# Euler's Formula

 

The proof Euler's formula, using Taylor series , is presented.

Referring to the tables of mathematical formulas , the Taylor series of $e^z$ is given by
$e^z = 1 + z + \dfrac{z^2} { 2! } + \dfrac{z^3} { 3! } + \dfrac{z^4} { 4! } + \dfrac{z^5} { 5! } + \dfrac{z^6} { 6! } + \dfrac{z^7} { 7! } ... + \dfrac{z^n}{n!} + ...$

Let $z = i x$, where $x$ is a real number and $i = \sqrt {-1}$ is the imaginary unit.
$e^{(ix)} = 1 + {(ix)} + \dfrac{{(ix)}^2} { 2! } + \dfrac{{(ix)}^3} { 3! } + \dfrac{{(ix)}^4} { 4! } + \dfrac{{(ix)}^5} { 5! } + \dfrac{{(ix)}^6} { 6! } + \dfrac{{(ix)}^7} { 7! } ...$

Take the $i$ from inside the brackets, $e^{(ix)}$ may be written as
$e^{(ix)} = 1 + i x + i^2 \dfrac{x^2} { 2! } + i^3 \dfrac{x^3} { 3! } + i^4 \dfrac{x^4} { 4! } + i^5 \dfrac{x^5} { 5! } + i^6 \dfrac{x^6} { 6! } + i^7 \dfrac{x^7} { 7! } ... + i^n \dfrac{x^n}{n!} + ...$

Note that
$i^2 = - 1$
$i^3 = i^2 i = - i$
$i^4 = i^3 i = 1$
$i^5 = i^4 i = i$
$i^6 = i^5 i = - 1$
$i^7 = i^6 i = -i$ ...
and so on.

Therefore $e^{(ix)}$ may be written as
$e^{(ix)} = 1 + i x - \dfrac{x^2} { 2! } - i \dfrac{ x^3} { 3! } + \dfrac{x^4} { 4! } + i \dfrac{x^5} { 5! } - \dfrac {x^6} { 6! } - i \dfrac{x^7} { 7! } ...$

Let us group the terms making the real part together and the terms making the imaginary together and rewrite $e^{(ix)}$ as
$e^{(ix)} = ( 1 - \dfrac{x^2} { 2! } + \dfrac{x^4} { 4! } - \dfrac {x^6} { 6! } ...) + i ( x - \dfrac{ x^3} { 3! } + \dfrac{x^5} { 5! } - \dfrac{x^7} { 7! } ...) \qquad (I)$

Referring to the
tables of mathematical formulas , the Taylor series of $\sin x$ and $\cos x$ are given by
$\sin x = x - \dfrac{x^3}{ 3!} + \dfrac{x^5}{ 5!} - \dfrac{ x^7}{ 7!} + ...$

$\cos x = 1 - \dfrac{x^2}{2!} + \dfrac{x^4}{4!} - \dfrac{x^6}{6!} + ...$

Note that the real part of $e^{(ix)}$ in $(I)$ is given by
$1 - \dfrac{x^2} { 2! } + \dfrac{x^4} { 4! } - \dfrac {x^6} { 6! } ...$ which is the Taylor series of $\cos x$

and the imaginary part of $e^{(ix)}$ in $(I)$ is given by
$x - \dfrac{ x^3} { 3! } + \dfrac{x^5} { 5! } - \dfrac{x^7} { 7! } ...$ which is the Taylor series of $\sin x$

and hence the Euler formula
$e^{ix} = \cos x + i \sin x$