“Class” of functions whose inverse, where defined, is the same “class”

Please excuse the amateurish use of the term "class", I don't know what the exact term is for what I'm looking for.

Anyway, details.

I'm asking specifically about real-valued functions on the real domain ($\mathbb{R}\to\mathbb{R}$). To keep things simple, let's assume that the function is defined on some interval of interest, and is continuous and strictly monotonic in that interval, so that there is an inverse function that's also continuous and monotonic.

I am looking for a "class" of functions where the inverse is of the same "class". By "class", I mean a set of functions with a finite number parameters that, if you changed them, the function would still be in the same "class". (An obvious example of what I mean by a "class" is the polynomials: you can change the coefficients but the function is still a polynomial.) Again I apologize if this is omitting a detail or if there's a nice little word for this that I don't know.

I know of a few examples of "classes" that meet these criteria, including:

• Linear functions
• Piecewise linear functions

Polynomials, of course, do not fit this criteria in general: the inverse of a polynomial is generally not a polynomial. I don't think rational functions do either, but I'm not sure.

For the record, I am asking partly out of curiosity, and partly because I have a nice application in mind. I have a application where I need a function that can approximate a curve with perfect round-tripping ($f(f'(x))=x$, exactly). We're using piecewise linear approximation for now, but it's desirable to be smooth as well.

Thanks.

• I like this question! If nobody provides a satisfactory answer within two days I'll add a bounty to it. – Harambe Nov 1 '17 at 1:42
• Monotonic increasing or decreasing? Does it matter? – Somos Nov 3 '17 at 0:16

Their domains aren't generally all of $\Bbb R$, but the set of Moebius transformations, functions of the form $$f(x) = \frac{ax + b}{cx + d} \quad \text{with}\quad ad - bc \neq 0,$$ are a wonderful group of functions whose inverses are also of that form (and subsume linear functions, by letting $c = 0$). They're also called "linear fractional transformations" (I thought it was conventional to call them linear fractional transformations when you were only considering real number inputs, but evidently I'm wrong about that).

Generally people allow complex inputs for Moebius transformations.

• I'm not sure how much of a deal-breaker it is that their domains aren't all of $\Bbb R$. Perhaps I should have asked first... – pjs36 Nov 1 '17 at 1:26
• Thanks for the answer. You'll notice in the question I spoke of "interval of interest", so not a deal breaker unless the particular interval if interest happens to be all real numbers. – Carl Banks Nov 1 '17 at 22:10
• I just had a thought. What about a composition of Moebius transformations. Subject to the limitations I mentioned (they are defined in an interval of interest and are continuous and monotonic in that interval--so no poles in the interval), will its inverse be a composition of Moebius transformations? Would that be generally true? Intuitively it seems like it would be, but I haven't thought it out yet. – Carl Banks Nov 7 '17 at 0:17
• Yep! I coyly referred to the set of Moebius functions as a group, and I did mean it in the mathematical sense -- compositions of Moebius transformations and inverses of Moebius transformations are also Moebius transformations. The two interact by $(f \circ g)^{-1} = g^{-1} \circ f^{-1}$ (this is true if $f$ and $g$ are elements of any group, not just the group of Moebius transformations). – pjs36 Nov 7 '17 at 0:23
• I see. Unfortunately being a real group kind of defeats the purpose of composition for me; I was hoping that by composing functions one could create a more complex function, and if it still fit the criteria then that'd be a great answer. I suppose it could still work (the family of functions doesn't have to be a group to satisfy my criteria, I don't think). But it looks doubtful that anyone will come up with a whammy, something that could be used to approximate an arbitrary monotonic curve, so take your best answer. – Carl Banks Nov 8 '17 at 18:42

The biggest family is of course: $\{ \textrm{all invertible functions} \}$.

In addition to pjs's family, another small family i could think of is the family:

$\{f_{c,d}(x) = cx^{d} | c, d \in \mathbb{R}^\times \}$

Of course, in this case: $f_{c,d}^{-1} = f_{c^{(d^{-1})}, d^{-1}}$.

• Actually, {all invertible functions} would not meet the criteria, which required "class" to vary over finite number of parameters. I put that criterion in there precisely to disqualify smart aleck answers, and also that no one could claim "infinite polynomials" as the answer. – Carl Banks Nov 5 '17 at 21:09
• @CarlBanks what do you mean by infinite polynomials: formal power series? This will not meet the criterion because there is no guarantee that inverses will exist. I think that finite number of parameters is a unnatural specification, because in math, i feel one is often interested in things with infinitely many parameters: e.g formal power series, vectors in infinite dimensional spaces etc.. who knows there might be some interesting things to say with infinitely many parameters. For example, you have a maxinal family, whereas you cannot have such a notion with only finitely many parameters. – Maithreya Sitaraman Nov 6 '17 at 1:56
• This also brings rise the idea of the order on the families by inclusion, which is also an interesting thing to study. – Maithreya Sitaraman Nov 6 '17 at 1:58
• but yes, if you want to restrict the notion of class to finitely many parameters you are absolutely correct of course.. – Maithreya Sitaraman Nov 6 '17 at 2:03
• Correct, power series it the proper term. And you can assume that that inverse exists since that was another one of the ground rules: assume this is over an interval of interest where the function is continuous and strictly monotonic. (I don't know if the inverse can be represented as another power series--but I would guess there is some subset of power series where that is true.) ---- As for why the finite restrictions, besides what I already mentioned, there is also this: (applications) tag. – Carl Banks Nov 6 '17 at 23:53

