A function f is said to be periodic if it repeats its nature after certain interval and that interval is called period of the function. If T be the period of f(x), then f(x + T) = f(x). For example f(x) = tan x is periodic with period `pi`. Then find the period of f(x) that satisfies the following condition.
(i) f(x + 1) - f(x + 3) = 0
(ii) f(x + 3) + f(x) = 0
(iii) f(x-1) + f(x+3) = f(x+1) + f(x + 5)
(iv) `f(x + p) = 1+{(1-f(x-p))^(2n+1)}^((1)/(2n+1)), n in N`

Explanation

(i) We have,
`f(x + 1) - f(x+3) = 0`
`rArr f(x+1) = f(x+3) = f(x+1+2)`
Let us put x-1 for x, we get
f(x-1 + 1) = f(x - 1 + 3)
`rArr f(x) = f(x + 2)`
which shows that the period of f(x) is 2.
(ii) We have,
f(x) + f(x+3) = 0
`rArr f(x) = -f(x+3)`
`= --f(x+3+3))`
`= f(x+6)`
The period f(x) is 6.
(iii) We have,
f(x-1) + f(x+3) = f(x+1) + f(x+5)
Let us put x+2 in place of x.
f(x+1) + f(x+5) = f(x+3)+f(x+7)
Adding them, we get
`f(x-1) = f(x+7)`
`rArr f(x-1) = f(x - 1 + 8)`
`rArr f(x) = f(x + 8)`
The period of f(x) is 8.
(iv) We have,
`f(x+p) = 1 + {(1-f(x-p))^(2n+1)}((1)/(2n+1)), n in N`
`rArr f(x+p)-1)^(2n+1) = (1-f(x-p))^(2n+1)`
`rArr f(x+p) - 1= 1- f(x-p)`, as f(x) is a real valued function.
`rArr f(x+p) + f(x-p) = 2`
`rArr f(x+3p) + f(x+p) = 2`
which on subtraction gives ltbtgt f(x+3p) - f(x-p) = 0
write x+p for x, we get
f(x+4p) = f(x)
`rArr f(x) = f(x + 4p)`
which shows that f(x) is periodic with period 4p.