Taylor series of i[x]
#7
(01/23/2017, 07:38 AM)mike3 Wrote: 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.

Yes, we have to xor the indeces to get a new one of the multiplication, but the method to determine the sign is unkown, I could not find any useful formula neither in that Arxiv paper nor anywhere.
I have written a program in PARI/gp which can xor two rational numbers, and yes, your speculation is correct, so i[1/3] i[1/5] = sgn() i[2/5] and i[1/2] i[1/3] = sgn() i[5/6].
For instance, here is the result of my programme code:
https://www.dropbox.com/s/5rcwawcfyisvcf...).bmp?dl=0
y = 1 xor x
And as you mentioned at the whole numbers the function is problematic, but it just has a cut-off point like sgn(x) at 0 or d|x|/dx at 0. It is on my graph you can see above.
And it is really simple thing to extrapolate onto complexes and interpolate onto reals. But this is not my question.
My question is that how we can determine the sign of multiplication of base units. I feel, intuitively, that these "signs" will be have a form like exp(ia). So these will be complex signs, if I can say it.
Now we are closer than ever before. I believe it is expansionable to Taylor series, but this is just my opinion.
Xorter Unizo
Reply


Messages In This Thread
Taylor series of i[x] - by Xorter - 01/12/2017, 04:50 PM
RE: Taylor series of i[x] - by sheldonison - 01/12/2017, 08:50 PM
RE: Taylor series of i[x] - by Xorter - 01/13/2017, 05:26 PM
RE: Taylor series of i[x] - by Xorter - 01/13/2017, 07:13 PM
RE: Taylor series of i[x] - by Xorter - 01/14/2017, 10:14 AM
RE: Taylor series of i[x] - by mike3 - 01/23/2017, 07:38 AM
RE: Taylor series of i[x] - by Xorter - 02/26/2017, 11:10 AM
RE: Taylor series of i[x] - by Xorter - 03/01/2017, 03:06 PM
RE: Taylor series of i[x] - by Xorter - 03/04/2017, 09:40 AM
RE: Taylor series of i[x] - by Xorter - 04/06/2017, 03:42 PM
RE: Taylor series of i[x] - by Xorter - 04/11/2017, 12:18 PM
RE: Taylor series of i[x] - by Xorter - 07/10/2017, 04:07 PM
RE: Taylor series of i[x] - by Xorter - 02/20/2018, 09:55 PM

Possibly Related Threads…
Thread Author Replies Views Last Post
  Divergent Series and Analytical Continuation (LONG post) Caleb 54 57,686 03/18/2023, 04:05 AM
Last Post: JmsNxn
  Discussion on "tetra-eta-series" (2007) in MO Gottfried 40 41,706 02/22/2023, 08:58 PM
Last Post: tommy1729
Question Tetration Asymptotic Series Catullus 18 22,393 07/05/2022, 01:29 AM
Last Post: JmsNxn
Question Formula for the Taylor Series for Tetration Catullus 8 13,776 06/12/2022, 07:32 AM
Last Post: JmsNxn
  Calculating the residues of \(\beta\); Laurent series; and Mittag-Leffler JmsNxn 0 3,728 10/29/2021, 11:44 PM
Last Post: JmsNxn
  Trying to find a fast converging series of normalization constants; plus a recap JmsNxn 0 3,562 10/26/2021, 02:12 AM
Last Post: JmsNxn
  Reducing beta tetration to an asymptotic series, and a pull back JmsNxn 2 6,866 07/22/2021, 03:37 AM
Last Post: JmsNxn
  Perhaps a new series for log^0.5(x) Gottfried 3 10,779 03/21/2020, 08:28 AM
Last Post: Daniel
  An explicit series for the tetration of a complex height Vladimir Reshetnikov 13 43,634 01/14/2017, 09:09 PM
Last Post: Vladimir Reshetnikov
  Complaining about MSE ; attitude against tetration and iteration series ! tommy1729 0 5,821 12/26/2016, 03:01 AM
Last Post: tommy1729



Users browsing this thread: 2 Guest(s)