Timeline for Norms of the Dirichlet kernel
Current License: CC BY-SA 4.0
10 events
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| Jun 26, 2019 at 12:19 | history | edited | Bazin | CC BY-SA 4.0 |
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| Jun 26, 2019 at 9:20 | comment | added | Bazin | Let us continue this discussion in chat. | |
| Jun 26, 2019 at 9:18 | comment | added | Bazin | @RajeshDachiraju In the first place, I am unhappy with the standard references which do not provide a true equivalence with a proof much more complicated than the one I suggest (and wrote). Next, in an asymptotic expansion, all terms carry some information and could be interesting. As I wrote before, implementing an integration-by-parts method with a singular amplitude may be weird and that example is the simplest and most natural which I could think of. | |
| Jun 26, 2019 at 1:47 | comment | added | Rajesh D | Understood, thats why I said "related" and not "duplicate". What I don't understand is your motivation. Is your motivation is to see if Euler-Maclaurin is useful here? or do you really have the application of this result in any larger problem? Normally, one is interested in the asymptotic result and don't want to look at other smaller terms. | |
| Jun 25, 2019 at 19:43 | comment | added | Bazin | @RajeshDachiraju In fact it seems to be a very natural example to study the classical Euler-Maclaurin formula in a situation where the amplitude has pointwise singularities because of the $\vert \sin t\vert$. It is interesting to see that in spite of that inconvenience, an asymptotic expansion can still be reached. Moreover, I insist on the fact that a direct Euler-Maclaurin style proof is simpler and more precise than the proof provided in that answer. | |
| Jun 25, 2019 at 19:39 | comment | added | Bazin | @RajeshDachiraju In the answer mentioned in your comment, the method which is presented does not even provide an equivalent in my sense. However to get that is not very difficult: you may write for $\theta\in [0,π/2]$, $sin\theta=\theta/(1+\theta^2 \sigma(\theta))$ and thus $(sin\theta)^{-1}=\theta^{-1}(1+\theta^2 \sigma(\theta))$. The term $1$ in the parenthesis gives you the main term, you only have to prove that the contribution of the other term is asymptotically smaller, which is not difficult. | |
| Jun 25, 2019 at 19:20 | history | edited | YCor |
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| Jun 25, 2019 at 16:27 | comment | added | Rajesh D | Possibly related : math.stackexchange.com/q/27517/2987 | |
| Jun 25, 2019 at 16:25 | comment | added | Rajesh D | If I may ask, why are you specifically interested in the rate of convergence? | |
| Jun 25, 2019 at 15:53 | history | asked | Bazin | CC BY-SA 4.0 |