座標平面上に, ベクトル \(\overrightarrow{u _ 1} , \overrightarrow{u _ 2} , \cdots , \overrightarrow{u _ n}\) がある. 座標平面上のベクトル \(\overrightarrow{p}\) のうち, \(\textstyle\sum\limits _ {k=1}^n k \left| \overrightarrow{p} -\overrightarrow{u _ k} \right|^2\) を最小にするものを \(\overrightarrow{v}\) とし, そのときの最小値を \(m\) とする. 次の問いに答えよ.

1. (1)　\(\overrightarrow{v}\) を \(\overrightarrow{u _ 1} , \overrightarrow{u _ 2} , \cdots , \overrightarrow{u _ n}\) を用いて表せ.

　以下, \(\overrightarrow{u _ k} = \left( \cos \dfrac{k \alpha}{n} , \sin \dfrac{k \alpha}{n} \right) \ ( k = 1, 2, \cdots , n )\) の場合を考える, ただし, \(\alpha\) は正の定数とする.

2. (2)　\(\displaystyle\lim _ {n \rightarrow \infty} \left| \overrightarrow{v} \right|^2\) を求めよ.

3. (3)　\(\displaystyle\lim _ {n \rightarrow \infty} \dfrac{m}{n^2}\) を求めよ.

【 解 答 】

(1)

\[\begin{align} \textstyle\sum\limits _ {k=1}^n k \left| \overrightarrow{p} -\overrightarrow{u _ k} \right|^2 & = \textstyle\sum\limits _ {k=1}^n k \left( \left| \overrightarrow{p} \right|^2 -2 \overrightarrow{p} \cdot \overrightarrow{u _ k} + \left| \overrightarrow{u _ k} \right|^2 \right) \\ & = \dfrac{n(n+1)}{2} \left| \overrightarrow{p} \right|^2 -2 \left( \textstyle\sum\limits _ {k=1}^n k \overrightarrow{u _ k} \right) \cdot \overrightarrow{p} +\textstyle\sum\limits _ {k=1}^n k \left| \overrightarrow{u _ k} \right|^2 \\ & = \dfrac{n(n+1)}{2} \left| \overrightarrow{p} -\dfrac{2}{n(n+1)} \textstyle\sum\limits _ {k=1}^n k \overrightarrow{u _ k} \right|^2 \\ & \qquad +\textstyle\sum\limits _ {k=1}^n k \left| \overrightarrow{u _ k} \right|^2 -\dfrac{2}{n(n+1)} \left| \textstyle\sum\limits _ {k=1}^n k \overrightarrow{u _ k} \right|^2 \quad ... [1] \end{align}\] よって \[ \overrightarrow{v} = \underline{\dfrac{2}{n(n+1)} \textstyle\sum\limits _ {k=1}^n k \overrightarrow{u _ k}} \]

(2)

(1) の結果と条件より \[\begin{align} \left| \overrightarrow{v} \right|^2 & = \dfrac{4}{n^2 (n+1)^2} \left\{ \left( \textstyle\sum\limits _ {k=1}^n k \cos \dfrac{k \alpha}{n} \right)^2 +\left( \textstyle\sum\limits _ {k=1}^n k \sin \dfrac{k \alpha}{n} \right)^2 \right\} \\ & = \underline{\dfrac{4}{\left( 1 +\frac{1}{n} \right)^2}} _ {[2]} \left\{ \left( \underline{\dfrac{1}{n} \textstyle\sum\limits _ {k=1}^n \dfrac{k}{n} \cos \dfrac{k \alpha}{n}} _ {[3]} \right)^2 +\left( \underline{\dfrac{1}{n} \textstyle\sum\limits _ {k=1}^n \dfrac{k}{n} \sin \dfrac{k \alpha}{n}} _ {[4]} \right)^2 \right\} \end{align}\] ここで, [2] ～ [4] について, \(n \rightarrow \infty\) とすれば \[\begin{align} [2] & \rightarrow 4 , \\ [3] & \rightarrow \displaystyle\int _ 0^1 x \cos \alpha x \, dx \\ & = \left[ \dfrac{x \sin \alpha x}{\alpha} \right] _ 0^1 -\dfrac{1}{\alpha} \displaystyle\int _ 0^1 \sin \alpha x \, dx \\ & = \dfrac{\sin \alpha}{\alpha} +\dfrac{1}{\alpha} \left[ \dfrac{\cos \alpha x}{\alpha} \right] _ 0^1 \\ & = \dfrac{\alpha \sin \alpha +\cos \alpha -1}{\alpha^2} , \\ [4] & \rightarrow \displaystyle\int _ 0^1 x \sin \alpha x \, dx \\ & = \left[ -\dfrac{x \cos \alpha x}{\alpha} \right] _ 0^1 +\dfrac{1}{\alpha} \displaystyle\int _ 0^1 \cos \alpha x \, dx \\ & = -\dfrac{\cos \alpha}{\alpha} +\dfrac{1}{\alpha} \left[ \dfrac{\sin \alpha x}{\alpha} \right] _ 0^1 \\ & = \dfrac{-\alpha \cos \alpha +\sin \alpha}{\alpha^2} \end{align}\] よって \[\begin{align} \displaystyle\lim _ {n \rightarrow \infty} \left| \overrightarrow{v} \right|^2 & = 4 \left\{ \left( \dfrac{\alpha \sin \alpha +\cos \alpha -1}{\alpha^2} \right)^2 +\left( \dfrac{-\alpha \cos \alpha +\sin \alpha}{\alpha^2} \right)^2 \right\} \\ & = \underline{\dfrac{4}{\alpha^4} \left( \alpha^2 -2 \alpha \sin \alpha -2 \cos \alpha +2 \right)} \end{align}\]

(3)

(1) の結果より \[\begin{align} \left| \overrightarrow{v} \right| & = \dfrac{2}{n(n+1)} \left| \textstyle\sum\limits _ {k=1}^n k \overrightarrow{u _ k} \right| \\ \text{∴} \quad \left| \textstyle\sum\limits _ {k=1}^n k \overrightarrow{u _ k} \right| & = \dfrac{n(n+1)}{2} \left| \overrightarrow{v} \right| \end{align}\] これを用いれば, [1] より \[\begin{align} m & = \textstyle\sum\limits _ {k=1}^n k \left| \overrightarrow{u _ k} \right|^2 -\dfrac{2}{n(n+1)} \left| \textstyle\sum\limits _ {k=1}^n k \overrightarrow{u _ k} \right|^2 \\ & = \dfrac{n(n+1)}{2} \left( 1 -\left| \overrightarrow{v} \right|^2 \right) \end{align}\] よって, (2) の結果も用いて \[\begin{align} \displaystyle\lim _ {n \rightarrow \infty} \dfrac{m}{n^2} & = \displaystyle\lim _ {n \rightarrow \infty} \dfrac{1 +\frac{1}{n}}{2} \left( 1 -\left| \overrightarrow{v} \right|^2 \right) \\ & = \underline{\dfrac{\alpha^4 -4 \alpha^2 +8 \alpha \sin \alpha +8 \cos \alpha -8}{2 \alpha^4}} \end{align}\]