問題2の各値を求める問題です。 1. $\Gamma(6)$

1. 問題の内容

問題2の各値を求める問題です。

1. $\Gamma(6)$

2. $\frac{\Gamma(6)}{\Gamma(3)}$

3. $\frac{\Gamma(3.5)}{\Gamma(1.5)}$

4. $\frac{\Gamma(4)\Gamma(3.5)}{\Gamma(5.5)}$

5. $B(\frac{3}{2}, 2)$

2. 解き方の手順

1. $\Gamma(6)$

$\Gamma(n+1) = n\Gamma(n)$ の公式を利用します。

$\Gamma(6) = 5\Gamma(5) = 5 \cdot 4\Gamma(4) = 5 \cdot 4 \cdot 3\Gamma(3) = 5 \cdot 4 \cdot 3 \cdot 2\Gamma(2) = 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1\Gamma(1) = 5! = 120$

2. $\frac{\Gamma(6)}{\Gamma(3)}$

$\Gamma(6) = 120$ であることと、$\Gamma(3) = 2! = 2$ であることから、$\frac{\Gamma(6)}{\Gamma(3)} = \frac{120}{2} = 60$

3. $\frac{\Gamma(3.5)}{\Gamma(1.5)}$

$\Gamma(n+1) = n\Gamma(n)$ を利用して、

$\Gamma(3.5) = 2.5\Gamma(2.5) = 2.5 \cdot 1.5\Gamma(1.5)$

$\frac{\Gamma(3.5)}{\Gamma(1.5)} = \frac{2.5 \cdot 1.5\Gamma(1.5)}{\Gamma(1.5)} = 2.5 \cdot 1.5 = \frac{5}{2} \cdot \frac{3}{2} = \frac{15}{4} = 3.75$

しかし、画像では異なる計算をしているようです。

$\Gamma(3.5) = \frac{5}{2} \Gamma(\frac{5}{2}) = \frac{5}{2} \frac{3}{2} \Gamma(\frac{3}{2})$

$\Gamma(1.5) = \frac{1}{2} \Gamma(\frac{1}{2})$

$\frac{\Gamma(3.5)}{\Gamma(1.5)} = \frac{\frac{5}{2} \frac{3}{2} \Gamma(\frac{3}{2})}{\frac{1}{2} \Gamma(\frac{1}{2})} = \frac{15}{2} \frac{\Gamma(\frac{3}{2})}{\Gamma(\frac{1}{2})}$

$\Gamma(\frac{3}{2}) = \frac{1}{2} \Gamma(\frac{1}{2})$

$\frac{\Gamma(3.5)}{\Gamma(1.5)} = \frac{15}{4}$

4. $\frac{\Gamma(4)\Gamma(3.5)}{\Gamma(5.5)}$

$\Gamma(4) = 3! = 6$

$\Gamma(3.5) = \frac{5}{2} \Gamma(\frac{5}{2}) = \frac{5}{2} \frac{3}{2} \Gamma(\frac{3}{2}) = \frac{15}{4} \Gamma(\frac{3}{2}) = \frac{15}{4} \frac{\sqrt{\pi}}{2}$

$\Gamma(5.5) = \frac{945}{32} \sqrt{\pi}$

$\frac{\Gamma(4) \Gamma(3.5)}{\Gamma(5.5)} = \frac{6 \frac{15}{4} \frac{\sqrt{\pi}}{2}}{\frac{945}{32} \sqrt{\pi}} = \frac{6 \cdot 15 \cdot 32}{4 \cdot 2 \cdot 945} = \frac{2880}{7560} = \frac{4}{10.5} = \frac{8}{21}$

5. $B(\frac{3}{2}, 2)$

$B(x,y) = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$ の公式を利用します。

$B(\frac{3}{2}, 2) = \frac{\Gamma(\frac{3}{2})\Gamma(2)}{\Gamma(\frac{3}{2}+2)} = \frac{\Gamma(\frac{3}{2})\Gamma(2)}{\Gamma(\frac{7}{2})} = \frac{\frac{1}{2}\Gamma(\frac{1}{2}) \cdot 1!}{\frac{5}{2}\Gamma(\frac{5}{2})} = \frac{\frac{1}{2}\sqrt{\pi}}{\frac{5}{2} \frac{3}{2}\frac{1}{2}\sqrt{\pi}} = \frac{\frac{1}{2}}{\frac{15}{8}} = \frac{1}{2} \cdot \frac{8}{15} = \frac{4}{15}$

3. 最終的な答え

1. $\Gamma(6) = 120$

2. $\frac{\Gamma(6)}{\Gamma(3)} = 60$

3. $\frac{\Gamma(3.5)}{\Gamma(1.5)} = \frac{15}{4}$

4. $\frac{\Gamma(4)\Gamma(3.5)}{\Gamma(5.5)} = \frac{8}{21}$

5. $B(\frac{3}{2}, 2) = \frac{4}{15}$