# 数学代写|离散数学作业代写discrete mathematics代考|MATH200

## 数学代写|离散数学作业代写discrete mathematics代考|Solved Problems

1. Determine whether the sequence $\left{f_n\right}={3 n}$ is a solution of the recurrence relation: $f_n=2 f_{n-1}-f_{n-2}$, for $n=2,3,4, \ldots$
Solution.
Suppose $f_n=3 n$. Then for $n \geq 2$,
\begin{aligned} f_n &=2 f_{n-1}-f_{n-2} \ &=2[3(n-1)]-3(n-2) \quad \text { since } f_n=3 n \ &=6 n-6-3 n+6=3 n \end{aligned}
$\therefore \quad\left{f_n\right}$, where $f_n=3 n$, is a solution of the recurrence relation.
2. Show that the sequence $\left{f_n\right}$ is a solution of the recurrence relation $f_n=-3 f_{n-1}+4 f_{n-2}$ if $f_n=2(-4)^n+3$
Solution.
\begin{aligned} f_n &=-3 f_{n-1}+4 f_{n-2} \ &=-3\left[2(-4)^{n-1}+3\right]+4\left[2(-4)^{n-2}+3\right] \ &=-6(-4)^{n-1}-9+8(-4)^{n-2}+12 \ &=-6(-4)^{n-1}+8(-4)^{n-2}+3 \ &=-6(-4)^{n-1}-2(-4)^{n-1}+3 \ &=2(-4)^n+3 \end{aligned}
$\therefore f_n=2(-4)^n+3$ is a solution of the recurrence relation.
Now, we discuss about a class of recurrence relations known as linear recurrence relations with constant coefficients.

## 数学代写|离散数学作业代写discrete mathematics代考|Linear Recurrence Relation

A recurrence relation of the form
$$a_0 f_n+a_1 f_{n-1}+a_2 f_{n-2}+\cdots+a_k f_{n-k}=f(n)$$
where $a_i$ ‘s are constants, is called a linear recurrence relation with constant coefficients. The recurrence relation (2.4) is known as a $k^{\text {th }}$-order recurrence relation, provided both $a_0$ and $a_k$ are non-zero.

Note: The phrase ” $k^{\text {th }}$-order” means that each term in the sequence depends only on the $k$ previous terms.
Example 1:
Consider the Fibonacci sequence defined by the recurrence relation $f_n=f_{n-1}+f_{n-2}, n \geq 2$ and the initial conditions $f_0=0$ and $f_1=1$. The recurrence relation is called a second-order relation because $f_n$ depends on the two previous terms of $f_n$.
Example 2:
Consider the recurrence relation $f(k)-5 f(k-1)+6 f(k-2)=4 k+10$ defined for $k \geq 2$, together with the initial conditions $f(0)=\frac{7}{3}$ and $f(1)=5$. Clearly, it is a second-order linear recurrence relation.

# 离散数学代写

## 数学代写|离散数学作业代写discrete mathematics代考|Solved Problems

1. 判断是否顺序 \left{f_n!right $}={3 \mathrm{n}}$ 是递归关系的解: $f_n=2 f_{n-1}-f_{n-2}$ ，为了 $n=2,3,4, \ldots$
解决方案。
认为 $f_n=3 n$. 然后为 $n \geq 2$ ，
$$f_n=2 f_{n-1}-f_{n-2} \quad=2[3(n-1)]-3(n-2) \quad \text { since } f_n=3 n=6 n-6-3 n+6=3 n$$
解决方案。
$$f_n=-3 f_{n-1}+4 f_{n-2} \quad=-3\left[2(-4)^{n-1}+3\right]+4\left[2(-4)^{n-2}+3\right]=-6(-4)^{n-1}-9+8(-4)^{n-2}$$
$\therefore f_n=2(-4)^n+3$ 是递归关系的解。
现在，我们讨论一类称为常系数线性递推关系的递推关系。

## 数学代写|离散数学作业代写discrete mathematics代考|Linear Recurrence Relation

$$a_0 f_n+a_1 f_{n-1}+a_2 f_{n-2}+\cdots+a_k f_{n-k}=f(n)$$

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