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 Infinite tetration and superroot of infinitesimal bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 01/30/2008, 05:09 PM Ivars Wrote:Barrow said (not exact quote): if x^x=1/2 has no solutions in real and complex numbers, it may lead to creation of new numbers just to fill that gap. Has anyone already tried to define such numbers? But it has a solution in the complex numbers? $\frac{1}{2}=x^x$ $\ln\left(\frac{1}{2}\right)=x\ln(x)=e^yy$ $W\left(\ln\left(\frac{1}{2}\right)\right)=y=\ln(x)$ $x=e^{W\left(\ln\left(\frac{1}{2}\right)\right)}$ Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 01/30/2008, 07:28 PM bo198214 Wrote:Ivars Wrote:Barrow said (not exact quote): if x^x=1/2 has no solutions in real and complex numbers, it may lead to creation of new numbers just to fill that gap. Has anyone already tried to define such numbers? But it has a solution in the complex numbers? $\frac{1}{2}=x^x$ $\ln\left(\frac{1}{2}\right)=x\ln(x)=e^yy$ $W\left(\ln\left(\frac{1}{2}\right)\right)=y=\ln(x)$ $x=e^{W\left(\ln\left(\frac{1}{2}\right)\right)}$ He actually wrote he does not know if it has. "There is no square superroot of 1/2, " . And it was Bromer in 1987 article. So it has solution in complex numbers. x= e^W(W(-ln2/2)). Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 01/30/2008, 08:21 PM I was not able to wait too long: So finally we have first ill defined definitions-they has to be ill defined as they include infinities so that we can build logical structures out of them defying Godel. 1) i is undiferrentiated structureless infinitesimal(s) (or -i, or both- they are NOT equal.) 2) the phase state of undifferenntiated, structureless infinitesimal is perfect chaos: no structures, just motion 3) the entropy of primordial see of such infinitesimal(s) is -i*lnI=-pi/2 4) It is maximal negative entropy there can be 5) The direction of development is such that infinitesimals organize in structures reducing negative entropy ( increasing share of entropy of structures moving in infinitesimal sea. 6) When there is no structures, just motion, any temperature is absolute 0, as is any kinetic energy . 7) What are the first structures remains to be seen, but surely interaction of i with itself as there is nothing else in a way that increases entropy from -pi/2 When there are structures, they move in infinitesimal sea, and there is temperature corresponding to kinetic energy of such structures. 9) between each level of structures of infintisemals, SCALES are increasing infinitely, speed is decreasing infinitely compared to previous scale. 10) Imaginary continuos time which works in these scales appears as first structures appear. Before that, when there is only motion, there is no TIME. 11) The time scales of imaginary time get infinitely longer with each scale, still being infintiely small 12) Infinitesimal behaviour in SQ EQUALS continuous mathematics- which may be missing at least tetration operation to give full picture, but probably all hyperoperations 13) As structures are created, entropy gets less negative, temperature and kinetic energy grows- 14) There is a threshold point where 1/2 of initial negative entropy is reduced by thermal motion of created structures of ALL scales so far that infinitely bigger structure witn infinitely larger scale can not be created anymore. So the next scale is finitely larger, time gets real and discrete, speed - finite. The limiting entropis are -pi/2, -pi/4(leading to sqrt(2) - long live Pythagoras! - 0, +pi/4, pi/2 4 phase regions, 1 still missing. 15) Quantum mechanics works in 2scales on BOTH sides of this phase transition- from here the ambiquity it has. K-vortex operates below this when it is just oscillating, slipping differential, above when it gets a grip and spin aligns with the axis that caused that grip to appear. 16) From this point onwards, thermal motion increases faster than creation of structures- so increased complexity of structures is possible only in limited space, not all SQ domain- so we start to see aggregation of structures in particles, etc etc. That is how far i have gotten. All this is connected to the prime number theorem, so that: in SQ, lni/i > entropy , entropy being -pi/2, -pi/3 etc, - probably, i representing a structure in each scale in such formula , so here the internal complexity of infinitesimal structures is not obvious- math lacks this dimension. On attempt to register it was backtetrating hyperdimensional volume -i and writing each dimension as dI , so that step to next dimension is -I+di, and it is possible to think about integration.differentiation along this dimension of infinitesimals. dI is basically a change in infinitesimal structure expressed as change in hypervolume. In SQ, lni/i is bigger than number of "primes", primes representing structures built. in limit case, phase transition between continuous/discrete, number of primes counted from both sides has a property lnx/x=pi(x) above SQ, pi(x) < lnx/x above SQ, the entropy of structures is pi(x) while the entropy of chaos left is lnx/x . x = infinity at phase transition point, so the bigger local structures are built in qbove quantum space, the bigger the difference between lnx/x and pi(x) , until, at last discrete scale, ln2/2 < 1 or ln1/1<1 . if the state of perfect order is possible to reach, it would have only structure and no motion, so entropy would be pi/2. As we know/ there should be 4 phases of Aether - one of which is perfect chaos, another structured chaos, another chaotic structures, another perfect order. Each has a well defined number , entropy. 