Timeline for Smooth out numerical fluctuation at known point
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 17, 2022 at 23:24 | history | became hot network question | |||
| Oct 17, 2022 at 16:01 | vote | accept | Frotaur | ||
| Oct 17, 2022 at 15:48 | answer | added | user293787 | timeline score: 2 | |
| Oct 17, 2022 at 15:41 | answer | added | Bob Hanlon | timeline score: 3 | |
| Oct 17, 2022 at 15:40 | comment | added | Frotaur | Indeed, that seems to be the case. May I ask how you arrived at this value ? Indeed my function is actually dependent on several parameters, so I would need to do that systematically (Edit : I found it, it is simply simplify. I will try that out if it works generally) | |
| Oct 17, 2022 at 15:35 | comment | added | user293787 | Your expression is equal to (1690 (20-13 s)^2)/(100+33 s)*1/(20 Sqrt[77+70 s]-26 s Sqrt[77+70 s]+7 Sqrt[1000+330 s])^2 if I did not make a mistake. In this way of writing it, there is no cancellation between -1 and 0. | |
| Oct 17, 2022 at 15:29 | comment | added | Frotaur | I tried, but that doesn't help. Indeed, the cancelling of the divergence is not a simple one such as $x^2/x$. While the two pieces of the functions diverge in the same and opposite way at $sp$, these factors differ away from sp. In other words, only the first (singular) term in an expansion around $sp$ cancels, but this does not hold for all s. (Original message was deleted, it was suggesting the use of "Together". I will leave my comment as I think it is an important clarification). | |
| Oct 17, 2022 at 15:20 | history | asked | Frotaur | CC BY-SA 4.0 |