2.3.7 So­lu­tion dot-g

Ques­tion:

Show that the func­tions $e^{4{\rm i}{\pi}x}$ and $e^{6{\rm i}{\pi}x}$ are an or­tho­nor­mal set on the in­ter­val 0 $\raisebox{-.3pt}{$\leqslant$}$ $x$ $\raisebox{-.3pt}{$\leqslant$}$ 1.

An­swer:

You need to show that both func­tions are nor­mal­ized, $\left\vert\left\vert e^{4{\rm i}{\pi}x}\right\vert\right\vert$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1 and $\left\vert\left\vert e^{6{\rm i}{\pi}x}\right\vert\right\vert$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1, and that they are mu­tu­ally or­thog­o­nal, $\left\langle{e}^{4{\rm i}{\pi}x}\vert e^{6{\rm i}{\pi}x}\right\rangle$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0. Work each out in turn (don't for­get to take com­plex con­ju­gate of the first func­tion in the in­ner prod­ucts):

$$\vert\vert e^{4{\rm i}\pi x}\vert\vert = \sqrt{\langle e^{4{... ...^{4{\rm i}\pi x}{ \rm d}x} = \sqrt{\int_0^1 1 { \rm d}x} = 1$$

$$\vert\vert e^{6{\rm i}\pi x}\vert\vert = \sqrt{\langle e^{6{... ...^{6{\rm i}\pi x}{ \rm d}x} = \sqrt{\int_0^1 1 { \rm d}x} = 1$$

$$\langle e^{4{\rm i}\pi x}\vert e^{6{\rm i}\pi x}\rangle = \i... ...}x = \frac{1}{2{\rm i}\pi} e^{2{\rm i}\pi x}\Big\vert _0^1 = 0$$

(Since the Euler for­mula shows that $e^{{\rm i}2\pi}$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1.)