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using sinh(x) ?
#1
im thinking about using asymt of exp(x) to solve tetration.

perhaps not the best one , but sinh(x) comes to my mind.

if we take the half-iterate of sinh(x) by using taylor series , we get a good approximation of the half-iterate of exp(x) for x large.

if this good approximation is analytic in say [e^e,e^e^e^e] we could use that interval and take logs or exp of it to compute the half-iterate for [-oo,+oo] up to a relatively high precision.

in fact , i assume , we can choose our precision if the good approximation is entire , by using bigger intervals and taking logs of it ( as somewhat done above ).

we might even consider the mittag-leffler expansion to avoid problems with non-analytic issues ( thus taking mittag-leffler expansion of the formal taylor series )

thats basicly the idea , but as said maybe sinh(x) isnt the best , on the other hand its probably the easiest.

furthermore some want - and me too actually - that half-iterates have non-negative (2+n)th derivatives and strictly positive zeroth , first and second derivatives.

that last restriction troubles me , especially when i try to get closer to exp(x) than sinh(x) by using function that have non-negative derivatives ... getting conditions on a particular n'th derivate isnt hard but the whole problem is more troublesome ( maybe someone knows a solution to this ? )

i didnt mention that all the above ofcourse is about functions going from R -> R , its obvious , but i just mention it to avoid potential confusion.

also every approximation of exp(x) should equal x at x = 0 only and be larger than id(x).

hope its clear.

maybe you considered this once too ?


greetings my fellow euh ... tetrationalists

regards

tommy1729

ps : showing that all derivatives are positive can sometimes be easy , but in general its hard , see for example " li's criterion " for the RH.
maybe its easy here and i missed a trivial thing ... ( im getting old ? :p )
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#2
(02/28/2010, 12:22 AM)tommy1729 Wrote: im thinking about using asymt of exp(x) to solve tetration.

perhaps not the best one , but sinh(x) comes to my mind.

if we take the half-iterate of sinh(x) by using taylor series , we get a good approximation of the half-iterate of exp(x) for x large.

no replies ...

is there a problem with the half-iterate of sinh(x) ? does its taylor series radius = 0 ?
is the inner radius of the mittag-leffler = 0 ?

...
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#3
Quote:is there a problem with the half-iterate of sinh(x) ? does its taylor series radius = 0 ?
is the inner radius of the mittag-leffler = 0 ?

Its not really a problem, but indeed the convergence radius of the fractional iterates would be 0. This is usually the case with parabolic ( f'(p)=1, f(p)=p ) development. But its not really a problem as you can use the Levy iteration formula (which though may be too slow) or the Ecalle formulas for parabolic iteration.
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#4
ok , i found my silly mistake !!

i wrote sinh(x) but i meant 2*sinh(x) !!

so in the OP replace sinh(x) with 2*sinh(x).

and then we dont have a parabolic fixpoint !

and we do have an asympt.

sorry !

regards

tommy1729
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#5
(03/09/2010, 01:31 PM)tommy1729 Wrote: so in the OP replace sinh(x) with 2*sinh(x).

and then we dont have a parabolic fixpoint !

and we do have an asympt.

And what do you do with this asymptote then?
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#6
(03/10/2010, 11:38 AM)bo198214 Wrote:
(03/09/2010, 01:31 PM)tommy1729 Wrote: so in the OP replace sinh(x) with 2*sinh(x).

and then we dont have a parabolic fixpoint !

and we do have an asympt.

And what do you do with this asymptote then?

i take the half-iterate of it by using taylor series expanded at 0.

f(x) is a taylor series , f(f(x)) = 2*sinh(x).

f(x) is then a good asympt 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))

something like that i think.

regards

tommy1729
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#7
(03/11/2010, 12:50 AM)tommy1729 Wrote:
(03/10/2010, 11:38 AM)bo198214 Wrote:
(03/09/2010, 01:31 PM)tommy1729 Wrote: so in the OP replace sinh(x) with 2*sinh(x).

and then we dont have a parabolic fixpoint !

and we do have an asympt.

And what do you do with this asymptote then?

i take the half-iterate of it by using taylor series expanded at 0.

f(x) is a taylor series , f(f(x)) = 2*sinh(x).

f(x) is then a good asympt 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))

something like that i think.

regards

tommy1729

all clear ?
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#8
i think it works for the real line.

however in the neighbourhood of the 2 fixed points of exp(x) it will fail.

i think so because for fixed point y :

log log f exp exp y => log log f(y) => equals y only if f(y) = y

so i assume a "button-like" region of correctness.
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#9
Actually Paul Lévy [1] showed how to obtain an iteration of e^x if we have an iteration of e^x-1.
Say is an Abel function of e^x-1, then



is an Abel function of . This should also work for beta being the Abel function of .

This approach is actually equivalent to the "change of base" approach we considered here on the forum, also Walker [2] used a similar method. (But I am in the moment to lazy to detail how exactly they imply each other.) It is still open whether it is analytic, but it is proven to be infinitely differentiable in [2].

[1] Lévy, P. (1927). Sur l'itération de la fonction exponentielle. C. R., 184, 500–502.
[2] Walker, P. L. (1991). Infinitely differentiable generalized logarithmic and exponential functions. Math. Comput., 57(196), 723–733.
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#10
(05/31/2010, 05:52 AM)bo198214 Wrote: Actually Paul Lévy [1] showed how to obtain an iteration of e^x if we have an iteration of e^x-1.
Say is an Abel function of e^x-1, then



is an Abel function of . This should also work for beta being the Abel function of .

This approach is actually equivalent to the "change of base" approach we considered here on the forum, also Walker [2] used a similar method. (But I am in the moment to lazy to detail how exactly they imply each other.) It is still open whether it is analytic, but it is proven to be infinitely differentiable in [2].

[1] Lévy, P. (1927). Sur l'itération de la fonction exponentielle. C. R., 184, 500–502.
[2] Walker, P. L. (1991). Infinitely differentiable generalized logarithmic and exponential functions. Math. Comput., 57(196), 723–733.

but is that equivalent to my solution ?
do we get the same half-iterate for exp(x) ?

thanks for the post and references.

regards

tommy1729
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