行列 \(A = \dfrac{1}{2} \left( \begin{array}{cc} 0 & -1 \\ 1 & -1 \end{array} \right)\) に対して, 座標空間の点 \(\text{P} _ n\) の座標 \(( a _ n , b _ n , c _ n ) \ ( n = 1, 2, 3, \cdots )\) を, \(( a _ 1 , b _ 1 , c _ 1 ) = ( 1, 0, 0 )\) . \[ \left( \begin{array}{c} a _ {n+1} \\ b _ {n+1} \end{array} \right) = A \left( \begin{array}{c} a _ n \\ b _ n \end{array} \right) , \ c _ {n+1} = c _ n +\sqrt{a _ n b _ n} \quad ( n= 1, 2, 3, \cdots ) \] で定める.

1. (1)　\(A^3\) を求めよ.

2. (2)　点 \(\text{P} _ 2 , \text{P} _ 3 , \text{P} _ 4\) の座標を求めよ.

3. (3)　点 \(\text{P} _ n\) の座標を求めよ.

【 解 答 】

(1)

\[\begin{align} A^2 & = \dfrac{1}{4} \left( \begin{array}{cc} 0 & -1 \\ 1 & -1 \end{array} \right) \left( \begin{array}{cc} 0 & -1 \\ 1 & -1 \end{array} \right) = \dfrac{1}{4} \left( \begin{array}{cc} -1 & 1 \\ -1 & 0 \end{array} \right) , \\ A^3 & = \dfrac{1}{8} \left( \begin{array}{cc} -1 & 1 \\ -1 & 0 \end{array} \right) \left( \begin{array}{cc} 0 & -1 \\ 1 & -1 \end{array} \right) = \underline{\dfrac{1}{8} \left( \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right)} \end{align}\]

(2)

\[\begin{align} \left( \begin{array}{c} a _ 2 \\ b _ 2 \end{array} \right) & = A \left( \begin{array}{c} a _ 1 \\ b _ 1 \end{array} \right) \\ & = \dfrac{1}{2} \left( \begin{array}{cc} 0 & -1 \\ 1 & -1 \end{array} \right) \left( \begin{array}{c} 1 \\ 0 \end{array} \right) = \dfrac{1}{2} \left( \begin{array}{c} 0 \\ 1 \end{array} \right) , \\ c _ 2 & = c _ 1 +\sqrt{a _ 1 b _ 1} = 0 \end{align}\] よって, \(\text{P} _ 2 \ \underline{\left( 0 , \dfrac{1}{2} , 0 \right)}\) . \[\begin{align} \left( \begin{array}{c} a _ 3 \\ b _ 3 \end{array} \right) & = A^2 \left( \begin{array}{c} a _ 1 \\ b _ 1 \end{array} \right) \\ & = \dfrac{1}{4} \left( \begin{array}{cc} -1 & 1 \\ -1 & 0 \end{array} \right) \left( \begin{array}{c} 1 \\ 0 \end{array} \right) = \dfrac{1}{4} \left( \begin{array}{c} -1 \\ -1 \end{array} \right) , \\ c _ 3 & = c _ 2 +\sqrt{a _ 2 b _ 2} = 0 \end{align}\] よって, \(\text{P} _ 3 \ \underline{\left( -\dfrac{1}{4} , -\dfrac{1}{4} , 0 \right)}\) . \[\begin{align} \left( \begin{array}{c} a _ 4 \\ b _ 4 \end{array} \right) & = A^3 \left( \begin{array}{c} a _ 1 \\ b _ 1 \end{array} \right) \\ & = \dfrac{1}{8} \left( \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right) \left( \begin{array}{c} 1 \\ 0 \end{array} \right) = \dfrac{1}{8} \left( \begin{array}{c} 1 \\ 0 \end{array} \right) , \\ c _ 4 & = c _ 3 +\sqrt{a _ 3 b _ 3} = \dfrac{1}{4} \end{align}\] よって, \(\text{P} _ 4 \ \underline{\left( \dfrac{1}{8} , 0 , \dfrac{1}{4} \right)}\) .

(3)

(1) の結果より, \(A^3 = \dfrac{1}{8} E\) なので

1. 1*　\(n = 3m-2 \ ( m = 1, 2, \cdots )\) のとき \[ \left( \begin{array}{c} a _ {3m-2} \\ b _ {3m-2} \end{array} \right) =\dfrac{1}{8^{m-1}} \left( \begin{array}{c} a _ 1 \\ b _ 1 \end{array} \right) =\dfrac{1}{8^{m-1}} \left( \begin{array}{c} 1 \\ 0 \end{array} \right) \]

2. 2*　\(n = 3m-1 \ (m = 0, 1, 2, \cdots )\) のとき \[ \left( \begin{array}{c} a _ {3m-1} \\ b _ {3m-1} \end{array} \right) =\dfrac{1}{8^{m-1}} \left( \begin{array}{c} a _ 2 \\ b _ 2 \end{array} \right) =\dfrac{1}{8^{m-1}} \left( \begin{array}{c} 0 \\ \dfrac{1}{2} \end{array} \right) \]

3. 3*　\(n = 3m \ ( m=1, 2, \cdots )\) のとき \[ \left( \begin{array}{c} a _ {3m} \\ b _ {3m} \end{array} \right) =\dfrac{1}{8^{m-1}} \left( \begin{array}{c} a _ 3 \\ b _ 3 \end{array} \right) = -\dfrac{1}{8^{m-1}} \left( \begin{array}{c} \dfrac{1}{4} \\ \dfrac{1}{4} \end{array} \right) \]

また \[\begin{align} c _ {3m+3} & = c _ {3m+2} +\sqrt{a _ {3m+2} b _ {3m+2}} = c _ {3m+2} \\ & = c _ {3m+1} +\sqrt{a _ {3m+1} b _ {3m+1}} = c _ {3m+1} \\ & = c _ {3m} +\sqrt{a _ {3m} b _ {3m}} = c _ {3m} +\dfrac{1}{4 \cdot 8^{m-1}} \end{align}\] したがって, \(m \geqq 2\) のとき \[\begin{align} c _ {3m} & = c _ 3 +\textstyle\sum\limits _ {k=1}^{m-1} \dfrac{1}{4 \cdot 8^{k-1}} \\ & = \dfrac{1}{4} \cdot \dfrac{1 -\frac{1}{8^{m-1}}}{1 -\frac{1}{8}} \\ & = \dfrac{2}{7} \left( 1 -\dfrac{1}{8^{m-1}} \right) \end{align}\] \(c _ 3 = 0\) なので, \(m = 1\) のときも満たしている.
以上より, \(m= 1, 2, \cdots\) に対して \(\text{P} _ n\) の座標は \[ \underline{\left\{ \begin{array}{ll} \left( \dfrac{1}{8^{m-1}} , 0 , \dfrac{2}{7} \left( 1 -\dfrac{1}{8^{m-1}} \right) \right) & \left( n=3m-2 \text{のとき} \right) \\ \left( 0 , \dfrac{1}{2 \cdot 8^{m-1}} , \dfrac{2}{7} \left( 1 -\dfrac{1}{8^{m-1}} \right) \right) & \left( n=3m-1 \text{のとき} \right) \\ \left( -\dfrac{1}{4 \cdot 8^{m-1}} , -\dfrac{1}{4 \cdot 8^{m-1}} , \dfrac{2}{7} \left( 1 -\dfrac{1}{8^{m-1}} \right) \right) & \ \left( n=3m \text{のとき}\right) \end{array} \right.} \]