TITLE: What is the upper bound of $R\underbrace{(3,3,3, \ldots,3)}_\text{$k$ times}$?
QUESTION [9 upvotes]: Just some context: $R\underbrace{(3,3,3, \ldots,3)}_\text{$k$ times}\leq n$, means that any colouring of a complete graph, $k_n$, on $n$ vertices or more with $k$ colours must contain a monochromatic triangle.
Claim: $R\underbrace{(3,3,3, \ldots,3)}_\text{$k$ times} \leq 3k!$
Proof:
Consider the minimum number of monochromatic edges connected to an arbitrary vertex, $V$. At least $\big\lceil{\frac{3k!-1}{k}}\big\rceil$ edges will be monochromatic. $\big\lceil{\frac{3k!-1}{k}}\big\rceil$ =$\frac{3k!}{k}$ when $k>1$. Therefore, vertex $V$ must be connected to $3(k-1)!$ monochromatic edges.
By trying to avoid a monochromatic triangle of color $k_1$, we will be forced into creating a monochromatic triangle of color {$k_2$, $k_3$, $\ldots$, or $k_i$}. If we connected any of the vertices $\{1,2,3, \ldots, 3(k-1)!\}$ with a blue edge, a blue monochromatic triangle would be formed, to avoid this, the remaining $3(k-1)!$ vertices must be colour with the remaining $k-1$ colours.
We can then consider a subgraph of the remaining vertices $\{1,2,3,\ldots,3(k-1)!\}$ and colour it with the remaining $k-1$ colours. We can focus on a new vertex, $V'$, within the subgraph.
Vertex $V'$ will be connected to at least $\displaystyle{\frac{3(k-1)!}{k-1}}$ monochromatic edges. Which equals $3(k-2)!$ edges.
We can continue this process until we consider the subgraph of remaining vertices $3(k-(k-1))!$ and colour it with the remaining $k-(k-1)$ colours. This means that this final subgraph has 3 vertices and has to be coloured with 1 colour. Therefore, a monochromatic triangle is unavoidable. Therefore, $R\underbrace{(3,3,3, \ldots,3)}_\text{$k$ times} \leq 3k!$.
I managed to get it down to $3k!$, I was wondering if anyone had better approximations?

REPLY [15 votes]: All questions about Ramsey numbers for small graphs should be first checked in Staszek Radziszowki's amazing frequently updated survey.  On page 40 we find the upper bound $(e-\frac16)k!+1\approx 2.55 k!$, proved by Xu Xiaodong, Xie Zheng and Chen Zhi in a paper published in Chinese.


TITLE: Tensors = matrices + covariance/contravariance?
QUESTION [2 upvotes]: I have read several topics on tensors but it is still not clear to me. Tensors are different from matrices because they contain additional information about how do they transform.
To fully specify a $n\times n$ matrix, one need to specify its $n^2$ components.
My question is: what do someone need to specify to fully define a tensor ?
Is it true to say that, to fully define a 2D tensor, one need to provide its $n^2$ components plus whether the tensor is covariant or contravariant along each dimension ? (or is there something else) ?

REPLY [4 votes]: Usually, a matrix is thought of a representation of a linear operator: a map that takes a vector and spits out another vector.  Say $A$ is some linear operator and $v$ is some vector, then $A(v)$ is the output vector.
An equivalent way of looking at it, however, is to say that there is a map $B$ that takes two vectors $v, w$ and spits out a scalar, given by $B(v,w) = A(v) \cdot w$, say.  Such a map is what is usually described in the literature when talking about tensors.
Where do contravariance and covariance come in?  Well, the above idea of a tensor is actually a bit of a cheat.  There might not be an inner product; we might not be able to freely convert between vectors and covectors using it.  So instead of saying that $B$ takes two vectors as arguments, let $B$ be a map taking one vector $v$ and a covector $\alpha$ instead, so that $B(v, \alpha) = \alpha(A(v))$.
(You'll note that, if there is a way to convert from vectors to covectors, then any tensor acting on $p$ vectors and $q$ covectors could be converted to one that acts on $p+q$ vectors, for instance.)
A general tensor could take any number of vector or covector arguments, or a mix of the two in any number.
In physics, it's common to look at the components of a tensor with respect to some basis--rather than supply whatever vectors or covectors that might be relevant to a problem, we supply a set of basis vectors and covectors instead, so we need only remember the coefficients.  If $e_i$ is the $i$th basis vector and $e^j$ is the $j$th basis covector, then $B(e_i, e^j) = {B_i}^j$ takes us from the more math-inclined definition of a tensor to the more familiar form to a physicist.


