Blogrolling On IV
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.
$xn J^2 + xy J + n = J$
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.
posted by Kea | 3:00 PM
3 Comments:
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!
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...
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.
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