functional equation

Dimitris X

Find all functions $f: \mathbb{R^*} \to \mathbb{R^*}$ satisfying the functional relation

$f(f(x)-x)=2x,\forall x\in \mathbb{R^*}$

**Posted:** Sun Nov 08, 2009 7:53 am

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ΠΑΙΡΝΩ ΤΑΜΠΕΛΑ ΚΑΙ ΕΓΩ ΤΟΥ ΕΘΝΙΚΟΥ ΠΡΟΔΟΤΗ ΑΦΙΕΡΩΜΕΝΟ ΚΑΙ ΑΥΤΟ ΣΕ ΚΑΘΕ ΔΟΥΛΟ ΠΑΤΡΙΩΤΗ.....

reason

hi!
First we consider the function g such that: $g(x)=f(x)-x$ so the f.e become:

$g(g(x))+g(x)=2x$ $\forall x\in{\mathbb{R^{*}}}$

then we consider the sequence $(x_{n})$ define by: $x_{0}=x$ and $g(x_{n})=x_{n+1}$ so if we solve this equation: $x^{2}+x-2=0$ we find $1$ and $-2$ ==>

$x_{n}=\alpha+(-2)^{n}\beta ;(\alpha,\beta)\in{\mathbb{R}}$ with $x=\alpha+\beta$ ,then for $n=1$ we find: $g(x)=\alpha- 2\beta$ ==> $f(x)=x+\alpha-2\beta$ $(*)$ .

We have: $f(f(x)-x)=2x$ so by using $(*)$ we find: $\beta=0$ ==> $f(x)=2x \forall x\in{\mathbb{R^{*}}}$

**Posted:** Sun Nov 08, 2009 11:37 am

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$hamza$

mavropnevma

I fail to see why from the star (*) relation follows $\beta = 0$ (other than it's convenient to be so).

**Posted:** Sun Nov 08, 2009 12:12 pm

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Listen to REMBETIKA for decoding the handle.

reason

we have $f(f(x)-x)=2x$ and $f(x)=x+\alpha-2\beta$ so:
$f(x)-x=x+\alpha-2\beta-x=\alpha-2\beta$ and $f(\alpha-2\beta)=2(\alpha-2\beta)$ ,we have $x=\alpha+\beta$ too so we find : $2(\alpha-2\beta)=2(\alpha+\beta)$ ==> $\beta=0$ ..

**Posted:** Sun Nov 08, 2009 12:24 pm

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$hamza$

mavropnevma

I still don't see why $f(\alpha-2\beta)=2(\alpha-2\beta)$ .

**Posted:** Sun Nov 08, 2009 12:40 pm

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Listen to REMBETIKA for decoding the handle.

reason

we have : $f(x)=x+\alpha-2\beta$ so $f(\alpha-2\beta)=\alpha-2\beta+\alpha-2\beta$ ==> $f(\alpha-2\beta)=2(\alpha-2\beta)$

**Posted:** Sun Nov 08, 2009 12:46 pm

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$hamza$

mavropnevma

I don't think you can do that. Let's go back to where you started solving the recurrence relatiion: you started with $x_0 = x$ , etc. Then the coefficients $\alpha$ and $\beta$ depend on the value of the initial term, so in fact you have $\alpha_x$ and $\beta_x$ .

Now, you got $f(x) = x + \alpha_x - 2\beta_x$ , and you're not allowed to replace $x$ with any value you want, respectively $\alpha_x - 2\beta_x$ , because for this value you may have different coefficients $\alpha$ and $\beta$ , in fact $\alpha_{\alpha_x - 2\beta_x}$ and $\beta_{\alpha_x - 2\beta_x}$ .

My point is that coefficients $\alpha$ and $\beta$ are not "universal", over all $x$ .

**Posted:** Sun Nov 08, 2009 1:11 pm

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Listen to REMBETIKA for decoding the handle.

reason

yes,I undrestand your point, but I think that can correct my solution:
we have:

$g(g(x)) + g(x) = 2x$

$g(x) = x - 3\beta_{x}$

$g(g(x)) = g(x) - 3\beta_{g(x)}$

then we conclude that: $- 3(2\beta_{x} - \beta_{g(x)}) = 0$ ==> $2\beta_{x} = \beta_{g(x)}$ , then by using this result we can find that: $g(g(x)) + g(x) = 2x$ <==> $2x - 12\beta_{x} = 2x$ ==> $\beta_{x} = 0$ so $g(x) = x$ ==> $f(x) = 2x \forall x\in{\mathbb{R^{*}}}$

I hope that will be correct.

**Posted:** Sun Nov 08, 2009 1:45 pm

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$hamza$

Vincent Gilbert · P versus NP Offline Joined: 11 Aug 2009 Posts: 35

First , $f(x)-x \in R^* \forall x \in R^*$
$g(x)=f(x)-x \Rightarrow g: R^* \rightarrow R^* , g(g(x))+g(x)=2x$
LEt's consider the sequence : $u_0=x_0; u_{n+1}=g(x_n)$
$\Rightarrow u_n>0 (*) \forall n \in N ; u_(n+1)+u_n=2u_(n-1)$
$\Rightarrow u_n=\alpha+(-2)^n .\beta$
* $\beta <0 \Rightarrow u_{2n}<0$ when $n$ is large enough (Contradition (*))
* $\beta>0\Rightarrow u_{2n+1}<0$ when $n$ is large enough (Contradition (*))
Hence : $\beta=0$
$\Rightarrow u_1=u_0=x_0$
$\Rightarrow g(x_0)=x_0$
$\Rightarrow f(x)=2x \forall x \in R^*$

**Posted:** Sun Nov 08, 2009 5:43 pm

Mashimaru

**Posted:** Mon Nov 09, 2009 6:21 pm

Vincent Gilbert · P versus NP Offline Joined: 11 Aug 2009 Posts: 35

You're right

**Posted:** Tue Nov 10, 2009 2:52 am

Mashimaru

I think there should be some additional condition to this functional equation because of the following reasons:

1. It is clearly that we just need to solve an equivalent problem:
Find every function $g: \mathbb{R}^*\to\mathbb{R}^*$ satisfies: $g(g(x)) + g(x) = 2x,\forall x\in\mathbb{R}$ .

