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01/12/2017, 04:50 PM
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?
Xorter Unizo
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01/12/2017, 08:50 PM
(This post was last modified: 01/12/2017, 09:22 PM by sheldonison.)
(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?
I'm not familiar with Sedenion, but it is an abstract algebra concept, not a complex analytic function, right? So by definition, unless there is some mapping to a complex function, then it would not have a Taylor series...
- Sheldon
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(01/12/2017, 08:50 PM)sheldonison Wrote: I'm not familiar with Sedenion, but it is an abstract algebra concept, not a complex analytic function, right? So by definition, unless there is some mapping to a complex function, then it would not have a Taylor series...
Well, I believe it is differentialable and if it is, it will have an exact Taylor series.
We just need to get its derivative. I could start to determine it, but I could not finish it, just look at it:
\( di_x / dx = lim (i_{x+h}-i_x)/h = i_x lim (1-i_x / i_{x+h})/h \), where h approaches to 0.
So I know that its derivative will have multiplicative part as i[x] and another thing which is questionable. What is lim (1 - i[x]/i[x+h])/h.
I think it is absolutely solvable ... but how?
Xorter Unizo
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01/13/2017, 07:13 PM
(This post was last modified: 01/13/2017, 07:43 PM by Xorter.
Edit Reason: I forgot subtituting x=0.
)
(01/12/2017, 08:50 PM)sheldonison Wrote: I'm not familiar with Sedenion, but it is an abstract algebra concept, not a complex analytic function, right? So by definition, unless there is some mapping to a complex function, then it would not have a Taylor series...
Ah, yes, and I could recognise something interesting.
I tried to take i[x] into the Carleman matrix and to square it with itself, it seems successfully, maybe. Well, I got that the new Carleman matrix 2nd coloumn have I[1,0], I[1,1], I[1,2], ... where
\( I_{1,N} = 1/N! \sum_{k=1}^{\infty} i^{(N)k}_x i^{(k)}_x / k! |_{x=0} \)
So
\( i_x ^{o2} = i_{i_x} = \sum_{k=0}^{\infty} I_{1,k} x^k =
= \sum_{k=0}^{\infty} i^{(k)}_x / k! + x \sum_{k=0}^{\infty} i'^{k}_x i^{(k)}_x / k! + x^2 \sum_{k=0}^{\infty} i''^{k}_x i^{(k)}_x / k! + ... |_{x=0} \)
Now I feel we are closer than before.
Xorter Unizo
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01/14/2017, 10:14 AM
(This post was last modified: 01/14/2017, 06:05 PM by Xorter.)
Here is another interesting fact:
The Taylor series of i[x] is the following:
\( i_x = \sum_{k=0}^{\infty} i^{(k)}_x x^k / k! = i_x + x i'_x + x^2 1/2 i''_x + x^3 1/6 i'''_x + ... \)
and
\( i_x^{o2} = i_{i_x} = \sum_{k=0}^{\infty} i^{(k)}_x / k! + x \sum_{k=0}^{\infty} i'^k_x i^{(k)}_x / k! + x^2 \sum_{k=0}^{\infty} i''^k_x i^{(k)}_x / k! + ... |_{x=0} \)
So i[i[x]] has a part as 1 looking like this:
\( i_{i_x} = 1 + x \sum_{k=0}^{\infty} i'^k_x i^{(k)}_x / k! + x^2 \sum_{k=0}^{\infty} i''^k_x i^{(k)}_x / k! + ... |_{x=0} \)
Therefor di[x]/dx cannot be 0, then what can it be?
Xorter Unizo
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01/23/2017, 07:38 AM
(This post was last modified: 01/23/2017, 08:24 AM by mike3.)
(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.
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(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
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03/01/2017, 03:06 PM
(This post was last modified: 03/01/2017, 10:42 PM by Xorter.)
Okey, here is a website: http://jwbales.us/ where you can calculate the sign by a programme. I checked the source of it, and I take the important part of it out. You can read it at the bottom.
I have made an own java html code, and it works. So we have a recursive formula for the signs. The next question is how to make just one formula and how to interpolate (to reals) and extrapolate (to complexes).
