差分

:$$\begin{array}{l}

z^1=&0+z\\

\end{array}$$

:$$z^n=\displaystyle-\left[\sum_{k=0}^{\lfloor (n-2)/2\rfloor}\binom{n-k-2}{k}(-1)^k(2\cos\theta)^{n-2k-2}\right]+\left[\sum_{k=0}^{\lfloor (n-1)/2\rfloor}\binom{n-k-1}{k}(-1)^k(2\cos\theta)^{n-2k-1}\right]z$$

==導出==