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My question is about existence of a non-trivial solution of the functional equation $f(f(f(x)))=-x$ where $f$ is a continuous function defined on $\mathbb{R}$. Also, what about the general one $f^n(x)=-x$ where $f^n$ is understood in the sense of composition of functions and $n$ is odd. Eventually, is there some theory about continuous solutions of $f^n(x)=g(x)$ where $g$ is a fixed continuous function. I tried a research here but all what I found was about $f^2(x)=g(x)$, i.e, "square root" in the sense of composition. Thanks.

EDITED : I was looking for a non-trivial solution with $n$ odd, sorry for the inconvenience. I reformulated my question. Thanks for the last poster who showed that the unique continuous solution to $f^n(x)=-x$ where $n$ is odd is the trivial one.

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6  
$f(x)=-x$ is a rather obvious example for your first question –  Henry Nov 16 '11 at 23:33
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@Henry: it works obviously for all odd $n$ ;) –  Damian Sobota Nov 16 '11 at 23:52
    
Are there examples of a function other than $f(r)=-r$ that does this? –  josh Nov 17 '11 at 0:35

3 Answers 3

up vote 7 down vote accepted

The only solution is $f(x)=-x$ for all odd $n$.

We have $f(0)=0$ because if $f(0)=a$, then $f(f(a)))=0$ and $-a=f(f(f(a)))=f(0)=a$, so $a=0$.

Since $f^{2n}(x)=x$ for all $x$, the function $f$ cycles sets of at most $2n$ elements $\{\pm a_1,\pm a_2,\ldots,\pm a_n\}$. Fix one of these cycles, and assume without loss of generality that all the $a_i$ are nonzero, and denote by $a>0$ the least positive number in this cycle. Then $f((0,a))$ is an open interval with the endpoints $f(0)=0$ and $f(a)=b$. Assume that $b\neq \pm a$. Since $|b|>|a|$, the interval $f((0,a))$ must contain either $a$ or $-a$, which means there is an $x, |x|<a$ such that $f(x)=a$ or $f(x)=-a$, which is a contradiction, because $f$ is a bijection and $\pm a$ are the images of some $a_i, |a_i|\geq a$. Thus it must hold that $|f(a)|=a$ for all $a$, and consequently $f(a)=-a$ for all $a$.

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Thank you, this is what I was searching for. The last pending question is about $f^n(x)=g(x)$ (see my first post, I edited it). –  Dijkschneier Nov 17 '11 at 13:54
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I think $f^n=g$ is too broad a question. Already functional square roots are difficult. I suppose you've already seen the MO thread mathoverflow.net/questions/17614/solving-ffxgx ? A poster in it recommends the book "Iterative functional equations", but I don't know anything about the book or any general theory. –  Samuel Nov 17 '11 at 15:44

As Henry points out in the comments, when $n$ is odd $f(x)=-x$ obviously works.

When $n$ is even, there is no such function. To see this, note the following properties must hold:

  • $f$ is a bijection, and hence stricly monotonic.
  • $f(0)=0$. Indeed, if $f(0)=a$, then $f^n(a) = f^{n+1}(0) = f(f^n(0)) = f(0) = a$, so $a=0$ since $0$ is the only fixed point of $-x$.
  • From the previous two properties, we must have that $f$ is either increasing, in which case $f(x)$ has the same sign as $x$ for every $x$, or decreasing, in which case $f(x)$ has the opposite sign as $x$ for every $x$. Either way, $f^n(x)$ and $x$ share the same sign, and so $f^n(x) \neq -x$.
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I have seen an example of a piecewise continuous function $f$ satisfying $f^2(x)=-x$. It's best described with a picture, but I don't know how to make those in this forum. Instead, I can describe it for nonnegative $x$, and then I'll ask you to give it odd symmetry.

EDIT: I did indeed mess up the formula. I think this is what I wanted. \begin{align} f(0) &= 0 \\ f(x) &= x+1 &&\mbox{for $x\in(2n-2,2n-1]$}\\ f(x) &= 1-x &&\mbox{for $x\in(2n-1,2n]$}\\ \end{align}

Again, this is for nonnegative $x$ only. Extend to negative $x$ with odd symmetry.

Sketch this for $n=1,2,3$ and you'll get the idea. Assuming I haven't messed up the formula, you get a graph that reminds me of TIE fighters from Star Wars.

EDIT: I uploaded a picture - much easier than I thought.

Graph of $f$

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Why the downvote? If you are going to downvote you should leave a comment. The OP's original question did not specify $n$ to be odd. I know this function is not continuous but I thought the OP would find it interesting. –  alex.jordan Nov 17 '11 at 17:42
    
I wasn't the downvote, but if you're not going to answer the question, you should be explicit about that in the first sentence of the post (or at least very quickly). –  Eric Stucky Mar 9 '13 at 0:29
    
@Eric Hi Eric, in an early version of this question, OP asked about $f^n(x)=-x$ without specifying $n$ odd. The question has changed since it was first asked, and I think this answer applies to the original question. But this is from a while back. –  alex.jordan Mar 9 '13 at 3:10
    
Oh, that was a while ago. Sorry for dragging it up for no good reason :/ –  Eric Stucky Mar 9 '13 at 6:10

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