Let #f# such that #f:RR->RR# and for some positive #a# the equation #f(x+a)=1/2+sqrt(f(x)+f(x)^2)# holds for all #x#. Prove that the function #f(x)# is periodic?
3 Answers
If f is periodic, with period a,
Rationalizing,
between two neighboring points.
There is no such function
Explanation:
The conditions cannot be satisfied, so there is no such function
Given:
#f(x+a) = 1/2 + sqrt(f(x)+f(x)^2)#
Note first that
Given that
#sqrt(f(x) + f(x)^2) > sqrt(f(x)^2) = f(x)#
So:
#f(x+a) > f(x) + 1/2#
Hence:
#f(x-a) < f(x) - 1/2#
So if
#f(-na) < f(0) - n/2 < f(0) - (f(0) - 1/2) = 1/2#
-- contradiction.
Making
so we can observe that