ていへんかけるたかさわるに

勉強している中で次のような関数を考えた.
\begin{equation*}
\DeclareMathOperator{\Ar}{Ar}
\DeclareMathOperator{\Arccos}{Arccos}
\DeclareMathOperator{\Arcsin}{Arcsin}
\DeclareMathOperator{\Arr}{Arr}
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\varphi(x,y) = (\cos(2\arccos(x)), \sin(2\arcsin(y))).
\end{equation*} この関数は単位円
\begin{equation*}
S^1 = \left\{ (x,y) \mid x^2+y^2=1 \right\} = \left\{ (\cos\theta, \sin\theta) \mid 0 \le \theta < 2\pi \right\}
\end{equation*} からそれ自身への関数で, 単位円上を点 $(x,y)$ が一周するときに点 $\varphi(x,y)$ が単位円上を 2 周する.
\begin{align}
\varphi(\cos\theta, \sin\theta)
& = (\cos(2\arccos(\cos\theta)), \sin(2\arcsin(\sin\theta))) \label{phi(cos(theta),sin(theta))}\tag{1} \\
& = (\cos(2\theta), \sin(2\theta)). \notag
\end{align} ただし逆三角関数 $\arccos$, $\arcsin$ は多価関数なので式 (\ref{phi(cos(theta),sin(theta))}) には曖昧さがある. これをきちんと計算してみる.

まず, $\Arccos$, $\Arcsin$ をそれぞれ閉区間 $[0,\pi]$, $\left[-\frac{1}{2}\pi,\frac{1}{2}\pi\right]$ に主値をとる逆三角関数とする. つまり $\Arccos, \Arcsin : [-1,1] \rightarrow \Real$ で,
\begin{equation*}
0 \le \Arccos(x) \le \pi, \quad
-\frac{1}{2}\pi \le \Arcsin(x) \le \frac{1}{2}\pi, \quad
(-1 \le x \le 1).
\end{equation*} が成り立つ.

$\Arccos$, $\Arcsin$ の値については, $(x,y) = (\cos\theta,\sin\theta)$ と表わし $n$ を任意の整数としたとき,
\begin{align}
\Arccos(\cos\theta) &=
\begin{cases}
\theta - n\pi &
(n\pi \le \theta < (n+1)\pi; \qq\text{$n$ は偶数}), \\
(n+1)\pi - \theta &
(n\pi \le \theta < (n+1)\pi; \qq\text{$n$ は奇数}),
\end{cases}
\\
& \notag \\
\Arcsin(\sin\theta) &=
\begin{cases}
\theta - n\pi &
\left(\frac{2n-1}{2}\pi \le \theta < \frac{2n+1}{2}\pi;
\qq\text{$n$ は偶数}\right), \\
n\pi - \theta &
\left(\frac{2n-1}{2}\pi \le \theta < \frac{2n+1}{2}\pi;
\qq\text{$n$ は奇数}\right).
\end{cases}
\label{ex:Arcsin-value}
\end{align} となっている. 具体的に $n=0,1,2,3,4$ について $\Arccos$, $\Arcsin$ の値を書き下してみると以下のようになる.
\begin{align}
\Arccos(\cos\theta) & =
\begin{cases}
\theta & (0 \le \theta < \pi; n=0), \\
2\pi - \theta & (\pi \le \theta < 2\pi; n=1), \\
\theta - 2\pi & (2\pi \le \theta < 3\pi; n=2), \\
4\pi - \theta & (3\pi \le \theta < 4\pi; n=3), \\
\theta - 4\pi & (4\pi \le \theta < 5\pi; n=4).
\end{cases}
\label{ex:Arccos-value2}\tag{2}
\\
& \notag \\
\Arcsin(\sin\theta) & =
\begin{cases}
\theta
& \left(-\frac{1}{2}\pi \le \theta < \frac{1}{2}\pi; n=0\right), \\
\pi - \theta
& \left(\frac{1}{2}\pi \le \theta < \frac{3}{2}\pi; n=1\right), \\
\theta - 2\pi
& \left(\frac{3}{2}\pi \le \theta < \frac{5}{2}\pi; n=2\right), \\
3\pi - \theta
& \left(\frac{5}{2}\pi \le \theta < \frac{7}{2}\pi; n=3\right), \\
\theta - 4\pi
& \left(\frac{7}{2}\pi \le \theta < \frac{9}{2}\pi; n=4\right).
\end{cases}
\label{ex:Arcsin-value2}\tag{3}
\end{align}

