sequence compare to quadratic equation

wolfric

If i once understood why you could do this I've forgotten. Can anyone link me (or even better explain themselves) how you link these two?

A(n)= Aa(n-1) + Ba(n-2)

t^2 -AT -B =0

and a(n) =Cs(n) +Dt(n) n=> 0

I did this for the leaving cert... can't believe I've forgotten so soon

Fremen

Hm, I'm not sure I completely understand your notation, but I think I know what you're getting at. I don't know if this is the standard proof you learned in Leaving cert, but here goes.

If [latex]u_n[/latex] converges to a limit [latex]L[/latex], and [latex]u_n[/latex] satisfies the recurrence relation

[latex]u_n - A u_{n-1} - B_{n-2}= 0[/latex]
then [latex]u_n[/latex] is a solution of

[latex]x^2 - Ax - B = 0[/latex]

Here's why. If the sequence converges, the terms get closer and closer. At the limit (with a little hand-waving which can be made rigorous), we have

[latex]\frac{u_n}{u_{n-1}} = \frac{u_{n-1}}{u_{n-2}}[/latex].

In other words,

[latex]u_n = \frac{u_{n-1}^2}{u_{n-2}}[/latex].

Subbing into the recurrence relation and dividing across by [latex]u_{n-2}[/latex] we have

[latex]\frac{u_{n-1}^2}{u_{n-2}^2} - A\frac{u_{n-1}}{u_{n-2}} - B= 0[/latex].

Now, let [latex] x = \frac{u_{n-1}}{u_{n-2}}[/latex].
and we're done.

Edit: brought everything over to the left hand so I don't have to write plus signs. Can't work out how to do them, "+" isn't working.