Book edition 7th

Author(s) Kenneth H. Rosen

Pages 808 pages

ISBN 9780073383095

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Short Answer

What is the generating function for $\{a_{k}\}$ , where $a_{k}$ is the number of solutions of $x_{1} + x_{2} + x_{3} + x_{4} = k$ when $x_{1}, x_{2}, x_{3}$ , and $x_{4}$ are integers with $x_{1} \geq 3$ , $1 \leq x_{2} \leq 5, 0 \leq x_{3} \leq 4$ , and $x_{4} \geq 1 ?$
Use your answer to part (a) to find $a_{7}$ .

(a)The total generating function is $\frac{x^{5} {(1 + x + x^{2} + x^{3} + x^{4})}^{2}}{{(1 - x)}^{2}}$ .

(b) $a_{7} = 10$

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Use Generating Function:

Generating function for the sequence $a_{0}, a_{1}, \dots, a_{k}, \dots$ of real numbers is the infinite series;

G (x) = a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{k} x^{k} + \dots = \sum_{k = 0}^{+ \infty} a_{k} x^{k}

Extended binomial theorem;

role="math" localid="1668610635640" ${(1 + x)}^{u} = \sum_{k = 0}^{+ \infty} (\begin{matrix} u \\ k \end{matrix}) x^{k}$

Since $x_{1} \geq 3$ the series representing $x_{1}$ should then contain only terms with a power of at least 3:

$x^{3} + x^{4} + x^{5} + \dots$

Since $1 \leq x_{2} \leq 5$ the series representing $x_{2}$ should then contain only terms with a power of at least 1 and at most 5:

$x + x^{2} + x^{3} + x^{4} + x^{5}$

Since $0 \leq x_{3} \leq 4$ the series representing $x_{3}$ should then contain only terms with a power of at least 0 and at most 4

$1 + x + x^{2} + x^{3} + x^{4}$

Since $x_{4} \geq 1$ the series representing $x_{4}$ should then contain only terms with a power of at least 1:

$x + x^{2} + x^{3} + \dots$

Step 2: The generating function is the product of the series for each variable:

$(1 + x + x^{2} + x^{3} + x^{4}) (x + x^{2} + x^{3} + x^{4} + x^{5}) (x^{3} + x^{4} + x^{5} + \dots) = x {(1 + x + x^{2} + x^{3} + x^{4})}^{2} (x^{3}) (x) {(1 + x + x^{2} + \dots)}^{2} = x^{5} {(1 + x + x^{2} + x^{3} + x^{4})}^{2} {(1 + x + x^{2} + \dots)}^{2} = x^{5} {(1 + x + x^{2} + x^{3} + x^{4})}^{2} {(\sum_{k = 0}^{+ \infty} x^{k})}^{2} = x^{5} {(1 + x + x^{2} + x^{3} + x^{4})}^{2} \cdot {(\frac{1}{1 - x})}^{2} = \frac{x^{5} {(1 + x + x^{2} + x^{3} + x^{4})}^{2}}{{(1 - x)}^{2}}$

b).

Step 3: Use Extended binomial theorem:

$\frac{x^{5} {(1 + x + x^{2} + x^{3} + x^{4})}^{2}}{{(1 - x)}^{2}} = x^{5} \cdot {(\sum_{k = 0}^{4} x^{k})}^{2} \cdot \frac{1}{{(1 - x)}^{2}} = x^{5} \cdot {(\frac{1 - x^{5}}{1 - x})}^{2} \cdot \frac{1}{{(1 - x)}^{2}} = x^{5} \cdot \frac{{(1 - x^{5})}^{2}}{{(1 - x)}^{4}} = x^{5} \cdot {(1 + (- x^{5}))}^{2} \cdot (1 + (- x))^{- 4}$

By further simplification:

localid="1668668833805" $\frac{x^{5} {(1 + x + x^{2} + x^{3} + x^{4})}^{2}}{{(1 - x)}^{2}} = x^{5} \cdot \sum_{m = 0}^{+ \infty} (\begin{matrix} 2 \\ m \end{matrix}) {(- x^{5})}^{m} . \sum_{k = 0}^{+ \infty} (\begin{matrix} - 4 \\ k \end{matrix}) {(- x)}^{k} = x^{5} \cdot \sum_{m = 0}^{+ \infty} (\begin{matrix} 2 \\ m \end{matrix}) {(- 1)}^{m} x^{5 m} . \sum_{k = 0}^{+ \infty} (\begin{matrix} - 4 \\ k \end{matrix}) {(- 1)}^{k} x^{k} = x^{5} \cdot \sum_{m = 0}^{+ \infty} b_{m} x^{5 m} \cdot \sum_{k = 0}^{+ \infty} c_{k} x^{k}$

Let localid="1668611635844" $b_{m} = (\begin{matrix} 2 \\ m \end{matrix}) (- 1)^{m}$ and $c_{k} = (\begin{matrix} - 4 \\ k \end{matrix}) (- 1)^{k}$

Step 4: The coefficient of x7a7 if m=0 and k=2 :

The coefficient of $x^{7}$ is the sum of the coefficients for each possible combination of localid="1668612574555" $m$ and $k$ :

localid="1668612490312" $a_{7} = b_{0} c_{2} = (\begin{matrix} 2 \\ 0 \end{matrix}) (- 1)^{0} (\begin{matrix} - 4 \\ 2 \end{matrix}) (- 1)^{2} = 10$

Hence, $a_{7} = 10$ .

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Discrete Mathematics and its Applications

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Short Answer

Step by Step Solution

Step 1: Use Generating Function:

Step 2: The generating function is the product of the series for each variable:

Step 3: Use Extended binomial theorem:

Step 4: The coefficient of x7a7 if m=0 and k=2 :

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Discrete Mathematics and its Applications

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Short Answer

What is the generating function for ak, where ak is the number of solutions of x1+x2+x3+x4=k when x1,x2,x3, and x4are integers with x1≥3, 1≤x2≤5,0≤x3≤4, and x4≥1?Use your answer to part (a) to find a7.

Step by Step Solution

Step 1: Use Generating Function:

Step 2: The generating function is the product of the series for each variable:

Step 3: Use Extended binomial theorem:

Step 4: The coefficient of x7a7 if m=0 and k=2 :

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