Mathematics, 16.07.2019 16:30 sswd
Consider the sequence defined recursively by a1=β2, a2=2, an+1=8anβ1β2an. we can use matrix diagonalization to find an explicit formula for an. a. find a matrix that satisfies [anan+1]=m[anβ1an]
Answers: 1
![\left[\begin{array}{ccc}a_{n+1}\\a_n\\\end{array}\right] =M \left[\begin{array}{ccc}a_n\\a_{n-1}\end{array}\right]](/tpl/images/0096/9623/8920e.png)
is ![M^{n-1}=\left[\begin{array}{ccc}2&-4\\1&0\end{array}\right]^{-1}\left[\begin{array}{ccc}2^{n-1}&0\\0&(-1)^{n-1}\end{array}\right]\left[\begin{array}{ccc}2&-4\\1&1\end{array}\right]](/tpl/images/0096/9623/9fad8.png)
.![\left[\begin{array}{ccc}a_3\\a_2\\\end{array}\right] =\left[\begin{array}{ccc}-2&8\\1&0\end{array}\right] \left[\begin{array}{ccc}a_2\\a_1\end{array}\right]](/tpl/images/0096/9623/7396e.png)
: ![\left[\begin{array}{ccc}a_{n+1}\\a_n\\\end{array}\right] =\left[\begin{array}{ccc}-2&8\\1&0\end{array}\right] \left[\begin{array}{ccc}a_n\\a_{n-1}\end{array}\right]](/tpl/images/0096/9623/41999.png)
![M=\left[\begin{array}{ccc}-2&8\\1&0\end{array}\right]](/tpl/images/0096/9623/ce0c9.png)
![M=\left[\begin{array}{ccc}2&-4\\1&0\end{array}\right]^{-1}\left[\begin{array}{ccc}2&0\\0&-4\end{array}\right]\left[\begin{array}{ccc}2&-4\\1&1\end{array}\right]](/tpl/images/0096/9623/a6aa5.png)
-th term pair is captured by
is given by the dot product of the second row of 
with 
.
. It has eigenvalues 
, where
, where
is the matrix whose columns are 
and 
and 
is the diagonal matrix whose non-zero entries are the eigenvalues of 
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Consider the sequence defined recursively by a1=β2, a2=2, an+1=8anβ1β2an. we can use matrix diagonal...
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