Show that set is an orthonormal set in PC(0,l)

[Shackleford]

Show that set \{ (2/l)^{1/2} sin(n-1/2)(\pi x/l)\}^\infty_1 is an orthonormal set in PC(0,l).
Of course, I need to show that <\phi_m, \phi_n> = \delta_{mn}
&lt;\phi_m, \phi_n&gt; = (2/l) \int_0^l sin(m-1/2)(\pi x/l) sin(n-1/2)(\pi x/l)\, dx<br /> \\ = 1/l \int_0^l cos(m-n)(\pi x/l) cos(m+n+1)(\pi x/l)\, dx<br /> \\ = \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<br /> \\ = \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<br />

I used the trig identity:
2 sin \theta sin \psi = cos(\theta - \psi) - cos(\theta + \psi)

 

[mathman]

What is your question? The expression you have (you should check it out) = 0 for $m\ne n$ and =1 for $m=n$.

 

[Shackleford]

What is your question? The expression you have (you should check it out) = 0 for $m\ne n$ and =1 for $m=n$.

Thanks for the reply. When I looked at it, I could swear it's zero in both cases. Let me check again.

 

[Shackleford]

<br /> \\ = \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<br />

When you evaluate at x = l and x = 0, the argument for sine is either zero or a multiple of pi. When m = n, there's an indeterminate form in the common term.

 

[mathman]

When m=n, I suggest you don't use the cos identity, but use the original form, where the integral is $2/l\int_0^lsin^2((m-1/2)(\pi x/l))dx$.

I think you have some errors in the derivation. You have the product of cos terms, rather than the difference, although this seems to have been corrected for the next line. When you integrated the cos (m+n+!)...) term, you divided by m-n.