period of f(x)

[stuart clark]

If be a real value function which satisfies

and

. then period of   is

[senior]

f(x+3/2) - f(x+1/2) = f(x+1) - f(x) --> T = 1/2;
however, if you are asking for fundamental period, it needs further proving, I guess.

[stuart clark]

Yes i am asking for fundamental period.

[senior]

f(x+3/2) - f(x+1/2) = f(x+1) - f(x) --> f(k) - f(k-1) = f(1) - f(0) = M and
the difference between f(k) and f(k-1) is increasing as we move toward infinity. So M must be zero to satistfy |f(x)| <= 2
Then, f(x) = f(x-1) for all x. And T = 1.
Sorry for previous answer.

[stuart clark]

Thanks Moderator

please I want more explanation for last post.

[senior]

We have f(x+3/2) - f(x+1/2) = f(x+1) - f(x),
Let M = f(1) - f(0), Then for x=0, we have f(3/2) - f(1/2) = f(1) - f(0) = M
So, we move 1/2 forward, and now have f(3/2) - f(1/2) = M
x = 1/2 --> we have f(2) - f(1) = f(3/2) - f(1/2) = M, we moved 1/2 forward again.
x = 1 --> we have f(5/2) - f(3/2) = f(2) - f(1) = M ....
f(x+3/2) - f(x+1/2) = M always...
So, if f(x+3/2) =  f(x+1/2) + M and if M is non-zero, eventually |f(x)| will be greater than 2. So, M = 0, and we have
f(x+3/2) =  f(x+1/2) which  means T = 1