TITLE: A question about metric spaces that are compact and denumerably infinite
QUESTION [1 upvotes]: Let S be any compact and denumerably infinite metric space and let d be the metric of S. We shall say that S satisfies condition C, if there exists at least one infinite sequence-------------p(1),p(2),.....,p(n),....of points of S such that (1) each point of S occurs once and only once in the sequence (2) the series d(p(1),p(2))+d(p(2),p(3))+.....+d(p(n),p(n+1))+....is convergent. If S satisfies condition C, an infinite version of the travelling salesman problem can be posed for the points of S. S always contains at least one limit point, since otherwise it could be covered by an infinite set of pairwise disjoint open balls, each containing a point of S-an impossibility if S is compact. It is easy to show that S cannot satisfy condition C if it contains more than one limit point. My question is: If S contains one and only one limit point, does S then necessarily always satisfy condition C?

REPLY [1 votes]: I'll show a counterexample in Euclidean $\ (\mathbb R^2\ \,d)$:

THEOREM There exists a countable compact $\ X\subseteq\mathbb R^2,\ $ which has exactly one limit point (hence $\ X\ $ is infinite), and such that every route $\ (p(1)\ p(2)\ \ldots)\ $ of $\ X\ $ has infinite length.

Here we understand that $\ (p(1)\ p(2)\ \ldots)\ $ being a route of $\ X\ $ means that all points $\ p(t)\ $ are different, and
$$ \{p(1)\ p(2)\ \ldots\}\ =\ X$$
and the length of such route is $\ \sum_{k=1}^\infty d(p(k)\ p(k+1))$.
PROOF   Let $\ O:=(0\ 0)\in\mathbb R^2.\ $ Let variables $\ k\ n\ $ run only through natural numbers $\ 1\ 2\ \ldots\ .\ $ Define
$$ X\ :=\ \{O\}\ \cup\ \left\{(\frac 1k\ \frac 1n)\ :\ \max(k\ n)\le 2\cdot\min(k\ n)\right\} $$
Consider $\ \ \delta(x) := \min_{y\in X\setminus\{x\}} d(x\ y)\ \ $ for every $\ x\in\mathbb R^2\setminus\{O\}.\ $ This minimum is well defined and satisfies:
$$ \delta\left(\left(\frac 1k \frac 1n\right)\right)\ \ge\ \frac 1{\max(k\ n)\cdot(\max(k\ n)+1)}$$
There are exactly $\ 2\cdot n+1\ $ points $\ \left(\frac 1r\ \frac 1s\right)\in X\ $ such that $\ \min(r\ s) = n.\ $ For each $\ 2\cdot n+1\ $ of such points $\ x\ $ we have $\ \delta(x)\ge \frac 1{2\cdot n\cdot(2\cdot n+1)}.\ $
Now consider an arbitrary route $\ (p(1)\ p(2)\ \ldots)\ $ of $\ X.\ $ For each of the $\ 2\cdot n+1\ $ points $\ p(t)\ $ such that it is one of points $\ x\ $ from above, we get:
$$ d(p(t)\ p(t+1))\ \ge\ \frac 1{2\cdot n\cdot(2\cdot n+1)}$$
Thus the sum of these $\ 2\cdot n + 1\ $ distances is at least $\ \frac 1{2\cdot n}.\ $ Thus (with a harmless abuse of notation):
$$\sum_{t=1}^\infty d(p(t)\ p(p(t+1))\ \gt \sum_{n=1}^\infty \frac 1{2\cdot n}\ \ge\ \infty$$
END of PROOF


TITLE: Justifying $\log{\frac{1}{P_{X}(x)}}$ as the measure of self information
QUESTION [1 upvotes]: I was reviewing self information and then came to realize that there is one idea that I have that I believe should be wrong but don't know why.
Let self-information associated with a random variable realization be:
$$I_{X}(x) = \log{\frac{1}{P_{X}(x)}}$$
The main idea is that realizing rare events contains more information that realizing event that you knew would occur for sure, being because rare events tell us more information about the distribution than what we already knew about it.
However, for extremely low values of $P_{X}(x)$ I find that justification (or intuition) for defining $I_{X}(x)$ the way it was just weird or wrong.
For example, say that I was trying to learn a distribution by observing values coming from it every so often and applying some learning algorithm or sampling technique. Then, an event that is extremely rare happens and I update what I have learned so far about the distribution taking into account the rare event. However, since the event is so rare, we update what we have learned and if the learning procedure makes sense, I should that it should not change its estimate of the distribution too much, because otherwise it wouldn't be consistent that this event is actually rare. Therefore, that means for me that we could have just not seen the event and it would have been just as good off. We might never see that event again anyway. So event with low probability actually didn't tell us that much about that distribution itself.
This intuition is exactly the opposite to how the definition of information is justified. Does someone know why my argument does not make sense? Is it just for events that basically have a probability so close to zero that they are basically zero?

REPLY [3 votes]: There is a good example on page 71 of David MacKay's book Information Theory, Inference, and Learning Algorithms (link to legal download page). Imagine a game of battleships on a chessboard where there is only one ship and 63 empty squares. As we ask "is it on square x?" for each of the squares, the answer is probably "no", so a "no" answer gives us little information (the figures are in the book itself). If the answer is "yes" then we suddenly get a lot of information (game over). In any case the sum of all the self-informations is always six bits.
If you could ask questions like "Is it in the top half?" then the answer to each question would have information content of one bit. In this case we would get to the answer more quickly than by asking about individual squares, but the sum of all the self-informations is still exactly six bits.


TITLE: find power of orthonormal matrix to get identity matrix
QUESTION [1 upvotes]: I have this orthonormal matrix:
$ Q =\frac{1}{9} \left(\begin{matrix}
4 & 8 & -1 \\
-4 & 1 & -8 \\
-7 & 4 & 4
\end{matrix}\right)$
If I calculate $Q^4$, I get the identity matrix. Is there a way to calculate the power $k$ of $Q^k$ to get the identity?

REPLY [1 votes]: By computing $Q^TQ=I_3$ we see that $Q$ is orthogonal. Hence $Q$ is similar to
$$R_\theta=\begin{pmatrix}\pm1&0&0\\
0&\cos\theta&-\sin\theta\\
0&\sin\theta&\cos\theta
\end{pmatrix}$$
Let $C_i$ the $i$-th column of $Q$. We have $C_1\wedge C_2=C_3$ so $Q$ is a rotation i.e. positive orthogonal matrix, hence $+1$ is an eigenvalue and from the trace of $Q$ we get $\cos\theta=0$ and so $\sin\theta=\pm1$. Now we see that $R_\theta^4=I_3$ hence $Q^4=I_3$.


text:
\section{Related Work} \label{s:related}

The problem of speeding up block propagation has received considerable attention in the blockchain community: for example using block compression \cite{}. There are also cryptographic and coding techniques to reduce the message overhead in block forwarding.
For example, in Graphene a combination of Bloom filters and IBLTs are used to compact the messages needed for syncing the local copies of blockchain across the \pp network. In Velocity~\cite{chawla2019velocity},  on the other hand, fountain codes are used for decentralized delivery of blocks and subsequent reconstruction at the peer.

However, optimizing the \pp topology itself for faster delivery in blockchains has not received as much attention. While much work has been done in traditional

An exception is Kadcast~\cite{rohrer2019kadcast} which proposes using the Kademlia's structured overlay, and UDP-based block forwarding, for

a more efficient propagation.
Many structured overlays for efficient content storage and lookup, including with good locality properties~\cite{rowstron2001pastry}, have been proposed in the past but they are not explicitly optimized for broadcast traffic~\cite{lua2005survey}.
Indeed Bitcoin, and other popular blockchains, do not require communication primitives beyond simple broadcasting of messages, and this is a key reason why an unstructured random topology has sufficed there. For example, the Bitcoin p2p network does not support sending messages targeted to a particular peer (unicast).

Some prior work for blockchains has focused on building dedicated high-speed low-latency distribution networks, such as bloXroute~\cite{klarman2018bloxroute}, Falcon, FIBRE.

These solutions are not fully decentralized as using them requires the peers to place trust on the distribution network in the middle.
The routes are also susceptible to man-in-the-middle attack.

config: arxiv


TITLE: Can particle physics be represented as an algebra?
QUESTION [2 upvotes]: Possibly the most useful thing anyone could tell me about particle physics:
Naively, one could try and make an algebra by enumerating all the types of particles and defining equivalence relationships (some rules), for example
Objects:
object type 1: photon
object type 2: electron
object type 3 ...
object type n
Rules:
rule 1: a photon is equivalent to an electron and a positron
rule 2: a proton is equivalent to three quarks
rule ....
rule n
Then one could study the properties of these relationships as a mathematical structure containing objects that represent particles and rules that define equivalence - i.e. as an algebra. I assume such an algebra would be, if formulated correctly, represented by Feynman diagrams (for some particles at least). Does such an algebra have a name? Is there more than one? Is this at all connected to the idea behind stating that the standard model is U(1) x SU(2) x SU(3)?
By an algebra I understand a closed and consistent collection of mathematical objects and relationships.

