Timeline for Orthonormal basis and decay
Current License: CC BY-SA 3.0
24 events
| when toggle format | what | by | license | comment | |
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| Aug 11, 2017 at 12:57 | vote | accept | Zinkin | ||
| Aug 11, 2017 at 6:17 | comment | added | Zinkin | @ChristianRemling the way the question was initially phrased was that $f$ only overlapped with $h_0$ and $h_1$, it had zero overlap with all other $h_i$. Thus, the projection of $f$ onto the span of the $h_i$ was equal to the projection onto the span of $h_0$ and $h_1$, no? | |
| Aug 11, 2017 at 0:26 | comment | added | Christian Remling | I think your edits completely killed the question: $h_1$ is trivially a linear combination of $g_0,g_1$ only, simply because the initial pieces of the $g$'s always have the same linear span as those of the $h$'s, just by how Gram-Schmidt works, and all the extra information is completely irrelevant for this. Your previous version looked much more interesting, where you ask this question about a new function $f$ rather than $h_1$. | |
| Aug 10, 2017 at 21:30 | comment | added | user69109 | @MateuszKwaśnicki sure, I was just curious. | |
| Aug 10, 2017 at 21:29 | comment | added | Mateusz Kwaśnicki | @Tokoyo: As usual, it took me more than a couple of minutes. | |
| Aug 10, 2017 at 21:27 | answer | added | Mateusz Kwaśnicki | timeline score: 1 | |
| Aug 10, 2017 at 21:11 | comment | added | user69109 | @MateuszKwaśnicki so do you have now a proof for the statement? | |
| Aug 10, 2017 at 20:19 | comment | added | Mateusz Kwaśnicki | Of course I messed this up again, $\langle h_0, g_1\rangle = 0$. I was thinking about $\langle h_1, g_0\rangle$. Sorry. | |
| Aug 10, 2017 at 20:08 | comment | added | Mateusz Kwaśnicki | @Zinkin: Although $\langle h_0, g_1\rangle$ is non-zero, $h_0$ is indeed not interesting. But so is $h_1$! I will write up an answer in a couple of minutes. By the way, $\langle h_0, h_1\rangle = \langle h_1, h_2\rangle$, if I understand correctly. | |
| Aug 10, 2017 at 17:50 | history | edited | Zinkin | CC BY-SA 3.0 |
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| Aug 10, 2017 at 16:40 | history | edited | Zinkin | CC BY-SA 3.0 |
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| Aug 10, 2017 at 15:55 | history | edited | Zinkin | CC BY-SA 3.0 |
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| Aug 10, 2017 at 15:48 | history | edited | Zinkin | CC BY-SA 3.0 |
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| Aug 10, 2017 at 14:31 | comment | added | Zinkin | @MateuszKwaśnicki not quite, since $\langle h_0,g_n\rangle =\delta_{0,n}$ but if we answer this question for $h_1$ as well, then I think we have it. Thank you for your remark, I will adapt the question accordingly. | |
| Aug 10, 2017 at 14:25 | history | edited | Zinkin | CC BY-SA 3.0 |
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| Aug 10, 2017 at 13:55 | history | edited | Zinkin | CC BY-SA 3.0 |
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| Aug 10, 2017 at 12:55 | comment | added | Mateusz Kwaśnicki | If I understand correctly, you do not really need $f$ and the projection: you ask how fast do the inner products $\langle h_0, g_n\rangle$ go to zero, right? | |
| Aug 10, 2017 at 12:51 | history | edited | Zinkin | CC BY-SA 3.0 |
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| Aug 10, 2017 at 8:12 | history | edited | Zinkin | CC BY-SA 3.0 |
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| Aug 10, 2017 at 8:02 | comment | added | Dirk | Your formula for $g_2$ does not seem right. | |
| Aug 10, 2017 at 7:28 | history | edited | Zinkin | CC BY-SA 3.0 |
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| Aug 10, 2017 at 5:58 | history | edited | Zinkin | CC BY-SA 3.0 |
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| Aug 10, 2017 at 0:18 | history | edited | Zinkin | CC BY-SA 3.0 |
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| Aug 9, 2017 at 23:50 | history | asked | Zinkin | CC BY-SA 3.0 |