We consider 2 * sinh(x).
We take the half-iterate of it by using taylor series expanded at 0.
Thus f(x) is a taylor series , f(f(x)) = 2*sinh(x).
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To do so we use something equivalent to the koenigs function. But we prefer something that is more suitable for taking derivatives, doing calculus and analysis etc.
Koenigs function is analytic and it works, at least it converges numerically to an analytic solution.
See https://en.wikipedia.org/wiki/Koenigs_function
Many equivalent fixpoint methods exist that accelerate the convergeance and is a more suited for taking some derivatives.
In particular setting up equations for the taylor coefficients or using not just the first linear approximation at the fixpoint but a polynomial approximation ( the taylor polynomial at 0 of some higher degree ).
But again it is equivalent to the Koenigs function , NOT another solution. So it has the same nice properties.
Details is a subtopic in dynamical systems and beyond this topic.
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Clearly f(x) is then a good asymptote for the half iterate of exp(x) for large x.
Let g(g(x)) = exp(x)
Then g(x) =
Lim k-> oo log log log log ...( k times ) (f( e^e^e^...( k times ) ^ x))
Clearly this can be generalized to tetration.
For a real 'r' to compute exp^[r] ( ^[] is iteration ) replace f(x) with (2*sinh(x))^[r] in the above formula.
Now you have ( for x and r being real ) exp^[r](x) aka tetration. It turns out this function is continuous and infinitely differentiable on the real axis (C^oo for fixed r and variable x). And it is probably analytic.
By analogue for a realvalued b we can compute (b^(x))^[r] and find the same properties as long as b > exp(1/2).
All of this has been formally proven for b > exp(2) but Im working on the smaller bases too.
( One of the main reasons is that for b > exp(2) , 2*sinh(2x) has only one fixpoint (=0) whereas for exp(1/2) < b < exp(2) you have 3 fixpoints : zero and two nonreal complex conjugates )
Can this method be improved ?
YES !
*** The exp(x) - exp(-3/5 x) method ***
Using f(x) = exp(x) - exp(-3/5 x) instead of 2sinh for the 2sinh method.
You might wonder why.
well the higher derivatives go into agreement with exp(x) for re(x) large unlike with 2sinh(x).
We also get the fixpoint at 0 and the function looks to go to exp in a more smooth way.
All the derivatives are positive at the origin and for x > 0.
This resembles exp(x) better thus in a way.
The solution is probably analytic and certainly well-behaved.
It also satisfies the semi-group isomo property. In fact it satisfies a monoid isomo property.
More precisely we have exp[a+b](x) = exp[a](exp[b](x)) = exp[b](exp[a](x)) for all real a,b > 0.
That property is a uniqueness condition !!
Notice that
exp(x) - exp( - 1/3 x) does NOT work since it has a fixpoint at some negative x.
This method is thus a consensus and compromise between using the fixpoint at 0 for exp(x) - 1 and 2sinh(x).
exp(x) - 1 is for base exp(1/e) however and is parabolic, so this method is more 2sinh type.
Notice that although this method is considered an improvement it is probably exactly the same function as with the 2sinh method. This follows from the consideration that the 2sinh method also probably satisfies the monoid isomo property which is afterall a uniqueness condition. But it has benefits numerically and in terms of calculus.
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Strongly related is the section " fake function theory ".
Also check out the sublinks.
***
All the above is for tetration base e. Other bases and ideas are being investigated.
Below is a link to my profile at the tetration forum.
At the moment I only accept the tetration forum as a good source for tetration although you can find some info and links on wiki , I do however not accept the wiki page.
http://math.eretrandre.org/tetrationforum/member.php?action=profile&uid=47
or
https://tetrationforum.org/member.php?action=profile&uid=47
(also posted at https://tetrationforum.org/showthread.php?tid=1750 on 05/16/2023, 11:30 PM )
Thank you for your intrest.
tommy1729
update :
another way is this one :
f(x) = exp(x) - exp(-3/5 x)
f^[z](x) = lim f^[m]( ( (exp(x) - exp(-3/5 x))/x )^z f^[-m](x) )
( notice the signs of m are switched because f(x) is repelling here. The inverse of f(x) has no simple closed form hence the reason )
All in all it is not all that different, but the viewpoint and insight makes all the difference.
Keep in mind to use some common sense just as with the other methods.
For instance if x is large and m is small and z is large , this probably does not give a good approximation yet and requires more work and/or iterations and/or analytic continuation.
As a consequence I write :
exp^[z](x) = lim ln^[m] ( f^[2m]( ( (exp(x) - exp(-3/5 x))/x )^z f^[-2m](exp^[m](x)) ) )
For more standard functions this method also works for instance :
f(x) = x/( 1 + ln(1+x) )
f^[z](x) = lim f^[-m]( (1 + ln(1+x))^z f^[m](x) )