489
回編集
差分
編集の要約なし
:$$\displaystyle z^n=\left[\sum_{k=0}^{\lfloor (n-1)/2\rfloor}\binom{n-k-1}{k}(-1)^kr^{n-2k-1}l^{k}\right]z-\left[\sum_{k=0}^{\lfloor (n-2)/2\rfloor}\binom{n-k-2}{k}(-1)^kr^{n-2k-2}l^{k}\right]l$$
:$$\begin{array}{l}z^1=&z$$\\:$$z^2=&rz-l$$\\:$$z^3=&(r^2-l)z-rl$$\\:$$z^4=&(r^3-2rl)z-(r^2-l)l$$\\:$$z^5=&(r^4-3r^2l+l^2)z-(r^3-2rl)l$$\\:$$&\quad\quad\quad\quadvdots\\vdots$$:$$z^n=&A_{n}z-A_{n-1}\quad\begin{cases}
A_0=0\\
A_1=1\\
A_{k}=(A_{k-1})r-(A_{k-2})l
\end{cases}\end{array}$$
:$$\displaystyle z^n=\left[\sum_{k=0}^{\lfloor (n-1)/2\rfloor}\binom{n-k-1}{k}(-1)^k(2\cos\theta)^{n-2k-1}\right]z-\sum_{k=0}^{\lfloor (n-2)/2\rfloor}\binom{n-k-2}{k}(-1)^k(2\cos\theta)^{n-2k-2}$$
:$$\begin{array}{l}z^1&=z$$&\\:$$z^2&=rz-1&=(2\cos\theta)z-1$$\\:$$z^3&=(r^2-1)z-r&=[(2\cos\theta)^2-1]z-(2\cos\theta)$$\\:$$z^4&=(r^3-2l2r)z-(r^2-1)&=[(2\cos\theta)^3-2(2\cos\theta)]z-[(2\cos\theta)^2-1]$$\\:$$z^5&=(r^4-3r^2+1)z-(r^3-2r)&=[(2\cos\theta)^4-3(2\cos\theta)^2+1]z-[(2\cos\theta)^3-2(2\cos\theta)]$$\\:$$&\quad\quad\quad\quadvdots\\vdots$$:$$z^n&=A_{n}z-A_{n-1}\quad\begin{cases}
A_0=0\\
A_1=1\\
A_{k}=(A_{k-1})r-(A_{k-2})
\end{cases}&\end{array}$$
==導出==