11.3: Solving through Iteration
- Page ID
- 83459
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Given a recursively defined sequence \(\{a_k\}\text{,}\) we can “unravel” the recursive definition to determine an explicit formula for the general term \(a_k\) which involves only the index \(k\text{.}\)
Solve the recurrence relation from Example 11.2.1.
Solution
The sequence in the example was defined recursively by \(h_0 = 100\) and
\begin{align*} h_k & = \dfrac{3}{4} \, h_{k-1} \text{,} & k & \ge 1 \text{.} \end{align*}
We can apply this formula to every term in the sequence, except for the first, using the pattern “each term is three-quarters of the previous term.” That is,
\begin{align*} h_k & = \dfrac{3}{4} \, h_{k-1} \text{,} & h_{k-1} & = \dfrac{3}{4} \, h_{k-2} \text{,} & h_{k-2} & = \dfrac{3}{4} \, h_{k-3} \text{,} & & \ldots\text{.} \end{align*}
Therefore, for \(k \ge 1\text{,}\) we can calculate
\begin{align*} h_k & = \dfrac{3}{4}\,h_{k-1} \\ & = \dfrac{3}{4}\left(\dfrac{3}{4}\,h_{k-2}\right) = \left(\dfrac{3}{4}\right)^2 h_{k-2}\\ & = \left(\dfrac{3}{4}\right)^2\left(\dfrac{3}{4}\,h_{k-3}\right) = \left(\dfrac{3}{4}\right)^3 h_{k-3}\\ & \vdots \\ & = \left(\dfrac{3}{4}\right)^k h_0 = \left(\dfrac{3}{4}\right)^k (100). \end{align*}
(Note that this formula is also valid for \(k=0\text{.}\))
We can verify our formula by substituting it into the original recurrence relation:
\begin{equation*} \text{RHS} = \dfrac{3}{4}\,h_{k-1} = \dfrac{3}{4}\,\left(\dfrac{3}{4}\right)^{k-1} h_0 = \left(\dfrac{3}{4}\right)^k h_0 = h_k = \text{LHS}\text{.} \end{equation*}
We could also prove our formula is correct by induction.
Solve the recurrence relation \(a_k = r a_{k-1}\) for \(k\ge 1\text{,}\) where \(r\) is a constant and the first term \(a_0\) is arbitrarily chosen.
- Compare.
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This example generalizes the previous example.
Solution
Through iteration, we obtain
\begin{equation*} a_k = r a_{k-1} = r (r a_{k-2}) = r(r(r a_{n-3})) = \cdots = r^k a_0\text{.} \end{equation*}
Solve the recurrence relation from Example 11.2.2.
Solution
The sequence in the example was defined recursively by \(a_0 = 1\) and
\begin{align*} a_k & = ka_{k-1} \text{,} & k & \ge 1 \text{.} \end{align*}
Therefore, for \(k \ge 1\text{,}\) we have
\begin{align*} a_k & = k a_{k-1} \\ & = k ( (k-1) a_{k-2} ) \\ & = k ( (k-1) ( (k-2) a_{k-3} )) \\ & \vdots \\ & = k(k-1)(k-2)\cdots (k - (k-1)) a_{k-k} \text{.} \end{align*}
Simplifying this last expression leads to
\begin{equation*} a_k = k(k-1)(k-2)\cdots 1 \cdot a_0 = k! \text{.} \end{equation*}
Note that the formula \(a_k = k!\) is also valid for \(k=0\) when we adopt the convention \(0! = 1\text{.}\)
Verify that the formula in the solution to the above worked example satisfies the recurrence relation.
Solve the recurrence relation \(a_0 = 1\text{,}\) \(a_1 = \dfrac{1}{2}\text{,}\) and
\begin{align*} a_k & = \dfrac{k}{2(k-1)}\, a_{k-1} \text{,} & k & \ge 2 \text{.} \end{align*}
Solution
Iterating, we obtain
\begin{align*} a_k & = \dfrac{k}{2(k-1)}\, a_{k-1} \\ & = \dfrac{k}{2(k-1)} \left( \dfrac{k-1}{2(k-2)}\, a_{k-2} \right) = \dfrac{k}{2^2(k-2)} \, a_{k-2}\\ & = \dfrac{k}{2^2(k-2)} \left( \dfrac{k-2}{2(k-3)}\, a_{k-3} \right) = \dfrac{k}{2^3(k-3)} \, a_{k-3}\\ & \vdots \\ & = \dfrac{k}{2^{k-1} (k-(k-1)}) \, a_{k-(k-1)} \text{.} \end{align*}
Simplifying this last expression, we obtain
\begin{equation*} a_k = \dfrac{k}{2^{k-1}} \, a_1 = \dfrac{k}{2^k} \text{.} \end{equation*}