Complex Exponential — Jemoka Knowledge Base

Recall that Euler’s Equation exists:

\begin{equation} f(x) = e^{i k \omega x} = \cos (k\omega x) + i \sin(k\omega x) \end{equation}

and, for \omega = \frac{2\pi}{L}, this is still L periodic! Next up, we make an important note:

\begin{equation} e^{ik\omega x}, e^{-i k \omega x} \end{equation}

is linearly independent over x. inner product over complex-valued functions recall all of the inner product properties. Now, for functions periodic over [0,L] (recall we have double this if the function is period over [-L, L]:

\begin{equation} \langle f, g \rangle = \frac{1}{L} \int_{0}^{L} f(x) \overline{g(x)} \dd{x} \end{equation}

similar to all other inner products, \langle f,f \rangle = 0 IFF f = 0, and \langle f,g \rangle = 0 implies that f and g are orthogonal. complex exponentials are orthonormal For L > 0, and \omega = \frac{2\pi}{L}, consider:

\begin{equation} \langle e^{ik_{1} \omega x}, e^{ik_{2} \omega x} \rangle \end{equation}

Importantly, we have the property that: \langle e^{ik_{1} \omega x}, e^{ik_{2} \omega x} \rangle = 0 if k_1 \neq k_2 \langle e^{ik_{1} \omega x}, e^{ik_{2} \omega x} \rangle = 1 if k_1 = 1