The recursive computation is perfectly possible. Starting with we obtain...
It seems that the 'negative sequence' replies, with alternating signs, the 'positive sequence'...
Kind regards
Let the Fibonacci sequence Fn be defined by its recurrence relation (1) Fn=F(n-1)+F(n-2) for n>=3. Show that there is a unique way to extend the definition of Fn to integers n<=0 such that (1) holds for all integers n, and obtain an explicit formula for the terms Fn with negative indices n.
I am completely stuck on how to prove this by induction. I figured that Fn = Fn+2-Fn+1, but i don't know how to apply induction to this to prove it for all integers n
You have that . This is a homogenous linear recurrence relation. That said, it's solution is of the form . Where are the solutions the auxiliary polynomial and are determined by initial conditions.
It turns out that where is the golden ratio.
This is clearly a canonical way to extend to