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Searching for an asymptotic to exp[0.5]
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.

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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.

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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

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Messages In This Thread
RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 09/03/2015, 10:31 PM

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