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 Infinite Pentation (and x-srt-x) bo198214 Administrator Posts: 1,417 Threads: 92 Joined: Aug 2007 05/31/2011, 10:29 PM (This post was last modified: 05/31/2011, 10:31 PM by bo198214.) (04/11/2009, 09:16 AM)andydude Wrote: If we interpolate these points, one would except the interpolation to diverge for infinite pentation, and x-superroot-x, but the interpolating polynomial of these integer points do seem to converge, Now I doubt about the interpolation method. If the interpolation of the self-tetra-root would yield a valid function, then shoud the interpolation of the simple self-root $n^{1/n}$ also converge to the self-root $x^{1/x}$ on $x>0$. But this seems not to be the case. An interpolation polynomial of degree 400 (401 sample points) still has a negative value at 0.25.     And if we compare the values it seems that the negativity gets rather worse: 101 points: $f(0.25)\approx -0.235$ 201 points: $f(0.25)\approx -0.330$ 301 points: $f(0.25)\approx -0.364$ 401 points: $f(0.25)\approx -0.378$ So it really looks as if the interpolation (even if it converges) does not converge to $x^{1/x}$ which is positive everywhere. So I would conclude that the interpolation of the self-tetra-root also does not converge to a self-tetra-root, even if it converges. Catullus Fellow Posts: 154 Threads: 31 Joined: Jun 2022 06/20/2022, 05:37 AM (This post was last modified: 06/24/2022, 02:13 AM by Catullus.) (05/31/2011, 10:29 PM)bo198214 Wrote: And if we compare the values it seems that the negativity gets rather worse: 101 points: $f(0.25)\approx -0.235$ 201 points: $f(0.25)\approx -0.330$ 301 points: $f(0.25)\approx -0.364$ 401 points: $f(0.25)\approx -0.378$ So it really looks as if the interpolation (even if it converges) does not converge to $x^{1/x}$ which is positive everywhere.(-1)^(1/-1) = -1. Negative one is not positive. Sincerely: Catullus « Next Oldest | Next Newest »

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