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+--- Thread: Deriving tetration from selfroot? (/showthread.php?tid=129)
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RE: Deriving tetration from selfroot? - Ivars - 03/24/2008
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
RE: Deriving tetration from selfroot? - Ivars - 03/25/2008
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