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I am very interested in functions $\gamma:\mathbb R\to\mathbb R$ with the following property: $$\gamma^2(x)=x$$ One form of a function satisfying this is $$f(x)=\frac{a-x}{1+bx}$$ Which has the property $f^2(x)=x$. Infinitely many more functions with this property can be obtained by finding some other injective function $g$, its inverse $g^{-1}$, and then composing $g, g^{-1},$ and $f$ as follows: $$g^{-1}\circ f\circ g$$ However, I am not very interested in involutory functions of this form, since they seem to all be ripoffs of the general form that I already stated.

In fact, it seems that all involutory functions can be put in the form $$g^{-1}\circ f\circ g$$ for some $g$, and for some $a,b$. I can't find any counterexamples, but I don't know how to prove it either. It seems to me that the best way to approach this would be to set up some kind of differential equation like $$(f'\circ f)(x)=\frac{1}{f'(x)}$$ But I have absolutely no idea how I might show that any involutory function can be put in the aforementioned form.

Any ideas?

NOTE: I'm sure there are some elaborate piecewise-defined answers that can destroy my conjecture. However, I can't expect people to know what I mean when I ask to prove this for all "reasonable" functions - so I will establish some stricter restrictions on $\gamma$. The function must be expressible using some finite composition of these functions and their inverses: $$\phi_1(x,a)=x+a$$ $$\phi_2(x,a)=ax$$ $$\phi_3(x,a)=x^a$$ $$\phi_4(x,a)=a^x$$ For example, $x^2+x+1$ can be expressed as $$\phi_1(\phi_3(x,2),\phi_1(x,1))$$

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  • $\begingroup$ what is the role played by $n$? $\endgroup$ – Francesco Polizzi Aug 12 '17 at 16:45
  • $\begingroup$ Whoops, sorry. I had $\gamma^n$ but decided to change it to $2$. I'll fix it! $\endgroup$ – Franklin Pezzuti Dyer Aug 12 '17 at 16:46
  • $\begingroup$ The answer will very likely depend on your setting. Just sets and functions? Spaces and continuous maps? Groups and homomorphisms? $\endgroup$ – Randall Aug 12 '17 at 16:46
  • $\begingroup$ @Randall I just changed my question to specify that $\gamma$ maps reals to reals. $\endgroup$ – Franklin Pezzuti Dyer Aug 12 '17 at 16:48
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    $\begingroup$ @Nilknarf If you don't require continuity any involutive bijection can do e.g. the transposition $t_{a,b}$ which exchanges only $a\not=b$. $\endgroup$ – Duchamp Gérard H. E. Aug 12 '17 at 16:51
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There is a general approach using power series. Suppose that $g(x) = b_1x+b_2x^2+b_3x^3+\dots\quad$ and we require that $x+f(x)=g(-xf(x))\;$ for some power series $f(x)$. The connection with $g(x)$ implies $f(f(x))=x.\quad$ Solving for the coefficients of $f(x)$ term by term gives the expansion $$f(x)=-x+b_1x^2-b_1^2x^3+(b_1^3+b_2)x^4-(b_1^4+3b_1b_2)x^5+\dots$$ which is the general form of involution with fixed point $0$ unlike your $f(x)=(a-x)/(1+bx)$ where $f(0)=a$ and $f(a)=0$ with $a\neq 0$. However, If $a=0$ then $$f(x)=-x/(1+bx)=-x+bx^2-b^2x^3+b^3x^4+\dots$$ which is the case where $g(x)=bx$.

You might ask for involutions $f(x)=x+a_2x^2+\dots\;$ but the only example is $f(x)=x$.

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Here are two families of involutory functions (A and B):

A) $$f_k(x)=\dfrac{kx}{\sqrt{x^2-k^2}} \ \ \ \text{for any} \ k>0 \tag{1}$$

(one can check that $f_k(f_k(x))=x$).

enter image description here

Fig. 1 : The curves of functions $f_k$ (see equation (1)) are symmetrical with respect to the angle bissector of the first quadrant (in red with equation $y=x$).

The only drawback is that, instead of having only one forbidden point, like for your functions, there is a whole forbidden interval : $$I_k=[-k,k],$$

(that can be made arbitrarily short).

Besides, nothing prevents to set for example $f_k(x)=x$ when $x \in I_k$...


It's worth telling how I found function $f$.

I thought in polar coordinates, to the following function $$\text{at first, to : } \ r=\dfrac{1}{\sin(2 \theta)}, \ \ \text{then, to : } \ r=\dfrac{2k}{\sin(2 \theta)}\tag{2}$$

Why that ? For two reasons :

  • Because its graphical representation is symmetrical with respect to line bissector $y=x$ : we have the same value of $r$ for $\theta=\dfrac{\pi}{4}-\alpha$ and $\theta=\dfrac{\pi}{4}+\alpha$ ;

  • Because, $\sin(2 \theta)$ placed in the denominator will generate (horizontal and vertical) infinite branches, i.e., an "infinite" $r$ for all $\theta=k \pi/2$.

Squaring (2) gives :

$$r^2=\dfrac{4k^2}{(2 \sin \theta \cos \theta)^2} \ \ \iff \ \ (r \sin \theta \ r\cos \theta)^2=k^2r^2\tag{3}$$

It's time to switch to cartesian coordinates, with conversion formulas :

$$x=r \cos \theta, \ \ y=r \sin \theta$$

we can transform (3) into :

$$x^2y^2=k^2(x^2+y^2)\tag{4}$$

It is enough now to extract $y$ from (4) and take the square root to obtain (1).


B) On the model of (4), we can consider functions defined in an implicit way like this :

$$x^3y^3=k^3(x^3+y^3)\tag{5}$$

or with a cartesian equation :

$$g_k(x)=\dfrac{kx}{\sqrt[3]{x^3-k^3}}\tag{6}$$

They possess a superiority over functions $f_k$ : they are defined everywhere but for value $x=k$.

enter image description here

Fig. 2 : Some curves of functions $g_k$ ($k=0.5, 1, 1.5, 2$) with different colors (see equation (6)) Their horizontal and vertical asymptotes (not represented here) are $x=k$ and $y=k$.

Remark : one can evidently build other involutive functions based on the same recipes...

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You want an involution $h$ for which no invertible $g$ satisfies $hg=gf$. In fact if $h$ is the identity function this is equivalent to $g=gf$, which by invertibility implies $f$ is the identity function. This fails for any $a,\,b$.

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