinity as possible, that is:

$A = \lim\limits_{n \rightarrow \infty} P\left(1 + \dfrac{r}{n}\right)^{nT}$

$= P \lim\limits_{n \rightarrow \infty} \left(1 + \dfrac{r}{n}\right)^{\frac{n}{r} \times rT}$

$= P \left\{\lim\limits_{n \rightarrow \infty} \left(1 + \dfrac{r}{n}\right)^{\frac{n}{r}}\right\} ^ {rT}$

As $n \rightarrow \infty$, $\dfrac{n}{r} \rightarrow \infty$, therefore,

$A = P \left\{\lim\limits_{\frac{n}{r} \rightarrow \infty} \left(1 + \dfrac{r}{n}\right)^{\frac{n}{r}}\right\} ^ {rT} = Pe^{rT}$

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IV. More General Applications:

From this example, the major takeaway should be the situation for which this limit is applicable. The application of this approach is dependent on two basic conditions:

- The growth (or decay) of any entity which is growing (or decaying) continuously over a period of time

- The growth (or decay) rate at any instant is linearly proportional to the current number (or quantity) of that entity.

Given these two conditions, this limit can be used for analysis of many things, like population growth of a biological organism, radioactive decay, fluid flow, chemical reactions, etc.

Let's say $Q_0$ be the initial quantity of the entity $E$, whose growth we are estimating, and $E$ grows at a rate which at any instant is proportional to the quantity of $E$ at that instant, during the time period that we are estimating for.

The growth rate $\Delta Q$ is linearly proportional to the the quantity $Q$ at that time, that is:

$\dfrac{\Delta Q}{\Delta t} \propto Q \Rightarrow \dfrac{\Delta Q}{\Delta t} = kQ$ ($-k$ for negative growth or decay)

and $T$ is the total time period for which we are estimating the growth, then applying similar concept,

$Q_T = Q_0e^{\pm kT}$ ($\pm$ for growth or decay resp.)

We will learn about these applications in much more details when we cover integration.