## New answers tagged ca.classical-analysis-and-odes

1

The only criterion I know is based on a Theorem in the second book of Dunford and Schwartz, see Theorem X1.6.29 and following. If the resolvent of an Hilbert-Schimdt operator satisfies some decay estimates on some rays dividing the complex plane, then the span of the generalized eigenfuntions is dense. The theorem generalizes to operators having trace-class ...

fa.functional-analysis real-analysis ca.classical-analysis-and-odes operator-theory sp.spectral-theory

5

In the paper The Green Function for uniformly elliptic equations Manuscripta Mathematica Gruter & Widman proved - among other things - pointwise estimates about the gradient of the Green function under the Dini continuity assumption on the leading coefficients. This is in section 3 Theorem 3.3

17

One way to get the claimed value of the given integral, $J$ in notation below, is by starting from the standard relation
$$
\begin{aligned}
\zeta(s)
&= \frac 1{\Gamma(s)}\int_0^\infty \frac {x^{s-1}}{e^x-1}\; dx\ ,\qquad
\text{ so }\\
\zeta'(s)
&=
\frac\partial{\partial s}
\left(\
\frac 1{\Gamma(s)}\int_0^\infty \frac {x^{s-1}}{e^x-1}\; dx
\ \...

18

I don't have a published source for this integral, but if need be you could refer to the following derivation:
$$\int_0^\infty \frac{x^2\ln{x}}{e^x-1}\,dx=\int_0^\infty x^2 e^{-x}\ln x\sum_{k=0}^\infty e^{-kx}$$
$$=\sum_{k=0}^\infty\frac{3-2\gamma-2 \ln (k+1)}{(k+1)^3}$$
$$=(3-2\gamma)\zeta(3)+2\zeta^\prime(3).$$
The integral over $x^2 e^{-(k+1)x}\ln x$ ...

0

Check Computing a family of reproducing kernels for
statistical applications, 1996, by Christine Thomas-Agnan

1

Certainly not, if $C$ is supposed to be real. For instance, suppose that $\rho_1$ and $\rho_2$ are probability densities such that $\rho_1\rho_2=0$. Then the left-hand side of your inequality is $\infty$, whereas its right-hand side is $2C$.

1

Perhaps the most elementary way to see that $f$ is not continuous at zero is to lower bound the sum at $x=\pi/k$. Note that summands coming from $n_1=2\ell k+s$ and $n_2=n_1+k$ together have non-negative contribution to the sum when $s=1,\dots k$ (and $\ell\in \mathbb{Z}_{\geq 0}$), so the sum is actually lower bounded by
$$
\sum_{n=1}^{2k}\frac{\sin{(n\cdot ...

10

This function is (on the real line, at least) the product of
$$ \exp( \mu^2 \sum_{i=1}^\infty |z_i|^2 - 2 \mu \Re(\sum_{i=1}^\infty z_i)) \quad (1)$$
and the Hadamard type product
$$ \prod_{i=1}^\infty E_1( 2 \mu\Re z_i - \mu^2 |z_i|^2) \quad(2)$$
where $E_1$ is the first elementary factor
$$ E_1(z) := (1-z) \exp(z).$$
The expression (1) is clearly entire in ...

0

This not an answer to my question (I accepted Iosif's answer) but additional information based on my most recent research, that may be helpful to the reader interested in these processes. It is too long for an update of my question.
For simplicity, we use the notation $F(x)=F_0(x)$ and $F(x-k)=F_k(x)$.
Theorem A
Regardless of the distribution $F_k$, if $b-a$ ...

pr.probability ca.classical-analysis-and-odes stochastic-processes probability-distributions sequences-and-series

4

$\newcommand{\1}{\mathbf 1}\newcommand{\ep}{\varepsilon}\newcommand{\tr}{\operatorname{tr}}$The min-max value is $\sqrt n$.
Indeed, take any real $n\times n$ matrix $H$ with $|\det H|=1$. By the singular value decomposition,
\begin{equation}
H=U^TDV,
\end{equation}
where $U$ and $V$ are some orthogonal matrices and $D$ is the diagonal matrix with ...

