Why $1/(1-x) \neq 1/(-x+1)$ in long division for generating functions?

Question

If you do a long division of $\frac{1}{1-x}$ and $\frac{1}{-x+1}$ to obtain a formal series of generating functions, you'll get different results:

$$ \frac{1}{1-x} = 1 + x + x^2 + x^3 + ... $$ $$ \frac{1}{-x+1} = -\frac{1}{x}-\frac{1}{x^2}-\frac{1}{x^3}-... $$

But we expect in both cases to get $g(x) = 1 + x + x^2 + x^3 +...$

I do know that one can show that generating functions, corresponding to $\frac{1}{1-x}$ and $\frac{1}{-x+1}$ are the same by considering $(1-x)g(x) = (-x+1)g(x) = 1$ identity, but it is quite disturbing for me that long polynomial division, which I used to trust, does not work. Moreover, the former fraction, $\frac{1}{-x+1}$ feels more natural during long division process, because it has the highest term going first.

So, what is the reason of why I get different results? (for example, maybe summation operation is not commutative in this kind of things so we have $(1-x) \neq (-x+1)$)

Answer 1

In the ring of formal power series, $x$ is "smaller" than $1$, not larger. The most significant (i.e. "highest") term of $1-x$ is $1$, not $-x$.

The valuation in the ring of formal power series is the opposite of the one you're used to in terms of polynomial degree.

Answer 2

The first series holds when $|x| < 1$. The second holds when $|x| > 1$.

The concept behind this is analytic continuation.

Answer 3

Long division does not give an infinite series. It gives a finite series plus a remainder (and that leads to an infinite series if the remainder tends to $0$).

If you write down finitely many terms plus the remainder it turns out they're the same. (For the first series the remainder tends to $0$ if $|t|<1$, while for the second the remainder tends to $0$ if $|t|>1$..)