168 lines
3.9 KiB
TeX
Executable File
168 lines
3.9 KiB
TeX
Executable File
\section{Fibonacci}
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\definition{}
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The \textit{Fibonacci numbers} are defined by the following recurrence relation:
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\begin{itemize}
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\item $f_0 = 0$
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\item $f_1 = 1$
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\item $f_n = f_{n-1} + f_{n-2}$
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\end{itemize}
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\problem{}
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Let $F(x)$ be the generating function that corresponds to the Fibonacci numbers. \par
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Find the generating function of $0, f_0, f_1, ...$ in terms of $F(x)$. \par
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Call this $G(x)$.
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\begin{solution}
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\begin{equation*}
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G(x) = xF(x)
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\end{equation*}
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\end{solution}
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\vfill
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\problem{}
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Find the generating function of $0, 0, f_0, f_1, ...$ in terms of $F(x)$.
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Call this $H(x)$.
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\begin{solution}
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\begin{equation*}
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H(x) = x^2F(x)
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\end{equation*}
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\end{solution}
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\vfill
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\problem{}
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Calculate $F(x) - G(x) - H(x)$ using the recurrence relation that
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we used to define the Fibonacci numbers.
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\begin{solution}
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\begin{align*}
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F(x) - G(x) - H(x)
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&=~ f_0 + (f_1 - f_0)x + (f_2 - f_1 - f_0)x^2 + (f_3 - f_2 - f_1)x^3 + ... \\
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&=~ f_0 + (f_1 - f_0)x \\
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&=~ x
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\end{align*}
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\end{solution}
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\vfill
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\pagebreak
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\problem{}
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Using the problems on the previous page, find $F(x)$ in terms of $x$.
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\begin{solution}
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\begin{align*}
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x
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&=~ F(x) - G(x) - H(x) \\
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&=~ F(x) - xF(x) - x^2F(x) \\
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&=~ F(x)(1-x-x^2)
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\end{align*}
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So,
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\begin{equation*}
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F(x) = \frac{x}{1-x-x^2}
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\end{equation*}
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\end{solution}
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\vfill
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\definition{}
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A \textit{rational function} $f$ is a function that can be written as a quotient of polynomials. \par
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That is, $f(x) = \frac{p(x)}{q(x)}$ where $p$ and $q$ are polynomials.
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\problem{}
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Solve this equation for $F(x)$, expressing it as a rational function.
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\begin{solution}
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\begin{align*}
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F(x)
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&=~ \frac{-x}{x^2+x-1} = \frac{-x}{(x-a)(x-b)} \\
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&=~ \frac{1 - \sqrt{5}}{2\sqrt{5}}\frac{1}{x-a} + \frac{-1 - \sqrt{5}}{2\sqrt{5}}\frac{1}{x-b}
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\end{align*}
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where
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\begin{equation*}
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a = \frac{-1 + \sqrt{5}}{2} ;~~
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b = \frac{-1 - \sqrt{5}}{2}
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\end{equation*}
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\end{solution}
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\vfill
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\pagebreak
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\problem{}<pfd>
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Now that we have a rational function for $F(x)$, \par
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find a closed-form expression for its coefficients.
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\vspace{2mm}
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Do this using \textit{partial fraction decomposition:} \par
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We can break up a rational function $\frac{p(x)}{(x-a)(x-b)}$ as follows:
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\begin{equation*}
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F(x) = \frac{p(x)}{(x-a)(x-b)} = \frac{c}{x-a} + \frac{d}{x-b}
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\end{equation*}
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where $c$ and $d$ are constants.
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\begin{solution}
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\begin{align*}
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F(x)
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&=~
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\left(\frac{1 - \sqrt{5}}{2\sqrt{5}}\right)\left(\frac{-1}{a}\right)\left(\frac{1}{1-\frac{x}{a}}\right)
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+ \left(\frac{-1 - \sqrt{5}}{2\sqrt{5}}\right)\left(\frac{-1}{b}\right)\left(\frac{1}{1-\frac{x}{b}}\right) \\
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&=~
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\left(\frac{1}{\sqrt{5}}\right)\left(\frac{1}{1-\frac{x}{a}}\right)
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+ \left(\frac{-1}{\sqrt{5}}\right)\left(\frac{1}{1-\frac{x}{b}}\right) \\
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&=~
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\frac{1}{\sqrt{5}}\left(1 + \frac{x}{a} + \left(\frac{x}{a}\right)^2 + ...\right)
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- \frac{1}{\sqrt{5}}\left(1 + \frac{x}{b} + \left(\frac{x}{b}\right)^2 + ...\right)
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\end{align*}
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\end{solution}
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\vfill
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\problem{}
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Using problems from the introduction and \ref{pfd}, find an expression
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for the coefficients of $F(x)$ (and this, for the Fibonacci numbers).
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\begin{solution}
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\begin{align*}
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f_0 &= \frac{1}{\sqrt{5}} - \frac{1}{\sqrt{5}} = 0 \\
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f_1 &= \frac{1}{a\sqrt{5}} - \frac{1}{b\sqrt{5}} = 1 \\
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f_n &= \frac{1}{\sqrt{5}}\left(\frac{1}{a^n} - \frac{1}{b^n}\right)
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= \frac{1}{\sqrt{5}}\left(
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\left(\frac{1 + \sqrt{5}}{2}\right)^n
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- \left(\frac{1-\sqrt{5}}{2}\right)^n
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\right)
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\end{align*}
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\end{solution}
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\vfill
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\pagebreak
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\problem{Bonus}
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Repeat the method of recurrence, generating function,
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partial fraction decomposition, and geometric series
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to find a closed form for the following sequence:
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\begin{equation*}
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a_0 = 1 ;~~ a_{n+1} = 2a_n + n
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\end{equation*}
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\hint{
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When doing partial fraction decomposition with a
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denominator of the form $(x-a)^2(x-b)$,
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you may need to express your expression as a sum of three fractions:
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$\frac{c}{(x-a)^2} + \frac{d}{x-a} + \frac{e}{x-b}$.`'
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}
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\vfill
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\pagebreak |