Question

Let A_0=1 and a_1=1 use generating functions to solve the recurrence equation a_(n+2)= a_(n+1)+ 2a_n for n>=0

Answer 1

so i tried to work this out and i started with my
G(x)=a_o+a_1x+a_2X^2+...
-1X(G(x)
-2X^2 G(x)
and got (1-x-2X^2) G(x) = 1-2x^2

G(x)= 1-2x^2 / (1-x-x^2)

but i am not sure if it right or where to go from there... and help would be great thanks :)

Answer 2

i was considering a homogenous and aprticular solution similar to diffy Qs

Answer 3

a_(n+2)= a_(n+1)+ 2a_n

a_(n+2) - a_(n+1) - 2a_n = 0

for some reason, the subs can go exponential

r^(n+2) - r^(n+1) - 2r^n = 0

r^n (r^2 - r - 2) = 0 when r = whatever the qf gives us i spose

Answer 4

r = -1, 2 .... but then it gets fuzzy again

Answer 5

\[a_n=C2^n+D(-1)^n\]

use \(a_0=1,a_1=1\) to find \(C,D\)

Answer 6

I think this is a nice recipe
http://www.cs.columbia.edu/~cs4205/files/CM2.pdf

Answer 7

You have to do some manipulation so you can use \[A(x)=\sum_{n=0}^{\infty}a_nx^n \]
Like at one point I was using:
\[\sum_{n=0}^{\infty}a_{n+2}x^{n}=x^{-2}\sum_{n=0}^{\infty}a_{n+2}x^{n+2}=x^{-2}(A(x)-a_2x^2-a_1x^1-a_0)\]

Answer 8

Or You can call big A big G. whatever

Answer 9

a 67 page recipe eh ... not quite rachel ray approved

Answer 10

if my head was not about to explode id love to read thru it better :)

Answer 11

page 10-11

Answer 12

I made a error

Answer 13

\[\sum_{n=0}^{\infty}a_{n+2}x^{n}=x^{-2}\sum_{n=0}^{\infty}a_{n+2}x^{n+2}=x^{-2}(A(x)-a_1x^1-a_0) \]