# Tetration Forum

Full Version: Searching for an asymptotic to exp[0.5]
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I think

Let s = exp^[1/2]

For suffic Large positive real x

Exp(X + DC) > S( [1 + C / ( x - AC )] s(x) ) > exp(x + BC).

Were A , B , C and D are real constants.
And A,B,D depend on C.

Also suff Large depends on C too ;

X >> (a^2 + b^2 + c^2)^4

At least i think.

The inspiration and arguments came from fake function theory , hence no new thread.

Regards

Tommy1729
Update on previous post

Im able to prove

For sufficiently Large x >> C :

S( [1 + C/x^2] s(x) ) = exp(x + o(1))

However this was without fake function theory and not so elegant imho.

o is little-o notation here.

Regards

Tommy1729
The master
First i considered fake function theory.
Then An elementary argument.

Those 2 led to the last 2 Posts.

However a simple argument is

S(x [1 + 1/t(x)] ) > s(x) [1 + 1/t(x)].

Omg

So simple.

Probably wrote similar stuff before , so my apologies.

However

S( x [1 + 1/u(x)] ] < s(x) [1 + 1/t(x)]

Remains An issue and the Ideas from the last few posts remain intresting.

Regards

Tommy1729

In the context of TPID 17

1) does this conjecture even hold ??
We have waay to little supporting examples.

We have the sqrt(n) factor in semi-exp and exp but for instance double exponential is not confirmed.

For instance for SUM x^n / n^(n^n).

Im Hoping you guys can help.

--- assuming its true ---

I noticed

Min ( f^2) = min ( f )^2

Yet (1 + f)^2 = 1 + 2f + f^2.

therefore

A_n = min ( f(x)/x^n )

Is improved by b_n = A_n + 2 sqrt A_n.

That is only a minor improvement.

I tried to use the same arguments and as you might have guessed

Replacing ^2 and sqrt by say ^3 and cuberoot.

However it seems to violate ...

I even hesitated to post this because intuition can get you in trouble here.

Clearly there is something missing here.

Yes i know , we approximate the LHS in

P(x) < min f

With a polynomial degree n , so if we take ^(n/2) we end Up with the largest naively possible root.
Because O(x^2)^(n/2) = O(x^n)
We cant ALLOW growth smaller then O (x^2) since this violates the conditions f>0,f ' > 0 , f '' > 0.

But that does not explain enough.

It does show the upperbound factor

< O (n/2)

But that is very close from O ( ln(n) sqrt(n) ).

All intuïtieve logic fails.

However i believe we can repeat the b_n argument and thus arrive at the improved

C_n = a_n + 2 sqrt a_n + 2 sqrt(a_n + 2 sqrt a_n) + ...

~~ a_n + 2 ln(n) sqrt(a_n).

But that is still far from the desired
C ( a_n ln(n+1) sqrt(n) )

Its getting weird , I know.

Regards

Tommy1729
Ok trying to clarity the previous post.

G (x) ~ g(0) + f(x)^2 = g0 + g1 x + ...

(Taylor expansion)

F(x) = f1 x + f2 x^2 + ...

From this we can say gn = f1 f#n-1# + f2 f#n-2# + ... + (f#n/2#)^2.
However since (min [f / x^(n/2) ]) ^2 = min[f^2 / x^n] ,
gn is wrongly estimated as gn ~ (f#n/2#)^2.

Assuming

Ass 1

f#k# f#n-k# < ( f#n/2# )^2

( notice this depends on the convergeance speed of fn ! )

we thus get the improved estimate

g'n ~< (n/2) (f#n/2#)^2.

Hence a correcting factor upper bound : n/2.

---

Well that is the idea.

Handwaving informal and sketchy ... Yes i admit.
But still.

Issues ??

1) n is not even.
2) repeating the argument ... As in (q^2)^2. Leading to arbitrary correcting factors ??
3) similar to 2); replacing ^2 with other functions such as ^5 , also leading to arbitrary correcting factors.

Solutions to 1) 2) and 3) are considered but not formal.

I hope 1) 2) and 3) make clear what i meant with intuition failure.

Also the correcting factor n/2 is far from sqrt ( ln(n) n ).

---

Here i considered the n th Taylor polynomials with sqrt.

Clearly i Cannot meaningfully take anything beyond ^(2/n) since

( X^2 )^(n/2) = x^n.

---

Hope this clarifies a bit.

Sorry for the late reply , but this subject is tricky !

Since Ass 1 depends on converg speed im not even sure that TPID 17 is correct.
On the other hand it holds for semi-exp and exp and sheldon believes in a very similar variant.

