05/14/2014, 11:42 PM

Ok after spending some time looking at the posts I take back most of my criticism. I did not have much time too think or read about this lately.

Btw sheldon edited his posts a few times, which makes them better naturally.

If we are talking in the context of growth approximation , rather than approximation of exp^[1/2] then it turn out my ideas and sheldons are similar and consistant with eachother and by themselves.

the tommy sheldon conjecture

1/a_n = O( exp(n * (ln^[1/2](n))^B ) )

For some real B > 0.

The conjecture made by sheldon that motivates his use of taking the derivative is true.

Hence h_n is justified.

The main difference between sheldon and me is then that i take ln^[1/2] or its " fake " and sheldon uses the h_n.

Although the h_n is more accurate ( since his derivative was justified ) it does not say how fast h_n " grows ".

So its basicly

ln[1/2](n) VS h_n

( I do not yet understand where his second derivative comes from , is this the same logic that goes from a_n to b_n ?? Also his integrals need more study and/or explaination. Forgive me if im a bit behind )

Some remarks first :

Its not so clear that

growth ( f(x) ) <=> growth ( exp^[1/2] ) <=> growth ( a_n )

is an equivalence relationship.

Just to avoid jumping into conclusions.

More importantly , we have encountered h_n before on this forum !

I do not know exactly where and how but I remember it.

That might be usefull.

Another remark of yet to determine value :

We could use the difference operator too :

a_n < exp( exp^[0.5](t_n) - n t_n )

where t_n is the inverse of exp^[1/2](x) - exp[1/2](x-1).

---

I Consider it important we only looked at

a_n < stuff

But a good boundary means

stuff < a_n < stuff

and the stuff < a_n is not adressed yet ...

---

BUT LETS CONTINUE :

ln[1/2](n) VS h_n

take the inverse on both sides

exp^[1/2](n) VS D exp^[1/2](n)

Now the key is too notice :

(exp^[1/2](n))^(1-o(1)) < D exp^[1/2](n) < (exp^[1/2](n))^(1+o(1))

Hence 50 % of the tommy sheldon conjecture has been proven, more specific :

1/a_n < O( exp(n * (ln^[1/2](n))^(1+o(1)) ) )

( B = 1 for the < part )

...

So it comes down (in part) to finding asymptotics to D exp^[1/2](n).

We know D exp^[1/2](n) = exp^[1/2](n) / ( n^( 1-o(1) ) )

from which the above follows easily.

regards

tommy1729

Btw sheldon edited his posts a few times, which makes them better naturally.

If we are talking in the context of growth approximation , rather than approximation of exp^[1/2] then it turn out my ideas and sheldons are similar and consistant with eachother and by themselves.

the tommy sheldon conjecture

1/a_n = O( exp(n * (ln^[1/2](n))^B ) )

For some real B > 0.

The conjecture made by sheldon that motivates his use of taking the derivative is true.

Hence h_n is justified.

The main difference between sheldon and me is then that i take ln^[1/2] or its " fake " and sheldon uses the h_n.

Although the h_n is more accurate ( since his derivative was justified ) it does not say how fast h_n " grows ".

So its basicly

ln[1/2](n) VS h_n

( I do not yet understand where his second derivative comes from , is this the same logic that goes from a_n to b_n ?? Also his integrals need more study and/or explaination. Forgive me if im a bit behind )

Some remarks first :

Its not so clear that

growth ( f(x) ) <=> growth ( exp^[1/2] ) <=> growth ( a_n )

is an equivalence relationship.

Just to avoid jumping into conclusions.

More importantly , we have encountered h_n before on this forum !

I do not know exactly where and how but I remember it.

That might be usefull.

Another remark of yet to determine value :

We could use the difference operator too :

a_n < exp( exp^[0.5](t_n) - n t_n )

where t_n is the inverse of exp^[1/2](x) - exp[1/2](x-1).

---

I Consider it important we only looked at

a_n < stuff

But a good boundary means

stuff < a_n < stuff

and the stuff < a_n is not adressed yet ...

---

BUT LETS CONTINUE :

ln[1/2](n) VS h_n

take the inverse on both sides

exp^[1/2](n) VS D exp^[1/2](n)

Now the key is too notice :

(exp^[1/2](n))^(1-o(1)) < D exp^[1/2](n) < (exp^[1/2](n))^(1+o(1))

Hence 50 % of the tommy sheldon conjecture has been proven, more specific :

1/a_n < O( exp(n * (ln^[1/2](n))^(1+o(1)) ) )

( B = 1 for the < part )

...

So it comes down (in part) to finding asymptotics to D exp^[1/2](n).

We know D exp^[1/2](n) = exp^[1/2](n) / ( n^( 1-o(1) ) )

from which the above follows easily.

regards

tommy1729