Timeline for Coefficients from Stone–Weierstrass versus Fourier transform
Current License: CC BY-SA 4.0
23 events
| when toggle format | what | by | license | comment | |
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| Apr 29, 2022 at 9:57 | history | edited | Jukka Kohonen | CC BY-SA 4.0 |
Weierstrass and other typos
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| Aug 25, 2010 at 13:18 | vote | accept | Dorian | ||
| S Aug 25, 2010 at 3:08 | vote | accept | Dorian | ||
| Aug 25, 2010 at 13:18 | |||||
| S Aug 25, 2010 at 3:07 | vote | accept | Dorian | ||
| S Aug 25, 2010 at 3:08 | |||||
| Aug 25, 2010 at 3:07 | vote | accept | Dorian | ||
| S Aug 25, 2010 at 3:07 | |||||
| Aug 24, 2010 at 22:02 | answer | added | Mike Hall | timeline score: 5 | |
| Aug 24, 2010 at 20:50 | comment | added | Yemon Choi | Could someone (preferably the original author) please edit the last paragraph so that it makes more sense? As it stands it is making false statements, and it would help us to give a useful answer to the question if the question were phrased more accurately. | |
| Aug 24, 2010 at 19:00 | answer | added | Helge | timeline score: 7 | |
| Aug 24, 2010 at 18:43 | answer | added | Otis Chodosh | timeline score: 3 | |
| Aug 24, 2010 at 17:25 | history | edited | Pete L. Clark | CC BY-SA 2.5 |
corrected spelling
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| Aug 24, 2010 at 16:18 | comment | added | Andrea Ferretti |
The set $\{ e^{inx} \}$ is trivially not dense in $L^1(0,1)$. What you probably mean is that their linear combinations (trigonometric polynomials) are. Still, this does not give a priori a development as a Fourier series, since the coefficients of the trigonometric polynomial approximation to a given function $f$ may not stabilize.
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| Aug 24, 2010 at 14:23 | comment | added | Otis Chodosh | Oh, I guess one way to do it is to prove that $\{e^{inx}\}$ seperates points and use en.wikipedia.org/wiki/… . | |
| Aug 24, 2010 at 14:19 | history | edited | Dorian | CC BY-SA 2.5 |
typo
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| Aug 24, 2010 at 14:11 | comment | added | Otis Chodosh | Sorry, I really mean, dense in $C(\mathbb{S}^1)$ functions. | |
| Aug 24, 2010 at 14:10 | comment | added | Otis Chodosh | I think @Dorian is referring to a corollary of Stone-Weierstrass which is apparently that $e^{inx}$ is a dense set in $C([0,1])$. See en.wikipedia.org/wiki/Stone-weierstrass#Applications_2. I can't remember how this proof goes off of the top of my head though. | |
| Aug 24, 2010 at 14:04 | comment | added | Peter Humphries | The Stone-Weierstrass theorem says for every continuous function $f$ and for every $\varepsilon > 0$ there exists a trigonometric polynomial $p$ such that $|f(x) - p(x)| < \varepsilon$ for all $x \in [0,1]$; that is, density w.r.t. the sup norm. Now $C([0,1])$ is dense in $L^1([0,1])$ w.r.t. the $L^1$ norm, but this is a completely different matter (and it's obviously a much weaker statement): there exist $L^1$ functions without close approximations by continuous functions w.r.t. the sup norm --- take the normal example (Dirichlet's monster) of $f(x) = 1$ on the rationals and $0$ otherwise. | |
| Aug 24, 2010 at 13:55 | history | edited | Dorian | CC BY-SA 2.5 |
I improved the title and changed a typo in my teX notation for $e^{ikx}$.
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| Aug 24, 2010 at 13:54 | comment | added | Otis Chodosh | The coefficients are the same. If you work in $L^2([0,1])$ instead, its easier to see. It all comes from the fact that Fourier coefficients are unique. If $\sum c_n e^{inx} = \sum d_n e^{inx}$ then $\sum (c_n - d_n) e^{inx} = 0$. Inner product with $e^{imx}$ and you get $c_n = d_n$. Its just another way to prove that Fourier series converge in $L^2$ (and thus $L^1$) | |
| Aug 24, 2010 at 13:47 | comment | added | Otis Chodosh | you need to change your exponents to $e^{ikx}$ so they display properly.. | |
| Aug 24, 2010 at 13:38 | comment | added | Dorian | What you're saying is false I think unless you're misunderstanding me. I'm working on say L^1[0,1] (any compact set will suffice). I can approximate $1$ as well as I'd like for instance by functions $\sum c_k sin(kx)$ for instance on $[0,2pi]$. I'm not talking about pointwise convergence just $L^1$ convergence. | |
| Aug 24, 2010 at 13:31 | history | edited | Dorian | CC BY-SA 2.5 |
added 23 characters in body
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| Aug 24, 2010 at 13:27 | comment | added | Thierry Zell | Your question might be; in which $L^1$? All your functions are $2\pi$-periodic, so you cannot approximate anything that isn't. | |
| Aug 24, 2010 at 13:13 | history | asked | Dorian | CC BY-SA 2.5 |