45405 費氏數列與 $\frac{z}{1-z-z^2}$ 的洛朗級數

貳、費氏數列的冪級數與一般式

\begin{align*} \lim_{n\to\infty}\frac{F_{n+1}}{F_n}=&\lim_{n\to\infty}\dfrac{\Big(\dfrac{1+\sqrt{5}}{2}\Big)^{n+1}-\Big(\dfrac{1-\sqrt{5}}{2}\Big)^{n+1}} {\Big(\dfrac{1+\sqrt{5}}{2}\Big)^n-\Big(\dfrac{1-\sqrt{5}}{2}\Big)^n}\\ =&\lim_{n\to\infty}\dfrac{\Big(\dfrac{1+\sqrt{5}}{2}\Big) -\dfrac{1-\sqrt{5}}{2}\Big(\dfrac{1-\sqrt{5}}{1+\sqrt{5}}\Big)^n}{1-\Big(\dfrac{1-\sqrt{5}}{1+\sqrt{5}}\Big)^n}=\dfrac{1+\sqrt{5}}{2}, \end{align*}

\begin{align*} \lim_{n\to\infty}\left|\dfrac{F_{n+1}z^{n+1}}{F_nz^n}\right|\lt1,\ &\hbox{則}\ \sum_{n=0}^\infty F_nz^n\ \hbox{收斂,}\\ \lim_{n\to\infty}\left|\dfrac{F_{n+1}z^{n+1}}{F_nz^n}\right|\gt1,\ &\hbox{則}\ \sum_{n=0}^\infty F_nz^n\ \hbox{發散。} \end{align*}

\begin{align*} \dfrac{\sqrt{5}+1}{2}\cdot |z|\lt1,\ &\hbox{則}\ \sum_{n=0}^\infty F_nz^n\ \hbox{收斂,}\\ \dfrac{\sqrt{5}+1}{2}\cdot |z|\gt1,\ &\hbox{則}\ \sum_{n=0}^\infty F_nz^n\ \hbox{發散。} \end{align*}

\begin{align*} |z|\lt\dfrac{\sqrt{5}-1}{2},\ &\hbox{則}\ \sum_{n=0}^\infty F_nz^n\ \hbox{收斂,}\\ |z|\gt\dfrac{\sqrt{5}-1}{2},\ &\hbox{則}\ \sum_{n=0}^\infty F_nz^n\ \hbox{發散。} \end{align*}

\begin{align*} \left|\lim\limits_{n\to\infty} F_nz^n\right|=&\lim\limits_{n\to\infty} \left|F_nz^n\right|=\lim\limits_{n\to\infty} F_n\left|z^n\right|\\ =&\lim_{n\to\infty} \dfrac 1{\sqrt 5} \Big[\Big(\dfrac{1+\sqrt{5}}{2}\Big)^n-\Big(\dfrac{1-\sqrt{5}}{2}\Big)^n\Big]\cdot \Big(\dfrac{\sqrt{5}-1}{2}\Big)^n\\ =&\lim_{n\to\infty} \dfrac 1{\sqrt 5}\Big[1-(-1)^n\Big(\dfrac{\sqrt{5}-1}{2}\Big)^{2n}\Big]=\dfrac 1{\sqrt{5}}, \end{align*}

\begin{align*} \sum\limits_{n=0}^\infty F_nz^n =&\sum\limits_{n=0}^\infty \dfrac 1{\sqrt 5} \Big[\Big(\dfrac{1+\sqrt{5}}{2}\Big)^n-\Big(\dfrac{1-\sqrt{5}}{2}\Big)^n\Big]z^n\\ =&\sum\limits_{n=0}^\infty \dfrac 1{\sqrt 5}\Big[\Big(\dfrac{1+\sqrt{5}}{2}z\Big)^n-\Big(\dfrac{1-\sqrt{5}}{2}z\Big)^n\Big]\\ =&\dfrac 1{\sqrt 5}\Big[\sum\limits_{n=0}^\infty \Big(\dfrac{1+\sqrt{5}}{2}z\Big)^n-\sum\limits_{n=0}^\infty \Big(\dfrac{1-\sqrt{5}}{2}z\Big)^n\Big]\\ =&\dfrac 1{\sqrt 5}\left( \dfrac{1}{1-\dfrac{1+\sqrt 5}{2}z}-\dfrac{1}{1-\dfrac{1-\sqrt 5}{2}z} \right)\\ =&\dfrac{z}{1-z-z^2}. \end{align*}

參、$\dfrac{z}{1-z-z^2}$的洛朗級數

\begin{align*} f(z)=&\sum\limits_{n=0}^\infty a_n(z-z_0)^n+\sum_{n=1}^\infty \frac{b_n}{(z-z_0)^n},\qquad (R_1\lt|z-z_0|\lt R_2)\\ \hbox{其中} a_n=&\frac {1}{2\pi i}\int_C\frac{f(z)}{(z-z_0)^{n+1}}dz,\qquad (n=0,1,2,\ldots)\\ \hbox{且} b_n=&\frac {1}{2\pi i}\int_C\frac{f(z)}{(z-z_0)^{-n+1}}dz,\qquad (n=1,2,\ldots). \end{align*}

