21

More generally, if $1\le k\le N-1$ is an integer, where $N$ is a positive interger,
$$S_{N,k} := \sum_{n=0}^\infty\biggl( \frac{1}{(N n+N-k)^2} + \frac{1}{(N n+k)^2}
\biggr) = \frac{\pi^2}{N^2\sin^2(\pi k/N)}.$$
Your sum is $S_{10,1}-S_{10,3}$.

16

Denoting $H_0=0$, we have $$\sum_{n=1}^\infty \frac{H_n}{n(n+1)2^n}=\sum_{n=1}^\infty \left(\frac1n-\frac1{n+1}\right)\frac{H_n}{2^n}=\sum_{n=1}^\infty \frac{1}n\left(\frac{H_n}{2^n}-\frac{H_{n-1}}{2^{n-1}}\right)\\=\sum_{n=1}^\infty\frac1{n^22^n}-\sum_{n=1}^\infty\frac{H_n}{(n+1)2^{n+1}}.$$
It is well known that $\sum_{n=1}^\infty\frac1{n^22^n}=\frac{\pi^2}{...

15

The answer is yes if $\epsilon<1$, and no when $\epsilon\geq 1$.
This follows from Carleman's quasianalyticity criterion, see for example, Hormander, Analysis of linear partial differential operators, Vol. I, Chap I, Section 1, Theorem 1.3.8.
(Carleman's original proof used Complex Analysis, and it was reproduced in the books on the subject. Hormander has ...

12

This is just a standard application of the Baire category theorem.
Proposition: Suppose that $X$ is a topological space where the Baire category theorem holds and $g_{n}:X\rightarrow[0,\infty]$ for each natural number $n$. Suppose that $\overline{\lim}_{n}\sup_{I}g_{n}\geq 1$ for each non-empty open set $I$. Then there is a point $x\in X$ such that $\...

9

An elementary non-existence proof may be of interest.
Let $f:[a,b]\to\mathbb{R}$ an increasing, continuous, and not absolutely continuous function: I claim there exists a point $c\in[a,b]$ where the Dini derivative $D^*f(c):=\limsup_{x\to c}\frac{f(x)-f(c)}{x-c}$ is infinite. For the proof, we may assume w.l.o.g. that $f$ is strictly increasing (we may just ...

9

No, there is not. This follows by an extended version of Lebesgue's differentiation theorem, which asserts that the derivative of $f(x) = \mu((-\infty, x))$ is infinite almost everywhere with respect to the singular part of $\mu$.
To be specific: Let $\nu$ be the Lebesgue measure and
$$ D_\mu \nu(x) = \lim_{t \to 0^+} \frac{\nu([x-t, x+t])}{\mu([x-t, x+t])} =...

9

This is true. This was proved by Hardy and Littlewood and the proof is reproduced in Zygmund's Trigonometric Series (which I don't have access to at the moment).
(Contradicting my prior "answer")
The answer to this question is negative. Such counterexamples are known as "Pauli partners" and are studied in, among other places, the quantum ...

9

This is asking for the value of an $L$-function of an even Dirichlet character $\chi$ at a positive even integer, and these have known values. It is analogous to the explicit expressions for the Riemann zeta-function at positive even integers (definitely not at positive odd integers!), but instead of the values being rational multiples of powers of $\pi$, ...

9

You seem to have a typo in your value of $a$ - evaluating the sum as is yields something more like 1.76. Converting to an integral,
$$
\lim_{n\rightarrow \infty } \frac{1}{n^2 } \sum_{i=1}^{n} \sum_{j=1}^{n} \frac{1}{\sqrt{(i/n)^2 + (j/n)^2 } } =\\ \int_{0}^{1} dx \int_{0}^{1} dy\, \frac{1}{\sqrt{x^2 +y^2 } } = 2\ln (1+\sqrt{2} )
$$
On the other hand,
$$
\...

7

First some heuristics, before constructing the complete answer -
this looks a bit more transparent if one considers
$$
A^2 = \begin{pmatrix} -\partial_{x}^{2} +x^2 & 1 \\ 1 & -\partial_{x}^{2} +x^2 \end{pmatrix}
$$
Then, denoting the standard harmonic oscillator eigenfunctions (i.e., the eigenfunctions of $-\partial_{x}^{2} +x^2 $ with eigenvalues $\...

7

Put $\mu:=\alpha-1$; then $\sum^{\infty}_{n=1}\frac{ x^{n-1}}{(n-1)!n^\alpha} = x^{-1}\sum^{\infty}_{n=1}\frac{ x^n}{n!n^\mu}=x^{-3/2}I_\mu(x)$ for the function $I_\mu(x)$ given in Johannes Trost’s answer to Asymptotic expansion of $\sum\limits_{n=1}^{\infty} \frac{x^{2n+1}}{n!{\sqrt{n}} }$, with a complete asymptotic series.