Thanks for trying to understand what I'm asking, with your input I can aim to make this more comprehensible. So I guess I'm looking for a formal structure which reflects particle reactions which is abstracted away from spatial, temporal and energetic considerations as far as is possible (is it at all possible?).

REPLY [1 votes]: What you want is an algebraic relation on particles which tells you which ones are allowed to turn into which other ones. These algebraic relations exist, and they are the conservation laws. The conservation laws come in two types--- discrete and continuous. The full list of exact conservation laws is known, and the exact ones are these:

Energy/Momentum
Angular momentum
Electric charge
CPT

In addition, there is one more conserved quantity, which is algebraically more complicated

Strong color charge

This one is only applicable at short distances, and it tells you which collections of quarks and gluons can turn into which other collections of quarks and gluons. All objects are neutral at long distances.
In addition to these laws, there are two more nearly exact conservation laws:

Generational lepton number (number of electrons+neutrinos of each generation)
Baryon number (the number of end-stable protons plus neutrons)

There are further conservation laws which are not at all exact, but are preserved by the electromagnetic and strong interactions at low energies. These are parity, and charge conjugation.
All processes which obey the exact conservation laws are allowed in principle, although they might have vanishingly small probability amplitude.


TITLE: Can a black hole absorb linear momentum?
QUESTION [2 upvotes]: I know that black holes maintain the linear momentum of the object that created it. But if I throw a baseball into a black hole, will it absorb that momentum as well?
I think the answer has to be "yes" because momentum needs to be conserved, but I am having trouble getting my head around how a black hole can react to momentum.

REPLY [2 votes]: Black hole will absorb the momentum and start to move with a constant speed.
A black hole solution can be transformed to any other inertial frame via a global Lorentz transformation, so it can move with any speed through space.


text:
\begin{document}

	\maketitle

	\begin{abstract}
		Let $ (G_n)_{n=0}^{\infty} $ be a non-degenerate linear recurrence sequence with power sum representation $ G_n = a_1(n) \alpha_1^n + \cdots + a_t(n) \alpha_t^n $. In this paper we will prove a function field analogue of the well known result that in the number field case, under some non-restrictive conditions, for $ n $ large enough the inequality $ \abs{G_n} \geq \left( \max_{j=1,\ldots,t} \abs{\alpha_j} \right)^{n(1-\varepsilon)} $ holds true.
	\end{abstract}

	\section{Introduction}

	Let $ (G_n)_{n=0}^{\infty} $ be a non-degenerate linear recurrence sequence with power sum representation $ G_n = a_1(n) \alpha_1^n + \cdots + a_t(n) \alpha_t^n $.
	This expression makes sense for a sequence $ (G_n)_{n=0}^{\infty} $ taking values in any field $ K $; the characteristic roots $ \alpha_i $ as well as the coefficients of the polynomials $ a_i $ then lie in a finite extension $ L $ of $ K $. In this paper $ K $ is either a number field or a function field in one variable of characteristic zero.
	The condition non-degenerate means in the number field case that no ratio $ \alpha_i/\alpha_j $ for $ i \neq j $ is a root of unity and in the function field case that no ratio $ \alpha_i/\alpha_j $ for $ i \neq j $ is contained in the field of constants.
	In the number field case it is well known that, if $ \max_{j=1,\ldots,t} \abs{\alpha_j} > 1 $, then for any $ \varepsilon > 0 $ the inequality
	\begin{equation}
		\label{p3-eq:ineqofinterest}
		\abs{G_n} \geq \left( \max_{j=1,\ldots,t} \abs{\alpha_j} \right)^{n(1-\varepsilon)}
	\end{equation}
	is satisfied for every $ n $ sufficiently large.

	The purpose of this paper is to prove an analogous result in the case of a function field in one variable of characteristic zero. Firstly, we will prove a theorem which states an inequality for an arbitrary valuation in the splitting field $ L $ of the characteristic polynomial belonging to the linear recurrence sequence.
	Secondly, we will derive a corollary for the special case of polynomial power sums. In this special case the inequality takes a form very similar to \eqref{p3-eq:ineqofinterest}.
	At this point we remark the following: Theorem 1 in \cite{fuchs-petho-2005} already implies that under some non-degeneracy conditions the degree of polynomials in a linear recurrence sequence with polynomial roots cannot be bounded and therefore must grow up to infinity as $ n $ does. But this theorem does not say how fast it must grow. In the present paper we will give a bound depending on $ n $ for the minimal possible degree of $ G_n $.

	Since the number field case is often mentioned, e.g. in \cite{vanderpoorten-1984} or in \cite{evertse-1984} or more recently in \cite{akiyama-evertse-petho-2017} and \cite{bugeaud-kaneko-2019}, but it is not that easy to access a proof of it (cf. \cite{vanderpoorten-schlickewei-1982} or the formulation in \cite{vanderpoorten-1989}), we give a complete proof based on results of Evertse and Schmidt in the appendix.
	By doing so we contribute to the goal that well known facts should be fully accessible with proof following van der Poorten's wise statement in \cite{vanderpoorten-1989} that ``all too frequently, [that] the well known is [often] not generally known, let alone known well''.

	\section{Results and notations}

	Throughout the paper we denote by $ K $ a function field in one variable over $ \CC $. By $ L $ we usually denote a finite algebraic extension of $ K $.
	For the convenience of the reader we give a short wrap-up of the notion of valuations that can e.g. also be found in \cite{fuchs-heintze-p2}:
	For $ c \in \CC $ and $ f(x) \in \CC(x) $, where $ \CC(x) $ is the rational function field over $ \CC $, we denote by $ \nu_c(f) $ the unique integer such that $ f(x) = (x-c)^{\nu_c(f)} p(x) / q(x) $ with $ p(x),q(x) \in \CC[x] $ such that $ p(c)q(c) \neq 0 $. Further we write $ \nu_{\infty}(f) = \deg q - \deg p $ if $ f(x) = p(x) / q(x) $.
	These functions $ \nu : \CC(x) \rightarrow \ZZ $ are up to equivalence all valuations in $ \CC(x) $.
	If $ \nu_c(f) > 0 $, then $ c $ is called a zero of $ f $, and if $ \nu_c(f) < 0 $, then $ c $ is called a pole of $ f $, where $ c \in \CC \cup \set{\infty} $.
	For a finite extension $ K $ of $ \CC(x) $ each valuation in $ \CC(x) $ can be extended to no more than $ [K : \CC(x)] $ valuations in $ K $. This again gives up to equivalence all valuations in $ K $.
	Both, in $ \CC(x) $ as well as in $ K $ the sum-formula
	\begin{equation*}
		\sum_{\nu} \nu(f) = 0
	\end{equation*}
	holds, where the sum is taken over all valuations in the considered function field.
	Moreover, valuations have the properties $ \nu(fg) = \nu(f) + \nu(g) $ and $ \nu(f+g) \geq \min \set{\nu(f), \nu(g)} $ for all $ f,g \in K $.
	Each valuation in a function field corresponds to a place and vice versa.
	The places can be thought as the equivalence classes of valuations.
	For more information about valuations and places we refer to \cite{stichtenoth-1993}.

	For any power sum $ G_n = a_1(n) \alpha_1^n + \cdots + a_t(n) \alpha_t^n $ with $ a_j(n) = \sum_{k=0}^{m_j} a_{jk} n^k $ and any valuation $ \mu $ (in a function field $ L/K $ containing the $ \alpha_j $ and the coefficients of the $ a_j $) we have the trivial bound
	\begin{align*}
		\mu(G_n) &= \mu\left( a_1(n) \alpha_1^n + \cdots + a_t(n) \alpha_t^n \right)
		\geq \min_{j=1,\ldots,t} \mu(a_j(n) \alpha_j^n) \\
		&\geq \min_{j=1,\ldots,t} \mu(a_j(n)) + \min_{j=1,\ldots,t} \mu(\alpha_j^n) \\
		&\geq \min_{j=1,\ldots,t} \min_{k=0,\ldots,m_j} \mu(a_{jk} n^k) + n \cdot \min_{j=1,\ldots,t} \mu(\alpha_j) \\
		&= \min_{\substack{j=1,\ldots,t \\ k=0,\ldots,m_j}} \mu(a_{jk}) + n \cdot \min_{j=1,\ldots,t} \mu(\alpha_j)
		= \widetilde{C} + n \cdot \min_{j=1,\ldots,t} \mu(\alpha_j).
	\end{align*}
	Our main result is now the following theorem which gives a bound in the other direction:

	\begin{mythm}
		\label{p3-thm:funcfieldcase}
		Let $ (G_n)_{n=0}^{\infty} $ be a non-degenerate linear recurrence sequence taking values in $ K $ with power sum representation $ G_n = a_1(n) \alpha_1^n + \cdots + a_t(n) \alpha_t^n $. Let $ L $ be the splitting field of the characteristic polynomial of that sequence, i.e. $ L = K(\alpha_1, \ldots, \alpha_t) $. Moreover, let $ \mu $ be a valuation on $ L $. Then there is an effectively computable constant $ C $, independent of $ n $, such that for every sufficiently large $ n $ the inequality
		\begin{equation*}
			\mu(G_n) \leq C + n \cdot \min_{j=1,\ldots,t} \mu(\alpha_j)
		\end{equation*}
		holds.
	\end{mythm}

