I am considering the following problem:
Suppose you are given a function $u: C \rightarrow C$, find a function $g$ such that $g(g) = u$ (Let's assume that such a function exists). And by "find", I mean give a series representation of $g$ in terms of known special functions as well as values of $u$.
This is a special case of (the Inverse Operator Problem):
Suppose you are given a function $u: C \rightarrow C$, find a function $g$ such that for some metafunctional $L$ $L(g) = u$. Which is solved by giving a series representation of $L^{(-1)}$
Which is a generalization of (the Inverse Function Problem):
Suppose you are given a number $x \in C$, find a number $t$ such that for some function $f$ we have that $f(t) = C$. This is solved by giving a series representation of $f^{-1}$ which can be done by the Lagrange inversion theorem given knowledge of the values of $f$ and it's derivatives.
So now I was wondering, how does one generalize the techniques of Lagrange to the operator problem I had given?
Here's the meta-plan I set up:
create: Linear Meta-Meta Functionals L
Express some Meta-Functional M as a series using L and a suitable basis
Then attempt lagrange Inversion Theorem trick, by determining a chain rule for L
However I've gotten stuck on point 2.
Some notation:
functions will be of the form $\text{symbol}(x)$ example $f(x)$, $A(x)$, $e^x+2^x$
meta-functions will be of the form $\text{symbol}(f)$ example $L(f)$, $f'+\frac{1}{f^2 + f(x+1)} $
meta-meta-functionls will be of the form $\text{symbol}(L)$ example $O(L)$, $L(f(x+1)-f(x)) + L^2$
Binding notation: The expression $(U)_{\alpha \leftarrow \beta}$ indicates to evaluate the expression U and then substitute every instance of $\alpha$ with $\beta$. This will be used to avoid ambiguity on operators.
Consider $O(L) = \frac{\partial L}{\partial f}$ this isn't well defined for all meta functions but it doesn't happen to be defined for $L(f) = f(f)$ for which it takes on the value $$O(L) = O(f(f)) = f'(f)$$
Furthermore its null-space is the set of all functions, and its linear which gives rise to the following
$$ O(A(x)) = 0$$ $$ O(A(x)f) = A(x)$$ $$ O(\frac{1}{2}A(x)f^2) = A(x)f$$
etc... which is the naturally way a Taylor series is generated. Thus we have that:
$$ f(f) = f(f)_{f \leftarrow g} + O[f(f)]_{f \leftarrow g} (f - g) + \frac{1}{2}O^2[f(f)]_{f \leftarrow g}(f - g)^2+ ... \frac{1}{n!} O^n[f(f)]_{f \leftarrow g}(f- g)^n + ... $$
Over some radius of convergence. This becomes:
$$ f(f) = g(g) + g'(g)(f - g) + \frac{1}{2}g''(g)(f-g)^2 + \frac{1}{6}g'''(g)(f-g)^3 + ... $$
Except there's a slight problem. It's obvious that we already know
$$g(f) = g(g) + g'(g)(f - g) + \frac{1}{2}g''(g)(f-g)^2 + \frac{1}{6}g'''(g)(f-g)^3 + ... $$
What has happened here is we have generated a meta-taylor series for the meta function $f(f)$ with radius of convergence 0.
I want another operator that actually gives me some non-zero radius of convergence. Because only once I have such a series representation, can I then progress to create the machinery for a generalized Lagrange inversion theorem.
