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Help understanding this approximation

by Ryuzaki
Tags: approximation
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Ryuzaki
#1
Jun19-14, 12:27 PM
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In a paper that I'm reading, the authors write:-

[itex]N_e \approx \frac{3}{4} (e^{-y}+y)-1.04[/itex] ------------ [itex](4.31)[/itex]

Now, an analytic approximation can be obtained by using the expansion with respect to the inverse number of "e-foldings" ([itex]N_e[/itex] is the number of "e-foldings"). For instance, eq. [itex](4.31)[/itex] yields:-

[itex]e^y = \dfrac{3}{4N_e} - \dfrac{9ln(N_e)}{16(N_e)^2} -\dfrac{0.94}{(N_e)^2} + O(\dfrac{ln^2(N_e)}{(N_e)^3})[/itex]

Can anyone tell me how this approximation is done? I'm not familiar with the $O$ notation either. What does it mean? How do the authors arrive at that expression?

If anyone should require it, the original paper can be found here: https://arxiv.org/pdf/1001.5118.pdf?...ication_detail
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pasmith
#2
Jun19-14, 01:35 PM
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Quote Quote by Ryuzaki View Post
In a paper that I'm reading, the authors write:-

[itex]N_e \approx \frac{3}{4} (e^{-y}+y)-1.04[/itex] ------------ [itex](4.31)[/itex]

Now, an analytic approximation can be obtained by using the expansion with respect to the inverse number of "e-foldings" ([itex]N_e[/itex] is the number of "e-foldings"). For instance, eq. [itex](4.31)[/itex] yields:-

[itex]e^y = \dfrac{3}{4N_e} - \dfrac{9ln(N_e)}{16(N_e)^2} -\dfrac{0.94}{(N_e)^2} + O(\dfrac{ln^2(N_e)}{(N_e)^3})[/itex]

Can anyone tell me how this approximation is done?

It's an asymptotic expansion. Finding these is more of an art than a science. Hinch is a good introduction.

I'm not familiar with the $O$ notation either. What does it mean?
See http://en.wikipedia.org/wiki/Big_O_notation.


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