How to prove the periodic problem of two functions? If f (x) is an odd function and the equation f (a + x) = f (A-X) holds for all x ∈ R, it is proved that the period of F (x) is 4a If f (x) is symmetric with respect to (a, Y0) and x = B, it is proved that the period of F (x) is 4 (B-A)

How to prove the periodic problem of two functions? If f (x) is an odd function and the equation f (a + x) = f (A-X) holds for all x ∈ R, it is proved that the period of F (x) is 4a If f (x) is symmetric with respect to (a, Y0) and x = B, it is proved that the period of F (x) is 4 (B-A)

1. It is known that f (a + x) = f (A-X), because f (x) is an odd function, so f (A-X) = - f [- (A-X)], the second formula is substituted into the first formula
F (a + x) = - f [- (A-X)], the deformation is
f(x+a)= -f(x-a) ……………… ①
The form of the following formula is
f(x+2a)= f[(x+a)+a]= -f[(x+a)-a]= -f(x) ……………… ②
There are two forms of the model
F (x + 4a) = f [(x + 2a) + 2A] = - f (x + 2a)
f(x+4a)= f(x)
So the period of function f (x) is 4a
2. Because f (x) is symmetric about point (a, Y0), f (a + x) = - f (A-X)
Because f (x) is symmetric with respect to x = B, f (B + x) = f (b-X)
Replace X of the first formula with X-B to get f (a + X-B) = - f (a + b-X)
If x in the second formula is replaced by x-a, f (B + x-a) = f (a + b-X)
The sum of the two formulas leads to
f[x+(b-a)]= - f[x-(b-a)] ……………… ①
The form of the following formula is
f[x+2(b-a)]= f[x+(b-a)+ (b-a)]= -f[x+(b-a)- (b-a)]= -f(x) ……………… ②
There are two forms of the model
F [x + 4 (B-A)] = f [x + 2 (B-A) + 2 (B-A)] = - f [x + 2 (B-A)]
f[x+4(b-a)]= f(x)
So the period of function f (x) is 4 (B-A)