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Invoking Eq. (56), we can remove the a, b, and α dependence: 

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The solution set of cos(( k + 3 / 2) θ ) / cos(( k + 1 / 2) θ ) < − 1 is 

and the solution set of sin(( k + 1) θ ) / sin( kθ ) < − 1 is 

In (70), k = n − j and j varies from 1 to n − 1 so k can take any value between 1 and n − 1. It is not hard to show that the intersection of the sets (70) having k between 1 and $^{n}$− 1 is (( $^{n}$− 1) π/n, π ). In (69), j varies from 1 to n so k takes values between 0 and $^{n}$− 1. The intersection of (( n − 1) π/n, π ) with all the sets (69) having k between 0 and n − 1 is ((2 n − 1) π/ (2 n + 1) , π ). Thus a necessary condition for all positive time intervals is that θ be in the range ((2 n − 1) π/ (2 n + 1) , π ). 

Eq. (55) may be inverted as x = √ y exp iφ, if cos( φ ) = − b (1 + y ) / (2 a √ y ) = − b cos( β/ 2) /a. We have φ ∈ ℜ if 0 ≤ cos $^{2}$( β/ 2) ≤ a $^{2}$/b$^{2}$. Equivalently, we may write the condition for capture as: 

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The left inequality follows from Eq. (60) so any solution to Eq. (60) in the interval (2 n − 1) π/ (2 n + 1) ≤ θ ≤ π satisfying 

<img src='content_image/332090.jpg'>

corresponds to allowed capture. Let Q be defined by Q ( θ ) = cos $^{2}$(( n +1 / 2) θ ) / cos $^{2}$( θ/ 2) . On the interval under consideration, Q increases monotonically in θ, from 0 at the left endpoint to (2 n + 1) 2 at the right endpoint. From Eq. (60), we can show that b = aQ/ (1 + a − aQ ) . When this is used in the inequality we have: 

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