(06/28/2014, 11:16 PM)tommy1729 Wrote: I wanted to remark that the "fake function" idea is generalizable.
We can find any entire asymptotic with nonnegative derivatives by simply using a rising integer function T(n) :
\( f(x) = \sum_{n = 1}^{\infty} a_{T(n)} x^{T(n)} \; < \; \exp^{0.5}(x) \)
\( \forall {x \gt 0 } \; \; a_{T(n)} x^{T(n)} \; < \; \exp^{0.5}(x) \)
\( \forall {x \gt 0 } \; \; \log(a_{T(n)} x^{T(n)}) \; < \; \log(\exp^{0.5}(x)) \)
\( \forall {x \gt 0 } \; \; \log(a_{T(n)}) + {T(n)}\log(x) \; < \; \log(\exp^{0.5}(x)) \)
etc.
This simple idea / equation is very powerfull.
The fundamental theorem of fake function theory.
In combination with the post about the inverse gamma function we could for instance find the asymptotic :
\( f(x) = \sum_{n = 1}^{\infty} \frac{x^{T(n)}} {T(n) !} \)
with the method above.
To stay in the spirit of the exp.
regards
tommy1729
To combine 2 ideas of me we can use this to find an alternative asymptotic half-iterate F(z) of exp(z) such that F(z) = F(-z).
Or even the additional F^[2](z) = F^[2](-z) if we take F(0) = 0.
We do that by defining F(z) = a0 + a1 x + a3 x^3 + a5 x^5 + ...
In other words we take T(n) = 2n + 1 and then solve like described above.
The main difference with 2sinh^[0.5](z) is that we get a different region of almost agreement with exp(z) or even 2sinh(z) after 2 iterations.
Also , though related , notice the derivatives of 2sinh^[0.5](z) are not all positive.
Btw we should also get the similar hadamard product for this F(z) as well as for 2sinh^[0.5].
But how do the plots look like ? And where are the zero's ?
This requires some additional work.
An intresting lemma for the 2sinh if true would be this :
A := 2 pi
2sinh^[0.5](z) = a0 + a1 z + a2 z^2 + ...
g(z) := |a0| + |a1| z + |a2| z^2 + ...
Conjecture : |g(z)| < 1 + A 2sinh^[0.5](|z|)^A
Stronger and weaker variants are also of intrest.
regards
tommy1729
