A periodic function

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A periodic function

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periodic function

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Source: IMO 1968 B2

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Rushil

#1 h Nov 4, 2005, 7:09 AM • 4 Y

Y by Adventure10, ImSh95, Mango247, Mango247

Let $f$ be a real-valued function defined for all real numbers, such that for some $a>0$ we have $f(x+a)={1\over2}+\sqrt{f(x)-f(x)^2}$ for all $x$ .
Prove that $f$ is periodic, and give an example of such a non-constant $f$ for $a=1$ .

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#2 h Nov 5, 2005, 9:09 AM • 4 Y

Y by Adventure10, Adventure10, ImSh95, Mango247

i think you can solve by plugging x->x+a,x->x+2a etc.

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Davron

#3 h Jun 20, 2006, 5:33 AM • 3 Y

Y by Adventure10, ImSh95, Mango247

Directly from the equality given: f(x+a) ≥ 1/2 for all x, and hence f(x) ≥ 1/2 for all x.

So f(x+2a) = 1/2 + √( f(x+a) - f(x+a)2 ) = 1/2 + √f(x+a) √(1 - f(x+a)) = 1/2 + √(1/4 - f(x) + f(x)2) = 1/2 + (f(x) - 1/2) = f(x). So f is periodic with period 2a.

We may take f(x) to be arbitrary in the interval [0,1). For example, let f(x) = 1 for 0 ≤ x < 1, f(x) = 1/2 for 1 ≤ x < 2. Then use f(x+2) = f(x) to define f(x) for all other values of x.

kalva

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#4 h Aug 3, 2011, 9:48 AM • 2 Y

Y by ImSh95, Adventure10

Since $\sqrt{f(x)-(f(x))^{2}}\geq 0$ , $f(x+a)\geq \frac{1}{2}$ and $f(x)\geq \frac{1}{2}$ .
Squaring the first equation, we obtain:
$(f(x+a))^{2}-f(x+a)+\frac{1}{4}=f(x)-(f(x))^{2}$ , which is equivalent to: $|f(x)-\frac{1}{2}|=\sqrt{f(x+a)-(f(x+a))^{2}}$ . (*)

Plugging $x+a$ instead of $x$ into the original equation, we get $f(x+2a)-\frac{1}{2}=\sqrt{f(x+a)-(f(x+a))^{2}}$ . Backing this into (*) and applying the condition $f(x)\geq \frac{1}{2}$ , we get: $f(x)-\frac{1}{2}=f(x+2a)-\frac{1}{2}$ , or equivalently $f(x)=f(x+2a)$ .

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DottedCaculator

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DottedCaculator

#5 h Aug 24, 2021, 2:11 PM • 2 Y

Y by centslordm, ImSh95

Solution

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ZETA_in_olympiad

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ZETA_in_olympiad

#6 h May 12, 2022, 12:33 PM • 1 Y

Y by ImSh95

The functional equation is equivalent to $f(x+a)=\sqrt{f(x)(1-f(x))}+1/2 \implies (f(x+a)-1/2)^2=1/4-(f(x)-1/2)^2 \implies f(x+2a)=f(x).$ Hence $f$ is periodic.

For example, $f(x)=\frac{1}{2}(1+|\cos \frac{\pi x}{2}|),$ which is seen to work. And thus we're done.

This post has been edited 1 time. Last edited by ZETA_in_olympiad, May 12, 2022, 12:34 PM

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#7 h May 12, 2022, 2:44 PM • 1 Y

Y by ImSh95

A way of aproaching this problem is to simply substitute $x=x+a$ .
$\therefore f(x+2a)=\frac{1}{2}+\sqrt{f(x+a)-f(x+a)^2}$
Now substitute $f(x+a)={1\over2}+\sqrt{f(x)-f(x)^2}$
$\therefore f(x+a)-f(x+a)^2={1\over2}+\sqrt{f(x)-f(x)^2} - ({1\over2}+\sqrt{f(x)-f(x)^2})^{2}={1\over4}-f(x)+f(x)^2=(f(x)-\frac{1}{2})^2.$
$\therefore f(x+2a)=\frac{1}{2}+\sqrt{(f(x)-\frac{1}{2})^2}=\frac{1}{2}+|f(x)-\frac{1}{2}|.$
Note that $\sqrt{a^2}=|a|\neq a$ for negative $a$ .
So we must prove $f(x)\geq\frac{1}{2} \forall x\in \mathbb{R}$ to complete our proof.
Luckily, this is easy as $\sqrt a\geq 0 \forall a\in\mathbb{R} \Rightarrow f(x+a)\geq\frac{1}{2} \Rightarrow f(x)\geq \frac{1}{2} \forall x\in \mathbb{R}$ .
$\therefore f(x+2a)=f(x) \forall x\in \mathbb{R}$ and we are done!
So, $f(x)$ is periodic with period $2a$ .

Finding an example is pretty elusive, but the periodicity should give us the hint of using trignometry.
But $f(x)\geq\frac{1}{2} \forall x\in \mathbb{R}.$ and there is no such condition on trignometric functions, we may also have to use the modulus function.
Combining these two along with the bound on $f(x)$ we get that the function may be of the form $f(x)=\frac{1}{2}+|g(x)|$ where $g(x)$ may be a trignometric function which contains $x$ .
After fiddling around with it a little bit we see that $f()=\frac{1}{2}+\frac{1}{2}|cos(\frac{\pi x}{2})|$ satisfies the condition, completing the 2nd part of the problem!
Hope you liked it!!! :thumbup:

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