TITLE: Use of Fresnel lenses
QUESTION [0 upvotes]: Major camera manufacturers are now using Fresnel lens elements in high end telephoto lenses to reduce weight. I can understand the use of such lenses in light concentration applications but how can they perform to the high standard of photographic imaging? I would think that the facets would produce stray light, thus reducing image contrast, among other possible consequences.

REPLY [0 votes]: @diffycue linked to an advertisement for a Nikon lens that uses "phase Fresnel lens elements."
Here's what Nikon's own web site has to say:

when there is a strong light source within the frame or when light enters the lens from outside of the frame, ring-shaped colored flare may occur according to shooting conditions. This phenomenon can be minimized with “PF Flare Control” to be included inCapture NX-D (ver.1.1.0 or later).

In other words, the facets do produce stray light, and the stray light does reduce the image contrast and, it does have other consequences; but all of those consequences are predictable, and digital image processing software is able to (mostly?) undo them.
I suppose that the DSP is performed in the camera if the camera exports a .jpeg file (or other "common" image format), and it most likely is performed by software running on a workstation that imports "RAW" images from the camera.

P.S., This actually is a new twist on an idea that's been out there for a few years: My own camera, a Fujifilm X-T30, uses lenses that are not corrected for certain geometric distortions. The in-camera software knows about the distortion and corrects for it when saving .jpeg files, and software on my workstation does the same if I import the RAW files.


TITLE: Why is classical mechanics determinism based on position and momentum only and not forces and scattering rules?
QUESTION [6 upvotes]: Consider a closed system (say a box) of $n$ particles. There is a well-known idiom/meme/law in classical mechanics that says that the position and momentum of those $n$ particles is all that is needed to determine the future and past configuration of the system. (A minor question, does this idiom have a name?)
Why do we not consider two other pieces of information to be as important as the position and momentum of the $n$ particles?

The force field (force as a function of space, and preferably of time too).
The particle interaction rules (When two particles touch each other, do they whiz by or do they scatter? if they scatter what are the rules for determining angles?)

REPLY [5 votes]: The information about the forces is just as important in predicting the time evolution of a classical mechanical system.
When people say that the initial position and momenta are "all" that's needed to predict the evolution of the configuration of the system, they do not mean this in the strict sense that you're indicating.  In particular, the statement they're really making is is along the lines of the following:

For a given set of interactions (forces) between particles in a system, and for a given set of initial positions and momenta of all of the particles, one can (modulo some somewhat pathological examples) determine the positions and momenta of the particles for all times using Newton's Second Law for each particle:
  \begin{align}
  \mathbf F_i = m\ddot {\mathbf x}_i
\end{align}
  where $\mathbf F_i$ is the net force on particle $i$.

Notice that if the forces are unknown, then one can't even write down Newton's Second Law, so there are no equations of motion to solve for the time evolution of the system and, in that case, the initial data are essentially useless.

REPLY [1 votes]: An experimentalist's answer is that the dynamics of all systems are described by second order differential equations, either classical or quantum mechanical.
The classical case is called deterministic because all one needs for describing the trajectories in space and time  of the particles in the system are solutions of the differential equations . The forces and fields entering the  problem exist in the setups of the differential equations.
Once one has the solutions to the equations, the general  form of the  trajectories, the  values  of the position and momentum at one time, t=t', define uniquely  the evolution of the system, basically by construction.
In the quantum mechanical case even though the  differential equations of the dynamics of the problem are second degree, and the potentials are included,  the solutions define probabilities for observing the evolution of the system,  not definite trajectories, and thus  the system cannot be deterministic in the classical sense.


TITLE: Showing that two regular expressions represent complementary regular languages over {0,1}
QUESTION [1 upvotes]: How do up you show that two that the regular expressions, such as $(01+1)^*$ and $(0+1)^*\left(0 + 00(0+1)^*\right)$ represent complementary regular languages over $\{0,1\}$? I'm trying to do some problems from my textbook (An Intro to Formal Lang and Automata by Linz) but I'm stuck on this problem.
Any help would be great, thanks.

