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Jan
17
comment an exponential sum involving quadratics
@r9m In this wiki page, there is a formula $g(a;p)=(\frac{a}{p})g(1;p)$. But when a=0, LHS=p but RHS=0.
Jan
17
comment an exponential sum involving quadratics
@JyrkiLahtonen BTW, to be precise $c=-\frac{b^2}{4a}$. Here $\frac{1}{4a}$ means the multiplicative inverse of $4a$ in the filed $\mathbb{Z}_q$.
Jan
17
comment an exponential sum involving quadratics
@JyrkiLahtonen Let me rephrase your idea. The case $b=0$ is known, so we want to reduce the matter to this case. Since we are working on the field $\mathbb{Z}_q$, we can do all the +,-*,/. Write $ak^2+bk=a(k+l)^2+c$ for some $l$ and $c$. By a change of variable, we can drop $l$. $c$ depends on both $a$ and $b$. So we can see that $b$ only contributes to a modulation and it does not affect the magnitude. Am I correct?
Jan
16
accepted equivalence of any two polynomials of same degree
Jan
16
asked equivalence of any two polynomials of same degree
Jan
15
accepted on an exponential inequality
Jan
15
asked an oscillatory integral with two parameters
Jan
15
accepted Does every non-commutative ring have the same numbers of left ideals and right ideals?
Jan
15
accepted $f^2+2f+1$ is a polynomial implies that $f$ is a polynomial
Jan
15
asked on an exponential inequality
Jan
15
asked $\{x:|f(x)|>5\}$ is the union of at most $2m$ intervals on each of which $f$ is monotone
Jan
14
asked construction of a partition function
Jan
13
asked an exponential sum involving quadratics
Jan
11
asked inserting absolute value in Hilbert transform and a discrete version of Hilbert transform
Jan
10
awarded  Yearling
Jan
8
answered If $\nu$ is a complex measure, then $L^1(\nu) = L^1(|\nu|)$
Jan
6
answered (complex measures) $d\nu=d\lambda +f\,dm \Rightarrow d|\nu|=d|\lambda| +|f|\,dm$ and $f\in L^1(\nu)\Rightarrow f\in L^1(|\nu|)$
Jan
5
comment the (2,2,1) boundedness of a “product” operator
I'm not sure about the boundedness. I suspect it is true and I'm trying to prove it. You may also try to disprove it.
Jan
5
comment the (2,2,1) boundedness of a “product” operator
I want to show that $T$ is strong (2,2,1) bounded. The coefficients $C(j,k)$ make the problem hard.
Jan
2
asked the (2,2,1) boundedness of a “product” operator