34 lines
1.5 KiB
TeX
34 lines
1.5 KiB
TeX
\input{decls.tex}
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\title{Polynomials}
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\begin{document}
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\maketitle
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\begin{itemize}[leftmargin=10em]
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\item[Polynomial in $\C$] \quad $\forall n \in \N, a_{i \in [0, n]} \in \C, a_n \neq 0. \quad P(x) = \sum_{i=0}^{n} a_{i}x^i$
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\item[Degree] \quad $n$
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\item[Leading Coefficient] \quad $a_n$
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\item[Monic Polynomial] \quad Polynomial with $a_n = 1$
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\end{itemize}
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The Abel-Ruffini theorem states that there exists a degree-5 polynomial with roots that cannot be expressed with $+ - * / \surd$
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\section*{Rational Root Theorem}
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\[
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\forall \text{ polynomials } P(x) \text { with } \forall i. a_i \in \Z.\quad \forall \frac{p}{q} \in \Q.~P\paren{\frac{p}{q}} = 0 \implies p|a_0 \land q|a_n \\ % x|y x divides y
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\]
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This means that monic polynomials have no rational non-integral roots.
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\section*{Polynomial Division}
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\[\forall P(x), D(x).~ \exists Q(x), R(x).~ P(x) = D(x)Q(x) + R(x), \operatorname{degree}(R) < \operatorname{degree}(D)\]
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\section*{Remainder Theorem}
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\[\forall P(x), c. P(c) = 0 \iff \paren{x - c}|P(x)\]
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Proof:
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\begin{align*}
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\text{Given that }& P(c) = 0: \\
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& \exists Q(x), R(x).~P(x) = Q(x)(x - c) + R(x) & \text{Division of polynomials} \\
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& P(c) = Q(c)(c - c) + R(c) & \\
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& 0 = Q(c)(0) + R(c) & \\
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& R(c) = 0 & \\
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& \forall x.~R(x) = 0 & \text{As $D(x)$ has degree $1$, $R(x)$ must have degree $0$} \\
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& \therefore~(x - c)|P(x) \\
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\text{Given that }& (x - c)|P(x) \\
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& \exists Q(x).~P(x) = Q(x)(x - c) & \text{Division of polynomials, remainder $0$} \\
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& \therefore~P(c) = Q(c)(c - c) = 0\cdot Q(c) = 0
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\end{align*}
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\end{document} |