description: “Full derivation of Euler’s Formula” slug: euler-formula public: true

Euler’s Formula

Euler’s Formula

eix=cos⁡(x)+isin⁡(x)e^{ix} = \cos(x) + i\sin(x)

where:

  • ee is Euler’s number (~2.718), the base of natural logarithms,
  • ii is the imaginary unit (i2=−1i^2 = -1),
  • xx is a real number (often thought of as an angle in radians).

Properties

  1. Bridge between algebra and geometry
  • The exponential function eixe^{ix} traces out points on the unit circle in the complex plane as xx changes.
  • This makes a direct link between exponential growth/decay and circular motion.
  1. Trigonometry becomes exponential
  • With this formula, sine and cosine can be expressed in terms of exponentials:
cos⁡(x)=eix+e−ix2,sin⁡(x)=eix−e−ix2i\cos(x) = \frac{e^{ix} + e^{-ix}}{2}, \quad \sin(x) = \frac{e^{ix} - e^{-ix}}{2i}
  • These are known as Euler’s identities.

Euler’s Identity

By plugging in x=πx = \pi into Euler’s formula:

eiπ+1=0e^{i\pi} + 1 = 0

This equation connects five of the most fundamental constants in mathematics: e,i,π,1,0e, i, \pi, 1, 0. Many mathematicians consider it the most beautiful equation ever written.

Proof

  • Euler’s Formula can be proved with exe^x, sinxsinx, cosxcosx, and Maclaurin’s Series.

1. f(x)=exf(x) = e^x

  • First, check the recurrence relation of exe^x’s derivatives when x=0x = 0
f(1)(x)=f(2)(x)=f(3)(x)=…=f(n)(x)=exf(1)(0)=f(2)(0)=f(3)(0)=…=f(n)(0)=1\begin{align*} &f^{(1)}(x) = f^{(2)}(x) = f^{(3)}(x) = \ldots = f^{(n)}(x) = e^x \\ &f^{(1)}(0) = f^{(2)}(0) = f^{(3)}(0) = \ldots = f^{(n)}(0) = 1 \\ \end{align*}
  • Then, expand exe^x with Maclaurin’ Series
∴ex=f(0)+f(1)(0)1!x+f(2)(0)2!x2+…+f(n)(0)n!xn+…=1+11!x+12!x2+…+1n!xn+…=1+x1!+x22!+…+xnn!+…\begin{align*} \therefore e^x &= f(0) + \frac{f^{(1)}(0)}{1!}x + \frac{f^{(2)}(0)}{2!}x^2 + \ldots + \frac{f^{(n)}(0)}{n!}x^n + \ldots \\ &= 1 + \frac{1}{1!}x + \frac{1}{2!}x^2 + \ldots + \frac{1}{n!}x^n + \ldots \\ &= 1 + \frac{x}{1!} + \frac{x^2}{2!} + \ldots + \frac{x^n}{n!} + \ldots \\ \end{align*}

2. f(x)=sin(x)f(x) = sin(x)

  • Now, check the recurrence relation of sin(x)sin(x)’s derivatives when x=0x = 0
f(1)(x)=cos(x)f(2)(x)=−sin(x)f(3)(x)=−cos(x)f(4)(x)=sin(x)\begin{align*} &f^{(1)}(x) = cos(x) \quad f^{(2)}(x) = -sin(x) \\ &f^{(3)}(x) = -cos(x) \quad f^{(4)}(x) = sin(x) \\ \end{align*} f(1)(0)=1f(2)(0)=0f(3)(0)=−1f(4)(0)=0\begin{align*} &f^{(1)}(0) = 1 \quad f^{(2)}(0) = 0 \\ &f^{(3)}(0) = -1 \quad f^{(4)}(0) = 0 \\ \end{align*}
  • Then, expand sin(x)sin(x) with Maclaurin’s Series as well
∴sin(x)=f(0)+f(1)(0)1!x+f(2)(0)2!x2+…+f(n)(0)n!xn+…=0+11!x−13!x3+…+(−1)(n−1)1(2n−1)!x(2n−1)+…=x1!+−x33!+…+(−1)(n−1)x(2n−1)(2n−1)!+…\begin{align*} \therefore sin(x) &= f(0) + \frac{f^{(1)}(0)}{1!}x + \frac{f^{(2)}(0)}{2!}x^2 + \ldots + \frac{f^{(n)}(0)}{n!}x^n + \ldots \\ &= 0 + \frac{1}{1!}x - \frac{1}{3!}x^3 + \ldots + (-1)^{(n -1)}\frac{1}{(2n - 1)!}x^{(2n - 1)} + \ldots \\ &= \frac{x}{1!} + \frac{-x^3}{3!} + \ldots + \frac{(-1)^{(n - 1)}x^{(2n-1)}}{(2n-1)!} + \ldots \\ \end{align*}

