Typos
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@ -4,7 +4,7 @@
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Say we have a sequence $a_0, a_1, a_2, ...$. \par
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Say we have a sequence $a_0, a_1, a_2, ...$. \par
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The \textit{generating function} of this sequence is defined as follows:
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The \textit{generating function} of this sequence is defined as follows:
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\begin{equation*}
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\begin{equation*}
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A(x) = \sum_{n=0}^\infty a_nx^n = a_0 + a_1x + a_2x^2 + a_3x+3 + ...
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A(x) = \sum_{n=0}^\infty a_nx^n = a_0 + a_1x + a_2x^2 + a_3x^3 + ...
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\end{equation*}
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\end{equation*}
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Under some circumstances, this sum does not converge, and thus $A(x)$ is undefined. \par
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Under some circumstances, this sum does not converge, and thus $A(x)$ is undefined. \par
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@ -44,6 +44,7 @@ Assuming $|x| < 1$, show that
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\begin{equation*}
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\begin{equation*}
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\frac{1}{1-x} = 1 + x + x^2 + x^3 + ...
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\frac{1}{1-x} = 1 + x + x^2 + x^3 + ...
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\end{equation*}
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\end{equation*}
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\hint{use some clever algebra. What is $x \times (1 + x + x^2 + ...)$? }
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\begin{solution}
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\begin{solution}
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Let $S = 1 + x + x^2 + ...$ \par
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Let $S = 1 + x + x^2 + ...$ \par
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@ -50,7 +50,7 @@ we used to define the Fibonacci numbers.
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\pagebreak
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\pagebreak
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\problem{}
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\problem{}<fibo>
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Using the problems on the previous page, find $F(x)$ in terms of $x$.
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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{solution}
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@ -77,7 +77,7 @@ A \textit{rational function} $f$ is a function that can be written as a quotient
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That is, $f(x) = \frac{p(x)}{q(x)}$ where $p$ and $q$ are polynomials.
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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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\problem{}
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Solve this equation for $F(x)$, expressing it as a rational function.
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Solve the equation from \ref<fibo> for $F(x)$, expressing it as a rational function.
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\begin{solution}
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\begin{solution}
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\begin{align*}
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\begin{align*}
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@ -98,20 +98,20 @@ Solve this equation for $F(x)$, expressing it as a rational function.
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\pagebreak
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\pagebreak
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\problem{}<pfd>
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\definition{}
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Now that we have a rational function for $F(x)$, \par
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\textit{Partial fraction decomposition} is an algebreic technique that works as follows: \par
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find a closed-form expression for its coefficients.
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If $p(x)$ is a polynomial and $a$ and $b$ are constants,
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we can rewrite the rational function $\frac{p(x)}{(x-a)(x-b)}$ as follows:
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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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\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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\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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\end{equation*}
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where $c$ and $d$ are constants.
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where $c$ and $d$ are constants.
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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 using partial fraction decomposition.
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\begin{solution}
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\begin{solution}
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\begin{align*}
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\begin{align*}
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F(x)
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F(x)
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