# $\DeclareMathOperator{\arcsinh}{arcsinh}w =\arcsinh( 1 + 2 \arcsinh( 1 + 2^2 \arcsinh ( 1 + 2^{2^2} \arcsinh( 1 + 2^{2^{2^2}} \arcsinh( 1 + \dotsm$

In the context of positive reals consider

$$w= \arcsinh( 1 + 2 \arcsinh( 1 + 2^2 \arcsinh ( 1 + 2^{2^2} \arcsinh( 1 + 2^{2^{2^2}} \arcsinh( 1 + \dotsm$$

Now consider a real $A > w$

Then the iterations

$A_0 = A$

$A_1 = (\sinh(A) - 1)/2$

$A_2 = (\sinh(A_1) - 1)/ 2^2$

$...$

grow superexponentially ! (Tetration)

Reals $B$ in $[0,w[$ do not give a sequence $B_n$ ( same iterations as above ) that grows to infinity.

So the interesting things are

1) the value of $w$

2) How fast does the sequence

$w_0 = w$

$w_1 = (\sinh(w_0) - 1 ) / 2$

$...$

grow ?

Could it be slower than superexponential ?

Consider the analogue

$$3 = \sqrt{1 + 2 \sqrt{ 1 + 3 \sqrt{ 1 + 4 \sqrt {...}}}}$$

Where the iterations $c_1 = 3, c_n = ( c_{n-1}^2 - 1) / n$ give the sequence $3,4,5,6,7,...$ rather than a double exponential growth ( as expected at first ).

Unfortunately tetration-like ideas tend to give Numbers too large for computation, If we are not careful. So we need a trick or some theory probably.

I tried to compute $w$ and arrived at the estimate

$$w = 2.613\,022\,592\,281\,8\!\dots$$

This Number might be wrong but this is my guess. Also the number seems familiar but that may be my imagination.

Is there an efficient way to compute $w$ ?

Are my digits correct ?

And the main question again :

How fast does $w_n$ grow ??

But this is in the context of reals only.

• Sorry, I modified the title after the body, and the macro to do that should have been set at the beginning of the title. It's fixed now. – Bernard Jan 31 '18 at 22:37
• I noticed Thanks – mick Jan 31 '18 at 22:42
• B.t.w., the official name of this function is asinh, or (in old style) Argsinh, or even in very old style, Argsh since sinh was denoted sh. – Bernard Jan 31 '18 at 22:47
• I have seen arcsinh used too. But Thanks for the info. In Dutch it is boogsinh :) – mick Jan 31 '18 at 22:50
• What does boog mean? – Bernard Jan 31 '18 at 22:55