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 Infinite tetration and superroot of infinitesimal Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 07/05/2008, 11:26 AM We can use these twisting or rotation properties of hyperoperations to get a glimpse what in x[z]y y dimensional space is: We know that ( i may be mistaken in these analogies): x^x=x[3]x=x[4]2 So according to my conventions, in 2 dimensional space, this is edge x TWISTED 4 times by pi/2, or 2pi, or a TWISTED line, related to torsion of line. In x dimensional space, it is edge x twisted 3 times via pi/2. So one twist is removed from line and attributed to space itself. Hopefully this can be extended to lower operations like multiplication and addition and zeration to see their geometrical meaning. Negation as inverse twists and imagination as rotation (?) of edges will follow. Rotations ( imaginations) will lead in case of straight angles to a kind of Hilbert space filling curve with edge x in n or other number of dimensions. Ivars Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 07/10/2008, 12:26 PM (This post was last modified: 07/10/2008, 12:33 PM by Ivars.) Well I hope Henryk will not punish me for these wild conjectures. This is the place I can store them and hope to get some feedback. I studied what i could for 2 weeks and understood little, got lost almost totally but also understood that most books on whatever start with the same- real vector spaces , and then go on developing the ideas. Most who write today about quaternions or not seem not to have studied Hamiltons Elements of quaternions since he gives there much more than just vectors to be used for 3D rotations in visualisation. He develops quaternion mathematics- functions, series, calculus, links to geometry, etc. It is not mentioned anywhere in new books even about quaternions. Even first explanator of quaternions Tait did not use most of that. But that is another story to be looked at. Briefly, if we compare number types geometrically with straight line elements: Scalars are lentgh of line, or ordinary numbers. They also may measure the translational motion of one end into other. Vectors are directed ordinary numbers, obtaine by inverse summation : A-B . e.g 7-5 is a directed number 2 and is a vector. While -2 is the same vector in opposite direction. Normal imaginary unit have been related to these a Amplitude*phase. Than , by same logic we must (?) have numbers that are obtained by inverse multiplication, i.e. division. It seems quaternions are these, as Grassmann or Clifford defined his algebra from existance multiplicative inverse, of which there are only 4 known. For some reason, it involves imaginary space of 3D and is related to rotations in space. What is rotated in imaginary space around origin is ordinary imaginary number I- point on imaginary 2-sphere- which is the same as I for normal complex numbers, but generally not the same as pure quaternion axis i,j,k - none of them. x^2+1=0 has infinitely many solutions- +- I -is -antipodal points on this imaginary 2-sphere in pure quaternion space. Then, next we may have numbers obtained by inverse exponentiation, or log. They may be twisted (?) numbers. Or what else could we do to an imaginary line segment (I have not figured out how and why) . e.g. Because octonions are non-associative, as is not exponetiation usually ( x^y is not y^x usually). Twisted numbers may oscillate? Then, we move to 4th operation, or tetration. What numbers we may obtain by inverse tetration? The inversion of infinite tetration is self root. Self root may denote looping numbers. But for self root to be inverse of tetration, we need infinite tetrations. In the middle, as long as number of tetrations is finite, we can get inverse by various order superrots as long as we tetrate one and the same variable ( number) . We may need another set of numbers or we may be able to survive with combinations of existing. Basically what we can add to number as imaginary line after translating, directing, rotating, twisting it is what? It may be further internal twists of the imaginary line? if that is related to sedenions, what could it mean geometrically and why it is not anymore divisor algebra? The reason I mention this in tetration forum is that it seems to me that numbers have internal structure, which may well be imaginary since we have no idea about it when we look at number line. It only reveals itself under proper operations. So number line looks (perhaps) as a rope which has internal degrees of freedom, brought out under fast enough operations applicated enough or proper amount of times. This imaginary structure at least at integer operation numbers may be related to imaginary numbers and their geometric developments into quaternions, octonions, sedenions etc. ... Ivars Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 07/11/2008, 09:27 AM I would like to start with an possible physical interpretation of I as an imaginary dimension. If You have a directed number on a line- vector AB in 1D, you can inverse it by turning it inside out via its origin- at first You contract it to 0 (point) while STORING continuously information about what it was - both size and direction- in