Deriving tetration from selfroot?
#11
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
#12
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




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