Skip to main content

All Questions

Filter by
Sorted by
Tagged with
4 votes
2 answers
305 views

Generalization of van der Corput's estimate on oscillatory integrals

Question: Given exponents $0<\alpha<\beta$ and an interval $[a,b]\subset(0,\infty)$ are there constants $C,d>0$ such that for any $\lambda_1,\lambda_2\in\mathbb{R}$, $$\left|\int_a^be(\...
Joel Moreira's user avatar
  • 1,701
4 votes
1 answer
297 views

Proving bounds on analytic functions using only the Taylor expansion

I wonder if there is a general method for obtaining bounds on an analytic function using only its Taylor expansion (not using its special properties such as satisfying a good differential equation, ...
Mostafa - Free Palestine's user avatar
3 votes
1 answer
113 views

maximum likelihood estimation of X is better than that of f(X)?

Consider a binary variable $C$ with $\Pr(C=0)=\Pr(C=1)=0.5$. Consider a random vector $X \in \mathbb{R}^d$, characterized by $C$, with PDF $p_m(x)$, $m\in\{0,1\}$. Define a maximum likelihood (ML) ...
Jeff's user avatar
  • 482
2 votes
0 answers
122 views

Comparing the truncated $\ell^{1}$-norm of polynomial coefficients with the supremum norm on the unit disc

Let $p=a_{0}+a_{1}z+\ldots+a_{n}z^{n}$ be a polynomial. Consider the following truncated $\ell^{1}$-seminorm of the coefficients of $p$: $$\|p\|_{\ell^{1},\text{trun.}}:=\sum_{k=1}^{n}|a_{k}|=\|p-a_{0}...
Calculix's user avatar
  • 321
1 vote
0 answers
66 views

Parameter estimation of a Taylor expansion

Let $a,b$ two real numbers, $\theta$ a real parameter and suppose that you have an analytic function of the form: $$ f_\theta(x)\triangleq \sum_{k\in\mathbb{N}}a_k(\theta)x^k \quad\forall x\in[a,b], $$...
NancyBoy's user avatar
  • 393
0 votes
1 answer
51 views

Strict positive type function on hypersurface also of positive type in neighborhood?

Let $u\in C^\infty(\mathbb{R}^n\times\mathbb{R}^n)$ be symmetric and of strictly positive type on some hypersurface $S \subset \mathbb{R}^n$ diffeomorphic to $\{0\}\times\mathbb{R}^{n-1}$. This means ...
megggs's user avatar
  • 13