- #1
Ryuzaki
- 46
- 0
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?origin=publication_detail
[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?origin=publication_detail