Continuum sum = Continuum product tommy1729 Ultimate Fellow Posts: 1,742 Threads: 382 Joined: Feb 2009 08/21/2013, 11:27 PM One of my earliest teenage conjectures was this : In the context of nonnegative real numbers consider A = A A + B = AB A + B + C = ABC ... It was clear that A could not be 0. A = 2 works fine. 2 = 2 2 + 2 = 2*2 So to go to the next variable D we need to solve 2 + 2 + D = 2*2*D or simpler 4 + D = 4D 4 = 3D D = 4/3 And basicly to go further we need to solve recursively D0 + D1 = D0*D1 D0 = (D0 - 1)D1 D1 = D0/(D0 - 1) Now this clearly is an iterated MÃ¶bius transform which we understand well. Notice the fixpoint X = X/(X - 1) X - 1 = 1 => X = 2. We ignore the solution 0 here because as said in the beginning 0 "fails". This is basicly the solution for "discrete sum = discrete product" Now the conjecture is " Continuum sum = Continuum product " Is given by the continuum iteration of X/(X-1). Basicly just handwaving and saying what is true for positive integers should be true for positive reals. *** I called equations involving continuum sums and/or continuum products " chaotic equations " or "chaos equations ". I was also aware of the idea that f ' (x) = continuum product [ f(x) ] had tetration as a solution. I tried to find some books about " chaos equations " but failed to find any. I assumed I would learn about it later , which never happened. Since I found the sinh method I forgot about it as it appeared the " best solution ". At that time I did not know what complex analytic meant hence it seemed perfect. Unfortunately this somewhat made me forget about these " chaos equations ". I also somewhat forgot about those " chaos equations " by the failure of finding closed forms solutions and the fear that all solutions would be as nonstandard as tetration. And ofcourse the lack of serious theory or tools to work with this kind of stuff. I spoke to a few people about this and I think I mentioned it at sci.math but that did not help much. Here and Now seems right to mention this stuff again. Problem is these " chaos equations " require more than just a method to compute a continuum sum ; probably comparable to being able to do differentiation and solving PDE. regards tommy1729 JmsNxn Ultimate Fellow Posts: 1,057 Threads: 121 Joined: Dec 2010 08/22/2013, 04:01 PM I was thinking about making the transformation from continuum sum to continuum product and encountered the following problem. The continuum sum can be written as a triple integral. If $f(y)$ is defined on the line $\Re(s) = \sigma$ And as its imaginary part goes to infinity it grows slower than $Ce^{-|\Im{s}|\frac{\pi}{2}}$ decays to zero. $\sum_a^b f(y)\, \sigma y = \frac{1}{2 \pi i}\Big[ \frac{1}{\Gamma(s)}\int_0^{\infty}e^{-t} \int_0^{\infty}u^{s-1} \int_{\sigma-i \infty}^{\sigma + i\infty}\Gamma(y)f(y)(t-u)^{-y}\,dydudt \,\,\Big]_{s=a}^{{s=b}$ Now the immediate problem is that I think mellin inversion will not work on functions like the logarithm, or any function that gives complex values to the logarithm. and this is how we go from the continuum sum to the continuum product. Therefore, we'd have to find a more general way of inverting the mellin transform that could work on functions like the logarithm. Or we could do some algebra and prove the result that $\frac{d}{ds} \sum_1^s f(y) \, \sigma y = \sum_1^s \frac{df(y)}{dy}\,\sigma y$ As well as the same result for the integral. With that we can get an expression for the factorial as: $C \Gamma(s) = e^{\int_1^s \psi(t) dt}$ Where psi is the di gamma function and can be shown to equal: $\psi(s) - C = \sum_1^{s-1} \frac{\sigma y}{y}$ « Next Oldest | Next Newest »

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