show-that-the-signals-in-exercises-3-6-form-a-basis-for-the-solution-set-of-the-accompanying-differe

Question

Show that the signals in Exercises $3-6$ form a basis for the solution set of the accompanying difference equation. 
The signals and equation in Exercise 1

Nikolaos Alexandrakis · Accepted Answer

wick of these two signals from problems 03 to decay in minor street like a. And we need to see you too, so that the V&#x27;s two signals form a basis for the solution set off the other line. This could be gruesome. We will apply like a sorority test, which you can find it at the end off things to four. Seven from the book. And we consider the casserole schematics off the system, which is this one. And the girls are the best. Says that if these Matics isn&#x27;t very people for at least one K then without Hlynur its independence. Um, so we try for simplicity toe study The turbulent off these mavericks four k equals zero. That is the definite off the mavericks one city well minus three on these evaluates not my arms. Three mind must say with equals minus six, which is other than zero. So this mattress is in vegetable and we have lena have independence off the solution. Um, this fucking it&#x27;s consistent with theory 17 from Bates 2 50 that says that this solution off the often and for their described the Anamika discreet occasion forms sodium Asian funds to two dimensional solutions

Michael Jacobsen · Answer

in this video, we have a second order difference equation that&#x27;s provided here. And our goal is to find a basis for H where h is a subspace of this space of all signals. That is the solution to this difference equation. To start out, we have two solutions that we&#x27;re going to verify. We have for the 1st 1 that y que is equal to negative three to the power of K. What&#x27;s substitute this solution into the difference equation? To verify that it is indeed a solution. First, why K plus two would be negative three to the power of K plus two plus six times why K plus one is negative three to the power of K plus one plus nine times y que is nine times negative three to the power of K. Now, for this expression we have so far we can factor out negative three to the power of K And then what we&#x27;re left with is negative three squared plus six times negative three to the power one plus night. It&#x27;s all together. This is going to result in negative three to the Power k. This will be multiplying the group where we have negative three squared for a 96 by negative three for a negative 18 plus nine, and this results in zero altogether. So we have one solution so far, and keep in mind that are difference. Equation is of second order. That implies that subspace h is of dimension to so if we can find another solution, will be ready to describe the basis for age. So let&#x27;s consider next y que equals k to the K times Negative three to the power of K For this particular signal. If we substituted into the difference equation here we obtain the following. Why K Plus Two is going to be K plus two times negative three to the power of K plus two. Then we have plus six times why K plus one which is going to be a K plus one times negative three to the power of K plus one. Then lastly, we have plus nine times y que where y que is given to be K times Negative three to the Power K. If this equals zero, then we&#x27;ll have a second solution and will be ready to describe the basis. Let&#x27;s use the same strategy as before and factor out negative three to the Power K. This time we&#x27;re going to be left with K plus two times negative three squared, plus six times K plus one times negative three to the power of one. I&#x27;ll just write as negative three, then plus nine times K. So let&#x27;s simplify inside the group first to see what we&#x27;re dealing with. We have first from here ah value of nine which will distribute into this group giving us nine K plus 18. Next. Considered the six and the negative three. Multiplying them together gives us negative 18 which will distribute into this group that gives us naked of 18 K minus a team. Then copy down Plus nine K. Now all together we have, Let&#x27;s see here two copies of nine K with negative 18 K that cancels out than 18 minus 18 produces zero. So this entire quantity is euro, and we&#x27;ve just verified that the signal is a solution. And we previously verified that the other signal isas Well, now we&#x27;re ready to state our conclusion. Let&#x27;s capture the fact that we have a second order difference equation From that we can say since the dimension of age is equal to two. And since let&#x27;s use negative three to the power K notation and que times negative three to the power of K are not multiples of each other. We have that negative three of the power of K and K times negative Three of the power of K is a basis four h so sometimes refer to these two particular symbols signals as to be a fundamental set of solutions to this difference equation that we started with. Now that we have a basis, our solution is complete.

