/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q5E Show that the signals in Exercis... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Q5E (page 191)

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

\({{\bf{3}}^k}\), \(k{\left( { - {\bf{3}}} \right)^k}\) \({y_{k + {\bf{2}}}} + {\bf{6}}{y_{k + {\bf{1}}}} + {\bf{9}}{y_k} = {\bf{0}}\)

Short Answer

Expert verified

\({3^k}\) and \(k{\left( { - 3} \right)^k}\) form a basis of the solution set for the difference equation.

Step by step solution

01

Check for \({\left( { - {\bf{3}}} \right)^k}\)

Substitute \({y_k} = {\left( { - 3} \right)^k}\) in thedifference equation \({y_{k + 2}} + 6{y_{k + 1}} + 9{y_k} = 0\).

\(\begin{aligned} {y_{k + 2}} + 6{y_{k + 1}} + 9{y_k} &= {\left( { - 3} \right)^{k + 2}} + 6{\left( { - 3} \right)^{k + 1}} + 9{\left( { - 3} \right)^k}\\ &= {\left( { - 3} \right)^k}\left[ {{{\left( { - 3} \right)}^2} + 6\left( { - 3} \right) + 9} \right]\\ &= {\left( 3 \right)^k}\left[ {9 - 18 + 9} \right]\\ &= 0\end{aligned}\)

02

Check for \(k{\left( { - {\bf{3}}} \right)^k}\)

Substitute\({y_k} = k{\left( { - 3} \right)^k}\)in the difference equation\({y_{k + 2}} + 6{y_{k + 1}} + 9{y_k} = 0\).

\(\begin{aligned} {y_{k + 2}} + 6{y_{k + 1}} + 9{y_k} &= \left( {k + 2} \right){\left( { - 3} \right)^{k + 2}} + 6\left( {k + 1} \right){\left( { - 3} \right)^{k + 1}} + 9k{\left( { - 3} \right)^k}\\ &= {\left( { - 3} \right)^k}\left[ {\left( {k + 2} \right){{\left( { - 3} \right)}^2} + 6\left( {k + 1} \right)\left( { - 3} \right) + 9k} \right]\\ &= {\left( 3 \right)^k}\left[ {9k + 18 - 18k - 18 + 9k} \right]\\ &= 0\end{aligned}\)

Thus, for all\(k\),\({\left( { - 3} \right)^k}\)is in the solution set H.

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, \({3^k}\) and \(k{\left( { - 3} \right)^k}\) forms a basis of the solution set for the difference equation.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

(M) Determine whether w is in the column space of \(A\), the null space of \(A\), or both, where

\({\mathop{\rm w}\nolimits} = \left( {\begin{array}{*{20}{c}}1\\2\\1\\0\end{array}} \right),A = \left( {\begin{array}{*{20}{c}}{ - 8}&5&{ - 2}&0\\{ - 5}&2&1&{ - 2}\\{10}&{ - 8}&6&{ - 3}\\3&{ - 2}&1&0\end{array}} \right)\)

Suppose a \({\bf{5}} \times {\bf{6}}\) matrix A has four pivot columns. What is dim Nul A? Is \({\bf{Col}}\,A = {\mathbb{R}^{\bf{3}}}\)? Why or why not?

Use coordinate vector to test whether the following sets of poynomial span \({{\bf{P}}_{\bf{2}}}\). Justify your conclusions.

a. \({\bf{1}} - {\bf{3}}t + {\bf{5}}{t^{\bf{2}}}\), \( - {\bf{3}} + {\bf{5}}t - {\bf{7}}{t^{\bf{2}}}\), \( - {\bf{4}} + {\bf{5}}t - {\bf{6}}{t^{\bf{2}}}\), \({\bf{1}} - {t^{\bf{2}}}\)

b. \({\bf{5}}t + {t^{\bf{2}}}\), \({\bf{1}} - {\bf{8}}t - {\bf{2}}{t^{\bf{2}}}\), \( - {\bf{3}} + {\bf{4}}t + {\bf{2}}{t^{\bf{2}}}\), \({\bf{2}} - {\bf{3}}t\)

Question: Exercises 12-17 develop properties of rank that are sometimes needed in applications. Assume the matrix \(A\) is \(m \times n\).

14. Show that if \(Q\) is an invertible, then \({\mathop{\rm rank}\nolimits} AQ = {\mathop{\rm rank}\nolimits} A\). (Hint: Use Exercise 13 to study \({\mathop{\rm rank}\nolimits} {\left( {AQ} \right)^T}\).)

Let \(B = \left\{ {\left( {\begin{array}{*{20}{c}}{\bf{1}}\\{ - {\bf{4}}}\end{array}} \right),\,\left( {\begin{array}{*{20}{c}}{ - {\bf{2}}}\\{\bf{9}}\end{array}} \right)\,} \right\}\). Since the coordinate mapping determined by B is a linear transformation from \({\mathbb{R}^{\bf{2}}}\) into \({\mathbb{R}^{\bf{2}}}\), this mapping must be implemented by some \({\bf{2}} \times {\bf{2}}\) matrix A. Find it. (Hint: Multiplication by A should transform a vector x into its coordinate vector \({\left( {\bf{x}} \right)_B}\)).
See all solutions

Recommended explanations on Math Textbooks

Pure Maths

Read Explanation

Probability and Statistics

Read Explanation

Decision Maths

Read Explanation

Calculus

Read Explanation

Geometry

Read Explanation

Theoretical and Mathematical Physics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.