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Question

# a. Sketch a phase portrait for the dynamical system $\dot x(t+1)=A \dot x(t)$ where$A=\left[\begin{array}{ll} 2 & 1 \\ 3 & 2 \end{array}\right].$b. In his paper "On the Measurement of the Circle", the great Greek mathematician Archimedes (c. 280-210 B.C.) uses the approximation $\frac{265}{153} < \sqrt{3} < \frac{1351}{780}$ to estimate $\cos \left(30^{\circ}\right).$ He does not explain how he arrived at these estimates. Explain how we can obtain these approximations from the dynamical system in part a. c. Without using technology, explain why $\frac{1351}{780}-\sqrt{3} < 10^{-6}.$ d. Based on the data in part b, give an underestimate of the form p/q of $\sqrt{3}$ that is better than the one given by Archimedes.

Solution

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$\textbf{a.}$ We solve:

$\det (A-\lambda I)=0$

$\begin{vmatrix}2-\lambda&1\\3&2-\lambda\end{vmatrix}=0$

$(2-\lambda )^2-3=0$

$\lambda ^2-4\lambda +1=0$

$\lambda _{1,2}=\dfrac{4\pm \sqrt{16-4}}{2}=2\pm \sqrt{3}$

Thus,

$B=\begin{bmatrix}\dfrac{1}{\sqrt{3}}&0\\0&-\dfrac{1}{\sqrt{3}}\end{bmatrix}.$

Now, for $\lambda _{1,2}=2\pm \sqrt{3}$, we solve:

$(A-\lambda I)x=0$

$\begin{bmatrix}\mp \sqrt{3}&1\\3&\mp \sqrt{3}\end{bmatrix}\begin{bmatrix}x_1\\x_2\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}$

$\mp \sqrt{3}x_1+x_2=0,\ 3x_1\mp \sqrt{3}x_2=0$

We can choose eigenvectors

$v_{1,2}=\begin{bmatrix}\pm \dfrac{1}{\sqrt{3}}\\1\end{bmatrix}$

, thus

$S=\begin{bmatrix}\dfrac{1}{\sqrt{3}}&-\dfrac{1}{\sqrt{3}}\\1&1\end{bmatrix}.$

Therefore,

$A^t=SB^tS^{-1}=\dfrac{1}{2}\begin{bmatrix}(2-\sqrt{3})^t+(2+\sqrt{3})^t&-\dfrac{\sqrt{3}}{3}((2-\sqrt{3})^t-(2+\sqrt{3})^t)\\-\sqrt{3}((2-\sqrt{3})^t-(2+\sqrt{3})^t)&(2-\sqrt{3})^t+(2+\sqrt{3})^t\end{bmatrix}.$

So, for

$x_0=\begin{bmatrix}x_1\\x_2\end{bmatrix}$

, we have

\begin{align*}A^tx_0&=\dfrac{1}{2}\begin{bmatrix}(2-\sqrt{3})^t+(2+\sqrt{3})^t&-\dfrac{\sqrt{3}}{3}((2-\sqrt{3})^t-(2+\sqrt{3})^t)\\-\sqrt{3}((2-\sqrt{3})^t-(2+\sqrt{3})^t)&(2-\sqrt{3})^t+(2+\sqrt{3})^t\end{bmatrix}\begin{bmatrix}x_1\\x_2\end{bmatrix}\\&=\dfrac{1}{2}\begin{bmatrix}(2-\sqrt{3})^t(x_1-\dfrac{\sqrt{3}}{3}x_2)+(2+\sqrt{3})^t(x_1+\dfrac{\sqrt{3}}{3}x_2)\\(2-\sqrt{3})^t(-\sqrt{3}x_1+x_2)+(2+\sqrt{3})^t(\sqrt{3}x_1+x_2)\end{bmatrix}.\end{align*}

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