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another infinite composition gaussian method clone - Printable Version +- Tetration Forum (https://tetrationforum.org) +-- Forum: Tetration and Related Topics (https://tetrationforum.org/forumdisplay.php?fid=1) +--- Forum: Mathematical and General Discussion (https://tetrationforum.org/forumdisplay.php?fid=3) +--- Thread: another infinite composition gaussian method clone (/showthread.php?tid=1693) |
another infinite composition gaussian method clone - tommy1729 - 01/20/2023 another infinite composition gaussian method clone Yeah nothing huge to mention. But just another to add to the list of gaussian method , beta method and similar methods. we want f(s+1) = exp( t(s) * f(s) ) and t(s) strictly rises from 0 at s = -oo to 1 at s = + oo in a fast way. The gaussian method had t(s) = (1 + erf(s))/2. The beta has t(s) = 1/(1 + exp(-s)). the incomplete gamma method had t(s) = 1/(1 + inc.gamma(-s)) etc You get the idea. So we want to construct a t(s). We need something that looks like erf(s) or tanh(s) Let c = 2/pi Then such a candidate is the " special function " : c tt(x) where tt(x) is the " tommy tan " function ( which I invented many decades ago as a teenager, and somewhat inspired me to do math ... on the other hand I forgot about it mainly lol ) see here where my friend mick asks about the value c : https://math.stackexchange.com/questions/191008/a-curious-limit-for-frac-pi2 tt(x) = sum_(n>0) (-1)^[n+1] x^(2n-1) [(2n)! ln(2n)]^(-1) or for the tex fans : \[ tt(x) = \sum_{n>0} (-1)^{[n+1]} x^{2n-1} [(2n)! \ln(2n)]^{-1}\\ \] ( notice tt(-x) = - tt(x) ) then t(s) becomes (1 + c tt(s))/2. NOTICE that c tt(s) is entire ! Now this function has been resurrected and given a life purpose , it might be nice to investigate its properties. *** I have been thinking/dreaming about an addition function formula tt(a+b) = ... or an asymptotical addition formula. But maybe that is just a dream not worth persuing. *** ( and ofcourse there are " fake function ideas " related to this but i have mentioned this already too often ) regards tommy1729 RE: another infinite composition gaussian method clone - JmsNxn - 01/23/2023 Honestly, I'm super thankful that my ideas have affected you so much, Tommy. But I'd be much more interested in a discussion of the family of solutions. Let's take: \[ T[t](s+1) = e^{t(s) T(s)}\\ \] Where the functions: \(t\) form a quasi space of functions such that: \[ \sum_{j=1}^\infty |t(s-j)| < \infty\\ \] And: \[ T[t](s) = \Omega_{j=1}^\infty e^{t(s-j)z}\,\bullet z\\ \] But the "addition" on this space is \(t_0 \oplus t_1 = \frac{t_0(s) + t_1(s)}{2}\), and \(t(\infty) = 1\). Both mine and your methods, and all of your methods belong to this space. I'm pretty fucking confident nothing is going to work for \(b=e\), it might for others. But base \(e\) is just so fucking volatile. And I'm confident both our methods fail at holomorphy in \(\mathbb{C}\).I'd be very surprised at this point if any \(t\) produces holomorphic functions. And if we do find one, it'll be partitioned as \(\Im s \to \pm \infty\) that \(t \to L^{\pm}\). Or something like that.... We can only find a representation for Kneser in this manner. We're not going to break things.... RE: another infinite composition gaussian method clone - tommy1729 - 01/24/2023 I am more optimistic |