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Apr 22 at 17:32 history edited Nemo CC BY-SA 4.0
added reference
Feb 18 at 2:27 comment added Iosif Pinelis Also, why the Gauss sums don't depend on $j$? I have just tried $n=5$, and then the sums do depend on $j$.
Feb 18 at 2:25 comment added Iosif Pinelis Can you detail the second equality in your first proof (where $f$ first appears) and the use of the Poisson summation formula (?) in your second proof, where $\sum_{n=0}^a$ first appears?
Feb 16 at 19:34 history edited Nemo CC BY-SA 4.0
thanks to Iosif Pinelis' observations, a typo is corrected, and a missing phase factor inserted
Feb 16 at 18:49 comment added Iosif Pinelis Sorry for the delayed response. No, I don't have a better answer, and I don't think I ever implied anything like that. Concerning the first line of your first proof, I think there is at least one typo there. Concerning the first line of your second proof, I think the factor $(-1)^{-1/4}$ is missing on right-hand side there; hopefully, this does not affect the final result.
Feb 15 at 7:12 history edited Nemo CC BY-SA 4.0
added proof without using Poisson summation
Feb 15 at 5:28 history edited Nemo CC BY-SA 4.0
deleted 38 characters in body
Feb 14 at 20:33 comment added Iosif Pinelis I think the formula is correct, but the manipulations with the delta functions need detalization/justification. I think answers are written not only for the OP, but also for their other readers -- so, they should be detailed enough in any case.
Feb 14 at 6:07 history edited Nemo CC BY-SA 4.0
added 666 characters in body; added 3 characters in body; added 1 character in body
Feb 14 at 5:53 history edited Nemo CC BY-SA 4.0
added info on calculation of the Gauss sum
Feb 14 at 2:26 comment added Iosif Pinelis Previous comment continued: Mathematica does not know this identity even for $f=1$.
Feb 13 at 22:24 comment added Iosif Pinelis Can you explain how the "(well known?) fact" is proved?
Feb 13 at 7:16 history answered Nemo CC BY-SA 4.0