REPLY [1 votes]: $(01+1)^*$ represents the language consisting precisely of those words in which every $0$ (if there is one) is immediately followed by a $1$. This excludes all words containing $00$ and all words that end in $0$, and nothing else. An easy way to see this is to realize that $(01+1)^*$ describes the language the same language that you get if you start with $\{1,2\}^*$ (corresponding to the regular expression $(2+1)^*$) and replace each $2$ in every word by $01$. The resulting word clearly must end in $1$ and clearly cannot contain $00$, but it’s not hard to show that it contains everything else.
I assume that your other regular expression is supposed to be $(0+1)^*[0+00(0+1)^*]$; the two $(0+1)^*$ expressions allow the generation of arbitrary strings, so the language is everything of the form $u0$ with $u\in\{0,1\}^*$ together with everything of the form $u00v$ with $u,v\in\{0,1\}^*$ $-$ in other words, everything that either ends in $0$ or contains $00$, precisely the complement of the first language.


TITLE: Prove that $A \subseteq B$ and $B$ is an ideal of $R$.
QUESTION [1 upvotes]: Let $R$ be a commutative ring with unity, let $A$ be an ideal of $R$ and let $a \in R$. Define $B = \{b \in R ~|~ ab \in A \}$.
Prove that $A \subseteq B$ (where $A$ does not have to be the proper subset of $B$) and $B$ is an ideal of $R$.
$\textbf{My Attempt:}$
Prove of $A \subseteq B$:
Since, $A$ is an ideal of $R$.
Then, by definition of ideal, $A$ is a subring $R$ and $A \subseteq R$.
Also, by definition of ideal, $A$ is an ideal of $R$, if for all $r \in R$ and for all $x \in A$ both $rx$ and $xr$ are in $A$.
Since, $R$ is a commutative ring with unity. Then, $A$ is a commutative ring with unity as well.
So, $rx = xr$ is in $A$.
Since, $A \subseteq R$ then by def of subset, for all $x \in A$ $\implies$ $x \in R$.
So, for all $r \in R$ and for all $x \in R$ such that $rx = xr$ is in $A$.
Since, we defined $B = \{ b \in R ~|~ ab \in A \}$. So, $rx = xr$ is also in $B$.
So, this proves that elements that in $A$ is also in $B$. Which means that $A \subseteq B$.
Prove of $B$ is an ideal of $R$:
Let $y \in B$ and since we let $a \in R$
Since, we define $B = \{ b \in R ~|~ ab \in A \}$. So, $y \in R$ and $ay \in A$.
Since, $A \subseteq R$. So, $ay \in R$.
So, we have that $B \subseteq R$.
Then, we need to apply the ideal test.
ideal test case (1):
Let $u,v \in B$. WTP: $u - v \in B$.
Since, $u, v \in B$. Then, $u, v \in R$ and $au \in A$ and $av \in A$.
Since, $R$ is a commutative ring with unity.
Then, by def of commutative ring, $u - v \in R$ as well.
Since, $A$ is an ideal of $R$. Then, by definition of ideal, $A$ is a subring $R$.
Which means $A$ is a commutative ring with unity as well.
Since, $au \in A$ and $av \in A$.
So, $au - av = a(u - v)$ which $a(u - v) \in A$.
Since, we prove that if $u, v \in B$ both $u - v \in R$ and $a(u - v) \in A$.
So, $u - v \in B$ which means ideal test case (1) passed.
ideal test case (2):
Let $\beta \in B$ and $\gamma \in R$. WTP: both $\gamma\beta$ and $\beta\gamma$ are in $B$.
Since, $\beta \in B$. Then, $\beta \in R$ and $a\beta \in A$.
Since, both $R$ and $A$ are commutative ring with unity. So, $\gamma \beta = \beta \gamma \in R$.
Since, $A$ is an ideal of $R$ and $\gamma \in R$ and $a\beta \in A$.
So, $a\beta\gamma = a\gamma\beta \in A$.
So, we proved that both $\gamma\beta$ and $\beta\gamma$ are in $B$. Which means ideal test case (2) passed.
Therefore, we proved that $B$ is an ideal of $R$.
$\textbf{Questions:}$
I don't feel my prove of both $A \subseteq B$ (where $A$ does not have to be the proper subset of $B$) and $B$ is an ideal of $R$ are quite right. Are there any other ways to prove better than mine ? Or are there anyting I can fixes for my prove to make it better ?

