連分数の収束子と漸化式
連分数の収束子と漸化式
数列\(\left(a_{n}\right)_{n\in\mathbb{N}_{0}},\left(b_{n}\right)_{n\in\mathbb{N}_{0}}\)があるとき、連分数\(\left[\left(a_{0},b_{0}\right);\left(a_{1},b_{1}\right),\left(a_{2},b_{2}\right),\cdots,a_{n}\right]\)について次が成り立つ。
\[ \begin{cases} p_{-1}=1\\ p_{0}=a_{0}\\ p_{n}=a_{n}p_{n-1}+b_{n-1}p_{n-2} & n\in\mathbb{N} \end{cases} \] \[ \begin{cases} q_{-1}=0\\ q_{0}=1\\ q_{n}=a_{n}q_{n-1}+b_{n-1}q_{n-2} & n\in\mathbb{N} \end{cases} \] である。
行列で表すと、
\[ \left(\begin{array}{c} p_{n}\\ q_{n} \end{array}\right)=\left(\begin{array}{cc} p_{n-1} & p_{n-2}\\ q_{n-1} & q_{n-2} \end{array}\right)\left(\begin{array}{c} a_{n}\\ b_{n-1} \end{array}\right) \] となる。
この\(\frac{p_{n}}{q_{n}}\)を収束子という。
\(\forall n\in\mathbb{N}_{0},b_{n}=r\)のときは、
\[ \left[a_{0};a_{1},a_{2},\cdots,a_{n}\right]_{r}=\frac{p_{n}}{q_{n}} \] となる。
\[ \frac{p_{n}}{q_{n}}-\frac{p_{n-1}}{q_{n-1}}=\frac{\left(-1\right)^{n+1}r^{n}}{q_{n}q_{n-1}} \] となる。
\[ \frac{p_{n}}{q_{n}}=a_{0}+\sum_{k=1}^{n}\frac{\left(-1\right)^{k+1}r^{k}}{q_{k}q_{k-1}} \] となる。
\[ \left[\left(a_{0},b_{0}\right);\left(a_{1},b_{1}\right),\cdots,\left(a_{k},b_{k}\right),\cdots a_{n}\right]=\left[\left(a_{k},b_{k}\right),\cdots a_{n}\right]p_{k-1}+b_{k-1}p_{k-2} \]
\[ \frac{p_{k}}{p_{k-1}}=\left[\left(a_{k},b_{k-1}\right);\left(a_{k-1}b_{k-2}\right),\cdots\left(a_{2},b_{1}\right),\left(a_{1},b_{0}\right),a_{0}\right] \]
\[ \frac{q_{k}}{q_{k-1}}=\left[\left(a_{k},b_{k-1}\right);\left(a_{k-1}b_{k-2}\right),\cdots\left(a_{2},b_{1}\right),a_{1}\right] \]
数列\(\left(a_{n}\right)_{n\in\mathbb{N}_{0}},\left(b_{n}\right)_{n\in\mathbb{N}_{0}}\)があるとき、連分数\(\left[\left(a_{0},b_{0}\right);\left(a_{1},b_{1}\right),\left(a_{2},b_{2}\right),\cdots,a_{n}\right]\)について次が成り立つ。
(1)漸化式
\[ \left[a_{0};\left(a_{1},b_{1}\right),\left(a_{2},b_{2}\right),\cdots,\left(a_{n},b_{n}\right)\right]=\frac{p_{n}}{q_{n}} \] が成り立つような数列\(p_{n},q_{n}\)は\[ \begin{cases} p_{-1}=1\\ p_{0}=a_{0}\\ p_{n}=a_{n}p_{n-1}+b_{n-1}p_{n-2} & n\in\mathbb{N} \end{cases} \] \[ \begin{cases} q_{-1}=0\\ q_{0}=1\\ q_{n}=a_{n}q_{n-1}+b_{n-1}q_{n-2} & n\in\mathbb{N} \end{cases} \] である。
行列で表すと、
\[ \left(\begin{array}{c} p_{n}\\ q_{n} \end{array}\right)=\left(\begin{array}{cc} p_{n-1} & p_{n-2}\\ q_{n-1} & q_{n-2} \end{array}\right)\left(\begin{array}{c} a_{n}\\ b_{n-1} \end{array}\right) \] となる。