imaginary dimension I. After that You release this information from imaginary dimension gradually until You get - AB on the same line. To perform the same inversion by rotation, You have to create a picture of this imaginary dimension in plane with linear real number scale ( |AB*I|= |AB|) - and than we get complex plane, where the inversion is performed by rotation via this imaginary dimension I which we have to append perpendicularly to our real line at least for the time we perform the inversion which maintains all information required until we have rotated the initial vector by Pi or -Pi. This implies that the process of inversion happens linearly ( so that as we reduce lenght of AB by dx the value in imaginary dimension grow as c*dx. This linearity leads to sin/cos relations in complex plane and also to exponetial dependance on imaginary angle I use below. Question- does it have to be- is it linear? The question is I a unit of this imaginary dimension I or just the dimension with arbitrary units needed to perform "physical" actions like inversion of real space dimension via itself? As an Example, one can take an infinitesimally thin shell elastic tube and inverse it by rolling it backwards on itself. During this rolling, the information about the length of the tube would be rolled in the very small torus it creates, while the information about direction would be retained in orientation of torus and relative Length of tube rolled in very thin torus and not rolled in yet or rolled out already. This would work equally to I as long as it works with infinitesimal shell perhaps of imaginary thickness,which I have to check in literature. In this case this torus would play the role of imaginary dimension relative to our Real dimension we started with. If, however, we would start with imaginary dimension and try to inverse it, we would have 2 imaginary dimensions, of which one would store information about the inversion of other. The trick with quaternions seems to be that to keep such system closed in some sense, as Complex numbers which are algebraically closed- any algebraic equation in complex numbers leads only to complex roots- which helps greatly with series expansions - so that one imaginary dimension stores information about the inversion of other , while the other keep information about the inversion of former- seems kind of difficult because we get a double torus - but is it impossible? in 3 Imaginary dimensions , we would have 3 toruses keeping information about each others inversions VIA ITSELF + also relative orientations. According to quaternions, they can not do it without the help of a scalar- or is it so? I have to try to picture this in more detail. Next thing to ask is is there a natural metric on these imaginary toruses etc because it obviosly can not use real numbers to store this spatial information (it can if we ascribe real numbers to such dimension- like a+b*i+c*j+d*j, where a,b,c,d = real) - how does imaginary dimension store length since length is measured relative to another real line segment which is not given in imaginary space dimension. Ivars Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 10/16/2008, 10:39 AM (This post was last modified: 10/16/2008, 10:59 AM by Ivars.) Ivars Wrote:When looking at x^y as hypercube in dimension y with edge x, I have made implicit assumption that the ANGLE in which each edge of such hypercube creates with another is 90 degrees in any dimension y, just by analogy, or Pi/2, as in general definitions of hypercube. However, this angle may be another,may be negative, may be imaginary. This leads to interpretation of real etc. extensions of hyperoperation number z in x[z]y as angle between such edges x of a hypercube in dimension y either directly, or via some exponential relation like log (pi/2)^n/log (pi/2) = n for integer z, and z*log pi/2/log (pi/2) =z for all other. , or n= log (e^n*I*pi/2)/log I , z= log (e^z*I*pi/2)/log I. In expression : x[z]y x- edge of hypervolume in y dimensions y- dimensions of hypervolume with edge x z- is related to varying angle between edges x of hypervolume in y dimensions, perhaps via logarithm or some trigonometric function. A meaningful definition should recover usual angles pi/2 in case of hypercube in integer dimensions. The geometric interpretation of hyperoperation in general and its real, complex extensions is related to projective geometry in imaginary form extended to non-integer, negative and imaginary dimensions. To give a more concrete test example: I[I]I is a hypervolume with edge I in I dimensions with angle between imaginary edges (e^(I*I*pi/2) = e^(-pi/2) I[-I]I is a hypervolume with edge I in I dimensions with angle between imaginary edges e^pi/2. In I dimensional space, such angle might have a meaning. Ivars For example in I dimensional space, angles (?) e^(-pi/2) and e^(pi/2) are both solutions of equation: z^(2*I)+1=0 which is equivalent to : (z^I+I)*(z^I-I)=0 and has solutions: z1=e^pi/2 z2=e^-pi/2 It is easy to see that these solutions are such that z1*z2=1 or z1= 1/z2. This is involution, and z1 , z2 are 2 real fixed points of involution on imaginary line=dimension I. All REAL circles in projective plane have in common circular points