2

This is not true. Indeed, suppose that $X_k=X_{s;k}=k+sZ_k$, where $s\downarrow0$ and the $Z_k$'s are any iid random variables (r.v.'s).
To obtain a contradiction, suppose that, for the random Borel measure $\mu_s$ over $\mathbb R$ defined by $\mu_s(B):=\sum_{k\in\mathbb Z}1(X_{s;k}\in B)$, the distribution of the random variable (r.v.) $\mu_s(B)$ is Poisson ...

pr.probability ca.classical-analysis-and-odes stochastic-processes probability-distributions sequences-and-series

3

Hjalmar Rosengren gave a nice formula for $f_{m,n}(x)$ based on the theory of hypergeometric series. Here I provide a direct elementary proof of the same based on generating series.
Let us introduce the differential operator
$$D_r:=\frac{1}{r!}\frac{\partial^r}{\partial x^r}.$$
Then
$$\sum_{k=0}^\infty\binom{n+k}{k}x^k=D^n\left(\frac{1}{1-x}\right)=\frac{1}{(...

3

With proofs abound, I would like to record yet another instance of application due to the Wilf-Zeilberger techniques. The aim is to prove the identity shown above (from GH from MO, Hjalmar Rosengren):
$$\sum_r\binom{m+n+k-r}{k-r}\binom{m}r\binom{n}r\binom{m+k}k^{-1}\binom{n+k}k^{-1}=1.$$
The mechanical process begins with letting (suppressing other variables)...

23

See Meyer, Walter; Kay, David C., A convexity structure admits but one real linearization of dimension greater than one, J. Lond. Math. Soc., II. Ser. 7, 124-130 (1973). ZBL0271.52002.
Theorem 4. If $V$ and $W$ are real vector spaces with $\dim V>1$, and $f:V\to W$ is a one-to-one mapping which preserves convexity, then $f$ is either linear (if $f(0)=0$) ...

6

This is Gauss' hypergeometric function
$F(n+1,m+1,1;x)$. You can then apply the huge theory of hypergeometric functions to derive further expressions. For instance, Euler's transformation formula gives the alternative expression
$$\frac 1{(1-x)^{m+n+1}}\,F(-m,-n,1;x)=\frac 1{(1-x)^{m+n+1}}\sum_{k=0}^{\min(m,n)}\binom mk\binom nk x^k$$
for the same series as ...

1

Come on, for the first property just use the fact that
$$ \int_{-n}^n K(x) dx = \sum_{j = -n+1}^{n-1} \phi(j) + \tfrac12(\phi(-n)+\phi(n)) $$
has a finite limit as $n \to \infty$, together with convergence of
$$ \biggl| \int_{-a}^a K(x) dx - \int_{-\lfloor a\rfloor}^{\lfloor a\rfloor} K(x) dx \biggr| \leqslant |\phi(-\lfloor a\rfloor-1)| + |\phi(-\lfloor a\...

1

$\newcommand\R{\mathbb R}\newcommand{\Z}{\mathbb{Z}}\newcommand{\ep}{\epsilon}\newcommand{\fl}[1]{\lfloor#1\rfloor}$The answer is yes to each of your two questions.
Let $a_n:=\phi(n)$. Then
\begin{equation*}
K(x)=\sum_{n\in\Z}a_n R(x-n).
\end{equation*}
Note that for all $j\in\Z$ we have $K(j)=a_j$ and $K$ linear (or, more exactly, affine) on the ...

5

Rohrlich has conjectured that the multiplicative relations in $\mathbb{C}^\times / \overline{\mathbb{Q}}^\times$ between values of $\Gamma$ at rational numbers are generated by the multiplication formula and the reflection formula. In conceptual terms, Lang says that $\Gamma$ is an odd punctured distribution on $\mathbb{Q}/\mathbb{Z}$, and that conjecturally,...

5

Since $\sin^2 u=(1-\cos 2u)/2$ and $\sinh^2 u=(\cosh 2u-1)/2$, after the substitution $x=\sqrt{1-h^2}$ the two integrals respectively reduce to the integrals
$$\int_0^1\frac{x \cos x\,dx}{\sqrt{1-x^2}}\quad\text{and}\quad
\int_0^1\frac{x \cosh x\,dx}{\sqrt{1-x^2}}.$$
In turn, integrating by parts, we get
$$\int_0^1\frac{x \cos x\,dx}{\sqrt{1-x^2}}
=1-\int_0^...

1

I think you can to look at the geometry of jet bundles, the equivalence problem a la "E. Cartan". See R. Gardner: The Equivalence Problem and Application (I don't remember well the title of the book, sorry), the work of D. C. Spencer...
Indeed, I think, all is about the characteristics, and so, ultimately, about the Curvature... In smooth category, ...

dg.differential-geometry ca.classical-analysis-and-odes sg.symplectic-geometry curvature contact-geometry

2

There is an implementation around (Pari/GP; in the tetration-forum) which claims to have a Kneser-implementation. It is a bit difficult to handle, so I'll show here a simpler version (essentially polynomial) of a tetration-function which seems to approximate that Kneser function when the polynomial's order is increased.
With a polynomial of order 32 I got ...