	For the special case of a linear recurrence sequence of complex polynomials having complex polynomials as characteristic roots we get the following lower bound for the degree of the $ n $-th member of the sequence:

	\begin{mycorol}
		\label{p3-corol:polypowersum}
		Let $ (G_n)_{n=0}^{\infty} $ be a non-degenerate linear recurrence sequence of polynomials in $ \CC[x] $ with power sum representation $ G_n = a_1(n) \alpha_1^n + \cdots + a_t(n) \alpha_t^n $ such that $ \alpha_1, \ldots, \alpha_t \in \CC[x] $. Then there is an effectively computable constant $ C $, independent of $ n $, such that for every sufficiently large $ n $ the inequality
		\begin{equation*}
			\deg G_n \geq n \cdot \max_{j=1,\ldots,t} \deg \alpha_j - C
		\end{equation*}
		holds.
	\end{mycorol}

	It would be interesting to also prove a function field variant of Corollary 3.1 in \cite{akiyama-evertse-petho-2017}. However, because of Lemma 3.1 therein, which is based on Dirichlet's classical approximation theorem, we are not (yet) able to prove such a statement.

	In the proof given in the next section we are going to apply the subsequently given special case of Theorem 1 in \cite{fuchs-petho-2005}:
	\begin{myprop}
		\label{p3-prop:luorbounded}
		Let $ K $ be as above and $ L $ be a finite extension of $ K $ of genus $ \gfr $. Let further $ \alpha_1, \ldots, \alpha_d \in L^* $ with $ d \geq 2 $ be such that $ \alpha_i / \alpha_j \notin \CC^* $ for each pair of subscripts $ i,j $ with $ 1 \leq i < j \leq d $. Moreover, for every $ i = 1, \ldots, d $ let $ \pi_{i1}, \ldots, \pi_{ir_i} \in L $ be $ r_i $ linearly independent elements over $ \CC $. Put
		\begin{equation*}
			q = \sum_{i=1}^{d} r_i.
		\end{equation*}
		Then for every $ n \in \NN $ such that
		\begin{equation*}
			\set{ \pi_{il} \alpha_i^n : l=1,\ldots,r_i, i=1,\ldots,d }
		\end{equation*}
		is linearly dependent over $ \CC $, but no proper subset of this set is linearly dependent over $ \CC $, we have
		\begin{equation*}
			n \leq C = C(q, \gfr, \pi_{il}, \alpha_i : l=1,\ldots,r_i, i=1,\ldots,d).
		\end{equation*}
	\end{myprop}

	The proof will also make use of height functions in function fields. Let us therefore define the height of an element $ f \in L^* $ by
	\begin{equation*}
		\Hc(f) := - \sum_{\nu} \min \setb{0, \nu(f)} = \sum_{\nu} \max \setb{0, \nu(f)}
	\end{equation*}
	where the sum is taken over all valuations in the function field $ L / \CC $.
	Observe that for every $ z \in L \setminus \CC $ we have $ \Hc(z) = \sum_{\nu} \max \setb{0, \nu(z)} = \sum_{P} \max \setb{0, \nu_P(z)} = \deg \sum_{P} \max \setb{0, \nu_P(z)} P = \deg (z)_0 = [L:\CC(z)] = \deg_{\CC}(z) $, by Theorem I.4.11 in \cite{stichtenoth-1993}, where we have used that all places have degree one since we are working over $ \CC $; therefore instead of the height one can use $ \deg_{\CC}(z) = [L:\CC(z)]$ (as in \cite{zannier-2008}).
	Additionally we define $ \Hc(0) = \infty $.
	This height function satisfies some basic properties that are listed in the lemma below which is proven in \cite{fuchs-karolus-kreso-2019}:

	\begin{mylemma}
		\label{p3-lemma:heightproperties}
		Denote as above by $ \Hc $ the height on $ L/\CC $. Then for $ f,g \in L^* $ the following properties hold:
		\begin{enumerate}[a)]
			\item $ \Hc(f) \geq 0 $ and $ \Hc(f) = \Hc(1/f) $,
			\item $ \Hc(f) - \Hc(g) \leq \Hc(f+g) \leq \Hc(f) + \Hc(g) $,
			\item $ \Hc(f) - \Hc(g) \leq \Hc(fg) \leq \Hc(f) + \Hc(g) $,
			\item $ \Hc(f^n) = \abs{n} \cdot \Hc(f) $,
			\item $ \Hc(f) = 0 \iff f \in \CC^* $,
			\item $ \Hc(A(f)) = \deg A \cdot \Hc(f) $ for any $ A \in \CC[T] \setminus \set{0} $.
		\end{enumerate}
	\end{mylemma}

	Furthermore we will use the following function field analogue of the Schmidt subspace theorem. A proof can be found in \cite{zannier-2008}:
	\begin{myprop}[Zannier]
		\label{p3-prop:functionfieldsubspace}
		Let $ F/\CC $ be a function field in one variable, of genus $ \gfr $, let $ \varphi_1, \ldots, \varphi_n \in F $ be linearly independent over $ \CC $ and let $ r \in \set{0,1, \ldots, n} $. Let $ S $ be a finite set of places of $ F $ containing all the poles of $ \varphi_1, \ldots, \varphi_n $ and all the zeros of $ \varphi_1, \ldots, \varphi_r $. Put $ \sigma = \sum_{i=1}^{n} \varphi_i $. Then
		\begin{equation*}
			\sum_{\nu \in S} \left( \nu(\sigma) - \min_{i=1, \ldots, n} \nu(\varphi_i) \right) \leq \bino{n}{2} (\abs{S} + 2\gfr - 2) + \sum_{i=r+1}^{n}  \Hc(\varphi_i).
		\end{equation*}
	\end{myprop}

	\section{Proofs}

	\begin{proof}[Proof of Theorem \ref{p3-thm:funcfieldcase}]
		Denote the coefficients of the polynomial $ a_j(n) \in L[n] $ by $ a_{j0}, a_{j1}, \ldots, a_{jm_j} $ where $ m_j $ is the degree of $ a_j(n) $. So
		\begin{equation*}
			a_j(n) = \sum_{k=0}^{m_j} a_{jk} n^k.
		\end{equation*}

		First assume that the recurrence sequence is of the shape $ G_n = a_1(n) \alpha_1^n $. Using Lemma \ref{p3-lemma:heightproperties} we get
		\begin{align*}
			\mu(G_n) &= \mu(a_1(n)) + n\mu(\alpha_1) \leq \Hc(a_1(n)) + n\mu(\alpha_1) \\
			&\leq \sum_{k=0}^{m_1} \Hc(a_{1k}n^k) + n\mu(\alpha_1) = \sum_{k=0}^{m_1} \Hc(a_{1k}) + n\mu(\alpha_1).
		\end{align*}

		Thus from now on we can assume that $ t \geq 2 $.
		Let $ \pi_{j1}, \ldots, \pi_{jk_j} $ be a maximal $ \CC $-linear independent subset of $ a_{j0}, a_{j1}, \ldots, a_{jm_j} $. Then we can write the sequence in the form
		\begin{equation*}
			G_n = \sum_{j=1}^{t} \left( \sum_{i=1}^{k_j} b_{ji}(n) \pi_{ji} \right) \alpha_j^n
		\end{equation*}
		with polynomials $ b_{ji}(n) \in \CC[n] $.
		Since $ a_j(n) $ is not the zero polynomial, there is for each $ j $ at least one index $ i $ such that $ b_{ji}(n) $ is not the zero polynomial.
		Without loss of generality we can assume that no $ b_{ji}(n) $ is the zero polynomial since otherwise we can throw out all zero polynomials and renumber the remaining terms. It does not matter whether all $ \pi_{ji} $ occur in the sum or not.
		Moreover we assume that $ n $ is large enough such that $ b_{ji}(n) \neq 0 $ for all $ j,i $.

