Rational operators (a {t} b); a,b > e solved

[tommy1729]

So, now, I'm going to define the function f(b), which returns the base which has the fixed point of "b".
\( f(e)=\eta \), since e is the fixed point of b=\( \eta \)
\( f(2)=\sqrt{2} \), since 2 is the lower fixed point of b=sqrt(2)
\( f(3)=1.442249570 \), since 3 is the upper fixed point of this base
\( f(4)=\sqrt{2} \), since 4 is the upper fixed point of b=sqrt(2)
\( f(5)=1.379729661 \), since 5 is the upper fixed point of this base
I don't know how to calculate the base from the fixed point, but that's the function we need, and we would like the function to be analytic!

Oh, Sheldon seems to be quite tired from all the calculation and discussion.
Sheldon, give your self some time to rest!
Your function is \( f(z)=z^{1/z} \)

About those fixpoints and bases.

Bo is correct ofcourse but i wanted to add how to find the other fixpoint.

I end with a joke because i did NOT show which T is the correct one : it is the SMALLEST > 1 to be precise ...
Analytic continuation is hard for a proof of that minimal property. The convex nature and fast growth of exp type functions is intuitive but also IMHO unconvincing / informal / weak.
So no satisfying proof. Perhaps food for thought.

( certainly possible ! )

Anyway here it is ( and T = t )


X^1/x = y^1/y

Ln(x)/ x = ln(y)/y

Ln(x)/x = ln(T x)/ T x

T ^ 1/T x^1/T = x

T x = x^t

X^(t - 1) = t

Ofcourse new similar problem

t1 ^ (1/(t1 - 1)) = t2 ^ (1/(t2 - 1))

Regards

Tommy1729
The master