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 Spiral Numbers tommy1729 Ultimate Fellow Posts: 1,370 Threads: 335 Joined: Feb 2009 11/19/2015, 10:20 PM (This post was last modified: 11/20/2015, 12:57 PM by tommy1729.) Spiral Numbers The idea is simplest when thinking in terms of polar coordinates. For a,c > 0 and b,d real , the complex numbers satisfy (a,b) (c,d) = (ac , b + d mod 2 pi) The idea of spiral numbers is (a,b)(c,d) = (ac , b + d) So far for products. The sum for spiral numbers is defined by X + Y = ln( exp(X) exp(Y) ). So it comes down to finding a good ln and exp. My guess is exp(a,b) = ( exp(a + ab) , e b) Where |*| is the absolute value. And the ln is just the inverse. For X^Y we use exp( ln X * Y ). I wonder how the algebra works out. Is this a good idea ? I wonder what you think. Regards Tommy1729 tommy1729 Ultimate Fellow Posts: 1,370 Threads: 335 Joined: Feb 2009 11/20/2015, 01:13 PM I edited post 1 with a different exp and ln. Im not sure if it is ok now. The big questions seem to be 1) is there a -1 and An i ? 2) do we have the distributive property ? It seems the only sqrt of 1 = 1. ( no solution x in x^2 = 1 apart 1 ). However it seems we have An additive inverse of 1. This suggests (-1)^2 =\= 1 ! So it Cannot be true that -1 exists * in the usual sense *. Since -1 is *weird* this rises questions about i. -- It seems that by the above and the fact that spiral numbers are not iso to complex numbers and does not contain them, The spiral numbers are closer to the reals then to the complex. Regards Tommy1729 tommy1729 Ultimate Fellow Posts: 1,370 Threads: 335 Joined: Feb 2009 02/29/2016, 10:56 PM One of the most intresting ways to continue is this Z1,z2 are complex. R1,r2 are real. (Z1,r1) + (z2,r2) = (z1 + z2,[r1 + r2]\2). This way we have commutative and associative Sum and product. Also we have the distributive property , no zero-divisors and algebraic closure. There exist other ways to define the Sum in a Nice way , but now we have the complex Numbers as a subset ( r1 = r2 = 0 ). ----- One alternative is (Z1,r1) + (z2,r2) = (z1 + z2,ln(exp(r1) + exp(r2))). This is also distributive ! The connection to hyperoper is clear now. ----- Regards Tommy1729 The master tommy1729 Ultimate Fellow Posts: 1,370 Threads: 335 Joined: Feb 2009 02/29/2016, 11:12 PM The alternative relates to my generalized distributive property. Funny though , so do variants more in the style of the " standard Tommy spiral Numbers " . Regards Tommy1729 Open your Mind Neo. marraco Fellow Posts: 93 Threads: 11 Joined: Apr 2011 03/01/2016, 12:38 AM (This post was last modified: 03/01/2016, 12:39 AM by marraco.) (11/19/2015, 10:20 PM)tommy1729 Wrote: Is this a good idea ? I wonder what you think. Why the name Spiral numbers? A good idea for what purpose? Are you aiming to make a field with fractional dimension? Your space has a geometrical representation? (I'm a visual person) I have the result, but I do not yet know how to get it. Gottfried Ultimate Fellow Posts: 766 Threads: 119 Joined: Aug 2007 03/01/2016, 12:30 PM (This post was last modified: 03/01/2016, 03:33 PM by Gottfried.) "Log-polar" (see also wikipedia https://en.wikipedia.org/wiki/Log-polar_coordinates ) representation might fill the prerequisites a tiny bit better. For instance the regular tetration with a complex fixpoint, curving around the fixpoint can then be approximated by linear interpolation and that interpolation agrees better and better with the regular tetration if the coordinate is translated into vicinity of the fixpoint using the functional equation. I have a picture of this in my comparision "5 methods for interpolation" ( http://go.helms-net.de/math/tetdocs/Comp...ations.pdf ) posted here earlier. Gottfried Helms, Kassel tommy1729 Ultimate Fellow Posts: 1,370 Threads: 335 Joined: Feb 2009 03/01/2016, 01:27 PM Intresting Gottfried. I was aware of the wiki , but perhaps a link to your work ? However my numbers are not log-polar since they are not iso to complex. Since you are a matrix expert ; how about a matrix representation for my spiral Numbers ? Also are my spiral Numbers iso to The ring R(x^3) or the group ring R+(C_4) ?? Although my Numbers have no zero-divisors there might still be a connection. Notice (0,r) * (1,-r) = (0,0). So if we do not consider (0,r) as 0 and (0,0) as zero , we get a sort of zero-divisor. The concept of zero is complicated here. X a = X for all a does not exist , but a - a is always (0,0). --- Notice that the spiral numbers keep track of additions in collatz if designed so. Regards Tommy1729 Gottfried Ultimate Fellow Posts: 766 Threads: 119 Joined: Aug 2007 03/01/2016, 04:16 PM (03/01/2016, 01:27 PM)tommy1729 Wrote: Intresting Gottfried. I was aware of the wiki , but perhaps a link to your work ? I just included the link in my previous reply Gottfried Helms, Kassel tommy1729 Ultimate Fellow Posts: 1,370 Threads: 335 Joined: Feb 2009 03/01/2016, 09:53 PM Correction : the spiral Numbers do not have associative addition. The alternative definition is , but they are iso to 2 copies of the complex plane. ( iso to (C,C) or (C,R) because exp( ln a + ln b ) iso a + b ) I guess this rules out the matrix representation for the normal spiral Numbers. Regards Tommy1729 tommy1729 Ultimate Fellow Posts: 1,370 Threads: 335 Joined: Feb 2009 03/01/2016, 10:15 PM The name spiral number comes from the analogue of polar coördinates. Multiplication is identical just without the mod 2 pi for the angle r. Visually this means that multiplication is on a spiral rather than on a plain or circle ( as with the complex Numbers ). Hence " spiral numbers ". To have a meaningful connection between addition and multiplication , I required the distributive law. Overview of typical distributive Numbers with invertible operators and a unit (?) 1) commutative and associative Rings , grouprings. Iso to copies of R and C ( when algebraicly closed ). 2) noncommutative and associative Matrices. 3) commutative and nonassociative Tommy's spiral Numbers 4) noncommutative and nonassociative No intresting cases known unless anticommutative (Lie). ??? Not sure how this gets us " fractional dimension " or " new Numbers for tetration ". Feel Free to correct or improve. Regards Tommy1729 « Next Oldest | Next Newest »

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