		Consider as a next step the set
		\begin{equation*}
			M := \set{\pi_{ji} \alpha_j^n : i=1,\ldots,k_j, j=1,\ldots,t}.
		\end{equation*}
		We intend to apply Proposition \ref{p3-prop:luorbounded}.
		If $ M $ is linearly dependent over $ \CC $, then we choose a minimal linear dependent subset $ \widetilde{M} $ of $ M $, i.e. a linearly dependent subset $ \widetilde{M} $ with the property that no proper subset of $ \widetilde{M} $ is linearly dependent. Let $ \widetilde{G_n} $ be the linear recurrence sequence associated with this subset $ \widetilde{M} $, that is
		\begin{equation*}
			\widetilde{G_n} = \sum_{j=1}^{s} \left( \sum_{i=1}^{\widetilde{k_j}} b_{ji}(n) \pi_{ji} \right) \alpha_j^n
		\end{equation*}
		for $ s \leq t $ and after a suitable renumbering of the summands. Since $ \pi_{j1}, \ldots, \pi_{jk_j} $ are $ \CC $-linear independent we have $ s \geq 2 $.
		Applying Proposition \ref{p3-prop:luorbounded} to
		\begin{equation*}
			\widetilde{M} := \set{\pi_{ji} \alpha_j^n : i=1,\ldots,\widetilde{k_j}, j=1,\ldots,s}
		\end{equation*}
		gives an upper bound for $ n $. Thus for $ n $ large enough this subset $ \widetilde{M} $ of $ M $ cannot be linearly dependent.
		Because of the fact that there are only finitely many subsets of $ M $, therefore for $ n $ large enough the set $ M $ must be linearly independent.

		We assume from here on that $ n $ is large enough such that $ M $ is linearly independent.
		For each fixed $ n $ we have $ b_{ji}(n) \in \CC^* $. Thus the set
		\begin{equation*}
			M' := \set{b_{ji}(n) \pi_{ji} \alpha_j^n : i=1,\ldots,k_j, j=1,\ldots,t}.
		\end{equation*}
		is linearly independent over $ \CC $ and contains for each $ j=1,\ldots,t $ at least one element.
		Let $ S $ be a finite set of places of $ L $ containing all zeros and poles of $ \alpha_j $ for $ j=1,\ldots,t $ and of the nonzero $ a_{ji} $ for $ j=1,\ldots,t $ and $ i=1,\ldots,m_j $ as well as $ \mu $ and the places lying over $ \infty $.
		Now applying Proposition \ref{p3-prop:functionfieldsubspace} yields
		\begin{equation*}
			\sum_{\nu \in S} \left( \nu(G_n) - \min_{\substack{j=1,\ldots,t \\ i=1,\ldots,k_j}} \nu\left( b_{ji}(n) \pi_{ji} \alpha_j^n \right) \right) \leq \bino{\sum_{j=1}^{t} k_j}{2} (\abs{S} + 2\gfr - 2) =: C_1
		\end{equation*}
		and since each summand in the sum on the left hand side is non-negative
		\begin{equation*}
			\mu (G_n) - \min_{\substack{j=1,\ldots,t \\ i=1,\ldots,k_j}} \mu\left( b_{ji}(n) \pi_{ji} \alpha_j^n \right) \leq C_1.
		\end{equation*}
		Therefore for all $ j_0=1,\ldots,t $ and $ i_0=1,\ldots,k_{j_0} $ we get
		\begin{align*}
			\mu (G_n) &\leq C_1 + \min_{\substack{j=1,\ldots,t \\ i=1,\ldots,k_j}} \mu\left( b_{ji}(n) \pi_{ji} \alpha_j^n \right) \\
			&\leq C_1 + \mu\left( b_{j_0i_0}(n) \pi_{j_0i_0} \alpha_{j_0}^n \right) \\
			&= C_1 + \mu\left( \pi_{j_0i_0} \right) + n \mu\left( \alpha_{j_0} \right) \\
			&\leq C_1 + \max_{\substack{j=1,\ldots,t \\ i=0,\ldots,m_j ,\ a_{ji} \neq 0}} \mu(a_{ji}) + n \mu\left( \alpha_{j_0} \right) \\
			&\leq C_1 + \max_{\substack{j=1,\ldots,t \\ i=0,\ldots,m_j ,\ a_{ji} \neq 0}} \Hc(a_{ji}) + n \mu\left( \alpha_{j_0} \right) \\
			&= C_2 + n \mu\left( \alpha_{j_0} \right).
		\end{align*}
		Since this holds for all $ j_0=1,\ldots,t $ we have
		\begin{equation*}
			\mu (G_n) \leq C_2 + n \cdot \min_{j=1,\ldots,t} \mu(\alpha_j).
		\end{equation*}
	\end{proof}

	\begin{proof}[Proof of Corollary \ref{p3-corol:polypowersum}]
		We can apply Theorem \ref{p3-thm:funcfieldcase} with $ L = K = \CC(x) $ and $ \mu = \nu_{\infty} $. This yields
		\begin{align*}
			-\deg G_n &= \nu_{\infty} (G_n) \leq C + n \cdot \min_{j=1,\ldots,t} \nu_{\infty} (\alpha_j) \\
			&= C - n \cdot \max_{j=1,\ldots,t} \deg \alpha_j
		\end{align*}
		which immediately implies the inequality in question.
	\end{proof}

	\section*{Appendix. The number field case}

	In this appendix we will give a proof for the following theorem:
	\begin{mythm}
		\label{p3-thm:numfieldcase}
		Let $ (G_n)_{n=0}^{\infty} $ be a non-degenerate linear recurrence sequence taking values in a number field $ K $ and let $ G_n = a_1(n) \alpha_1^n + \cdots + a_t(n) \alpha_t^n $ with algebraic integers $ \alpha_1,\ldots,\alpha_t $ be its power sum representation satisfying $ \max_{j=1,\ldots,t} \abs{\alpha_j} > 1 $. Denote with $ \abs{\cdot} $ the usual absolute value on $ \CC $. Then for any $ \varepsilon > 0 $ the inequality
		\begin{equation*}
			\abs{G_n} \geq \left( \max_{j=1,\ldots,t} \abs{\alpha_j} \right)^{n(1-\varepsilon)}
		\end{equation*}
		is satisfied for every $ n $ sufficiently large.
	\end{mythm}

	From here on $ K $ will denote a number field.
	In the proof we will need three auxiliary results which are listed below. The first one is a result of Schmidt and can be found in \cite{schmidt-1999}:
	\begin{mythm}
		\label{p3-thm:zerobound}
		Suppose that $ (G_n)_{n \in \ZZ} $ is a non-degenerate linear recurrence sequence of complex numbers, whose characteristic polynomial has $ k $ distinct roots of multiplicity $ \leq a $. Then the number of solutions $ n \in \ZZ $ of the equation
		\begin{equation*}
			G_n = 0
		\end{equation*}
		can be bounded above by
		\begin{equation*}
			c(k,a) = e^{(7k^a)^{8k^a}}.
		\end{equation*}
	\end{mythm}
	The second one is a result of Evertse. The reader will find it as Theorem 2 in \cite{evertse-1984}. We use the notation
	\begin{equation*}
		\norm{\xbf} = \max_{\substack{k=0,\ldots,t \\ i=1,\ldots,D}} \abs{\sigma_i(x_k)}
	\end{equation*}
	with $ \set{\sigma_1, \ldots, \sigma_D} $ the set of all embedings of $ K $ in $ \CC $ and $ \xbf = (x_0,x_1,\ldots,x_t) $.
	Moreover, we denote by $ \Oc_K $ the ring of integers in $ K $:
	\begin{mythm}
		\label{p3-thm:valuationprodineq}
		Let $ t $ be a non-negative integer and $ S $ a finite set of places in $ K $, containing all infinite places. Then for every $ \varepsilon > 0 $ a constant $ C $ exists, depending only on $ \varepsilon, S, K, t $ such that for each non-empty subset $ T $ of $ S $ and every vector $ \xbf = (x_0,x_1,\ldots,x_t) \in \Oc_K^{t+1} $ with
		\begin{equation*}
			x_{i_0} + x_{i_1} + \cdots + x_{i_s} \neq 0
		\end{equation*}
		for each non-empty subset $ \set{i_0,i_1,\ldots,i_s} $ of $ \set{0,1,\ldots,t} $ the inequality
		\begin{equation*}
			\left( \prod_{k=0}^{t} \prod_{\nu \in S} \norm{x_k}_{\nu} \right) \prod_{\nu \in T} \norm{x_0 + x_1 + \cdots + x_t}_{\nu} \geq C \left( \prod_{\nu \in T} \max_{k=0,\ldots,t} \norm{x_k}_{\nu} \right) \norm{\xbf}^{-\varepsilon}.
		\end{equation*}
		is valid.
	\end{mythm}
	Furthermore, we will need the following lemma which also can be found in \cite{evertse-1984}:
	\begin{mylemma}
		\label{p3-lemma:valprodbound}
		Suppose $ K $ is a number field of degree $ D $, let $ f(X) \in K[X] $ be a polynomial of degree $ m $ and $ T $ a non-empty set of primes on $ K $. Then there exists a positive constant $ c $, depending only on $ K,f $ such that for all $ r \in \ZZ $ with $ r \neq 0 $ and $ f(r) \neq 0 $ it holds that
		\begin{align*}
			c^{-1} \abs{r}^{-Dm} &\leq \left( \prod_{\nu} \max \setb{1, \norm{f(r)}_{\nu}} \right)^{-1} \leq \prod_{\nu \in T} \norm{f(r)}_{\nu} \\
			&\leq \prod_{\nu} \max \setb{1, \norm{f(r)}_{\nu}} \leq c \abs{r}^{Dm}.
		\end{align*}
	\end{mylemma}

