23 votes
Accepted

A finite alternating sum

I have obtained a formula for the generating function of your sequence. Let $S_n$ be defined as in the quesion. We extend the definition to $n = 0$ by demanding $0^0 = 0$, hence $S_0 = 0$. Consider $...
WhatsUp's user avatar
  • 3,232
18 votes

A finite alternating sum

WhatsUp's generating function allows to get an asymptotics: we have $$\sum S_n t^n=\frac{t}{e^{t/e}-t},\sum S_n e^nt^n=\frac{t}{e^{t-1}-t}.$$ Denote $t-1=x$, then $$\frac{t}{e^{t-1}-t}=\frac{1+x}{e^x-...
Fedor Petrov's user avatar
17 votes

exponential sum over variety

This is a very general sort of problem and you will get very different answers depending on exactly which $V, f,g$ you need. The least savings you could ask for is $\sqrt{p}$ savings over the ...
Will Sawin's user avatar
  • 135k
17 votes
Accepted

Deligne's theorem on exponential sums

Yes, smoothness is equivalent to the gradient being nonzero for every $x \in \overline{\mathbb F}_q^n \setminus \{0\}$. I would define smoothness of the hypersurface defined by $Q_d$ as the condition ...
Will Sawin's user avatar
  • 135k
16 votes
Accepted

Why are Deligne-type exponential sum estimates so hard to use?

There are a lot of subtle reasons such exponential sums can fail to exhibit square-root cancellation. First let me comment on two reasons suggested in your answer: (1) trying to have an explicit ...
Will Sawin's user avatar
  • 135k
13 votes
Accepted

Arguments of exponential sums

There are at least 3 ways of seeing this - an elementary way, via the determinants of $\ell$-adic sheaves, and via Katz's calculus of finite field hypergeometric functions. Elementary: Subtraction ...
Will Sawin's user avatar
  • 135k
11 votes

Mean square estimate for the Kloosterman sums

You cannot have estimates like (*) for any $\theta<1$. Fouvry and Michel showed that (see Theorem 1.2 there) $$ \sum_{c\le x} |S(m,n;c)|/\sqrt{c} \gg_k \frac{x}{\log x} (\log \log x)^k, $$ for ...
Lucia's user avatar
  • 43.3k
10 votes
Accepted

Does anyone recognize this exponential sum?

Yes, these sums occur as the arithmetic part of the Fourier expansion of period kernels $\sum_{ad-bc=1}(a\tau+b)^{-k}(c\tau+d)^{-k}$, the analytic part being J-Bessel functions. The derivation is not ...
Henri Cohen's user avatar
  • 11.5k
10 votes
Accepted

The exploration of the asymptotic behavior of a simple sum. $\sum_{k=1}^{\infty} (k^{1/k} - 1)$

The OP asks the question whether the series $S =\sum_{k=2}^{\infty}(k^{1/k} -1)$ diverges. The answer is that the series diverges, because $$k^{\frac{1}{k}}-1 = \exp\left(\frac{\ln k}{k}\right)-1>\...
Carlo Beenakker's user avatar
10 votes

A basic estimate of exponential sums

The main idea is to apply Poisson summation formula. It allows to replace $a/q$ by $\lfloor q/a\rfloor$ and then repeat the procedure using continued fraction expansion of $a/q$. This idea belongs to ...
Alexey Ustinov's user avatar
9 votes
Accepted

An exponential sum over squares

Actually, $|\sum_{M < n \leq N} e(x/n^2)| \sim c \sqrt{x}$, where $c \approx 0.016151690 + 0.0738060263i$. To see this, write $$\sum_{M < n \leq N} e(x/n^2) = \sum_{M < n \leq \epsilon \sqrt{...
George Shakan's user avatar
9 votes

counting points on unit sphere mod p

Since this old question has resurfaced, let me sketch two ways to prove the stated formulæ using algebraic geometry: The first way is fairly elementary. Let us stick for definiteness with the number ...
Gro-Tsen's user avatar
  • 29.9k
9 votes

Upper bound an integral with exponential function

The integral in question can be rewritten as $$ \begin{aligned} I&:=\frac1{\sqrt n}\,\int_{-a\sqrt n}^{(1-a)\sqrt n} e^{-u^2}\Big(1-\exp\Big\{-\frac{u^4/n}{1-u^2/n}\Big\}\Big)\,du \\ &\le\...
Iosif Pinelis's user avatar
9 votes

Estimates for certain double-Kloosterman sums

I can do it for $q$ prime, and thus $q$ squarefree, by a long but elementary manipulation followed by the Weil bound. Until the end, the elementary manipulations will work for $q$ arbitrary. Taking ...
Will Sawin's user avatar
  • 135k
9 votes
Accepted

Cancellation in a very rapidly oscillating exponential sum

I doubt that one is able to get as far as $T = \exp(\log^{2-\varepsilon} x)$ with Weyl differencing. Standard Weyl differencing arguments, such as that in Theorem 8.4 of Iwaniec, Henryk; Kowalski, ...
Terry Tao's user avatar
  • 108k
9 votes
Accepted

Does there exist some irrational $x,\alpha$ so that this Weyl sum is $o(\sqrt N)$?

