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 Infinite tetration and superroot of infinitesimal jaydfox Long Time Fellow Posts: 440 Threads: 31 Joined: Aug 2007 12/20/2007, 07:25 PM Quote:My only problem with this is that I have a deep-seated hatred of Cantor's numbers. I'll save that talk for later. Heh, guess I'm not alone... ~ Jay Daniel Fox Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 12/20/2007, 09:02 PM Then I better even do not start with them. I read a little aboout ordinals, cardinals etc. after Andy's post and I get the same feeling as dealing with limits -I just do not get it.Can it not be simpler? Infinitesimals though feel rather easy and intuitive, as well as graded infinities and different scales and rotations of infinitesimals. jaydfox Long Time Fellow Posts: 440 Threads: 31 Joined: Aug 2007 12/20/2007, 09:08 PM Well, I'm sure everyone has his own way of thinking of infinities and infinitesimals that is unorthodox. Personally, I think of "Big Oh" limits of functions taken to infinity, but using only a power of x. For example, let's say I have a function f(x), and I want to find a power of x that represents the big oh limit. Well, f(x)=x^2+x will have a power of 2. For f(x)=3x^2-x^1.999, it's still 2. But what about f(x)=ln(x)*x^2? It's big oh limit will be greater than 2, but smaller than all real numbers greater than 2. I consider this 2 plus an infinitesimal. YMMV. ~ Jay Daniel Fox Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 12/20/2007, 09:50 PM In my notation, it would be (as infinity is 1/i for grade i infinitesimals): f(1/i) = ln(1/i)*(1/i)^2 = -lni* (-i)^2 = (-i)^3 * pi/2 So its big oh limit would be 3, but pi/2 would appear. This means this limit will be transcending the squared infinity because of extra -i*pi/2. What does it mean to transcend infinity that is created by 1/infinitesimal I do not know yet. Maybe it is like exausting infinity, since pi continues beyond any infinity in length.... --i*pi/2 is obviously not a usual infinitesimal just multiplied by a finite real because -i^2 * e^-ipi/2 = -k where -k is infinity perpendicular to -i ^2 (not in complex plane, but perpendicular to it). I would say big oh limit for this function is 2 + (1 turned by 90 degrees). At complex plane, it may look like 2+ infinitesimal, but that would be just a projection of -k on complex plane ( the 3rd i than is infinitesimal multiplied by -pi/2). Well, not exactly like this, of course.... andydude Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 12/21/2007, 02:08 AM We have mentioned lots of different kinds of infinities and infinitesimals, just to mention two others: differentials (dx) and singularities (f(x) = infinity). Differentials are probably the most standard way of looking at infinitesimals, and singularities are probably the most standard way of looking at infinities. And of course, both of these things can appear "in different orders". Andrew Robbins Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 12/21/2007, 05:49 PM (This post was last modified: 12/21/2007, 05:55 PM by Ivars.) Sure, but what is dx^(1/dx) or f(x)^(1/f(x)) from Your example? I know from Euler that ln((1+dx)^(1/dx))= 1 so (1+dx)^(1/dx) = e. But what about dx^(1/dx)? Ivars andydude Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 12/21/2007, 08:06 PM Ivars Wrote:But what about dx^(1/dx)? Simple calculus would state that: $\frac{d}{dx}\left(x^{1/x}\right) = x^{1/x}(1+\ln(x))\frac{1}{x^2}$ so moving the dx part, we obtain: $d\left(x^{1/x}\right) = x^{1/x}(1+\ln(x))\frac{1}{x^2}dx$ is this what you are talking about? If not, then I suppose that $d\left(x^{1/x}\right) \ne dx^{1/{dx}}$ which means we are no closer to knowing the answer, but since you said $(1+dx)^{1/dx} = e$ then that means that: $dx^{1/dx} = \sum_{k=0}^{1/dx}\left({1/dx \atop k}\right)(-1)^{1/dx} (1+dx)^k = \sum_{k=0}^{1/dx}\frac{1}{k!}(-1)^{1/dx}$ which means $dx^{1/dx}$ is indeterminate, since $(-1)^{\infty}$ is indeterminate. Andrew Robbins jaydfox Long Time Fellow Posts: 440 Threads: 31 Joined: Aug 2007 12/21/2007, 08:40 PM For complex z, the function z^(1/z) has an essentialy singularity at z=0. For real x, using the primary branch of roots for positive x, the function x^(1/x) goes to 0 from the right. But it's indeterminate from the left, because we're taking a negative number to various fractional and even irrational powers, which forces complex numbers and never gives us a consistent complex argument, even though the modulus consistently goes to infinity. Either way, dx^(1/dx) must be indeterminate. ~ Jay Daniel Fox Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 12/21/2007, 10:41 PM Yes I see, thank You for answers, but: As an infinitesimal, dx in this expression is always either only positive, or only negative . And never 0. But smaller than any real (bigger if negative) . That is the definition of infinitesimal in non-standard analysis-in hyperreal set. The theory that works in calculus without using limits, so I guess application of limits in this case is misleading , but I do not know what to apply. dx in this case is not a limit dx->0 as real, but an extension of real set, or, rather set in which reals are just subset. It is a different type of number, so to say- infinitesimal. So is 1/dx = 1/infinitesimal= definition of infinity (also belongs to hyperreals). http://en.wikipedia.org/wiki/Non-standard_analysis jaydfox Long Time Fellow Posts: 440 Threads: 31 Joined: Aug 2007 12/21/2007, 11:00 PM If you want to treat the infinitesimal as always positive, then it's simply the logical end of the sequence dx, dx^2, dx^3, ..., dx^(1/dx). In ordinal notation, it's 1/(omega^omega), and omega^omega is a rather small ordinal in the hierarchy of countable infinities. (Not that I particularly like the hierarchy.) ~ Jay Daniel Fox « Next Oldest | Next Newest »

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