REPLY [2 votes]: It’s correct, but it’s much wordier than necessary. For instance, the proof that $A\subseteq B$ could be reduced to this:

Let $x\in A$. $A$ is an ideal, so $ax\in A$, and that by definition means that $x\in B$. Thus, $A\subseteq B$.

In fact it can even be reduced to this:

$aA\subseteq RA=A$, so $A\subseteq B$.

Here $RA=\{rx:r\in R\text{ and }x\in A\}$. $RA\subseteq A$ because $A$ is an ideal, and $RA=A$ because $R$ is a ring with unity.
The proof that $B$ is an ideal can also be drastically shortened.

Let $b_0,b_1\in B$. Then $a(b_0-b_1)=ab_0-ab_1$, where $ab_0,ab_1\in A$. $A$ is a subring of $R$, so $ab_0-ab_1\in A$, and therefore $b_0-b_1\in B$, i.e., $B$ is closed under subtraction.
Now let $b\in B$ and $r\in R$. Then $a(br)=(ab)r$, where $ab\in A$, and $A$ is an ideal of $R$, so $(ab)r\in A$ as well. Thus, $br\in B$, and since $R$ is commutative, $rb$ is also in $B$, and we’ve shown that $B$ is an ideal of $R$.

Indeed, if I were writing it up for myself, I’d shorten it even further:

If $b_0,b_1\in B$, then $a(b_0-b_1)=ab_0-ab_1\in A-A=A$, so $b_0-b_1\in B$. If $b\in B$ and $r\in R$, then $a(br)=(ab)r\in Ar\subseteq A$, so $rb=br\in B$. Thus, $B$ is an ideal.

When you’re just learning, it’s generally better to err on the side of giving more detail than necessary, but it’s also a good idea to work on cutting out parts that really can safely be left to the reader: including too much detail makes it harder to read. With practice one learns to strike an appropriate balance for a given audience.


text:
\begin{document}
\maketitle
\begin{abstract}

  The analysis of games played on graph-like
  structures is of increasing importance due to the prevalence of
  social networks, both virtual and physical, in our daily life.
  As well as being relevant in computer science, mathematical
  analysis and computer simulations of such distributed games are
  vital methodologies in economics, politics and epidemiology,
  amongst other fields. Our contribution is to give compositional
  semantics of a family of such games as a well-behaved mapping,
  a strict monoidal functor, from a category of open graphs
  (syntax) to a category of open games (semantics). As well as
  introducing the theoretical framework, we identify some
  applications of compositionality.

\end{abstract}

\section{Introduction}
\subfile{1-introduction}
\section{Network games}~\label{sec:games_on_graphs}
\subfile{2-gamesongraphs}
\section{Open games}~\label{sec:opengames}
\subfile{3-opengames}
\section{Open graphs}~\label{sec:graphs}
\subfile{4-opengraphs}
\section{Games on graphs via functorial semantics}~\label{sec:def_functor}
\subfile{5-functorialsemantics}
\section{Examples}~\label{sec:example}
\subfile{6-examples}
\section{Conclusions}
\subfile{7-conclusions}
\bibliography{ms}