TITLE: How to solve an hyperbolic Angle Side Angle triangle?
QUESTION [2 upvotes]: If from an hyperbolic triangle $ \triangle ABC$ the angels  $\angle A$, $\angle B$ and side $c$ are given. (ASA triangle, a side and both adjacent angles are given)
How can I calculate the remaining angles and sides? (or at least one of them)
In Euclidean geometry you start with calculating $\angle C$ by substracting  $\angle A$ and  $\angle B$ from a straight angle, but in hyperbolic geometry that obviously doesn't work.

REPLY [2 votes]: You can use the cosine rule for hyperbolic triangles:
$\sin A \sin B \cosh c  = \cos C + \cos A \cos B$ or
$\cos C = \sin A \sin B \cosh c - \cos A \cos B $
and then use the law of sines to find the other two sides:
$\frac{\sin C}{\sinh c}=\frac{\sin B}{\sinh b}=\frac{\sin A}{\sinh a}$
Sources:
http://en.wikipedia.org/wiki/Hyperbolic_triangle
http://www.maths.gla.ac.uk/wws/cabripages/hyperbolic/hypertrig.html


TITLE: What is the simplest proof that the density of primes goes to zero?
QUESTION [62 upvotes]: By density of primes, I mean the proportion of integers between $1$ and $x$ which are prime. The prime number theorem says that this is asymptotically $1/\log(x)$.
I want something much weaker, namely that the proportion just goes to zero, at whatever rate. And I want the easiest proof possible.
The simplest proof I know uses estimates involving the binomial coefficient $\binom{2n}{n}$, but the argument still feels a bit involved.
Does anyone know an even simpler proof that the density of primes goes to zero?

REPLY [2 votes]: Terry Tao's proof can be slightly reworked to give a little more — that $\mu^\ast(P) = 0$ for every arithmetic upper quasi-density $\mu^\ast$ on $\mathbb N$, where $P$ is the set of primes. And a refinement of the unnumbered theorem in Terry Tao's post is Theorem 1 in [R.C. Buck, The Measure Theoretic Approach to Density, Amer. J. Math. 68 (1946), No. 4, 560-580], whose proof is as simple as it could be. (To apply Buck's theorem, one has to know that the series of reciprocals of all primes is divergent, but this is in turn as easy to prove as the divergence of the harmonic series, see [J.A. Clarkson, On the series of prime reciprocals, Proc. Amer. Math. Soc. 17 (1966), 541]).

An arithmetic upper quasi-density (on $\mathbb N$) is a $[0,1]$-valued function $\mu^\ast$ defined on the whole power set of $\mathbb N$ such that, for all $X, Y \subseteq \mathbb N$ and $h, k \in \mathbb N^+$, the following hold:

$\mu^\ast(X) \le \mu^\ast(\mathbb N) = 1$.

$\mu^\ast(X \cup Y) \le \mu^\ast(X) + \mu^\ast(Y)$.

$\mu^\ast(k \cdot X + h) = k^{-1} \mu^\ast(X)$ , where $k
    \cdot X + h := \{kx + h \colon x \in X\}$.

An arithmetic upper density is then a $\subseteq$-monotonic arithmetic upper quasi-density; and an arithmetic [quasi-]density is the restriction $\mu$ of an arithmetic upper [quasi-]density $\mu^\ast$ to the family of all sets $X \subseteq \mathbb N$ such that $\mu^\ast(X) = 1 - \mu^\ast(\mathbb N \setminus X)$. We refer to such a restriction as the arithmetic [quasi-]density induced by $\mu^\ast$.
Examples of arithmetic upper densities are the upper asymptotic (or upper natural) density (link to Wiki.en), the upper Banach density (link to MathWorld), the upper logarithmic density (link to OEIS), the upper Pólya density, the upper analytic (or upper Dirichlet) density (link to OEIS), and the upper Buck density. The arithmetic density induced by the upper asymptotic density is the asymptotic density, that induced by the upper Banach density is the Banach density, etc.
All of this and a bit more is discussed in joint work with Paolo Leonetti, where arithmetic upper [quasi-]densities are just called upper [quasi-]densities. In particular, Corollary 3.4 in [Leonetti & T., On small sets of integers, Ramanujan J. 57 (2021), 275-289] has a proof of the statement from the first paragraph of this answer (that $\mu^\ast(P) = 0$ for every arithmetic upper quasi-density).


TITLE: Calculating the residues of $\prod_{i=1}^N\frac{1}{a_is+1}$
QUESTION [1 upvotes]: I have a generic expression
$$\prod_{i=1}^N \frac{1}{a_is+1}$$
where $a_i$ are constants and some of the $a_i$ elements necessarily not the same. The thing I want to obtain is I want to express this product
$$ \prod_{i=1}^N \frac{1}{a_is+1} = \sum_{k=1}^t\sum_{\ell=1}^{v_t} \frac{\rho_{k,\ell}}{(a_ks+1)^j}$$
where $\sum_{k=1}^tv_t=N$. I have found that the elements $\rho_{k,\ell}$ can be obtained as
$$ \rho_{k,\ell}= \frac 1 {(v_k-\ell)!a_\ell^{v_k-\ell} } \frac {\partial^{v_k-\ell}}{\partial s^{v_k -\ell}} \left[ \prod_{m=1,m\neq k}^t \frac 1 {1 + a_ks} \right]$$
but I could not perform the derivations to achieve $\rho_{k,l}$ constants. If you help me, I'll be so glad. Thank you!

