2.3.5 So­lu­tion dot-e

Ques­tion:

Ver­ify that $\sin(x)$ is not a nor­mal­ized func­tion on the in­ter­val 0 $\raisebox{-.3pt}{$\leqslant$}$ $x$ $\raisebox{-.3pt}{$\leqslant$}$ $2\pi$, and nor­mal­ize it by di­vid­ing by its norm.

An­swer:

$$\vert\vert\sin(x)\vert\vert = \sqrt{\left\langle\vphantom{\s... ...ngle } = \sqrt{\int_0^{2\pi} \sin^2(x){ \rm d}x} = \sqrt{\pi}$$

Since $\vert\vert\sin(x)\vert\vert$ is not one, $\sin(x)$ is not a nor­mal­ized func­tion on 0 $\raisebox{-.3pt}{$\leqslant$}$ $x$ $\raisebox{-.3pt}{$\leqslant$}$ $2\pi$. If you di­vide by its norm, i.e. by $\sqrt{\pi}$, how­ever,

$$\vert\vert\sin(x)/\sqrt{\pi}\vert\vert = \sqrt{\int_0^{2\pi}... ...qrt{\pi})(\sin(x)/\sqrt{\pi}){ \rm d}x} = \sqrt{\pi /\pi} = 1$$