	\begin{proof}[Proof of Theorem \ref{p3-thm:numfieldcase}]
		Since the characteristic roots $ \alpha_j $ of $ G_n $ are algebraic integers we can find a nonzero integer $ z $ such that $ z a_j(n) \alpha_j^n $ are algebraic integers for all $ j=1,\ldots,t $ and all $ n \in \NN $.
		Denote by $ L $ the splitting field of the characteristic polynomial of the sequence $ G_n $, i.e. $ L = K(\alpha_1,\ldots,\alpha_t) $.
		Choose $ S $ as a finite set of places in $ L $ containing all infinite places as well as all places such that $ \alpha_1, \ldots, \alpha_t $ are $ S $-units.
		Let $ \mu $ be such that $ \norm{\cdot}_{\mu} = \abs{\cdot} $ is the usual absolute value on $ \CC $. In particular we have $ \mu \in S $. Further define $ T = \set{\mu} $.

		As $ G_n $ is non-degenerate the sequence $ \widetilde{G_n} = zG_n $ is non-degenerate, too. Therefore applying Theorem \ref{p3-thm:zerobound} yields for $ n $ large enough
		\begin{equation*}
			z a_{j_1}(n) \alpha_{j_1}^n + \cdots + z a_{j_s}(n) \alpha_{j_s}^n \neq 0
		\end{equation*}
		for each non-empty subset $ \set{j_1,\ldots,j_s} $ of $ \set{1,\ldots,t} $.
		Thus we can apply Theorem \ref{p3-thm:valuationprodineq} and get
		\begin{equation*}
			\left( \prod_{j=1}^{t} \prod_{\nu \in S} \norm{z a_j(n) \alpha_j^n}_{\nu} \right) \abs{zG_n} \geq C \max_{j=1,\ldots,t} \abs{z a_j(n) \alpha_j^n} \norm{z\xbf}^{-\varepsilon}
		\end{equation*}
		for $ \xbf = (a_1(n) \alpha_1^n, \ldots, a_t(n) \alpha_t^n) $.
		Assuming without loss of generality that $ \abs{\alpha_1} = \max_{j=1,\ldots,t} \abs{\alpha_j} $ and using that $ z $ is a fixed integer as well as that the $ \alpha_j $ are $ S $-units, we can rewrite this as
		\begin{align}
			\left( \prod_{j=1}^{t} \prod_{\nu \in S} \norm{a_j(n)}_{\nu} \right) \abs{G_n} &\geq C_1 \max_{j=1,\ldots,t} \abs{a_j(n) \alpha_j^n} \norm{\xbf}^{-\varepsilon} \nonumber \\
			&\label{p3-eq:appproof1}\geq C_1 \abs{a_1(n) \alpha_1^n} \norm{\xbf}^{-\varepsilon} \\
			&= C_1 \abs{a_1(n)} \abs{\alpha_1}^n \norm{\xbf}^{-\varepsilon}. \nonumber
		\end{align}

		In preparation for the next step note that there exists a positive constant $ A $ such that
		\begin{equation*}
			\max_{\substack{j=1,\ldots,t \\ i=1,\ldots,D}} \abs{\sigma_i(\alpha_j)} \leq A \cdot \abs{\alpha_1}.
		\end{equation*}
		We decompose $ \varepsilon = \gamma \cdot \delta $ with small $ \delta $ and $ A^{\gamma} \leq \abs{\alpha_1} $.
		Using this we get the estimates
		\begin{align*}
			\norm{\xbf} &= \max_{\substack{j=1,\ldots,t \\ i=1,\ldots,D}} \abs{\sigma_i(a_j(n) \alpha_j^n)} = \max_{\substack{j=1,\ldots,t \\ i=1,\ldots,D}} \abs{\sigma_i(a_j(n)) \sigma_i(\alpha_j)^n} \\
			&\leq \max_{\substack{j=1,\ldots,t \\ i=1,\ldots,D}} \abs{\sigma_i(a_j(n))} \cdot \max_{\substack{j=1,\ldots,t \\ i=1,\ldots,D}} \abs{\sigma_i(\alpha_j)}^n \\
			&\leq C_2 n^m \cdot \max_{\substack{j=1,\ldots,t \\ i=1,\ldots,D}} \abs{\sigma_i(\alpha_j)}^n \leq C_2 n^m A^n \abs{\alpha_1}^n
		\end{align*}
		with $ m = \max_{j=1,\ldots,t} \deg a_j $, and
		\begin{equation*}
			\norm{\xbf}^{\varepsilon} \leq C_3 n^{m\varepsilon} A^{\gamma n\delta} \abs{\alpha_1}^{n\varepsilon} \leq C_3 n^{m\varepsilon} \abs{\alpha_1}^{n(\varepsilon+\delta)}.
		\end{equation*}
		Now we insert this into inequality \eqref{p3-eq:appproof1} which gives us
		\begin{align*}
			\left( \prod_{j=1}^{t} \prod_{\nu \in S} \norm{a_j(n)}_{\nu} \right) \abs{G_n} &\geq C_4 \abs{a_1(n)} \abs{\alpha_1}^n n^{-m\varepsilon} \abs{\alpha_1}^{-n(\varepsilon+\delta)} \\
			&\geq C_5 n^{-m\varepsilon} \abs{\alpha_1}^{n(1-\varepsilon-\delta)}.
		\end{align*}
		Applying Lemma \ref{p3-lemma:valprodbound} to the product in the brackets on the left hand side gives the bound
		\begin{equation*}
			\prod_{j=1}^{t} \prod_{\nu \in S} \norm{a_j(n)}_{\nu} \leq \prod_{j=1}^{t} C_6^{(j)} n^{Dm} \leq C_7 n^{tDm}.
		\end{equation*}
		Altogether for $ n $ large enough the inequality
		\begin{equation*}
			\abs{G_n} \geq C_8 n^{-tDm-m\varepsilon} \abs{\alpha_1}^{n(1-\varepsilon-\delta)}
		\end{equation*}
		holds.
		Hence, recalling $ \abs{\alpha_1} = \max_{j=1,\ldots,t} \abs{\alpha_j} $, for $ n $ large enough we get
		\begin{equation*}
			\abs{G_n} \geq \left( \max_{j=1,\ldots,t} \abs{\alpha_j} \right)^{n(1-\widetilde{\varepsilon})}.
		\end{equation*}
		This proves the theorem.
	\end{proof}


TITLE: Why should SUSY be expected naturally?
QUESTION [7 upvotes]: In the last 40 years (approximately) people have been "discovering", "rediscovering" and "studying" SUSY as a powerful tool and "symmetry principle".
Question:
What if SUSY is not realized in Nature at the end? Is SUSY the only path to "relate" fermions and bosons or what else? Remark: SUSY has not been discovered yet, so keep you totally conservative. What if there is no SUSY?
Bonus:
What are the merits of SUSY? What are its main issues? I do know some answers to this, but I think it could very enlightening if we "listed" pros and contras of current supersymmetric theories to see where we are NOW.

