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Searching for an asymptotic to exp[0.5]
Back to basics

In addition to post 17,18 notice that

D exp^[1/2](x) = D exp^[1/2] (exp(x) ) * exp(x) / exp^[1/2](exp(x)).

That follows from ln exp^[a] ( exp(x) ) = exp^[a](x) and the chain rule for derivatives.

By induction / recursion this gives a nice way ( product ) to compute the derivative.

This strenghtens the conclusions from post 17 , 18 and shows that

1 + o(1) <<_n 2.

( smaller after only a few iterations n )

We conclude by noting that the Taylor T

T = Sum_{K=4}^{oo} d_k x^k

With d_k = exp( - k^2 )

grows slower then exp^[1/2](x) , yet faster then any polynomial.

T has growth 0 , like exp( ln^2 (x) ) and similar.

Did we meet T yet ?? I believe T was a fake of Some elementairy like exp( ln^2 ) or such ...
This again leads to the desire of inverse fake or its related integral transforms ...

---

Regards

Tommy1729


Reply
I request complex plots of f(x) = fake exp^[1/3](x) , f(f(x)) and f(f(f(x))).

Like sheldon did for fake exp^[1/2](x) in one of the early posts in this thread.

It is very important !!

( potentially new results/conjectures based on those plots ! )

Regards

Tommy1729

Ps make sure to make backups of this website/content ? Bo ?
Reply
Im currently into this 

f(x) = integral from 1 to +oo [ t^x g(t) dt ]

and the related

f(x) = a_0 + a_1* 2^x + a_2* 3^x + a_3 4^x + ...

( for suitable f(x) )

This is ofcourse similar to finding taylor series and fake function theory ( so far ).

The idea floating around of finding approximate entire dirichlet series with a_n >=0 is ofcourse tempting.

However maybe the inequality

a_n < min ( f(x)/n^x )

might be less efficient here ??

What do you guys think ?

NOTICE the integral transform is NOT the mellin transform.

Anyone knows an inverse integral transform for this ?

I think f(x) = gamma(x,1) + (constant) might make an interesting case ...

regards

tommy1729
Reply
(07/18/2021, 11:47 PM)tommy1729 Wrote: Im currently into this 

f(x) = integral from 1 to +oo [ t^x g(t) dt ]

and the related

f(x) = a_0 + a_1* 2^x + a_2* 3^x + a_3 4^x + ...

( for suitable f(x) )

This is ofcourse similar to finding taylor series and fake function theory ( so far ).

The idea floating around of finding approximate entire dirichlet series with a_n >=0 is ofcourse tempting.

However maybe the inequality

a_n < min ( f(x)/n^x )

might be less efficient here ??

What do you guys think ?

NOTICE the integral transform is NOT the mellin transform.

Anyone knows an inverse integral transform for this ?

I think f(x) = gamma(x,1) + (constant) might make an interesting case ...

regards

tommy1729

As a small example :

 integral from 1 to +oo [ t^x g(t) dt ]

with g(t) = exp(- ln(t)^2 ) 

equals :

(1/2) * ( erf((x+1)/2) +1) *  sqrt(pi) * exp( (1/4)* (x+1)^2 ).

I find this fascinating.

btw g(t) reminds me again of the binary partition function.

Differentiation under the integral sign (dx) will probably be a usefull trick too.

More math must exist.

regards

tommy1729
Reply
(07/21/2021, 05:11 PM)tommy1729 Wrote:
(07/18/2021, 11:47 PM)tommy1729 Wrote: Im currently into this 

f(x) = integral from 1 to +oo [ t^x g(t) dt ]

and the related

f(x) = a_0 + a_1* 2^x + a_2* 3^x + a_3 4^x + ...

( for suitable f(x) )

This is ofcourse similar to finding taylor series and fake function theory ( so far ).

The idea floating around of finding approximate entire dirichlet series with a_n >=0 is ofcourse tempting.

However maybe the inequality

a_n < min ( f(x)/n^x )

might be less efficient here ??

What do you guys think ?

NOTICE the integral transform is NOT the mellin transform.

Anyone knows an inverse integral transform for this ?

I think f(x) = gamma(x,1) + (constant) might make an interesting case ...

regards

tommy1729

As a small example :

 integral from 1 to +oo [ t^x g(t) dt ]

with g(t) = exp(- ln(t)^2 ) 

equals :

(1/2) * ( erf((x+1)/2) +1) *  sqrt(pi) * exp( (1/4)* (x+1)^2 ).

I find this fascinating.

btw g(t) reminds me again of the binary partition function.

Differentiation under the integral sign (dx) will probably be a usefull trick too.

More math must exist.

regards

tommy1729

Ofcourse im not trying directly to solve the integral transform for exp^[0.5].
What I am trying here is to solve for standard functions ... and then use them.

As in TT(exp^[0.5](s)) = TT (sum over standard functions) = sum over TT (standard functions)

Where TT stands for " tommy-transform " the integral transform mentioned above.

If this is the best strategy , I do not know.  It just felt natural.

regards

tommy1729
Reply


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