Question

Solve the recurrence relation \(2a_n = 7a_{n-1} - 3a_{n-2}; a_0 = a_1 = 1 \) please see here for more info http://math.stackexchange.com/questions/1068309/solve-the-recurrence-relation-2a-n-7a-n-1-3a-n-2-a-0-a-1-1

Answer 1

okay

Answer 2

PLEASE

Answer 3

there is something of an exponent based method that I recollect.

let an = x^n and solve for x^n

Answer 4

\[2x^n=7x^{n-1}-3x^{n-2}\]
\[2x^n-7x^{n-1}+3x^{n-2}=0\]
\[x^{n-2}(2x^2-7x+3)=0\]
something along thise lines

Answer 5

that's what i did.

Answer 6

seems close but no cigar and i don't know what the hell else to do so

Answer 7

solve the quadratic:

(x-6/6)(x-1/6)

(x-1)(6x-1) = 0 when x=1 or 1/6

which we can then use in working out an explicit formula

Answer 8

my factoring seems off

Answer 9

ok i thought you divided the x^n version by x^n-2 not multiplied

Answer 10

so that you solve for 2x^2 - 7x - 3

Answer 11

i factored out x^n-2 to get a quadratic (x^2)

Answer 12

oh

Answer 13

ok yea so roots are 1/2 and 3

Answer 14

4x = 7 +- sqrt(49-4(2)(3))

4x = 7 +- 5

4x = 2 or 12

x = 1/2 or 3 yes

Answer 15

since an = x^n i believe we work out a system of equations that fit with a combination of the 2 solutions

Answer 16

therefore the constants in the solution are 4/7 and 3/7 which works for a_0 but not a_1 or a_2, for a_2 it gives 4 which is 2a_2 .

Answer 17

it's supposed to be a_n = b(1/2)^n + d(3)^n

Answer 18

so then b+d = 1 = b(1/2) + d(3); b = 4/7, d = 3/7

Answer 19

doens't work

Answer 20

an = (7an-1 - 3an-2)/2

a0 = 1
a1 = 1

a2 = (7(1)-3(1))/2 = 2
a3 = (7(2) -3(1))/2 = 11/2

Answer 21

my computer is going real slow

Answer 22

sok sok i think i know what happened

Answer 23

the wolf gets us to
\[a_n = \frac15 2^{-n} (6^n+4)\]
so we have something to guide us to

Answer 24

(2^-n 2^n 3^n + 2^-n 2^2) / 5

an = 3^n/5 + 4(1/2)^n/5

Answer 25

yeah here's the thing though i messed up initially because i used what i got for x which was -1/2 and -3. then i messed up again not fixing my whole solution when i just fixed those to positive. but why did i get -1/2 and -3 when the roots are 1/2 and 3. even quadratic equation calculators say what i got on paper, negatives. why are they positive?

Answer 26

b and d are actually 4/5 and 1/5

Answer 27

a0 = A(3^0) + B(1/2)^0 = 1
a1 = A(3^1) + B(1/2)^1 = 1

A + B = 1
3A + B/2 = 1


-3A -6B/2 = -3
3A + B/2 = 1
--------------
-5B/2 = -2

B = 4/5
A = 1/5

Answer 28

slow computer here, but can you type up your exponent approach?

Answer 29

2an = 7an-1 - 3an-2

let an = x^n

2x^n = 7x^(n-1) - 3x^(n-2)

2x^n - 7x^(n-1) + 3x^(n-2) = 0

if we factor out an x^n we end up with

x^n [2 - 7x^(-1) + 3x^(-2)] = 0

which isnt all that conventional to me so lets factor out the x^-2

x^n-2 [2x^2 - 7x + 3] = 0

Answer 30

i get positive roots so im not sure how you approached it