I, -I. Obviously (??) , ALL IMAGINARY circles in projective plane have in common 2 real points (in angle space? or e^( angle) space) ?) e^pi/2, e^-pi/2. Infinite tetration transforms one of these real common points of Imaginary circles (e^pi/2) into both common points of real circles (+I, -I) since h( e^pi/2) = a) h( I^(1/I) = I b) h((1/I)^I) =-I But it does not transform the other common point of imaginary circles, e^-pi/2, into anything so interesting, as h( e^-pi/2) = 0,47541.. Ivars Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 03/01/2009, 12:58 PM (This post was last modified: 03/02/2009, 05:23 PM by Ivars.) GFR Wrote:Actually, the beautiful formula I propose is: ssqrt(x) = ln(x) / W(ln(x)), which, for x = 1/2, gives: ssqrt(1/2) = ln(1/2) / W(ln(1/2)) = 0.26289282802173525.. + 0.4996694356833174.. i Perfectly coherent with Henryk's formula. GFR Since 0^0 = 1 then $ssqrt(1) = 0$ The reason for 0^0=1 is explained here ( it is not 100% based on real number limits, actually, not at all, its based on symbolic calculations ( binomial theorem): D.Knuth 2 notes on Notations Quote:Evidently Libri’s main purpose was to show that unlikely functions can be expressed in algebraic terms, somewhat as we might wish to show that some complex functions can be computed by a Turing Machine. “Give me the function 0^0^x , and I’ll give you an expression for [x divides m].” Most mathematicians agreed that 0^0 = 1, but Cauchy [5, page 70] had listed 0^0 together with other expressions like 0/0 and ∞−∞ in a table of undefined forms. Libri’s justification for the equation 0^0 = 1 was far from convincing, and a commentator who signed his name simply “S” rose to the attack [45]. August Mobius [36] defended Libri, by presenting his former professor’s reason for believing that 0^0 = 1 (basically a proof that limx→0+ x^x = 1). Mobius also went further and presented a supposed proof that limx→0+ f(x)^g(x) = 1 whenever limx→0+ f(x) = limx→0+ g(x) = 0. Of course “S” then asked [3] whether Mobius knew about functions such as f(x) = e−1/x and g(x) = x. (And paper [36] was quietly omitted from the historical record when the collected works of Mobius were ultimately published.) The debate stopped there, apparently with the conclusion that 0^0 should be undefined. But no, no, ten thousand times no! Anybody who wants the binomial theorem to hold for at least one nonnegative integer n must believe that 0^0 = 1, for we can plug in x = 0 and y = 1 to get 1 on the left and 0^0 on the right. The number of mappings from the empty set to the empty set is 0^0. It has to be 1. On the other hand, Cauchy had good reason to consider 0^0 as an undefined limiting form, in the sense that the limiting value of f(x)^g(x) is not known a priori when f(x) and g(x) approach 0 independently. In this much stronger sense, the value of 0^0 is less defined than, say, the value of 0+0. Both Cauchy and Libri were right, but Libri and his defenders did not understand why truth was on their side. If we use the formula: $ssqrt(y) = \ln(y) / W(\ln(y))$ for y=1 $ssqrt(1) = \ln(1) / W(\ln(1))$ $\ln(1) = 0$ $W(0) = 0$ Then $ssqrt(1) = 0/0$ Which not what x^x=y as x=0 says. It says$ssqrt(1)=0$ Differentiating at 1 according to l'Hopitals rule gives ( I hope I did not make a mistake): $ssqrt(1) = ln(1)/W(0) + ln(1) =0/0 + 0$ No big help here. May be if we divide the powerseries for ln(1+x) and W(x) near 0 we get clear 0, since in these new powerseries the coefficient at $x^0$ would be 0, only powers of 0 remain, so the resulting series will equal 0 at x=0. $x^{(x^x)} = 1$ then as $0^0=1$, $x^1=1$ in this case 3rd superroot of y=1 is NOT=0 , but 1. 4th superroot is again ln(1) =0 , 5th is 1 or ( perhaps) ln(e) etc. So here even/odd superroots of 1 clearly divide, and it is consistent with formula for ssqrt(1). Ivars bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 03/01/2009, 01:22 PM (This post was last modified: 03/01/2009, 01:24 PM by bo198214.) Ivars Wrote:Question: What about other superroots of 1? 3rd, 4th, n-th, x-th, zth etc? If: The answer to this is really easy, you just need to look at the graph. It is well known that $\lim_{x\to 0} {^n x}$ is 0 for odd $n$ and 1 for even $n$. It follows that the solution of ${^n x}=1$ is only 1 for odd $n$ and $\{0,1\}$ for even $n$. Quote:$x^{(x^x)} = 1$ then as $0^0=1$, $x^0=1$ in this case 3rd superrot of y=1 is also x=0 $0^{0^0}=0^1=0$. (though note that statements like $0^0=1$ can be quite wrong, for example when considering $\left(e^{-x^2}\right)^{1/x}$. If you naively set $\lim_{x\to\infty}=0^0=1$ this is wrong because $\lim_{x\to \infty} e^{-x^2 * x} = \lim_{x\to\infty} e^{-x} = 0$.) Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 03/01/2009, 01:29 PM (This post was last modified: 03/01/2009, 02:32 PM by Ivars.) Hi bo I edited my post a bit, there was that obvious mistake with 3rd superroot. Yes $0^0=1$ is naively set-but is it not provable from GFR , your formulas for superroot involving Lambert function? As Knuth says, it is not based on real number limits only. More like an agreement needed for symbolic machine calculations using binomial theorem etc. Ivars « Next Oldest | Next Newest »

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