Euler's Formula

Derivation of Euler Formula

The derivation of Euler's formula involves concepts from calculus, power series, and complex numbers. Consider the power series expansions of the exponential function \( e^x \), the cosine function \( \cos(x) \), and the sine function \( \sin(x) \). The power series expansions for these functions are: \[ e^x = 1 + x + \dfrac{x^2}{2!} + \dfrac{x^3}{3!} + \dfrac{x^4}{4!} + \dfrac{x^5}{5!} + \cdots \] \[ \cos(x) = 1 - \dfrac{x^2}{2!} + \dfrac{x^4}{4!} - \dfrac{x^6}{6!} + \cdots \] \[ \sin(x) = x - \dfrac{x^3}{3!} + \dfrac{x^5}{5!} - \dfrac{x^7}{7!} + \cdots \] Now, consider what happens when we substitute \( ix \) into the series expansions of \( e^x \): \[ e^{ix} = 1 + ix + \dfrac{(ix)^2}{2!} + \dfrac{(ix)^3}{3!} + \dfrac{(ix)^4}{4!} + \dfrac{(ix)^5}{5!} + \cdots \] \[ = 1 + ix - \dfrac{x^2}{2!} - i\dfrac{x^3}{3!} + \dfrac{x^4}{4!} + i\dfrac{x^5}{5!} - \cdots \] Write \( e^{ix} \) as a complex number of the standard form \( Re + i Im \): \[ e^{ix} = \left(1 - \dfrac{x^2}{2!} + \dfrac{x^4}{4!} \cdots \right) + i \left( x - \dfrac{x^3}{3!} + \dfrac{x^5}{5!} \cdots \right) \] Notice that the even-powered \( \left(1 - \dfrac{x^2}{2!} + \dfrac{x^4}{4!} \cdots \right) \) terms form the series expansion of \( \cos(x) \), and the odd-powered \( \left( x - \dfrac{x^3}{3!} + \dfrac{x^5}{5!} \cdots \right) \) terms form the series expansion of \( \sin(x) \). Thus, combining them, we can write: \[ \Large \color{red} {e^{ix} = \cos(x) + i\sin(x)} \]

Identities and Formulas Related to Euler Formula

Euler's Identity

When \( x = \pi \), Euler's formula becomes: \[ e^{i\pi} = \cos(\pi) + i\sin(\pi) \] Which may be written as \[ \Large \color{red} { e^{i\pi} + 1 = 0} \] Notice that the above equation combines the five most important constants in mathematics: \( e \), \( \pi \), \( i \), \( 1 \), and \( 0 \).

De Moivre's Formula

The extension Euler's formula for any integer power \( n \) : \[ (e^{i x})^n = (\cos(x) + i\sin(x))^n \] Using exponential rule \[ (e^{i x})^n = e^{i n x} = \cos (n x) + i \sin (n x) \] Hence Demivre's formula \[ \Large \color{red} {(\cos(x) + i\sin(x))^n = \cos (n x) + i \sin (n x)} \] This formula is incredibly useful for calculating powers of complex numbers.

Euler's Formula for \( \sin \) and \( \cos \)

Start with the expansions of \[ e^{ix} = \cos(x) + i\sin(x) \] and \[ e^{-ix} = \cos(-x) + i\sin(- x) = \cos(x) - i\sin(x) \] Adding and subtracting the left and right sides, we obtain \[ e^{ix} + e^{-ix} = 2 \cos(x) \] and \[ e^{ix} - e^{-ix} = 2 i\sin(x) \] Solve for \( \cos(x) \) and \( \sin(x) \) to obtain \[ \Large \color{red} { \sin(x) = \dfrac{e^{ix} - e^{-ix}}{2i} } \] \[ \Large \color{red} { \cos(x) = \dfrac{e^{ix} + e^{-ix}}{2} } \] These identities relate the trigonometric functions \( \sin(x) \) and \( \cos(x) \) to the exponential function \( e^{ix} \). These identities have applications in various fields of mathematics, physics, and engineering. They provide deep insights into the connections between exponential growth, periodic motion, and complex numbers.

Trigonometric Identities and Euler's Formula

We present an example on how to use Euler's formula to proove trigonometric identities.

Example

Proove the trigonometric identity \( \sin(A + B) = \sin A \cos B + \cos A \sin B \)

Solution

Use the identity \( \sin(x) = \dfrac{e^{ix} - e^{-ix}}{2i} \) to write \[ \sin(A+B) = \dfrac{e^{i(A+B)} - e^{-i(A+B)}}{2i} \quad (I)\] Use the exponential property \( e^{x+y} = e^x e^y \) to rewrite \( e^{i(A+B)} - e^{-i(A+B)} \) as \[ e^{i(A+B)} - e^{-i(A+B)} = e^{iA} e^{iB} - e^{-iA} e^{-iB} \] We now use Euler's formula to expand the terms on the right side \[ e^{i(A+B)} - e^{-i(A+B)} = (\cos A + i \sin A)(\cos B + i \sin B) - (\cos (-A) + i \sin(- A))(\cos(- B) + i \sin (-B) )\] Use the identities \( \cos (-A) = \cos A \) and \( \sin(-A) = - \sin A \) to rerite the above as \[ e^{i(A+B)} - e^{-i(A+B)} = (\cos A + i \sin A)(\cos B + i \sin B) - (\cos A - i \sin A)(\cos B - i \sin B) \] Expand the right side and simplify \[ e^{i(A+B)} - e^{-i(A+B)} = 2 i \sin A \cos B + 2 i \cos A \sin B \] Substitute in (I) above to obtain \[ \sin(A+B) = \dfrac{2 i \sin A \cos B + 2 i \cos A \sin B}{2i} \] Simplify to obtain the very well known trigonometric formula: \[ \Large \color{red} { \sin(A+B) = \sin A \cos B + \cos A \sin B } \]