16
$\begingroup$

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 3x3 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 2x2 rotation matrix.

  2. Prove the action property (that composing these series yields a new series of the same form, with parameters respecting 2x2 matrix multiplication), [this is where i'm stuck hence this question]

  3. Look to now construct series that respect the action of 3x3 matrix multiplication, perhaps inspired by the completion of (2).

$\endgroup$
4
$\begingroup$

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 are $$ f(x) = \omega + \omega^2 y - y^2 - \omega y^3 -\omega^2 y^4 + y^5 + O(y^6) \, = \frac{\omega + \omega^2 y - y^2} {(1 + y^3)}. $$ $$ g(x) = \omega - \omega y + y^2 + \omega^2 y^3 - \omega y^4 + y^5 + O(y^6) \, = \omega + \frac{y}{\omega(\omega+y)}. $$ The radius of convergence for both series is $\,1\,$ centered at $\, \omega \,$ and includes $\, 0<x<1. \,$ Check by composition that $$ f(f(x)) = g(x), \, f(g(x)) = x, \, g(g(x)) = f(x). $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.