Science:Math Exam Resources/Courses/MATH307/April 2010/Question 04 (c)

MATH307 April 2010

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Question 04 (c)

Consider the following recursion relation: $x_{n+1}=ax_{n}+bx_{n-1}$ where $n\geq 1$, and $a,b$ $\epsilon$ $\mathbb {R}$ Assume that $\displaystyle \lambda _{1}\neq \lambda _{2}$. What subspace must the initial vector ${\begin{bmatrix}x_{1}\\x_{0}\end{bmatrix}}$ belong to for the sequence $\{x0,x1,x2,...\}$ to decay, i.e. have $\lim _{n\to \infty }x_{n}$ Set the appropriate condition on b or a, and identify the subspace.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint

Checking a solution serves two purposes: helping you if, after having used the hint, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

Solution

Knowing that $\lambda _{1}\neq \lambda _{2}$ we’ll proceed after having solved for $a$ So the matrix we’ll be working with is: $A_{n}={\begin{bmatrix}1-b&b\\1&0\end{bmatrix}}^{n}$ We’re going to diagonalize this matrix so that we can get an isolated equation for $x_{n}$ which is what we want to decay. Since, based on part b), what the eigenvalues of $A$ are after having solved for $a$ (eigenvalues are: $1$, $-b$), it’s easy to put our matrix into our desired format: $A_{n}={\frac {1}{b+1}}{\begin{bmatrix}1&-b\\1&1\end{bmatrix}}{\begin{bmatrix}1&0\\0&-b^{n}\end{bmatrix}}{\begin{bmatrix}1&b\\-1&1\end{bmatrix}}$ so our new relation from part a) is: ${\begin{bmatrix}x_{n+1}\\x_{n}\end{bmatrix}}$ = ${\frac {1}{b+1}}$${\begin{bmatrix}1&-b\\1&1\end{bmatrix}}{\begin{bmatrix}1&0\\0&-b^{n}\end{bmatrix}}{\begin{bmatrix}1&b\\-1&1\end{bmatrix}}$${\begin{bmatrix}x_{1}\\x_{0}\end{bmatrix}}$ Let’s work through what our $A_{n}$ matrix is, having multiplied our matrices: =${\frac {1}{b+1}}$${\begin{bmatrix}1&-b\\1&1\end{bmatrix}}{\begin{bmatrix}1&0\\0&-b^{n}\end{bmatrix}}{\begin{bmatrix}1&b\\-1&1\end{bmatrix}}$ =${\frac {1}{b+1}}$${\begin{bmatrix}1&b^{n+1}\\1&-b^{n}\end{bmatrix}}{\begin{bmatrix}1&b\\-1&1\end{bmatrix}}$ =${\frac {1}{b+1}}$${\begin{bmatrix}1-b^{n+1}&b+b^{n+1}\\1+b^{n}&b-b^{n}\end{bmatrix}}$ And so our general equation becomes: ${\begin{bmatrix}x_{n+1}\\x_{n}\end{bmatrix}}$ = ${\frac {1}{b+1}}$${\begin{bmatrix}1-b^{n+1}&b+b^{n+1}\\1+b^{n}&b-b^{n}\end{bmatrix}}$${\begin{bmatrix}x_{1}\\x_{0}\end{bmatrix}}$ from this system, we can see what $x_{n}$ must be: $x_{n}={\frac {(1+b^{n})x_{1}+(b-b^{n})x_{0}}{b+1}}$ Now that we know what $x_{n}$ is based on our choice of variables, let’s solve what we set out to do, i.e. $\lim _{n\to \infty }x_{n}=0$ so $\lim _{n\to \infty }$ ${\frac {(1+b^{n})x_{1}+(b-b^{n})x_{0}}{b+1}}$ Now comes what our choice in $b$ must be, we need these $b$ terms to disappear completely, thus we need them to tend towards zero. So assume $\left|b\right|<1$, Now as $n\to \infty$, we’re left with: ${\frac {(1)x_{1}+(b)x_{0}}{b+1}}$ which we want to be zero. We come again to our choice of $b$, Let’s assume that $b=0$ which implies that our choice of $x_{0}$ is arbitrary and that $x_{1}=0$, which makes sense as in this case $x_{0}$ will not be used at all because of our choice in $b$. Continuing along, we see that where ${\vec {x}}={\begin{bmatrix}0\\x_{0}\end{bmatrix}}$ and where our choice of $b=0$ that our initial vector belongs to the nullspace! We can see that as even if we assume that $b\neq 0$ then if we assume that $x_{1}=x_{0}$ then the initial vector is simple the zero vector! Now we come to the assumption, what if $b\neq 0$ and $x_{1}\neq x_{0}$, well again we’re left with: ${\frac {(1)x_{1}+(b)x_{0}}{b+1}}$ And now the only way we can get this to be zero is if $x_{1}+bx_{0}=0$ Which would mean that $x_{1}=-bx_{0}$ So we come to the conclusion that our initial vector is ${\begin{bmatrix}-bx_{0}\\x_{0}\end{bmatrix}}$ for $0\leq \left|b\right|<1$