$F(z)=\dfrac{z}{1-z-z^2}$, 其中 $z$ 為複變數, 將 $F(z)$ 進行部分分式分解, 可得 $$F(z)=\dfrac{z}{1-z-z^2}=\dfrac 1{\sqrt 5}\Bigg( \dfrac{1}{1-\dfrac{1+\sqrt 5}{2}z}-\dfrac{1}{1-\dfrac{1-\sqrt 5}{2}z} \Bigg),$$ 得知解析函數 $F(z)$ 在複數平面上只有兩個奇異點 $z_1=\dfrac{-1+\sqrt{5}}{2}$, $z_2=\dfrac{-1-\sqrt{5}}{2}$, 我們將 $F(z)$ 在 $z=0$ 為中心的開圓盤及環狀區域做洛朗展開, 得到以下定理 7。

1. 當 $|z|\lt\dfrac{\sqrt 5-1}{2}$, $F(z)=\sum\limits_{n=0}^\infty F_nz^n$。
2. 當 $\dfrac{\sqrt 5-1}{2}\lt|z|\lt\dfrac{\sqrt 5+1}{2}$, $F(z)=-\dfrac 1{\sqrt 5} \Bigg[\sum\limits_{n=1}^\infty \Big(\dfrac{\sqrt{5}-1}{2}\cdot \dfrac 1z\Big)^n+\sum\limits_{n=0}^\infty \Big(\dfrac{1-\sqrt{5}}{2}\cdot z\Big)^n\Bigg]$。
3. 當 $|z|\gt\dfrac{1+\sqrt 5}{2}$, $F(z)=\sum\limits_{n=1}^\infty (-1)^n\dfrac{F_n}{z^n}$。

1. 當 $|z|\lt\dfrac{\sqrt 5-1}{2}$,

\begin{align*} F(z)=&\dfrac{z}{1-z-z^2}=\dfrac 1{\sqrt 5}\Bigg( \dfrac{1}{1-\dfrac{1+\sqrt 5}{2}z}-\dfrac{1}{1-\dfrac{1-\sqrt 5}{2}z} \Bigg)\\ =&\dfrac 1{\sqrt 5}\Bigg(\sum_{n=0}^\infty\Big(\frac{\sqrt 5+1}{2}\cdot z\Big)^n-\sum_{n=0}^\infty\Big(\frac{1-\sqrt 5}{2}\cdot z\Big)^n\Bigg)\\ =&\dfrac 1{\sqrt 5}\left(\sum_{n=0}^\infty\Bigg(\Big(\frac{\sqrt 5+1}{2}\cdot z\Big)^n-\Big(\frac{1-\sqrt 5}{2}\cdot z\Big)^n\Bigg)\right)\\ =&\sum_{n=0}^\infty \dfrac 1{\sqrt 5}\left(%\sum_{n=0}^\infty \Big(\frac{\sqrt 5+1}{2}\Big)^n-\Big(\frac{1-\sqrt 5}{2}\Big)^n\right)z^n,\qquad\quad\hbox{(由引理1)}\\ =&\sum_{n=0}^\infty F_nz^n. \end{align*}

2. 當 $\dfrac{\sqrt 5-1}{2}\lt|z|\lt\dfrac{\sqrt 5+1}{2}$,

\begin{align*} F(z)=&\dfrac{z}{1-z-z^2}=\dfrac 1{\sqrt 5}\Bigg(\dfrac{1}{1-\dfrac{1+\sqrt 5}{2}z}-\dfrac{1}{1-\dfrac{1-\sqrt 5}{2}z}\Bigg)\\ =&\dfrac 1{\sqrt 5}\Bigg(\frac{1}{\dfrac{1+\sqrt 5}{2}z\Big(\dfrac 2{1+\sqrt 5}\cdot \dfrac 1z-1\Big)} -\dfrac 1{1-\dfrac{1-\sqrt 5}{2}\cdot z}\Bigg)\\ =&\dfrac 1{\sqrt 5}\left(-\frac{1}{\Big(\dfrac{1+\sqrt 5}{2}z\Big)} \sum_{n=0}^\infty \Big(\dfrac 2{1+\sqrt 5}\cdot \dfrac 1z\Big)^n -\sum_{n=0}^\infty\Big(\dfrac{1-\sqrt 5}{2}\cdot z\Big)^n\right)\\ =&\dfrac {-1}{\sqrt 5}\left(\sum_{n=1}^\infty \Big(\dfrac {\sqrt 5-1}2\cdot \dfrac 1z\Big)^n +\sum_{n=0}^\infty\Big(\dfrac{1-\sqrt 5}{2}\cdot z\Big)^n\right). \end{align*}

