I have the the following problem from my Fourier analysis book:
Show that $\{ (2/l)^{1/2}\sin(n-\frac{1}{2})(\pi x/l) \}_1^{\infty}$
is an orthonormal set in $PC(0,l)$, i.e. class of piecewise continuous
functions on the interval $[0,l]$.
The dot product between two functions $\phi_n$, $\phi_m$ $\in PC(0,l)$ is defined as:
$$\langle\phi_n, \phi_m\rangle = \int_0^l\phi_n\, \overline{\phi_m}\:dx.$$
So my problem is to show that:
$$\langle\phi_n, \phi_m\rangle = \left\{\begin{array}{lr}
1 & : m=n\\
0 & : m\neq n
\end{array}\right\}.$$
My problem is that I get confused with the complex conjugate $\overline{\phi_m}$. What is the complex conjugate of $$\phi_m = \frac{\sqrt{2}\sin\left( \frac{m\pi x}{l}-\frac{\pi x}{2l}\right)}{\sqrt{l}} \;\;\;(= \overline{\phi_m}\;\;?)$$
Should I use the identity $$\sin\theta=\frac{e^{i\theta}-e^{-i\theta}}{2i}$$
in this problem?
Thnx for any hints =)
Best Answer
$$\langle \phi_n , \phi_m \rangle = \frac{2}{l}\int^l_0 \left[\sin\left( \frac{(2n-1)\pi x}{2l}\right)\sin\left( \frac{(2m-1)\pi x}{2l}\right)\right]\;dx $$
$$=\frac{1}{l}\int^l_0 \left[ \cos\left( \frac{(n-m)\pi x}{l}\right)-\cos\left( \frac{(n+m-1)\pi x}{l}\right) \right]\;dx \;\;\;\text{(by the product-to-sum rule)}$$
when $n=m$, the integral evaluates to:
$$=\frac{1}{l}\int^l_0 \left[ 1-\cos\left( \frac{(2n-1)\pi x}{l}\right) \right]\;dx =\frac{1}{l}\left[ x-\frac{l}{(2n-1)\pi}\sin\left(\frac{(2n-1)\pi x}{l}\right)\right]^l_0 = \frac{1}{l}\cdot l = 1$$
when $n\neq m$, the integral evaluates to:
$$=\frac{1}{l}\left[ \frac{l}{(n-m)\pi }\sin\left( \frac{(n-m)\pi x}{l}\right)-\frac{l}{(n+m-1)\pi }\sin\left( \frac{(n+m-1)\pi x}{l}\right) \right]^l_0 =\frac{1}{l}\cdot 0=0$$