TITLE: What is meant by 2D fourier transform of an image?
QUESTION [7 upvotes]: I have some questions about this interesting concept I came across about 2D Fourier transform, please...
Firstly, the Fourier transform of a 1D signal (such as a sound recording) is as follows:

The first picture is a graph of the real sound file, and the second picture is the sorted frequency bins of the analysed original recording. Although the the graphs are often plotted next to each other, the frequency domain graph really is not at all related to space dimensions, as it has no relations at all to the physical shape of the sound waves (or images file) etc. It is just a histogram of the strength of the different frequencies found after analysis, from bin 1, 2, 3... into a chart.
But, when we come to the 2D Fourier transform for images, suddenly I have trouble even picturing what this might possibly mean?

What is meant by the Fourier transform of a 2D signal? Do we take many 1D Fourier charts in the x-direction as before and do another meta Fourier transform in the y-direction on these frequency charts?
Since frequency charts are not even about spatial relationships, but really about histograms of tabulated frequencies, what does the frequency picture even represent? How to calculate the color value for each of the point $(x,y)$ in the frequency image, and what does the $x$-axis represent, what does the $y$-axis represent in this frequency image?
EDIT: Forward Fourier transform results

REPLY [2 votes]: I'll give you a mathematical expression to augment Gilbert's answer. The Fourier transform in 1D is given by
$$\hat f(k)=\int\mathrm dx \, e^{ikx}f(x).$$
There are ofcourse other conventions but I chose a convenient one. You can then consider a (grayscale) image as a 2D function $f(x,y)$ which gives the intensity of the image at every point $(x,y)$. The Fourier transform in 2D is given by
$$\hat f(k_x,k_y)=\int\mathrm dx\,\mathrm dy \, e^{i(k_xx+k_y y)}f(x,y).$$
The output is, just like $f(x,y)$, a two dimensional function. So the output is again an image! This image will generally be complex so to show this image often the absolute value is taken of the output. Also the log is often taken to bring out details with low intensity. It looks like this happened for your images as well but I'm not sure.
For implemnting this on images you can have a look at the discrete Fourier transform. Note that depending on the implementation the $(0,0)$ might end up either in the corners or in the center of your image. If you want to use the discrete Fourier transform a lot you should always use a library/predefined function because there exists an algorithm to compute the discrete Fourier transform called the Fast Fourier Transform which, like the name implies, is much faster.


TITLE: Proof of Group Permutations
QUESTION [0 upvotes]: Prove that $\sigma\tau = \sigma\mu$ and $\tau\sigma = \mu\sigma$ in $S_n$, then $\tau = \mu$.
Workings:
Proof:
Since $\sigma\tau = \sigma\mu$
Then for $\tau$
$\tau = e\tau$
$\tau = (\sigma^{-1}\sigma)\tau$
$\tau = \sigma^{-1}(\sigma\tau)$
$\tau = \sigma^{-1}(\sigma\mu)$
$\tau = (\sigma^{-1}\sigma)\mu$
$\tau = e\mu$
$\tau = \mu$
Since $\tau\sigma = \mu\sigma$
Then for $\tau$
$\tau = \tau e$
$\tau = \tau(\sigma\sigma^{-1})$
$\tau = (\tau\sigma)\sigma^{-1}$
$\tau = (\mu\sigma)\sigma^{-1}$
$\tau = \mu(\sigma\sigma^{-1})$
$\tau = \mu e$
$\tau = \mu$
Therefore $ \tau = \mu$
I'm wondering if this correct. Any help will be appreciated.

REPLY [1 votes]: Those proofs look correct. It is more common to take the given equation and multiply both sides, either on the left or on the right, by $\sigma^{-1}$. But your proofs are equivalent to that method.
The problem seems to be badly stated, since only one of the two hypotheses is needed to get the conclusion, not both of them. Are you sure there was the word "and" between the two hypotheses rather than the word "or"?


TITLE: Cauchy problem $y' = y + 10\sin y, y(0) = a$
QUESTION [2 upvotes]: I have a Cauchy problem:
 $y' = y + 10\sin y, y(0) = a$
It is need to find $\frac{\partial y}{\partial a} \Big|_a$ where $a = 0$.
I don't know what to do here. Could you help me please?
Thank you.