\appendix
\section{Proofs for Section~\ref{sec:graphs}}~\label{app:graphs}
\subfile{A1-graphs}
\section{Proofs for Sections~\ref{sec:def_functor}}~\label{app:def_functor}
\subfile{A2-functorialsemantics}
\end{document}

config: arxiv


TITLE: Inverse for $1-zb(z)$
QUESTION [0 upvotes]: I need to find the inverse of $1-zb(z)$ with $b(z)=\sum_{n=0}^{\infty}b_nz^n$.
I have tried several approaches where I among other things have tried using the methods in my calculus book but nothing seems to quite work.
What would be the best approach to find this inverse (and what is it)?
(I'm not necessarily asking for the full solution - just the approach and the result so I can deduce the answer myself and know that I got the right answer)

REPLY [1 votes]: Consider the case where $b_1 = 1$ and $b_j = 0$ for all other $j$. Then your function is $f(z) = 1 - z^2$. Since $f(1) = f(-1) = 0$, this function doesn't have an inverse.
Since the problem of finding the inverse cannot be solved for this specific case, it also can't be solved for the general case. You're going to need to tell us more about the coefficients for us to be of any use.


TITLE: Is it true that if $(b)\subset (c) \subset R$, then $(c)/(b)\simeq R/(a)$ as $R$-modules, where $a\cdot c = b$?
QUESTION [1 upvotes]: Is it true that if $(b)\subset (c) \subset R$, $b\neq 0$, then $(c)/(b)\simeq R/(a)$, where $a\cdot c = b$?
$R$ is ring and $(b)$, $(c)$ are its ideals, $(c)/(b)$  is ideal of $R/(b)$ by third theorem of isomorphism for rings.
We can construct homomorphism of $R$-modules  $\varphi: (c)/(b) \rightarrow R/(a)$, such as $\varphi(x\cdot c+(b)) = x+(a)$. It's correct construction, because if $x\cdot c + (b) = y\cdot c+ (b)$, then $(x-y)\cdot c \in (b)$, then $(x-y) \in (a)$. I think it's also easy to prove that it is isomorphism.

REPLY [1 votes]: Your map is not well-defined for the reason you state. More precisely if $(x - y) c \in (b)$, then you cannot simply conclude that $x - y \in (a)$.

For the moment, let us consider a more general situation: Let $I, J \subset R$ be ideals. Then, one defines the colon ideal $I : J$ as $$I : J := \{r \in R \mid rJ \subset I\}.$$

Back to your question:
From your choice of $a$, it is clear that $a \in (b : c)$ or $(a) \subset (b : c)$. But later on, you seem to assume that $(a) = (b : c)$.
But this may not be the case. As a simple counterexample, we can take $R = \Bbb Z$, $b = c = 0$, and $a = 43$. (This is also a counterexample to your proof.)
As a less trivial example, we may take $R = \Bbb Z/6 \times \Bbb Z$, $b = (0, 0)$, $c = (3, 0)$, $a = (2, 0)$.
This can be avoided in nice situations, though. More precisely, we can say the following:

Theorem. Let $R$ be an integral domain, and let $a, b, c \in R$ be such that $a \cdot c = b$ and $c \neq 0$. Then, $(b : c) = (a)$.

Proof. $(a) \subset (b : c)$ is clear. Conversely, let $x$ be such that $xc \in (b)$. Then, $xc = by$ for some $y \in R$. We can write $b = ac$ to conclude that $xc = acy$. Since $c \neq 0$, we may cancel it to conclude that $x = ay \in (a)$. $\qquad \square$

But instead of restricting ourselves to integral domains, we can modify the question to ask:
$$\text{Is } (c)/(b) \simeq R/(b : c)?$$

However, your question is still not true (even in the case of domains). As a counterexample, you may take $R = \Bbb Z$, $b = 4$, and $c = 2$. In this case, the question is equivalent to asking $$2\Bbb Z/4\Bbb Z \simeq \Bbb Z/2\Bbb Z.$$
However, the above is not true. Indeed, the product of any two elements in the left ring is equal to $0$, which is not true for the right ring.

