Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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`