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@jemoka / Jemoka Knowledge Base / raw/concept/kbhcomplex_exponential.md
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--- title: "Complex Exponential" source: https://www.jemoka.com/posts/kbhcomplex_exponential/ --- 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 \(k_1 = 1\)