Book edition 5th

Author(s) Otto Bretscher

Pages 442 pages

ISBN 9780321796974

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Short Answer

Three holy men (let’s call them Anselm, Benjamin, and Caspar) put little stock in material things; their only earthly possession is a small purse with a bit of gold dust. Each day they get together for the following bizarre bonding ritual: Each of them takes his purse and gives his gold away to the two others, in equal parts. For example, if Anselm has 4 ounces one day, he will give 2 ounces each to Benjamin and Caspar.
**(a) If Anselm starts out with 6 ounces, Benjamin with 1 ounce, and Caspar with 2 ounces, find formulas for the amounts a(t), b(t), and c(t) each will have after t distributions.
Hint: The vector $[\begin{matrix} 1 \\ 1 \\ 1 \end{matrix}], [\begin{matrix} 1 \\ - 1 \\ 0 \end{matrix}] and [\begin{matrix} 1 \\ 0 \\ - 1 \end{matrix}]$ , and will be useful.
(b) Who will have the most gold after one year, that is, after 365 distributions?**

The solutions are,

$a (t) = 3 . 1^{t} + 2 . {(- \frac{1}{2})}^{t} + {(- \frac{1}{2})}^{t} = 3 + 3 (- \frac{1}{2}) b (t) = 3 - 2 . (- \frac{1}{2}) (a) c (t) = 3 - (- \frac{1}{2})$

(b) Benjamin

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Solving for (a)

This is dynamical system, with,

$A = [\begin{matrix} 0 & \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & 0 & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{2} & 0 \end{matrix}]$

This matrix’s eigenvectors are,

$v_{1} = [\begin{matrix} 1 \\ 1 \\ 1 \end{matrix}], v_{2} = [\begin{matrix} 1 \\ - 1 \\ 0 \end{matrix}], v_{3} [\begin{matrix} 1 \\ 0 \\ - 1 \end{matrix}] λ_{1} = 1, λ_{2, 3} = - \frac{1}{2} .$

With the corresponding eigenvalues being .

We also have

$x_{0} = [\begin{matrix} 6 \\ 1 \\ 2 \end{matrix}] = 3 [\begin{matrix} 1 \\ 1 \\ 1 \end{matrix}] + 2 [\begin{matrix} 1 \\ - 1 \\ 0 \end{matrix}] + [\begin{matrix} 1 \\ 0 \\ - 1 \end{matrix}]$

Therefore,

role="math" localid="1659583142610" $a (t) = 3 . 1^{t} + 2 . {(- \frac{1}{2})}^{t} + {(- \frac{1}{2})}^{t} = 3 + 3 {(- \frac{1}{2})}^{t} b (t) = 3 - 2 . {(- \frac{1}{2})}^{t} (a) c (t) = 3 - {(- \frac{1}{2})}^{t}$

Step 2: Solving for (b)

We have,

$a (365) = 3 - 3 {(\frac{1}{2})}^{365}, b (365) = 3 + 2 {(\frac{1}{2})}^{365}, c (365) = 3 + {(\frac{1}{2})}^{365}$

So Benjamin will have the most ounces of gold.

Hence final solution is,

(a)

(b) Benjamin

$a (t) = 3 . 1^{t} + 2 . {(- \frac{1}{2})}^{t} + {(- \frac{1}{2})}^{t} = 3 + 3 {(- \frac{1}{2})}^{t} b (t) = 3 - 2 . {(- \frac{1}{2})}^{t} (a) c (t) = 3 - {(- \frac{1}{2})}^{t}$

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Linear Algebra With Applications

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