REPLY [1 votes]: Before we answer the question we also look somewhat closer at OP's expressions which contain some (technical) inaccuracies and which should be discussed beforehand.
We consider the product $\prod_{i=1}^s\frac{1}{a_is+1}$ where the numbers $a_i$ are not necessarily distinct. In fact we have precisely $t$ pairwise different elements which we denote as
\begin{align*}
\{a_1,a_2,\ldots,a_N\}=\{b_1,b_2,\ldots,b_t\}\qquad\text{with}\qquad |\{b_1,b_2,\ldots,b_t\}|=t
\end{align*}
Each $b_k, 1\leq k\leq t$ occurs in $\{a_1,a_2,\ldots,a_N\}$ with multiplicity $\nu_k$, so that $\sum_{k=1}^t\nu_k=N$.
We obtain the following representation
\begin{align*}
 \prod_{i=1}^N\frac{1}{a_is+1} = \prod_{k=1}^t\left(\frac{1}{b_ks+1}\right)^{\nu_k}\tag{1}
 \end{align*}
Selecting $k_0, 1\leq k_0\leq t$ we have a pole $-\frac{1}{b_{k_0}}$ of order $\nu_{k_0}$ and a Laurent series expansion
\begin{align*}
\prod_{k=1}^t\left(\frac{1}{b_ks+1}\right)^{\nu_k}&=\sum_{l=-\nu_{k_0}}^{-1}\alpha_l\left(s-\frac{1}{b_{k_0}}\right)^l+H(s)\\
&=\sum_{l=-\nu_{k_0}}^{-1}\alpha_l\left(-\frac{1}{b_{k_0}}\right)^l\left(b_{k_0}s+1\right)^l+H(s)
\end{align*}
where the right-hand sum is the principal part of the Laurent series and $H(s)$ is a holomorphic function. Multiplication with $(b_{k_0}s+1)^{\nu_{k_0}}$ gives
\begin{align*}
\prod_{{k=1}\atop{k\ne k_0}}^t\left(\frac{1}{b_ks+1}\right)^{\nu_k}
&=\sum_{l=-\nu_{k_0}}^{-1}\alpha_l\left(-\frac{1}{b_{k_0}}\right)^l\left(b_{k_0}s+1\right)^{l+\nu_{k_0}}+H(s)\left(b_{k_0}s+1\right)^{\nu_{k_0}}\\
&=\sum_{l=1}^{\nu_{k_0}}\underbrace{\alpha_{-l}\left(-b_{k_0}\right)^l}_{\rho_{k_{0},l}}\left(b_{k_0}s+1\right)^{\nu_{k_0}-l}+\tilde{H}(s)\tag{2}\\
\end{align*}
with $l$ substituted by $-l$ in the last line.

From (2) we derive OP's representation for $\rho_{k_{0},l_{0}}, 1\leq l_0\leq \nu_{k_{0}}$ as
\begin{align*}
\color{blue}{\rho_{k_{0},l_{0}}}
&\color{blue}{=\frac{1}{\left(\nu_{k_{0}}-l_0\right)!b_{k_{0}}^{\nu_{k_{0}}-l_0}}\,
\frac{d^{\nu_{k_{0}}-l_0}}{ds^{\nu_{k_{0}}-l_0}}\prod_{{k=1}\atop{k\ne k_{0}}}^t\left(b_ks+1\right)^{-\nu_{k}}}\tag{3}\\
\end{align*}
Using $\mu=\nu_{k_{0}}-l_0$ as short-hand we obtain:
\begin{align*}
\color{blue}{\rho_{k_{0},l_{0}}}
&=\frac{1}{\mu!b_{k_0}^{\mu}}\,
\frac{d^{\mu}}{ds^{\mu}}\prod_{{k=1}\atop{k\ne k_{0}}}^t\left(b_ks+1\right)^{-\nu_{k}}\\
&=\frac{1}{\mu!b_{k_{0}}^{\mu}}\,
\sum_{q_1+\cdots+q_{k_0}^{\prime}+\cdots+q_t=\mu}\binom{\mu}{q_1,\ldots,q_{k_0}^{\prime},\ldots,q_t}\prod_{{k=1}\atop{k\ne k_{0}}}^t\frac{d^{q_k}}{ds^{q_k}}\left(b_ks+1\right)^{-\nu_{k}}\tag{4}\\
&=\frac{1}{\mu!b_{k_{0}}^{\mu}}\,
\sum_{q_1+\cdots+q_{k_0}^{\prime}+\cdots+q_t=\mu}\binom{\mu}{q_1,\ldots,q_{k_0}^{\prime},\ldots,q_t}\\
&\qquad\qquad\qquad\qquad\cdot\prod_{{k=1}\atop{k\ne k_{0}}}^t\left(-\nu_k\right)^{\underline{q_k}}b_k^{q_k}\left(b_ks+1\right)^{-\nu_{k}-q_k}\tag{5}\\
&\,\,\color{blue}{=\frac{(-1)^{\mu}}{\mu!b_{k_{0}}^{\mu}}\,
\sum_{q_1+\cdots+q_{k_0}^{\prime}+\cdots+q_t=\mu}\binom{\mu}{q_1,\ldots,q_{k_0}^{\prime},\ldots,q_t}}\\
&\qquad\qquad\qquad\qquad
\color{blue}{\cdot\prod_{{k=1}\atop{k\ne k_{0}}}^t\frac{(\nu+q_k-1)!}{(\nu-1)!}b_k^{q_k}\left(b_ks+1\right)^{-\nu_{k}-q_k}}\tag{6}\\
\end{align*}

