using sinh(x) ?

[tommy1729]

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 )

[tommy1729]

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 ?

...

[bo198214]

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.

[tommy1729]

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

[bo198214]

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?

[tommy1729]

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

[tommy1729]

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 ?

[tommy1729]

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.

[bo198214]

Actually Paul Lévy [1] showed how to obtain an iteration of e^x if we have an iteration of e^x-1.
Say \( \beta \) is an Abel function of e^x-1, then

\( \alpha(x)-\alpha(x_0)=\lim_{n\to\infty} \beta(\exp^{[n]}(x)) - \beta(\exp^{[n]}(x_0)) \)

is an Abel function of \( e^x \). This should also work for beta being the Abel function of \( 2\sinh(x) \).

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.

[tommy1729]

Actually Paul Lévy [1] showed how to obtain an iteration of e^x if we have an iteration of e^x-1.
Say \( \beta \) is an Abel function of e^x-1, then

\( \alpha(x)-\alpha(x_0)=\lim_{n\to\infty} \beta(\exp^{[n]}(x)) - \beta(\exp^{[n]}(x_0)) \)

is an Abel function of \( e^x \). This should also work for beta being the Abel function of \( 2\sinh(x) \).

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