REPLY [6 votes]: There is a standard paradigm for thinking about the new physics that lies beyond the standard model, at higher and higher energies: weak-scale supersymmetry, grand unification, string theory. The purpose of weak-scale supersymmetry is to stabilize weak-scale physics (i.e. everything we know about) against quantum corrections. Grand unification can explain various features of the standard model, such as the assortment of representations and hypercharges. String theory offers a unique and subtle model of quantum gravity, and it contains supersymmetry and grand unification for free.
So it's a very compelling paradigm, but also a very broad one. It has been the work of decades to explore all the variations on this theme - there might be tens of thousands of models in the literature which fit this paradigm - and people continue to discover new theoretical possibilities every month. Conceptual and mathematical progress occurs too, alongside the exploration of concrete models. For example, the idea of a string-theory landscape, and the way it dovetails with eternal inflation, is a new perspective on the meaning of this "theory space" that people have been exploring, mostly at the level of field theory.
But of course theoretical work isn't confined to this paradigm. Numerous deviations from it are also being explored. We also continue to get new data about the world, both positive (new phenomena) and negative (confirmation of standard-model predictions), which has its impact on the viability and plausibility of all those thousands of specific models. And finally, there is that constant conceptual progress which incrementally clarifies our theoretical options, and which sometimes transforms our idea of what a particular theory means.
One feature of the present day is that the idea of weak-scale supersymmetry is under pressure, because only the Higgs has turned up, and no undeniable deviations from the standard model are being seen, and many possible deviations are not being seen. However, there are many possibilities, and the majority of the LHC's work still lies ahead. So I think the most we can say is that there is a shift in the theoretical centre. In the absence of a theoretical or experimental revolution, models of weak-scale supersymmetry will continue to be considered for years to come, but there will also be more attention to alternatives, such as high-scale supersymmetry and no supersymmetry (including string theory without supersymmetry). If we continue to see nothing but standard model at the LHC, there may be a shift in theoretical focus, from predicting masses of superpartners, to explaining the detailed features of the standard model (such as fermion mass ratios). Under such circumstances, the chief appeal of a model may become how many such features it can explain, and how well.
It seems clear that physics will continue to advance, even if not in ways that any one person managed to foresee. Even if we were to see nothing but standard model on Earth, there would still be the universe to explain. It would be a new era for fundamental physics if its primary empirical guidance were to become astrophysics, but if it happens, it happens, and methods and models will adjust accordingly. And meanwhile, that steady conceptual and mathematical progress always holds out the prospect of transforming how everything looks, even without further empirical input. Perhaps the string landscape will be completely mapped thanks to a generation-long effort; perhaps someone from one of the "deviant" streams of research will find a starting point which magically explains all the features of the standard model that are presently just assumptions.
Supersymmetry is there, or it isn't. Either way, I expect the human race to find out eventually, through a combination of empirical and theoretical advances, and as just one aspect of finding out the general truth about how physics works. It may show up undeniably at the LHC when it powers up to higher energies. Or we may already be seeing it as dark matter, in which case proving that dark matter is superparticles rather than something else may be a much more demanding task, than just turning the LHC up to maximum. Or supersymmetry may be confined to ultra-high energies, or not there at all, in which case there will have to be profound theoretical advances before we can understand the situation.


TITLE: $f$ is bounded by $M$ on $[a, b]$ and if the restriction of $f$ to every interval $[c, b]$ where $c$ in $(a, b)$ is Riemann integrable
QUESTION [3 upvotes]: If $f$ is bounded by $M$ on $[a, b]$ and if the restriction of $f$ to every interval $[c, b]$ where $c$ in $(a, b)$ is Riemann integrable, then $f$ is Riemann integrable and that $\int _c^b f \to \int_a^b f$ as $c \to a+$.
I am stuck with this problem.

REPLY [1 votes]: Proof. Since $f$ is bounded on $[a,b]$, there exists an $M>0$ such that $|f(x)|≤M$ for all $x∈[a,b]$. Choose an $\varepsilon>0$. Define $δ_1=\min\{\varepsilon/2M,b-a\}$. Choose $c∈(a,b)$ such that $a<c<a+δ_1$. Consider a tagged partition $\dot{\mathcal{P}}_1=\{([x_{i-1},x_i],t_i)\}_{i=1}^n$ of $[a,b]$ with $\Vert \dot{\mathcal{P}}_1\Vert<δ_1$. Then
\begin{equation}
\left|S(f|_{[a,c]}; \dot{\mathcal{P}}_1)\right|≤\sum_{i=1}^n|f(t_i)|(x_i-x_{i-1})≤M\sum_{i=1}^n(x_i-x_{i-1})=M(c-a)<\frac{\varepsilon}{2} \tag{1}.
\end{equation}
This proves that the restriction of $f$ to $[a,c]$ is Riemann integrable with
\begin{equation}
\lim_{c\rightarrow a^+} \int_a^c f=0.
\end{equation}
Since $f∈\mathcal{R}[c,b]$, there exists a $δ_2>0$ such that for any tagged partition $\dot{\mathcal{P}}_2$ of $[c,b]$ with $\Vert \dot{\mathcal{P}}_2\Vert<δ_2$,
\begin{equation}
\left|S(f|_{[c,b]};\dot{\mathcal{P}}_2)-∫_c^b f\right|<\frac{\varepsilon}{2}.
\end{equation}
Define $\delta=\min\{\delta_1,\delta_2\}$. Consider a tagged partition $\dot{\mathcal{P}}$ of $[a,b]$ as $\mathcal{P}=\mathcal{P}_1 \cup \mathcal{P}_2$ with the same tags as in $\dot{\mathcal{P}}_1$ and $\dot{\mathcal{P}}_2$. Then $\Vert \dot{\mathcal{P}}\Vert<δ$ and
\begin{equation}
\left|S(f;\dot{\mathcal{P}})-S(f|_{[c,b]} ;\dot{\mathcal{P}}_2)\right|=\left|S(f|_{[a,c]};\dot{\mathcal{P}}_1)\right|<\frac{\varepsilon}{2}.
\end{equation}
Now it follows from the triangle inequality that
\begin{equation}
\left|S(f;\dot{\mathcal{P}})-∫_c^b f\right|≤\left|S(f;\dot{\mathcal{P}})-S(f|_{[c,b]};\dot{\mathcal{P}}_2)\right|+\left|S(f|_{[c,b]};\dot{\mathcal{P}}_2)-∫_c^b f\right|<\varepsilon.
\end{equation}
This proves that $f∈\mathcal{R}[a,b]$. By Additivity Theorem,
\begin{equation}
∫_a^b f=∫_c^b f+∫_a^c f
\end{equation}
and eq. $(1)$, one obtains
\begin{equation}
∫_a^b f=\lim_{c→a^+}∫_c^b f.\blacksquare
\end{equation}


TITLE: Conditional version of Bayes formula
QUESTION [0 upvotes]: I'm doing a course on Advanced Probability and struggle hard with the topic of conditional expectations. There is one exercise in the Book of Klenke, which makes me wonder:
Let $(\Omega,\mathcal{A},\mathbb{P})$ be a probability space and $\mathcal{F}$ a sub-$\sigma$-algebra of $\mathcal{A}$. Prove the conditional version of Bayes' formula: for $A\in\mathcal{A}$ and $F\in \mathcal{F}$: $$\mathbb{P(F|\mathcal{A})}= \frac{\int_F\mathbb{P}(A|\mathcal{F})d\mathbb{P}}{\int_\Omega\mathbb{P}(A|\mathcal{F})d\mathbb{P}}.$$ How can one prove that? As far as I know it holds $\mathbb{P}(A|\mathcal{F})=\mathbb{E}(1_{A}|\mathcal{F})$, so I would put that in. But then I have no clue of where to go. Are there some probability experts? Thank you in advance.

REPLY [0 votes]: I suppose you want to show if $A\in\mathcal A, B\in \mathcal F$ where $ \mathcal F\subset \mathcal A$ is a sub sigma algebra then,

$\mathbb P(B\mid A)=\frac{\int _B \mathbb P(A\mid \mathcal F)d\mathbb P}{\int \mathbb P(A\mid \mathcal F)d\mathbb P}$.

Suppose $\mathbb P(A)>0$. Now $$\frac{\int _B \mathbb P(A\mid \mathcal F)d\mathbb P}{\int \mathbb P(A\mid \mathcal F)d\mathbb P}=\frac{\int_B E(1_A\mid \mathcal F)d\mathbb P}{\int E(1_A\mid \mathcal F)d\mathbb P}=\frac{\int E(1_{A\cap B}\mid \mathcal F)d\mathbb P}{E(1_A)}=\frac{E(E(1_{A\cap B}\mid \mathcal F))}{\mathbb P(A)}=\frac{E(1_{A\cap B})}{\mathbb P(A)}=P(B\mid A).$$ We use the definition of the conditional probability in the first step, second step is since $B\in\mathcal F$, i.e $1_B$ is measurable with resepct to $\mathcal F$, further we use in the second and third the definition of a conditional expectation.
Hope that will help you.


TITLE: The least number which leaves remainders 2, 3, 4, 5 and 6 on dividing by 3, 4, 5, 6 and 7 is
QUESTION [1 upvotes]: Problem - The least number which leaves remainders 2, 3, 4, 5 and 6 on dividing by 3, 4, 5, 6 and 7 is?
Solution - Here 3-2 = 1, 4-3 = 1, 5-4 = 1 and so on.
So required number is (LCM of 3, 4, 5, 6, 7) - 1 = 419
My confusion -
I didn't get the solution of this. Please explain how by subtracting the conclusion was drawn that the number will be the (LCM - 1)?

REPLY [0 votes]: You can see by inspection that -1 works. Then any other solution must be congruent to -1 mod lcm(3,4,5,6,7) = 420 and hence the least positive solution is 419.


