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Chapter 4: Q32E (page 191)

What is the order of the following difference equation? Explain your answer\({y_{k + 3}} + {a_1}{y_{k + 2}} + {a_2}{y_{k + 1}} + {a_3}{y_k} = 0\).

Short Answer

Expert verified

If \({a_3} = 0\) and \({a_2} \ne 0\), the order is 2.

If \({a_3} \ne 0\), the order is 3.

If \({a_3} = {a_2} = 0\) and \({a_1} \ne 0\), the order is 1.

If \({a_3} = {a_2} = {a_1} = 0\), the order is 0.

Step by step solution

01

Discuss the attributes of a difference equation

The given difference equation is \({y_{k + 3}} + {a_1}{y_{k + 2}} + {a_2}{y_{k + 1}} + {a_3}{y_k} = 0\). The order of a difference equation depends on the values of the coefficients \({a_1}\), \({a_2}\), and \({a_3}\).

02

Discuss the cases corresponding to the values of the coefficients

If \({a_3} = 0\) and \({a_2} \ne 0\), then the difference equation can be reduced to \({y_{k + 2}} + {a_1}{y_{k + !}} + {a_2}{y_k} = 0\), which implies that its order is 2.

If \({a_3} \ne 0\), then the difference equation cannot be reduced further, which implies that its order is 3.

If \({a_3} = {a_2} = 0\) and \({a_1} \ne 0\), then the difference equation can be reduced to \({y_{k + 1}} + {a_1}{y_k} = 0\), which implies that its order is 1.

If all the coefficients are zero, that is, \({a_3} = {a_2} = {a_1} = 0\), then the difference equation does not possess any signal value, which implies that its order is 0.

03

Draw a conclusion

If \({a_3} = 0\) and \({a_2} \ne 0\), the order is 2.

If \({a_3} \ne 0\), the order is 3.

If \({a_3} = {a_2} = 0\) and \({a_1} \ne 0\), the order is 1.

If \({a_3} = {a_2} = {a_1} = 0\), the order is 0.

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Most popular questions from this chapter

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

20. \(A = \left( {\begin{array}{*{20}{c}}{.8}&{ - .3}&0\\{.2}&{.5}&1\\0&0&{ - .5}\end{array}} \right),B = \left( {\begin{array}{*{20}{c}}1\\1\\0\end{array}} \right)\).

(M) Let \(H = {\mathop{\rm Span}\nolimits} \left\{ {{{\mathop{\rm v}\nolimits} _1},{{\mathop{\rm v}\nolimits} _2}} \right\}\) and \(K = {\mathop{\rm Span}\nolimits} \left\{ {{{\mathop{\rm v}\nolimits} _3},{{\mathop{\rm v}\nolimits} _4}} \right\}\), where

\({{\mathop{\rm v}\nolimits} _1} = \left( {\begin{array}{*{20}{c}}5\\3\\8\end{array}} \right),{{\mathop{\rm v}\nolimits} _2} = \left( {\begin{array}{*{20}{c}}1\\3\\4\end{array}} \right),{{\mathop{\rm v}\nolimits} _3} = \left( {\begin{array}{*{20}{c}}2\\{ - 1}\\5\end{array}} \right),{{\mathop{\rm v}\nolimits} _4} = \left( {\begin{array}{*{20}{c}}0\\{ - 12}\\{ - 28}\end{array}} \right)\)

Then \(H\) and \(K\) are subspaces of \({\mathbb{R}^3}\). In fact, \(H\) and \(K\) are planes in \({\mathbb{R}^3}\) through the origin, and they intersect in a line through 0. Find a nonzero vector w that generates that line. (Hint: w can be written as \({c_1}{{\mathop{\rm v}\nolimits} _1} + {c_2}{{\mathop{\rm v}\nolimits} _2}\) and also as \({c_3}{{\mathop{\rm v}\nolimits} _3} + {c_4}{{\mathop{\rm v}\nolimits} _4}\). To build w, solve the equation \({c_1}{{\mathop{\rm v}\nolimits} _1} + {c_2}{{\mathop{\rm v}\nolimits} _2} = {c_3}{{\mathop{\rm v}\nolimits} _3} + {c_4}{{\mathop{\rm v}\nolimits} _4}\) for the unknown \({c_j}'{\mathop{\rm s}\nolimits} \).)

In Exercise 1, find the vector x determined by the given coordinate vector \({\left( x \right)_{\rm B}}\) and the given basis \({\rm B}\).

1. \({\rm B} = \left\{ {\left( {\begin{array}{*{20}{c}}{\bf{3}}\\{ - {\bf{5}}}\end{array}} \right),\left( {\begin{array}{*{20}{c}}{ - {\bf{4}}}\\{\bf{6}}\end{array}} \right)} \right\},{\left( x \right)_{\rm B}} = \left( {\begin{array}{*{20}{c}}{\bf{5}}\\{\bf{3}}\end{array}} \right)\)

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?

Let \({M_{2 \times 2}}\) be the vector space of all \(2 \times 2\) matrices, and define \(T:{M_{2 \times 2}} \to {M_{2 \times 2}}\) by \(T\left( A \right) = A + {A^T}\), where \(A = \left( {\begin{array}{*{20}{c}}a&b\\c&d\end{array}} \right)\).

  1. Show that \(T\)is a linear transformation.
  2. Let \(B\) be any element of \({M_{2 \times 2}}\) such that \({B^T} = B\). Find an \(A\) in \({M_{2 \times 2}}\) such that \(T\left( A \right) = B\).
  3. Show that the range of \(T\) is the set of \(B\) in \({M_{2 \times 2}}\) with the property that \({B^T} = B\).
  4. Describe the kernel of \(T\).
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