この\(\frac{p_{n}}{q_{n}}\)を収束子という。
\(\forall n\in\mathbb{N}_{0},b_{n}=r\)のときは、
\[ \left[a_{0};a_{1},a_{2},\cdots,a_{n}\right]_{r}=\frac{p_{n}}{q_{n}} \] となる。
(2)隣接関係式
\[ \frac{p_{n}}{q_{n}}-\frac{p_{n-1}}{q_{n-1}}=\frac{\left(-1\right)^{n+1}}{q_{n}q_{n-1}}\prod_{k=0}^{n-1}b_{k} \] \(\forall n\in\mathbb{N}_{0},b_{n}=r\)のときは、\[ \frac{p_{n}}{q_{n}}-\frac{p_{n-1}}{q_{n-1}}=\frac{\left(-1\right)^{n+1}r^{n}}{q_{n}q_{n-1}} \] となる。
(3)和公式
\[ \frac{p_{n}}{q_{n}}=a_{0}+\sum_{k=1}^{n}\frac{\left(-1\right)^{k+1}}{q_{k}q_{k-1}}\prod_{j=0}^{k-1}b_{j} \] \(\forall n\in\mathbb{N}_{0},b_{n}=r\)のときは、\[ \frac{p_{n}}{q_{n}}=a_{0}+\sum_{k=1}^{n}\frac{\left(-1\right)^{k+1}r^{k}}{q_{k}q_{k-1}} \] となる。
(4)
\(1\leq k\leq n\)とする。\[ \left[\left(a_{0},b_{0}\right);\left(a_{1},b_{1}\right),\cdots,\left(a_{k},b_{k}\right),\cdots a_{n}\right]=\left[\left(a_{k},b_{k}\right),\cdots a_{n}\right]p_{k-1}+b_{k-1}p_{k-2} \]
(5)
\(k\in\mathbb{N}\)とする。\[ \frac{p_{k}}{p_{k-1}}=\left[\left(a_{k},b_{k-1}\right);\left(a_{k-1}b_{k-2}\right),\cdots\left(a_{2},b_{1}\right),\left(a_{1},b_{0}\right),a_{0}\right] \]
(6)
\(k\in\mathbb{N}\)とする。\[ \frac{q_{k}}{q_{k-1}}=\left[\left(a_{k},b_{k-1}\right);\left(a_{k-1}b_{k-2}\right),\cdots\left(a_{2},b_{1}\right),a_{1}\right] \]
(1)
\(n=0\)のとき、\(\frac{p_{0}}{q_{0}}=\frac{a_{0}}{1}=a_{0}\)となるので成り立つ。\(n=k\)のとき成り立つと仮定すると、
\begin{align*} \left[\left(a_{0},b_{0}\right);\left(a_{1},b_{1}\right),\left(a_{2},b_{2}\right),\cdots,a_{k+1}\right] & =\left[\left(a_{0},b_{0}\right);\left(a_{1},b_{1}\right),\left(a_{2},b_{2}\right),\cdots,\left(a_{k-1},b_{k-1}\right),\left[\left(a_{k},b_{k}\right);a_{k+1}\right]\right]\\ & =\left[\left(a_{0},b_{0}\right);\left(a_{1},b_{1}\right),\left(a_{2},b_{2}\right),\cdots,\left(a_{k-1},b_{k-1}\right),\left[\left(a_{k}+\frac{b_{k}}{a_{k+1}}\right)\right]\right]\\ & =\left[\left(a_{0},b_{0}\right);\left(a_{1},b_{1}\right),\left(a_{2},b_{2}\right),\cdots,\left(a_{k-1},b_{k-1}\right),\left[\left(\frac{a_{k}a_{k+1}+b_{k}}{a_{k+1}}\right)\right]\right]\\ & =\left[\left(a_{0},b_{0}\right);\left(a_{1},b_{1}\right),\left(a_{2},b_{2}\right),\cdots,\left(a_{k-1},b_{k-1}\right),\frac{a_{k}a_{k+1}+b_{k}}{a_{k+1}}\right]\\ & =\frac{p_{k}'}{q_{k}'}\\ & =\frac{a_{k}'p_{k-1}'+b_{k-1}'p_{k-2}'}{a_{k}'q_{k-1}'+b_{k-1}'q_{k-2}'}\\ & =\frac{a_{k}'p_{k-1}+b_{k-1}p_{k-2}}{a_{k}'q_{k-1}+b_{k-1}q_{k-2}}\\ & =\frac{\frac{a_{k}a_{k+1}+b_{k}}{a_{k+1}}p_{k-1}+b_{k-1}p_{k-2}}{\frac{a_{k}a_{k+1}+b_{k}}{a_{k+1}}q_{k-1}+b_{k-1}q_{k-2}}\\ & =\frac{\left(a_{k}a_{k+1}+b_{k}\right)p_{k-1}+a_{k+1}b_{k-1}p_{k-2}}{\left(a_{k}a_{k+1}+b_{k}\right)q_{k-1}+a_{k+1}b_{k-1}q_{k-2}}\\ & =\frac{a_{k+1}\left(a_{k}p_{k-1}+b_{k-1}p_{k-2}\right)+b_{k}p_{k-1}}{a_{k+1}\left(a_{k}q_{k-1}+b_{k-1}q_{k-2}\right)+b_{k}q_{k-1}}\\ & =\frac{a_{k+1}p_{k}+b_{k}p_{k-1}}{a_{k+1}q_{k}+b_{k}q_{k-1}}\\ & =\frac{p_{k+1}}{q_{k+1}} \end{align*} となるので\(n=k+1\)のときも成り立つ。
ここで\(p_{k}',q_{k}'\)は\(\left(a_{0},\left(a_{1},b_{1}\right),\left(a_{2},b_{2}\right),\cdots,\left(a_{k-1},b_{k-1}\right),\frac{a_{k}a_{k+1}+r}{a_{k+1}}\right)\)に対しての\(p_{k},q_{k}\)である。
故に数学的帰納法より\(n\in\mathbb{N}\)に対して題意は成り立つ。
(2)
\begin{align*} \frac{p_{n}}{q_{n}}-\frac{p_{n-1}}{q_{n-1}} & =\frac{p_{n}q_{n-1}-p_{n-1}q_{n}}{q_{n}q_{n-1}}\\ & =\frac{1}{q_{n}q_{n-1}}\left|\begin{array}{cc} p_{n} & q_{n}\\ p_{n-1} & q_{n-1} \end{array}\right|\\ & =\frac{1}{q_{n}q_{n-1}}\left|\left(\begin{array}{cc} a_{n} & b_{n-1}\\ 1 & 0 \end{array}\right)\left(\begin{array}{cc} p_{n-1} & q_{n-1}\\ p_{n-2} & q_{n-2} \end{array}\right)\right|\\ & =\frac{1}{q_{n}q_{n-1}}\left(-b_{n-1}\right)\left|\left(\begin{array}{cc} p_{n-1} & q_{n-1}\\ p_{n-2} & q_{n-2} \end{array}\right)\right|\\ & =\frac{1}{q_{n}q_{n-1}}\left|\begin{array}{cc} p_{0} & q_{0}\\ p_{-1} & q_{-1} \end{array}\right|\prod_{k=1}^{n}\frac{\left|\left(\begin{array}{cc} p_{k} & q_{k}\\ p_{k-1} & q_{k-1} \end{array}\right)\right|}{\left|\left(\begin{array}{cc} p_{k-1} & q_{k-1}\\ p_{k-2} & q_{k-2} \end{array}\right)\right|}\\ & =\frac{1}{q_{n}q_{n-1}}\left|\begin{array}{cc} a_{0} & 1\\ 1 & 0 \end{array}\right|\prod_{k=1}^{n}\left(-b_{k-1}\right)\\ & =\frac{1}{q_{n}q_{n-1}}\left(-1\right)\left(-1\right)^{n}\prod_{k=1}^{n}b_{k-1}\\ & =\frac{\left(-1\right)^{n+1}}{q_{n}q_{n-1}}\prod_{k=0}^{n-1}b_{k} \end{align*}(3)
