(01/12/2017, 04:50 PM)Xorter Wrote: I have been interested in that what the taylor series of i[x] is for long years.
There are infinity base units from i[0]=1 through i[pi] to i[10000...]. Their multiplication with each other usually is -1, but we can be sure only when x is an integer bigger than 0.
Here is a multiplication table for base units from i[0] to i[15]: https://en.wikipedia.org/wiki/Sedenion
(So we xor their indexes: n^m where ^ is the xor binary operator giving somehow a sign ... unfortunately I do not know what the form for this is.)
But the biggest question is that how we can get its taylor series. Any idea?
Well you haven't really defined terms but I take it by "base units" you mean the imaginary units that appear in the Cayley-Dickson construction and want to see if it would make sense to define those for non-integer indices somehow.
Well, the XOR bit actually provides what would probably be the most natural way: if \( e_x \) is a base unit with some non-negative real number \( x \) which is not a dyadic fraction (i.e. we declare an uncountable set of base units \( B = \{ e_x : x \in \mathbb{R}\ \mathrm{and}\ \ x \ge 0 \} \)), then that \( x \) has a unique binary expansion and if we have two of those units we could define \( e_x e_y = \mathrm{sign}(x, y) e_{x\ \mathrm{xor}\ y} \) where the XOR operation is just XORing the infinite binaries of \( x \) and \( y \) and \( \mathrm{sign} \) is some unknown generalization of the signing for when \( x \) and \( y \) are integer. Dyadic fractions are a problem though, because they have two different binary expansions, one ending in endless 0s and another in endless 1s. But we could say that the integers can all be considered as such, and the XOR we use there corresponds to doing it with representations ending in all 0s, thus the most natural rule would then be to just resolve them by taking their binaries to end in all 0s.
Thus we'd get that, say, \( e_{1/3} e_{1/5} \) should be \( e_{2/5} \) times some unknown sign, and \( e_{1/2}e_{1/3} \) is \( e_{5/6} \) times some unknown sign. Not sure of the sign bit yet but I'm sure there is some sort of method which can be used to generate it, and that would paint the way toward its extension.
These functions will, of course, be non-continuous when viewed as functions of the indices, so will not be analytic and have no Taylor series. But this is a rather different kind of interpolation/generalization animal than tetration, so you can't expect the same rules to apply. I also suspect that were one to craft an analytic interpolation it would be rather cumbersome and highly arbitrary, and not have any neat algebraic properties, like this might. Unlike tetration, we actually have algebra rules that point toward a generalization, like in exponentiation (e.g. \( e^{x + y} = e^x e^y \) -- that's real-number "e" not unital "e"). So in some sense, this is actually a much easier problem.
What I'm saying is that Taylor series are probably the wrong way to go about this, and the answer is really a lot simpler.
ADD!: I was googling "sedenions xor" and got this
https://arxiv.org/pdf/math/0011260.pdf
where they seem to touch on this a bit and suggests the indexing is arbitrary. This would make sense -- you can throw an isomorphism on top. Which means we could relabel the elements to make the indices follow some other rule. But what that really goes to show is the real "meat" here is in the signs. IOW, still not close. But I really don't think Taylor series are of any use here.
