another infinite composition gaussian method clone tommy1729 Ultimate Fellow Posts: 1,859 Threads: 402 Joined: Feb 2009 01/20/2023, 12:00 AM (This post was last modified: 01/20/2023, 12:13 AM by tommy1729.) 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...r-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 JmsNxn Ultimate Fellow Posts: 1,176 Threads: 123 Joined: Dec 2010 01/23/2023, 02:35 AM (This post was last modified: 01/23/2023, 02:37 AM by JmsNxn.) 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.... tommy1729 Ultimate Fellow Posts: 1,859 Threads: 402 Joined: Feb 2009 01/24/2023, 12:53 AM I am more optimistic « Next Oldest | Next Newest »

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