## 数学代写|离散数学作业代写discrete mathematics代考|Recurrence Relations obtained from Solutions

Before giving an algorithm for solving a recurrence relation, we will examine a few recurrence relations that arise from certain closed form expressions. The procedure is illustrated by the following examples.

1. Form the recurrence relation given $f_n=3 \cdot 5^n, n \geq 0$.
Solution.
If $n \geq 1$, then
\begin{aligned} f_n=3 \cdot 5^n &=3 \cdot 5 \cdot 5^{n-1} \ &=5 \cdot 3 \cdot 5^{n-1} \ &=5 f_{n-1} \end{aligned}
$\therefore$ The recurrence relation is $f_n=5 f_{n-1}$ with $f_0=3$.
2. Find the recurrence relation satisfying $y_n=A(3)^n+B(-2)^n$.
Solution.
Given $y_n=A(3)^n+B(-2)^n$.
\begin{aligned} \therefore \quad y_{n+1} &=A(3)^{n+1}+B(-2)^{n+1}=3 A(3)^n-2 B(-2)^n \ y_{n+2} &=A(3)^{n+2}+b(-2)^{n+2}=9 A(3)^n+4 B(-2)^n \end{aligned}
Eliminating $A$ and $B$ from the above equations,
$$\left|\begin{array}{ccc} y_n & 1 & 1 \ y_{n+1} & 3 & -2 \ y_{n+2} & 9 & 4 \end{array}\right|=0$$
Expanding along column 1,
$$\begin{array}{ll} & y_n(12+18)-y_{n+1}(4-9)+y_{n+2}(-2-3)=0 \ \text { or } & 30 y_n+5 y_{n+1}-5 y_{n+2}=0 \ \text { or } & 6 y_n+y_{n+1}-y_{n+2}-0 \ \text { or } & y_{n+2}-y_{n+1}-6 y_n=0 \end{array}$$

## 数学代写|离散数学作业代写discrete mathematics代考|Solving Linear Homogenous Recurrence Relations

Consider a linear homogenous recurrence relation of degree $k$ with constant coefficients
$$f_n=a_1 f_{n-1}+a_2 f_{n-2}+\cdots+a_k f_{n-k}$$
where $a_1, a_2, \ldots, a_k$ are real numbers and $a_k \neq 0$. The hasic approach for solving linear homogenous recurrence relations is to look for solutions of the form $f_n=r^n$, where $r$ is a constant. Note that $f_n=r^n$ is a solution of the recurrence relation $f_n=a_1 f_{n-1}+a_2 f_{n-2}+\cdots+a_k f_{n-k}$ if and only if
$$r^n=c_1 r^{n-1}+c_2 r^{n-2}+\cdots+c_k r^{n-k} .$$

When both sides of this equation are divided by $r^{n-k}$ and the right-hand side is subtracted from the left, we obtain
$$r^k-c_1 r^{k-1}-c_2 r^{k-2}-\cdots-c_{k-1} r-c_k=0 .$$
Consequently, the sequence $\left{f_n\right}$ with $f_n=r^n$ is a solution if and only if $r$ is a solution of this last equation.

# 离散数学代写

## 数学代写|离散数学作业代写discrete mathematics代考|Recurrence Relations obtained from Solutions

1. 形成给定的递归关系 $f_n=3 \cdot 5^n, n \geq 0$.
解决方案。
如果 $n \geq 1$ ，然后
$$f_n=3 \cdot 5^n=3 \cdot 5 \cdot 5^{n-1} \quad=5 \cdot 3 \cdot 5^{n-1}=5 f_{n-1}$$
$\therefore$ 递归关系是 $f_n=5 f_{n-1}$ 和 $f_0=3$.
2. 找到递归关系满足 $y_n=A(3)^n+B(-2)^n$.
解决方案。
鉴于 $y_n=A(3)^n+B(-2)^n$.
$$\therefore \quad y_{n+1}=A(3)^{n+1}+B(-2)^{n+1}=3 A(3)^n-2 B(-2)^n y_{n+2} \quad=A(3)^{n+2}+b(-2)^{n+2}=9 A(3)^n$$
消除 $A$ 和 $B$ 从上面的方程式，
沿着第 1 列展开，
$$y_n(12+18)-y_{n+1}(4-9)+y_{n+2}(-2-3)=0 \text { or } 30 y_n+5 y_{n+1}-5 y_{n+2}=0 \text { or } 6 y_n+y_{n+1}$$

## 数学代写|离散数学作业代写discrete mathematics代考|Solving Linear Homogenous Recurrence Relations

$$f_n=a_1 f_{n-1}+a_2 f_{n-2}+\cdots+a_k f_{n-k}$$

$f_n=a_1 f_{n-1}+a_2 f_{n-2}+\cdots+a_k f_{n-k}$ 当且仅当
$$r^n=c_1 r^{n-1}+c_2 r^{n-2}+\cdots+c_k r^{n-k} .$$

$$r^k-c_1 r^{k-1}-c_2 r^{k-2}-\cdots-c_{k-1} r-c_k=0 .$$

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