I must be missing something.
The importance of the decending chain condition for the derivatives perhaps ??

---

Maybe i know someone who could help us here.

---

Pls inform me if its much simpler then i think but i guess it is not.

Regards

Tommy1729

About issues 2) and 3) and in general ...

Suppose we want the correcting factors c_n for f(x).

Since a_n depends on the truncated fake Taylor polynomial of degree n ,
At best we can take ^(2/n) < as explained before >.

But there is another upperbound.

F(x)^q needs to satisfy the fundamental conditions

D^a [F(x)^q] > 0 for a E (0,1,2).

In particular a = 2.

This places An upper bound on q INDEP of n but DEP on the values and rate of descent of the a_n.

And this is the balance we look for

:

the faster a_n descends , the larger q is.
And vice versa.

There we cannot " repeat the argument " as much as we want , nor choose any m-th root we want ( or other function ).

This gives hope for proving results of type

Correcting factors ~< O ( ( ln(n) n)^gamma ).

gamma ~ 1/2 is then close to a proof of TPID 17.

I call Q = 1/q the power level of f(x).

Guess this clarifies alot.

Im not sure how this relates to sheldon's integrals , Hadamard products and zeration [ min,- algebra ] yet.
Although I have Some Ideas ...

Regards

Tommy1729
However Some functions have a power level of infinity.
And this makes it nontrivial to even decide if this strategy is helpful.

More investigation is neccessary.
Unfortunately i lack time.

For example exp(x) has a power level of infinity.
Exp(x)^a = exp(a x).

The analogue idea of using semi-logarithms instead of sqrt comes to mind.

Another idea is to write

F(x) = exp(a(x)) b(x)
With b(x) growing slower than exp.

Then repeat if necc with a(x) until we get functions a*(x),b(x) that grow slower then exp , and then use the power level tricks on them.

[ this assumes F(x) grows slower then some power tower exp^[k](x). ]

Regards

Tommy1729
To give An example that should work.

F(x) = Sum a_n x^n

With a_n = exp(-n^2)

To compute the fake a_n we consider

F_n(x) = Sum^n a_i x^i

Now we solve

(Alpha x^2)^[b] = a_n x^n

So alpha is close to a_n ^ (log_2(n-1))^(-1).

Or alpha ~ exp( - n^2 / log_2(n-1) )

And b ~ ln(n-1)/ln(2).

--

Max_x Integral_0^x exp(- t^2) exp(- (x-t)^2 ) dt =

Exp(- x^2/2 ) C.

Where C is a constant ( upper bound constant with respect to x ).

Therefore the correcting factor for taking sqrt is C.
And thus the correcting factor for taking r th root is C^log_2{r}.
Or r^( ln{C}\ln(2) ).

From the computation of b we get max{r} ~ 2^b.

Therefore D_n = (n-1)^( ln{C}\ln(2) )

Where D_n is An upper bound on the (final) correcting factor for a_n.

--

So we should have

Exp(- n^2) =< D_n min( f(x) / x^n ).

--

Something like that.

It looks a bit like using convolution and fourrier analysis , but its different.

Regards

Tommy1729

Let f(x) be a real-entire function with all derivatives > 0 and f(0) >= 0.

Let C be the Cauchy constant ( 1/ 2 pi i ).

The taylor coëfficiënts are given by the contour integral



C *

The estimate from fake function theory,
Min (f(x) x^{-n}) can also be given by a contour integral

Let g(x,n) = f(x) / x^n.

then



Min (f(x) x^{-n}) = C *

So the correcting factors are given by

Cor(n) = \ =

So the question becomes to estimate , bound or simplify \.

Not sure how to proceed here.

But now we have a reformulation in terms of more standard calculus ; in terms of (contour) integration.

I call this the " ratio formulation " and TPID 17 can be expressed in it.

Im aware I did not mention alot of related things such as the specification of the contours , numerical methods , Laplace etc etc.

Certainly special cases can be solved but a general idea is missing.

I was able to prove / disprove the expressibility in similar cases , but contour integration is a bit trickier then " ordinary " integrals.

Ideal would be if we could express this ratio as a single contour.

But im not sure if that is possible.

While considering that, the idea of

" contour derivative " \

Comes to mind.

For Some of you - or most - this was already clear I assume.

But for completeness I make this post.

Also Sheldon has similar ideas and I am not sure how exactly they relate ...

Regards

Tommy1729
I think Mick did a good job posting the problem on MSE.