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F(f(x))=exp(exp(x)). - 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: F(f(x))=exp(exp(x)). (/showthread.php?tid=1555) Pages:
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RE: F(f(x))=exp(exp(x)). - Gottfried - 07/01/2022 Try this one: Code: g(x) =2*( 1 -0.693147180560*(x/2-1) - 1.32551766575*(x/2-1)^2 - 2.50705892304*(x/2-1)^3 - 4.77252348338*x^4 With a convergence range of about \(1/2 \) this should give the "dual function" (as I once christened this) for \( f(x) = b^x \) with \(b= \sqrt 2 \) to the effect that \(g(g(x))=f(f(x)) \) but \( g(x) \ne f(x) \) A short inspection, (using Euler-summing of the series when (not-too-much) out of convergence) gave \[ g(0) = 2.46791405115 \qquad g(g(0)) = \sqrt 2 \] while, of course, \[ f(0) = 1 \qquad f(f(0)) = \sqrt 2 \] Note, that the value \(2.46791405115 \) is exactly the value in my picture (recently posted here) of imaginary iteration of \(b^x\) where \( b= \sqrt 2\) and \( f(0)=1 \) is iterated with imaginary height . The power series of the function \(g(x) \) was created using the Schroeder-procedere as mentioned in my first answer with the help of diagonalizing the Carlemanmatrix for \(t(x) = 2^x-1 \). Gottfried RE: F(f(x))=exp(exp(x)). - JmsNxn - 07/01/2022 (07/01/2022, 10:30 PM)Gottfried Wrote: Note, that the value \(2.46791405115 \) is exactly the value in my picture (recently posted here) of imaginary iteration of \(b^x\) where \( b= \sqrt 2\) and \( f(0)=1 \) is iterated with imaginary height . Don't you just love the parallel of \(\sqrt{-1} = i\)! RE: F(f(x))=exp(exp(x)). - Gottfried - 07/01/2022 (07/01/2022, 10:39 PM)JmsNxn Wrote:(07/01/2022, 10:30 PM)Gottfried Wrote: Note, that the value \(2.46791405115 \) is exactly the value in my picture (recently posted here) of imaginary iteration of \(b^x\) where \( b= \sqrt 2\) and \( f(0)=1 \) is iterated with imaginary height . Yeppo !
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