341(f(x+p) -1)2 = 1-(f(x)-1)2
Hence f(x) = 1 + \sqrt{2 f(x+p) - f^2(x+p)}
Hence f(x-p) = 1 + \sqrt{2 f(x) - f^2(x)}
Thus we have f(x-p) = f(x+p) so that 2p is a period of f(x)
if f(x) belogs [1,2] when x belongs R and for a fixd positive real no P. f(x+p)=1+√2f(x)-f(x)^2.
prove that f(x) is periodic.
-
UP 0 DOWN 0 0 5
5 Answers
341
62THe question has been discussed a couple of times eureka.. I am sure you must have seen.. Read Prophet sir's thread...
That is the question..