Euler's Formula

Prerequisites: Exponential Function, Trigonometric Functions, Imaginary Numbers. Full path (4 pages)

Best explanations

Concrete Intuitive Understanding Of Euler’s Formula

Intuitive
Concrete
Abstract
Why is it interesting?

Intuitive

Think of complex numbers as points in a plane, with real numbers along the horizontal axis and imaginary numbers along the vertical axis. Multiplying by the imaginary unit turns any point by a quarter turn around the origin.

Ordinary exponential growth means the rate of change is proportional to the current value and points in the same direction: the more you have, the faster you grow, and you grow outward. Now imagine a quantity whose rate of change is the imaginary unit times its current value. The change always points a quarter turn away from the current position, so it is always perpendicular to the line from the origin. Motion that is always perpendicular to the radius does not move the point closer or further away. It goes around in a circle at constant speed.

Euler's formula says exactly this. Starting at 1 and following this "sideways growth" for some time takes us the same angle around the unit circle. The horizontal position of that point is the cosine of that angle and the vertical position is the sine of that angle. After a time pi we have gone half way around and arrive at minus one.

This is why oscillations and waves in physics are so often written with complex exponentials. A rotating point in the plane, seen from the side, moves back and forth like a pendulum, and rotations are much easier to calculate with than sines and cosines.

Concrete

For a great explanation, see Intuitive Understanding Of Euler’s Formula by Kalid Azad
See also Easy Trig Identities With Euler’s Formula by Kalid Azad

The exponential function has the property that it is its own derivative.

$\frac{d(e^x)}{dx} = e^x \qquad (1)$

The is-its-own-derivative implies the following Taylor series:

$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \cdots \qquad (2)$

2-cycle

We also know two functions that are each other's derivative; a derivation 2-cycle.

$\sinh(x) = x +{\frac {x^{3}}{3!}}+{\frac {x^{5}}{5!}}+{\frac {x^{7}}{7!}}+\cdots \qquad (3)$

$\cosh(x) = 1+{\frac {x^{2}}{2!}}+{\frac {x^{4}}{4!}}+{\frac {x^{6}}{6!}}+\cdots \qquad (4)$

formula for the hyperbolic sine:

$\sinh(x) = \frac{e^x - e^{-x}}{2} \qquad (5)$

formula for the hyperbolic cosine:

$\cosh(x) = \frac{e^x + e^{-x}}{2} \qquad (6)$

We can count taking the derivative of the exponential function as a 1-cycle. For the $\sinh$ and $\cosh$ the split is achieved by using the cyclic property of successive multiplication of a factor of -1. With -1 you get a 2-cycle.

4-cycle

The next step is a sequence of two functions that are each others derivative, but after the second derivation the resulting function is the negative of the original, so the total cycle is a 4-cycle.

The Taylor series expansion of the trigonometric functions follows from the 4-cycle derivatives property:

$\sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots \qquad (7)$

$\cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots \qquad (8)$

At this point it is of course helpful to devise a compact notation for the above Taylor series. The $\sin$ and $\cos$ can be expressed in terms of the exponential function by using a factor that has the property of repeating in a 4-cycle, with sign alteration along the way. Interestingly, while we can connect this 4-cycle factor to other areas of mathematics, for the purpose of this demonstration making this connection is not a necessity.

We can pretend to invent a novel entity for this, we can call this factor the 'inventive number', abbreviated 'i'.

For this invented factor the following 4-cycle is defined:
$i^1 = i$
$i^2 = -1$
$i^3 = -i$
$i^4 = 1$

The formulas for the sine and cosine then are:

$\sin(x) = \frac{e^{ix} - e^{-ix}}{2i} \qquad (9)$

$\cos(x) = \frac{e^{ix} + e^{-ix}}{2} \qquad (10)$

Euler's formula follows from (9) and (10):

$e^{ix} = \cos(x) + i\sin(x) \qquad (11)$

The key points:

The connection between the exponential function and the trigonometric functions is the property of getting the same function back after taking the derivative. The difference is whether the original function comes back right away, or in a 2-cycle pattern, or in a 4-cycle pattern.

