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Jan 6, 2021 at 18:15 comment added Yemon Choi Can I just point out that this definition of "Fourier coefficient" looks a bit strange for $f(x)=\sin(x)$, which last time I looked was a perfectly reasonable $2\pi$-periodic smooth function...
Jan 6, 2021 at 11:55 comment added Yemon Choi Are you only interested in even functions $f$, i.e. those satisfying $f(x)=f(2\pi -x)$? If not then your definition of "Fourier coefficient" might not be appropriate
Dec 28, 2020 at 16:44 vote accept spaceman
Dec 28, 2020 at 14:30 history edited YCor CC BY-SA 4.0
fixed typo, changed tag
Dec 26, 2020 at 19:11 answer added Salcio timeline score: 3
Dec 25, 2020 at 9:23 history became hot network question
Dec 25, 2020 at 3:48 answer added Iosif Pinelis timeline score: 25
Dec 24, 2020 at 19:41 comment added Fedor Petrov @Itay but it's higher derivatives are quite large in a sense.
Dec 24, 2020 at 19:40 comment added spaceman Hi @Itay, I'm not necessarily looking for a smoothness condition, but just another general condition for this to hold, e.g. a sufficient decay, has to satisfy condition (X) etc. type of condition. The smoothness assumption was put in place to simplify matters.
Dec 24, 2020 at 19:11 comment added Itay I doubt there will be such a condition in terms of smoothness. Given any smooth function $f$ and a positive integer $n$, the function $f- \hat{f}(n) cos(nx)$ has the same Fourier coefficients as $f$, besides the $n$th, which would be 0.
Dec 24, 2020 at 18:15 history asked spaceman CC BY-SA 4.0