(\ref{ex:Arccos-value2}), (\ref{ex:Arcsin-value2}) に基いて関数 $A_1, A_2 : S^1 \rightarrow [0,2\pi)$ を
\begin{align}
A_1(\cos\theta,\sin\theta) & =
\begin{cases}
\Arccos(\cos\theta) & (0 \le \theta < \pi), \\
2\pi - \Arccos(\cos\theta) & (\pi \le \theta < 2\pi),
\end{cases}
\label{def:A_1}\tag{4} \\
& \notag \\
A_2(\cos\theta,\sin\theta) & =
\begin{cases}
\Arcsin(\sin\theta) & \left(0 \le \theta < \frac{1}{2}\pi \right), \\
\pi - \Arcsin(\sin\theta) & \left(\frac{1}{2}\pi \le \theta < \frac{3}{2}\pi \right), \\
\Arcsin(\sin\theta) + 2\pi & \left(\frac{3}{2}\pi \le \theta < 2\pi \right).
\end{cases}
\label{def:A_2}\tag{5}
\end{align} と定義する. こうすると
\begin{equation}
\label{ex:Arccos,Arcsin}\tag{6}
A_1(\cos\theta) = \theta, \quad A_2(\sin\theta) = \theta
\end{equation} となり, $\theta$ を $0$ から $2\pi$ に変化させたときに, それに連れて $A_1$, $A_2$ も $0$ から $2\pi$ まで同様に変化する.

ここで, あらためて関数 $\varphi : S^1 \rightarrow S^1$ を
\begin{equation*}
\varphi(x,y) = (\cos(2 A_1(x,y), \sin(2 A_2(x,y))
\end{equation*} と定義する.

$A_1$, $A_2$ の上記の性質 (\ref{ex:Arccos,Arcsin}) から, $\theta$ を $0$ から $2\pi$ まで変化させたとき, $\varphi(\cos\theta,\sin\theta)$ は単位円上を 2 周する. すなわち
\begin{equation*}
\varphi(\cos\theta,\sin\theta) = \varphi(\cos(\theta+\pi),\sin(\theta+\pi))
\end{equation*} が成り立つ.

式 (\ref{phi(cos(theta),sin(theta))}) は, 区間 $[0,2\pi)$ を 4 つの区間
\begin{equation*}
\textstyle \big[0,\frac{1}{2}\pi \big), \big[\frac{1}{2}\pi,\pi \big), \big[\pi,\frac{3}{2}\pi \big), \big[\frac{3}{2}\pi,2\pi \big),
\end{equation*} に分けて, 各区間について $A_1$, $A_2$ の定義を用いて計算すれば導ける. たとえば, $\frac{3}{2}\pi \le \theta < 2\pi$ のとき,
\begin{equation*}
\frac{5}{2}\pi \le \theta + \pi < 3\pi
\end{equation*} である. (\ref{ex:Arccos-value2}), (\ref{ex:Arcsin-value2}), (\ref{def:A_1}), (\ref{def:A_2}) から
\begin{align*}
A_1(\cos\theta, \sin\theta)
& = 2\pi - \Arccos(\cos\theta)
= 2\pi - (2\pi - \theta)
= \theta, \\
A_2(\cos\theta, \sin\theta)
& = \Arcsin(\sin\theta) + 2\pi
= (\theta - 2\pi) + 2\pi
= \theta, \\
A_1(\cos(\theta + \pi), \sin(\theta + \pi))
& = \Arccos(\cos(\theta + \pi)) \\
& = (\theta + \pi) - 2\pi \\
& = \theta - \pi, \\
A_2(\cos(\theta + \pi), \sin(\theta + \pi))
& = \pi - \Arcsin(\sin(\theta + \pi)) \\
& = \pi - (3\pi - (\theta + \pi)) \\
& = \theta - \pi.
\end{align*} よって
\begin{align*}
\varphi(\cos\theta, \sin\theta)
& = (\cos(2 A_1(\cos\theta,\sin\theta)), \sin(2 A_2(\cos\theta,\sin\theta))) \\
& = (\cos(2\theta), \sin(2\theta)). \\
\varphi(\cos(\theta+\pi), \sin(\theta+\pi))
& = (\cos(2 A_1(\cos(\theta+\pi),\sin(\theta+\pi))), \\
& \hspace{4em} \sin(2 A_2(\cos(\theta+\pi),\sin(\theta+\pi)))) \\
& = (\cos(2(\theta - \pi)), \sin(2(\theta - \pi))) \\
& = (\cos(2\theta), \sin(2\theta))).
\end{align*} となり, (\ref{phi(cos(theta),sin(theta))}) が成り立っている.