search
Search
Login
Unlock 100+ guides
menu
menu
web
search toc
close
Comments
Log in or sign up
Cancel
Post
account_circle
Profile
exit_to_app
Sign out
search
keyboard_voice
close
Searching Tips
Search for a recipe:
"Creating a table in MySQL"
Search for an API documentation: "@append"
Search for code: "!dataframe"
Apply a tag filter: "#python"
Useful Shortcuts
/ to open search panel
Esc to close search panel
to navigate between search results
d to clear all current filters
Enter to expand content preview
icon_star
Doc Search
icon_star
Code Search Beta
SORRY NOTHING FOUND!
mic
Start speaking...
Voice search is only supported in Safari and Chrome.
Navigate to

Intuitive Guide on Euler's Number

schedule Aug 12, 2023
Last updated
local_offer
Tags
mode_heat
Master the mathematics behind data science with 100+ top-tier guides
Start your free 7-days trial now!

This guide requires that you are familiar with the concept of compound interests. If not, please consult our guide on compound interests.

Motivating example

Recall that the formula to compute compound interest:

$$\begin{align*} A=P\Big(1+\frac{r}{n}\Big)^{nt} \end{align*}$$

Where:

  • $A$ is the final amount in the bank account after $t$ years.

  • $P$ is the principal, that is, the starting amount in the bank account.

  • $r$ is the interest rate (e.g. $0.1$).

  • $n$ is the number of compounds per year.

  • $t$ is the number of years elapsed.

Now, consider a simple case with the following assumptions:

  • we initially start with one dollar in the bank account, that is, $P=1$.

  • the interest rate is $100\%$, that is, the bank will double our money at every compound period. This would mean $r=1$.

  • we are interested in our bank balance after one year, that is, $t=1$.

For this simple case, our formula for compound interest becomes:

$$\begin{align*} A=\Big(1+\frac{1}{n}\Big)^{n} \end{align*}$$

The only variable here is $n$, which is the number of compounds per year. For instance, $n=1$ for annual compounds, whereas $n=4$ for quarterly compounds. Let's now compute $A$ for different values of $n$ like so:

$n$

$A$

$n$

$A$

1 (annually)

2.0000

52 (weekly)

2.6926

2 (bi-annually)

2.2500

365 (daily)

2.7146

4 (quarterly)

2.4414

8760 (hourly)

2.7181

12 (monthly)

2.6130

525600 (per min)

2.7183

Note that the values of $A$ are all rounded to four decimal places. We can see that as $n$ increases, the value of $A$ also increases, but the amount of increase slows down. For instance, increasing $n$ from $1$ to $2$ gives us an additional $\$0.25$, but increasing $n$ from $365$ to $8760$ only gives us around $\$0.003$ additionally.

As a result, $A$ seems to be converging to some constant of around $2.718$; this constant is the famous Euler's constant $e$. The larger the $n$, the closer we approach $e$, hence it makes sense to define $e$ as the limit of $A$ as $n$ tends to infinity.

Example.

Formal definition of Euler's constant

Euler's constant $e$ is defined as follows:

$$e=\lim_{n\to\infty} \Big[\Big(1+\frac{1}{n}\Big)^n\Big]$$

Note that $e$ is irrational:

$$e=2.71828...$$

Let's now expand on the formal definition of Euler's constant and express $e^\lambda$ as a limit where $\lambda$ is some scalar number.

Theorem.

Expressing the power of Euler's constant as limits

If $\lambda\in\mathbb{R}$, then:

$$e^\lambda= \lim_{n\to\infty}\Big(1+\frac{\lambda}{n}\Big)^n$$

Proof. From the basic rules of taking powers, we have that:

$$\Big(1+\frac{\lambda}{n}\Big)^{n} =\Big[\Big(1+\frac{\lambda}{n}\Big)^{n/\lambda}\Big]^\lambda $$

Now, let's take the limit as $n$ tends to infinity:

$$\begin{equation}\label{eq:yTGF12ryu589R5LxGty} \lim_{n\to\infty}\Big(1+\frac{\lambda}{n}\Big)^{n} = \lim_{n\to\infty}\Big[\Big(\Big(1+\frac{\lambda}{n}\Big) ^{n/\lambda}\Big)^\lambda\Big] \end{equation}$$

We now want to show that the right-hand side is equal to $e^\lambda$. Let's define a new variable:

$$k=\frac{n}{\lambda}$$

Notice how $k$ also tends to infinity as $n$ tends to infinity. Rewriting the right-hand side of \eqref{eq:yTGF12ryu589R5LxGty} in terms of $k$ gives:

$$\lim_{k\to\infty}\Big[\Big(\Big(1+\frac{1}{k}\Big) ^{k}\Big)^\lambda\Big]$$

Using the power rule of limits, we can take the power of $\lambda$ outside the limit:

$$\begin{equation}\label{eq:PxSqujAPFRRw0WW1Wlc} \begin{aligned}[b] \lim_{k\to\infty} \Big[\Big(\Big(1+\frac{1}{k}\Big)^{k}\Big)^\lambda\Big] &=\Big[ \lim_{k\to\infty} \Big(\Big(1+\frac{1}{k}\Big)^{k}\Big)\Big]^\lambda\\ &=e^\lambda\\ \end{aligned} \end{equation}$$

Substituting \eqref{eq:PxSqujAPFRRw0WW1Wlc} into \eqref{eq:yTGF12ryu589R5LxGty} gives us the desired identity:

$$e^\lambda= \lim_{n\to\infty}\Big(1+\frac{\lambda}{n}\Big)^n$$

This completes the proof.

robocat
Published by Isshin Inada
Edited by 0 others
Did you find this page useful?
thumb_up
thumb_down
Comment
Citation
Ask a question or leave a feedback...
thumb_up
0
thumb_down
0
chat_bubble_outline
0
settings
Enjoy our search
Hit / to insta-search docs and recipes!