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 | |
|---|---|---|---|---|---|
| Nov 9, 2019 at 18:15 | comment | added | Atom | Wait, did you mean time translation? | |
| Nov 9, 2019 at 18:15 | comment | added | knzhou | @Atom You need linearity in the independent variable ($x$) and translational invariance in the dependent variable ($t$), which is precisely what we have. | |
| Nov 9, 2019 at 18:12 | comment | added | Atom | But $F(x)=-kx$ isn’t translationally invariant, is it? | |
| Nov 9, 2019 at 18:05 | comment | added | knzhou | @Atom By "diagonal" I mean the general intuition that a problem can be decomposed into independent parts, each of which can be solved easily independently. That is useful for the same reason diagonalizing a matrix is. Fourier series allow you to do this for any system of linear translationally invariant differential equations. | |
| Nov 9, 2019 at 17:48 | comment | added | Atom | That’s clear. But where does the ‘diagonal’ come into picture? | |
| Nov 9, 2019 at 17:43 | comment | added | knzhou | @Atom Now take a more complicated problem, like $F = - kx - \epsilon x^3$ for small $\epsilon$. The solution is almost a sinusoid, and in fact you can write it as $\sum_n a_n \cos(\omega_0 n t)$ where the terms rapidly fall off in size. Again, trying to do this with a Taylor series would be disastrous. | |
| Nov 9, 2019 at 17:41 | comment | added | knzhou | @Atom The simplest example is the harmonic oscillator, taught in every first physics class, $F = - kx$. You probably saw this solved by "guessing the solution was $\cos(\omega t)$". That's a cheap way of saying that the problem is trivially solved by Fourier series; you don't even need infinitely many terms, you get the whole thing with one term. Imagine trying to solve it by setting $x(t) = \sum_n a_n t^n$. | |
| Nov 9, 2019 at 17:33 | comment | added | Atom | Can you please explain your last paragraph in some detail? I don’t know how a problem is rendered “diagonal”. | |
| Nov 9, 2019 at 17:26 | history | answered | knzhou | CC BY-SA 4.0 |