Timeline for Can phase significantly concentrate a function's spectrum?
Current License: CC BY-SA 3.0
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| Apr 13, 2017 at 12:57 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Jul 1, 2015 at 0:36 | history | edited | Yemon Choi | CC BY-SA 3.0 |
deleted 81 characters in body
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| Jul 1, 2015 at 0:34 | comment | added | Yemon Choi | @WillSawin Good point. I'll update my answer | |
| Jul 1, 2015 at 0:05 | comment | added | Will Sawin | @DustinG.Mixon For any function $f$, if we let $f_n$ be $f$ composed with multiplication by $n$ then there is a simple formula for $\hat{f}_n$. | |
| Jun 30, 2015 at 23:58 | comment | added | Dustin G. Mixon | @WillSawin - I have a lot more intuition about the Fourier transform of $\sin(2\pi n\cdot)$ than I do about $f_n$. Perhaps you conflated the two? | |
| Jun 30, 2015 at 22:09 | comment | added | Will Sawin | Doesn't $||f_n||_{A(\mathbb R/\mathbb Z)} = ||f_1||_{A(\mathbb R/\mathbb Z)}$ because the Fourier transofrm of $||f_n||$ is the Fourier transform of $||f_1||$ with extra zeroes? | |
| Jun 30, 2015 at 14:14 | comment | added | Dustin G. Mixon | Thanks! I'm not sure I agree with your gut, though. Taking $x_a(t):=\operatorname{sin}(2\pi at/n)$, simulations suggest that $\inf_{a,n}\|F_nx_a\|_1/\|F_n|x_a|\|_1=\pi/4$, where $F_n$ denotes the DFT over $\mathbb{Z}/n\mathbb{Z}$. | |
| Jun 30, 2015 at 13:35 | history | answered | Yemon Choi | CC BY-SA 3.0 |