(2)より、\begin{align*} \frac{p_{n}}{q_{n}} & =\frac{\left(-1\right)^{n+1}}{q_{n}q_{n-1}}\prod_{k=0}^{n-1}b_{k}+\frac{p_{n-1}}{q_{n-1}}\\ & =\frac{p_{0}}{q_{0}}+\sum_{k=1}^{n}\left(\frac{p_{k}}{q_{k}}-\frac{p_{k-1}}{q_{k-1}}\right)\\ & =a_{0}+\sum_{k=1}^{n}\frac{\left(-1\right)^{k+1}}{q_{k}q_{k-1}}\prod_{j=0}^{k-1}b_{j} \end{align*} となるので与式は成り立つ。
(4)
\begin{align*} \left[\left(a_{0},b_{0}\right);\left(a_{1},b_{1}\right),\cdots,\left(a_{k},b_{k}\right),\cdots a_{n}\right] & =\left[\left(a_{0},b_{0}\right);\left(a_{1},b_{1}\right),\cdots,\left[\left(a_{k},b_{k}\right),\cdots a_{n}\right]\right]\\ & =\left[\left(a_{k},b_{k}\right),\cdots a_{n}\right]p_{k-1}+b_{k-1}p_{k-2} \end{align*}(5)
\begin{align*} \frac{p_{k}}{p_{k-1}} & =\frac{a_{k}p_{k-1}+b_{k-1}p_{k-2}}{p_{k-1}}\\ & =a_{k}+b_{k-1}\frac{p_{k-2}}{p_{k-1}}\\ & =\left[\left(a_{k},b_{k-1}\right);\frac{p_{k-1}}{p_{k-2}}\right]\\ & =\left[\left(a_{k},b_{k-1}\right);\left(a_{k-1}b_{k-2}\right),\cdots\left(a_{2},b_{1}\right),\frac{p_{1}}{p_{0}}\right]\\ & =\left[\left(a_{k},b_{k-1}\right);\left(a_{k-1}b_{k-2}\right),\cdots\left(a_{2},b_{1}\right),\left(a_{1},b_{0}\right),\frac{p_{0}}{p_{-1}}\right]\\ & =\left[\left(a_{k},b_{k-1}\right);\left(a_{k-1}b_{k-2}\right),\cdots\left(a_{2},b_{1}\right),\left(a_{1},b_{0}\right),a_{0}\right] \end{align*}(6)
\begin{align*} \frac{q_{k}}{q_{k-1}} & =\frac{a_{k}q_{k-1}+b_{k-1}q_{k-2}}{q_{k-1}}\\ & =a_{k}+b_{k-1}\frac{q_{k-2}}{q_{k-1}}\\ & =\left[\left(a_{k},b_{k-1}\right);\frac{q_{k-1}}{q_{k-2}}\right]\\ & =\left[\left(a_{k},b_{k-1}\right);\left(a_{k-1}b_{k-2}\right),\cdots\left(a_{2},b_{1}\right),\frac{q_{1}}{q_{0}}\right]\\ & =\left[\left(a_{k},b_{k-1}\right);\left(a_{k-1}b_{k-2}\right),\cdots\left(a_{2},b_{1}\right),a_{1}\right] \end{align*}
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(*)簡単な連分数展開
\[
\left[x;x,x,\cdots\right]=\frac{1}{2}\left(x+\sqrt{x^{2}+4}\right)
\]
連分数と最大公約数
\[
\gcd\left(p_{n},q_{n}\right)=1
\]
連分数の定義
\[
\left[a_{0};a_{1},a_{2},a_{3},\cdots\right]:=a_{0}+\frac{1}{a_{1}+\frac{1}{a_{2}+\frac{1}{a_{3}+\cdots}}}
\]
連分数の性質
\[
\left[\left(a_{0},b_{0}\right);\left(a_{1},b_{1}\right),\cdots,a_{n}+c\right]=\left[\left(a_{0},b_{0}\right);\left(a_{1},b_{1}\right),\cdots,\left(a_{n},b_{n}\right),\frac{b_{n}}{c}\right]
\]