11.2: Recurrence Relations
- Page ID
- 83458
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)a sequence \(\{a_k\}\) from a set \(A\text{,}\) where \(a_0,a_1,\ldots ,a_{K-1}\) are defined explicitly, and for \(k\ge K\text{,}\) the term \(a_k\) is defined in terms of some (or all) of the previous terms in the sequence \(a_0, \, a_1, \, \ldots , \, a_{k-1}\)
for a recursively-defined sequence, the formula that defines the general term \(a_k\) recursively in the previous terms \(a_0, \, a_1, \, \ldots , \, a_{k-1}\)
A ball is dropped from a height of 100 cm. On each bounce, it returns to \(75\%\) of its previous height.
Let \(h_k\) be the height in centimetres after the \(k^{th}\) bounce. Then \(h_0 = 100\) and the recurrence relation is
\begin{align*} h_k & = \dfrac{3h_{k-1}}{4} \text{,} & k & \ge 1 \text{.} \end{align*}
The terms of the sequence are
\begin{equation*} 100, \, 75, \, 56.25, \, 42.1875, \, \ldots \text{.} \end{equation*}
Set \(a_0 = 1\text{,}\) and let \(a_k = k a_{k-1}\) for \(k \ge 1\text{.}\) Then the terms of the sequence are
\begin{equation*} 1, \, 1, \, 2, \, 6, \, 24, \, 120, \, \ldots \text{.} \end{equation*}
The sequence
\begin{equation*} 0, \, 1, \, 1, \, 2, \, 3, \, 5, \, 8, \, 13, \, 21, \, 34, \ldots \end{equation*}
can be defined recursively by \(a_0 = 0\text{,}\) \(a_1 = 1\text{,}\) and
\begin{align*} a_k & = a_{k-1} + a_{k-2} \text{,} & k & \ge 2 \text{.} \end{align*}
Define a sequence \(\{A_k\}\) from \(\mathscr{P}(\mathbb{N})\) recursively as follows. Let \(A_0 = \emptyset\text{,}\) and take the recurrence relation to be
\begin{align*} A_k & = A_{k-1} \cup \{k\} \text{,} & k & \ge 1 \text{.} \end{align*}
Then the terms of the sequence are
\begin{equation*} \emptyset, \, \{1\}, \, \{1,2\}, \, \{1,2,3\}, \, \ldots , \, \{1, 2, \ldots , k \}, \, \ldots \text{.} \end{equation*}