7.6: Exercises for Linear Recurrences
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Exercises
Find a basis for the space \(V\) of sequences \([x_{n})\) satisfying the following recurrences, and use it to find the sequence satisfying \(x_{0} = 1\), \(x_{1} = 2\), \(x_{2} = 1\).
- \(x_{n+3} = -2x_{n} + x_{n+1} + 2x_{n+2}\)
- \(x_{n+3} = -6x_{n} + 7x_{n+1}\)
- \(x_{n+3} = -36x_{n} + 7x_{n+2}\)
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b. \(\{[1), [2^n), [(-3)^n)\}\); \(x_n = \frac{1}{20}(15 + 2^{n+3} + (-3)^{n+1})\)
In each case, find a basis for the space \(V\) of all sequences \([x_{n})\) satisfying the recurrence, and use it to find \(x_{n}\) if \(x_{0} = 1\), \(x_{1} = -1\), and \(x_{2} = 1\).
- \(x_{n+3} = x_{n} + x_{n+1} - x_{n+2}\)
- \(x_{n+3} = -2x_{n} + 3x_{n+1}\)
- \(x_{n+3} = -4x_{n} + 3x_{n+2}\)
- \(x_{n+3} = x_{n} - 3x_{n+1} + 3x_{n+2}\)
- \(x_{n+3} = 8x_{n} - 12x_{n+1} + 6x_{n+2}\)
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- \(\{[1), [n), [(-2)^n)\}\); \(x_n = \frac{1}{9}(5 -6n + (-2)^{n+2})\)
- \(\{[1), [n), [n^2)\}\); \(x_n = 2(n - 1)^2 - 1\)
Find a basis for the space \(V\) of sequences \([x_{n})\) satisfying each of the following recurrences.
- \(x_{n+2} = -a^{2}x_{n} + 2ax_{n+1}\), \(a \neq 0\)
- \(x_{n+2} = -abx_{n} + (a + b)x_{n+1}\), \((a \neq b)\)
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b. \(\{[a^{n}), [b^{n})\}\)
In each case, find a basis of \(V\).
- \(V = \{[x_{n}) \mid x_{n+4} = 2x_{n+2} - x_{n+3}\), for \(n \geq 0\}\)
- \(V = \{[x_{n}) \mid x_{n+4} = -x_{n+2} + 2x_{n+3}\), for \(n \geq 0\}\)
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b. \([1, 0, 0, 0, 0, \dots)\), \([0, 1, 0, 0, 0, \dots)\),
\([0, 0, 1, 1, 1, \dots)\), \([0, 0, 1, 2, 3, \dots)\)
Suppose that \([x_{n})\) satisfies a linear recurrence relation of length \(k\). If \(\{\mathbf{e}_{0} = (1, 0, \dots, 0), \\ \mathbf{e}_{1} = (0, 1, \dots, 0), \dots, \mathbf{e}_{k-1} = (0, 0, \dots, 1)\}\) is the standard basis of \(\mathbb{R}^k\), show that
\[x_{n} = x_{0}T(\mathbf{e}_{0}) + x_{1}T(\mathbf{e}_{1}) + \cdots + x_{k-1}T(\mathbf{e}_{k-1}) \nonumber \]
holds for all \(n \geq k\). (Here \(T\) is as in Theorem .)
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Add texts here. Do not delete this text first.
Show that the shift operator \(S\) is onto but not one-to-one. Find \(\text{ker }S\).
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Find a basis for the space \(V\) of all sequences \([x_{n})\) satisfying \(x_{n+2} = -x_{n}\)
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By Remark 2,
\[\def\arraystretch{1.5} \begin{array}{l} \left[i^n + (-i)^n\right) = \left[2, 0, -2, 0, 2, 0, -2, 0, \dots \right) \\ \left[i(i^n - (-i)^n)\right) = \left[0, -2, 0, 2, 0, -2, 0, 2, \dots \right) \end{array} \nonumber \]
are solutions. They are linearly independent and so are a basis.