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I am supposed to find period of function following the below relation if exists, for domain belonging to all real numbers.

$$f(x+T) = 1 + \left(2f(x)-f(x)^2\right)^{\frac12}$$

I just took $1$ to other side and squared and then differentiated to get $f(x+T) = f(-x)$ which indicates that period is $2T$.

Query: I don't think differentiation is a good way, because it is not mentioned that $f(x)$ is differentiable. Can someone please help in clarifying the underlying ideas or suggesting a altogether different approach.

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    $\begingroup$ Just some thoughts: Take $g(x) = f(x) - 1$ then $g(x+T) = \sqrt{1 - g(x)^2}$. Note that $g$ is always positive and from $g^2(x) + g(x+T)^2 = 1$ we should be able to parametrize $g(x) = |\sin(\theta(x))|$ so that $g(x+T) = |\cos(\theta(x))|$ where we require $\theta(x+T) = \theta(x) + \pi \mod 2\pi$. One solution would be $f(x) = 1 + |\sin(x)|$ with $T = \pi$. $\endgroup$
    – Winther
    Commented Jul 31, 2017 at 17:47
  • $\begingroup$ you mean $g(x) = f(x) -1$. Okay in that case what you wrote is correctAnd I am not used to parametrization hence no comments on that part.And the answer is 2T so how does $pi$ related to 2T $\endgroup$
    – vishal mishra
    Commented Jul 31, 2017 at 17:54
  • $\begingroup$ What I meant with that is if $a^2 + b^2 = 1$ then this means that $(a,b)$ is a point on the unit circle so there is an "angle" $\theta$ such that $a = \sin(\theta)$ and $b = \cos(\theta)$. (and yes there are some sign errors in the comment above + I probably messed up the $\theta$ relation, so I would double check this) $\endgroup$
    – Winther
    Commented Jul 31, 2017 at 18:04

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I think this shows it:

If you rearrange as you say:

$(f(x+T)-1)^2=2f(x)-f(x)^2$

$f(x+T)^2-2f(x+T)+1=2f(x)-f(x)^2$

$2f(x)−f(x)^2-f(x+T)^2+2f(x+T)=1$ (eq. 1)

Substituting $x+T$ for $x$:

$2f(x+T)−f(x+T)^2-f(x+2T)^2+2f(x+2T)=1$ (eq. 2)

Subtracting eq. 2 from eq. 1:

$2f(x)-f(x)^2-2f(x+2T)+f(x+2T)^2=0$

$2f(x)-f(x)^2=2f(x+2T)-f(x+2T)^2$

...so the period is $2T$.

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    $\begingroup$ Does your final expression enough to say that period is 2T. I thought if you get something like $f(x)=f(x+T)$ only then the period is T, what you have written is F(f(x))= F((x+T)). $\endgroup$
    – vishal mishra
    Commented Aug 3, 2017 at 6:19

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