Hold on we can do a bit better.
For base sqrt(2) = a using sheldon's y and z.
a^^0 is defined by
a^^0 ^ a^^0 = 2^4 = 4^2 = 16.
( 2 and 4 are a^^oo or the fixpoints )
So a^^0 = ssqrt(16).
From sheldon we then get
a^(y z) = 16.
subst a = sqrt(2) [ for example , this idea works for other bases too ]
=> y z = 8.
Now notice r = -(r+1) for r = -1/2.
So a^^(-1/2) ^ 2 = 8.
So a^^(-1/2) = sqrt( 8 ) = 2 sqrt(2).
Since we have a^^0 and a^^(-1/2) we actually have
a^^(K/2) for all integer K.
Maybe this idea can be extended to a unique analytic solution ...
¯\_(ツ)_/¯
Regards
tommy1729
The master
For base sqrt(2) = a using sheldon's y and z.
a^^0 is defined by
a^^0 ^ a^^0 = 2^4 = 4^2 = 16.
( 2 and 4 are a^^oo or the fixpoints )
So a^^0 = ssqrt(16).
From sheldon we then get
a^(y z) = 16.
subst a = sqrt(2) [ for example , this idea works for other bases too ]
=> y z = 8.
Now notice r = -(r+1) for r = -1/2.
So a^^(-1/2) ^ 2 = 8.
So a^^(-1/2) = sqrt( 8 ) = 2 sqrt(2).
Since we have a^^0 and a^^(-1/2) we actually have
a^^(K/2) for all integer K.
Maybe this idea can be extended to a unique analytic solution ...
¯\_(ツ)_/¯
Regards
tommy1729
The master
