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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

3 years ago

A rectangle has a length of 80 units and a width of 39 units. Find the length of the diagonal.

levacccp [35]

Equation = a² + b² = c²

a = 80
b = 39
c = Diagonal

a² + b² = c²
80² + 39² = c²
6400 + 1521 = c²
c² = 7921
c = √7921
c = 89

Answer = 89 units

6 0

4 years ago

Dalvin has a toy car collection of 200 toy cars. He keeps 194 of the toy cars on his wall. What percentage of Dalvin's toy car c

aleksklad [387]

Answer: 97%

Step-by-step explanation:

what is 194 out of 200 ? 97%

AND IM SO SORRY ABOUT RED DEV!L

5 0

2 years ago

5y-30=-5y+30 and also can you show the steps :}

KatRina [158]

5y-30=-5y+30

add 30 to both sides

5y = -5y+60

add 5y to both sides

10y = 60

divide both sides by 10

y = 6

hope this helps!

8 0

3 years ago

Stuck on question 1, can anyone help???

poizon [28]

'A' is the square root of 25. That's 5, so take A=5 with you
as you go to the next step.

B is A³. A³ means (A x A x A). We know that 'A' is 5, so 'B' is (5x 5 x 5) = 125 .
Take B=125 with you to the next step.

'C' is B - 25. We know that 'B' is 125. So C = (125 - 25) = 100 .
Take C=100 with you to the next step.

'D' is the square root of 'C'. We know that C=100, so D = √100 .
The square root of 100 is 10, so D=10.
Take D=10 with you to the next step .

'E' is D+39. We know that D=10. So E=(10+39) = 49 .
Take E=49 with you to the last step.

'F' is the square root of 'E'. We know that E=49.

7 0

3 years ago

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Can anyone let me use their Netflix account?Thank you!​

Can anyone let me use their Netflix account?Thank you!