Comment:

In (4) we use the general Leibniz rule to find the $q_k$-th derivative of the product. We use the prime symbol for $q_{k_0}^{\prime}$ to indicate $q_{k_0}$ is not part of the sum resp. multinomial coefficient.

In (5) and (6) we use falling factorials
\begin{align*}
\left(-\nu_k\right)^{\underline{q_k}}&=(-\nu)(-\nu-1)\cdots(-\nu-q_k+1)\\
&=(-1)^{q_k}\nu(\nu+1)\cdots(\nu+q_k-1)\\
&=(-1)^{q_k}\frac{(\nu+q_k-1)!}{(\nu-1)!}
\end{align*}


TITLE: Number theory problem, parity
QUESTION [1 upvotes]: Let $f(x)$ denote the number of (not necessarily distinct) prime factors of $x$. Let $n > 1$ be the smallest positive integer for which there are more $i$ with with $f(i)$ even than $f(i)$ odd in the range $1 \leq i \leq n$. Compute $n$ modulo 1000.

REPLY [2 votes]: I don't know any way to do this without just doing a lot of computing, and I don't see the "modulo 1000" thing as making it any easier. At http://oeis.org/A072203 there is a tabulation of "(Number of oddly factored numbers $\le n$) $-$ (number of evenly factored numbers $\le n$)" where "A number $m$ is oddly or evenly factored depending on whether $m$ has an odd or even number of prime factors [counting with multiplicity]." It says,
"Polya conjectured that $a(n) \ge 0$ for all $n$, but this was disproved by Haselgrove. Lehman gave the first explicit counterexample, $a(906180359) = -1$; the first counterexample is at $906150257$ (Tanaka)."
The Tanaka reference is Tanaka, M. "A Numerical Investigation on Cumulative Sum of the Liouville Function". Tokyo Journal of Mathematics 3:1 (1980), pp. 187-189.


TITLE: Determining the probability of multiple condtional events
QUESTION [0 upvotes]: (I hope I used the correct terminology, I am a social sciences guy, not a math guy, and I am a bit out of my element. Please be kind.)
I am trying to determine the cumulative probability of two finite events with three possible outcomes.
To put it in simple terms, what is the cumulative probability of failure, failure being a roll of a 1 or 2 on a six sided die.
IE: Individual rolls a die, and gets a 1 (failure), and then rolls a second die 3 (success...)
Individual 2 rolls a die, and gets a failure on both rolls, and individual 3 succeeds on both rolls.
How do I calculate the aggregate chance that nothing will happen?

REPLY [0 votes]: The answer depends on a couple of things that you haven't fully specified. First of all, the probability of each of the 3 outcomes, and secondly, whether the events are indepedent or not (i.e., whether the outcome of the first event in any way influences the outcome of the second).
In your example, these are answerable: the three outcomes (A = 1 or 2, B = 3 or 4, C = 5 or 6) are equally probable, and so each has a probability of $1/3$. And the outcome of the second roll is not affected by the first. There are 9 possible outcomes of the two rolls, each having probability = $\frac 1 3 \times\frac 1 3 = \frac 1 9$: $$AA, AB, AC, BA, BB, BC, CA, CB, CC$$
If we consider $A$ a failure, then note that there are $5$ combinations including $A$, so the probability of failure will be $\frac 5 9$.


