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
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