Abstract

The complex exponential $\exp: (\mathbb{C}, +) \to (\mathbb{C}^\times, \cdot)$ , defined by the everywhere convergent power series, is a surjective group homomorphism, $e^{z + w} = e^z e^w$ , with kernel $2\pi i\,\mathbb{Z}$ . Hence

$\mathbb{C}/2\pi i\,\mathbb{Z} \cong \mathbb{C}^\times.$

Restricted to the imaginary axis it gives the homomorphism $\mathbb{R} \to U(1) = \{z \in \mathbb{C} : |z| = 1\}$ , $\theta \mapsto e^{i\theta}$ , with kernel $2\pi\mathbb{Z}$ , so $U(1) \cong \mathbb{R}/2\pi\mathbb{Z}$ . Euler's formula identifies this map with $\theta \mapsto \cos\theta + i \sin\theta$ , the point at angle $\theta$ on the unit circle. In Lie-theoretic language, $i\mathbb{R} = \mathfrak{u}(1)$ is the Lie algebra of $U(1)$ and $i\theta \mapsto e^{i\theta}$ is its exponential map.

The proof of Euler's formula only uses $i^2 = -1$ . For any element $J$ of an associative real algebra with $J^2 = -1$ , splitting the exponential series into even and odd powers gives

$e^{\theta J} = \cos\theta\,\mathbb{1} + \sin\theta\, J.$

Examples are the matrix $J = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$ , for which $e^{\theta J}$ is the rotation matrix of $SO(2)$ , and any unit pure imaginary quaternion $u$ . For the Pauli matrices and a unit vector $\hat n$ one has $(\hat n \cdot \vec\sigma)^2 = \mathbb{1}$ , so $(-i\,\hat n \cdot \vec \sigma)^2 = -\mathbb{1}$ and

$\exp\!\left(-\frac{i\theta}{2}\,\hat n \cdot \vec\sigma\right) = \cos\frac{\theta}{2}\,\mathbb{1} - i \sin\frac{\theta}{2}\,\hat n \cdot \vec \sigma,$

which is the $SU(2)$ element representing a rotation by $\theta$ about $\hat n$ .

The functions $\theta \mapsto e^{in\theta}$ , $n \in \mathbb{Z}$ , are exactly the continuous homomorphisms $U(1) \to U(1)$ , the characters of the circle group. By the Peter-Weyl theorem they form an orthonormal basis of $L^2(U(1))$ with respect to the normalized measure $d\theta/2\pi$ , which is the statement that Fourier series converge in $L^2$ . The Fourier transform is the analogous expansion in the characters $x \mapsto e^{ikx}$ of the group $\mathbb{R}$ .

Why is it interesting?

Euler's formula is a deep result of complex analysis that shows a surprising relationship between seemingly unrelated notions:

$e^{ix} = \cos(x) + i \sin(x) \,$

where $e^{ix}$ denotes the exponential function and $\cos(x)$ , $\sin(x)$ are the usual trigonometric functions. If we evaluate this equation at $x= \pi$ , we get

$e^{i\pi } = \cos(\pi) + i \sin(\pi) = -1 -i 0 = -1 \, .$

This shows a deep relationship between the exponential function, the imaginary unit $i$ and $\pi$ . (Pi is the ratio between circumference and diameter shared by all circles.)

Like a Shakespearean sonnet that captures the very essence of love, or a painting that brings out the beauty of the human form that is far more than just skin deep, Euler's equation reaches down into the very depths of existence. Keith Devlin

[Euler's equation] is absolutely paradoxical; we cannot understand it, and we don't know what it means, but we have proved it, and therefore we know it must be the truth". Benjamin Peirce

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