TITLE: Prove $\lim_{n \rightarrow \infty} \frac{1}{n}\cdot \frac{3 + \frac{1}{n}}{4 - \frac{1}{n}} = 0 $
QUESTION [1 upvotes]: Prove $\lim_{n \rightarrow  \infty} \frac{1}{n}\cdot \frac{3 + \frac{1}{n}}{4 - \frac{1}{n}} = 0   $
Let $\epsilon > 0$ be arbitrary. I want to find $N$ such that $n \in \mathbb{N}$ guarantees $ \left | \frac{1}{n}\cdot \frac{3 + \frac{1}{n}}{4 - \frac{1}{n}} - 0  \right | < \epsilon$.
$\Leftrightarrow \frac{1}{n}\cdot \frac{3 + \frac{1}{n}}{4 - \frac{1}{n}} < \epsilon$
$\Leftarrow \frac{1}{n} \cdot \frac{4}{3} < \epsilon \Leftrightarrow n > \frac{4}{3} \cdot \frac{1}{\epsilon} \Rightarrow n \geq N$
Take $N = \left \lfloor \frac{4}{3} \cdot \frac{1}{\epsilon} \right \rfloor + 1$
In the above proof, I'm confused as to how $\frac{1}{n}\cdot \frac{3 + \frac{1}{n}}{4 - \frac{1}{n}} < \epsilon$
$\Leftarrow \frac{1}{n} \cdot \frac{4}{3} < \epsilon$. How did $\frac{3 + \frac{1}{n}}{4 - \frac{1}{n}}$ become $\frac{4}{3} $?

REPLY [0 votes]: Since $n \in \mathbb N \implies n \geq 1$, we know that:
$$
\frac{1}{n} \leq 1 \iff 3 + \frac{1}{n} \leq 4
$$
as well as that:
$$
\frac{-1}{n} \geq -1 \iff 4 - \frac{1}{n} \geq 3
$$
Now to make a fraction smaller, we can either decrease the numerator or increase the denominator. Thus, it follows that:
$$
\epsilon > \frac{1}{n} \cdot \frac{4}{3}
\geq \frac{1}{n} \cdot \frac{3 + \frac{1}{n}}{4 - \frac{1}{n}}
$$


TITLE: Meaning of $A^B$ where $A$ and $B$ are sets
QUESTION [4 upvotes]: I've tried to google this with a variety of search terms, but most results refer to the power set, which is $2^S$, not $A^B$.
Does it mean the set of all maps (functions) from the set $B$ to the set $A$? What is the definition of this notation?
Thank you for reading!

REPLY [1 votes]: As others have said, yes, $A^B$ means the set of functions from $B$ to $A$. I would add that, with this notation, we also have the "exponential law"
$$
(A^B)^C \cong A^{B \times C}
$$
for any sets $A, B, C$, where $B \times C$ is the Cartesian product of $B$ and $C$.


TITLE: When does a stochastic process have its sample paths a.s. in the reproducing kernel hilbert space (RKHS) induced by its covariance function?
QUESTION [2 upvotes]: Let $T$ be a compact metrizable space. Consider a centered second order measurable process $(X_t\colon t\in T)$ with continuous covariance function $c(t,s):= \mathbb{E}X_t X_s$.
Are there any known sufficient conditions ensuring that $t \mapsto X_t$ lies almost surely in the reproducing kernel Hilbert space $\mathcal{H}(c)$ generated by $c$? What about necessary conditions on that matter?
The question is motivated by the Karhunen-Loève representation for $X_t$ (endow $T$ with a strictly positive Borel measure $\nu$ for this to make sense),
\begin{equation}
\sup_{t\in T} \mathbb{E}\left(X_t - \sum_{j=1}^n\langle X_\cdot , \varphi_j\rangle_{L^2(\nu)}\varphi_j(t)\right)^2\rightarrow 0
\end{equation}
and the fact that $\{\sqrt{\lambda_j}\varphi_j\}$ is an orthonormal basis for $\mathcal{H}(c)$ so that one may wonder whether the (formal) series
\begin{equation}
\sum_{j=1}^\infty a_j\,\sqrt{\lambda_j}\varphi_j, \qquad a_j:=\lambda_j^{-1/2}\langle X_\cdot , \varphi_j\rangle_{L^2(\nu)}
\end{equation}
is a well defined element of $\mathcal{H}(c)$.
In the above, $(\lambda_j,\varphi_j)$ are the eigenvalues / eigenfunctions as given in Mercer's Theorem.

REPLY [3 votes]: The Gaussianity assumption mentioned in the answer of Joris is a particular case of a more general condition:

The family of random variables $\{\langle X, \varphi \rangle^2 \mid \varphi \in S\}$ is uniformly integrable, where $S$ is the set of linear functionals of the form $\langle X, \varphi \rangle := \sum_{k=1}^n \varphi_k X(t_k)$ whose $\mathcal{H}(c)^\ast$-norm is $\le 1$ (i.e. $\forall \xi \in \mathcal{H}(c): \sum_k \varphi_k \xi(t_k) \le \Vert \xi \Vert_{\mathcal{H}(c)}$).