REPLY [2 votes]: I interpret this question as follows, as indicated in my comments:  given the ordinary differential equation
$\dfrac{dy(x)}{dx} = y'(x) = y(x) + 10 \sin y(x) \tag{1}$
with
$y(0) = a, \tag{2}$
we can actually think of $y$ as a function $y(x, a)$ of the two variables $x$ and $a$; for each value of $a$, we obtain a unique trajectory $y(x, a)$ satisfying $y(0, a) = a$.  Bearing this in mind, and assuming for the moment that $y(x, a)$ is sufficiently differentiable as a function of both $x$ and $a$, it follows that the derivative $\partial y(x, a) / \partial a$ is a meaningful quantity; indeed, for fixed $a$ it metrics the varying rate of change of the trajectory $y(x, a)$ with respect to the initial point as we travel along an integral curve of (1).  For this reason, as well as a certain amount of experience I have had with similar problems, I assume the OP is asking for $(\partial y(x, a) \partial a)_{a = 0}$ as a function of $x$; indeed, if all that is wanted is
$(\dfrac{\partial y(x, a)}{\partial a})_{x = 0, a = 0} = (\dfrac{\partial y(0, a)}{\partial a})_{a = 0} = \dfrac{\partial a}{\partial a} = 1, \tag{3}$
there is not much question here.
To find $(\partial y(x, a) / \partial a)$ as a function of $x$ we may proceed as follows:
Since the function $y + 10 \sin y$ is analytic in $y$, the solution $y(x, a)$, is an analytic function of the two variables $x$ and $a$.  Therefore all partial derivatives of $y(x, a)$ in $x$ and $a$ exist and may be taken in any order; in particular
$\dfrac{\partial y(x, a)}{ \partial x \partial a} = \dfrac{\partial y(x, a)}{ \partial a \partial x}; \tag{4}$
for notational convenience, I introduce the convention
$y_a(x, a) = \dfrac{\partial y(x, a)}{\partial a}; \tag{5}$
then (4) reads
$\dfrac{\partial y_a(x, a)}{\partial x} = \dfrac{\partial}{\partial a} (\dfrac{\partial y(x, a)}{\partial x}); \tag{6}$
using (1), (6) may be written
$\dfrac{\partial y_a(x, a)}{\partial x} = \dfrac{\partial}{\partial a}(y(x, a) + 10 \sin y(x, a)); \tag{7}$
using the chain rule et al:
$\dfrac{\partial}{\partial a}(y(x, a) + 10 \sin y(x, a)) = y_a(x, a) + 10 y_a(x, a) \cos y(x, a)$
$= (1 + 10 \cos y(x, a))y_a(x, a); \tag{8}$
whence (7) becomes
$\dfrac{\partial y_a(x, a)}{\partial x} = (1 + 10 \cos y(x, a))y_a(x, a); \tag{9}$
of course since $y(0, a) = a$, we take the initial condition associated with this equation to be
$\dfrac{\partial y(0, a)}{ \partial a} = 1; \tag{10}$
(9) is a first order, one- dimensional, ordinary, linear, homogeneous equation for $y_a(x, a)$ with non-constant coefficient $1 + \cos y(x, a)$; it is common knowledge that the solution to (9) with any given $(\partial y(0, a) / \partial y)$ is
$\dfrac{\partial y(x, a)}{\partial a} = \dfrac{\partial y(0, a)}{\partial a} e^{\int_0^x(1 + 10 \cos y(s,a)) ds}; \tag{11}$
(11) is easily verified by direct differentiation; that it satisfies the requisite initial condition is also easily seen by simply setting $x = 0$.  I leave these (essentially) no-work details to be filled in by my readers.  Inspecting (11), we see that the integral occurring in the exponential in fact depends on the solution $y(x, a)$ to (1) starting at $y = a$: we need to know this solution to evaluate (11).  This piece of the puzzle is provided by the observation that the solution to (1) with $a = 0$, is in fact the constant solution $0$:
