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 Infinite tetration and superroot of infinitesimal bo198214 Administrator Posts: 1,386 Threads: 90 Joined: Aug 2007 02/01/2008, 09:29 PM (This post was last modified: 02/01/2008, 09:38 PM by bo198214.) Hey Gianfranco, open your eyes! Catastrophe theory will not solve your problems with multivalued functions, nor will the use of the term "multivalued function". I think I clarified in detail what terms are there and how to use them: 1. There is an analytic function (single valued) 2. If the function has only isolated singularities choose slices starting at the singularities for extending this function (uniquely/single valued) to the sliced complex plane. 3. Determine branches of the function by paths that cross the slices. If your original analytic function $f$ is not injective, you can invert it at a point with $f'(x_0)\neq 0$. That means there is a unique inversion $f^{-1}$ in a neighborhood of $(x_0,f(x_0))$. If you look at the branches of $f^{-1}$ you will find among them also the other values $x$ for which $f(x)=y$, i.e. for each $x$ with $f(x)=y$ there will be a branch ${f^{-1}}_k$ such that ${f^{-1}}_k(y)=x$. Or in other words for each $x$ there is a $k$ such that ${f^{-1}}_k(f(x))=x$. For example the function $\sqrt{x}$ (sliced as usual) has the branches $\sqrt{x}_k = (-1)^k \sqrt{x}$, i.e. at all only 2 branches that is $\sqrt{x}$ and $-\sqrt{x}$. And hence $\sqrt{x^2}$ is not always equal to $x$ but we have to choose the right branch for example $-\sqrt{(-2)^2}=-2$ and $\sqrt{2^2}=2$. In general we have merely $\sqrt{x^2}=|x|$. And no multivalued function consideration will save you from considering this. Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 02/01/2008, 10:00 PM If You think about infinite plane, and start to tear it apart. There is one plane until its teared in 2. Then You tear them again. If there was ever a single valued function mapping something to this plane, after a phase transtion ( exactly at moment when finally tear percolates - planes are separated-) there is suddenly a double valued function. As You continue along in the same way ( You can always tear an infinite plane by making infinite/percolation value cuts) after every phase transition ( and not before) You get more and multiple valued function when outputs move from one infinite plane to another. To me it seems natural to have multivalued functions since in nature everything happens via phase transitions and dividing. Actually, a single valued function would just work inside region between 2 phase transitions, so it will have no connection wiht evolution of the system. Which is pretty boring. Ivars GFR Member Posts: 174 Threads: 4 Joined: Aug 2007 02/01/2008, 11:00 PM (This post was last modified: 02/01/2008, 11:02 PM by GFR.) Dear Henryk! I went out of the main field, a little bit rooming around and, therefore, you said: bo198214 Wrote:Hey Gianfranco, open your eyes! Catastrophe theory will not solve your problems with multivalued functions, nor will the use of the term "multivalued function". I think I clarified in detail what terms are there and how to use them:Actually "Thom's Catastrophe Theory" is not (... yet) a consistent mathematical theory, but only a set of ideas, to be further developed. On the contrary, the expression "multivalued functions" covers (little bit) more official mathematical concepts. Nevertheless, both the supersquare root (for 1/Eta < x < 1) and the "infinite-tower" h (for 1 < b < Eta) seem to appear as two-alued "functions", for the analysis of which the existing mathematical tools are not completely available yet (don't be nervous, please, and try to accept my ... critical descriptive innocent language). The theory of analytic functions is very respectable and extremely serious. It is what it can be made available now, but this is only the present status of the affairs. Tomorrow, your little little children will perhaps have also other more efficient (and complicated) instruments. Unfortunately, our eyes will not be around to see it. You seemed to accept my way of presenting the inversion of function y = x^x (square tetration?), i.e. of x = ssqrt(y), by the use of the logical union of two branches of the Lambert Function (orders -1 and 0). I tried, then, to propose using the same methodology for representing the two h branches, which (together) represent the "inverse" of b = h^(1/h) = selfroot(h). For the moment, these are only practical "tricks" and I dared to imagine a future theory justifying them. I see and respect the role of analytic functions, as well as the exigencies put forward by the necessary serial developments. A good theory should be the most appropriate to "justifying the past