5th is missing- i think that is the one living inside that phase transition between SQ and quantum, may be wrongly. That is as far as I have got. Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 01/31/2008, 07:41 AM (This post was last modified: 01/31/2008, 09:24 AM by Ivars.) I have a question: Since we have h(e^(pi/2)) = h(i^(1/i)) = h( (1/i)^i) = -i ( or ,perhaps, also +i- but that is not so important at this stage), and we interpret it as Andrew did it: if -i is a hypervolume of length e^(pi/2) being tetrated infinitely - kind of building up higher dimensions of immeasureable ( transcendental or not?) edge e^pi/2, What would be the hypersurface (area) of that hypervolume -i ? Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 01/31/2008, 11:13 AM Well when I read the defintion of hypersurface in Wikipedia: In algebraic geometry, a hypersurface in projective space of dimension n is an algebraic set that is purely of dimension n − 1. It is then defined by a single equation F = 0, a homogeneous polynomial in the homogeneous coordinates. (It may have singularities, so not in fact be a submanifold in the strict sense.) Letting out the subtleties, the hypersurface of hypervolume -I must be dI= - I lnI = -pi/2 And, for hypervolume I -> dI = IlnI = pi/2. Because there is nothing else given, the only way to reduce infinite dimensionality of hypervolume I to (infinity -1 ) is by taking a differential of it. GFR Member Posts: 174 Threads: 4 Joined: Aug 2007 01/31/2008, 11:55 AM (This post was last modified: 01/31/2008, 12:01 PM by GFR.) Ivars Wrote:bo198214 Wrote:Ivars Wrote:Barrows said: "... if x^x=1/2 has no solutions in real and complex numbers ... "But ... : $\frac{1}{2}=x^x$ $\ln\left(\frac{1}{2}\right)=x\ln(x)=e^yy$ $W\left(\ln\left(\frac{1}{2}\right)\right)=y=\ln(x)$ $x=e^{W\left(\ln\left(\frac{1}{2}\right)\right)}$He actually wrote he does not know if it has. "There is no square superroot of 1/2, " . And it was Bromer in 1987 article. So it has solution in complex numbers. x= e^W(W(-ln2/2)).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 GFR Member Posts: 174 Threads: 4 Joined: Aug 2007 01/31/2008, 02:25 PM Hi Ivars! Concerning: Ivars Wrote:I have a question: Since we have h(e^(pi/2)) = h(i^(1/i)) = h( (1/i)^i) = -i (or, perhaps, also +i, but that is not so important at this stage), and we interpret it as Andrew did it: if -i is a hypervolume of length e^(pi/2) being tetrated infinitely - kind of building up higher dimensions of immeasureable (transcendental or not?) edge e^pi/2,With "rho" and "pi": Greek letters, let me try to put it like this: - pi = 3.1415592654.. : the famous constant universal ratio; -+/- i = sqrt(-1) : the two (constant!) square roots of -1; - rho = e^(pi/2) : a critical base for an "infinite tetrate"; - h = {-i, +i} : the imaginary conjugate units. Well, we have: rho#(+oo) = [e^(pi/2)]#(+oo) = {-i,+i} rho = i^(1/i) = i^(-i) = (-i)^(-1/i) = (1/i)^(i) = e^(pi/2) Concerning a possible mnemonical and amusing math-fiction model, we might say that -i and +i are the two possible symmetrical (conjugate) imaginary unit heights of a tower with base "rho" and an infinite number of stores. In other words, {-i, +i} are the two softly collapsed heights of an infinite tower with base (and not an "edge") rho. But, it is not reasonable to measure the actual "volume" of this kind of "towers" and, particularly, their external "surfaces". These are, in my opinion, label concepts without any specific mathematical meaning. GFR Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 01/31/2008, 03:56 PM GFR No not really shall we give up so fast! If we have created a hypervolume lets find its hypersurface. Then we can think what it means. -i^(-1/i) is a new one- but that is the right notation since it takes -i as it is - even better -i^( 1/-i) without doing any simplifications. Really, such small things are important. I try to never simplify an expression too early. What is the difference between edge and base e^(pi/2)? Andrew mentioned that if e^( pi/2) is one dimensional, it is lenght . Do you think e^(pi/2) is 2 dimensional in some sense, or something else? Ivars GFR Member Posts: 174 Threads: 4 Joined: Aug 2007 01/31/2008, 05:06 PM No, it is a real number. Like e is also a real number, even when it is the "base" of an exponential. I don't see why it would be 2-dimensional when it is the "base" of a tetrational expression. I don't have any prejudice idea about "edges", but the name of that object is the "base" of the tetrational expression. In y = b ^ x, b is the base, x is the (super)exponent and y is the result of b-tetra-x. For x-> +oo, y -> h, which we may call "the height of the infinite tower", for a kind of nice word-joke. Both you and me we know that we don't have any "tower" around. By the way, for my old-fashoned mind, every well simplified expression, if it is correctly done, is equivalent, or there is something wrong. GFR bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 02/01/2008, 12:30 AM (This post was last modified: 02/01/2008, 12:30 AM by bo198214.) 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 That reminds me to update the wikipedia tetration article as there is no formula for the square super root yet. And indeed your formula is also a solution: We first see that your formula $\frac{\ln(x)}{W(\ln(x))}$ is equal to $\frac{1}{h(1/x)}$ which is a solution to $y^y=x$: $\left(\frac{1}{h(1/x)}\right)^{\frac{1}{h(1/x)}}=\frac{1}{1/x}=x$ as $h$ is the inverse function of $x^{1/x}$. « Next Oldest | Next Newest »

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