First note that for any x,
(x+a)+(b-x) is always equal to a+b, so points (x+a, f(x+a)) and (b-x, f(b-x)) are symmetrical about x = (a+b)/2.
Note also that when X changes, x+a can run all over the domain of F, so F is symmetric about x = (a+b)/2.
2. If the function satisfies f(x+a) = f(x+b), the function has a period |a-b|, where a ≠ b.
The difference between this one and the last one is that x+a+x+b is not always equal to a constant, so there is no symmetry. But they will subtract.
Is a constant, and the function values of the two independent variables of the difference constant are the same, so we can talk about periodicity.
3. If the function satisfies f(x+a)+f(b-x) = c, then the function image is symmetric about this point ((a+b)/2, c/2).
It is proved that it is similar to 1, especially when c = 0, the function image is symmetric about ((a+b)/2,0).
More specifically, when a=b=c=0, the function image is symmetrical about the origin, so f is odd function.
4. If the function is symmetric (a≠b) about points (a, c), (b, c) and (b, c), then the period of the function is 2 |a-b|.
5. If the function is symmetrical about the point (a, c) and the straight line x = b (a≠b), then the period of the function is 4 |a-b|.