Added. What goes wrong? For the updated question, your map $\varphi$ is indeed well-defined.
However, it is not a homomorphism of rings. (It fails to be multiplicative, which is not surprising since you wouldn't expect $(xc)(yc) = (xy)c$.)
However, it is additive and $R$-linear. Thus, we may treat $\varphi$ as a map of $R$-modules. It is straightforward to note that $\varphi$ is injective since $$(x - y) \in (b : c) \Leftrightarrow (x - y) \cdot c \in (b).$$
Lastly, we check that it is onto. Let $r \in R$ be arbitrary. Then, we have $$\varphi(rc + (b)) = r + (a).$$
Thus, we have an isomorphism of $R$-modules.


TITLE: If surface tension uses perimeter and length to define, how can we use it to predict the surface a water droplet must take in 3D space?
QUESTION [1 upvotes]: Often we say a water droplet takes spherical shape in zero gravity due to surface tension. However, to my knowledge surface tension is related to the length of the boundary over which the surface of the liquid is formed; yet a sphere is a closed surface, hence it has no boundary line. So how do we apply the concept of surface tension here?

REPLY [0 votes]: There is a way of defining surface tension $\Sigma$ as a force per unit length on the boundary of a surface, but that turns out not to be the most useful way of thinking about $\Sigma$.  This is similar to how pressure $P$ can be conceived in mechanics as a force per unit area, but in thermodynamics, we find that it is usually more useful to think of pressure as the quantity that appears in the First Law of Thermodynamics, $dU=dQ-P\,dV$.  This makes it possible to redefine pressure as a partial derivative,
$$P=-\left(\frac{\partial U}{\partial V}\right)_{S}.$$
(The appearance of entropy $S$. here is because we need to specify that this is derivative is taken during an adiabatic process, with no heat transfer $dQ=T\,dS=0$.)
Surface tension plays the same role as $-P$ in a two-dimensional system like a surface.  The differential work associated with a surface tension is $dW=\Sigma\,dA$, where $dA$ is the differential change in surface area.  (The difference in sign between this equation and $dW=-P\,dV$ for a fluid arises from the fact that pressure wants to expand the volume of a fluid, but surface tension wants to shrink the area of a surface.  For a one-dimensional system, like a rope of spring, with a tension ${\cal T}$, stretched by a distance $d\ell$, the differential work is $dW={\cal T}\,d\ell$, which is nothing more than the Work-Energy Theorem.)  So we can think of the $\Sigma$ as the rate at which the energy increases as he increase the surface area adiabatically,
$$\Sigma=\left(\frac{\partial U}{\partial A}\right)_{S}.$$
With this notion of surface tension, it is easy to see that if you start with a spherical drop of water and start to deform it (while keeping the volume constant), the area will necessarily increase by an amount $dA$.  This is associated with an energy cost $\Sigma\,dA$, so the sphere (since it is the most compact configuration, with the smallest surface area to volume) is the lowest-energy, hence equilibrium, state of the drop.


TITLE: Find $f^{(2015)}(0)\space \text{ of }\space f(x)=\sin(\cos(x))$.
QUESTION [0 upvotes]: Find $f^{(2015)}(0)\space \text{  of  }\space f(x)=\sin(\cos(x))$.
I am pretty lost here. The hints say I should use the fact that the function is even, but I don't know how to.

REPLY [5 votes]: Note $f(x)=\sin{(\cos{x})}$ is  even function,
use this well know result
because if $f(x)=f(-x),x\in R$,then we have
$$f'(x)=-f'(-x)$$
Let $x=0$,$$f'(0)=0$$
simaler we have
$$f'''(x)=-f'''(-x)$$
$$f'''(0)=0$$
$$\cdots $$
then we have
$$f^{(2k+1)}(0)=0,k\in N^{+}$$

REPLY [4 votes]: Hint: What is the derivative of an even function? What about an odd?


