Recent content by Blouge

Hi, I have some questions about time travel. My understanding is that it is an accepted theoretical consequence of General Relativity. If the scenarios I'm describing are too ridiculous then perhaps analagous microscopic scenarios would be more palatable. Consider these scenarios: Scenario...

>The truth is that ... 0.999... = 1. This is true when dealing with the real number system. If you are dealing with the hyperreals, then the two numbers are different, and in fact their exact difference can be computed and expressed in terms of infinitesimals: 1 - 0.99999... = 1 - 0.9 sum(...

I'm not sure if it makes things more or less clear, but the surreal numbers and the hyperreals are infinitely more dense than the real numbers. Conceptually, what seems like a point in these number systems is actually a monad (a small interval full of points) when viewed at a finer scale.

>exp doesn't have a fixed point There are complex, not real, fixed points. If you take -1 (or some other starting point) and apply the ln function repeatedly, it converges to some complex number c for which exp(c) = c. You can find infinitely many other fixed points by applying (2Npi + ln)...

I see now. Thanks Hurkyl! For [-inf,a), f(x) = f0(x) = a - aa/x For [a,0), f(x) = f1(x) = exp(f0^-1(x)) = exp(aa / (a - x)) For [0,exp(a)), f(x) = f2(x) = exp(f1^-1(x)) = exp( a - aa / ln x ) For [exp(a), exp(exp(a))), f(x) = f3(x) = exp(f2^-1(x)) = ... etc... The only problem is that there's...

>Okay, so that's the piece of f defined on [-inf, -1). Now use the functional equation to figure out what f should be on [-1, 0)... I guess, for y on [-1,0) you can say that f(y) = exp(-1/(1+y)): f(f(x)) = exp(x) f(y) = exp(f^-1(y)) f(y) = exp(-1/(1+y))

>And f(x) = x+3 is not a bijection from [-inf, a) to [a, 0) for any choice of a. How about a = -1 and f(x) = -1 - 1/x? I don't see where that gets you. Going back to the power series idea (where the constant a i used is ln ln ln ... -1 ) d^2/(dx)^2 f(f(x)) = f''(x) f'(f(x)) + f'(x)^2...

>I claim it's not a problem: any choice works. Um, f(x) = x + 3 is not a good choice: x + 3 + 3 = exp(x)? I'm trying to find a choice that satisfies the equation.

>if f(x) has inverse, then you can get it in high school, I think. I'm trying to find out a more direct way of expressing the unknown function f(x) -- not focusing on a particular value of x or known function f. >choose f to be a bijection Yes, the whole problem is how to "choose" f that...

If f(f(x)) = exp(x), then what is f(x)? I don't think that f(x) can be an entire function. It would be nice to have f(x) be monotonically increasing for positive x. I tried to attack this problem by using a method in the book "Bypasses", and it works, but it involves numerical calculation of...

>Is it not possible to define in anyway such indeterminate quantities? Sure, as x approaches 0, sin x approaches 0 while sin x / x approaches 1. sin x / (x*x) approaches is "undefined" (i.e. it approaches plus or minus infinity) as real x approaches 0. So, 1 / (x*x) at x = 0 is a much larger...

Maybe it's because the infinite sums that start off involving small numbers like 1, 2, 3, .. or 1!, 2!, 3!, etc. yield the most interesting and general constants: e=1+1+1/2+1/6+1/24+1/120+... approx = 2 + 1/2 = 2.5 pi= 4 - 4/3 + 4/5 - 4/7 +... approx = 4 - 4/3 = 2.66666... phi = (1 + 1/phi) =...