# A Theoretical Framework to Determine RHP Zero Dynamics in Sequential Interacting Sub-Systems
## Appendix A  
Proof of Proposition 4.: Assume by contradiction: \(\alpha_{0}>\beta_{0}\Rightarrow\alpha_{k}<\beta_{k}\)\(\forall k=\overline{1,n}\). If \(T_{\alpha}=min\{\mathcal{C}\}\) and \(T_{\beta}=max\{\mathcal{D}\}\), then:  
\[\alpha_{k} > \prod_{i=1}^{n}\gamma_{i}T_{\alpha}^{k}\frac{(2n)!}{k!(2n-k)!}= \hat{\alpha}_{k}\] \[\beta_{k} < \prod_{i=1}^{n}(1-\gamma_{i})T_{\beta}^{k}\frac{n!}{k!(n-k)!}= \hat{\beta}_{k}\]  
resulting in \(\alpha_{k}<\beta_{k}\Rightarrow\hat{\alpha}_{k}<\hat{\beta}_{k}\), i.e.,  
\[\prod_{i=1}^{n}\gamma_{i}T_{\alpha}^{k}\frac{(2n)!}{k!(2n-k)!}<\prod_{i=1}^{n}( 1-\gamma_{i})T_{\beta}^{k}\frac{n!}{k!(n-k)!}\]From the hypothesis, we have \(\alpha_{0}>\beta_{0}\Rightarrow\prod\limits_{i=1}^{n}\gamma_{i}>\prod\limits_{i=1} ^{n}(1-\gamma_{i})\), then (A1) holds if and only if:  
\[T_{\alpha}^{k}\frac{(2n)!}{k!(2n-k)!} < T_{\beta}^{k}\frac{(n)!}{k!(n-k)!}\] \[T_{\alpha}^{k} < T_{\beta}^{k}\frac{n!}{k!(n-k)!}\frac{k!(2n-k)!}{(2n)!}\] \[T_{\alpha} < T_{\beta}\sqrt[k]{\frac{n!}{(n-k)!}\frac{(2n-k)!}{(2n)!}}\] \[T_{\alpha} < T_{\beta}\underbrace{\sqrt[k]{\frac{(2n-k)(2n-k-1)\cdots(2n-k-n+ 1)}{2n(2n-1)\cdots(2n-n+1)}}}_{p}\]  
To study the last inequality, the limits of \(p\) are analyzed. Thus, when \(k\to 1\), \(p\) becomes:  
\[\lim\limits_{k\to 1}p=\frac{n}{2n}=\frac{1}{2}\]  
and when \(k\to n\), we have:  
\[\lim\limits_{k\to n}p = \sqrt[n]{\frac{(2n-n)(2n-n-1)\cdots(2n-n-n+1)}{2n(2n-1)\cdots(2n- n+1)}}=\] \[= \sqrt[n]{\frac{n(n-1)\cdots 1}{2n(2n-1)\cdots(n+1)}}<\frac{1}{2}\ \ \forall n=2,3,..\]  
Since the second case is less restrictive then the first case, only \(k\to 1\) will be analyzed, obtaining:  
\[T_{\alpha}<\frac{T_{\beta}}{2}\Leftrightarrow 2T_{\alpha}<T_{\beta}\]  
However, from second premise given in Proposition 4, it results that the product of two elements from \(\mathcal{C}\) cannot be lower than an element from \(\mathcal{D}\).  
Hence, our assumption \(\alpha_{0}>\beta_{0}\Rightarrow\alpha_{k}<\beta_{k}\) is incorrect, resulting in \(\alpha_{0}>\beta_{0}\Rightarrow\alpha_{k}>\beta_{k}\).  
Proof of Proposition 5.: Firstly, we show that the function is strictly increasing in \(I=(0,1)\). Then, to prove the unique real solution is sufficient to check the different sign in the function, when \(\hat{\gamma}\) tends to the interval bound, finally, the values of the real solution are computed through algebraic calculations. The first derivative of \(\zeta_{0}(\hat{\gamma})\) is:  
\[\zeta_{0}^{{}^{\prime}}(\hat{\gamma})=n\hat{\gamma}^{n-1}-n(\hat{\gamma}-1)^{n -1}\]  
Since the hypothesis implies that \(n\) is an even number, it results that \(\zeta_{0}^{{}^{\prime}}(\hat{\gamma})>0\ \ \forall\hat{\gamma}\in I\). Thus, \(\zeta_{0}(\hat{\gamma})\) is strictly increasing in \(I\).  
In the second part, the limits of the function are verified:  
\[\lim\limits_{\hat{\gamma}\to 0^{+}}\zeta_{0}(\hat{\gamma}) = -1\] \[\lim\limits_{\hat{\gamma}\to 1^{-}}\zeta_{0}(\hat{\gamma}) = 1\]  
Thus, the function has different signs in the interval extremes and is monotonically increasing, resulting in that it has only one real solution independent of the value of \(n\).  
In the last part of the proof, we show that the unique real solution is equal to \(\hat{\gamma}=\frac{1}{2}\). Hence, a real number must be found, such that:  
\[\hat{\gamma}^{n}-(\hat{\gamma}-1)^{n} = 0\] \[\hat{\gamma}^{n} = (\hat{\gamma}-1)^{n}\ \ \forall n=2,4,\cdots\]  
Since in the hypothesis, \(n\) is assumed an even number and \(\hat{\gamma}\in(0,1)\), taking the \(n\)th root in the above equation, we have:  
\[\hat{\gamma} = -\hat{\gamma}+1\] \[\hat{\gamma} = \frac{1}{2}\]  
Furthermore, it can be proven that that other \(n-1\) complex solutions have all the same real part.