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Last updated on October 15, 2025
Euler’s number, denoted by 𝑒, is a fundamental mathematical constant introduced by Jacob Bernoulli in 1683. Later, Leonhard Euler studied it further, which is why it is named after him.
Euler's number, written as e, is an irrational and transcendental constant, with its value being approximately 2.71828. This value is approximate because the decimal expansion goes infinitely without repeating.
We see this value appear naturally in many areas of math and science while describing continuous growth or decay. The Euler's number helps describe any process that requires continuous and smooth change.
Euler’s number, e is defined by the following equation:
Euler’s formula for compound interest,
\(A = P^{rt}\)
Where,
FV stands for future value
PV represents present value of balance or sum
e is the mathematical constant
r is Interest rate being compounded, and
t is time in years
Euler’s form connects trigonometry and exponential functions helping in complex analysis. it provides an efficient framework unifying exponential and trigonometric expressions helping simplify mathematical computations.
For any value of x, the formula is given by:
\(e^ix = cos x + isin x\)
Here,
cos and sin represent the trigonometric ratios functions.
i is the imaginary unit, and
e is the base of the natural logarithm.
Geometrically, this formula can be visualized on a complex plane where \(e^{i𝜃} \) traces a unit circle as the angle θ is measured in radians.
Let us look at the approximate proof of this formula for better understanding.
Let’s start with the taylor series expansion of the exponential function \(e^x\):
\(e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \cdots \)
This expansion holds true for real as well as complex numbers.
Substituting x = iθ, where i is the imaginary unit (i2 = -1)
\(e^{i\theta} = 1 + i\theta + \frac{(i\theta)^2}{2!} + \frac{(i\theta)^3}{3!} + \frac{(i\theta)^4}{4!} + \cdots \)
Now, we group real and imaginary terms separately:
\(e^{i\theta} = \left(1 - \frac{\theta^2}{2!} + \frac{\theta^4}{4!} - \cdots \right) + i\left(\theta - \frac{\theta^3}{3!} + \frac{\theta^5}{5!} - \cdots \right) \)
The trigonometric series is
\(\cos \theta = 1 - \frac{\theta^2}{2!} + \frac{\theta^4}{4!} - \cdots \)
\(\sin \theta = \theta - \frac{\theta^3}{3!} + \frac{\theta^5}{5!} - \cdots \)
Now, substituting these expansions back to our expression for \(e^iθ\), we get
\(e^iθ = cosθ + isinθ\)
Hence, proved \(e^iθ = cosθ + isinθ\)
Euler's identity: From the above formula, we get \(e^ix = cosx + isinx\)
when x = π, this formula give the identity
\(e^{i π} = cos π + isin π\)
\(e^{i π} = -1 + i (0) \), because cos π = -1 and sin π = 0
\(e^{i π} = -1\) or \(e^{iπ} + 1 = 0\)
This is Euler's identity.
A polyhedron is a three-dimensional solid shape. It consists of flat faces and straight edges. Cubes, cuboids, prisms and pyramids are some examples of polyhedra. If a polyhedron does not intersect itself, its vertices, faces and edges follow a specific relationship.
According to Euler's formula, the sum of the number of vertices and faces is exactly two more than the number of edges. Mathematically, it can be expressed as:
\(F + V - E = 2\)
Here, F, V and E represent the number of faces, vertices and edges.
The value of e is approximately 2.718. Euler's number is mostly used to calculate the rate of change or growth, such as in finance, radioactive decay, and so on. Here are some examples
Example 1: Calculate the final amount when $100 is invested for 5 years at a 4% interest rate compounded continuously.
Solution: Euler's formula for compounding interest is
A = Pert
Given, \(P = 100\)
\( r = 0.04\)
\(t = 5\)
\(A = 100e^{0.04 \times 5} \)
\(= 100 × 1.2214\)
\( = 122.14\)
Therefore, the money in the account after 5 years is $122.14.
Example 2: Find the value of e when n = 3
Solution: Given n = 3,
\(\lim\limits_{x \to \infty} ( 1 + \frac1n)^n \)
\(e =\) \((1 + \frac13)^3 \) \(= 2.37037\)
This is an approximation; Euler's number e ≈ 2.71828.
Given below are a few tips and tricks, that help students better understand and apply the Euler number.
1. Biology: It is used to calculate the exponential growth and decay of organisms
2. Physics: Radioactive decay follows an exponential pattern modeled using Euler’s number.
3. Finance: Compound interest calculations in finance reveal growth and decline patterns, which support better risk management
4. Computer Science: It helps study complex algorithms in fields such as machine learning, computer graphics, optimization, and many more.
5. Weather: Euler’s number is used in studying weather changes, such as temperature changes over time, which involves exponential functions.
Calculate the final amount when $1000 is invested for 4 years at a 6% interest rate compounded continuously.
A = 1271.24
Using the formula A = Pert
A = Total money with interest
P = 1000
r = 0.06
t = 4
A = 1000 e0.06 × 4
A = 1000 e0.24
A = 1000 × 1.271249
A = 1271.24
Find the value of e when n = 5
2.48832
Given n = 5,
\(\lim\limits_{x \to \infty} ( 1 + \frac1n)^n\)
\((1 + \frac15)^5\) = 2.48832
Evaluate lim(n→∞) (1+3/n)^n
20.0855
We know that,
\(\lim\limits_{x \to \infty} ( 1 + \frac1n)^n\) = e,
Given \((1 + \frac3n)^n\)which is equivalent to,
\((1 + \frac3n)^n\) = \(1 + \frac1 {n/3}^{n/3 \times 3}\)= \(1 + \frac1 {n/3}^{n/3 }\)
The limit approaches,
\(\lim\limits_{x \to \infty} ( 1 + \frac3n)^n\) = e3
We know e = 2.71828, then, e3 = 2.718283 = 20.0855
Therefore, (1+3n)n = 20.08553
Calculate the final amount when $800 is invested for 9 years at a 6% interest rate compounded continuously.
1372.80
Using the formula A = Pert
A = Total money with interest
P = 800
r = 0.06
t = 9
A = 800 × e0.06 × 9
A = 800 × e0.54
A = 800 × 1.7160068
A = 1372.80
Sonia invests $2,000 in a savings account that earns 5% annual interest, compounded continuously. How much money will she have in her account after 6 years?
$2,699.72
Sonia’s investment grows with continuous compounding, which uses the formula: \(A = P e^{rt} \)
Where:
A = amount after t years
P = principal = 2000
r = annual interest rate in decimal = 0.05
t = time in years = 6
\(e ≈ 2.71828\)
Substituting the values, we get
\(A = 2000 \cdot e^{0.05 \cdot 6} = 2000 \cdot e^{0.3} \approx 2000 \cdot 1.34986 \approx 2699.72 \)
So, after 6 years, Sonia will have $2,699.72 in her account. Continuous compounding grows the money slightly faster than regular compounding because interest is added constantly.
Hiralee Lalitkumar Makwana has almost two years of teaching experience. She is a number ninja as she loves numbers. Her interest in numbers can be seen in the way she cracks math puzzles and hidden patterns.
: She loves to read number jokes and games.