Question

Solving recurrence relation?
a(n) - 6a(n-1) + 8a(n-2) = 0 , n > 2 , a(0) = 2 , a(1) = 4

Answer 1

is it
\[a ^{n} - 6a ^{(n-1)} +8a ^{(n-2)} = 0\]

Answer 2

yes but all of the exponents are on the bottom

Answer 3

you know the first 2 values, and thats all you need to determine the 3rd, 4th, 5th, etc ....

i would make a short list and see if a pattern develops

Answer 4

there was one other method that i believe converts a_(n-r) -> x^r and solves it somehow

Answer 5

an - 6an-1 + 8an-2 = 0

n=2

a2 - 6a1 + 8a0 = 0
a2 - 6(4) + 8(2) = 0 ; a2 = 8



n=3

a3 - 6a2 + 8a1 = 0
a3 - 6(8) + 8(4) = 0 ; a3 = 16


n=4

a4 - 6a3 + 8a2 = 0
a4 - 6(16) + 8(8) = 0 ; a4 = 60


etc...

Answer 6

a(0) is plugged into 6a(n-1)?

Answer 7

when n=2, we can solve for \(a_3\) by letting n=2

\[\Large a_n-6a_{n-1}+8a_{n-2}=0\]
\[\Large a_2-6a_{2-1}+8a_{2-2}=0\]
\[\Large a_2-6a_{1}+8a_{0}=0\]
we know the values of a0 and a1, 2 and 4
\[\Large a_2-6(4)+8(2)=0\]
solve for \(a_2\)

Answer 8

now im lost again lol

Answer 9

well a2 would be 8

Answer 10

yes, and then we know the value of 3 elements of the sequence, 2, 4, 8

a3 can be found using a2 and a1 in the same manner, but this is a long manner and can be tiresome

Answer 11

i see an explanation that i can try (i need the practice) using ar^(n-r) in place of a_(n-r)

Answer 12

r is prolly a bad chose for the base tho ...

\[\Large a_n-6a_{n-1}+8a_{n-2}=0\]
replace all the subs by r^(n-k); such that say; an-2 becomes ar^(n-2)
\[\Large ar^n-6ar^{n-1}+8ar^{n-2}=0\]factor out an ar^(n-2), since its the highest power
\[\Large ar^{n-2}(r^2-6r+8)=0\]and solve for r

Answer 13

the tiny alpha looked like an "a" to me ... it appears that "alpha" is just some constant

Answer 14

(r-4)(r-2), therefore r = 2 or 4

the general solution is therefore:\[C_0~2^n+C_1~4^n\]
since we know a0 and a1 to be 2 and 4 ... those cheeky monkeys!! we can solve for the constants

\[C_0~2^0+C_1~4^0=2\]\[C_0~2^1+C_1~4^1=4\]

Answer 15

lol....are you a teacher?

Answer 16

not officially, but i do love to learn new stuff. And this method was fairly new for me :)

Answer 17

sadly i have to head to work! to early for math problems for me lol

Answer 18

im already at work, lousy office job ..... drives me sane i tell you, SANE!!

Answer 19

its nice that openstudy saves your progress whenever you want to continue

Answer 20

have fun :)

Answer 21

hah thank for the help have a nice day!

Answer 22

if we continued the plug and play action ...

an - 6an-1 + 8an-2 = 0

n=2

a2 - 6a1 + 8a0 = 0
a2 - 6(4) + 8(2) = 0 ; a2 = 8



n=3

a3 - 6a2 + 8a1 = 0
a3 - 6(8) + 8(4) = 0 ; a3 = 16


n=4

a4 - 6a3 + 8a2 = 0
a4 - 6(16) + 8(8) = 0 ; a4 = 32

n=5

a5 - 6a4 + 8a3 = 0
a5 - 6(32) + 8(16) = 0 ; a5 = 64

n=6

a6 - 6a5 + 8a4 = 0
a6 - 6(64) + 8(32) = 0 ; a6 = 128

we get a sequence:

2,4,8,16,32,64,128, .... which is recognizable as powers of 2

a0 = 2 = 2^(0+1)
a1 = 4 = 2^(1+1)
...
an = 2^(n+1)


by solving the system of equation that used the power method


A(2^0) + B(4^0) = 2
A(2^1) + B(4^1) = 4


A + B = 2
2A + 4B = 4


-2A -2B = -4
2A +4B = 4
------------
2B = 0 ; B = 0 , A = 2

an = 2(2^n) = 2^(n+1) ; which matches :)