[2sinh] using exp(x) - exp(-3/5 x) tommy1729 Ultimate Fellow Posts: 1,924 Threads: 415 Joined: Feb 2009 05/16/2023, 11:30 PM Using f(x) = exp(x) - exp(-3/5 x) instead of 2sinh for the 2sinh method. You might wonder why. well the higher derivatives go into agreement with exp(x) for re(x) large unlike with 2sinh(x). We also get the fixpoint at 0 and the function looks to go to exp in a more smooth way. All the derivatives are positive at the origin and for x > 0. This resembles exp(x) better thus in a way. It also satisfies the semi-group isom property. Notice that  exp(x) - exp( - 1/3 x)  does NOT work since it has a fixpoint at some negative x. This method is thus a consensus and compromise between exp(x) - 1 and 2sinh(x). exp(x) - 1 is for base exp(1/e) however and is parabolic, so it is more 2sinh type. regards tommy1729 tommy1729 Ultimate Fellow Posts: 1,924 Threads: 415 Joined: Feb 2009 05/16/2023, 11:32 PM also it is still faster than the gaussian method. regards tommy1729 tommy1729 Ultimate Fellow Posts: 1,924 Threads: 415 Joined: Feb 2009 06/18/2023, 11:49 PM I realize a variant of this with parabolic fixpoint (probably at 0) is the key to solving some of my older post ideas. But not exp(x) - 1 that is not asymptotic enough. so maybe something like exp(x) - exp(- 3/5 x) + exp(- a x) - exp(- b x) But that gives me the system of equations (from taylor expansion ) - a + b + 8/5 = 1 a^2 / 2 - b^2 + 8/25 > 0 - a^3 + b^3/6 + 76/375 > 0 ... b = 1 + a - 8/5 ... I am unsure. This does not give all derivatives > 0 I think, so it kinda fails. There are also some additional conditions. I need to work on this. I will come back to this later. Sorry for being vague. Open to suggestions ! regards tommy1729 « Next Oldest | Next Newest »

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