$y(x, 0) = 0; \tag{12}$
(12) also readily checked by direct substitution into (1), (2). Since
$\cos y(x, 0) = \cos 0 = 1 \tag{13}$
for all $x$, we see that
$\dfrac{\partial y(x, 0)}{\partial x} = \dfrac{\partial y(0, 0)}{\partial a} e^{\int_0^x (1 + 10 \cos y(s, 0)) ds} = \dfrac{\partial y(0, 0)}{\partial a} e^{\int_0^x (1 + 10 \cos 0)ds}$
$= 1 e^{\int_0^x 11ds} = e^{11x}.  \tag{14}$
To summarize,
$(\dfrac{\partial y(x, a)}{\partial a})_{a= 0} = e^{11x}. \tag{15}$
Appendix:  Some Theoretical Considerations; References:
That unique trajectories of (1) do indeed exist follows from the Lipschitz continuity of the function $f(y)  = y + 10 \sin y$, which in turn may be seen to be a consequence of its differentiability; indeed
$\dfrac{df}{dy} = 1 + 10 \cos y, \tag{16}$
from which
$\vert \dfrac{df}{dy} \vert = \vert 1 + 10 \cos y \vert \le 1 + 10 \vert \cos y \vert \le 11 \tag{17}$
for all $y$; thus, for $y_2 \ge y_1$,
$\vert f(y_2) - f(y_1) \vert = \vert \int_{y_1}^{y_2} \dfrac{df}{dy} dy \vert \le \int_{y_1}^{y_2} \vert \dfrac{df}{dy} \vert dy \le \int_{y_1}^{y_2} 11 dy \le 11 \vert y_2 - y_1 \vert, \tag{18}$
while if $y_1 \ge y_2$ essentially the same calculation shows that
$\vert f(y_1) - f(y_2) \vert = \vert \int_{y_2}^{y_1} \dfrac{df}{dy} dy \vert \le \int_{y_2}^{y_1} \vert \dfrac{df}{dy} \vert dy \le \int_{y_2}^{y_1} 11 dy \le 11 \vert y_1 - y_2 \vert;\tag{19}$
from (19) and (20) we conclude that we always have
$\vert f(y_2) - f(y_1) \vert \le 11 \vert y_2 - y_1 \vert; \tag{20}$
which shows that $f(y)$ is possessed of a global Lipschitz constant $K = 11$.  From here the standard theory of existence and uniqueness takes over and we conclude that there is in fact a precisely one solution to (1) through any initial point $(x_0, y_0)$.  These essential facts are common knowledge; a detailed exposition, including proofs of all the relevant details, may be found in J. K. Hale's most excellent treatment, Ordinary Differential Equations (ISBN-13:  978-0-486-47212-9), Chapter I; see also the text of Hirsch, Smale and Devaney, Differential Equations, Dynamical Systems, and an Introduction to Chaos (ISBN O-12-349703-5), Chapter 17.  And again, the fact that the trajectory $y(x, a)$ is unique allows $y$ to be unabiguously defined as a function of $x$ and $a$.
But there's more.  $y + 10 \sin y$ is not merely Lipschitz continuous; it infinitely differentiable for all $y$, indeed (real) analytic.  From this it follows, by an extension of the theory of existence and uniqueness, that the solution $y(x, a)$ of (1) with $y(0, a) = a$ is itself infinitely differentiable in $x$ and $a$.  (Indeed, the solutions to $y'(x, a) = f(y, a)$ are $C^k$ whenever $f(y, a)$ is.)  In particular, $y(x, a)$ is a $C^2$ function of $x$ and $a$.  This fact underlies the method presented above to find that $(\partial y(x, a) / \partial a)_{a = 0} = e^{11x}$, since the differential equation for $(\partial y(x, a) / \partial a)$, i.e. (4), (7), (10), hinges on the validity of interchanging the order of the derivatives $\partial / \partial x$ and $\partial / \partial a$.  In addition to the above references, material on the differentiability of solutions, including all the relevant proofs, may be found in these notes.  End of Appendix.