2. By considering the sequence $\{u_n\}_{n=0}^{\infty}$ defined by $u_0 = x, u_{n+1} =g(u_n)$ , it is known that $u_n = \lambda_1 + \lambda_2 (-2)^n$ , thus if $\lambda_2\neq 0$ , the sequence $\{u_n\}_{n=0}^{\infty}$ takes both negative and positive values. Thus if we add the continuity of $g(x)$ , we can deduce that $\exists x: g(x)=0$ , contradiction. Moreover, if $g: \mathbb{R}^+\to\mathbb{R}^+$ or $g: \mathbb{R}^- \to \mathbb{R}^-$ , we can also find $g$ .

3. Still consider the sequence $\{u_n\}$ above, we see that the definition of $g$ is just in $\{u_n\}$ , therefore if we can point out a partion of $\mathbb{R}^*$ into such disjoint sequences like $\{u_n\}$ , i.e. sequences satisfies $u_{n+1} + u_n - 2u_{n-1} = 0$ then there is a function $g$ . This is just my sense that we can construct infinitely many functions $g$ satisfies the problem in this way. maybe

**Posted:** Tue Nov 10, 2009 4:55 am

Dimitris X

solution (Abhay Kumar Jha).

We begin with observation that $f(x)\ge 0,\forall x\in R^*$ .Define a new function $g$ on $R^*$ by setting $g(x) = f(x) - x$ .Since $f(x)\ge x$ ,we see that $g(g(x))$ is meaningful.A simple computation gives $g(g(x)) = f(g(x)) - g(x) = f(f(x) - x) - (f(x) - x) = 3x - f(x) = 2x - g(x).$

Thus we have to find $g: R^* \to R^*$ satisfying

$g(g(x)) + g(x) = 2x.$

Fix some $a$ in $R^*$ and define $u_n = g^n(a)$ where $g^n(x) = g(g^{n - 1}(x)).$ Then $< u_n >$ satysfies the recurrence relation $u_{n + 2} + u_{n + 1} + - 2u_n = 0$ for all $n \ge 0$ ,where $u_o = a$ .This diference equation for $u_n$ has auxiliary equation $x^2 + x - 2 = 0$ which has solutions $x = 1$ and $x = - 2$ .Using the theory of difference equations,the general solution of the difference equation is given by

$u_n = A(1)^n - B( - 2)^n$

for some constants $A$ and $B$ .However, we see that $g(x)\ge 0$ for all $x\in R^*$ ,and hence $u_n = g^n(a) \ge 0$ .Since $( - 2)^n$ alters sign as $n$ runs through natural numbers,we conclude that $B = 0$ .This forces $u_n = A$ for all $n$ and using initial condition,we can see that $2A = u_1 + u_2 = 2a$ .We thus obtain $A = a$ and hence $u_n = a$ for all $n$ .But then $g(a) = u_1 = a$ and we conclude that $f(a) = 2a$ .Since this is true for every $a\in R^*$ we arrive at the solution $f(x) = 2x$ .

P.s.1
I cant understand the solution very well because i'm not familiar with sequences and thats why i post this problem on unsolved,hoping for an elementary solution....

P.s.2
The book uses the notation $Ro$ and i guess he mean $R^*$ ......

**Posted:** Tue Nov 10, 2009 5:55 am

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ΠΑΙΡΝΩ ΤΑΜΠΕΛΑ ΚΑΙ ΕΓΩ ΤΟΥ ΕΘΝΙΚΟΥ ΠΡΟΔΟΤΗ ΑΦΙΕΡΩΜΕΝΟ ΚΑΙ ΑΥΤΟ ΣΕ ΚΑΘΕ ΔΟΥΛΟ ΠΑΤΡΙΩΤΗ.....

Mathias_DK

**Posted:** Tue Nov 10, 2009 6:17 am

Mashimaru

**Posted:** Wed Nov 11, 2009 2:08 am

FelixD

Maybe $f: \mathbb{R_+} \to \mathbb{R_+}$ ^^

**Posted:** Thu Nov 12, 2009 9:29 am

Mashimaru

**Posted:** Sat Nov 14, 2009 2:51 am

pco

**Posted:** Sat Nov 14, 2009 2:59 am

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Patrick

Mashimaru

**Posted:** Sat Nov 14, 2009 6:39 am

pco

Just take one basis $\{b_i\}$ of the $\mathbb Q$ -vector space $\mathbb R$

Then define :
$f_1(\sum q_ib_i)=\sum q_if(b_i)$ with $f(b_1)=2b_1$ and $f(b_i)=-b_i$ for all other elements of the basis. You got a solution.

$f_2(\sum q_ib_i)=\sum q_if(b_i)$ with $f(b_2)=2b_2$ and $f(b_i)=-b_i$ for all other elements of the basis. You got another solution.

$f_3(\sum q_ib_i)=\sum q_if(b_i)$ with $f(b_3)=2b_3$ and $f(b_i)=-b_i$ for all other elements of the basis. You got another solution.

And so on. Choosing randomly $f(b_i)$ as either $2b_i$ , either $-b_i$ , you build infinitely many different solutions to $f(f(x)-x)=2x$

**Posted:** Sat Nov 14, 2009 7:28 am

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Patrick