Mathematics: Consider the sequence defined recursively by a1=−2, a2=2, an+1=8an−1−2an. we can use matrix diagonalization to find an explicit formula for an. a. find a matrix that satisfies [anan+1]=m[an−1an]

Consider the sequence defined recursively by a1=−2,

Ответ:

ja815792

13.03.2022

The matrix satisfy the $\left[\begin{array}{ccc}a_{n+1}\\a_n\\\end{array}\right] =M \left[\begin{array}{ccc}a_n\\a_{n-1}\end{array}\right]$ is $M^{n-1}=\left[\begin{array}{ccc}2&-4\\1&0\end{array}\right]^{-1}\left[\begin{array}{ccc}2^{n-1}&0\\0&(-1)^{n-1}\end{array}\right]\left[\begin{array}{ccc}2&-4\\1&1\end{array}\right]$

Step-by-step explanation:

Given: Consider the sequence defined recursively by $a_1=-2,\;a_2=2,\;a_{n+1}=-2a_n+8a_{n-1}\;\rm{for},\; n\geq2$ .

According to question,

From the given values: $\left[\begin{array}{ccc}a_3\\a_2\\\end{array}\right] =\left[\begin{array}{ccc}-2&8\\1&0\end{array}\right] \left[\begin{array}{ccc}a_2\\a_1\end{array}\right]$

And generally we can write for $(n+1,\;n)$ : $\left[\begin{array}{ccc}a_{n+1}\\a_n\\\end{array}\right] =\left[\begin{array}{ccc}-2&8\\1&0\end{array}\right] \left[\begin{array}{ccc}a_n\\a_{n-1}\end{array}\right]$

So, here

$M=\left[\begin{array}{ccc}-2&8\\1&0\end{array}\right]$

So using diagonalise equation of the matrix using eigen values we can write: $M=S^{-1}DS$

$M=\left[\begin{array}{ccc}2&-4\\1&0\end{array}\right]^{-1}\left[\begin{array}{ccc}2&0\\0&-4\end{array}\right]\left[\begin{array}{ccc}2&-4\\1&1\end{array}\right]$

Which

$M^{n-1}=\left[\begin{array}{ccc}2&-4\\1&0\end{array}\right]^{-1}\left[\begin{array}{ccc}2^{n-1}&0\\0&(-1)^{n-1}\end{array}\right]\left[\begin{array}{ccc}2&-4\\1&1\end{array}\right]$

Therefore, the matrix satisfy the $\left[\begin{array}{ccc}a_{n+1}\\a_n\\\end{array}\right] =M \left[\begin{array}{ccc}a_n\\a_{n-1}\end{array}\right]$ is $M^{n-1}=\left[\begin{array}{ccc}2&-4\\1&0\end{array}\right]^{-1}\left[\begin{array}{ccc}2^{n-1}&0\\0&(-1)^{n-1}\end{array}\right]\left[\begin{array}{ccc}2&-4\\1&1\end{array}\right]$ .

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Ответ:

anondriap

13.03.2022

Mathematics

$\begin{cases}a_1=-2\\a_2=2\\a_{n+1}=-2a_n+8a_{n-1}&\text{for }n\ge2\end{cases}$

From the given starting values, we have

$\begin{bmatrix}a_3\\a_2\end{bmatrix}=\begin{bmatrix}-2&8\\1&0\end{bmatrix}\begin{bmatrix}a_2\\a_1\end{bmatrix}$

and more generally, the (n+1,n)

-th term pair is captured by

$\begin{bmatrix}a_{n+1}\\a_n\end{bmatrix}=\begin{bmatrix}-2&8\\1&0\end{bmatrix}\begin{bmatrix}a_n\\a_{n-1}\end{bmatrix}$

so that

$\mathbf M=\begin{bmatrix}-2&8\\1&0\end{bmatrix}$

We notice that

$\begin{bmatrix}a_{n+1}\\a_n\end{bmatrix}=\mathbf M\begin{bmatrix}a_n\\a_{n-1}\end{bmatrix}=\mathbf M^2\begin{bmatrix}a_{n-1}\\a_{n-2}\end{bmatrix}=\cdots=\mathbf M^{n-1}\begin{bmatrix}a_2\\a_1\end{bmatrix}$

which is to say that the explicit formula for a_n

is given by the dot product of the second row of $\mathbf M^{n-1}$ with $\begin{bmatrix}a_2&a_1\end{bmatrix}^\top$ .

First we diagonalize $\mathbf M$ . It has eigenvalues $\lambda$ , where

$\det(\mathbf M-\lambda\mathbf I)=\begin{vmatrix}-2-\lambda&8\\1&-\lambda\end{vmatrix}=0\implies\lambda_1=2,\lambda_2=-4$

with corresponding eigenvectors $\vec\eta$ , where

$(\mathbf M-2\mathbf I)\vec\eta_1=\mathbf0\implies\vec\eta_1=\begin{bmatrix}2\\1\end{bmatrix}$
$(\mathbf M+4\mathbf I)\vec\eta_2=\mathbf0\implies\vec\eta_2=\begin{bmatrix}-4\\1\end{bmatrix}$

Then we have

$\mathbf M=\mathbf S^{-1}\mathbf{DS}$

where $\mathbf S$ is the matrix whose columns are $\vec\eta_1$ and $\vec\eta_2$ and $\mathbf D$ is the diagonal matrix whose non-zero entries are the eigenvalues of $\mathbf M$ . So

$\mathbf M=\begin{bmatrix}2&-4\\1&1\end{bmatrix}^{-1}\begin{bmatrix}2&0\\0&-4\end{bmatrix}\begin{bmatrix}2&-4\\1&1\end{bmatrix}$

which gives

$\mathbf 
M^{n-1}=\begin{bmatrix}2&-4\\1&1\end{bmatrix}^{-1}\begin{bmatrix}2^{n-1}&0\\0&(-4)^{n-1}\end{bmatrix}\begin{bmatrix}2&-4\\1&1\end{bmatrix}$

From here, you would just need to determine the bottom row of the final matrix product.

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Ответ:

BeautyxQueen

15.11.2021

Mathematics

The first place after the decimal point is the tenths place, followed by the hundredths place, then the thousandths place. So the answer is:

Two hundred thirty five and six hundred seventy eight thousandths.

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Consider the sequence defined recursively by a1=−2, a2=2, an+1=8an−1−2an. we can use matrix diagonalization to find an explicit formula for an. a. find a matrix that satisfies [anan+1]=m[an−1an]

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