高専の数学解説

第17章 17.26

A=(0111)A=\begin{pmatrix}0&1\\1&1\end{pmatrix}とすると、与式より(anan+1)=An(a1a2)\begin{pmatrix}a_n\\a_{n+1}\end{pmatrix}=A^n\begin{pmatrix}a_1\\a_2\end{pmatrix}が求められるので

∣A−λE∣=∣−λ111−λ∣=(−λ)(1−λ)−1=λ2−λ−1|A-\lambda E|=\begin{vmatrix}-\lambda&1\\1&1-\lambda\end{vmatrix}=(-\lambda)(1-\lambda)-1=\lambda^2-\lambda-1

λ=1±1+42=12(1±5)\lambda=\frac{1\pm\sqrt{1+4}}{2}=\frac{1}{2}(1\pm\sqrt{5})

ここでα=12(1+5)\alpha=\frac{1}{2}(1+\sqrt{5})、β=12(1−5)\beta=\frac{1}{2}(1-\sqrt{5})とおくと

λ=12(1+5)\lambda=\frac{1}{2}(1+\sqrt{5})のとき

(−12(1+5)1112(1−5))⇒(1−1−12(1+5)112(1−5))⇒(1−1−12(1+5)0m)⇒(1−α−100)\begin{pmatrix}-\frac{1}{2}(1+\sqrt{5})&1\\1&\frac{1}{2}(1-\sqrt{5})\end{pmatrix}\Rightarrow\begin{pmatrix}1&-\frac{1}{-\frac{1}{2}(1+\sqrt{5})}\\1&\frac{1}{2}(1-\sqrt{5})\end{pmatrix}\Rightarrow\begin{pmatrix}1&-\frac{1}{-\frac{1}{2}(1+\sqrt{5})}\\0&m\end{pmatrix}\Rightarrow\begin{pmatrix}1&-\alpha^{-1}\\0&0\end{pmatrix}

m=12(1−5)−1−12(1+5)=1−5+42(1+5)=0m=\frac{1}{2}(1-\sqrt{5})-\frac{1}{-\frac{1}{2}(1+\sqrt{5})}=\frac{1-5+4}{2(1+\sqrt{5})}=0 これから P1=C(α−11)⇒C(1α)P_1=C\begin{pmatrix}\alpha^{-1}\\1\end{pmatrix}\Rightarrow C\begin{pmatrix}1\\\alpha\end{pmatrix}

λ=12(1−5)\lambda=\frac{1}{2}(1-\sqrt{5})のとき

(−12(1−5)1112(1+5))⇒(1−1−12(1−5)112(1+5))⇒(1−1−12(1−5)0n)⇒(1−β−100)\begin{pmatrix}-\frac{1}{2}(1-\sqrt{5})&1\\1&\frac{1}{2}(1+\sqrt{5})\end{pmatrix}\Rightarrow\begin{pmatrix}1&-\frac{1}{-\frac{1}{2}(1-\sqrt{5})}\\1&\frac{1}{2}(1+\sqrt{5})\end{pmatrix}\Rightarrow\begin{pmatrix}1&-\frac{1}{-\frac{1}{2}(1-\sqrt{5})}\\0&n\end{pmatrix}\Rightarrow\begin{pmatrix}1&-\beta^{-1}\\0&0\end{pmatrix}

n=12(1+5)−1−12(1−5)=1−5+42(1−5)=0n=\frac{1}{2}(1+\sqrt{5})-\frac{1}{-\frac{1}{2}(1-\sqrt{5})}=\frac{1-5+4}{2(1-\sqrt{5})}=0 これから P1=C(β−11)⇒C(1α)P_1=C\begin{pmatrix}\beta^{-1}\\1\end{pmatrix}\Rightarrow C\begin{pmatrix}1\\\alpha\end{pmatrix}

よってP=(11αβ)P=\begin{pmatrix}1&1\\\alpha&\beta\end{pmatrix} P−1=1β−α(β−1−α1)P^{-1}=\frac{1}{\beta-\alpha}\begin{pmatrix}\beta&-1\\-\alpha&1\end{pmatrix}

P−1AP=(α00β)P−1An−1P=(αn−100βn−1)An−1=P(αn−100βn−1)P−1P^{-1}AP = \begin{pmatrix} \alpha & 0 \\ 0 & \beta \end{pmatrix} \quad P^{-1}A^{n-1}P = \begin{pmatrix} \alpha^{n-1} & 0 \\ 0 & \beta^{n-1} \end{pmatrix} \quad A^{n-1} = P\begin{pmatrix} \alpha^{n-1} & 0 \\ 0 & \beta^{n-1} \end{pmatrix}P^{-1}

