Try the fastest way to create flashcards
Question

a. Sketch a phase portrait for the dynamical system x˙(t+1)=Ax˙(t)\dot x(t+1)=A \dot x(t) where

A=[2132].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 265153<3<1351780\frac{265}{153} < \sqrt{3} < \frac{1351}{780} to estimate cos(30).\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 13517803<106.\frac{1351}{780}-\sqrt{3} < 10^{-6}. d. Based on the data in part b, give an underestimate of the form p/q of 3\sqrt{3} that is better than the one given by Archimedes.

Solution

Verified
Step 1
1 of 4

a.\textbf{a.} We solve:

det(AλI)=0\det (A-\lambda I)=0

2λ132λ=0\begin{vmatrix}2-\lambda&1\\3&2-\lambda\end{vmatrix}=0

(2λ)23=0(2-\lambda )^2-3=0

λ24λ+1=0\lambda ^2-4\lambda +1=0

λ1,2=4±1642=2±3\lambda _{1,2}=\dfrac{4\pm \sqrt{16-4}}{2}=2\pm \sqrt{3}

Thus,

B=[130013].B=\begin{bmatrix}\dfrac{1}{\sqrt{3}}&0\\0&-\dfrac{1}{\sqrt{3}}\end{bmatrix}.

Now, for λ1,2=2±3\lambda _{1,2}=2\pm \sqrt{3}, we solve:

(AλI)x=0(A-\lambda I)x=0

[3133][x1x2]=[00]\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}

3x1+x2=0, 3x13x2=0\mp \sqrt{3}x_1+x_2=0,\ 3x_1\mp \sqrt{3}x_2=0

We can choose eigenvectors

v1,2=[±131]v_{1,2}=\begin{bmatrix}\pm \dfrac{1}{\sqrt{3}}\\1\end{bmatrix}

, thus

S=[131311].S=\begin{bmatrix}\dfrac{1}{\sqrt{3}}&-\dfrac{1}{\sqrt{3}}\\1&1\end{bmatrix}.

Therefore,

At=SBtS1=12[(23)t+(2+3)t33((23)t(2+3)t)3((23)t(2+3)t)(23)t+(2+3)t].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

x0=[x1x2]x_0=\begin{bmatrix}x_1\\x_2\end{bmatrix}

, we have

Atx0=12[(23)t+(2+3)t33((23)t(2+3)t)3((23)t(2+3)t)(23)t+(2+3)t][x1x2]=12[(23)t(x133x2)+(2+3)t(x1+33x2)(23)t(3x1+x2)+(2+3)t(3x1+x2)].\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*}

Create a free account to view solutions

Create a free account to view solutions

Recommended textbook solutions

Linear Algebra with Applications 5th Edition by Otto Bretscher

Linear Algebra with Applications

5th EditionISBN: 9780321796974 (2 more)Otto Bretscher
2,516 solutions
Linear Algebra and Its Applications 5th Edition by David C. Lay, Judi J. McDonald, Steven R. Lay

Linear Algebra and Its Applications

5th EditionISBN: 9780321982384 (3 more)David C. Lay, Judi J. McDonald, Steven R. Lay
2,070 solutions
Elementary Linear Algebra 11th Edition by Howard Anton

Elementary Linear Algebra

11th EditionISBN: 9781118473504Howard Anton
2,932 solutions
Elementary Linear Algebra 12th Edition by Anton Kaul, Howard Anton

Elementary Linear Algebra

12th EditionISBN: 9781119406778Anton Kaul, Howard Anton
3,078 solutions

More related questions

1/4

1/7