Proving $\frac{1}{1-x} \circ \frac{1}{1-x} = 1 - \frac{1}{x}$ from a series point of view

Question

It's an elementary exercise in grade school algebra that

$$ \frac{1}{1-\frac{1}{1-x}} = 1 - \frac{1}{x} $$

However from the series point of view it's not at all obvious. There are two different series expressions for $\frac{1}{1-x}$ which are

$$ \sum_{k=0}^{\infty} x^k = 1 + x+ x^2 + ... \ |x| < 1 $$ $$ - \sum_{k=1}^{\infty} \frac{1}{x^k} = -\frac{1}{x} - \frac{1}{x^2} - \frac{1}{x^3} ... \ |x| > 1 $$

and attempting to compose yields troubles: (there are 4 cases to analyze here)

$$ 1+ (1+x+x^2 ... ) + (1 + x + x^2 + ...)^2 + ... $$

This leads to coefficient blow up, and even with using zeta function values to renormalize infinities it leads to an expression that seems meaningless (or I should say, is very "difficult" to interpret).

$$ 1 + (-\frac{1}{x} - \frac{1}{x^2} ... ) + (-\frac{1}{x} - \frac{1}{x^2} ...)^2 ... $$

actually simplifies to the correct expression $1- \frac{1}{x}$ (so series can confirm the identity for: $|x|<1, 1 > |x-1|$)

$$ - \frac{1}{1 + x + x^2 ... } - \frac{1}{(1 + x + x^2 ... )^2 } ... $$

Is again intractable without referencing the geometric series formula.

$$ - \frac{1}{ - \frac{1}{x} - \frac{1}{x^2} ... } - \frac{1}{(-\frac{1}{x} - \frac{1}{x^2} ... )^2} ... $$

Is even more horrific (I nicknamed this expression Harmonic Hell).

My worry here is only 1 of these 4 compositions could be simplified into the correct target expression, how to correctly manipulate the other 3 to yield the target expression, after all there are specific domain, range combinations that I am losing when I only consider any one of these pairs of series (yet the expression $1 - \frac{1}{x}$ is true globally).

Motivation

This is part of a toy problem: Action of 3x3 invertible matrices on $\mathbb{C}$? where I began to wonder if it was possible to find an action of $3\times 3$ matrices on the complex plane.

My program of research was the following:

1. Interpret mobius transformations as literally pairs of laurent series which accept a quadruple of parameters $a,b,c,d$ corresponding to elements of a $2\times 2$ rotation matrix.

2. Prove the action property (that composing these series yields a new series of the same form, with parameters respecting $2\times 2$ matrix multiplication), [this is where i'm stuck hence this question]

3. Look to now construct series that respect the action of $3\times 3$ matrix multiplication, perhaps inspired by the completion of (2).

Answer 1

For your problem with power series for $\, f(x) := 1/(1-x), \,\, g(x) := 1-1/x, \,$ you would like to center each power series about the same number when the functions are composed. In this case the common center is $\, \omega, \,$ a primitive sixth root of unity because $\, \omega = f(\omega) = g(\omega) \,$ is a fixed point of both $\,f\,$ and $\,g.\,$ Thus, let $\, y := x-\omega \,$ be the local variable. Check that the two power series expansions in powers of $\, y\,$ are $$ f(x) = \omega + \omega^2 y - y^2 - \omega y^3 -\omega^2 y^4 + y^5 + O(y^6) \, = \frac{\omega + \omega^2 y - y^2} {(1 + y^3)}, \tag1$$ $$ g(x) = \omega - \omega y + y^2 + \omega^2 y^3 - \omega y^4 + y^5 + O(y^6) \, = \omega + \frac{y}{\omega(\omega+y)}. \tag2$$ The radius of convergence for both series is $\,1\,$ centered at $\, \omega \,$ and includes $\, 0<x<1. \,$ Check with composition the following equations: $$ f(f(x)) = g(x), \,\, f(g(x)) = x, \,\, g(f(x)) = x, \,\, g(g(x)) = f(x). \tag3$$