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

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

As well, the sine function can be written as

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

and cosine as

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

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

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:

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

if we add these together, we have

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

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