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I am trying to solve the functional equation:

Find all functions $f:\mathbb R^2\rightarrow \mathbb R$ such that for all $\left \{x_1,x_2,...,x_{2016} \right \}\subset \mathbb R$: $$f(f(f(...(f(x_1,x_2),x_3),...),x_{2016}))=x_1+x_2+...+x_{2016}$$

My work so far:

$f(x,y)=f(0+0+..+0+x,y)$

$f(f(f(...(f(0,0),0),...),x),y)=0+0+...+0+x+y=x+y$

I need help here.

Roman83
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  • Was this from a math contest? – almagest May 04 '16 at 09:04
  • @almagest: Yes, in 2007. I changed in 2007 for 2016. I think the solution is not affected – Roman83 May 04 '16 at 09:08
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    I suggest citing the source then! – flawr May 04 '16 at 09:17
  • @Roman83 It probably doesn't change anything, but it might also turn out to be relevant that, for instance, $2016$ is even while $2007$ isn't. – Arthur May 04 '16 at 10:19
  • @Arthur: Thank you. I understood. – Roman83 May 04 '16 at 10:34
  • Your picture of the text contains 17 lines, 14 are unrelated to the problem. You should change this. Also you didn't add a reference for your problem, e.g., which contest. You "work so far" shows not much effort. It maybe important that n=2007 and not n=2016. But you can start with n=3. This makes the structure of the problem simpler. Maybe this brings some insight after some work. – miracle173 May 04 '16 at 11:35
  • @Roman83: If you don't know the solution to this problem, it is absolutely unacceptable to change $2007$ to $2016$ like you did. You may have turned an easy question into an impossible one, which is just a waste of time for everybody. – TonyK May 04 '16 at 16:53
  • @TonyK: I got it. I do so I will not – Roman83 May 04 '16 at 17:03

2 Answers2

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Considering the outer most iteration (the one with $x_{2016}$), we obviously have $f(x,y) = g(x)+y$ for some function $g$, so:

$\begin{align*}x_1+x_2+\ldots+x_{2015}+x_{2016}&= f(f(f(\ldots f(x_1,x_2),\ldots,x_{2016}) \\ &= g(g(\ldots g(x_1)+x_2 \ldots )+x_{2015})+x_{2016} \\ \implies x_1+x_2+\ldots +x_{2015}\qquad \quad\,&=g(g(\ldots g(x_1)+x_2 \ldots )+x_{2015}) \end{align*}$

Setting $x_1 = \dots = x_{2014} =0$ and $c:=g(\ldots g(x_1)+x_2 \ldots ) = g(\ldots g(0)+0 \ldots )$ we get

$y = g(c+y)\;\forall y$

We subsitute $y$ with $t-c$ and get

$t-c = g(t)\;\forall t \qquad (*)$

Pluggin this equation back in the original results in

$\begin{align*}x_1+x_2+\ldots+x_{2016} &= f(f(f(\ldots f(x_1,x_2),\ldots,x_{2016}) \\ &\overset{(*)}{=} (x_1-c)+(x_2-c)+\ldots+(x_{2015}-c)+x_{2016} \\ &= x_1+x_2+\ldots +x_{2016} - 2015c\end{align*}$

Therefore $c=0$ and so $g(t) = t$. And this implies $f(x,y)=x+y \;\forall x,y$.

flawr
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  • I cant see that it is ovious (from the outer most iteration) that f(x,y)=g(x)+y. How can you proof this? – miracle173 May 04 '16 at 21:09
  • yes, of course. But it is still not obvious. If you can prove that you can find $x_1$,...,$x_{2015}$ such that $f(f(f(\ldots f(x_1,x_2),\ldots,x_{2015})=x $ for an arbitrary $x$ than we are done. But how can you prove this? – miracle173 May 04 '16 at 21:38
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So lets assume $f$ is a real function an

$$f(f(f(\ldots f(f(x_1,x_2),x_3)\ldots),x_{2016})= x_1+x_2+\ldots + x_{2016} \tag{1}$$

We now apply

$$f(.,x_{2017}) \tag{2}$$

on the LHS and RHS of $(1)$ and we get

$$f( f(f(f(\ldots f(f(x_1,x_2),x_3)\ldots),x_{2016}) ,x_{2017}) = f(x_1+x_2+\ldots + x_{2016},x_{2017}) \tag{3}$$

If we change the variable names in $(1)$ , $x_{i}$ by $x_{i+1}$, we get

$$f(f(f(\ldots f(f(x_2,x_3),x_4)\ldots),x_{2017})= x_2+x_3+\ldots + x_{2017} \tag{4}$$

In $(4)$ now we substitute $x_2$ by $f(x_1,x_2)$ and get

$$f(f(f(\ldots f(f( f(x_1,x_2) ,x_3),x_4)\ldots),x_{2017})= f(x_1,x_2) +x_3+\ldots + x_{2017} \tag{5}$$

The LHS of $(3)$ and $(5)$ are the same. So the RHS of $(3)$ and $(5)$ must be equal and we get

$$f(x_1+x_2+\ldots + x_{2016},x_{2017}) = f(x_1,x_2) +x_3+\ldots + x_{2017} \tag{6}$$

Here we set $x_1=0$, $x_2=0$, ..., $x_{2015}=0$ and get

$$f(x_{2016},x_{2017})=f(0,0)+x_{2016}+x_{2017} \tag{7}$$

We can use $(7)$ to evaluate the nested function expression of the LHS of $(1)$. You can start with the outer or with the inner expression. Finally you will get

$$f( f(f(f(\ldots f(f(x_1,x_2),x_3)\ldots),x_{2016}) ,x_{2017})= 2015 f(0,0)+x_1+x_2+x_3+\cdots+x_{2016} \tag{8}$$

The RHS of $(1)$ and $(8)$ are equal, so ist must be

$$2015f(0,0)=0 \tag{9}$$

From $(9)$ and $(7)$ we see that

$$f(x,y)=x+y \tag{10}$$

as expected.

miracle173
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