If

Find the general term xn.

ringearV
2021-08-16
Answered

Discrete mathematics

If${x}_{1}=2,{x}_{n}=4{X}_{n-1}-4n\mathrm{\forall}n\ge 2$ .

Find the general term xn.

If

Find the general term xn.

You can still ask an expert for help

Nicole Conner

Answered 2021-08-17
Author has **97** answers

Step 1

Given recurrence relation is

${x}_{n}=4{x}_{n-1}-4n$ , ${x}_{1}=2$

Rewriting given recurrence relation, we have

${x}_{n}-4{x}_{n-1}=-4n$

Associated homogeneous recurrence relation is

${x}_{n}-4{x}_{n-1}=0$

Auxiliary equation is

$m-4=0$

$\Rightarrow m=4$

$\Rightarrow$ Complementary solution is

$x}_{n}^{c}=C{\left(4\right)}^{n$ C is arbitrary constant.

Step 2

To find particular solution:

the non-homogeneous function is$Q\left(n\right)=-4n=0{\left(1\right)}^{n}-4n{\left(1\right)}^{n}$

$\Rightarrow$ particular solution will have form

${x}_{n}^{p}=A+Bn$

$\Rightarrow {x}_{n-1}^{p}=A+B(n-1)=A+Bn-B$

Substituting in given recurrence relation

$A+Bn=4(A+Bn-B)-4n$

$\Rightarrow A+Bn-4A-4Bn+4B=-4n$

$\Rightarrow (-3A+4B)+(-3B)n=-4n$

$\Rightarrow -3A+4B=0,-3B=-4$ by comparing coefficients of 1&n.

$\Rightarrow B=\frac{4}{3}$ and $3A=4B=4\times \frac{4}{3}$

$\Rightarrow A=\frac{16}{9}$ and $B=\frac{4}{3}$

Hence particular solution is

${x}_{n}^{p}=A+Bn=\frac{16}{9}+\frac{4}{3}n$

$\Rightarrow {x}_{n}^{p}=\frac{16}{9}+\frac{4}{3}n$

Step 3

General solution is

$x}_{n}={x}_{n}^{c}+{x}_{n}^{p$

${x}_{n}=C{\left(4\right)}^{n}+\frac{16}{9}+\frac{4}{3}n$

Given that,${x}_{1}=2$

$\Rightarrow 2=C{\left(4\right)}^{1}+\frac{16}{9}+\frac{4}{3}\times 1$

$\Rightarrow 2=4C+\frac{16+12}{9}$

Given recurrence relation is

Rewriting given recurrence relation, we have

Associated homogeneous recurrence relation is

Auxiliary equation is

Step 2

To find particular solution:

the non-homogeneous function is

Substituting in given recurrence relation

Hence particular solution is

Step 3

General solution is

Given that,

asked 2021-07-28

Let A, B, and C be sets. Show that

asked 2021-08-18

Discrete Mathematics Basics

1) Determine whether the relation R on the set of all Web pages is reflexive, symmetric, antisymmetric, and/or transitive, where$(a,b)\in R$ if and only if

I) everyone who has visited Web page a has also visited Web page b.

II) there are no common links found on both Web page a and Web page b.

III) there is at least one common link on Web page a and Web page b.

1) Determine whether the relation R on the set of all Web pages is reflexive, symmetric, antisymmetric, and/or transitive, where

I) everyone who has visited Web page a has also visited Web page b.

II) there are no common links found on both Web page a and Web page b.

III) there is at least one common link on Web page a and Web page b.

asked 2021-08-02

Suppose that A is the set of sophomores at your school and B is the set of students in discrete mathematics at your school. Express each of these sets in terms of A and B.

a) the set of sophomores taking discrete mathematics in your school

b) the set of sophomores at your school who are not taking discrete mathematics

c) the set of students at your school who either are sophomores or are taking discrete mathematics

Use these symbols:$\cap \cup$

a) the set of sophomores taking discrete mathematics in your school

b) the set of sophomores at your school who are not taking discrete mathematics

c) the set of students at your school who either are sophomores or are taking discrete mathematics

Use these symbols:

asked 2021-08-15

How many elements are in the set
{ 0, { { 0 } }?

asked 2022-07-03

11, Find the least integer n such that f(x) is 0( nx ) for each

of these functions.

a) f(x) = 2x + (logx)^10

b) f(x) = (x^4 + 5logx) / (x^4 + 10)

12/A sequence of pseudorandom numbers is generated as follows

x0 = 4

x_i = ( 6x_i–1 + 5 ) mod 13 if i > 0

Find x6

asked 2022-09-04

Prove that ${2}^{n+1}>(n+2)\cdot \mathrm{sin}(n)$ for all positive integers n.

Proof: If P(n) represents the given proposition.

(1) Basic Step: P(n) for $n=1$ is

$${2}^{1+1}>(1+2)\cdot \mathrm{sin}(1)$$

$${2}^{2}>3\cdot \mathrm{sin}(1)$$

$$4>3\cdot \mathrm{sin}(1)$$

Since $\mathrm{sin}(n)\le 1$ and $1<4/3$ and so $\mathrm{sin}(n)<4/3$ by putting $n=1$, we obtain

$$\mathrm{sin}(1)<4/3\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}4>3\cdot \mathrm{sin}(1)$$

Which is true.

(2) Induction Step: Suppose that P(n) is true for $n=k$, i.e. Let

$${2}^{k+1}>(k+2)\cdot \mathrm{sin}(k)$$

To prove that P(n) to be true for $n=k+1$

$${2}^{(k+1)+1}={2}^{k+1}2=2\cdot {2}^{k+1}>2(k+2)\cdot \mathrm{sin}(k)$$

How to prove P(n) to be true for $n=k+1$. I got stuck here. Would appreciate for your assistance. Also review my proof if there is any mistake while writing.

Proof: If P(n) represents the given proposition.

(1) Basic Step: P(n) for $n=1$ is

$${2}^{1+1}>(1+2)\cdot \mathrm{sin}(1)$$

$${2}^{2}>3\cdot \mathrm{sin}(1)$$

$$4>3\cdot \mathrm{sin}(1)$$

Since $\mathrm{sin}(n)\le 1$ and $1<4/3$ and so $\mathrm{sin}(n)<4/3$ by putting $n=1$, we obtain

$$\mathrm{sin}(1)<4/3\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}4>3\cdot \mathrm{sin}(1)$$

Which is true.

(2) Induction Step: Suppose that P(n) is true for $n=k$, i.e. Let

$${2}^{k+1}>(k+2)\cdot \mathrm{sin}(k)$$

To prove that P(n) to be true for $n=k+1$

$${2}^{(k+1)+1}={2}^{k+1}2=2\cdot {2}^{k+1}>2(k+2)\cdot \mathrm{sin}(k)$$

How to prove P(n) to be true for $n=k+1$. I got stuck here. Would appreciate for your assistance. Also review my proof if there is any mistake while writing.

asked 2022-03-18

2x ≡ 6(mod 4)