漸化式 - 例題
4.特殊解型
ex) a_{n+1} = 4a_n - 3, a_1 = 2 \\
\begin{array}{l} \\
特殊解\alphaを求める。\\
\alpha = 4 \alpha - 3 \\
\therefore \alpha = 1 \\
\\
a_{n+1} - 1 = 4(a_n - 1) とおける \\
a_n - 1 = 4 \cdot 4^{n-1} \\
a_n = (a_1 - 1) \cdot 4^{n-1} + 1 \\
a_n = 4^{n-1} + 1 \\
\end{array} \\
5.指数型
ex) a_{n+1} = 3a_n +4 \cdot 5^n, a_1 = 2 \\
\begin{array}{l} \\
両辺を5^{n+1}で割ると \\
\frac{a_{n+1}}{5^{n+1}} = \frac{3}{5}\frac{a_n}{5^n} + \frac{4}{5} \\
特殊解\alpha は、\alpha = \frac{3}{5}\alpha + \frac{4}{5} より、\alpha = 2 \\
\frac{a_{n+1}}{5^{n+1}} - 2 = \frac{3}{5}(\frac{a_n}{5^n} - 2) \\
\frac{a_n}{5^n} - 2 = (\frac{a_1}{5^1} - 2)(\frac{3}{5})^{n-1} \\
\frac{a_n}{5^n} - 2 = (\frac{2}{5} - 2)(\frac{3}{5})^{n-1} \\
\frac{a_n}{5^n} = - \frac{8}{5}(\frac{3}{5})^{n-1} + 2 \\
a_n = -8 \cdot 3^{n-1} + 2 \cdot 5^n
\end{array} \\
6.Ｎ次式型
ex) a_{n+1} = 2a_n + n^2-2n+3, a_1 = -3 \\
\begin{array}{l} \\
a_{n+1} + \alpha (n+1)^2 + \beta (n+1) + \gamma = 2(a_n + \alpha n^2 + \beta n + \gamma) \\
\Leftrightarrow a_{n+1} = 2a_n + \alpha n^2 + (-2\alpha + \beta)n -\alpha -\beta + \gamma \\
これを元の漸化式と比較すると\\
\left\{
\begin{array}{l}
\alpha &= 1\\
-2\alpha + \beta &= -2 \\
-\alpha -\beta + \gamma &= 3 \\
\end{array}
\right \}
\therefore \alpha = 1, \beta = 0, \gamma = 4 \\
よって\\
a_{n+1} + (n+1)^2 + 4 = 2(a_n + n^2 + 4) \\
a_n + n^2 + 4 = (a_1 + 1^2 + 4) \cdot 2^{n-1} \\
a_n + n^2 + 4 = (-3 + 1^2 + 4) \cdot 2^{n-1} \\
a_n = 2^n -n^2 - 4 \\
\end{array} \\