133 lines
4.2 KiB
TeX
133 lines
4.2 KiB
TeX
\documentclass{article}
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\input{../../preamble}
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\makecode{../..}
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\begin{document}
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\header{Sequences}{}
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\tableofcontents
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\section{Summations}%
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\hyperlabel{sec:summations}
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\subsection{\verified{Arithmetic Series}}%
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\hyperlabel{sub:sum-arithmetic-series}
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Let $(a_i)_{i \geq 0}$ be an arithmetic sequence with common difference $d$.
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Then for some $n \in \mathbb{N}$,
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\begin{equation}
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\hyperlabel{sub:sum-arithmetic-series-eq1}
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\sum_{i=0}^n a_i = \frac{(n + 1)(a_0 + a_n)}{2}.
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\end{equation}
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\code{Common/Real/Sequence/Arithmetic}
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{Real.Arithmetic.sum\_recursive\_closed}
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\begin{proof}
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Let $(a_i)_{i \geq 0}$ be an arithmetic sequence with common difference $d$.
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By definition, for all $k \in \mathbb{N}$,
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\begin{equation}
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\hyperlabel{sub:sum-arithmetic-series-eq2}
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a_k = (a_0 + kd).
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\end{equation}
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Define predicate $P(n)$ as "identity \eqref{sub:sum-arithmetic-series-eq1}
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holds for value $n$."
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We use induction to prove $P(n)$ holds for all $n \geq 0$.
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\paragraph{Base Case}%
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Let $k = 0$.
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Then $$\sum_{i=0}^k a_i = a_0 = \frac{2a_0}{2} =
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\frac{(k + 1)(a_0 + a_k)}{2}.$$
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Therefore $P(0)$ holds.
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\paragraph{Induction Step}%
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Assume induction hypothesis $P(k)$ holds for some $k \geq 0$.
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Then
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\begin{align*}
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\sum_{i=0}^{k+1} a_i
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& = \sum_{i=0}^k a_i + a_{k+1} \\
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& = \frac{(k + 1)(a_0 + a_k)}{2} + a_{k+1}
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& \text{induction hypothesis} \\
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& = \frac{(k + 1)(a_0 + (a_0 + kd))}{2} + (a_0 + (k + 1)d)
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& \eqref{sub:sum-arithmetic-series-eq2} \\
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& = \frac{(k + 1)(2a_0 + kd)}{2} + (a_0 + (k + 1)d) \\
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& = \frac{(k + 1)(2a_0 + kd) + 2a_0 + 2(k + 1)d}{2} \\
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& = \frac{2ka_0 + k^2d + 4a_0 + kd + 2kd + 2d}{2} \\
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& = \frac{(k + 2)(2a_0 + kd + d)}{2} \\
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& = \frac{(k + 2)(a_0 + a_0 + (k + 1)d)}{2} \\
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& = \frac{(k + 2)(a_0 + a_{k+1})}{2}
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& \eqref{sub:sum-arithmetic-series-eq2} \\
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& = \frac{((k + 1) + 1)(a_0 + a_{k+1})}{2}.
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\end{align*}
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Thus $P(k)$ implies $P(k + 1)$ holds true.
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\paragraph{Conclusion}%
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By mathematical induction, it follows for all $n \geq 0$, $P(n)$ is true.
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\end{proof}
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\subsection{\verified{Geometric Series}}%
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\hyperlabel{sub:sum-geometric-series}
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Let $(a_i)_{i \geq 0}$ be a geometric sequence with common ratio $r \neq 1$.
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Then for some $n \in \mathbb{N}$,
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\begin{equation}
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\hyperlabel{sub:sum-geometric-series-eq1}
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\sum_{i=0}^n a_i = \frac{a_0(1 - r^{n+1})}{1 - r}.
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\end{equation}
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\code{Common/Real/Sequence/Geometric}
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{Real.Geometric.sum\_recursive\_closed}
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\begin{proof}
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Let $(a_i)_{i \geq 0}$ be a geometric sequence with common ratio $r \neq 1$.
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By definition, for all $k \in \mathbb{N}$,
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\begin{equation}
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\hyperlabel{sub:sum-geometric-series-eq2}
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a_k = a_0r^k.
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\end{equation}
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Define predicate $P(n)$ as "identity \eqref{sub:sum-geometric-series-eq1}
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holds for value $n$."
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We use induction to prove $P(n)$ holds for all $n \geq 0$.
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\paragraph{Base Case}%
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Let $k = 0$.
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Then $$\sum_{i=0}^k a_i = a_0 = \frac{a_0(1 - r)}{1 - r} =
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\frac{a_0(1 - r^{k+1})}{1 - r}$$
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Therefore $P(0)$ holds.
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\paragraph{Induction Step}%
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Assume induction hypothesis $P(k)$ holds for some $k \geq 0$.
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Then
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\begin{align*}
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\sum_{i=0}^{k+1} a_i
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& = \sum_{i=0}^k a_i + a_{k+1} \\
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& = \frac{a_0(1 - r^{k+1})}{1 - r} + a_{k + 1}
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& \text{induction hypothesis} \\
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& = \frac{a_0(1 - r^{k+1})}{1 - r} + a_0r^{k + 1}
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& \eqref{sub:sum-geometric-series-eq2} \\
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& = \frac{a_0(1 - r^{k+1}) + a_0r^{k+1}(1 - r)}{1 - r} \\
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& = \frac{a_0(1 - r^{k+1} + r^{k+1}(1 - r))}{1 - r} \\
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& = \frac{a_0(1 - r^{k+1} + r^{k+1} - r^{k+2})}{1 - r} \\
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& = \frac{a_0(1 - r^{k+2})}{1 - r} \\
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& = \frac{a_0(1 - r^{(k + 1) + 1})}{1 - r}.
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\end{align*}
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Thus $P(k)$ implies $P(k + 1)$ holds true.
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\paragraph{Conclusion}%
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By mathematical induction, it follows for all $n \geq 0$, $P(n)$ is true.
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\end{proof}
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\end{document}
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