f(x) + f(x+1) = f(x+2)
does not determine a unique minimum period for f. In fact you have not even
indicated the domain of f, for all we know it could be Z or R. I would usually
assume that we were talking about functions on R for this type of question
In fact the recurence does not even ensure that f is periodic.
Now it may be the case that if f is a periodic solution (on either Z or R) then
there is a minimum period that the solution can have, but that is not the question
Perhaps it would help if you could tell us what class you are taking, what material (ie. chapter heading) this question originated from, what you were recently talking about in class, etc. Questions don't arise from a vacuum and evidently you are supposed to have a way to solve this thing. So if we know the topic it came from, maybe we can help you.
After wasting rather more time on this than I think its worth, I can tell
you there are no periodic solutions to the recurrence:
for f a function over Z, and the same proof shows that there aslo cannot
be a periodic solution for f a function over R either.
The solutions to this recurrence are generalised Fibinacci numbers and their
absolute values grow without bound.
fibonacci is only fr int rt???
2nd thin a fn need not hav sumthin lik sin cos root powers
it can be defined for 1 n 2 n 3 n4such as it follows dat eqn
and more dan dat .....if i say f(1)=-1 and f(2)=3 then f(3)=2 dat says it is not an ever increasin or decreasin fn
A second order linear recurence like the one in question has a solution uniquely
determined at all integer values of x by the values of f(1) and f(2).
f(n+2) = f(n+1) + f(n) ...(1)
and use a trial solution f(n)=a^n, then:
a^2 - a - 1=0.
This has roots (1+sqrt(5))/2 and (1-sqrt(5))/2, so any solution of (1) can be
f(n) = A[(1+sqrt(5))/2]^n + B[(1-sqrt(5))/2]^n
and values of A and B can be found to satisfy any values we require for f(1)
Now (1+sqrt(5))/2 = g ~=1.62, and (1-sqrt(5))/2 ~= 0.62, so as n becomes
large f(n) ~ A g^n. So the absolute value of f(n) goes to infinity as n goes
to infinity. Thus there is no periodic function over the integers Z that satisfies
the recurence (1).
Now if f is a function over R which satisfies (1), at the very least f(x), x an
integer must be bounded, but we have seen that it cannot be, so there are
no functions over R which are periodic solutions of (1). (there is a get out clause
here in that we could allow f to have singularities, which would then allow the
function to increase without bound at integer values of its argument but with
a forrest of infinities in each unit interval - as far as I can tell).
(Aside: there are always periodic solutions for functions over Z/nZ, in the
case of Z/2Z, they are all of period 3)