# Odd-order Butterworth filter gain at cutoff frequency

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.