Euler's identity

Euler's identity, sometimes called Euler's equation, is this equation:

$e^{i\pi }+1=0$

It features the following mathematical constants:

• $\pi$, pi

  $\pi \approx 3.14159$

• $e$, Euler's Number

  $e\approx 2.71828$

• $i$, imaginary unit

  $i=\sqrt{-1}$

It also features three of the basic mathematical operations: addition, multiplication and exponentiation.

Euler's identity is named after the Swiss mathematician Leonard Euler. It is not clear that he invented it himself.

Respondents to a Physics World poll called the identity "the most profound mathematical statement ever written", "uncanny and sublime", "filled with cosmic beauty" and "mind-blowing".

Mathematical proof of Euler's Identity using Taylor Series

Many equations can be written as a series of terms added together. This is called a Taylor series.

The exponential function $e^x$ can be written as the Taylor series

$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}\cdots =\sum _{k=0}^{\mathrm{\infty }}\frac{x^n}{n!}$

As well, the sine function can be written as

$\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}\cdots =\sum _{k=0}^{\mathrm{\infty }}\frac{(-1{)}^n}{(2n+1)!}{x}^{2n+1}$

and cosine as

$\cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}\cdots =\sum _{k=0}^{\mathrm{\infty }}\frac{(-1{)}^n}{(2n)!}{x}^{2n}$

Here, we see a pattern take form. $e^x$ seems to be a sum of sine and cosine's Taylor series, except with all of the signs changed to positive. The identity we are actually proving is $e^{ix}=\cos (x)+i\sin (x)$.

So, on the left side is $e^{ix}$, whose Taylor series is $1+ix-\frac{x^2}{2!}-\frac{i{x}^3}{3!}+\frac{x^4}{4!}+\frac{i{x}^5}{5!}\cdots$

We can see a pattern here, that every second term is i times sine's terms, and that the other terms are cosine's terms.

On the right side is $\cos (x)+i\sin (x)$, whose Taylor series is the Taylor series of cosine, plus i times the Taylor series of sine, which can be shown as:

$(1-\frac{x^2}{2!}+\frac{x^4}{4!}\cdots )+(ix-\frac{i{x}^3}{3!}+\frac{i{x}^5}{5!}\cdots )$

if we add these together, we have

$1+ix-\frac{x^2}{2!}-\frac{i{x}^3}{3!}+\frac{x^4}{4!}+\frac{i{x}^5}{5!}\cdots$

Therefore,

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

Now, if we replace x with $\pi$, we have:

$e^{i\pi }=\cos (\pi )+i\sin (\pi )$

Since we know that $\cos (\pi )=-1$ and $\sin (\pi )=0$, we have:

• $e^{i\pi }=-1$
• $e^{i\pi }+1=0$

which is the statement of Euler's identity.

Related pages

• −1 (number)
• Euler's formula