1.

Solve these recurrence relations together with the initial conditions given. a) an= an−1+6an−2 for n ≥ 2, a0= 3, a1= 6 b) an= 7a

8_murik_8 [283]

Answer:

a) 3/5·((-2)^n + 4·3^n)
b) 3·2^n - 5^n
c) 3·2^n + 4^n
d) 4 - 3 n
e) 2 + 3·(-1)^n
f) (-3)^n·(3 - 2n)
g) ((-2 - √19)^n·(-6 + √19) + (-2 + √19)^n·(6 + √19))/√19

Step-by-step explanation:

These homogeneous recurrence relations of degree 2 have one of two solutions. Problems a, b, c, e, g have one solution; problems d and f have a slightly different solution. The solution method is similar, up to a point.

If there is a solution of the form $a[n]=r^n$ , then it will satisfy ...

$r^n=c_1\cdot r^{n-1}+c_2\cdot r^{n-2}$

Rearranging and dividing by $r^{n-2}$ , we get the quadratic ...

$r^2-c_1r-c_2=0$

The quadratic formula tells us values of r that satisfy this are ...

$r=\dfrac{c_1\pm\sqrt{c_1^2+4c_2}}{2}$

We can call these values of r by the names r₁ and r₂.

Then, for some coefficients p and q, the solution to the recurrence relation is ...

$a[n]=pr_1^n+qr_2^n$

We can find p and q by solving the initial condition equations:

$\left[\begin{array}{cc}1&1\\r_1&r_2\end{array}\right] \left[\begin{array}{c}p\\q\end{array}\right] =\left[\begin{array}{c}a[0]\\a[1]\end{array}\right]$

These have the solution ...

$p=\dfrac{a[0]r_2-a[1]}{r_2-r_1}\\\\q=\dfrac{a[1]-a[0]r_1}{r_2-r_1}$

_____

Using these formulas on the first recurrence relation, we get ...

a)

$c_1=1,\ c_2=6,\ a[0]=3,\ a[1]=6\\\\r_1=\dfrac{1+\sqrt{1^2+4\cdot 6}}{2}=3,\ r_2=\dfrac{1-\sqrt{1^2+4\cdot 6}}{2}=-2\\\\p=\dfrac{3(-2)-6}{-5}=\dfrac{12}{5},\ q=\dfrac{6-3(3)}{-5}=\dfrac{3}{5}\\\\a[n]=\dfrac{3}{5}(-2)^n+\dfrac{12}{5}3^n$

__

The rest of (b), (c), (e), (g) are solved in exactly the same way. A spreadsheet or graphing calculator can ease the process of finding the roots and coefficients for the given recurrence constants. (It's a matter of plugging in the numbers and doing the arithmetic.)

_____

For problems (d) and (f), the quadratic has one root with multiplicity 2. So, the formulas for p and q don't work and we must do something different. The generic solution in this case is ...

$a[n]=(p+qn)r^n$

The initial condition equations are now ...

$\left[\begin{array}{cc}1&0\\r&r\end{array}\right] \left[\begin{array}{c}p\\q\end{array}\right] =\left[\begin{array}{c}a[0]\\a[1]\end{array}\right]$

and the solutions for p and q are ...

$p=a[0]\\\\q=\dfrac{a[1]-a[0]r}{r}$

__

Using these formulas on problem (d), we get ...

d)

$c_1=2,\ c_2=-1,\ a[0]=4,\ a[1]=1\\\\r=\dfrac{2+\sqrt{2^2+4(-1)}}{2}=1\\\\p=4,\ q=\dfrac{1-4(1)}{1}=-3\\\\a[n]=4-3n$

__

And for problem (f), we get ...

f)

$c_1=-6,\ c_2=-9,\ a[0]=3,\ a[1]=-3\\\\r=\dfrac{-6+\sqrt{6^2+4(-9)}}{2}=-3\\\\p=3,\ q=\dfrac{-3-3(-3)}{-3}=-2\\\\a[n]=(3-2n)(-3)^n$

_____

Comment on problem g

Yes, the bases of the exponential terms are conjugate irrational numbers. When the terms are evaluated, they do resolve to rational numbers.

6 0

2 years ago

How to find the measure of angle. If the measure is be and the angle is 60dgree

Sphinxa [80]

Answer:

i dont under your question. Do you mean on triangles or complamentary or supplementary angles it depends

6 0

3 years ago

Read 2 more answers

I'm trying to pass my performance task and finish high school and I have no idea what it really is asking me to do, rather how t

ElenaW [278]

Answer:

x

8

−

256

Rewrite

x

8

as

(

x

4

)

2

.

(

x

4

)

2

−

256

Rewrite

256

as

16

2

.

(

x

4

)

2

−

16

2

Since both terms are perfect squares, factor using the difference of squares formula,

a

2

−

b

2

=

(

a

+

b

)

(

a

−

b

)

where

a

=

x

4

and

b

=

16

.

(

x

4

+

16

)

(

x

4

−

16

)

Simplify.

Tap for more steps...

(

x

4

+

16

)

(

x

2

+

4

)

(

x

+

2

)

(

x

−

2

)

Step-by-step explanation:

3 0

3 years ago

Find f(g(-4)) and g(f(-4)) using the functions below. f(x) = 2x + 7 g(x) = x2 + 3x - 1

daser333 [38]

Answer:

f(g(-4)) = 13.

g(f(-4)) = -3.

Step-by-step explanation:

g(-4) = (-4)^2 + 3(-4) - 1

= 3.

so f(g(-4))

= f(3)

= 2(3) + 7

= 13.

f(-4) = 2(-4) + 7

= -1.

so g (f(-4))

= g(-1)

= (-1)^2 - 3 - 1

= -3.

8 0

3 years ago

A stone is thrown at an angle of 30° above the horizontal from the top edge of a cliff with an initial speed of 12 m/s. a stop w

stellarik [79]

To solve this problem you must apply the proccedure shown below:
1. You must apply the following formula:
yo=-(Voy)t+gt²/2
2. You have that:
Voy=VoxSin(30°)
Voy=(12 m/s)(Sin(30°))
Voy=6 m/s
t=5.6 s
g=9.8 m/s²
2. When you susbtitute these values into the formula, you obtain:
yo=-(6 m/s)(5.6 s)+(9.8 m/s^2)(5.6 s)²/2
yo=120.06 m
Therefore, the answer is: 120.06 m

3 0

3 years ago