We can bring this a little further by taking x=phi=1,61803399...Golden Mean and using the property that 1/phi = phi-1
BTW interesting to note that multiplication of true and inverse self roots of phi (This is not generally true for other numbers):
(phi^(1/phi))*((1/phi)^phi))= 1/phi = phi-1= h((phi)^(1/phi)) = 0,61803399 . One more interesting selfroot is:
(i^(1/i))*((1/i)^i))= (i^(1/i))^2=(e^(pi/2))^2=e^(pi)=23,14069263
so:
h((I/phi)^(phi/I)) = h(I*(1/phi)^(phi/I))=h((I*(phi-1))^(phi/I)) = phi/I = -I*phi=-((1/phi)+1)*I = -((I/phi) +I)
If we multiply/divide by 2, since I/2=sin (I*ln(phi)) we get:
h((I/phi)^(phi/I))) = (phi/2)/(I/2) = (phi/2)/(sin (I*ln(phi)))
Also:
h((1/phi)^phi) = h(phi-1)^phi) = phi = 1,61803399
BTW interesting to note that multiplication of true and inverse self roots of phi (This is not generally true for other numbers):
(phi^(1/phi))*((1/phi)^phi))= 1/phi = phi-1= h((phi)^(1/phi)) = 0,61803399 . One more interesting selfroot is:
(i^(1/i))*((1/i)^i))= (i^(1/i))^2=(e^(pi/2))^2=e^(pi)=23,14069263
so:
h((I/phi)^(phi/I)) = h(I*(1/phi)^(phi/I))=h((I*(phi-1))^(phi/I)) = phi/I = -I*phi=-((1/phi)+1)*I = -((I/phi) +I)
If we multiply/divide by 2, since I/2=sin (I*ln(phi)) we get:
h((I/phi)^(phi/I))) = (phi/2)/(I/2) = (phi/2)/(sin (I*ln(phi)))
Also:
h((1/phi)^phi) = h(phi-1)^phi) = phi = 1,61803399

