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Searching for an asymptotic to exp[0.5]
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
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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
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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

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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
Reply
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

Reply
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
Reply
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
Reply
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

Reply
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

[1]


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

[2]

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


So the correcting factors are given by

Cor(n) = [1]\[2] =




So the question becomes to estimate , bound or simplify [1]\[2].

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 " [1]\[2]

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


Here is the link :

http://math.stackexchange.com/questions/...r-dx-n-n-0

No votes or reactions yet.

Regards

Tommy1729
Reply


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