some comments 2021

[tommy1729]

consider y = ln ln ln ... a1^a2^a3^ ... ^x.

Then when we put that into an equation we get

exp(exp(exp(...y) = a1^a2^a3^...^x.

We know both sides are nonzero so we can take ln on both sides

exp(exp(...y) = ln(a1)* a2^a3^...^x.

So what if ln(a1)*a2^a3^...^x = 1 ??

Lets assume y exists again.
Then the Lhs again has a log. AND THAT LOG IS NOT 0.
So we get a different branch.

So solving this equation is just a matter of taking the correct branches all the time ???

THAT seems too optimistic !?

Keep in mind that choosing the branches must be consistant with analyticity as well !

Taking ln one more time :

exp(...y) = ln(ln(a1)) + ln(a2) * a3^...^x.

How do we know y exists again ??

Here is a toy idea :

ln(ln(a1)) + ln(a2)*a3^..^x = a3^..^x'

This switch from x to x' makes us able to repeat the above.

Ofcourse this assumes an x' exists !!

Im aware I mentioned 2 things in the previous post about simplifying a log (analytic continuation , lower multiplicity ) and boundaries, but Im still confused by this.

For real numbers this all works nicely. 
But for complex Im confused.

So I started thinking.

We could go the other direction ; THE INVERSE WAY ;

LET f(x) be a given function with a taylor series.

Now for a chosen set a_n we can inductively define :

f(x)  = a_0 + exp( a_1 + exp( a_2 + exp( a_3 + ...

As long as it converges and as long as f(x) - a_0 , ln(f(x) - a_0) - a_1,... are nonzero.

This is in a sense the analogue of taylor series , infinite compositions and power towers.

So I think there must be a way to make this all formal.

regards

tommy1729

Tom Marcel Raes