TITLE: Counting complex solutions on a disk.
QUESTION [0 upvotes]: I am looking for an example of an analytic map $f$ on the complex domain for which there exist $r>0$ such that for $\gamma=\{z:|z|=r\}$, a.s. for $\theta \in [0,2\pi)$ we have that
$$\int_{f(\gamma)} \frac{1}{2 \pi i} \frac{dw}{w} < \left( f(\gamma)e^{-2\pi i \theta}\cap \mathbb{R}_{+} \right).$$

REPLY [1 votes]: For $r=1$, the function $f(z) = (z+\sqrt2)^4$ works nicely. The path integral equals $0$, while the count on the right-hand side is always at least 1. (In fact, if you reduce $\sqrt2$ slightly, you can get the count to be at least 2 always.)

To deduce how I came up with this function, consider it as the composition of $z\mapsto z+\sqrt2$ with $z\mapsto z^4$, and think about what each map does to the complex plane.


TITLE: How do I find the longitude of the subsolar point?
QUESTION [0 upvotes]: On December 21, 2018 at 2:23pm PST I want to be standing as directly under the sun as humanly possible. Obviously the latitude of that point will be the Tropic of Capricorn. Assuming I have the right ascension for the sun at that time how would I know what line of longitude to stand on? Please use simple non-technical terms if possible.

REPLY [0 votes]: In case someone stumbles across this question and wants to know the math here's what I've been able to figure out.
All right ascensions(RA) are distances, in time, east of the vernal equinox given in hours, minutes, and seconds. The location of the vernal equinox depends on the standard used when giving the RA. For instance J2000 uses the location of the equinox at noon on Januar 1, 2000. The equinox is the imaginary line formed by the intersection of the equatorial plane and the ecliptic plane on the side of the planet where the sun was moving from below the equator to above it.
To begin finding the longitude of an object you need to compute the GMST or GAST for the date and time the measurement was given. You then subtract the GMST from the RA and multiply the result by 15 to get the longitude of the object.


TITLE: Asymptotics of integrals with respect to asymptotically equivalent distribution functions
QUESTION [2 upvotes]: Let us assume that $f\colon[0,\infty)\to[0,\infty)$ is a non-decreasing continuous function with
$$ \frac{f(t)}{t}\xrightarrow{t\to\infty}1. \tag{A}\label{A}$$
We then have the following elementary Lemma:

For all $k\in\mathbb N_0$ we have
  $$ \frac{\int_0^t s^k df(s)}{\int_0^t s^k ds}\xrightarrow{t\to\infty}1. $$

For the sake of completeness, a detailed elementary proof is given below which makes explicit use of the very simple nature of the integrand. However, it seems plausible that such a property should hold for much more general functions in place of $s^k$. In particular, the following statement seems reasonable to me.

Conjecture: If $h\colon[0,\infty)\to[0,\infty)$ is measurable and
  $$ \liminf_{t\to\infty}h(t)>0$$
  (and maybe something else), then we have
  $$ \frac{\int_0^t h(s) df(s)}{\int_0^t h(s) ds}\xrightarrow{t\to\infty}1. $$

Since $h$ cannot "lose mass at infinity", this property would give meaning to the idea that if a distribution function of a measure $\mu$ on $\mathcal B([0,\infty))$ is asymptotically equivalent to the identity (which is the distribution function of Lebesgue's measure), then "$\mu$ is asymptotically equivalent to Lebesgue's measure".
Of course, further natural steps would be to replace Lebesgue's measure by any other locally finite measure $\nu$ on $\mathcal B([0,\infty))$ with a continuous distribution function and then compare $\mu$ and $\nu$ asymptotically. Then one could ask if continuity can be dropped. But for the moment, I guess the above conjecture is nice enough.
Does anyone know a reference where something like this is stated (and proved)? Or does anyone have an idea how to prove it? Thanks a lot in advance!

Proof of the elementary Lemma: Using the Stieltjes product formula, we can write
\begin{align*}
 \frac{\int_0^t s^k df(s)}{\int_0^t s^k ds}&=\frac{k+1}{t^{k+1}} \int_0^t s^k df(s) \\
    &=\frac{k+1}{t^{k+1}} \left( t^kf(t)-\int_0^t f(s)ds^k \right) \\
 &=(k+1)\frac{f(t)}{t}-\frac{k+1}{t^{k+1}} \int_0^t f(s)ks^{k-1}ds \\
 &=(k+1)\frac{f(t)}{t}-k-\frac{k(k+1)}{t^{k+1}} \int_0^t (f(s)-s)s^{k-1}ds.
\end{align*}
By \eqref{A}, the first summand converges to $k+1$ for $t\to\infty$, so it remains to prove that
$$g(t):=\frac{k+1}{t^{k+1}} \int_0^t (f(s)-s)s^{k-1}ds$$
vanishes for $t\to\infty$. Let $\epsilon>0$. According to \eqref{A}, there is some $t_0\in(0,\infty)$ such that
$$\left|\frac{f(s)}{s}-1\right|\le\epsilon \quad \text{for all $s\in[t_0,\infty)$.}$$
Then
\begin{align*}
|g(t)|&\le \frac{k+1}{t^{k+1}} \int_0^{t_0} |f(s)-s|s^{k-1}ds + \frac{k+1}{t^{k+1}} \int_{t_0}^t (s\epsilon)s^{k-1}ds \\
&\le \frac{k+1}{t^{k+1}} \int_0^{t_0} |f(s)-s|s^{k-1}ds + \epsilon
\end{align*}
for all $t\in[t_0,\infty)$. As the integral in the last step is finite and does not depend on $t$, this shows that
$$\limsup_{t\to\infty}|g(t)|\le\epsilon,$$
which completes the proof, since $\epsilon$ was chosen arbitrarily.

REPLY [1 votes]: Here are some counter-examples: Let $f(t) = t+\sin t$. Then clearly $f(t)/t \to 1$ as $t \to \infty$.

If $h(s) = e^s$, then
$$ \frac{\int_{0}^{t}h(s)\,f(\mathrm{d}s)}{\int_{0}^{t}h(s)\,\mathrm{d}s} = 1 + \frac{\cos t+\sin t}{2} + \mathcal{O}(e^{-t}). $$

If $h(s) = 2+\cos s$, then
$$ \frac{\int_{0}^{t}h(s)\,f(\mathrm{d}s)}{\int_{0}^{t}h(s)\,\mathrm{d}s} = \frac{5}{4} + \mathcal{O}\left(\frac{1}{t}\right). $$


TITLE: Prove that f(x) can have any value between A and B
QUESTION [0 upvotes]: I need to prove the following statement:
Let f be a bounded and continuous function in the interval (a,b). if $\lim_{x\to a+}f(x)=A$ and $\lim_{x\to b-}f(x)=B$ and $A\neq B$ then f can get any value between and A and B in the interval (a,b).
What I did:
Let $ g(x) =
\begin{cases}
f(x),  & \text{$a<x<b$} \\
A, & \text{x=a}  \\
B, & \text{x=b} \\
\end{cases}$
. We can see that $g(x)$ is a continuous function in the the interval [a,b]. Therefore, I can use the Intermediate value theorem to prove that for any value between A and B, there exists $a<c<b$ such that g(c) is equal to that value, but in the interval (a,b) , $g(x)= f(x)$, so f(x) can get any value between A and B in the interval (a,b).
My Questions:
1)Can i define a function that is divided to 3 cases?
2)I didn't use the fact that f(x) is bounded, and that seems weird. what is wrong with my proof?
3) I am sure there must be a better way to prove it because the next statement i need to prove the same just for $A=\infty$ and $B=-\infty$(f(x) isn't bounded) and my trick won't work there, that's why I have no idea how to prove it.

REPLY [0 votes]: Your proof for the first problem is right.
Let's see if we can make sense of the second problem. Let's say that we are trying to show that $f$ takes on the value $C$. Via the property of limits, we know we can find some $x_1$ so that $f(x_1) > C$. Also we know we can find some $x_2$ so that $f(x_2) < C$. Consider the interval $[x_1, x_2]$ (or $[x_2,x_1]$ if you accidentally found these values so that $x_2 < x_1$). Then you can apply IVT to $f$ restricted to this interval to get the desired result.


TITLE: Finding a system of linear equations from solutions
QUESTION [2 upvotes]: Is there a simple way to find the system of linear equations given the solutions?
For example, find a system with 2 equations and 3 variables that has solutions (1, 4, -1) and (2, 5, 2).
I get:
x1 - x2 = -3
3x1 - x3 = 4

The method I used to get this answer was guessing and checking, but surely there is a simpler and more efficient way to go about it?
Also, are there other equations that would fit this solution set?
Thanks!