In the Gaussian case this condition is satisfied because the $L^2(\Omega,\mathsf{P})$ norm of the linear functionals $\langle X, \varphi \rangle$ (which is the same as $\Vert \varphi \Vert_{\mathcal{H}(c)^\ast}$) bounds an $L^p(\Omega,\mathsf{P})$ norm for $p>2$. More generally, it is also satisfied for processes that depend polynomially on a Gaussian system, or on Bernoulli random signs, or basically on anything where hypercontractivity works.
In general the condition above is enough to conclude that the sample paths of $X$ do not lie in $\mathcal{H}(c)$ (unless it's finite-dimensional).
To prove this, note that by the Lebesgue's theorem, uniform integrability implies that the $L^2$ and $L^0$ topologies (the latter is convergence in probability) coincide on the space of linear functionals of $X$. Note that the $L^0$ topology won't change if we change the probability measure $\mathsf{P}$ by an equivalent measure $\mathsf{Q}$.
Now argue towards a contradiction. Let $X$ be a random element of $\mathcal{H}(c)$. Then $\Vert X \Vert_{\mathcal{H}(c)}^2$ is a well-defined, finite random variable, so there exists a measure $\mathsf{Q}$, equivalent to $\mathsf{P}$, such that $\mathsf{E}_{\mathsf{Q}} \Vert X \Vert_{\mathcal{H}(c)}^2 < \infty$. On the other hand, $\mathsf{E}_{\mathsf{Q}} \Vert X \Vert^2_{\mathcal{H}(c)}$ is the trace of the covariance operator of $X$ in $\mathcal{H}(c)$ (computed for the measure $\mathsf{Q}$), so the covariance must be trace class, and therefore compact. Thus there is a sequence of linear functionals $\varphi_n \in S$, such that $\Vert \varphi_n \Vert_{\mathcal{H}(c)^\ast} = 1$, and $\langle \varphi_n, X \rangle \to 0$ in $L^2(\mathsf{Q})$. Thus we also have $\langle \varphi_n, X \rangle \to 0$ in $L^0(\mathsf{P})$, and, by uniform integrability, also in $L^2(\mathsf{P})$. This contradicts the fact that $\Vert \langle \varphi, X \rangle \Vert_{L^2(\mathsf{P})} = \Vert \varphi \Vert_{\mathcal{H}(c)^\ast} = 1$.


TITLE: Can it be decidable for any polynomials to have the intersecting point?
QUESTION [5 upvotes]: Give system of polynomials$$P_1(x_1,x_2,\dots,x_n)=0,$$$$\vdots,$$$$P_k(x_1,x_2,\dots,x_n)=0$$
Can it be decidable for thoses polynomials to have the intersecting point ?

REPLY [3 votes]: Given an ideal $I$ of $k[x_1,\ldots,x_n]$ with $k$ an algebraically closed field of characteristic zero, you can effectively compute a $Gr\ddot{o}bner$ basis for it. Checking whether $Z(I)$ is empty is equivalent to checking whether $1$ is in the  $Gr\ddot{o}bner$ basis (by Hilbert's Nullstellensatz). Checking whether $1$ is in the  $Gr\ddot{o}bner$ basis is a trivial check.
Maybe the same works in characteristic $>0$ as well.
As André Nicolas says the problem is much more difficult over non-algebraically closed fields, even if you restrict to just smooth projective curves.


TITLE: $f(0)=f(1)=0$, $f(x)=\frac{f(x+h)+f(x-h)}{2}$ implies $f(x)=0$ for $[0, 1]$
QUESTION [2 upvotes]: Question: Suppose $f$ is continuous on $[0, 1]$ with $f(0)=f(1)=0$. For $\forall x\in (0, 1)$, there $\exists h>0$ with $0\le x-h<x<x+h\le1$ such that  $f(x)=\frac{f(x+h)+f(x-h)}{2}$. Show that $\forall x\in(0, 1), f(x)=0$.
I tried to prove that $f$ is differentiable on $(0, 1)$ using the fact that $\frac{f(x+h)-f(x)}{h}=\frac{f(x)-f(x-h)}{h}$, but I realized that not all $h$ holds the equation, but there exists a particular $h$ in every $x$. So, this is not a proper approach.
I also thought about the concavity of $f$. Since $\forall x\in(0, 1),\exists h>0$ with $0\le x-h<x<x+h\le1$ such that
$$f(x)=f\left(\frac{x-h}{2}+\frac{x+h}{2}\right)\ge {1\over2}f\left(x-h\right)+{1\over2}f\left(x+h\right)$$
, which implies $f$ is concave downward, and
$$f(x)=f\left(\frac{x-h}{2}+\frac{x+h}{2}\right)\le {1\over2}f\left(x-h\right)+{1\over2}f\left(x+h\right)$$
, which implies $f$ is concave upward.
Two facts might imply that $f$ is constant, which in turn $\forall x\in[0, 1], f(x)=0$ since $f(0)=f(1)=0$. Is this approach correct?
I thought it has to be more precise, so I wanted to use the second derivative. But I actually failed to prove that $f$ is differentiable. Could you please give me some ideas about the question? Thanks a lot.