An−1=1β−α(11αβ)(αn−100βn−1)(β−1−α1)=1β−α(αn−1βn−1αnβn)(β−1−α1)A^{n-1} = \frac{1}{\beta - \alpha}\begin{pmatrix} 1 & 1 \\ \alpha & \beta \end{pmatrix}\begin{pmatrix} \alpha^{n-1} & 0 \\ 0 & \beta^{n-1} \end{pmatrix}\begin{pmatrix} \beta & -1 \\ -\alpha & 1 \end{pmatrix} = \frac{1}{\beta - \alpha}\begin{pmatrix} \alpha^{n-1} & \beta^{n-1} \\ \alpha^n & \beta^n \end{pmatrix}\begin{pmatrix} \beta & -1 \\ -\alpha & 1 \end{pmatrix}

=1β−α(αn−1β−βn−1α−αn−1+βn−1αnβ−βnα−αn+βn)= \frac{1}{\beta - \alpha}\begin{pmatrix} \alpha^{n-1}\beta - \beta^{n-1}\alpha & -\alpha^{n-1} + \beta^{n-1} \\ \alpha^n\beta - \beta^n\alpha & -\alpha^n + \beta^n \end{pmatrix}

これから

(anan+1)=1β−α(αn−1β−βn−1α−αn−1+βn−1αnβ−βnα−αn+βn)(a1a2)=1β−α(αn−1β−βn−1α−αn−1+βn−1αnβ−βnα−αn+βn)(11)\begin{pmatrix} a_n \\ a_{n+1} \end{pmatrix} = \frac{1}{\beta - \alpha}\begin{pmatrix} \alpha^{n-1}\beta - \beta^{n-1}\alpha & -\alpha^{n-1} + \beta^{n-1} \\ \alpha^n\beta - \beta^n\alpha & -\alpha^n + \beta^n \end{pmatrix}\begin{pmatrix} a_1 \\ a_2 \end{pmatrix} = \frac{1}{\beta - \alpha}\begin{pmatrix} \alpha^{n-1}\beta - \beta^{n-1}\alpha & -\alpha^{n-1} + \beta^{n-1} \\ \alpha^n\beta - \beta^n\alpha & -\alpha^n + \beta^n \end{pmatrix}\begin{pmatrix} 1 \\ 1 \end{pmatrix}

=1β−α(αn−1β−βn−1α−αn−1+βn−1αnβ−βnα−αn+βn)= \frac{1}{\beta - \alpha}\begin{pmatrix} \alpha^{n-1}\beta - \beta^{n-1}\alpha - \alpha^{n-1} + \beta^{n-1} \\ \alpha^n\beta - \beta^n\alpha - \alpha^n + \beta^n \end{pmatrix}

=1β−α(αn−1(β−1)+βn−1(1−α)αn(β−1)+βn(1−α))= \frac{1}{\beta - \alpha}\begin{pmatrix} \alpha^{n-1}(\beta - 1) + \beta^{n-1}(1 - \alpha) \\ \alpha^n(\beta - 1) + \beta^n(1 - \alpha) \end{pmatrix}

ここで β−α=−5\beta - \alpha = -\sqrt{5}、β−1=−α\beta - 1 = -\alpha、1−α=β1 - \alpha = \beta から

=15(αn−βnαn+1−βn+1)= \frac{1}{\sqrt{5}}\begin{pmatrix} \alpha^n - \beta^n \\ \alpha^{n+1} - \beta^{n+1} \end{pmatrix}

よって an=15(αn−βn)a_n = \frac{1}{\sqrt{5}}(\alpha^n - \beta^n)

τ=lim⁡n→∞an+1an=lim⁡n→∞αn+1−βn+1αn−βn=lim⁡n→∞α−βnαnβ1−βnαn=α−0⋅β1−0=α\tau = \lim_{n \to \infty}\frac{a_{n+1}}{a_n} = \lim_{n \to \infty}\frac{\alpha^{n+1} - \beta^{n+1}}{\alpha^n - \beta^n} = \lim_{n \to \infty}\frac{\alpha - \frac{\beta^n}{\alpha^n}\beta}{1 - \frac{\beta^n}{\alpha^n}} = \frac{\alpha - 0 \cdot \beta}{1 - 0} = \alpha

(lim⁡n→∞βnαnは分母が大きいのでn→∞では0となる)(\lim_{n \to \infty}\frac{\beta^n}{\alpha^n}は分母が大きいので n \to \infty では 0 となる)

解説: r31bn1z