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Special functions, orthogonal polynomials, harmonic analysis, ordinary differential equations (ODE's), differential relations, calculus of variations, approximations, expansions, asymptotics.
2
votes
Accepted
Dense subset for $C_0(\mathbb{R})$
The functions $f_a$ considered are (up to a multiplicative constant) the Fourier transforms of even tent functions. The real linear combinations of even tent functions yield all even continuous and pi …
3
votes
Upper bound for an inverse Laplace transform
Partial answer. If $g$ is known to be a non-negative function, Karameta Tauberian theorem relates the behavior of $G = \int_0^\cdot g$ at infinity (resp. at 0) to the behavior of $\mathcal{L}g$ at $0$ …
1
vote
Growth of the "cube of square root" function
For $-1<h<1$,
$$(1+h)^{3/2}+(1-h)^{3/2} = 2 \sum_{n=0}^{+\infty}{3/2 \choose 2n}h^{2n},$$
where
$${3/2 \choose 2n} = \prod_{k=1}^ {2n} \frac{5/2-k}{k}.$$
For $n \ge 1$, since $(-1)^{2n-2}=1$, we get
$ …
6
votes
Accepted
Existence of weird complex norms
Consider the norm $\Vert\cdot\Vert$ on $\mathbb{C}^2$ defined by $\Vert z \Vert^2 := |z_1|^2+|z_1+z_2|^2$. Then
$$\left\Vert \left(\begin{array}{c}1 \\ -1 \end{array}\right) \right\Vert^2 = 1,$$
$$\le …
5
votes
Rigorous estimates on roots of function
If my computations are correct, there is a root of the form $x=1+\sin^2(\frac\theta2)$ with:
$$\theta \in \left(\frac{k\pi}N,\frac{k\pi}N+\frac\pi{2N}\right) \text{ if }k < \frac{N}3-\frac12$$
$$\thet …
2
votes
Does there exist a continuous function $f(x)$ such that $f(0)=0$ and $0<\lim_{n\to\infty}\pr...
Possible way to find such a function $f$
Since
$$\prod_{k=1}^n \frac{ke}{n} = n!\Big(\frac{e}{n}\Big)^n \sim \sqrt{2\pi n} \text{ as } n \to +\infty,$$
it is sufficient to find a continuous function $ …
7
votes
Forcing the uniqueness of a solution of an ODE
This question would be possibly at a better place on MathStack Exchange.
Yet, once the statement of the question is corrected (the functions $y_n$ need to be defined on $\mathbb{R_+}$ and not only on …
0
votes
A counterexample to: $\frac{1-f(x)^2}{1-x^2}\le f'(x)$ — revisited
I complete the reformulation given by Andrea Marino and give another counterexample.
First, the inequality of the beginning can be written
$$\forall x \in (0,1), \quad \frac{f'(x)}{1-f^2(x)} \ge \frac …