Im having a hard time trying to solve this...
if $ (x_{k})_{k=0}^\infty $ is a causal succession such as
$$ Z(x_{k}) = X(z) \parallel z \parallel > R $$
prove that $$ Z( {{x_{k+2}}}) = Z^{2}X(z)-z^{2}x_{0}-zx_{1}, \parallel z \parallel > R$$
Can anybody help me?
The problem essential states that \begin{align} \sum_{k=0}^{\infty} x_{k} \ z^{-k} = X(z) \end{align} and asks to find a similar result for $x_{k+2}$. It can be seen that \begin{align} \sum_{k=0}^{\infty} x_{k+2} \ z^{-k} &= \sum_{k=2}^{\infty} x_{k} \ z^{-k+2} \\ &= z^{2} \left( \sum_{k=0}^{\infty} x_{k} \ z^{-k} - x_{0} z^{o} - x_{1} z^{-1} \right) \\ &= z^{2} X(z) - x_{0} z^{2} - x_{1} z. \end{align} Hence \begin{align} Z(x_{k+2}) = z^{2} X(z) - x_{0} z^{2} - x_{1} z. \end{align}
In a similar manor it can be shown that \begin{align} Z(x_{k+1}) = z X(z) - x_{0} z. \end{align}