An answer might be the family of all functions locally defined by power series of the form $$f(x) := -x + c_2 x^2 + c_3 x^3 + c_4 x^4 + c_5 x^5 + c_6 x^6 + c_7 x^7 + c_8 x^8 + O(x^9)$$ where $\; c_3 = -c_2^2,\; c_5 = 2c_2^4-3c_2c_4,\; c_7 = -13c_2^6+18c_2^3c_4-2c_4^2-4c_2c_6\;$ and they have the property that $\;f(f(x))=x.\;$ That is, each function is its own inverse. The coefficients of the even powers of $x$, the $c_2,c_4,c_6,\dots,$ are arbitrary while $c_3,c_5,c_7,\dots,\;$ are polynomials in the even power coefficients.

Each function is decreasing in an interval about the origin, but using linear or linear fractional transformations, they can be adjusted to fit a purpose. For example, take $\;a f(c(x-d))+b.$

Some examples of such functions $f$ are: $f(x) := -x/(1-x),\; f(x) := x-1+\sqrt{1-4x},\; f(x) := (-1+\sqrt{1-4x^3})/(2x^2).$

We now give a way to construct such functions. Let $\;g(x) := a_1x +a_2x^2 + a_3x^3 + O(x^4)\;$ where $\;x + f(x) = g(x f(x)).\;$ Now, starting with $\;f(x) = -x + g(x f(x)),\;$ this can be iterated to $f(x) = -x + g(x (-x + g(x f(x)))) = -x + g(x (-x + g(x (-x + \dots)))).\;$ Given a function $\;g(x),\;$ with $\;g(0)=0,\;$ we can compute $\;f(x)\;$ iteratively such that $\;f(f(x))=x.$ Explicitly, $$f(x) = -x + (-a_1)x^2 + (-a_1^2)x^3 + (-a_1^3+a_2)x^4 + (-a_1^4+3a_1a_2)x^5 + O(x^6).$$

Notice that, since $f(x)$ is determined by $g(x)$, which can be any power series with $\;g(0)=0,\;$ we can let it be a polynomial. For example, if $\;g(x) := -x^2,\;$ then $\;f(x) = -(1-\sqrt{1-4x^3})/(2x^2).\;$ If $\;g(x):=x,\;$ then $\;f(x)=-x/(1-x).$

Answering my own question, much later. The application I needed this for is still an issue, and I finally thought of something that I think will work. (Wait, what, you mean I want to use math for something? Weird.)

I realized that there is a pretty generic way to obtain a function that meets my requirements, though I can't say I have a proof that it can work in general for any desired interval of interest even if the function is monotonic in that interval.

Suppose we have a real-valued function $$F(x,y;A,B,C,D,...)$$, defined on some region of the $$x,y$$-plane in terms of a finite number of real-valued parameters $$A,B,C,D,...$$, such that there exists another set of real-valued parameters $$Z,Y,X,W,...$$ for which $$F(x,y;A,B,C,D,...)=F(y,x;Z,Y,X,W,...)$$. In other words, a function of $$x$$ and $$y$$ where you can swap $$x$$ and $$y$$ and the function will be of the same except for the value of the parameters. I'm sure there's a spiffy name for this like incomplete symmetry or something like that.

Now, if the equation $$F(x,y;A,B,C,D,...)=0$$ has a general closed-form solution for $$y$$ over some area where the curve is single-valued and monotonic in terms of $$x$$, then it will have the same general closed-form solution for $$y$$, except for the value of the parameters.

Thus, one can obtain a function that meets the requirements I set out.

Example. The following equation is the general form of an equation representing conic sections in the real plane:

$$Ax^2+Bxy+Cy^2+Dx+Ey+F=0$$

The general solution to this equation for $$x$$, over a region where the curve is single-valued and monotonic in terms of $$y$$, is:

$$y=Zx^2+Yx+X+W\sqrt{x^2+Vx+U}$$

Likewise, the general solution to this equation for $$y$$ is:

$$x=Zy^2+Yy+X+W\sqrt{y^2+Vy+U}$$

Thus the function $$f(x)=Zx^2+Yx+X+W\sqrt{x^2+Vx+U}$$, in a region where it's monotonic, has an inverse is of the same form.

The only thing is, I don't think this necessarily holds for arbitrary intervals, even if $$f$$ is monotonic in that interval and thus invertible. However, as a practical matter, for my use case, I think it will suffice for a simple reason: I can patch several of these together if I need to. If, for example, there is a point where the parameters have to change, I can put a break at that point.

So my general answer for this is piecewise patched conics, the inverse of which is also piecewise patched conics.