TITLE: Is my answer correct?
QUESTION [2 upvotes]: I'm trying to solve this question:

My solution:
Since $\varphi$ is continuous we have:
$C\text{ is convex}\implies C\text{ is connected}\implies \varphi(C)\text{ is connected}\implies \varphi(C) \text{ is an interval}$.
Am I right? I didn't use the fact the norm is induced by the inner product and in the chapter I'm studying the author haven't written yet about continuous function.
I found also the question a little bit weird, a set can have more than one infimum?
Thanks

REPLY [0 votes]: My first answer addressed the question you asked ("Is my proof correct?"). This one suggests how you might proceed with a correct proof:
Suppose that there were two points, $a, b$ whose squared distance to $p$ was minimal. Consider the function
$$
h(t) = \phi((1-t)a + tb) - \phi(a).
$$
Then $h(0) = h(1) = 0$. Can you evaluate $h(0.5)$, for instance? What can you say about the points $(1-t)a + tb$, for $t \in [0, 1]$?
Further hint: the algebra will be simplified a bit if you just assume that $p$ is the origin; for the non-origin case, first translate everything by $-p$ so that $p$ becomes the origin. Translation leaves distances invariant, so you've reduced the problem to a simpler case.


TITLE: About the proof of the theorem that assures that a surface is invariant with respect to a system of differential equations.
QUESTION [0 upvotes]: I have seen the next theorem in A Lotka-Volterra three species food chain,

Theorem 0.1: Let S be a smooth surface in $\mathbb{R}^3$ and
  \begin{equation}
\begin{cases}
&\frac{dx}{dt}=f(x,y,z),\\
&\frac{dy}{dt}=g(x,y,z),\\
&\frac{dz}{dt}=h(x,y,z),
\end{cases}
\end{equation}
  where $f$, $g$ and $h$ are continuously differentiable. Suppose that $\textbf{n}$
   is a normal vector to the surface S at $(x,y,z)$, and for all $(x,y,z)\in\mathbb{R}^3$ we have that
  \begin{equation*}
\textbf{n}\cdot\left(\frac{dx}{dt},\frac{dy}{dt},\frac{dz}{dt}\right)=0.
\end{equation*}
  Then, S is invariant with respect to the above system.

In the above reference, I have seen that they refer to V. Arnold's book "Ordinary Differential Equations" to search for the proof of this theorem, but I couldn't find it. Does someone know where could I find its proof? I don't mind if it is in Arnold's book or in another place.
Thank you very much for your help.

REPLY [1 votes]: Take any Calc III book and find a proof (or better prove it yourself) that if the surface is defined by the equation $u(x)=C$ then vector gradient $\nabla u(x)$ at the point $x$ is exactly the normal to this surface at this point.


TITLE: pure strategy equilibria and mixed strategy equilibria
QUESTION [1 upvotes]: Three players simultaneously pick a point on the interval $[0,1]$.
The player closest to the average of the three points wins $1$ dollar.
If there is a tie, then the dollar is split equally among them.
More formally, the players simultaneously choose strategies $ s_i ∈ [0,1]. $
The average of their choices is $ S = (s_1 + s_2 + s_3)/3. $
Player $i$’s payoﬀ function is
$ U_i(s_1 , s_2 , s_3) =
\begin{cases}
1/t,  & \text{if $i ∈ arg  min_j |s_j − S| $ } \\
0 & \text{otherwise}
\end{cases}$
where t is the number of players who tie (their choices are equally close to the average).
(a) What are the pure-strategy equilibria of this game?
(b) What are the mixed-strategy equilibria if the possible strategies are limited to playing $0$ or $1$, rather than $[0,1]$?
This question is about game theory, related with nash equilibrium. I have no idea where to start.

REPLY [1 votes]: (a)
Suppose all three choose the same number, $x$. The average is $x$ and the payoff is $\tfrac{1}{3}$. Is there a profitable deviation?
If one player deviates, he can affect the average and move it only third of the way towards his newly chosen value. The average is still closer to $x$ than to him, so he'll get $0$. Thus, for each $x$ in the segment, the strategy in which all players choose $x$ is a pure equilibrium.
Is there an equilibrium where 2 players choose the same $x$ and someone else chooses $y\neq x$? No - from the argument above, the $y$ player gets $0$ and he should switch to $x$.
Is there an equilibrium where all three players choose different numbers? No, since in such a scenario at least one of the players will get $0$ (the one that is far away from the average) and he has a profitable deviation - get closer. For example: choose the same as one of the others and get $0.5$ with him.
(b)
Suppose they choose 1 with probability $x,y,z$ (respectively). Write the expected payoff of player 3 when he chooses $1$ (as a function of $x,y$) and when he chooses $0$.
From indifference, they are suppose the same - this gives you an equation for $x,y$.
You can repeat the process to get $3$ equation with $3$ variables, and the solution is the desired equilibrium (again, $x=y=z=0$ and $x=y=z=1$ are pure equilibria, as stated in (a)).


TITLE: Problems with no known quantum advantage
QUESTION [10 upvotes]: I was wondering what the list of current natural computational problems is for which there is no known complexity advantage in using a quantum computer.
To start things off, I think computation of the edit distance is one for which the fastest known quantum algorithm seems to be the fastest known classical one. More tentatively, I would also suggest sorting as another problem for which there is no known quantum speedup (compared to the fastest known unit-cost word RAM algorithm).