3. 當 $|z|\gt\dfrac{1+\sqrt 5}{2}$,

\begin{align*} F(z)=&\dfrac{z}{1-z-z^2}=\dfrac 1{\sqrt 5}\Bigg(\dfrac{1}{1-\dfrac{1+\sqrt 5}{2}z}-\dfrac{1}{1-\dfrac{1-\sqrt 5}{2}z}\Bigg)\\ =&\dfrac 1{\sqrt 5}\Bigg(\frac{1}{\dfrac{1+\sqrt 5}{2}z\Big(\dfrac 2{1+\sqrt 5}\cdot \dfrac 1z-1\Big)} -\dfrac 1{\dfrac{1-\sqrt 5}{2}z\Big(\dfrac 2{1-\sqrt 5}\cdot \dfrac 1z-1\Big)}\Bigg)\\ =&\dfrac 1{\sqrt 5}\Bigg(-\frac{1}{\Big(\dfrac{1+\sqrt 5}{2}z\Big)} \sum_{n=0}^\infty \Big(\dfrac 2{1+\sqrt 5}\cdot \dfrac 1z\Big)^n +\frac{1}{\Big(\dfrac{1-\sqrt 5}{2}z\Big)} \sum_{n=0}^\infty \Big(\dfrac 2{1-\sqrt 5}\cdot \dfrac 1z\Big)^n\Bigg)\\ =&\dfrac 1{\sqrt 5}\left(-\sum_{n=0}^\infty \Big(\dfrac 2{1+\sqrt 5}\cdot \dfrac 1z\Big)^{n+1} +\sum_{n=0}^\infty\Big(\dfrac 2{1-\sqrt 5}\cdot \dfrac 1z\Big)^{n+1}\right)\\ =&\dfrac 1{\sqrt 5}\left(-\sum_{n=0}^\infty \Big(\dfrac {\sqrt 5-1}2\cdot \dfrac 1z\Big)^{n+1} +\sum_{n=0}^\infty\Big(-\dfrac {\sqrt 5+1}2\cdot \dfrac 1z\Big)^{n+1}\right)\\ =&\dfrac 1{\sqrt 5} \sum_{n=0}^\infty (-1)^{n+1}\left(\Big(\dfrac{\sqrt 5+1}{2}\cdot \dfrac 1z\Big)^{n+1} -\Big(\frac{1-\sqrt 5}{2}\cdot \dfrac 1z\Big)^{n+1}\right),\quad\hbox{(由引理1)}\\ =&\sum_{n=0}^\infty (-1)^{n+1} \dfrac {F_{n+1}}{z^{n+1}}=\sum_{n=1}^\infty (-1)^{n} \dfrac {F_{n}}{z^{n}}. \end{align*}

肆、冪級數的應用

\begin{align*} F(z)=&\sum_{n=0}^\infty F_nz^n=\frac z{1-z-z^2},\quad |z|\lt\frac{\sqrt 5-1}{2},\\ \hbox{得} &\sum_{n=1}^\infty F_nz^{n-1}=\frac 1{1-z-z^2},\quad |z|\lt\frac{\sqrt 5-1}{2}. \end{align*}

\begin{align} \sum_{n=1}^\infty \int_0^{z_1} F_nz^{n-1}dz=\int_0^{z_1}\frac 1{1-z-z^2}dz, \quad |z|\lt\frac{\sqrt 5-1}{2} ,\label{1} \end{align}

\begin{align} \dfrac 1{1-z-z^2}=\frac{\sqrt 5}{10}\Bigg( \dfrac{\sqrt 5-1}{1-\dfrac{1-\sqrt 5}{2}z}+\dfrac{\sqrt 5+1}{1-\dfrac{1+\sqrt 5}{2}z} \Bigg),\label{2} \end{align}

\begin{align} \int_0^{z_1}\frac 1{1-z-z^2}dz=&\frac{\sqrt{5}}{10}\Bigg( \int_0^{z_1}\dfrac{\sqrt 5-1}{1-\dfrac{1-\sqrt 5}{2}z}dz+\int_0^{z_1}\dfrac{\sqrt 5+1}{1-\dfrac{1+\sqrt 5}{2}z}dz \Bigg)\nonumber\\ =&\frac{\sqrt{5}}{10}\Bigg[(\sqrt 5-1) \Big(-\dfrac 2{1-\sqrt 5}\Big)Log\Big(1-\dfrac{1-\sqrt 5}{2}z_1\Big)\nonumber\\ &+(\sqrt 5+1)\Big(-\dfrac 2{1+\sqrt 5}\Big)Log\Big(1-\dfrac{1+\sqrt 5}{2}z_1\Big)\Bigg]\nonumber\\ =&\frac{\sqrt{5}}{10}\Bigg[2Log\Big(1-\dfrac {1-\sqrt 5}2z_1\Big)-2Log\Big(1-\dfrac{1+\sqrt 5}{2}z_1\Big)\Bigg] \nonumber\\ =&\frac{\sqrt{5}}{5}\Bigg[Log\Big(1-\dfrac {1-\sqrt 5}2z_1\Big)-Log\Big(1-\dfrac{1+\sqrt 5}{2}z_1\Big)\Bigg].\label{3} \end{align}

參考資料

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