Nikolaos Alexandrakis · Answer

we will. So that these two signals for a salute for, um a basis for the solution set off this underlined, uh, discreetly Grissom. First of all, we will apply the casserole test, which can be found at the end off Bates. So for seven in the book and we proceed by evaluating by calculating the castle automatics, which is these magics and consider the best 10 says that if there is one case, that&#x27;s the element of this matrix or in other words, of the sorority on is not zero. Um, that means that these two a store functions are linearly independent. We proceed by calculating the determinant for off this mother forgettable zero for simplicity. And we see that equals the detainment off this matics. That&#x27;s by substituting Terrier called zero. Um, the mathematics and the retirement equals five with this positive. So we have a limb our independence of these two solutions. Now, in order to verify that these two functions do soul of the clarification, we&#x27;ll just notice that five toe the K plus two course sign was okay. Plus toe I multiplied by pi over too. Like Woz, Kate by over two plus bye on by basic diplomatic identity is this Does this equals, don&#x27;t he? Five would be played by 53 k, which is 530.55 square multiplied by my nose because sinus Okay, bye over too. And so the minors gets out. And so the museum steps find and we proceed toe the games of the other signal in the exact same on the exemption away.

Nikolaos Alexandrakis · Answer

we will. So that the history signals are Who are you with? Something. So all of these things Critical ation forum and basis for the solution set of this occasion. Now we first keep in mind the theorem 17 from the book Bates Toe 50. Who says that the solution said off these discreet occasions Since it is off their three, it is three dimensional vector space. If it were so for their end, it would be an n dimensional vector space. So us in the biggest precise is we proceed the with psoriatic test and evaluate like a certain Patrick&#x27;s with this this one for Katie, plus one and keep lasto. And now we just need to find one. Okay, Uh, what was the best day? Who substitute was one in the bar? You? Because our young, which is the pregnant of these metrics and see if is not zero with me equals zero. He is inevitable, Andi. Then we&#x27;ll have. Then we will have dinner. Independence by the castle. Atticus gets on the phone at the end of hates to 47 eso week of one minus one on one, in here on zero and one and minus one. And, uh, to the five and five square, which is five. All right, so we just do some basic literal transformations, and, uh, we see that these metrics, we will put the equivalent in a parentis. Is this that land is These are them. Equals the government off the equivalent Mataric switches. If with is if we substructure the first column from the second, that is one minus one and one on the first column and then zero and then one minus one equals zero. And then we fell. Um, I miss fund doing here. Three on duh. Then five and five squared my loose minus one with vibrates at six. And mindless one, which is five minus one. It&#x27;s going for and solve this determined evaluates to the government off, please. Too happy to Matt. Tricks on this is minus 24 minus 12. It&#x27;s evaluates of minors 36. Now this is other than zero. And so we are done on the Mavericks is convertible for K equals zero. So by fear him by the basis field among aids stool drinking nine. We thought that see, these three function seems there&#x27;ll in early independent and, uh, they solve this this good system, they for use for basis for the solution set off this occasion

Michael Jacobsen · Answer

in this example, we have 1/3 order difference equation that&#x27;s provided its homogeneous since the right hand side is zero and were also provided with two different solutions Negative one of the K and three of the Power K. So what we&#x27;re going to do is let h denote the subspace of all signals s that contains a solution to this difference equation. And our objective is to determine if this will form a basis for H. Let&#x27;s start off with the following observation. The difference equation is off order three due to the three here by the sub script K plus three since its of order three. This implies that the dimension of the subspace h is three. So now we can say that are set negative one to the power of K and three to the power of K cannot form a basis. Four h. Let&#x27;s explain why that&#x27;s the case. We know since the dimension is equal to three, every basis of age must have exactly three vectors. But this particular set, whether it&#x27;s linearly, independent or not, only contains two different vectors, and so it could never form a basis. So we have to extend this set by finding 1/3 solution, and then and only then can we determine whether this is really a basis.

Problem

Show that the signals in Exercises $3-6$ form a b…

Problem 3 Easy Difficulty

Show that the signals in Exercises $3-6$ form a basis for the solution set of the accompanying difference equation.
The signals and equation in Exercise 1

Answer

the given signals $\left\{ 2 ^ { k } , ( - 4 ) ^ { k } \right\}$ are linearly independent.

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David C. Lay, Steven R. Lay, Judi J. McDonald

Linear Algebra and Its Applications

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the given signals $\left\{ 2 ^ { k } , ( - 4 ) ^ { k } \right\}$ are linearly independent.

Linear Algebra and Its Applications

Lily A.

Catherine R.

Heather Z.

Caleb E.

Vectors Intro

Vector Basics - Example 1

David C. Lay, Steven R. Lay, Judi J. McDonaldLinear Algebra and Its Applications

Lily A.

Catherine R.

Heather Z.

Caleb E.

David C. Lay, Steven R. Lay, Judi J. McDonald

Linear Algebra and Its Applications