## Proof: Euler's Equation

Euler's Equation states that: $e^{ix}=\cos x + i\sin x$ (where $$i=\sqrt{-1}$$ )

## Proof

If we consider the Teylor expansions of $$e^x$$, $$\cos x$$ and $$\sin x$$, we see that the latter two almost combine to make the former, with some differences in the signs of the terms:

$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\frac{x^5}{5!}+\frac{x^6}{6!}+\cdots$

$\hspace{0.5cm} \cos x=1\hspace{1cm}-\frac{x^2}{2!}\hspace{1.2cm}+\frac{x^4}{4!}\hspace{1.2cm}-\frac{x^6}{6!}\hspace{1cm}+\cdots$

$\sin x=\hspace{1cm}x\hspace{1.2cm}-\frac{x^3}{3!}\hspace{1.2cm}+\frac{x^5}{5!}\hspace{1.2cm}+\cdots \hspace{0.7cm}$

Notice however, that if we substitute $$ix$$ for $$x$$ in the Taylor expansion of $$e^x$$, we get the correct signs:

$e^{ix}=1+ix+\frac{(ix)^2}{2!}+\frac{(ix)^3}{3!}+\frac{(ix)^4}{4!}+\frac{(ix)^5}{5!}+\frac{(ix)^6}{6!}+\cdots$

so:

$e^{ix}=1+ix-\frac{x^2}{2!}-\frac{ix^3}{3!}+\frac{x^4}{4!}+\frac{ix^5}{5!}-\frac{x^6}{6!}+\cdots$

$\hspace{0.5cm} \cos x=1\hspace{1cm}-\frac{x^2}{2!}\hspace{1.2cm}+\frac{x^4}{4!}\hspace{1.2cm}-\frac{x^6}{6!}\hspace{1cm}+\cdots$

$\sin x=\hspace{1cm}x\hspace{1.2cm}-\frac{x^3}{3!}\hspace{1.2cm}+\frac{x^5}{5!}\hspace{1.2cm}+\cdots \hspace{0.7cm}$

i.e.    e^{ix}=cos x +i sin x

$Q.E.D.$