REPLY [1 votes]: The solution set of your system of equations is not simply $\{(1,4,-1), (2,5,2)    \}$.
For example, $(0,3,-4)$ is a solution. There are infinitely many others.
And we cannot obtain a system of two linear equations in three unknowns whose solution set is
$\{(1,4,-1), (2,5,2)    \}$.  For if the system has a solution, it has infinitely many.
As to your question about producing such a system in a systematic way, let the two equations be $a_1x+b_1y+c_1z=d_1$, $a_2x+b_2y+c_2z=d_2$. The conditions are
$a_1+4b_1-c_1=d_1$, $2a_1+5b_1+2c_1=d_1$, $a_2+4b_2-c_2=d_2$, $2a_2+5b_2+2c_2=d_2$.
We have $4$ equations in $8$ unknowns, so obviously a lot of slack.  One way to cut down on it is to decide that the $d_i$ will be $0$.
Choose say $a_1=0$, and solve for $b_1$ and $c_1$; Choose $b_2=0$, and solve for $a_2$ and $c_2$.


TITLE: $\int_a^b\frac{d}{dx}f(x,y)dy$ versus $\int_a^b\frac{\partial}{\partial x}f(x,y)dy$, which is correct?
QUESTION [4 upvotes]: For a two-variable function $f$, which symbol below is technically correct?

$$ \int_a^b\frac{d}{dx}f(x,y)dy ~~~~~~~ \text{versus} ~~~~~~~ \int_a^b\frac{\partial}{\partial x}f(x,y)dy$$

The first or the second? Or both? How about $\int_a^b\frac{\partial f}{\partial x}(x,y)dy$?
Notice that the point here is that for a two-variable function, can't we use $d/dx$? Must we use $\partial/\partial x$? You may say, there's a rule that whenever dealing with multivariable function, use $\partial$. However, notice that you would write $\frac{df(x,3)}{dx}$ instead of $\frac{\partial f(x,3)}{\partial x}$. Here $f$ itself is a two-variable function, but we're not concerning $f$ itself, but the expression $f(x,3)$ (just as you must have seen something like $\frac{d(\sin x+x^2)}{dx}$, imagine that it is the case that the nominatior $(\sin x+x^2)$ is replaced by the expression $f(x,3)$). So the rule that "whenever dealing with multivariable function, use $\partial$" does not make sense.
And on the other hand, when INSIDE the integral sign, the variable $y$ is somewhat already served as a constant (we say the variable is being binded), waiting to be integrate after the integrand is completely evaluated (just as in $\sum_{i=1}^{10}i^2$, the variable $i$ inside summation is binded).  So, the all story happen inside the integral sign, is only of one variable $x$. In this way, should we write $\int_a^b\frac{d}{dx}f(x,y)dy$?
PS: Feel free to comment or to answer. Any advice or experience is welcomed. :)

REPLY [3 votes]: For $f : \Omega \subseteq \mathbb{R}^2 \to \mathbb{R} : (x,y) \mapsto f(x,y)$ it does not really matter if we write $\displaystyle \frac{df}{dx}$ or $\displaystyle \frac{\partial f}{\partial x}$ for the derivative with respect to $x$. It is a matter of emphasis. We usually write $\displaystyle \frac{df}{dx}$ to emphasize we are deriving a function of one variable, namely $x$, and $\displaystyle \frac{\partial f}{\partial x}$ to emphasize we are deriving a function of more than one variable with respect to the variable named $x$. It is a question of making easier to the reader to know what is going on. It is much more common to use $\displaystyle \frac{\partial}{\partial x}$ in this case.
Now regarding to the technicality, $\displaystyle \int_a^b\frac{d}{dx}f(x,y)\ dy$ and $\displaystyle \int_a^b\frac{\partial}{\partial x}f(x,y)\ dy$ and $\displaystyle \int_a^b\frac{\partial f}{\partial x}(x,y)\ dy$ are not correct or incorrect but inaccurate.
Derivatives acts on functions, not on points. That is $\displaystyle \frac{\partial}{\partial x}$ acts on the function $f$ and not on the point $f(x,y)$. Furthermore we derive the function $f$ at the point $(p,q)$ with respect to $x$, we not derive the function $f$ at the point $(x,y)$ with respect to $x$. That is
$$ \frac{\partial f}{\partial x} : \Omega \to \mathbb{R} : (p,q) \mapsto \frac{\partial f}{\partial x}(p,q) $$
The latter one, $\displaystyle \frac{\partial f}{\partial x}(x,y)$, creates unnecessary ambiguity. To see that, set $x = 1$; then $\displaystyle \frac{\partial f}{\partial x}(x,y) = \frac{\partial f}{\partial 1}(1,y)$ or $\displaystyle = \frac{\partial f}{\partial x}(1,y)$ or $\displaystyle = \frac{\partial f}{\partial 1}(x,y)$?
Therefere, agreeing we want to avoid ambiguity we would write
$$ \int_a^b\frac{\partial f}{\partial x}(p,q)\ dq $$
However it also reasonable to write
$$ \int_a^b\frac{\partial f}{\partial x}(p,y)\ dy $$
because the derivative are not binding the variable $y$ in any way.
Let me know if you need further clarifications.


TITLE: Integration of rational functions by partial fractions
QUESTION [1 upvotes]: I have the following integral :
$$\int \frac{x^2+2x+3}{x^2-5x+6}~dx$$
Although I tried to solve it by partial fractions, I could not come up with an appropriate answer. My question is more about the technique which I should use here in order to evalue the integral. I understand that the denominator could be expressed as
$$(x-2)(x-3)$$
But then how to proceed with such a complicated numerator.

REPLY [4 votes]: First,
$$
x^2+2x+3=1\cdot (x^2-5x+6)+7x-3.
$$
Thus
$$
\frac{x^2+2x+3}{x^2-5x+6}=1+\frac{7x-3}{x^2-5x+6}.
$$
You know $x^2-5x+6=(x-2)(x-3)$. It is known that there exists $A,B$ such that
$$
\frac{7x-3}{x^2-5x+6}=\frac{A}{x-2}+\frac{B}{x-3}
$$
and so
$$
\frac{7x-3}{x^2-5x+6}=\frac{(A+B)x-(3A+2B)}{x^2-5x+6}.
$$
Thus we get $A=-11$ and $B=18$. Therefore
$$
\int\frac{x^2+2x+3}{x^2-5x+6}dx=x-11\ln|x-2|+18\ln|x-3|+C.
$$


TITLE: The word "representation" in the context of Lie groups
QUESTION [4 upvotes]: I'm still very new to learning about Lie groups, something I find particularly confusing is the use of the word representation in the context of Lie groups. Sources I've checked online go quite far over my head and tend to be quite mathematical, so I thought I would try to ask about this term specifically.
When we say a "representation" of a Lie group, I interpret this as meaning that given the abstract set of elements and group structures, there are several "things" that the elements of the set could be while still satisfying the axioms of being that particular Lie group. For instance, when we say that $U(1)$ "is the circle group", and is the set of complex numbers of modulus $1$ with the group operation defined as complex multiplication, I find it strange that we are identifying something entirely abstract (a particular Lie group) with something specific (the complex numbers and complex multiplication). Am I correct in saying that there exist other sets of elements related by a different group operation that are another "representation" of the $U(1)$ Lie group? So that the elements of this other representation are in a sense isomorphic to the set of complex numbers with modulus $1$?
Another example would be the $SU(2)$ group, my lectures define this as the "special unitary group of $2\times2$ matrices with determinant equal to $1$". This again seems like we are defining this Lie group to be the set of matrices satisfying those axioms, but to me a group is something entirely abstract. So just like the $U(1)$ group I just talked about, do there exist other mathematical objects paired with another mathematical operation that is in the same sense isomorphic to the set of matrices defined above? And this alternative set of elements whatever they may be is just another "representation" of the $SU(2)$ Lie group?
I hope this question makes sense, I have tried to be as broad with it as possible, I'm not necessarily singling out the $U(1)$ or $SU(2)$ groups as confusing, but trying to understand the general idea behind this.

REPLY [4 votes]: To a physicist, a "group representation" is a set of matrices (i.e elements of ${\rm GL}[{\mathbb F}]$ where $\mathbb F$ is a field, often  ${\mathbb F}={\mathbb C}$) having the same multiplication table as the abstract group. The representation is faithful if different group elements correspond to different matrices.  To a mathematician the representation is the map $D:G\to {\rm GL}[{\mathbb F}]$ with   $g\mapsto D(g)\in {\rm GL}[{\mathbb F}]$ rather than the set $\{D(g)\}$ of matrices. This difference in usage is occasionally confusing. This confusion is also enhanced because physicists often refer to the vector space on which the matrices act as the "representation."


TITLE: Detailed diagram with mathematical fields of study
QUESTION [17 upvotes]: Some time ago, I was searching for a detailed diagram with mathematical fields of study the nearest one I could find is in this file, second page.
I want something that shows information like: "Geometry leads to I topic, Geometry and Algebra leads do J topic and so on.
Can you help me?

REPLY [5 votes]: http://www.math-atlas.org/ love this page! It appears that is has not been updated in a long time. To date i have not found its equal