We are really close to have its Taylor series!
Code: function bitValue(number,bit){
var result = (number & (1<<bit)) >>> bit;
return result;
}
function twist(p,q){
var state="A0";
var pbit=0;
var qbit=0;
var result=1;
var p0=p;
var q0=q;
var pplus=1+p0;
var qplus=1+q0;
var stateA0 = [];
var stateA = [];
var stateB = [];
var stateNB = [];
var stateC = [];
var stateNC = [];
var loopIndex=Math.ceil(Math.LOG2E*Math.log((Math.max(pplus,qplus))));
var pXORq =eval(p0) ^ eval(q0);
stateA0[0] = "A0"
stateA0[1] = "A"
stateA0[2] = "B"
stateA0[3] = "NB"
stateA[0] = "A";
stateA[1] = "A";
stateA[2] = "C";
stateA[3] = "NC";
stateB[0] = "B";
stateB[1] = "NC";
stateB[2] = "B";
stateB[3] = "C";
stateNB[0] = "NB";
stateNB[1] = "C";
stateNB[2] = "NC";
stateNB[3] = "NB";
stateC[0] = "C";
stateC[1] = "NC";
stateC[2] = "NC";
stateC[3] = "NC";
stateNC[0] = "NC";
stateNC[1] = "C";
stateNC[2] = "C";
stateNC[3] = "C";
if (p < 0 || q < 0){
p=p0;
q=q0;
str = "Values of p and q cannot be negative!";
alert(str);
}
for (var i = 0; i < loopIndex; i++){
var j = loopIndex - i - 1;
var k = 2*bitValue(p,j)+bitValue(q,j);
if (state == 'C'){
state = stateC[k];
continue;
}
if (state == 'NC'){
state = stateNC[k];
continue;
}
if (state == 'B'){
state = stateB[k];
continue;
}
if (state == 'NB'){
state = stateNB[k];
continue;
}
if (state == 'A'){
state = stateA[k];
continue;
}
if (state == 'A0'){
state = stateA0[k];
continue;
}
}
if (state=="NB" || state=="NC") {
result="-1";
}
return result;
}
Xorter Unizo
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03/04/2017, 09:40 AM
(This post was last modified: 03/04/2017, 09:45 AM by Xorter.)
Okey, I programmed it and I got the next graph for y = Y(1;x)*(1 xor x):
https://www.dropbox.com/s/81rxif99w7j50u...D.bmp?dl=0
And this is my PARI/gp code:
Code: twist(p,q)={
L=ceil(1.4426950408889634*log((max(q+1,p+1))));
res=1;sta=1;
if(round(p)!=p||round(q)!=q,p*=2^60;q*=2^60;L+=60);
p=round(p);q=round(q);
T=[1,2,3,4;2,2,5,6;3,6,3,5;4,5,6,4;5,6,6,6;6,5,5,5];
for(i=0,L,j=L-i-1;k=2*bittest(p,j)+bittest(q,j)+1;sta=T[sta,k]);
if(sta==4||sta==6,return(-1),return(1));
}
xor(x,y)={return(bitxor(round(x*2^60),round(y*2^60))/2^60)}
I checked it in a really little interval (from 0 to 10^-8 ) and it does not look like something which is differentialable. So should I say it has not a derivative and Taylor series?
Anyway, can anyone determine the integral of it from 0 to x? It would be really interesting, I think so.
Xorter Unizo
Posts: 93
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Joined: Aug 2016
Okey, I have some news!
I got a little bit closer than before.
Let us get the length of the curve of points of the base units from i[exp(0)] to i[exp(2ipi)]:
In polar-coordinate system:
L = .5 [ sqrt(i[x]'^2 - 1) ] from 0 to 2pi
In cartesian coordinate system:
L = [ sqrt( i[exp(ix)]'^2 + 1 ) ] from 0 to 2pi
So
.5 [ sqrt(i[x]'^2 - 1) ] from 0 to 2pi = [ sqrt( i[exp(ix)]'^2 + 1 ) ] from 0 to 2pi
It looks a promising integral and differential equation.
What do you think, what should be the next step?
Xorter Unizo
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