\[A= \left( \begin{array}{ccc} y & 2 & 2x \\ x & y & 2 \\ 1 & x & y \end{array} \right) \]
.This matrix is constructed by building diagonals. First a 1 in the bottom left corner, above this a diagonal of
\[x\]
's (an approximation to \[ \sqrt[3]{2}\]
, which we guess), above this a diagonal of \[y\]
's (an approximation to \[\sqrt[3]{2^2}\]
, which we guess), above this a diagonal of 2's, and finally in the upper right, a \[2x\]
.If we take a non zero vector
\[\mathbf{v}\]
and repeatedly multiply this vector by \[A\]
. The components of \[A^n \mathbf{v|}\]
will tend to the ration \[2^{2/3} \colon 2^{1/3} \colon 1 \]
.Let our guess for
\[x= \sqrt[3]{2}\]
be 1 and let our guess for \[y=\sqrt[3]{2^2}\]
be 2. Then \[A= \left( \begin{array}{ccc} 2 & 2 & 2 \\ 1 & 2 & 2 \\ 1 & 1 & 2 \end{array} \right)\]
Let
\[\mathbf{v}= \begin{pmatrix}1\\1\\1\end{pmatrix}\]
.Then
\[A \mathbf{v}= \left( \begin{array}{ccc} 2 & 2 & 2 \\ 1 & 2 & 2 \\ 1 & 1 & 2 \end{array} \right) \begin{pmatrix}1\\1\\1\end{pmatrix} = \begin{pmatrix}6\\5\\4\end{pmatrix}\]
.\[A^2 \mathbf{v}= \left( \begin{array}{ccc} 2 & 2 & 2 \\ 1 & 2 & 2 \\ 1 & 1 & 2 \end{array} \right) \begin{pmatrix}6\\5\\4\end{pmatrix} = \begin{pmatrix}30\\24\\19\end{pmatrix}\]
.\[A^3 \mathbf{v}= \left( \begin{array}{ccc} 2 & 2 & 2 \\ 1 & 2 & 2 \\ 1 & 1 & 2 \end{array} \right) \begin{pmatrix}30\\24\\19\end{pmatrix} = \begin{pmatrix}146\\116\\92\end{pmatrix}\]
.\[A^4 \mathbf{v}= \left( \begin{array}{ccc} 2 & 2 & 2 \\ 1 & 2 & 2 \\ 1 & 1 & 2 \end{array} \right) \begin{pmatrix}146\\116\\92\end{pmatrix} = \begin{pmatrix}708\\562\\446\end{pmatrix}\]
.Hence
\[2^{2/3} \simeq \frac{708}{446}=1.587443946\]
and \[2^{1/3} \simeq \frac{562}{446}=1.260089686\]
.Compare these with the true values
\[2^{2/3} = 1.587401052\]
and \[2^{1/3} = 1.259921050\]
, both to 10 significant figures.To find an approximation to
\[5^{1/4}\]
use the matrix \[A= \left( \begin{array}{cccc} 4 & 5 & 5 & 10 \\ 2 & 4 & 5 & 5 \\ 1 & 2 & 4 & 5 \\ 1 & 1 & 2 & 4 \end{array} \right) \]
.The 1's in the second diagonal are an approximation to
\[5^{1/4}\]
, the 2's an approximation to \[5^{2/4}\]
, the 4's an approximation to \[5^{3/4}\]
.Let
\[\mathbf{v}= \begin{pmatrix}1\\1\\1\\1\end{pmatrix}\]
.Then
\[A \mathbf{v}= \left( \begin{array}{cccc} 4 & 5 & 5 & 10 \\ 2 & 4 & 5 & 5 \\ 1 & 2 & 4 & 5 \\ 1 & 1 & 2 & 4 \end{array} \right) \begin{pmatrix}1\\1\\1\\1\end{pmatrix} = \begin{pmatrix}24\\16\\12\\8\end{pmatrix}\]
.\[A^2 \mathbf{v}= \left( \begin{array}{cccc} 4 & 5 & 5 & 10 \\ 2 & 4 & 5 & 5 \\ 1 & 2 & 4 & 5 \\ 1 & 1 & 2 & 4 \end{array} \right) \begin{pmatrix}24\\16\\12\\8\end{pmatrix} = \begin{pmatrix}316\\212\\144\\96\end{pmatrix}\]
.\[A^3 \mathbf{v}= \left( \begin{array}{cccc} 4 & 5 & 5 & 10 \\ 2 & 4 & 5 & 5 \\ 1 & 2 & 4 & 5 \\ 1 & 1 & 2 & 4 \end{array} \right) \begin{pmatrix}316\\212\\144\\96\end{pmatrix} = \begin{pmatrix}4004\\2680\\1796\\1200\end{pmatrix}\]
.Hence
\[5^{3/4} \simeq \frac{4004}{1200}=3.336666667\]
, \[5^{2/4} \simeq \frac{2680}{1200}=2.233333333\]
and \[5^{1/4} \simeq \frac{1796}{1200}=2.1.496666667\]
.Compare these with the true values
\[5^{3/4} = 3.343701525\]
, \[5^{2/4} = 2.236067977\]
, and \[5^{1/4} = 1.495348781\]
, all to 10 significant figures.