Timeline for Why use Fourier series instead of Taylor?
Current License: CC BY-SA 4.0
36 events
| when toggle format | what | by | license | comment | |
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| Oct 27, 2023 at 21:50 | answer | added | Ján Lalinský | timeline score: 0 | |
| Oct 27, 2023 at 18:23 | answer | added | SargeATM | timeline score: -2 | |
| Feb 20, 2020 at 2:15 | review | Close votes | |||
| Feb 21, 2020 at 19:43 | |||||
| Nov 18, 2019 at 8:49 | vote | accept | Atom | ||
| Nov 18, 2019 at 8:44 | comment | added | Captain Emacs | @Atom I remembered to answer. See below. I hope the answer is helpful. | |
| Nov 17, 2019 at 21:35 | answer | added | Captain Emacs | timeline score: 5 | |
| Nov 13, 2019 at 7:39 | comment | added | Atom | @CaptainEmacs Done then! Will remind you next week! | |
| Nov 13, 2019 at 7:21 | comment | added | Captain Emacs | @Atom Ok, next week, if no-one else has answered this by then. | |
| Nov 12, 2019 at 18:21 | answer | added | march | timeline score: 7 | |
| Nov 12, 2019 at 5:59 | comment | added | Atom | @CaptainEmacs I beg thee to write an answer, please!! | |
| Nov 12, 2019 at 0:47 | comment | added | Captain Emacs | I do not have time to answer it in detail right now, but perhaps someone else can: a deep reason for using Fourier series is the fact that they represent a special decomposition that respects a symmetry operation on the set of functions you consider, in the standard case, the translations. This can be generalised to other group symmetries, and particularly beautifully so for compact or finite groups. There is a whole class of decompositions that emerges this way. The field is called "representation theory"/"Harmonic Analysis". | |
| Nov 11, 2019 at 22:48 | comment | added | user541686 | @Atom: Crucial question: Are you trying to eventually find the solution (closed-form), or are you trying to approximate it (computationally)? Taylor series can indeed make a ton of sense for the latter, but less so for the former. | |
| Nov 10, 2019 at 17:13 | answer | added | Lee Mosher | timeline score: 1 | |
| Nov 10, 2019 at 13:44 | comment | added | Atom | @The_Sympathizer Yes, the discussion here went off too far afield. | |
| Nov 10, 2019 at 13:35 | comment | added | The_Sympathizer | @Atom : It seems, then, you are also asking another question here, which is not "why use Fourier series", but rather "how could you discover the idea of a Fourier series if you did not know it before?" That's certainly a fascinating question, but not the one asked in the title! :) | |
| Nov 10, 2019 at 12:22 | answer | added | Mark Wildon | timeline score: 1 | |
| Nov 10, 2019 at 4:18 | vote | accept | Atom | ||
| Nov 18, 2019 at 8:49 | |||||
| Nov 9, 2019 at 23:15 | history | became hot network question | |||
| Nov 9, 2019 at 21:00 | history | tweeted | twitter.com/StackPhysics/status/1193272133394292736 | ||
| Nov 9, 2019 at 17:26 | answer | added | knzhou | timeline score: 7 | |
| Nov 9, 2019 at 17:22 | answer | added | tparker | timeline score: 20 | |
| Nov 9, 2019 at 17:20 | vote | accept | Atom | ||
| Nov 9, 2019 at 17:20 | |||||
| Nov 9, 2019 at 17:16 | comment | added | JMac | @Atom But for the entire function, a Fourier series will still approximate the periodic function better than a polynomial. The Fourier series repeats itself by it's own nature, just like the function you're modelling. The polynomials don't actually repeat themselves; you have to add higher and higher orders just to get that repetition. Even if the Taylor series models one period easier, the more periods you look at, the harder it is to model with Taylor series, while Fourier would be unchanged. | |
| Nov 9, 2019 at 17:11 | comment | added | Atom | @eyeballfrog Can you please give some reference? | |
| Nov 9, 2019 at 17:08 | comment | added | Atom | @JMac In my question, I never mentioned that I was talking about only sines or such functions, which have a finite Fourier series. For other, more general functions, the Fourier series is also an infinite series. | |
| Nov 9, 2019 at 17:06 | comment | added | Atom | @alephzero But the other, “more useful” tool can’t appear out of thin air. You’ll have to make them. One might be okay with just accepting the tools without motivation. But not me. | |
| Nov 9, 2019 at 17:02 | history | edited | Qmechanic♦ |
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| Nov 9, 2019 at 16:40 | comment | added | JMac | @Atom A Taylor series only models $\sin x$ or $\cos x$ exactly if you take an infinite expansion of the Taylor series. You can't model the infinite periodic nature of a sine wave without infinite terms in the polynomial. A Fourier series can perfectly converge on $sin$ or $cos$ without an infinite expansion. | |
| Nov 9, 2019 at 16:32 | comment | added | alephzero | @Atom the argument "suppose you only know Taylor" is just silly. If the only tool you have is a hammer, that doesn't mean a hammer is a useful tool for what you want to do. | |
| Nov 9, 2019 at 16:31 | comment | added | eyeballfrog | Everywhere the Taylor series converges uniformly, the Fourier series will as well. The difference is that the Fourier series for a periodic function also converges pointwise everywhere else. | |
| Nov 9, 2019 at 16:25 | comment | added | Atom | @JMac See the popular $\sin x$ and $\cos x$ series for starters. They “model” them exactly. | |
| Nov 9, 2019 at 16:12 | answer | added | The Photon | timeline score: 42 | |
| Nov 9, 2019 at 16:11 | comment | added | JMac | Depends quite a bit on the context. If I was trying to model the entire function with a series, I would know that a Taylor series is going to have issues though. Polynomial series can't really model periodic functions well. | |
| Nov 9, 2019 at 16:02 | comment | added | Atom | @JMac But suppose you didn’t have any hint that sine or cosine series can also approximate a function. Suppose that you only knew Taylor. Then would your intuition say the same? | |
| Nov 9, 2019 at 15:41 | comment | added | JMac | If it's a periodic function, wouldn't it make a lot more sense to use a periodic series? | |
| Nov 9, 2019 at 15:15 | history | asked | Atom | CC BY-SA 4.0 |