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
A. S. Adikesavan
Oct 23, 2016

If f is periodic, with period a,

#f(x+a)=f(x)=1/2+sqrt(f(x)+f(x)^2#

Rationalizing,

#f(x)=1/4, a constant#, and the period a becomes the distance

between two neighboring points.

George C.
Oct 23, 2016

There is no such function #f#

Explanation:

The conditions cannot be satisfied, so there is no such function #f#.

Given:

#f(x+a) = 1/2 + sqrt(f(x)+f(x)^2)#

Note first that #sqrt(...) >= 0# and hence #f(x) >= 1/2# for all #x in RR#

Given that #f(x) > 0#, we have #f(x) + f(x)^2 > f(x)^2# and hence:

#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 #n > ceil(2(f(0) - 1/2))# then:

#f(-na) < f(0) - n/2 < f(0) - (f(0) - 1/2) = 1/2#

-- contradiction.

Cesareo R.
Oct 23, 2016

Making #x = x+a#

#f(x+2a)=1/2+sqrt(f(x+a)-f(x+a)^2)=#
#=1/2+sqrt(1/2+sqrt(f(x)-f(x)^2)-(1/2+sqrt(f(x)-f(x)^2))^2)=#
#=1/2+sqrt(1/2+sqrt(f(x)-f(x)^2)-1/4-f(x)+f(x)^2-sqrt(f(x)-f(x)^2))=#
#=1/2+sqrt(1/4-f(x)+f(x)^2)=#
#=1/2+sqrt((1/2-f(x))^2)=#
#=1/2+abs(1/2-f(x))#

so we can observe that

#f(x) ge 1/2# so

#abs(1/2-f(x))=f(x)-1/2# and finally

#f(x+2a)=f(x)# so #f(x)# is periodic with period #2a#

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