Determine f(0) if given f(f(x))

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Determine f(0) if given f(f(x))

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Source: Baltic Way 2011

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WakeUp

#1 h Nov 6, 2011, 12:35 PM • 8 Y

Y by PhantomR, Number1, DeepanshuPrasad, Amir Hossein, Adventure10, Mango247, and 2 other users

Let $f:\mathbb{R}\to\mathbb{R}$ be a function such that
$f(f(x))=x^2-x+1$
for all real numbers $x$ . Determine $f(0)$ .

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15142 posts

#2 h Nov 6, 2011, 12:57 PM • 15 Y

Y by PhantomR, Number1, Pirkuliyev Rovsen, sayak38kolay, rajatgauss, numerology9, Nguyenhuyhoang, DeepanshuPrasad, HVishy, Centralorbit, Wizard0001, Adventure10, Mango247, and 2 other users

We have $f(f(0)) = f(f(1)) = 1$ .
Now, $f(x^2-x+1) = f(f(f(x))) = f(x)^2 - f(x) + 1$ , so $f(0)^2 - f(0) + 1 = f(1)^2 - f(1) + 1 = f(1)$ .
So $f(1)^2 - 2f(1) + 1 = 0$ , hence $\boxed{f(1)=1}$ . It follows $f(0)(f(0)-1) = 0$ .
Assume $f(0) = 0$ ; then $1=f(f(0)) = f(0) = 0$ , absurd. So $\boxed{f(0)=1}$ .

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#3 h Nov 9, 2011, 1:48 PM • 1 Y

Y by Adventure10

It's not end of this problem. It's necessary to show that such function exists.
But I heard that none of teams on Baltic Way showed this function and 10 out of 11 teams got 5/5 pts.

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pco

#4 h Nov 9, 2011, 2:06 PM • 2 Y

Y by Adventure10 and 1 other user

Swistak wrote:

It's not end of this problem. It's necessary to show that such function exists.
But I heard that none of teams on Baltic Way showed this function and 10 out of 11 teams got 5/5 pts.

I dont think showing the existence is demanded (the problem statement supposed existence)

But, if really required, it's not very difficult to build infinitely many solutions.

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#5 h Nov 20, 2011, 5:19 AM • 2 Y

Y by Vladimir_Djurica, Adventure10

it suffices to notice the following conclusion:
if $f(x_1)=f(x_2)$ ,then $x_1=x_2$ or $x_1+x_2=1$ .

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pco

#6 h Apr 5, 2013, 9:46 AM • 5 Y

Y by Pirkuliyev Rovsen, soad, Vladimir_Djurica, Adventure10, and 1 other user

pco wrote:

But, if really required, it's not very difficult to build infinitely many solutions.

I was required thru pm to give some examples. Here is one construction :

Let $g(x)=x^2-x+1$ . Notice that $g(1-x)=g(x)$ , $f(x)\ge x$ $\forall x$ and $g(x)$ is increasing over $[\frac 12, +\infty)$

1) Defining $f(x)$ over $[\frac 12, 1)$
===================
Let $a\in(\frac 12,\frac 34)$
Let the sequence $a_n$ defined as $a_0=\frac 12$ , $a_1=a$ and $a_{n+2}=g(a_n)$ $\forall n\ge 0$
$a_n$ is an increasing sequence whose limit is $1$
Let $h_0(x)$ be any bijection from $[a_0,a_1)\to [a_1,a_2)$
It's easy to build a sequence of bijections $h_n(x)$ from $[a_n,a_{n+1})\to[a_{n+1},a_{n+2})$ as $h_{n+1}(x)=g(h_n^{-1}(x))$ $\forall n\ge 0$

We can define $f(x)$ over $[\frac 12,1)$ as :
$\forall x\in [a_n,a_{n+1})$ : $f(x)=h_n(x)$
Obviously $f(f(x))=x^2-x+1$ $\forall x\in [\frac 12, 1)$

2) defining $f(x)$ for $x=1$
=================
Choose $f(1)=1$
Obviously $f(f(x))=x^2-x+1$ $\forall x\in [1,1])$

3) defining $f(x)$ over $(1,+\infty)$
====================
Let $b>1$ and $c\in(b,g(b))$

3.1) Defining $f(x)$ over $[b,+\infty)$

Let the sequence $b_n$ defined as $b_0=b$ , $b_1=c$ and $b_{n+2}=g(b_n)$ $\forall n\ge 0$
$b_n$ is an increasing sequence whose limit is $+\infty$
Let $k_0(x)$ be any bijection from $[b_0,b_1)\to [b_1,b_2)$
It's easy to build a sequence of bijections $k_n(x)$ from $[b_n,b_{n+1})\to[b_{n+1},a_{b+2})$ as $k_{n+1}(x)=g(k_n^{-1}(x))$ $\forall n\ge 0$

We can define $f(x)$ over $[b,+\infty)$ as :
$\forall x\in [b_n,b_{n+1})$ : $f(x)=k_n(x)$
Obviously $f(f(x))=x^2-x+1$ $\forall x\in [b,+\infty)$

3.2) Defining $f(x)$ over $(1,b)$

Let $g_{1+}$ be the restriction of $g(x)$ over $(1,+\infty)$ . $g_{1+}$ is a bijection from $(1,+\infty)\to(1,+\infty)$
Let the sequence $c_n$ defined as $c_0=c$ , $c_1=b$ and $c_{n+2}=g_{1+}^{-1}(c_n)$ $\forall n\ge 0$
$c_n$ is a decreasing sequence whose limit is $1$
It's easy to build a sequence of bijections $l_n(x)$ from $[c_{n+2},c_{n+1})\to[c_{n+1},c_n)$ as :
$l_0(x)=k_0^{-1}(g(x))$ $l_{n+1}(x)=l_n^{-1}(g(x))$ $\forall n\ge 0$

We can define $f(x)$ over $(1,b)$ as :
$\forall x\in [c_{n+2},c_{n+1})$ : $f(x)=l_n(x)$
Obviously $f(f(x))=x^2-x+1$ $\forall x\in (1,b)$

4) Defining $f(x)$ over $(-\infty,\frac 12)$
=========================
Just choose $f(x)=f(1-x)$

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#7 h Oct 13, 2020, 2:42 PM • 2 Y

Y by Mango247, Mango247

PCO how do come with such solutions that seems nearly impossible? what is the motivation?

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#8 h Apr 5, 2021, 1:54 PM

Y by

I have uploaded a video explaining this problem:
Link: Video Solution

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11388 posts

#9 h Apr 5, 2021, 1:57 PM

Y by

$f(f(f(x))=f(x)^2-f(x)+1=f(x^2-x+1)$
Setting $x=1$ , we obtain $f(1)^2-2f(1)+1=0$ , so $f(1)=1$ .
Setting $x=0$ , we get $f(0)^2-f(0)+1=1$ , so $f(0)\in\{0,1\}$
Set $x=0$ in the original equation; we get $f(f(0))=1$ , so $f(0)$ cannot be zero. Thus $\boxed{f(0)=1}$ .

This post has been edited 1 time. Last edited by jasperE3, Apr 5, 2021, 1:59 PM

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