2

Thanks Carlo for pointing out the solution @ math.stackexchange from Ron Gordon. Yes, this works. Indeed, one can use the same technique to solve my series for arbitrary positive integer $s$.
For this, start with the the contour pointed out by Carlo above, which vanishes:
$$
\oint_{C} dz \frac{1}{z^{2s+1} \sin(\pi z)\sin(\pi z (\sqrt{2}-1))}=0\,.
$$
The ...

1

$\newcommand{\Z}{\mathbb{Z}}\newcommand{\ep}{\epsilon}$Let $a_n:=\phi(n)$. Then
\begin{equation}
K(x)=\sum_{n\in\Z}a_n 1(n-1/2\le x<n+1/2).
\end{equation}
So, $K(x)=a_0=0$ if $1/2\le x<1/2$. So, for $\ep\in(0,1/2)$,
\begin{equation}
I_\ep:=\int_{1/\ep<|x|<\ep}K(x)\,dx=\int_{|x|<\ep}K(x)\,dx
=\sum_{n\in\Z}a_n J_n,
\end{equation}
...

1

For an account of integrability of linear ODE in the way that you describe, refer section 4 of the 2004 paper
A.G. Khovanskii, "On solvability and unsolvability of equations in explicit form"
Uspekhi Mat. Nauk. 59:4 69-146, English translation in Russian Math Surveys 59:4 661-736.
(Section 1.2 of the paper describes the "classical" ...

0

A follow-up of Kloeden and Platen is C.H. Skiadas, Exact Solutions of Stochastic Differential Equations (2010).

answered Nov 12 at 13:40

Carlo Beenakker

136k1212 gold badges329329 silver badges470470 bronze badges

3

Let me try to give a short argument that if LHS integral exists and equals $I=\int_{a}^b f(g(x))dg$ than $\int_{g(a)}^{g(b)} f(t)dt$ exists and equals $I$.
Without loss of generality $g(a)\leqslant g(b)$, and it suffices for any fixed $I_1>I$ to find a partition of $[g(a),g(b)]$ such that upper Darboux sum for $\int f$ is at most $I_1$ (then do the same ...

1

Make a sequence of functions of period $2\pi$ whose limit is of period $\pi$. Your map is discontinuous.

fa.functional-analysis ca.classical-analysis-and-odes probability-distributions harmonic-analysis periodic-functions

3

Looks like the answer is yes, they are equal, and perhaps you do not really need to assume the existence of the integral in the right-hand side, that is, $\int_{g(a)}^{g(b)} f(t) dt$. This is proved in the context of Henstock–Kurzweil integrals as Theorem 6.1 in [1].
The key lemma is that if $g(a) = g(b)$, then $\int_a^b f(g(x)) dg(x) = 0$. This is applied ...

1

Concerning Lemma 6: Clearly, this lemma is false if $F=\emptyset$ and, say, $u_n=nu$ for some nonzero $u\in H$. Assume therefore that $F\ne\emptyset$.
The inequality $\|u_n\|\le\|u_1\|+\|f\|$ will not always hold in this setting -- consider e.g. the case when $u_1=0$ and $u_n=2f\ne0$ for $n\ge2$.
However, since $F\ne\emptyset$, there is some $f_*\in F$. Then ...

fa.functional-analysis ca.classical-analysis-and-odes convex-optimization calculus-of-variations convex-analysis

1

Under the described hypotheses, the function $f: x \in \mathbf{R} \mapsto e^{-cx} \phi(x)$ would be decreasing, because $f'(x) = e^{-cx} (-c \phi(x) + \phi'(x)) \leq 0$ for all $x \in \mathbf{R}$. As $f(0) = 0$ and $f \geq 0$, one would have $f(x) = 0$ for all $x \geq 0$. For this reason a cutoff function cannot have $\phi' \leq c \phi$.
Here for ...

0

The answer is no.
Indeed, let $f:=\phi=\phi_\epsilon$. Let $a:=\sup\{x\colon f(x)=0\}$. Then $a$ is real, $f(a)=0$ and $f>0$ on the interval $(a,\infty)$. Without loss of generality, $a=0$, so that $f(0)=0$ and $f>0$ on the interval $(0,\infty)$.
Suppose now that for some real $c>0$ we have $f'\le cf$. Then $(\ln f)'\le c$ on $(0,\infty)$, whence $\...

