漸化式 - 例題３

## 7.階比型

```math
\require{cancel}
ex)　a_{n+1} = \frac{n+1}{n}a_n, a_1 = -3 \\
\begin{array}{l} \\
n \ge 2 のとき \\
\begin{array}{l} \\
a_n &= \frac{n}{n-1}a_{n-1} \\
&= \frac{n}{n-1}\frac{n-1}{n-2}a_{n-2} \\
&= \frac{n}{n-1}\frac{n-1}{n-2}\frac{n-2}{n-3}a_{n-3} \\
&= \vdots \\
&= \frac{n}{\cancel{n-1}}\frac{\cancel{n-1}}{\cancel{n-2}}\frac{\cancel{n-2}}{\cancel{n-3}} \cdots \frac{\cancel{2}}{1} a_1 \\
&= n a_1 \\
&= 3n \\
\end{array} \\
これは n = 1でも成り立つ \\
a_n = 3n \\
\end{array} \\
```

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## 8.次数相異型

```math
ex)　a_{n+1} = 3 \sqrt{a_n}, a_1 = 27 \\
\\
漸化式の形と初項の値から、すべての自然数 n についてa_n \gt 0 であるとわかる。 \\
\begin{array}{l} \\
a_{n+1} = 3 a_n^{\frac{1}{2}} \\
\log_3{a_{n+1}} = \log_3{3a_n^{\frac{1}{2}}} \\
\log_3{a_{n+1}} = \frac{1}{2}(\log_3{3a_n)} \\
\log_3{a_{n+1}} = \frac{1}{2}(\log_3{3} + \log_3{a_n}) \\
\log_3{a_{n+1}} = \frac{1}{2} + \frac{1}{2}(\log_3{a_n}) \Rightarrow 特殊解型 \\
\\
特殊解\alphaは、\alpha = \frac{1}{2} \alpha + 1 \rightarrow \alpha = 2 \\
\log_3{a_{n+1}} - 2 = \frac{1}{2}(\log_3{a_n} - 2) \\
\log_3{a_n} - 2 = (\log_3{a_1} - 2) (\frac{1}{2})^{n-1} \\
\log_3{a_n} - 2 = (\log_3{27} - 2) (\frac{1}{2})^{n-1} \\
\log_3{a_n} - 2 = (\frac{1}{2})^{n-1} \\
\log_3{a_n} = (\frac{1}{2})^{n-1} + 2 \\
a_n = 3^{(\frac{1}{2})^{n-1} + 2} \\
\end{array} \\
```

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## 9.分数型

```math
ex)　(q = 0 型) \\
a_{n+1} = \frac{3 a_n}{a_n + 2}, a_1 = \frac{1}{2} \\
\\
a_{n+1} = 0 だと仮定する。このときa_n = 0 となるが、 \\
これは a_{n+1} = a_n = a_{n-1} = \cdots = a_1 = 0 を意味し、a_1 = \frac{1}{2} に矛盾する。\\
よってすべての自然数nについてa_n \ne 0 \\
a_n \ne 0 より、逆数を取ると、\\
\begin{array}{l} \\

\frac{1}{a_{n+1}} = \frac{a_n + 2}{3a_n} \\
\frac{1}{a_{n+1}} = \frac{2}{3}\frac{1}{a_n} + \frac{1}{3} （特殊解型）\\
\\
特殊解\alphaは、\alpha = \frac{2}{3} \alpha + \frac{1}{3} \rightarrow \alpha = 1 \\
(\frac{1}{a_{n+1}} - 1) = \frac{2}{3}(\frac{1}{a_n} - 1)\\
\frac{1}{a_n} - 1 = (\frac{1}{a_1} - 1)(\frac{2}{3})^{n-1} \\
\frac{1}{a_n} - 1 = (\frac{2}{3})^{n-1} \\
\frac{1}{a_n} = (\frac{2}{3})^{n-1} + 1 \\
a_n = \frac{1}{(\frac{2}{3})^{n-1} + 1} \\
\end{array} \\
```

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```math
ex)　(q \ne 0 型) \\
a_{n+1} = \frac{4 a_n - 6}{a_n - 1}, a_1 = 5 \\
\\
\begin{array}{l} \\
特殊解\alpha は、\alpha = \frac{4 \alpha - 6}{\alpha - 1} \rightarrow \alpha^2 -5\alpha + 6 = 0 \\
(\alpha - 2)(\alpha - 3) = 0 よって \alpha = 2, 3 \\
\\
(a_{n+1} - 2) = \frac{4 a_n - 6}{a_n - 1} - 2 \\
= \frac{4 a_n - 6 - 2(a_n - 1)}{a_n - 1} \\
= \frac{2 a_n - 4}{a_n - 1} \\
= \frac{2(a_n - 2)}{(a_n - 2) + 1} \\
\\
a_{n+1} = 2 と仮定する。このときa_n = 2となるが、\\
これはa_{n+1} = a_n = \cdots = a_1 = 2 を意味し、a_1 = 5 と矛盾。\\
よってすべての自然数nについてa_n \ne = 2 \\
a_n \ne 2 より逆数をとると \\
\frac{1}{(a_{n+1} - 2)} = \frac{(a_n - 2) + 1}{2(a_n - 2)} \\
= \frac{1}{2}\frac{1}{a_n - 2} + \frac{1}{2} （特殊解型）\\
\\
特殊解\alphaは、\alpha = \frac{1}{2}\alpha + \frac{1}{2} \rightarrow \alpha = 1 \\
(\frac{1}{(a_{n+1} - 2)} - 1) = \frac{1}{2}(\frac{1}{a_n - 2} - 1) \\
\frac{1}{a_n - 2} - 1 = (\frac{1}{a_1 - 2} - 1)(\frac{1}{2})^{n-1} \\
\frac{1}{a_n - 2} - 1 = -\frac{2}{3}(\frac{1}{2})^{n-1} \\
\frac{1}{a_n - 2} = -\frac{2}{3}(\frac{1}{2})^{n-1} + 1 \\
a_n - 2 = \frac{1}{1 -\frac{2}{3}(\frac{1}{2})^{n-1}} \\
a_n = \frac{1}{1 -\frac{2}{3}(\frac{1}{2})^{n-1}} + 2 \\
\end{array} \\
```

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- $ 特殊解で q = 0 パターンに変形できる理由 $

```math
a_{n+1} - \alpha = \frac{p a_n + q}{ r a_n + s} - \alpha \\
a_{n+1} - \alpha = \frac{(p a_n + q) - \alpha(r a_n + s)}{ r (a_n - \alpha) + s + r \alpha} \\
a_{n+1} - \alpha = \frac{(p -\alpha r)a_n + q - \alpha s}{ r (a_n - \alpha) + s + r \alpha} \\
a_{n+1} - \alpha = \frac{(p -\alpha r)(a_n - \alpha) +\alpha(p - \alpha r) + (q - \alpha s)}{ r (a_n - \alpha) + s + r \alpha} \\
\\
ここで、\alpha(p - \alpha r) + (q - \alpha s) = 0 であればいい。\\
p \alpha + q = \alpha(r \alpha + s) \\
\therefore \alpha = \frac{p \alpha + q}{r \alpha + s} \\
つまり、\alpha = \frac{p \alpha + q}{r \alpha + s} とおいた特殊解\alphaを使用すれば、\\
q = 0の形式に変形できる。\\
```