and foreseeing the future". My wild ideas didn't completely cover the first part of the enlightened exigencies. Sorry about that. GFR bo198214 Administrator Posts: 1,386 Threads: 90 Joined: Aug 2007 02/02/2008, 12:16 AM GFR Wrote:A good theory should be the most appropriate to "justifying the past and foreseeing the future". My wild ideas didn't completely cover the first part of the enlightened exigencies. But as long as this theory isnt there I would prefer to discuss understandibly and in clearly defined terms (where possible). I mean our discussions were somewhat cloudy and several misconceptions occured with respect to multiple values. So it is quite preferable to have clear concepts here and now. GFR Member Posts: 174 Threads: 4 Joined: Aug 2007 02/02/2008, 03:31 PM Agreed! Let's try and use the most standard an clearest mathematical language as possible. The subjects are very complicated to deal with. Let us make an effort, all together. Otherwise, it will become impossible for anyone to follow and interact efficiently, with relevant interpretations and proposals. GFR Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 02/05/2008, 10:58 AM (This post was last modified: 02/05/2008, 11:01 AM by Ivars.) Since it is my beloved thread, I will allow this - without no mathematical language at all(allmost): Our learned and intuitive notion that mathematics are creatively limited are based on fact that the part of MATH that deals with OPERATIONS (motions) was essentially left undeveloped since infinity and divergence was forbidden and engineering took over "exact" sciences. In principle, You should be able to "differentiate" operation to get a new one by a set of rules that is not more complicated than normal math, just contra intuitive. You just need a space of operations, space of structures (which we know are set infinity apart from each other, but dynamically interacting) and some correspondence set between them by (infinite/finite?) set of observer functions. As there are 3 of them , there will always be scales at which they are not in balance, so uniqueness and certain unpredictability at certain scales will be present still. In this sence, 1/2 is a dynamical balance point between pure Operation (only motion) and pure Number ( only Structure) while Observer functions are in no way in symmetric balance- in fact, thay have infinite number of balances on both sidesof this balance. They are like ropes, roots, that hold this 1/2 balance in place, rooted in both Operation dominated (motion ,infinitesimal, SQ, imaginary ) and structure dominated ( number, above quantum, negative) spaces. Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 02/06/2008, 07:43 PM I was looking at these graphs: If You plot in polar coordinates (r,phi) following spirals : r= phi^(1/phi) , r= phi^(1/phi)^phi, r= phi^(1/phi)^phi^(1/phi) etc they all seem to converge to a circle with a radius e^(1/e) so that is just its area. But I have no software to check it, so it might be a mistake. May be it is smaller and has no connection to the problem. Anyway, it is a very strange graph crossing itself once at angle which seems not to be 5/2* pi - after one full rotation - but again, I have no software to check. With each rotation, the angle of crossing gets smaller, there also seems to be a limit angle of crossing, which seems bigger than pi/4 but that is after 5000 rotations -I can not simulate more. Now we think that infinity^(1/infinity) which was the beginning of this thread is not very well defined. Looking at the polar graph of r=phi^1/phi it does not seem so. First, there is well defined limit circle, I suppose, which has area. Second, there is a limit finite angle where outgoing spiral at phi< pi/2 cuts the inner limit circle. Third, there is the lenght of the spiral from origin ( which is also undefined, 0^1/0) to the point where it crosses limit cycle. So , in fact, phi^1/phi etc. defines at least 3 values, 2 of them 1 dimensional, which could be used as limits for any of limits we are looking for: e.g. infinitesimal^1/infinitesimal = lenght of the arc till crossing infinity ^1/infinity = angle of line from center to the crossing point, or, alternatively, tangent angle of the spiral at that point- in fact, there are 2 tangent angles-another to the circle at that point. Plus area of circle. if we change sign of r=-phi^(1/phi), spiral wounds the same way but with -pi phase offset. If we change also sign of exponent, r=-phi^(-1/phi) things spiral appears from infinity to fill the cycle from inside. in this case, there is no arc length, but 2 tangent angles + area is still there. The same with r=phi^(-1/phi), offset by pi. In both cases with negative exponents spirals turn anticlockwise , also in both cases with positive exponents. How ever, if we consider the limit cycle (ring) as the origin of spirals, spirals they wind in opposite directions, kind of squeezing the limit cycle in the middle. So why not define the limits using the finite tangent angles? Or their trigonometric functions, e.g tangens? The one of the outgoing/incoming spiral for infinitesimal^(1/infinitesimal), the one of the limit circle at crossing for infinity^1/infinity. There are options to define those limits , why no one is used? bo198214 Administrator Posts: 1,386 Threads: 90 Joined: Aug 2007 02/26/2008, 09:26 AM Ivars Wrote:r= phi^(1/phi) , r= phi^(1/phi)^phi, r= phi^(1/phi)^phi^(1/phi) etc ... Now we think that infinity^(1/infinity) which was the beginning of this thread is not very well defined. ... So why not define the limits using the finite tangent angles? Or their trigonometric functions, e.g tangens? The one of the outgoing/incoming spiral for infinitesimal^(1/infinitesimal), the one of the limit circle at crossing for infinity^1/infinity. There are options to define those limits , why no one is used? I dont understand why you invoke a polar graph here, the limits are very well defined and well-known: $\lim_{x\to\infty} x^{1/x} = 1$ $\lim_{x\to 0} x^{1/x} = 0$ You can already guess it when you look at the (cartesian) graph of the selfroot. $\lim_{x\to\infty} x^{(1/x)^x} = \lim_{x\to\infty} x^{1/x^x} = \lim_{x\to\infty} e^{\ln(x)/x^x}=e^{\lim_{x\to\infty}\ln(x)/x^x }=e^0=1$ $\lim_{x\to\infty} x^{(1/x)^{x^{1/x}}}=\exp\left(\ln(x)\frac{1 }{x^{x^{1/x}}}\right)$ Here we know that $x^{1/x}>1$ for $x>1$ and hence $x^{x^{1/x}}>x$ for $x>1$ and hence $\lim_{x\to\infty}\ln(x)/x^{x^{1/x}}=0$. Finally again: $\lim_{x\to\infty} x^{(1/x)^{x^{1/x}}}=1$ Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 02/26/2008, 09:31 PM bo198214 Wrote:I dont understand why you invoke a polar graph here, the limits are very well defined and well-known: $\lim_{x\to\infty} x^{1/x} = 1$ $\lim_{x\to 0} x^{1/x} = 0$ You can already guess it when you look at the (cartesian) graph of the selfroot. $\lim_{x\to\infty} x^{(1/x)^x} = \lim_{x\to\infty} x^{1/x^x} = \lim_{x\to\infty} e^{\ln(x)/x^x}=e^{\lim_{x\to\infty}\ln(x)/x^x }=e^0=1$ $\lim_{x\to\infty} x^{(1/x)^{x^{1/x}}}=\exp\left(\ln(x)\frac{1 }{x^{x^{1/x}}}\right)$ Here we know that $x^{1/x}>1$ for $x>1$ and hence $x^{x^{1/x}}>x$ for $x>1$ and hence $\lim_{x\to\infty}\ln(x)/x^{x^{1/x}}=0$. Finally again: $\lim_{x\to\infty} x^{(1/x)^{x^{1/x}}}=1$ Hi, I reread the beginning of the thread and it was dx^(1/dx) that was a problem, so lim x^1/x when x-> 0 is undefined in real numbers - that I understood from both jaydfox and Andy. Since in my understanding infinity is 1/infinitesimal, x-> infinity is perhaps complementary to first one. When You add another x making x^(1/x)^x You perhaps forcibly drive the expression into the limit You wish as exponent is higher level of infinitesimal than x is infinity. It does not prove (1/dx)^dx=1- these infinitesimals must be of same order. I chose polar coordinates because there You can see what infinity of self root looks like - it is a unit circle-all 4 expressions tend to it, though from different sides. See attached file. As to self root close to 0, dx^1/dx, 2 of selfroots seem to start at 0, 2-at infinity, depending on combination of signs, so this is still undecided for me. Attached Files Image(s)     bo198214 Administrator Posts: 1,386 Threads: 90 Joined: Aug 2007 02/26/2008, 10:23 PM (This post was last modified: 02/26/2008, 10:26 PM by bo198214.) Ivars Wrote:I reread the beginning of the thread and it was dx^(1/dx) that was a problem, so lim x^1/x when x-> 0 is undefined in real numbers - that I understood from both jaydfox and Andy. dx is used in the context of differentiation and integration. If you use it out of context, nobody knows what you mean. And if you additionally mix it into the theory of hyperreals, that doesnt simplify things. Limits however are well established in mathematics and $\lim_{x\downarrow 0} x^{1/x} = \lim_{x\downarrow 0} e^{\ln(x)/x} = 0$ because $\ln(x)/x \to -\infty$ for $x\downarrow 0$ ($x\downarrow 0$ means $x\to 0$ while $x>0$). There is nothing special about it. These are the other limits: $\lim_{x\downarrow 0} x^{1/x^x}= \lim_{x\downarrow 0} e^{\ln(x)/x^x} = 0$ because $\lim_{x\downarrow 0 } x^x = 1$. $\lim_{x\downarrow 0} x^{(1/x)^{x^{1/x}}} = 0$ because $\lim_{x\downarrow 0} x^{x^{1/x}} = 1$. Quote:I chose polar coordinates because there You can see what infinity of self root looks like - it is a unit circle What? The limit is a number and not a circle, how can a limit be a geometrical shape? « Next Oldest | Next Newest »

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