1

No, it is not correct. Simple counterexamples can be obtained as follows. Let $V_1=W$ and $V_2$ be any Banach spaces such that the identity is continuous but not surjective $V_2\to W$. Then $f$ defined by $(x,y)\mapsto x+y$ is even smooth with $\partial_2 f(x_0,y_0)$ injective $V_2\to W$ and there cannot be any $g,\Omega_1,\Omega_2$ as required since we ...

1

I asked Mathematica about the boundary behavior. First, the $P^{1/4}_{\nu } $ solution: We have, for $\epsilon \searrow 0$,
$$
(\sin \epsilon )^{1/4} P^{1/4}_{\nu } (\cos \epsilon ) = \frac{2^{1/4} }{\Gamma(3/4)} + O(\epsilon^{2} )
$$
and
\begin{eqnarray*}
(\sin (\pi -\epsilon ))^{1/4} P^{1/4}_{\nu } (\cos (\pi -\epsilon) ) &=& \frac{2^{3/4} \pi }{\...

1

What Maple tells you is that you could try a function of the form $Y(\theta)=(\sin\theta)^\frac14 y(\cos\theta)$. And indeed, when you do, you find that $y$ satisfies a Legendre equation, leading to the solutions provided. Your boundary condition in $y$ is a bit complicated at $x=\pm1$. Literally it is
$$
\lim_{x\to\pm1} \left(\frac14 \frac{x}{(1-x^2)^\...

0

No, we cannot.
Formally, $\varphi$ is the eigenfunction of the unbounded operator $L_s$ on $L^2(\Omega)$, defined initially by
$$ L_s u(x) = (-\Delta)^s u(x) \qquad \text{for } x \in \Omega , $$
where $u \in C_c^\infty(\Omega)$ (and it is understood that $u(x) = 0$ for $x \notin \Omega$), and then extended to an appropriate domain (e.g. by means of ...

fa.functional-analysis pr.probability ap.analysis-of-pdes ca.classical-analysis-and-odes elliptic-pde

0

As $g(t)=\log{t}$ is a strictly increasing function, what you really needs is just the maximum of $\frac{A}{B}.$
Also, after some experiments, it seems that, for pure imaginary numbers only, the maximum occurs when the imaginary part is equal to $\lambda$. That is (in your notation), when $a=0$ the maximum is obtained with $b=\lambda$. So "divide and ...

3

I find it useful to represent the Legendre function in terms of a hypergeometric function, using a formula from Wikipedia,
$$f(\theta)=\frac{ (1+\cos \theta)^{\mu/2} \, _2F_1\left(-\nu,\nu+1;1-\mu;\frac{1}{2} (1-\cos \theta)\right)}{\Gamma (1-\mu)\sin^\mu\theta(1-\cos \theta)^{\mu/2}}.$$
Then Mathematica gives me the small-$\theta$ expansion
$$f(\theta)=\...

answered Nov 5 at 21:35

Carlo Beenakker

136k1212 gold badges329329 silver badges470470 bronze badges

3

I think it is not possible to do much better than $\frac{1}{\sqrt n}$. More precisely, I believe the best $\alpha$ is $\frac{1}{\sqrt n}$ whenever $n$ is a power of $2$, and therefore (since $\alpha$ is non-decreasing in $n$) at most $\frac{\sqrt{2}}{\sqrt n}$ in general.
Indeed, consider $f(x)=\frac 1 2 \|A^{-1} x\|_2^2$ for a linear invertible map $A$ on $\...

fa.functional-analysis ca.classical-analysis-and-odes convex-optimization norms multivariable-calculus

12

The answer is yes indeed. It is a special case of Fox-H function, a variation of the confluent Fox-Wright $_{1}\Psi_{1}$ function (a generalization of the confluent hypergeometric function $_{1}F_{1}$) providing the inverse function. See a previous answer here for details and references. For this particular case solution is
(Setting $\alpha = a$), for $\...

1

If $a$ is rational, then the root, say $x_*$, of your equation is algebraic, and (say) Mathematica will find for you with any degree of accuracy. Otherwise, one can approximate $a$ by rational numbers.
Another way to get bounds on $x_*$ is to use a combination of the Newton and secant methods to bracket the root, as follows. For $a\in(0,1)$, using the ...

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