494
回編集
差分
編集の要約なし
:$$\begin{array}{l}
\begin{pmatrix}C_{n}&C_S_{n+1}\\S_C_{n+1}&S_{n+1}\end{pmatrix}\begin{pmatrix}1\\z\end{pmatrix}&=\begin{pmatrix}0&1\\-z\cdotp\bar{z}&z+\bar{z}\end{pmatrix}^n\begin{pmatrix}1\\z\end{pmatrix}\\
&=\begin{pmatrix}0&1\\-1&2\cos\theta\end{pmatrix}^n\begin{pmatrix}1\\z\end{pmatrix}
\end{array}$$
:$$\begin{array}{l}
z^1&=0+z&\\
z^2&=-1+rz\\&=-1+(2\cos\theta)z\\z^3&=-r+(-1+r^2)z\\&=-(2\cos\theta)+[(2\cos\theta)^2-1]z\\&=-(2\cos\theta)+[2\cos2\theta+1]z\\z^4&=-(-1+r^2)+(-2r+r^3)z\\&=-[(2\cos\theta)^2-1]+[(2\cos\theta)^3-2(2\cos\theta)]z\\&=-[2\cos2\theta+1]+[2(\cos\theta+\cos3\theta)]z\\z^5&=-(-2r+r^3)+(-3r^2+1+r^4)z\\&=-[(2\cos\theta)^3-2(2\cos\theta)]+[(2\cos\theta)^4-3(2\cos\theta)^2+1]z\\&=-[2(\cos\theta+\cos3\theta)]+[2(\cos2\theta+\cos4\theta)+1]z\\
&\quad\quad\quad\vdots\\
z^{2m}
&\displaystyle=-\left[1+2\sum_{k=1}^{m-1}\cos2k\theta\right]+\left[2\sum_{k=0}^{m-1}\cos(2k+1)\theta\right]z\\
z^{2m+1}
&\displaystyle=-\left[2\sum_{k=0}^{m-1}\cos(2k+1)\theta\right]+\left[1+2\sum_{k=1}^{m}\cos2k\theta\right]z\\
\end{array}$$