What is a function that satisfies the iterated functional equation $f(f(x))=-x$?

  • Context: MHB 
  • Thread starter Thread starter Jameson
  • Start date Start date
  • Tags Tags
    Functional
Click For Summary

Discussion Overview

The discussion revolves around finding a function \( f \) such that \( f(f(x)) = -x \), specifically for the case where \( f: \mathbb{R} \rightarrow \mathbb{R} \). Participants explore various forms of functions and their properties in relation to the given functional equation.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant notes the challenge of finding a function \( f \) that satisfies \( f(f(x)) = -x \) and mentions that common examples like \( f(x) = \frac{1}{x} \) do not apply.
  • Another participant proposes \( f(x) = ix \) and provides calculations to show it satisfies the equation, expressing curiosity about other potential solutions.
  • A different participant suggests that if \( f \) is of the form \( f(x) = ax + b \), then it must be either \( f(x) = ix \) or \( f(x) = -ix \), but later acknowledges the restriction to real functions.
  • One participant describes a piecewise function defined on disjoint intervals that satisfies the condition and extends it to all of \( \mathbb{R} \) using odd symmetry.
  • Another participant shares a complex piecewise function they developed, detailing how it maps various types of numbers and expressing uncertainty about whether all such functions follow similar patterns.
  • A participant references a similar problem found elsewhere, suggesting it may be useful for others in the discussion.

Areas of Agreement / Disagreement

Participants express varying approaches and potential solutions, but there is no consensus on a single function that satisfies the condition for all real numbers. Multiple competing views and methods remain present throughout the discussion.

Contextual Notes

Some participants note the importance of including all relevant information when posting problems, indicating that assumptions and restrictions on the function's domain are critical to finding solutions.

Jameson
Insights Author
Gold Member
MHB
Messages
4,533
Reaction score
13
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 [math]f(x)=\frac{1}{x}[/math] 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?
 
Last edited:
Mathematics news on Phys.org
How about:

$f(x)=ix$?
 
Last edited:
MarkFL said:
How about:

$f(x)=ix$?

$f(2)=2i$, $f(2i)=2i^2=-2$
$f(-2)=-2i$, $f(-2i)=-2i^2=2$

[math]f \left( \frac{1}{2} \right)= \frac{1}{2}i[/math], [math]f \left( \frac{1}{2}i \right) \frac{1}{2}i^2=-\frac{1}{2}[/math]

Yup, seems to work. Nice! I wonder if there are any others as well.
 
Jameson said:
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 $f(x)=-ix$.
 
Doh! I forgot my friend stipulated that f: R -> R.
 
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.:cool:
 
Jameson said:
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 $x\in\mathbb{R}$.
 
Jameson said:
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:

When posting the problem, please include all relevant information. This ensures that you get the best and most timely help possible.

<ducking&running&lmao> (Rofl)(Wave)(Beer)
 
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!
 

Attachments

Last edited:

Similar threads

Replies
3
Views
2K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 20 ·
Replies
20
Views
3K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 16 ·
Replies
16
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K