Now the function $J(1,y)$ really likes to be a complex number. Since it is always of norm 1 when the discriminant is non-positive, it is only real when $J = \pm 1$. Note that $J(1,y)$ may be generalised yet again using the rule

$xn J^2 + xy J + n = J$

which is also designed to give a norm 1 complex number at $x = 1$. In this case, the condition for $J(1,y) = \pm 1$ is

$y = 1 \pm Q$

for $n^2 = Q^2 – N^2$ where $Q$ and $N$ are whole numbers. When $n = 1$ this reduces to the previous case of $N = 0$, but a general Pythagorean triple gives a solution for $y$. Note that each such choice of $y$ leads to a new Abel series with coefficients somehow depending on $n$ and all of these series sum to $\pm 1$ although the substitution rule is different for each choice of $n$. Other choices for $y$ give roots of unity. For example, when $y = 6$ and $n = 3$ we find that $J = \frac{1}{6} [- 5 \pm \sqrt{11} i]$.

Update: Here is the original post on the Everything Seminar, with nice examples of series summing to -1.

### Like this:

Like Loading...

*Related*

## Matti Pitkanen said,

August 9, 2007 @ 5:26 am

Dea Kea,

I noticed that in Sums of divergent series posting the first sum was nothing but a 2-adic representation of -1 as 1+2+2^2+2^3+…. More general representation is (p-1)(1+p+p^2+…). Apparently they had not noticed the connection!

## Matt Noonan said,

August 9, 2007 @ 2:40 pm

Look again! This was the first way I tried to get people to devalue the notion of metric convergence, by pointing out that the real absolute value doesn’t see intersting details that the p-adic metrics do. In any case, there are plenty of series which are divergent in every completion of the rationals but still have reasonable sums, so it seems like the p-adics are only small portion of the big picture here.

That said, I like your idea of using the geometric series in categories with products and sums to get negative objects, with

-X = (X – 1)(1 + X + X^2 + …)

but it only seems to work if we can solve the equation Y + 1 = X for Y. Even then, it looks a little rocky: with FinSet and X = 1 = {0} you get

-1 = (1 – 1) (1 + 1 + 1 + …) = 0

which feels like a reflection of the pole at 1. It would be nice to patch this, but I think the tricks involved will be fundamentally algebraic, not metric or analytic. Just guessing, though: I haven’t had any luck defining negative sets yet…

## Kea said,

August 9, 2007 @ 8:19 pm

Thanks! My guess would be that negative objects are a fundamentally metrical concept (we are trying to do gravity here, after all). But I might be dreaming. Maybe -1 = 0 in FinSet because finite sets just don’t see negatives: that is, the negative objects are all forced to be empty sets.