Classification for involutory real infinite series
Out of curiosity I'm playing around with the concept of involutory functions. An involutory function (involution) is a function whose composition with itself is the identity function (i.e. ${f \circ f} = id$). In other words, involutions are functions that are their own inverses (i.e. $f = f^{-1}$).
Applying this concept to univariate real polynomials; it seems that all involutory polynomials have degree one(!) and belong to one of these two families:
- $id$: $id(x) = x$ (the identity function, trivially involutory)
- $p_1$: $p_1(x) = {-x + c}$, for some $c \in \mathbb{R}$
One well-known generalization of polynomials are infinite series, so that raises the question: Is there some pretty classification of involutory real infinite series, too?
Clarification in response to a comment: this is about "analytic functions which are involutory within some part of their region of convergence", but ideally from an infinite series perspective. If that perspective doesn't make sense, an explanation as to why it doesn't make sense would be nice.
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If we look at formal power series and ignore questions of convergence for now, we can take $f(x) = \sum_{i \ge 0} a_i x^i$. Then the question is which sequences of $a_i$ satisfy $$\sum_{j \ge 0} a_j \left(\sum_{i \ge 0} a_i x^i\right)^j = x$$
The case $a_0 \neq 0$ is awkward, because we immediately get the constraint $$f(f(0)) = \sum_{i \ge 0} a_i a_0^i = 0$$ which involves all of the coefficients. But if we focus on the case $a_0 = 0$ we get a series of finite constraints:
$$\begin{eqnarray*} a_1^2 &=& 1 \\ a_1^2 a_2 + a_1 a_2 &=& 0 \\ a_1^3 a_3 + 2 a_1 a_2^2 + a_1 a_3 &=& 0 \\ a_1^4 a_4 + 3 a_1^2 a_2 a_3 + a_2^3 + 2 a_1 a_2 a_3 + a_1 a_4 &=& 0 \\ a_1^5 a_5 + 4 a_1^3 a_2 a_4 + 3 a_1 a_2^2 a_3 + 3 a_1^2 a_3^2 + 2 a_2^2 a_3 + 2 a_1 a_2 a_4 + a_1 a_5 &=& 0 \\ &\vdots& \end{eqnarray*}$$Subcase $a_0 = 0, a_1 = 1$
We have $$\sum_{j \ge 2} a_j\left(x^j + \left(x + \sum_{i \ge 2} a_i x^i\right)^j\right) = 0$$ and it's easy to show by induction that $\forall j \ge 2: a_j = 0$. Uninteresting case.
Subcase $a_0 = 0, a_1 = -1$
By substituting from the earlier identities into the later ones we can simplify further. Note that half of the identities become trivial. NB I've also assumed characteristic not equal to 2 and halved both sides of all of the following identities:
$$\begin{eqnarray*} a_3 &=& -a_2^2 \\ a_5 &=& 2 a_2^4 - 3 a_2 a_4 \\ a_7 &=& -13 a_2^6 + 18 a_2^3 a_4 - 2 a_4^2 - 4 a_2 a_6 \\ a_9 &=& 145 a_2^8 - 221 a_2^5 a_4 + 50 a_2^2 a_4^2 + 35 a_2^3 a_6 - 5 a_4 a_6 - 5 a_2 a_8 \\ &\vdots& \end{eqnarray*}$$I haven't attempted to prove that the pattern of the even coefficients being free continues, but it seems plausible that there is a family of $|\mathbb{R}|^{|\mathbb{N}|}$ series, and that doesn't make me optimistic about a neat non-circular characterisation.
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