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Wikipedia gives the gain of an n-order Butterworth filter as
$$G^2(\omega)=\left |H(j\omega)\right|^2 = \frac {{G_0}^2}{1+\left(\frac{j\omega}{\omega_c}\right)^{2n}}$$
here. If you look at the gain at the cutoff frequency, the denominator becomes $(1+j^{2n})$, which blows up to infinity for odd $n$. Am I doing something wrong? Are Butterworth filters, or that formula, restricted to even $n$?
Congratulations, you've found an error on the wikipedia page on Butterworth filters. The squared magnitude of the frequency response of an $n^{th}$-order Butterworth low pass filter is
$$|H(j\omega)|^2=\frac{1}{1+\left(\frac{\omega}{\omega_c}\right)^{2n}}\tag{1}$$
There should be no imaginary unit $j$ in that formula.