Although I don't want to set a hard restriction, I am particularly interested in problems in NP and/or problems with no known efficient classical solution.

Following a suggestion of Juan Bermejo Vega  here is some further clarification.  I am interested in problems in NP for which there is currently no known big $O$ time complexity advantage at all if you use a quantum computer.
I am not focusing on cases where we can prove there can't be an advantage nor am I focusing on exponential speedups (i.e. polynomial would also be fine).  So far it seems the only two examples are the ones in my question which seems very surprising if it is really true.

REPLY [3 votes]: Recently, this paper in SODA 2018 shows a constant factor approximation algorithm for edit distance in quantum computers with subquadratic time. Note that, no constant factor approximation algorithm for edit distance in classical computers with subquadratic time is known, yet. Moreover, it is believed that no such algorithm exist in classical computers.


TITLE: How to convert an affine matrix to a homography matrix?
QUESTION [0 upvotes]: The title is quite self-explanatory, given an affine matrix, is there a way to convert it to a homography matrix?

REPLY [1 votes]: An affine transformation is a homography.  You can do nothing, and you're done.


TITLE: Is $u$ substitution nothing more than a change of coordinate but in $1$ dimension
QUESTION [0 upvotes]: Often while solving integrals, it becomes easier to solve it, by transforming into polar, spherical and other coordinate systems, from our standard cartesian coordinates.
My question is, is a simple $u$-substitution essentially the same thing, just in one dimension ? Is it nothing more than a simple change of coordinates, where we move from $x$ coordinate to $u$ coordinates ?
Moreover, is coordinate transformation or $u$ substitution, just a trivial change of basis, while integral transforms represent a standard change of basis. Can someone explain why one is considered trivial while the other is 'standard' ?

REPLY [3 votes]: Yes, integration by substitution is a special case of the following theorem:
$\newcommand{\d}{\mathop{}\!\mathrm{d}}$
Let $U,V\subset\mathbb{R}^n$ be open and $\Phi\colon U\to V$ a $C^1$-diffeomorphism. Then a measurable function $F\colon V\to \mathbb{R}$ is integrable iff $$G:=|\det\circ\d\Phi|\cdot(F\circ\Phi)$$ is integrable$^1$ and$$\int_V F=\int_U G$$
(Both integrals are defined with respect to the Lebesgue measure.) More generally, we can consider the Bochner integral and then this theorem also holds if $f$ is a function to a Banach space.

$^1$ Since $\phi\in C^1(U,V)$, $\d\Phi\in C(U,\mathbb R^{n\times n})$ and hence $|\det\circ\d\Phi|\in C(U,\mathbb R)$ is the composition of continuous functions.


TITLE: Proving that $\text{Var}(X)\cdot\text{Var}(Y)\geq\text{Cov}(X,Y)^2$ for real-valued random variables.
QUESTION [3 upvotes]: Consider nondegenerate real-valued random variables $X$ and $Y$ with finite mean and variance. We can prove the inequality
$$\text{Var}(X)\cdot\text{Var}(Y)-\text{Cov}(X,Y)^2\geq0$$
in a variety of ways:
(1) We can recognize left-hand side as the determinant of the variance-covariance matrix of the random vector $(X, Y)$, and then invoke the fact that variance-covariant matrices are positive semidefinite.
(2) The variance and covariance are invariant under shifting by a constant, so by mean-centering $X$ and $Y$, the inequality reduces to $E(X^2)E(Y^2)-E(XY)^2\geq0$, which follows from Cauchy-Schwarz for random variables.
(3) We can try to prove the inequality “directly” by unraveling the left-hand side as
$$\big(E(X^2)-E(X)^2\big)\big(E(Y^2)-E(Y)^2\big)-\big(E(XY)-E(X)E(Y)\big)^2$$
which upon simplifying gives (if I haven’t made an error)
$$\color{blue}{E(X^2)E(Y^2)-E(XY)^2}+\color{red}{2E(XY)E(X)E(Y)-E(X^2)E(Y)^2-E(Y^2)E(X)^2}$$
According to the argument in (2), using shift-invariance of variance and covariance, the red term should vanish, but I’m not seeing why. What algebraic cleverness am I missing?

