3

For example, the simple self-inverse function $y = 6 - x$ can be written as $x+y=6$, which is symmetric in $x$ and $y$. Less apparent (to me at least), $y = (x+1)/(1-x)$ can be expanded and written as $x + y = xy - 1$, which is also symmetric in $x$ and $y$, and must therefore be self-inverse. Can all self-inverse functions be expressed this way?

vitamin d
  • 5,783
user1153980
  • 1,121

2 Answers2

0

The answer is yes. First, notice that $f$ must satisfy $f:I\subset\mathbb{R}\to I\subset\mathbb{R}$, since it is its own inverse. The best way of justifying it is by looking at the $graph$ of the function, i.e., the set $$G = \{(x,f(x))\in\mathbb{R}^2:x\in I\}=\{(f(y),y)\in\mathbb{R}^2:y\in I\}$$ This equality holds, again, because $f$ is its own inverse. Therefore, the graph of $f$ is symmetric, that is to say, it satisfies $G=\{(y,x)\in\mathbb{R}^2:(x,y)\in G)\}$ (it is immediate to see it from the above equality).

Hence, if $G$ can be expressed as the set of points that follow an equation, $G=\{(x,y)\in\mathbb{R}^2:g(x,y)=0\}$, this equation $g(x,y)=0$ must be symmetric in the variables $(x,y)$ (again, it is immediate to see it since $G$ is symmetric)

If you have any problems seeing what I said, don't hesitate to write a comment.

Samuel M. A. Luque
  • 1,925
  • The argument here is either circular or only reducing the problem to proving that a set in $\mathbb{R}^2$ that is invariant by $s(x,y)=(x,y)$ can be defined as the zero set of symmetric functions? The claim in you second to last paragraph is not correct, since the $g(x,y)$ in the definition of $G$ can perfectly be one of the functions given in the question as examples. They are not symmetric. – plop Jun 10 '21 at 18:39
  • @plop My argument went as follows: $f$ is symmetric $\implies$ the graph of $f$ is a symmetric set in $\mathbb{R}^2$ $\implies$ the equation describing $G$ is symmetric in the variables (x,y). I'm sorry if I didn't make myself clear :( – Samuel M. A. Luque Jun 10 '21 at 18:55
  • What you are claiming is clear. The problem is that it is false. Take your $G$ defined as the zero set of $g(x,y)=y-\frac{1-x}{1+x}$. The function $g$ is not symmetric. For example $g(0,2)=1\neq \frac{1}{3}=g(2,0)$. – plop Jun 10 '21 at 19:12
  • Of course it is not, I'm saying that the equation $g(x,y)=0$ is symmetric, i.e., that the zeroes of $g$ are symmetric (and that is what OP asked). Did I state it otherwise? I am sorry if I did. – Samuel M. A. Luque Jun 10 '21 at 19:16
  • Take your time. I sense that you are going to start saying that what you mean by "the equation $g(x,y)=0$ being symmetric" is that the set of points that it defines is symmetric, which again is what the question is saying as hypothesis. So, in that case you are not saying anything. – plop Jun 10 '21 at 19:16
  • Let us continue this discussion in chat. – Samuel M. A. Luque Jun 10 '21 at 19:18
  • I apologize for any ambiguity. I realize that what I wanted to prove is strictly speaking not true. Consider x+y=6. What I intended was to say that if we swap x and y, we get a symbolically identical equation, which would be y+x=6, but y+x is symbolically different from x+y. It might be possible to rescue the original statement by allowing terms to be reordered in accordance with commutative, associative and distributive laws, but I am not confident that such a thing is possible. – user1153980 Jun 10 '21 at 23:31
  • @user1153980 Don't worry. Yes, you mean that the function $g(x,y)$ describing the graph of $f$ is symmetric ($g(x,y)=g(y,x)$, a totally unambiguous statement). But what really describes the graph of $f$ is the condition (or equation) $g(x,y)=0$ (in your case, $x+y-6=0$). In that case, $g$ is indeed symmetric. What I proved in my answer is that the zeroes of $g$ are symmetric. And since all we care about is the condition $g(x,y)=0$, i.e., the set of zeroes of $g$, we can always take a symmetric function to describe this (namely, $g(x,y)=1$ if $(x,y)\notin G$ and zero if they are in $G$. – Samuel M. A. Luque Jun 10 '21 at 23:55
  • @user1153980 Or, as plop pointed out, $g(x,y)=(x-f(y))(y-f(x))$ – Samuel M. A. Luque Jun 10 '21 at 23:59
-1

No, because most functions, even most self-inverse ones, cannot be expressed by any equation at all. We can choose any subset $A$ of $\Bbb R^{\gt 0}$ and define $$f_A(x)=\begin {cases} x&|x| \in A\\-x&|x| \not \in A \end {cases}$$ There are uncountably many subsets, each of which defines a self-inverse function. Most of them cannot be expressed as there are only countably many ways to express such a function.

Ross Millikan
  • 374,822
  • I would appreciate the downvoter saying what is wrong with this answer. I have a longtime frustration with people thinking every function has a simple formula. That is badly wrong. – Ross Millikan Jun 11 '21 at 02:55
  • Yes, your argument is wrong. All you are proving is that you cannot assign a formula to each symmetric set. But that is irrelevant to this question. Here the symmetric set (say of $\mathbb{R}^2$) is a quantified variable, you only need a single symbol, which gets a $\forall$ in front. For example, $f$, or your $A$. Based on that symbol the goal is to produce another symbol(or group of symbols), the formula, that then is required to satisfy the property of being symmetric. – plop Jun 11 '21 at 03:03
  • @plop: The question asks if every self-inverse function can be expressed this way, where this way implies a specific formula. – Ross Millikan Jun 11 '21 at 03:16
  • Yes, a specific formula, dependent on a variable $f$, since we are proving a statement for an arbitrary involution $f$. – plop Jun 11 '21 at 03:19
  • Compare to asking if we have a formula for the solutions of every quadratic equation over the reals. We wouldn't say "no, because not all real numbers are definable". We have the quadratic formula by having quantified variables for the coefficients and we give a formula in terms of those variables. – plop Jun 11 '21 at 03:25
  • @plop: but the language of analysis does not allow quantifying over sets of reals, only over reals. It is a first order theory. The quadratic equation works fine that way, but describing functions based on sets of reals does not. – Ross Millikan Jun 11 '21 at 03:41
  • Whether the statement can be written in some language and not another is fair game. The question didn't specify. That the answer cannot be expressed in a language that cannot express the question, is some information, not very interesting, though. It cannot be expressed in this language ${5}$ either. – plop Jun 11 '21 at 14:01