Thread Rating:
  • 2 Vote(s) - 4 Average
  • 1
  • 2
  • 3
  • 4
  • 5
Searching for an asymptotic to exp[0.5]
#87
(09/08/2014, 12:56 AM)tommy1729 Wrote: Dear Sheldon

I read your answer @stackexchange.

I assume your sequences are the derivatives and not the coefficients ?
Otherwise it seems divergent.

I intended to tell mick about the integral of (1-exp(x))/x to get a fake ln(x) or fake 1/x.

---
And then I started wondering about x fake(1/x) , exp(fake(ln(x)) - x etc.
---

I mentioned (1-exp(x))/x before.
However apparantly that was already given a long time at mathoverflow.

(as your link proves).

Its easy to get a fake sqrt from a fake ln.

So the main questions are these :

1) which is the best fake ln(x) ?
integral (1-exp(-x)/x) - C or C_2 * +fake(ln(x) exp(x)) exp(-x) ?

( I use +fake to refer to "our" method of fake functions ; with the positive derivatives )

2) Is there an analogue for the integral of (1 - exp(-x))/x to create a fake sqrt(x) ?
( I do NOT mean exp(fake(ln(x)/2) )

Answers to these will lead to new questions.

I want to remark that +fake(sqrt(x) exp(x)) exp(-x) is very likely a better approximation then exp(+fake(ln(x) exp(x)) exp(-x) / 2).

I also notice we can add variables to our ideas :

1) which is the best fake ln(x) ?
integral(1-exp(- A x)/x) - C or C_2 *+fake(ln(x) exp(B x)) exp(-B x) ?

Which leads to more intresting questions.

Anyways the most intresting is imho :

2) Is there an analogue for the integral of (1 - exp(-x))/x to create a fake sqrt(x) ?


regards

tommy1729

When I printed the numbers, they needed very very high precision for the entiref version, so the exponent isn't visible unless you scroll to the right. But it will converge, but computations are tricky with the entiref version due to precision loss as large numbers cancel each other out to generate the very small ln(x+1) assymptotic, so it is much more straightforward to work with the version, where entiref=f(x)exp(-x).

For exp^{0.5} the method with all positive derivatives woud be much much more compact, with very few Taylor series coefficients required to represent the function, and converges pretty well. If you generate a representation instead using exp(x)exp^{0.5)(x), then it will converge faster, but requires far more terms to do so, with something like 2n terms required for f(n). Also, it will only converge on the right hand side of the complex plane, whereas the all positive derivative version converges everywhere, and even at the negative real axis, it oscillates between zero and approximately 2x exp^{0.5}(x).
- Sheldon
Reply


Messages In This Thread
RE: Searching for an asymptotic to exp[0.5] - by sheldonison - 09/08/2014, 04:15 AM

Possibly Related Threads...
Thread Author Replies Views Last Post
  Merged fixpoints of 2 iterates ? Asymptotic ? [2019] tommy1729 1 586 09/10/2019, 11:28 AM
Last Post: sheldonison
  Another asymptotic development, similar to 2sinh method JmsNxn 0 2,558 07/05/2011, 06:34 PM
Last Post: JmsNxn



Users browsing this thread: 1 Guest(s)