Recurrence relation help?

Asked Feb 09 '14 at 22:26

Active Feb 10 '14 at 17:43

Viewed 20 times

$$t(n)=\begin{cases}n&\text{if }n=0,1,2,\text{ or }3\\t(n-1)+t(n-3)-t(n-4)&\text{otherwise.}\end{cases} $$ Express your answer as simply using the theta notation.

I don't know where to go with this.

$$t(n) - t(n-1) - t(n-3) + t(n-4) = 0$$

Is the characteristic polynomial $x^3 - x^2 - x +1 = 0$?

edited Feb 10 '14 at 17:43

asked Feb 09 '14 at 22:26

user127259

Hint: Use Master theorem – Bartosz Przybylski Feb 09 '14 at 22:37
@Bartek: Not for this one... – Raphael Feb 09 '14 at 22:42
Right, stupid me, its more like Fib numbers. – Bartosz Przybylski Feb 09 '14 at 22:47
Can you help me Raphael instead of sending me to that? – user127259 Feb 09 '14 at 23:42
There's a readable explanation of how to find solutions to these "linear homogeneous recurrence relations with constant coefficients" here. – Rick Decker Feb 10 '14 at 00:40
That's way more useful than Raphael's. Now I'm curious if my characteristic equation is correct? What throws me off is that mine jumps from n-1 to n-3. Does that mean anything? – user127259 Feb 10 '14 at 00:55
If mine is correct then my answers are (x-1)(x-1)(x+1)=0 – user127259 Feb 10 '14 at 00:57
The characteristic polynomial is $x^4-x^3-x-1$ (the $+t(n-4)$ should be $-t(n-4)$), which can be factored as $(x^2+1)(x^2-x-1)$. – Rick Decker Feb 10 '14 at 16:40
Why should it be -t(n-4) when It was negative to begin with and I moved it to the other side? My mistake I wrote the problem wrong and edited it now. Anyways I don't know how to work with that characteristic polynomial. Are the solutions all complex? – user127259 Feb 10 '14 at 17:42
Well working it out it gave me 2 real and 2 complex solutions. – user127259 Feb 10 '14 at 17:46

Recurrence relation help?

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