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Chapter 4: Q6E (page 191)

Show that the signals in Exercises 3-6 form a basis for the solution set of the accompanying difference equation.

\({{\bf{5}}^k}{\bf{cos}}\frac{{k\pi }}{{\bf{2}}}\), \({{\bf{5}}^k}sin\frac{{k\pi }}{{\bf{2}}}\), \({y_{k + {\bf{2}}}} + {\bf{25}}{y_k} = {\bf{0}}\)

Short Answer

Expert verified

\({5^k}\cos \frac{{k\pi }}{2}\)and\({5^k}\sin \frac{{k\pi }}{2}\)form the basis of the solution set for the difference equation.

Step by step solution

01

Check for \({{\bf{5}}^k}{\bf{cos}}\frac{{k\pi }}{{\bf{2}}}\)

Substitute\({y_k} = {5^k}\cos \frac{{k\pi }}{2}\)in the difference equation\({y_{k + 2}} + 25{y_k} = 0\).

\(\begin{aligned} {y_{k + 2}} + 25{y_k} &= {5^{k + 2}}\cos \frac{{\left( {k + 2} \right)\pi }}{2} + 25\left( {{5^k}\cos \frac{{k\pi }}{2}} \right)\\ &= {5^k}\left( {{5^2}\cos \frac{{\left( {k + 2} \right)\pi }}{2} + 25\cos \frac{{k\pi }}{2}} \right)\\ &= 25 \cdot {5^k}\left( {\cos \left( {\frac{{k\pi }}{2} + \pi } \right) + \cos \frac{{k\pi }}{2}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {\cos \left( {t + \pi } \right) = - \cos t} \right]\\ &= 25 \cdot {5^k}\left( 0 \right)\end{aligned}\)

02

Check for \({{\bf{5}}^k}{\bf{sin}}\frac{{k\pi }}{{\bf{2}}}\)

Substitute\({y_k} = {5^k}\sin \frac{{k\pi }}{2}\)in the difference equation\({y_{k + 2}} + 25{y_k} = 0\).

\(\begin{aligned} {y_{k + 2}} + 25{y_k} &= {5^{k + 2}}\sin \frac{{\left( {k + 2} \right)\pi }}{2} + 25\left( {{5^k}\sin \frac{{k\pi }}{2}} \right)\\ &= {5^k}\left( {{5^2}\sin \frac{{\left( {k + 2} \right)\pi }}{2} + 25\sin \frac{{k\pi }}{2}} \right)\\ &= 25 \cdot {5^k}\left( {\sin \left( {\frac{{k\pi }}{2} + \pi } \right) + \sin \frac{{k\pi }}{2}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {\sin \left( {t + \pi } \right) = - \sin t} \right]\\ &= 25 \cdot {5^k}\left( 0 \right)\end{aligned}\)

03

Check whether the solution is the basis of the difference equation

For all k, in n-dimensional vector space, the dimension of H is 2. So,\({5^k}\cos \frac{{k\pi }}{2}\)and\({5^k}\sin \frac{{k\pi }}{2}\)form the basis of the solution setfor the difference equation.

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Most popular questions from this chapter

Let \(A\) be any \(2 \times 3\) matrix such that \({\mathop{\rm rank}\nolimits} A = 1\), let u be the first column of \(A\), and suppose \({\mathop{\rm u}\nolimits} \ne 0\). Explain why there is a vector v in \({\mathbb{R}^3}\) such that \(A = {{\mathop{\rm uv}\nolimits} ^T}\). How could this construction be modified if the first column of \(A\) were zero?

[M] Let \(A = \left[ {\begin{array}{*{20}{c}}7&{ - 9}&{ - 4}&5&3&{ - 3}&{ - 7}\\{ - 4}&6&7&{ - 2}&{ - 6}&{ - 5}&5\\5&{ - 7}&{ - 6}&5&{ - 6}&2&8\\{ - 3}&5&8&{ - 1}&{ - 7}&{ - 4}&8\\6&{ - 8}&{ - 5}&4&4&9&3\end{array}} \right]\).

  1. Construct matrices \(C\) and \(N\) whose columns are bases for \({\mathop{\rm Col}\nolimits} A\) and \({\mathop{\rm Nul}\nolimits} A\), respectively, and construct a matrix \(R\) whose rows form a basis for Row\(A\).
  2. Construct a matrix \(M\) whose columns form a basis for \({\mathop{\rm Nul}\nolimits} {A^T}\), form the matrices \(S = \left[ {\begin{array}{*{20}{c}}{{R^T}}&N\end{array}} \right]\) and \(T = \left[ {\begin{array}{*{20}{c}}C&M\end{array}} \right]\), and explain why \(S\) and \(T\) should be square. Verify that both \(S\) and \(T\) are invertible.

Is it possible that all solutions of a homogeneous system of ten linear equations in twelve variables are multiples of one fixed nonzero solution? Discuss.

Find a basis for the set of vectors in\({\mathbb{R}^{\bf{2}}}\)on the line\(y = {\bf{5}}x\).

Question: Determine if the matrix pairs in Exercises 19-22 are controllable.

21. (M) \(A = \left( {\begin{array}{*{20}{c}}0&1&0&0\\0&0&1&0\\0&0&0&1\\{ - 2}&{ - 4.2}&{ - 4.8}&{ - 3.6}\end{array}} \right),B = \left( {\begin{array}{*{20}{c}}1\\0\\0\\{ - 1}\end{array}} \right)\).
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