I have a sequence $a_n$ such that $0 \leq a_n \leq \log n$, and I am considering $\sum_{n \leq X} a_n$. However, I prefer using smooth weights so I would like to approximate it with $\sum_{n \geq 1} a_n w(X)$.

I guess what I would like is a nice function $w$ such that $$ E(X) = | \sum_{n \leq X} a_n - \sum_{n \geq 1}a_n w(X) | $$ is "small" and that the derivative is "not too big". I know this is a bit vague, but I was wondering what $w$ is out there that I can achieve these two things as well as possible... Any comments would be appreciated.

PS alternatively I was curious about what I can expect, as in if I wanted $E(X) < X^{c}$, $0 < c<1$ then what is the best I can hope for for the upper bound of $|w'(x)|$? and vice versa. Thank you.

  • 1
    $\begingroup$ $\sum_{n \ge 1} a_n w_n(X)$. Nice, small, not too big depends on $(a_n)$. $\endgroup$ – reuns Dec 22 '17 at 16:52
  • 2
    $\begingroup$ Usually you smooth the sum with a term $w(n/X)$, not just $w(X)$, which, being independent of $n$, factors out of the sum. See Chapter 5 of Multiplicative Number Theory by Montgomery and Vaughan for examples. $\endgroup$ – Stopple Dec 22 '17 at 16:59

In Section 11.8 of Harman's book "Prime-Detecting Sieves", he introduced an infinitely differentiable function $\psi_1(t)$ on $\mathbb{R}$ such that $\psi_1(t)\in[0, 1]$ for all $t$ and $$ \psi_1(t)=\left\{ \begin{array}{lr} 1\text{ if }x-y+\Delta_1\le t\le x-\Delta_1,\\ 0\text{ if }t\not\in(x-y, x) \end{array} \right. $$ with $$ \psi^{(j)}(t)\ll_j\Delta_1^{-j}\text{ for }j=1, 2, ... $$ where $\Delta_1=yx^{-\eta}$ for some $\eta>0$. But the construction, which he gave in the appendix of the book, works for very general $x, y$ and $\Delta_1$. For example, you can choose $x=y$ and $\Delta_1$ to be a suitable power of $x$. As another example, in Fouvry and Iwaniec's paper "Gaussian Primes" (p. 257), they estimated the sum $$\sum_{\substack{n\equiv0\pmod d\\n\leq x}}a_n$$ by comparing it to $$\sum_{n\equiv0\pmod d}a_nf(n)$$ where $f$ is a suitable smooth function. Then they applied Poisson's summation to transform and estimate the latter sum (where the bounds on the derivatives of $f$ play an important role). And since $f(n)=1$ for most $n$ (say, for all $0\le n\le x^{1-\epsilon}$) and $|1-f(n)|\le1$ for all $n$, so the sum $$\sum_{\substack{n\equiv0\pmod d\\n\leq x}}a_n(1-f(n))$$ can be shown to be small.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.