- #1
Shackleford
- 1,656
- 2
Show that set [tex] \{ (2/l)^{1/2} sin(n-1/2)(\pi x/l)\}^\infty_1 [/tex] is an orthonormal set in PC(0,l).
Of course, I need to show that [tex] <\phi_m, \phi_n> = \delta_{mn} [/tex]
[tex] <\phi_m, \phi_n> = (2/l) \int_0^l sin(m-1/2)(\pi x/l) sin(n-1/2)(\pi x/l)\, dx
\\ = 1/l \int_0^l cos(m-n)(\pi x/l) cos(m+n+1)(\pi x/l)\, dx
\\ = \left( \frac 1 l \right) \frac {l} {\pi(m-n)} sin(m-n)(\pi x/l) \Big|_0^l - \left( \frac 1 l \right) \frac {l} {\pi(m-n)} sin(m+n+1)(\pi x/l)\ \Big|_0^l
\\ = \frac {1} {\pi(m-n)} sin(m-n)(\pi x/l) \Big|_0^l - \frac {1} {\pi(m-n)} sin(m+n+1)(\pi x/l)\ \Big|_0^l
[/tex]
I used the trig identity:
[tex] 2 sin \theta sin \psi = cos(\theta - \psi) - cos(\theta + \psi)[/tex]
Of course, I need to show that [tex] <\phi_m, \phi_n> = \delta_{mn} [/tex]
[tex] <\phi_m, \phi_n> = (2/l) \int_0^l sin(m-1/2)(\pi x/l) sin(n-1/2)(\pi x/l)\, dx
\\ = 1/l \int_0^l cos(m-n)(\pi x/l) cos(m+n+1)(\pi x/l)\, dx
\\ = \left( \frac 1 l \right) \frac {l} {\pi(m-n)} sin(m-n)(\pi x/l) \Big|_0^l - \left( \frac 1 l \right) \frac {l} {\pi(m-n)} sin(m+n+1)(\pi x/l)\ \Big|_0^l
\\ = \frac {1} {\pi(m-n)} sin(m-n)(\pi x/l) \Big|_0^l - \frac {1} {\pi(m-n)} sin(m+n+1)(\pi x/l)\ \Big|_0^l
[/tex]
I used the trig identity:
[tex] 2 sin \theta sin \psi = cos(\theta - \psi) - cos(\theta + \psi)[/tex]