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 Deriving tetration from selfroot? Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 03/24/2008, 10:26 PM (This post was last modified: 03/25/2008, 05:38 PM by Ivars.) As far as I understand selfroot and any root: Selfroot is x^(1/x): Selfroot = $x^{1/x}= e^{\ln(x^{(1/x})}=e^{(\ln(x)/x)}*e^{+-(2*\pi*I*{k/x)}$ For positive x, ${\ln(x)/x}$ is real, giving one real self root (k=0) $e^{(\ln(x)/x)}$, and infinite number of imaginary roots depending on ratio $2*{k/x}$. For negative x, ${\ln(x)/x}$ will be imaginary, of the form: ${\ln(x)/x}+-{((2m-1)/x)}*\pi*I$ so roots are: $e^{(\ln(x)/x)}*e^{-+{((2m-1)/x)}*\pi*I}*e^{+-(2*\pi*I*{k/x)}$, again infinite quantity in totality. m=0 gives: $e^{(\ln(x)/x)}*e^{+-{(1/x)}*\pi*I}*e^{+-(2*\pi*I*{k/x)}$, For imaginary and complex x, tomorrow, with mistakes here also hopefully corrected. Ivars Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 03/25/2008, 05:52 PM (This post was last modified: 03/25/2008, 06:02 PM by Ivars.) Selfroot is x^(1/x): Selfroot = $x^{1/x}= e^{\ln(x^{(1/x})}=e^{(\ln(x)/x)}*e^{+-(2*\pi*I*{k/x)}$ For Imaginary x, ${\ln(x)/x}$ will be imaginary, of the form: ${+-{(({1/2}+n)/x)}*\pi*I$ so roots are: $e^{+-{(({1/2}+n)/x)}*\pi*I }*e^{+-(2*\pi*I*{k/x)}$, again infinite quantity in totality. n=0 gives: $e^{+-{(({1/2}+n)/x)}*\pi*I}*e^{+-(2*\pi*I*{k/x)}$, If $x=I$ $I^{1/I}=e^{+{(({1/2}+n)/I)}*\pi*I }*e^{+-(2*\pi*I*{k/I)}$, $I^{1/I}=e^{+({1/2}+n)*\pi}*e^{+-(2*\pi*k)}$, For n=0, k=0 $I^{1/I}=e^{+({1/2})*\pi} = e^{\pi/2}$, Something wrong again; I will correct later. Ivars « Next Oldest | Next Newest »

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