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

Question

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.

Answer 1

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.

Answer 2

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: $f_{c,d}^{-1} = f_{c^{(d^{-1})}, d^{-1}}$.

Answer 3

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 $\;f(x)=-x/(1-x).$