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

e^{i\pi} + 1 = 0

It features the following mathematical constants:

i = \surd{-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.

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