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 The Generalized Gaussian Method (GGM) tommy1729 Ultimate Fellow Posts: 1,640 Threads: 366 Joined: Feb 2009 09/25/2021, 12:24 PM The Gaussian method can be easily generalized. suppose we use f(s) = exp( g(s) f(s-1) ) , then we are bounded in the sense that g(s) cannot grow to fast towards 1 as Re(s) goes to +oo. The reason is, if g(s) grows like O(exp(-exp(s)) ) then the (complex) argument (theta) gives us trouble. With erf(s) we are close to 1 + exp(-s^2) and because s^2 puts the imaginary line at 45° that is ok. With 1 + exp(-exp(s)) however the complex argument (theta) gives us issues. 1 + exp(-exp(s)) goes to 1 fast for positive real s , BUT because of the complex argument ( theta ) this does not hold for non-real s even if their real parts are large. So we look for functions g(s) between 1 + exp(-s^2) and 1 + exp(-exp(s)). This is cruxial to understand ! So how do we do that ? For starters it is also known that functions below O(exp(s)) can be completely defined by the value at 0 and its zero's. And we want the zero's to be close to the imag axis. This results in my generalized gaussian method. see pictures !! Regards Tom Marcel Raes tommy1729 Attached Files Image(s)         tommy1729 Ultimate Fellow Posts: 1,640 Threads: 366 Joined: Feb 2009 10/26/2021, 10:41 PM (This post was last modified: 10/28/2021, 12:11 PM by tommy1729.) The related integral above is quite complicated. So I came up with the following simplification. A different method but very similar. n are integers larger than 0. m is going to +infinity. $f(s)=e^{t(s)*f(s-1)}$ $t(s)=(J(s)+1)/2$ $J(s) =Erf(s*p_7(s))$ $p_7(s)=\prod(1+s^2/n^7)$ $sexp(s+s_e)=\ln^{[m]}f(s+m)$ This has similar properties as the other generalized gaussian method and it should be easier to implement. call it GGM2 or so. For bases other than e ; take the base e^b then we get  $f_b(s)=e^{b*t(s)*f(s-1)}$ $t(s)=(J(s)+1)/2$ $J(s) =Erf(s*p_7(s))$ $p_7(s)=\prod(1+s^2/n^7)$ $sexp_b(s+s_b)=\ln_b^{[m]}f_b(s+m)$ regards tommy1729 Tom Marcel Raes tommy1729 Ultimate Fellow Posts: 1,640 Threads: 366 Joined: Feb 2009 10/28/2021, 12:07 PM (This post was last modified: 10/28/2021, 12:12 PM by tommy1729.) (10/26/2021, 10:41 PM)tommy1729 Wrote: The related integral above is quite complicated. So I came up with the following simplification. A different method but very similar. n are integers larger than 0. m is going to +infinity. $f(s)=e^{t(s)*f(s-1)}$ $t(s)=(J(s)+1)/2$ $J(s) =Erf(s*p_7(s))$ $p_7(s)=\prod(1+s^2/n^7)$ $sexp(s+s_e)=\ln^{[m]}f(s+m)$ This has similar properties as the other generalized gaussian method and it should be easier to implement. call it GGM2 or so. For bases other than e ; take the base e^b then we get  $f_b(s)=e^{b*t(s)*f(s-1)}$ $t(s)=(J(s)+1)/2$ $J(s) =Erf(s*p_7(s))$ $p_7(s)=\prod(1+s^2/n^7)$ $sexp_b(s+s_b)=\ln_b^{[m]}f_b(s+m)$ regards tommy1729 Tom Marcel Raes A further idea is to generalize like this   for positive odd w ;  $t_w(s)=1+(J(s)-1)^w/2^w$ for instance w = 3 or w = 7. with w = 7 we get the case : n are integers larger than 0. m is going to +infinity. $f(s)=e^{t_w(s)*f(s-1)}$ $t_w(s)=1+(J(s)-1)^w/2^w$ $J(s) =Erf(s*p_7(s))$ $p_7(s)=\prod(1+s^2/n^7)$ $sexp(s+s_e)=\ln^{[m]}f(s+m)$ This has similar properties as the other generalized gaussian method and it should be easier to implement. call it GGM2 or so. For bases other than e ; take the base e^b then we get  $f_b(s)=e^{b*t_w(s)*f(s-1)}$ $t_w(s)=1+(J(s)-1)^w/2^w$ $J(s) =Erf(s*p_7(s))$ $p_7(s)=\prod(1+s^2/n^7)$ $sexp_b(s+s_b)=\ln_b^{[m]}f_b(s+m)$ Notice this latest new modifation does not change the range where we get close to 1 much , but is still getting faster to 1. regards tommy1729 Tom Marcel Raes ps : join " tetration friends " at facebook :p « Next Oldest | Next Newest »

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