6

The equivariant version of the implicit function theorem is the following.
Let $f: \mathbb{R}^p \times \mathbb{R}^n \to \mathbb{R}^m$ be a smooth function (possibly only defined on open neighborhoods) which is equivariant with respect to the action of a compact Lie group $G$ on $\mathbb{R}^n$ and $\mathbb{R}^m$. Let $(t_0, x_0)$ be such that $f(t_0, x_0) = ...

6

As Iosif said, in general the system you specified does not admit a solution. Here we will give a more pedestrian argument using only comparisons.
Monotonicity
Claim: if a solution exists, and $f(1) > 0$, then the function is monotonically decreasing; if $f(1) < 0$, then the function is monotonically increasing.
Proof: we will focus on the positive ...

5

First of all, your main equation contains $f(1)$ and therefore is not an ODE. Let us consider the ODE
\begin{equation*}
(r^2 - 2ar)f''(r) + 2(r-a) f'(r) - (4a + m(m+1))f(r) = p, \tag{1}
\end{equation*}
where $p$ is a real number; your equation corresponds to (1) with
\begin{equation}
p=-4af(1). \tag{2}
\end{equation}
The general real solution of (1) ...

5

Such a function does not exist.
Indeed, let $f\colon[a,b]\to\mathbb R$ be an increasing differentiable function. Then
$$f(x)-f(a)=(HK)\int_a^x f'(t)\,dt$$
for all $x$,
where $(HK)\int$ is the Henstock–Kurzweil integral. In particular, $f$ is Henstock–Kurzweil integrable on $[a,b]$.
Therefore and because $f'\ge0$, $f'$ is Lebesgue integrable on $[a,b]$, and ...

5

Let $b:=-\alpha\in\mathbb C$. We have to show that
$$f(x):=\sum^{\infty}_{N=1}\frac{N^b \, x^{N-1}}{\Gamma(N)}\sim e^{x}x^b \tag{1}$$
as $x\to\infty$.
Let $Y=Y_x$ denote a random variable with the Poisson distribution with parameter $x$. Using the idea of this answer, we have
$$f(x)=e^x x^b\, E\Big(\frac{Y+1}x\Big)^b. \tag{2}$$
Next, $\dfrac{Y+1}x\to1$ in ...

4

Let me add some thoughts to the already existing fine answers.
1. The answer is "no" as follows from Propositions 79 and 92 in Tao's lecture notes on differentiation theorems. Indeed, let $f:[a,b]\to\mathbb{R}$ be an increasing differentiable function. By the quoted propositions, $f'$ is absolutely integrable, and
$$f(x)=f(a)+\int_a^x f'(t)\,dt,\...

4

Since your equation is linear, this is sufficient. The convergence holds in the sense of distributions.
Notice the following: the limit, as a solution of the hyperbolic equation $u_t+u_x=0$ is unique, being actually $u(t,x)=u(0,x-t)$. In particular
$$\|u(t)\|_{L^2({\mathbb R})}\equiv\|u(0)\|_{L^2({\mathbb R})}.$$
Since on the other hand
$$\|u_\epsilon(t)\|_{...

3

The extended operator can be treated along similar lines as the $c=0$ case. One merely has to modify the algebra a little. Again, denote the standard harmonic oscillator eigenfunctions (i.e., the eigenfunctions of $-\partial_{x}^{2} +x^2 $ with eigenvalues $\lambda_{n} =2n+1$) as $\psi_{n} (x)$. Introduce also the standard raising and lowering operators $a^{\...

3

The inequality is scale invariant and holds for a ball of any radius. It follows by a standard argument that is the inductive step in what's known as Moser iteration.
The constant $C$ below can change from line to line but always depends only on the dimension. Let $B = B(0,r)$ and $2B = B(0,2r)$. Let $\chi$ be a smooth compactly supported function on $2B$ ...

3

This is not a full answer, but a pair of soft arguments suggesting that $A$ is dense in $[0, +\infty)$.
First Argument
Given any triple $(a,b,c)$, let $\displaystyle r(a,b,c)=\frac{c}{\text{rad}(abc)}$.
One can generate two new triples
$$t_1=(a(c+b),b^2,c^2)$$
$$t_2=(a^2,b(c+a),c^2)$$
with ratios
$$r(t_1)=r(a,b,c)\cdot\frac{c}{c+b}\cdot\frac{c+b}{\text{rad}(...