3. f(x)=cos(x)f(x) = cos(x)

  • Movign on, check the recurrence relation of cos(x)cos(x)’s derivatives when x=0x = 0
f(1)(x)=−sin(x)f(2)(x)=−cos(x)f(3)(x)=sin(x)f(4)(x)=cos(x)\begin{align*} &f^{(1)}(x) = -sin(x) \quad f^{(2)}(x) = -cos(x) \\ &f^{(3)}(x) = sin(x) \quad f^{(4)}(x) = cos(x) \\ \end{align*} f(1)(0)=0f(2)(0)=−1f(3)(0)=0f(4)(0)=1\begin{align*} &f^{(1)}(0) = 0 \quad f^{(2)}(0) = -1 \\ &f^{(3)}(0) = 0 \quad f^{(4)}(0) = 1 \\ \end{align*}
  • Then, expand cos(x)cos(x) with Maclaurin’s Series as well
∴cos(x)=f(0)+f(1)(0)1!x+f(2)(0)2!x2+…+f(n)(0)n!xn+…=1−12!x2+14!x4+…+(−1)n1(2n)!x2n+…=x1!+−x33!+…+(−1)nx2n(2n)!+…\begin{align*} \therefore cos(x) &= f(0) + \frac{f^{(1)}(0)}{1!}x + \frac{f^{(2)}(0)}{2!}x^2 + \ldots + \frac{f^{(n)}(0)}{n!}x^n + \ldots \\ &= 1 - \frac{1}{2!}x^2 + \frac{1}{4!}x^4 + \ldots + (-1)^n\frac{1}{(2n)!}x^{2n} + \ldots \\ &= \frac{x}{1!} + \frac{-x^3}{3!} + \ldots + \frac{(-1)^{n}x^{2n}}{(2n)!} + \ldots \\ \end{align*}

4. Bring everything together

∵ex=1+x1!+x22!+…+xnn!+…∴eiθ=1+iθ1!+(iθ)22!+…+(iθ)nn!+…eiθ=1+iθ1!−θ22!−iθ33!+θ44!+iθ55!−θ66!−iθ77!+…=cosθ+isinθ\begin{align*} \because e^x &= 1 + \frac{x}{1!} + \frac{x^2}{2!} + \ldots + \frac{x^n}{n!} + \ldots \\ \therefore e^{i\theta} &= 1 + \frac{i\theta}{1!} + \frac{(i\theta)^2}{2!} + \ldots + \frac{(i\theta)^n}{n!} + \ldots \\ e^{i\theta} &= 1 + \frac{i\theta}{1!} - \frac{\theta^2}{2!} - \frac{i\theta^3}{3!} + \frac{\theta^4}{4!} + \frac{i\theta^5}{5!} - \frac{\theta^6}{6!} - \frac{i\theta^7}{7!} + \ldots \\ &= cos\theta + isin\theta \end{align*}
  • Here we are, the Euler’s Formula
eiθ=cosθ+isinθe^{i\theta}= cos\theta + isin\theta