Recurrence Equation

[jinydu]

c(n+2) = -8c(n+1) - 16c(n), c(1) = -1, c(2) = 8

Given the above recurrence equation and starting conditions, the goal is to find the solution (that is, an equation that calculates c(n) without reference to previous terms).

The characteristic polynomial is:

n^2 = -8n - 16

n^2 + 8n + 16 = 0

n = -4 (repeated root)

What has made this problem difficult for me is the repeated root. Because of the repeated root, the solution given at

http://mathworld.wolfram.com/LinearR...eEquation.html

doesn't work because it involves dividing by zero.

In case it helps, here are some more terms:

c(0) = 0
c(1) = -1
c(2) = 8
c(3) = -48
c(4) = 256
c(5) = -1280
c(6) = 6144
c(7) = -28672

Also, the ratio between successive terms appears to be approaching some number close to -4.

Thanks

[FeLiNe]

I'm probably just missing something obvious somewhere, but what is wrong with eq. (11) on the Mathworld page? In the situation you present, A=-8, x₁=-1 and x₂=8 so the first term that starts with (Ax₁- x₂) is identical to zero and all that remains is the second term.

[jinydu]

I meant Equation 13 (I do know of a way to adjust the formula to take account of different starting terms, but that doesn't solve the problem of dividing by zero).

[wblipp]

Originally Posted by jinydu

What has made this problem difficult for me is the repeated root.

Repeated roots in difference equations are similar to repeated roots in differential equations. In this case the general solutions is

A*(-4)ⁿ + B*n*(-4)ⁿ

You pick A and B to match the first two terms.

One way to derive this is to take the two root solution with alpha = beta + epsilon and figure out what happens in the limit as epsilon goes to zero. You have
[(beta+epsilon)ⁿ - betaⁿ]/(beta+epsilon-beta)

=[betaⁿ + n*epsilon* beta^n-1 + epsilon²*?? - betaⁿ]/epsilon

= n*beta^n-1 + epsilon*??

Absobing a factor of beta into the constant, in the limit this is proportional to

n * betaⁿ

William

[jinydu]

Ok, that's the right answer (proven by induction).

Thanks

[Qbert]

Just for final clarification,
c(n)=(-1)^n*4^(-1 + n)*n
c(n)=(-1)**n*4**(-1 + n)*n

Is TeX form or even better MathML form possible to post?

[wblipp]

Originally Posted by Qbert

Just for final clarification,
c(n)=(-1)^n*4^(-1 + n)*n
c(n)=(-1)**n*4**(-1 + n)*n

Is TeX form or even better MathML form possible to post?

HTML superscript codes are supported as [sup]n[/sup], so you could write this as

c(n) = (-1)ⁿ4^n-1n