help me prove this

Let f(x+a) = 1/2 + √f(x) - (f(x))2

and 0<=f(x)<=1

Then prove that period of y=f(x) is 2a .

6 Answers

1708
man111 singh ·

$\textbf{Here $\mathbf{f(x+a)=\frac{1}{2}+\sqrt{f(x)-(f(x))^2}}$}$\\\\\\ \mathbf{\left(f(x+a)-\frac{1}{2}\right)^2=f(x)-(f(x))^2............................(1)}$\\\\\\ \textbf{Now $\mathbf{x\rightarrow x+a}$,We Get}$\\\\ \mathbf{\left(f(x+2a)-\frac{1}{2}\right)^2=f(x+a)-(f(x+a))^2............................(2)}$\\\\\\ $\mathbf{\left(f(x+2a)-\frac{1}{2}\right)^2=\frac{1}{2}+\sqrt{f(x)-(f(x))^2}-\frac{1}{4}-f(x)+(f(x))^2-\sqrt{f(x)-(f(x))^2}}\\\\\\ \mathbf{\left(f(x+2a)-\frac{1}{2}\right)^2=\left(f(x)-\frac{1}{2}\right)^2}$\\\\\\ \textbf{So $\mathbf{f(x+2a)-\frac{1}{2}=\pm \left(f(x)-\frac{1}{2}\right)}}$\\\\\\ \textbf{\boxed{\boxed{\mathbf{f(x+2a) = f(x)}$}} and $\mathbf{f(x+2a)+f(x)-1=0}$}

21
Shubhodip ·

thoughts for others (though well known)

give example of one such function

30
Ashish Kothari ·

I'm guessing. Fractional part of x?

1
jee12 ·

thanks a lot sir

21
Arnab Kundu ·

ashish, does that satisfy 0≤f(x)≤1?

30
Ashish Kothari ·

Okay I didnt notice there was condition on f(x).

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