REPLY [4 votes]: Since $f$ is continuous on a closed, bounded interval, it attains its bounds.
So, let $x^*\in [0,1]$ be a point such that $f(x^*)$ is a maximum on $[0,1].$ However, we know that $f(x^*)$ is the average of two points around it, which (since $f(x^*)$ is a maximum) must also be maxima, i.e.
$$\exists h_1>0\:\colon f(x^*+h_1)=f(x^*).$$
Repeating for $x^*+h_1,$ we form a sequence $x^*, x^* + h_1,x^*+h_1+ h_2,\dotsc.$ This sequence converges, since it is bounded above by 1.
Suppose it converges to some $L< 1.$ But then, since $f$ is continuous, $f(L)$ is also equal to $f(x^*)$ (by sequential continuity). So we can repeat the above process with $f(L)$ to obtain a new point $L+h$ at which $f$ also attains its maximum (a contradiction).
Hence $L=1$ and
$$f(x^*)=f(1)=0.$$
Repeating with the minimum of $f$ gives us that the maximum and minimum values attained by $f$ are both $0,$ and so $f$ is constant.


TITLE: Energy of the triplet states and the singlet states
QUESTION [0 upvotes]: When dealing with a system of two spin-$1/2$ particles, we use the singlet and triplet states as a common basis for the complete set of commuting observables $\{\hat S ^2,\hat S_z, \hat S_{z,1}, \hat S_{z,2}\}$, where $\hat {\mathbf S}=\hat {\mathbf S}_1+\hat {\mathbf S}_2$:
$$
\left.\begin{array}{ll}
|1,1\rangle & =\uparrow \uparrow \\
|1,0\rangle & =\frac{1}{\sqrt{2}}(\uparrow \downarrow+\downarrow \uparrow) \\
|1,-1\rangle & =\downarrow \downarrow \\
\end{array}\right\} \quad s=1 \quad \text { (triplet) } \\
\left.|0,0\rangle=\frac{1}{\sqrt{2}}(\uparrow \downarrow-\downarrow \uparrow)\right\} \quad s=0 \quad \text { (singlet) }
$$
It is said that the singlet state has lower energy than those states of the triplet, which I think makes sense since in the singlet state there are no unpaired particles, whereas in the triplet state the particles are unpaired. My question is: what is the energy associated to these states? And, are states of the triplet state all of the same energy?

REPLY [2 votes]: It's not necessarily the case, It depends on your Hamiltonian. It's true for the case of Hyperfine interaction, where in addition to the Coulomb interaction, there exists another interaction between the electron and proton in the Hydrogen atom. The Hamiltonian describing this interaction, which is due to  the magnetic moments of the two particles is,
$$H_{\text{hf}}=A\mathbf{S}_1\cdot \mathbf{S}_2\ \ \ (A>0)$$
$H_\text{hf}$ splits the ground state into two levels:
\begin{align*}
E_+&= -\text{Ry}+\frac{\hbar^2A}{4}\\
E_- &= -\text{Ry}-\frac{3\hbar^2 A}{4}
\end{align*}
and that corresponding states are triplets and singlet, respectively.