3

Your sum is not absolutely summable, but if you first sum on $y$ and use $\sum_{n\in\Bbb Z}1/(x+n)=\pi\cot(\pi x)$ (symmetrical sum) you get
$$\frac{\pi}{c}\sum_x\frac{\cot(\pi bx/c)}x=\frac{\pi}{c}\sum_{n=1}^{c-1}\cot(\pi bn/c)\sum_{m\in\Bbb Z}\frac{1}{mc+n}=\frac{\pi^2}{c^2}\sum_{n=1}^{c-1}\cot(\pi bn/c)\cot(\pi n/c)$$
using again the above formula for $\...

3

This is clearly false as stated, since a necessary condition is that $g(x)\to g(\xi)$ as $x\to\xi$ a. e. in the sphere, but if $g$ is merely $L^2$, this may well fail for every point.
The convergence holds in a weaker sense. For example, it is true that if $g_\rho(x)=g(\rho x)$ for $x\in\partial B_r(0)$ and $\rho<1$, then $g_\rho\to g$ a. e. and in $L^2$ ...

3

The polynomial $Q(y):=\frac43y-\frac13 {y^3}$ is increasing on the interval $[-1,1]$; it has fixed points $0$ and $\pm1$, and $\text{sgn }( Q(y)-y )=\text{sgn}y$. Thus the iterates of $Q$ starting from any $y\in [-1,1]\setminus\{0\}$ converge monotonically to $\text{sgn } y$ (in fact with exponential rate given by $Q'(\pm1)=\frac13$, and uniformly away from $...

3

This question has been studied in two papers of Peter Yuditskii and myself:
Zbl 1241.41005 (arXiv:1008.3765) and Zbl 1168.30020 (arXiv:math/0604324), where we determined the polynomial of best approximation to sgn(x), and the asymptotics of the error term. For the general case, take a linear combination of shifts sgn(x-a_j). There is no "Gibbs ...

answered Sep 18 at 12:31

Alexandre Eremenko

71.9k66 gold badges208208 silver badges344344 bronze badges

2

The answer is negative: If $f'(x) = 0$ "too often", then $R_f$ may fail to be equal to one almost everywhere.
Let $C$ be a fat Cantor set, let $I_n = (a_n, b_n)$ ($n \geqslant 2$) be the sequence of all finite components of the complement of $C$, and let $f$ be a differentiable function with the following properties:
$f(x) = 0$ for $x \in C$;
on ...

2

I'm pretty sure that you figured it out by now, but I'll post it for the sake of completeness. Any unimodal $\sqrt{p}$ can be approximated by a sum $\sum c_k\chi_{I_k}$ with $I_1\subset I_2\subset\dots$. Now just notice that the function $\int(\chi_{I_k})(\chi_{I_j})_t$ is non-increasing in $t\ge 0$ for any $k,j$.

2

Angeloni, Costarelli, and Vinti, in arXiv:1906.03021, study a discrete multivariate convolution
for a class of averaged product kernels of the form
$$\bar{\chi}_m(t_1,t_2,\ldots t_N)=\prod_{i=1}^N\left(\frac{1}{m}\int_{-m/2}^{m/2}\chi_i(t_i+v)\,dv\right).$$
The variation $V[f]$ of a function $f:\mathbb{R}^N\rightarrow \mathbb{R}$, defined by
$$V[f]=\int_{\...

answered Aug 28 at 18:00

Carlo Beenakker

133k1212 gold badges324324 silver badges461461 bronze badges

2

Reformulate
Supposing $f$ is a solution to your first formulation, with $f(1) = \lambda$. Let $\tilde{f}(r) = \lambda^{-1} f(r)$. Then $\tilde{f}$ solves the differential equation
$$ \tag{1} r(r-2a) \tilde{f}'' + 2(r-a) \tilde{f}' - m(m+1) \tilde{f} = \frac{4a^2}{r(r-2a)} (\tilde{f} - 1) + \frac{4a(1-2a)}{(1-a)r(r-2a)} \tilde{C} $$
with
$$\tilde{f}'(1) = \...

2

First integrate over $\theta_1,\theta_2$. Use the delta function representation (for $k\in\mathbb{R}$)
$$\int_{-\infty}^\infty e^{2\pi i k\theta}\,d\theta=\delta(k),$$
to evaluate
$$\int_{-\infty}^\infty \int_{-\infty}^\infty e^{2\pi i (v_1\cdot x)\theta_1+2\pi i(v_2\cdot x)\theta_2}\,d\theta_1 d\theta_2=\delta(v_1\cdot x)\delta(v_2\cdot x).$$
Next for the ...

answered Aug 27 at 13:12

Carlo Beenakker

133k1212 gold badges324324 silver badges461461 bronze badges

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