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\)
Section 11.2 Recurrence relations
recursively-defined sequence
a sequence \(\{a_k\}\) from a set \(A\text{,}\) where \(a_0,a_1,\dotsc,a_{K-1}\) 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 \(a_0, \, a_1, \, \dotsc, \, a_{k-1}\)
recurrence relation
for a recursively-defined sequence, the formula that defines the general term \(a_k\) recursively in the previous terms \(a_0, \, a_1, \, \dotsc, \, a_{k-1}\)
Example 11.2.1 . A bouncing ball.
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 \(\nth[k]\) bounce. Then \(h_0 = 100\) and the recurrence relation is
\begin{align*}
h_k \amp = \frac{3h_{k-1}}{4} \text{,} \amp k \amp \ge 1 \text{.}
\end{align*}
The terms of the sequence are
\begin{equation*}
100, \, 75, \, 56.25, \, 42.1875, \, \dotsc \text{.}
\end{equation*}
Example 11.2.2 . Factorial.
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, \, \dotsc \text{.}
\end{equation*}
Example 11.2.3 . Fibonacci sequence.
The sequence
\begin{equation*}
0, \, 1, \, 1, \, 2, \, 3, \, 5, \, 8, \, 13, \, 21, \, 34, \dotsc
\end{equation*}
can be defined recursively by \(a_0 = 0\text{,}\) \(a_1 = 1\text{,}\) and
\begin{align*}
a_k \amp = a_{k-1} + a_{k-2} \text{,} \amp k \amp \ge 2 \text{.}
\end{align*}
Example 11.2.4 . A sequence of sets.
Define a sequence \(\{A_k\}\) from \(\powset{\N}\) recursively as follows. Let \(A_0 = \emptyset\text{,}\) and take the recurrence relation to be
\begin{align*}
A_k \amp = A_{k-1} \cup \{k\} \text{,} \amp k \amp \ge 1 \text{.}
\end{align*}
Then the terms of the sequence are
\begin{equation*}
\emptyset, \, \{1\}, \, \{1,2\}, \, \{1,2,3\}, \, \dotsc, \, \{1, 2, \dotsc, k \}, \, \dotsc \text{.}
\end{equation*}
