I have a problem while calculating cutoff frequency, suppose we have these specs.
Firstly, I calculated the order of the filter and got $N=5.8858$ and round it up to get $N=6$.
Now I'm supposed to get $\Omega_c$. Using these equations:
\begin{cases}1+\left(\frac{0.2\pi}{\Omega_c}\right)^{2N} = \left(\frac{1}{0.89125}\right)^2 \quad&\quad (1)\\ 1+\left(\frac{0.3\pi}{\Omega_c}\right)^{2N} = \left(\frac{1}{0.17783}\right)^2\quad&\quad (2) \end{cases}
Now with $N=6$ and $T=1$, substituting in $(1)$
\begin{align} \left(\frac{0.2\pi}{\Omega_c}\right)^{12} &=\left(\frac{1}{0.89125}\right)^2 - 1 =0.25893\\ \implies (\Omega_c)^{12} &= \frac{{(0.2\pi)}^{12}}{0.25893}\\ &= 40.29 \end{align} But in the textbook it says $\Omega_c = 0.7032$, what I did wrong? Any help would be appreciated.