Prove that f(x) is periodic and find the period.?

Question

[ f(x + p) ]^2 = 1 - [ f(x) ]^2

Prove that f(x) is periodic and find the period in terms of p.

The period is NOT p.

f(x) is non constant, continuous and differentiable in the entire real domain.

Suppose f(x) = sin(x) and p = π/2?

ksoileau: I know for f(x) = sin(x) p = π/2, f(x+2p) = -f(x). Can you think of a periodic function where f(x+2p) = f(x)?
Other than |sin(x)| or |cos(x)|

Answer 1

f(x+p)^2+f(x)^2=1
f(x+2p)^2+f(x+p)^2=1
Subtracting, we get
f(x+2p)^2-f(x)^2=0
f(x+2p)^2=f(x)^2
So f(x)^2 is periodic with period 2p/k for some positive integer k.
Since f(x+2p)^2-f(x)^2=0,
we can say that
(f(x+2p)-f(x))*(f(x+2p)+f(x))
=0
for every x, so for every x either f(x+2p)=f(x) or f(x+2p)=-f(x) will hold. If for all x, f(x+2p)=f(x) holds then f is periodic with period 2p/k for some positive integer k. If for all x, f(x+2p)=-f(x) holds, then for all x f(x+4p)=f(x) and f is periodic with period 4p/k for some positive integer k. If for some x f(x+2p)=f(x) holds and for the rest of the x, f(x+2p)=-f(x) holds, that's where the fun starts. For example, suppose p is real but not rational and define f(x)=1/sqrt(2) if x is rational, otherwise define f(x)=-1/sqrt(2). Then f(x)^2=1/2 everywhere, so f(x+p)^2=1-f(x)^2 is satisfied for all x. Now choose any rational x. Clearly x+p is nonrational, so f(x+p)=-1/sqrt(2) yet f(x)=1/sqrt(2). So f is not periodic.


Dr D: Take f(x)=sin(2x). Then f(x+2p)=sin(2x+4p)
=sin(2x+2pi)=sin(2x)=f(x).

Answer 2

This is simply not true.

This aperiodic function of integer x and p = 1 satisfies the condition:

+1 0 -1 0 +1 0 +1 0 +1 0 -1 0 -1 0 -1 0 +1 0 +1




Non constant,
continuous and
differentiable in the entire real domain
all does not help.
It's easy to construct counter-example even in case of
infinitely many times differentiable function.
Show me _your proof, and I will tell you where it went wrong.

Answer 3

A function f is periodic if for f(x) and for all f(x + p) the value is the same. Since f(x + p)^2 = 1 - f(x)^2, this condition is satisfied (I skipped lots of detail...).

Don't know what the period is offhand.