Functiona Equation...?

Question

Functional Equation
f(x+p) = 1+( 2-3f(x)+3(f(x))^2-f(x)^3 )^1/3 .
How do I prove that f(x) is periodic given that p>0. ?

Answer 1

You don't, since it isn't necessarily true. There is something wrong with the phrasing of your question.
f(x) is not necessarily periodic, since
f(x+p) isn't necessarily dependent upon x or p.

Example: let
f(x) = f(x+p) = 1 + (1/2)^(1/3)
for all x and p. This f(x) is a solution of the functional equation.

f(x+p) = 1 + (1 + (1-f(x))^3)^(1/3)
= 1 + (1 + (- (1/2)^(1/3))^3)^(1/3)
= 1 + (1 - 1/2)^(1/3)
= 1 + (1/2)^(1/3)
= f(x)

The functional equation is solved, and the f(x) is a horizontal line, which is not periodic.

Also, you never indicated that the function has a certain domain. If we allow complex functions, there are many non-periodic functions which satisfy the equation. Using Dr D's derivation up to the point:
(h-1)^3 - (f-1)^3 = 0
(h-1)^3 = (f-1)^3
we notice that there are three cube roots of every number, so h-1 doesn't necessarily equal f-1. It could be that:

h-1 = (f-1) * [(-1 ± i√3)/2]
or h = f(x+2p) = (f(x) - 1) * [(-1 ± i√3)/2]
= [(-1 ± i√3)/2] * f(x) - [(-3 ± i√3)/2]

Answer 2

If f(x) is periodic with period p then you have to show that

f(x+p) = f(x)

f(x+p) = 1 + [2 - 3f(x) + 3f²(x) - f³(x)]^(1/3)
f(x+p) - 1 = [2 - 3f(x) + 3f²(x) - f³(x)]^(1/3)
[f(x+p) - 1]³ = 2 - 3f(x) + 3f²(x) - f³(x)
[f(x+p) - 1]³ = 1 + 1 - 3f(x) + 3f²(x) - f³(x)
[f(x+p) - 1]³ = 1 + [1 - f(x)]³

[f(x+p) - 1]³ - [1 - f(x)]³ = 1

Multiply both sides by [f(x+p) + 1]³ - [1 + f(x)]³

{[f(x+p) - 1]³ - [1 - f(x)]³}{[f(x+p) + 1]³ - [1 + f(x)³]} = [f(x+p) + 1]³ - [1 + f(x)]³

[f(x+p) - 1]^6 - [1 - f(x)]^6 = [f(x+p) + 1]³ - [1 + f(x)]³

Set x+p=x

[f(x) - 1]^6 - [1 - f(x)]^6 = [f(x) + 1]³ - [1 + f(x)]³
[f(x) - 1]^6 - (-1)^6[f(x)-1]^6 = 0
[f(x) - 1]^6 - [f(x)-1]^6 = 0

0=0

Apparently this shows f(x) is periodic but not necessarily with period p and most likely 2p as the exponents had to be doubled to show periodicity. Frankly DrD's method is more intuitive than mine.

Answer 3

Let g(x) = f(x+p) to simplify the writing.

g(x) = 1 + [1 + {1 - f(x) }^3 ]^(1/3)
rearranging
(g - 1)^3 = 1 + (1 - f)^3
(g-1)^3 + (f-1)^3 = 1 ... eqn (1)

Now let h(x) = f(x + 2p) = g(x + p)
By def'n h should have the same relationship to g as g does to f.
So (h - 1)^3 + (g-1)^3 = 1 ... eqn(2)

Eliminating g from equations 1 and 2:
(h-1)^3 - (f-1)^3 = 0
This can be factorized as the difference of two cubes:
(h - f) * [(h-1)^2 + (h-1)*(f-1) + (f-1)^2 ] = 0

If f and h are identically equal to 1, this solves the equation. However, as long as h = f, then that also satisfies the equation.
Apart from the trivial solution, f = 1 identically, the only real solution is : h(x) = f(x)

So f(x + 2p) = f(x)
f(x) is periodic with period, 2p.

Answer 4

If we had a - sign and 1 instead of 2
[1-f(x)]^3= 1-3f+3f^2-f^3 so your expression would be
f(x+p)=1-(1-f)=f(x)