# Iterated Functional Equation

#### Jameson

Staff member
I don't know what to title this but will change it if $f(f(x))$ has a name. Anyway, I need to find $f(x)$ such that $f(f(x))=-x$. My friend gave me this challenge question and I haven't been able to figure it out.

There are many examples where $f(f(x))=x$ for example $$\displaystyle f(x)=\frac{1}{x}$$ but that doesn't work for this one. I also tried a piecewise function but that didn't work either.

Lastly $f: \mathbb{R} \rightarrow \mathbb{R}$

Any ideas?

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#### MarkFL

Staff member

$f(x)=ix$?

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#### Jameson

Staff member

$f(x)=ix$?
$f(2)=2i$, $f(2i)=2i^2=-2$
$f(-2)=-2i$, $f(-2i)=-2i^2=2$

$$\displaystyle f \left( \frac{1}{2} \right)= \frac{1}{2}i$$, $$\displaystyle f \left( \frac{1}{2}i \right) \frac{1}{2}i^2=-\frac{1}{2}$$

Yup, seems to work. Nice! I wonder if there are any others as well.

#### Fernando Revilla

##### Well-known member
MHB Math Helper
Nice! I wonder if there are any others as well.
At least, if $f:\mathbb{C}\to \mathbb{C}$ has de form $f(x)=ax+b$, necessarily $f(x)=ix$ or $f(x)=-ix$.

#### Jameson

Staff member
Doh!!! I forgot my friend stipulated that f: R -> R.

#### MarkFL

Staff member
For a function of the form:

$f(x)=kx^n$ then $f(f(x))=k^{n+1}x^{n^2}$

and the only solution I can find is for $n=1$ leading to the example I gave. (I missed the other example cited by Fernando Revilla which comes from $k^2=-1\,\therefore\,k=\pm i$)

I bet there are other forms that might work though. I have to run now, but I will be thinking about this. #### Fernando Revilla

##### Well-known member
MHB Math Helper
Doh!!! I forgot my friend stipulated that f: R -> R.
The trick is to consider a set of disjoint intervals $(0,1],(1,2],(2,3],(3,4],\ldots$ covering $\mathbb{R}^+$. We define $f:\mathbb{R}^+\to \mathbb{R}$ in the following way

\begin{align}
f(0) &= 0 \\
f(x) &= x+1 &&\mbox{if $x\in(2k-2,2k-1]$}\\
f(x) &= 1-x &&\mbox{if $x\in(2k-1,2k]$}\\
\end{align}

We easily verify $f(f(x))=-x$. Now we can extend $f$ to $\mathbb{R}$ by odd symmetry i.e. if $x<0$ we define $f(x)=-f(-x)$ and the condition $f(f(x))=-x$ is verified for all $x\in\mathbb{R}$.

#### MarkFL

Staff member
Doh!!! I forgot my friend stipulated that f: R -> R.
Allow me to offer a bit of friendly advice when posting problems on math help sites:

<ducking&running&lmao>   #### rashtastic

##### New member
Hello all,

I don't recall where we originally found the problem but we certainly had some fun with it before I offered it to Jameson.

A similar problem was posted on stackoverflow, of all places.

Here was my lamentable first solution:

$f(x)=\left\{\begin{matrix} 0 & | &x=0 \\ -\frac{sign(x)}{|x|+1}{} & | & x\in \mathbb{Z} \setminus \{0\} \\ \frac{1}{x}-sign(x) & | & \frac{1}{x}\in \mathbb{Z}\\ \frac{1}{x} & | & |x|>1, x\notin\mathbb{Z}, \frac{1}{x}\notin\mathbb{Z}\\ -\frac{1}{x} & | & |x|<1, \frac{1}{x}\notin\mathbb{Z} \end{matrix}\right.$

There are three things going on:
1. 0 maps to 0 (I'll write 0 > 0), which it must do for all f
2. The integers and their reciprocals: 1>-1/2>-1>1/2>1> ...; 2>-1/3>-2>1/3>2>...
3. f maps everything else to its inverse or negative inverse: 3/5>-5/3>-3/5>5/3>3/5>...; pi>1/pi>-pi>-1/pi>pi>...

The pieces came from trying something and making different rules for the numbers it didn't work for.

After finding a few other solutions similar to Fernando's very smooth function, we wondered if all such f operated on these kinds of tricks, and the answer was an emphatic No. We could map pi^e > 2 > -pi^e > -2 > pi^e if we wanted, just as long as each nonzero pair (x,-x) was itself paired with another nonzero (y,-y), and those pairings of pairs could be whatever we liked. That thought led to the attached document. I hope you guys find it interesting and correct!

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