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

MATH307 April 2010

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[hide] 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?

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[show] Hint

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[show] 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 $0\leq \left|b\right|<1$