TITLE: Mistake in proof: Symmetric group acting on group ring
QUESTION [2 upvotes]: Let $G$ be finite group of order $n$ and $K$ be a field. The symmetric group on $G$, denoted by $S_G$ acts linearly on the group ring $K[G]$ by
$$\sigma\left(\sum_{g\in G}c_gg\right):=\sum_{g \in G}c_g\sigma(g) $$
The corresponding homomorphism
$$\rho:S_G\to GL(K[G])\cong GL_n(K)  $$
is, injective since $\rho(\sigma)=id$ implies $\sigma(g)=g$ for all $g\in G$.
Now if $K=\mathbb{F}_p$ this would imply that $n!=\# S_G$ divides $\# GL(K[G])=\#GL_n( \mathbb{F}_p)$, which is absolute nonsense since I didn't use that $G$ is a $p$-group or anything similar. Where did I make a mistake?

REPLY [2 votes]: There is no mistake and as JBL says in the comments you don't use anything about the group structure on $G$. Here is a simpler argument: $S_n$ acts $K$-linearly and faithfully on $K^n$ by permuting the coordinates, for any field $K$ (even any commutative ring). This gives a canonical embedding of $S_n$ into $GL_n(K)$ as the subgroup of permutation matrices which, when $K$ is a finite field $\mathbb{F}_q$, establishes that $n!$ divides the order
$$|GL_n(\mathbb{F}_q)| = q^{ {n \choose 2} } (q - 1)^n [n]_q!$$
of $GL_n(\mathbb{F}_q)$. Here
$$[n]_q! = \prod_{i=1}^n [i]_q = \prod_{i=1}^n \frac{q^i - 1}{q - 1}$$
is the $q$-factorial. It is by no means obvious at first glance, just looking at these formulas, that this divisibility ought to hold. So, how can we prove it directly? We can work one prime at a time. If $p$ is a prime recall that $\nu_p(n)$ is the greatest exponent $k$ such that $p^k \mid n$. Legendre's formula tells us that
$$\nu_p(n!) = \sum_{k \ge 1} \left\lfloor \frac{n}{p^k} \right\rfloor$$
so to prove the divisibility we need to show that $\nu_p(|GL_n(\mathbb{F}_q)|)$ is bigger than or equal to this for all primes $p$. I did this calculation recently here so we can steal it: if $\gcd(p, q) = 1$ then we get
$$\nu_p(|GL_n(\mathbb{F}_q)|) = \left\lfloor \frac{n}{d} \right\rfloor \nu_p(q^d - 1) + \sum_{k \ge 1} \left\lfloor \frac{n}{dp^k} \right\rfloor$$
where $d = \text{ord}_p(q)$ is the smallest positive integer such that $q^d \equiv 1 \bmod p$. In particular $d \le p-1$ and $\nu_p(q^d - 1) \ge 1$ which gives
$$\nu_p(|GL_n(\mathbb{F}_q)|) \ge \sum_{k \ge 0} \left\lfloor \frac{n}{(p-1)p^k} \right\rfloor$$
and the desired inequality is clear since $\frac{1}{p-1} > \frac{1}{p}$. If $p \mid q$ then $\nu_p(|GL_n(\mathbb{F}_q)|) \ge {n \choose 2}$ which is much larger than what we need.
This expression on the RHS can be interpreted as $\log_p$ of the order of the "Sylow $p$-subgroup" of $GL_n(\mathbb{Q})$, in the sense that every $p$-subgroup of $GL_n(\mathbb{Q})$ embeds into it (which means it controls how large a finite subgroup of $GL_n(\mathbb{Q})$ can be). See Serre's Bounds for the orders of the finite subgroups of $G(k)$ for much more on this.