I'm trying to find the closed form of the Fibonacci recurrence but, out of curiosity, in a particular way with limited starting information. I am aware that the Fibonacci recurrence can be solved fairly easily using the characteristic root technique (and its corresponding linear algebra interpretation):
http://discrete.openmathbooks.org/dmoi3/sec_recurrence.html
But what I'm wondering is if its possible to determine Fibonacci Recurrence's closed form using the following two theorems:
$\forall n \ge 2, \sum_{i=1}^{n-2} F_i = F_n - 2$
$\forall n \ge 1, F_n < ((1 + \sqrt 5)/2)^n $
I suspect that it may be possible because the 2nd theorem involves the "golden ratio" $\varphi = (1 + \sqrt 5)/2$, but I have no idea where I might start so some hints or a little insight would be appreciated. Both theorems are true and I can provide the proofs if necessary.
Im using the following definition of the Fibonacci Recurrence:
$F_0 = F_1 = 1$
$F_{i+1} = F_i + F_{i-1} , \forall i \ge 2$
To see that this does not work, note that your first relation quickly implies (for $n≥2$) $$F_n=F_{n-1}+F_{n-2}$$
which, of course, is the usual Fibonacci recursion. It also quickly shows that $F_2=2$.
Thus, to find a counterexample, we want initial conditions such that $F_0+F_1=2$ and for which the entire series satisfies the given inequality.
Take, for instance, $$F_0=\frac 12\quad \& \quad F_1=\frac 32$$
Standard methods show that, with those initial conditions, we get the closed formula $$F_n=\frac 12\times \left(\frac {1+\sqrt 5}2\right)^{n+1}+\;\;\frac 12\times \left(\frac {1-\sqrt 5}2\right)^{n+1}$$
But then simple numerical work establishes the desired inequality for modestly sized $n$ and for large $n$ the second term becomes negligible and the desired equality is easily shown for the first term.