Question

What is the generating function for $\left\{a_k\right\}$, 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 $x_1\ge 3$, $1\le x_2\le 5,0\le x_3\le 4$, and $x_4\ge 1?$

Use your answer to part (a) to find $a_7$.

Short answer

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

(b) $a_7=10$

Step-by-step solution

Use Generating Function:

Generating function for the sequence$a_0,a_1,\dots ,a_k,\dots$of real numbers is the infinite series;

$G\left(x\right)=a_0+a_{1}x+a_2{x}^2+\dots +a_k{x}^k+\dots =\sum _{k=0}^{+\infty }{a}_k{x}^k$

Extended binomial theorem;

role="math" localid="1668610635640" ${(1+x)}^u=\sum _{k=0}^{+\infty }\left(\begin{array}{c}u\\ k\end{array}\right)x^k$

Since $x_1\ge 3$ 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 $1\le x_2\le 5$ 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 $0\le x_3\le 4$ 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 $x_4\ge 1$ the series representing $x_4$should then contain only terms with a power of at least 1:

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

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

$$\begin{gathered} \left(1+x+x^2+x^3+x^4\right)\left(x+x^2+x^3+x^4+x^5\right)\left(x^3+x^4+x^5+\dots \right) \\ =x{\left(1+x+x^2+x^3+x^4\right)}^2\left(x^3\right)\left(x\right){\left(1+x+x^2+\dots \right)}^2 \\ =x^5{\left(1+x+x^2+x^3+x^4\right)}^2{\left(1+x+x^2+\dots \right)}^2 \\ =x^5{\left(1+x+x^2+x^3+x^4\right)}^2{\left(\sum _{k=0}^{+\infty }{x}^k\right)}^2 \\ =x^5{\left(1+x+x^2+x^3+x^4\right)}^2·{\left(\frac{1}{1-x}\right)}^2 \\ =\frac{x^5{\left(1+x+x^2+x^3+x^4\right)}^2}{{(1-x)}^2} \end{gathered}$$

b).

Use Extended binomial theorem:

$$\begin{gathered} \frac{x^5{\left(1+x+x^2+x^3+x^4\right)}^2}{{(1-x)}^2}=x^5·{\left(\sum _{k=0}^4{x}^k\right)}^2·\frac{1}{{(1-x)}^2} \\ =x^5·{\left(\frac{1-x^5}{1-x}\right)}^2·\frac{1}{{(1-x)}^2} \\ =x^5·\frac{{\left(1-x^5\right)}^2}{{(1-x)}^4} \\ =x^5·{\left(1+\left(-x^5\right)\right)}^2·(1+(-x){)}^{-4} \end{gathered}$$

By further simplification:

localid="1668668833805" $$\begin{gathered} \frac{x^5{\left(1+x+x^2+x^3+x^4\right)}^2}{{(1-x)}^2}=x^5·\sum _{m=0}^{ \\ +\infty }\left(\begin{array}{c}2\\ m\end{array}\right){\left(-x^5\right)}^m.\sum _{k=0}^{ \\ +\infty }\left(\begin{array}{c}-4\\ k\end{array}\right){\left(-x\right)}^k \\ =x^5·\sum _{m=0}^{ \\ +\infty }\left(\begin{array}{c}2\\ m\end{array}\right){\left(-1\right)}^m{x}^{5m}.\sum _{k=0}^{ \\ +\infty }\left(\begin{array}{c}-4\\ k\end{array}\right){\left(-1\right)}^k{x}^k \\ =x^5·\sum _{m=0}^{+\infty }{b}_m{x}^{5m}·\sum _{k=0}^{+\infty }{c}_k{x}^k \end{gathered}$$

Letlocalid="1668611635844" $b_m=\left(\begin{array}{c}2\\ m\end{array}\right)(-1{)}^m$and$c_k=\left(\begin{array}{c}-4\\ k\end{array}\right)(-1{)}^k$

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" $\mathrm{m}$and $k$:

localid="1668612490312" $$\begin{gathered} a_7=b_0{c}_2 \\ =\left(\begin{array}{c}2\\ 0\end{array}\right)(-1{)}^0\left(\begin{array}{c}-4\\ 2\end{array}\right)(-1{)}^2 \\ =10 \end{gathered}$$

Hence,$a_7=10$.