Timeline for Why use Fourier series instead of Taylor?
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| Nov 10, 2019 at 19:04 | comment | added | HerpDerpington | And to add to the second point - periodicity. | |
| Nov 10, 2019 at 4:18 | vote | accept | Atom | ||
| Nov 18, 2019 at 8:49 | |||||
| Nov 9, 2019 at 17:51 | comment | added | The Photon | @user45664, you're right that if the equation is linear, you can solve each term separately and then sum up the results. The advantage of using Fourier series is that each result for a series term will be orthogonal from the other results, making the final summing up simpler. | |
| Nov 9, 2019 at 17:45 | comment | added | knzhou | @Atom If you only care about having a "theoretical" solution, regardless of whether you can use it for anything, then this whole discussion makes no sense. For example, for literally any differential equation, you can solve it by saying the solution is the function $\text{Atom}(x)$, where the function is defined by being the solution to the differential equation. This is a closed form, final, exact theoretical solution to the problem. But you probably think it's completely absurd to solve a problem that way. Why? Because it's not good for anything. We always need a purpose in mind. | |
| Nov 9, 2019 at 17:39 | comment | added | Atom | Just one thing. When we talk about series solutions, do we really care (even at theoretical level) whether the infinite series is practically achievable? Cuz when we write the final solution (theoretically), we really are talking about the infinite series, not some truncation of it, aren’t we? | |
| Nov 9, 2019 at 17:37 | comment | added | Milan | "then you can consider each term on its own", this is indeed not unique for Fourier series. In linear systems, different terms will never interact. The point is (as you mentioned) that you can turn ODE's into algebraic equations and some PDE's into ODE's because complex exponentials are eigenfunctions of the differential operator. | |
| Nov 9, 2019 at 17:27 | comment | added | user45664 | What you say for Fourier series is true for any orthogonal basis--no?. I.e. "then you can consider each term on its own". | |
| Nov 9, 2019 at 17:20 | vote | accept | Atom | ||
| Nov 9, 2019 at 17:20 | |||||
| Nov 9, 2019 at 16:12 | history | answered | The Photon | CC BY-SA 4.0 |