Invoking Eq. (56), we can remove the a, b, and α dependence: 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: 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 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: 21 Q (1 + Q ) + a $^{(1}$− Q ) ≥ ( a + 1) sin $^{2}$( θ/ 2) . (73) sin $^{2}$( θ/ 2) ≤ b/ [( a + b )( a + 1)] (72) $^{1}$− a $^{2}$/b2 ≤ sin $^{2}$( β/ 2) ≤ 1 ⇔ ( b − a ) b ( a + 1) ≤ sin $^{2}$( θ/ 2) ≤ b ( a + b )( a + 1) . (71) k ⋃ l =1 ( lπ/ ( k + 1) , lπ/k ) . (70) k +1 ⋃ l =1 ((2 l − 1) π/ (2 k + 3) , (2 l − 1) π/ (2 k + 1)) . (69) t$_{2}$$_{j}$$_{+1}$ t$_{2}$$_{j}$ = − sin(( n − j + 1) θ ) sin(( n − j ) θ ) , j = 1 , 2 $^{,}$· · · ( n − 1) . (68)