(1) \(n\) を \(2\) 以上の自然数とするとき, 関数 \[ f _ n ( \theta ) = ( 1 +\cos \theta ) \sin^{n-1} \theta \] の \(0 \leqq \theta \leqq \dfrac{\pi}{2}\) における最大値 \(M _ n\) を求めよ.
(2) \(\displaystyle\lim _ {n \rightarrow \infty} \left( M _ n \right)^n\) を求めよ.
【 解 答 】
(1)
\[\begin{align} f'( \theta ) & = -\sin^n +( 1 +\cos \theta ) (n-1) \sin^{n-2} \theta \cos \theta \\ & = \sin^{n-2} \theta \left\{ \cos^2 \theta -1 +(n-1) \cos \theta ( 1 +\cos \theta ) \right\} \\ & = \sin^{n-2} \theta \left\{ \cos^2 \theta +(n-1) \cos \theta -1 \right\} \\ & = \sin^{n-2} \theta ( n \cos \theta -1 ) ( \cos \theta +1 ) \end{align}\] \(f'( \theta ) = 0\) をとくと \[ \theta = 0 , \ \cos \theta = \dfrac{1}{n} \] \(\cos \theta _ n = \dfrac{1}{n} \ \left( 0 \leqq \theta _ n \leqq \dfrac{\pi}{2} \right)\) とおけば, \(f( \theta )\) の増減は下表のとおり. \[ \begin{array}{c|ccccc} \theta & 0 & \cdots & \theta _ n & \cdots & \dfrac{\pi}{2} \\ \hline f'( \theta ) & 0 & + & 0 & - & \\ \hline f( \theta ) & & \nearrow & \text{最大} & \searrow & \\ \end{array} \] よって \[\begin{align} M _ n & = f( \theta _ n ) \\ & = \underline{\left( 1 +\dfrac{1}{n} \right) \left( \sqrt{1 -\dfrac{1}{n^2}} \right)^{n-1}} \\ \end{align}\]
(2)
\[\begin{align} \left( M _ n \right)^n & = \left( 1 +\dfrac{1}{n} \right)^n \left( 1 -\dfrac{1}{n^2} \right)^{\frac{n^2 -n}{2}} \\ & = \left( 1 +\dfrac{1}{n} \right)^n \left( 1 -\dfrac{1}{n^2} \right)^{-\frac{1}{2} (-n^2)} \left( 1 +\dfrac{1}{n} \right)^{-\frac{1}{2} n} \left( 1 -\dfrac{1}{n} \right)^{\frac{1}{2} (-n)} \\ & \rightarrow e \cdot e^{-\frac{1}{2}} \cdot e^{-\frac{1}{2}} \cdot e^{\frac{1}{2}} \quad ( \ n \rightarrow \infty \ \text{のとき} ) \\ & = e^{\frac{1}{2}} \end{align}\] よって \[ \displaystyle\lim _ {n \rightarrow \infty} \left( M _ n \right)^n = \underline{e^{\frac{1}{2}}} \]