The estimate $$ S_N(x) = S_N(x,\alpha) = o(N^{1/2}) \tag {1}$$ cannot hold when $x$ is irrational. Heuristically, the reason comes from the (approximate) modularity properties of $S_N(x,\alpha)$; this ...
Terry Tao's user avatar
  • 108k
8 votes
Accepted

The first case of the strong Littlewood conjecture

The following human-verifiable proof is in collaboration* with Fedja. Lemma $1$: We have the following for $0 \le x \le 3$: $$\frac{6204}{6750}x^2-\frac{8429}{60750}x^4+\frac{4475}{546750}x^6\le x.$$ ...
mathworker21's user avatar
8 votes
Accepted

Is the Lebesgue measure of the $x$ so that this exponential sum is $o(\sqrt{N})$ positive?

Ok, apologies if this is overkill, but this paper shows that for almost every (irrational) $\alpha \in \mathbb{R}$, only a measure $0$ set of $x \in \mathbb{R}$ satisfy $S_{N,\alpha}(x) = o(\sqrt{N})$ ...
mathworker21's user avatar
7 votes

Sums of twisted products of Kloosterman Sums

Your sum is $$\sum_{i=0}^2 (-1)^i \operatorname{tr}(\operatorname{Frob}_p, H^i_c( \mathbb A^1_{\overline{\mathbb F}_p}, \mathcal{K}\ell_2 (ab) \otimes \mathcal{K}\ell_2 (a(b+h)) \otimes \mathcal L_\...
Will Sawin's user avatar
  • 135k
7 votes

Hypotheses for exponent pairs

See Montgomery's book, Ten Lectures on the Interface Between Analytic Number Theory and Harmonic Analysis, specifically the examples on p.55. He considers variants of your sum, namely $\sum_{n \leq N}...
Matt Young's user avatar
  • 4,633
7 votes
Accepted

Moments of certain exponential sum

Yes. Without loss of generality, $X\geq 2$ is an integer. Then we have explicitly $$ v(\beta)=e\left(\frac{(X+1)\beta}{2}\right)\frac{\sin(\pi X\beta)}{\sin(\pi\beta)},\qquad\beta\not\in\mathbb{Z}.$$ ...
GH from MO's user avatar
  • 98.2k
7 votes
Accepted

Why are exponential sums so bad at solving this very easy problem?

The basic example of exponential sums in Number Theory is to count solutions to an equation such as $f(x_1,\cdots,x_n)\equiv 0$ $\mod p$ where $p$ is prime this is because $p$-adic solutions are a ...
Ivan Meir's user avatar
  • 4,782
7 votes
Accepted

Bound for sum of multiplicative character calculated over multivariate polynomial

Yes, there are several known bounds. The following statement, due to Katz, has quite strict conditions on $f$, but gives a very strong result. It is perhaps the simplest statement that gives such a ...
Will Sawin's user avatar
  • 135k
7 votes

On the upper-bound for a type of quintuple Kloosterman sums

The Newton polyhedron method for your sum is as follows: We first replace the quadratic character with an additive character and a simpler quadratic character using a Gauss sum, then introduce a ...
Will Sawin's user avatar
  • 135k
7 votes

On the estimate for the mixed 3-dimensional hyper-Kloosterman sum

This sum can be handled by elementary change of variables. Let $a = \overline{z} x$, $b=y$, and $c =\overline{xy}$. Then $x = \overline{bc}$, $y=b$, and $z=\overline{abc}$, so $(x,y,z) \to (a,b,c)$ ...
Will Sawin's user avatar
  • 135k
6 votes
Accepted

estimate for a sum of products of Weil's sum

Here is a different approach than Denis Chaperon de Lauzières's. Opening everything and using orthogonality of characters to remove the $x$ variable, we see that your sum is $1/p$ times $$\sum_{z_1,...
Will Sawin's user avatar
  • 135k
6 votes
Accepted

An Exponential Sum Restricted to Primes

No, this is definitely not true. First of all, there's no way we should expect better than square-root cancellation in an exponential sum without an incredible amount of structure, which the primes do ...
Greg Martin's user avatar
  • 12.7k
6 votes

Summing the infinite series $\sum_{k=0}^{\infty} \frac{x^k}{(k!)^2}$

The first way I try to solve questions like this is to "ask Maple" (or Mathematica). If you have access to, say, Maple, then you can type "sum(x^k/k!^2, k=0..infinity)" and it will report BesselI(0,...
Ryan O'Donnell's user avatar
6 votes
Accepted

Integral over an exponential sum with squares

The integral is equal to exactly one - just expand out the square and use orthogonality to get $$ I = \sum_{0\leq n,m<p} \int_0^1 e((n^2+m^2)t) \mathrm{d} t = \sum_{0\leq n,m<p}1_{n^2+m^2=0}=1.$...
Thomas Bloom's user avatar
  • 6,608
6 votes
Accepted

Can the following sum be counted or expressed in terms of special functions?

Probably the answer is negative. Your series is a restriction of the analytic function in two complex variables: $$F(\zeta,q)=\sum_{n=0}^\infty\frac{q^{n^2}}{n!}\zeta^n,\quad |q|\leq1,$$ obtained by ...
Alexandre Eremenko's user avatar

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