TITLE: Propagator in Path Integral Quantum Mechanism as Green Function of Schrodinger Equation
QUESTION [1 upvotes]: I'm studying in Ryder's book of QFT. I'm dealing with QM in the path integral approach and he is trying to prove that the propagator $K(x_f t_f;x_i t_i)$ is the Green function of the Schrodinger (S.) equation:
\begin{equation}
\frac{\hbar^2}{2m} \frac{d^2}{dx_f^2}\psi(x_f t_f) + i \hbar \frac{\partial}{\partial t_f} \psi(x_f t_f) =V(x_ft_f) \psi(x_ft_f) \quad (1)
\label{one}
\end{equation}
We have a generic wave function that satisfies S. equation $\psi$ and we can rewrite it as (eq. 5.27)
\begin{equation}
\psi(x_f t_f)= \phi(x_f t_f) - \frac{i}{\hbar}\int dx dt \; K_0(x_f t_f;x t)V(x,t)\psi(xt) \quad (2)
\label{eq:2}
\end{equation}
where $\phi(x_f t_f)$ is a free plane wave that therefore satisfies the free S. eq:
\begin{equation}
\frac{\hbar^2}{2m} \frac{d^2}{dx_f^2}\phi(x_f t_f) + i \hbar \frac{\partial}{\partial t_f} \phi(x_f t_f) =0 \quad (3)
\end{equation}
and $K_0$ is the free propagator. So plugging eq(1) into eq(2) and using eq (3) Ryder derives this
\begin{equation}
\frac{\hbar^2}{2m} \frac{d^2}{dx_f^2}K_0(x_f t_f;x t)+ i \hbar \frac{\partial}{\partial t_f} K_0(x_f t_f;x t) = \frac{-i}{\hbar} \delta(x_f-x)\delta(t_f-t)
\end{equation}
My problem is to get to this equation. Plugging (2) into (1) I can get as far as
\begin{equation}
\frac{-i}{\hbar}\int dx dt \; \left[\frac{\hbar^2}{2m} \frac{d^2}{dx_f^2} + i \hbar \frac{\partial}{\partial t_f}\right] ( K_0(x_f t_f;x t)) \cdot V(xt) \psi(xt) = V(x_ft_f) \phi(x_ft_f) - \frac{i}{\hbar} V(x_ft_f) \int dx dt K_0(x_f t_f;x t)V(x,t)\psi(xt)
\end{equation}
where I used (3) in the left hand side.

REPLY [0 votes]: I am sorry I do not have access to Ryder and would be reluctant to read it with you, anyway. I'll just point out that it is aggressively confusing to be calling $K_0$ the propagator, since it solves the inhomogeneous free TDSE, so it is a Green function, normally called G, that the three linked answers take pains to contrast to the propagator, the fundamental solution of the homogeneous equation.
It might, or might not, help you to note that, if, instead, you assumed it is a Green function,
\begin{equation}
\frac{\hbar^2}{2m} \frac{d^2}{dx_f^2}K_0(x_f t_f;x t)+ i \hbar \frac{\partial}{\partial t_f} K_0(x_f t_f;x t) = i\hbar \delta(x_f-x)\delta(t_f-t) \tag{5}
\end{equation}
then inserting (2) on the left-hand-side of (1), only, by virtue of (3), you obtain directly
\begin{equation}
\left(\frac{\hbar^2}{2m} \frac{d^2}{dx_f^2}  + i \hbar \frac{\partial}{\partial t_f} \right ) \psi(x_f t_f)\\  =  - \frac{i}{\hbar}\int dx dt \;      \left(\frac{\hbar^2}{2m} \frac{d^2}{dx_f^2}  + i \hbar \frac{\partial}{\partial t_f} \right )   K_0(x_f t_f;x t)V(x,t)\psi(xt)
\\ =  \int dx dt \;  \delta(x_f-x)\delta(t_f-t) V(x,t)\psi(xt)=V(x_ft_f) \psi(x_ft_f),
\end{equation}
so you obtain the right hand side of (1), that is you derive (1).
You might, or might not, choose to work backwards.


TITLE: Area of the region bounded by the curves $y=x^2$ & $y=x^4$
QUESTION [0 upvotes]: One of my last assignments for my first semester of calculus is a list of area, volume, and work problems using integrals, I worked one out, but can't find the answer anywhere, and just want to make sure I'm correct, here's my work:
Find the area of the region bounded by the curves $y=x^2$ and $y=x^4$
$x^4=x^2$
$x^4-x^2=0$
$x^2(x^2-1)=0$
$x=1, -1$
$\int_1^1(x^4-x^2)dx$
$\left[\frac{x^5}{5}-\frac{x^3}{3}\right]$ from $-1$ to $1$
$\left[\frac{(1)^5}{5}-\frac{(1)^3}{3}\right] - \left[\frac{(-1)^5}{5}-\frac{(-1)^3}{3}\right]$
$\left[\frac{1}{5}-\frac{1}{3}\right] - \left[\frac{-1}{5}+\frac{1^3}{3}\right]$
$\left[\frac{3-5}{15}-\frac{-3+5}{15}\right]$
$\left[\frac{-2}{15}+\frac{-2}{15}\right] = \frac{-4}{15}$
I double checked the definite integral online and it got $\frac{-4}{15}$ as well, so I guess that means I"m correct I'm just confused as to how an area can be negative unless I'm missing something obvious.
My prof also left a note that says "*Make sure you have the entire region", where "entire" is underlined, is there more to this problem than what I've done?

REPLY [1 votes]: I think you do indeed have the entire region. The only problem is that your integral should have been
$$\int_{-1}^1(x^2-x^4)dx$$
instead, because $x^2≥x^4$ in the region where $-1<x<1$.