using sinh(x) ?

[tommy1729]

Look.

Let f(z') = e - 0,2.
F(z') ~ exp^[1/2] ~ 2sinh^[1/2]
So 1 < z ' < e - 0,2.

Within radius 0,1 around z = z' :

| exp(f(z)) - f(exp(z)) | =< e^e

And

| exp(exp(f(z))) - f(exp(exp(z))) | =< e^e^e.


So

Ln | exp(f(z)) - f(exp(z)) | =< e.

And

Ln ln | exp( exp(f(z)) ) - f(exp(exp(z))) | =< e.

By induction we get

+1+

Ln^[n] | exp^[n] (f(z)) - f(exp^[n](z)) | < e + 1.

Therefore the m-test gives for +1+ : | m_1 | + | m_2 | + ... < e + 1.

Notice if | f(exp^[n](z)) | >= | exp^[n](f(z)) |
Then

Ln^[n] | 2 f(exp^[n](z)) | < e + 1 + ln(2).

So in this case the m-test for 2 sinh gives An upper bound :

| s1 | + |s2| + ... < e + 1 + ln(2).

So the m-test has passed it as analytic.


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Easy to generalize to e' > e and f(z '' ) = ~ e'.
And a radius 0,1 / e^(e^e').

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Easy to show that it holds at infinity too.

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Regards

Tommy1729