# Famous Theorems of Mathematics/Euler's equation

Euler's equation states that,

$\exp i\theta =i\sin \theta +\cos \theta$ From which, the famous identity,

$\exp i\pi +1=0$ may be deduced.

## Proof

Definition 1
$\exp \varphi =\sum _{n=0}^{\infty }{\frac {1}{n!}}\varphi ^{n}$
Definition 2
$\sin \varphi =\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{(2n+1)!}}\varphi ^{2n+1}$
Definition 3
$\cos \varphi =\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{(2n)!}}\varphi ^{2n}$

The following results will be assumed;

$\sin \pi =0$
$\cos \pi =-1$
Theorem 1
$\exp i\theta =\cos \theta +i\sin \theta$
Proof

By definition 1,

$\exp i\theta =\sum _{n=0}^{\infty }{\frac {1}{n!}}(i\theta )^{n}$

Observe that one may split the summation into two,

$\exp i\theta =\sum _{n=0}^{\infty }{\frac {1}{(2n)!}}(i\theta )^{2n}+\sum _{n=0}^{\infty }{\frac {1}{(2n+1)!}}(i\theta )^{2n+1}$

Evaluating,

$\exp i\theta =\sum _{n=0}^{\infty }{\frac {i^{2n}}{(2n)!}}\theta ^{2n}+\sum _{n=0}^{\infty }{\frac {i^{2n+1}}{(2n+1)!}}\theta ^{2n+1}$
$\exp i\theta =\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{(2n)!}}\theta ^{2n}+i\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{(2n+1)!}}\theta ^{2n+1}$

By definition 2 and 3,

$\exp i\theta =\cos \theta +i\sin \theta$  $\blacksquare$
Theorem 2
$\exp i\pi +1=0$
Proof

By theorem 1,

$\exp i\pi =\cos \pi +i\sin \pi$
$\exp i\pi =-1+0i$

Hence,

$\exp i\pi +1=0$  $\blacksquare$