# Euler's Identity - ${e}^{\mathrm{\pi i}}-1=0$

## Introduction

*e ^{πi}* + 1 = 0 is one of the more interesting
mathematical equations since it takes 3 different number concepts, the
natural log raised to pi times i, the imaginary constant. Simplifying
this expression leads to a simple answer of -1. Now I am pretty nerdy,
but this seems pretty amazing to me.

- Three of the most important numbers can be put in an expression to equal -1
- How a number raised to a power can be a negative
- How the proof is done with elementary functions

Because of all this, the concept of Math has intrigued the last few days, so I thought I would put my own explanation up on here and spread some math knowledge to all.

## Background

First off, I am going to explain what each of the numbers involved is.

*e* is the base of the natural logarithm. e can be expressed two
different ways. The typical definition is the following:

However, it can also be defined as:

$$e=\sum _{k=0}^{\infty}\frac{1}{k!}$$Both of these will lead to a series which goes on to infinity and will lead to an irrational number:

$$e=\mathrm{2.71828182845....}$$ *π* is the ratio of a circle's circumference to it's diameter.
This is an irrational number and is equal to:

*i* is defined as the following:

Because you cannot typically take the square root of a negative number, the idea of i or the imaginary digit was introduced. For example, the square root of -9 would then be 3i.

## Proof

Using the definition:

$${e}^{z}=\underset{x\to \mathrm{\infty}}{lim}{\left((,1,+,\frac{z}{x},)\right)}^{x}$$we get:

$${e}^{\mathrm{\pi i}}=1+\mathrm{ix}+{i}^{2}\frac{{x}^{2}}{\mathrm{2!}}+{i}^{3}\frac{{x}^{3}}{\mathrm{3!}}+{i}^{4}\frac{{x}^{4}}{\mathrm{4!}}+{i}^{5}\frac{{x}^{5}}{\mathrm{5!}}+\mathrm{...}$$Now using the definition of i:

$${e}^{\mathrm{\pi i}}=1+\mathrm{ix}-\frac{{x}^{2}}{\mathrm{2!}}-i\frac{{x}^{3}}{\mathrm{3!}}+\frac{{x}^{4}}{\mathrm{4!}}+i\frac{{x}^{5}}{\mathrm{5!}}-\mathrm{...}$$Rearraning the equation, we get:

$${e}^{\mathrm{\pi i}}=\left((,1,-,\frac{{x}^{2}}{\mathrm{2!}},+,\frac{{x}^{4}}{\mathrm{4!}},-,\mathrm{...},)\right)+i\left((,x,+,-,i,\frac{{x}^{3}}{\mathrm{3!}},+,i,\frac{{x}^{5}}{\mathrm{5!}},-,\mathrm{...},)\right)$$Using the following substitutions, from the Taylor Series:

$$\mathrm{cos}x=1-\frac{{x}^{2}}{\mathrm{2!}}+\frac{{x}^{4}}{\mathrm{4!}}-\mathrm{...}$$$$\mathrm{sin}x=x+-i\frac{{x}^{3}}{\mathrm{3!}}+i\frac{{x}^{5}}{\mathrm{5!}}-\mathrm{...}$$

We get:

$${e}^{\mathrm{ix}}=\mathrm{cos}x+i\mathrm{sin}x$$$${e}^{\mathrm{i\pi}}=\mathrm{cos}\pi +i\mathrm{sin}\pi $$

$${e}^{\mathrm{i\pi}}=\mathrm{-1}+i0$$

$${e}^{\mathrm{i\pi}}=\mathrm{-1}$$