Question

please help

Solve the 2nd order homogeneous linear recurrence relation with constant coefficients: f(n+2)-4f(n+1)+4f(n)=0 subject to the start-up conditions f(0)=3; f(1)=4

Answer 1

hey do you know how to do this?

Answer 2

if you could be kind enough to show me the work also i would greatly appreciate it...i have no idea how to do this

Answer 3

I am little rusty, but I think I will figure it out, hold on

Answer 4

no problem take your time :)

Answer 5

hi lalaly thaks for coming :)

Answer 6

thanks*

Answer 7

no problem:D

Answer 8

lalaly u left? =/

Answer 9

: f(n+2)-4f(n+1)+4f(n)=0

use characteristic equation

t^2 -4t +4

t=-2,-2

that's repeated root

we use the form
T(n)= C1 r^n + C2 n r^n
C1 -2^n + C2 n -2^n

now we have to solve for initial condition
C1 -2^0 + C2 *0 -2^0=3
C1=3

C1*-2+C2*-2=4
-6+C2*-2=4
-2 C2=10
C2=-5

T(n)=(3 -2^n + -5( n* -2^n)

Answer 10

lalaly u left? =/

Answer 11

so would i have to solve anything after that?

Answer 12

lalaly u left? =/

Answer 13

so if i write down every thing you have just written then then the problem should be complete right? the teacher shouldn't ask for more? sorry im just really lost in this class and i gotta pass this take home final to pass the class

Answer 14

let me check my work

Answer 15

\[C _{1} r ^{n} + C _{2} n r ^{n}\] is this what u meant to say? ...just trying to figure out where the sub numbers and exponents go

Answer 16

yes

Answer 17

hey so does it work out?
im also asking this question in another post just to see what other people get

Answer 18

should have been

T(n)=(3 2^n + -1( n* 2^n)

Answer 19

f(n+2)=4f(n+1)-4f(n)

f(2)= 4(f1)-4(f0)
f0=3
f1=4
f(2)= 16- 12=4

let test it on our equation by plugging in n=2

3(2)^2-1(2)(2)^2

12 -8=4

yes, it checks out