why is there no real tetration for b<1?

[bo198214]

Why hasn't this been considered for real extension?


for e^-e<b<1, even though the tetrational graph oscillates, it converges toward the infinite tetrate of b. what is the problem with this oscillation, and what's the problem with having complex values for real arguments?

There is already the very old thread Tetration below 1 which discusses the topic \( e^{-e}<b<1 \).
There also a picture is mentioned, which shows the complex values of the super-exponential to base \( 1/e \) for real arguments.

By extension, why can't you do real/complex tetration for all real bases, or all complex bases (except 0, maybe), for that matter? is the base involved in functions specifically for natural/real numbers in all of the known tetration extensions? or is it just because they would be a little too weird?

with complex exponentiation you have allways the problem of multiplicity. Thatswhy it is very useful to stay at the real axis to limit the number of solutions.

(-10)[4]4 seems to be -10. i suspect because of my calc's errors, it is a little off

No, computed with 2000 bits of precision, (-10)[4]3 deviates from 1 of order \( 10^{-10} \) in real and imaginary part.

for b = -e, after 23 random complex numbers.... (-e)[4]25 = -e (!)

Thats also not true computed with 20000 bits of precision.
(-e)[4]24 deviates from 1 of order \( 10^{-23} \).

for b = -exp(1/e), b[4]10 = b.

also not true. b[4]9 is close to 1, it deviates of order \( 10^{-22} \).

tried it on my calc, and iteration of (-.5)^x on x=1 results in oscillation: 1, -.5, -sqrt(2)i, 47.33965842+70.6208561i, 0, 1, -.5, ...

\( b^{b^{b^b}} \) is in the order of magnitude of \( 10^{-111} \) its not 0 (computed with 200000 bits of precision).