Given the following matrix : $$ A= \begin{bmatrix} 1 & 1 \\ 1 & 0 \\ \end{bmatrix} $$ and the sequences $a_n$, $b_n$, $ c_n$, $d_n$ such that $ A^n$= \begin{bmatrix} a_n& b_n \\ c_n & d_n \\ \end{bmatrix} I need to find the pair $(x,y)$ such that $a_{n+2}=xa_{n+1}+ya_n$ for every integer $n$. What do I need to do? Do I have to find a form for the matrix raised to the power $n$? Can I use Hamilton Cayley here? Could someone please explain in detail because I haven't worked with matrices and sequences until now. The problem also asks me the number of values for $n$ such that $b_n=c_n$ and the limit as $n$ goes to infinity of $\frac{a_n}{b_n}$.
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yes............. – Lola Jul 11 '17 at 15:34
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1Why don't you compute a few values of $A^n$ (and so also the sequences $(a_n)$ etc.) to see what kind of behaviour you get; you may even be able to guess the answer from this. – Angina Seng Jul 11 '17 at 15:37
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2Note that $A$ is one of the two matrices defining the matricial representation of the Stern-Brocot tree – G Cab Jul 11 '17 at 15:46
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1as well as of the matrix representation of Fibonacci Numbers – G Cab Jul 11 '17 at 15:53
1 Answers
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Note that
$$\begin{pmatrix} a_{n+1}&b_{n+1}\\c_{n+1}&d_{n+1} \end{pmatrix} = \begin{pmatrix} a_n&b_n\\c_n&d_n \end{pmatrix} \begin{pmatrix} 1&1\\1&0 \end{pmatrix} = \begin{pmatrix} a_n + b_n & a_n \\ c_n + d_n & c_n \end{pmatrix} $$
and
$$\begin{pmatrix} a_{n+2}&b_{n+2}\\c_{n+2}&d_{n+2} \end{pmatrix} = \begin{pmatrix} a_{n+1}&b_{n+1}\\c_{n+1}&d_{n+1} \end{pmatrix} \begin{pmatrix} 1&1\\1&0 \end{pmatrix} = \begin{pmatrix} 2a_n + b_n & a_n + b_n \\ 2c_n + d_n & c_n + d_n \end{pmatrix} $$
So,
$$ a_{n+2} = 2a_n + b_n$$ $$ a_{n+1} = a_n + b_n $$
Can you solve for $x$ and $y$ from here?
aras
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I tried solving it but I get stuck. Can you tell me what I should do furthermore? – Lola Jul 12 '17 at 17:24
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@Lola You want to find $x$ and $y$ such that $a_{n+2} = a_{n+1} + a_n$. You are given $a_{n+2} = 2a_n + b_n$ and $a_{n+1} = a_n + b_n$. Since you don't know $b_n$, it is a good idea to subtract these two equations so the $b_n$ terms cancel out. You are left with $a_{n+2} - a_{n+1} = a_n$. Rearranging this gives $a_{n+2} = a_{n+1} + a_n$. So, $x=y=1$. Did that help you? – aras Jul 12 '17 at 20:40