Suggest edit — Complex Exponential

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title: "Complex Exponential"
source: https://www.jemoka.com/posts/kbhcomplex_exponential/
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Recall that Euler&rsquo;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 \(L &gt; 0\), 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 \(k_1 = 1\)