REPLY [1 votes]: You have made no error in your calculations. However, you have misinterpreted the implication of shift invariance. It's not that the red term vanishes, but rather than if you replace $X$ and $Y$ in the blue portion with $X-\mathbb EX$ and $Y-\mathbb EY$ respectively, then when you simplify it out you will be left with the red and blue expressions together.
By the way, whenever $\textrm{Var}(Y)\not=0$ one has
$$
\textrm{Var} (X)\cdot \textrm{Var} (Y)-\textrm{Cov}(X,Y)^2=\frac{\textrm{Var}\bigl(\textrm{Var}(Y)\cdot X-\textrm{Cov}(X,Y)\cdot Y\bigr)}{\textrm{Var}(Y)}.
$$
I think this fits the bill of a "slick algebraic proof" of the type you are looking for.


TITLE: Bounds for the beta CDF
QUESTION [2 upvotes]: This question is closely related to a previous question that I asked here:
An inequality involving the beta distribution
Let $a,b$ be strictly positive integers, and let $F_{a,b}(x)$ denote the CDF for a Beta distribution with parameters $a$ and $b$: $$F_{a,b}(x) = \frac{\Gamma(a+b)}{\Gamma(a)\Gamma(b)}\int_0^x  \frac{ t^{a-1} (1-t)^{b-1} }{B(a,b)} dt $$
I am trying to understand the behavior of the function $$g(x)=\int_0^x\sqrt{F_{a,b}(t)} ~dt$$
Since there is not an algebraic expression for the integrand, I am wondering if there are any existing bounds for $F_{a,b}(\cdot)$ that would help me bound $g(x)$, for all $a$, $b$, and $x$?
In my previous question, it was answered that $$\frac{b}{a+b}\leq g(1)\leq\frac{2b}{a+b}$$
As one example, an easy observation is that if we fix $b=pa$ for some constant $p$ and let $a$ become large, then $F_{a,b}$ looks like a step function that jumps from $0$ to $1$ at the point $t = a/(a+b)$, and my integral inherits this same behavior.

REPLY [3 votes]: If $\min(a,b)\ge100$, then the mean of the beta distribution will be
$$m=\frac a{a+b},$$
and for its standard deviation we will have
\begin{equation}
    s\approx\sqrt{\frac{ab}{(a+b)^3}}=m\sqrt{\frac b{a(a+b)}}<m\sqrt{\frac1a}\le\frac m{10} \tag{0}
\end{equation}
and similarly
$$s\approx\sqrt{\frac{ab}{(a+b)^3}}=(1-m)\sqrt{\frac a{b(a+b)}}<(1-m)\sqrt{\frac1b}\le\frac{1-m}{10}.$$
So, the beta distribution will be approximated well by $N(m,s^2)$, with $s\lesssim\min(\frac m{10},\frac{1-m}{10})<\frac1{20}$, which latter is much less than $1$. So, for $x\in(0,1)$ and with $\Phi$ denoting the standard normal cdf, the integral in question, $g(x)$, will be approximated well by
\begin{equation}
    \int_0^x\sqrt{\Phi\Big(\frac{t-m}s\Big)}\,dt
=s \int_{-m/s}^{(x-m)/s}\sqrt{\Phi(z)}\,dz
\approx s\psi\Big(\frac{x-m}s\Big), \tag{1}
\end{equation}
where
\begin{equation}
    \psi(u):=\int_{-\infty}^u\sqrt{\Phi(z)}\,dz.
\end{equation}
The approximate equality in (1) holds because, by (0), $m/s$ is much greater than $1$.
So, to approximate $g(x)$, you only need to use the values of one function, $\psi$, of one real variable.
For the error $E(x):=g(x)-s\psi\big(\frac{x-m}s\big)$ of this approximation, here is the graph $\{(t,E(t))\colon0<t<1\}$ for $(a,b)=(100,500)$:


TITLE: Can the emf induced in a conductor passing through the magnetic field be explained via magnetic flux?
QUESTION [1 upvotes]: Let's say I have a conducting rod passing through a constant magnetic field. I get positive charges in the rod moving with velocity will experience a force (eg. upwards) and move up to the top of the rod and hence the potential difference between the top and bottom of the rod will induce an emf.
The thing is, Faraday's law also states that the induced emf is proportional to the rate of change in magnetic flux linkage. Here's what I don't get. Is flux linkage a property inherent in closed loops or can be used for rods as well? I know that rods technically do not have a 'number of turns' but they do have an area.
If so, then there will be magnetic flux through the rod, but as it moves through the field, the magnetic flux will remain the same, so there will be no induced emf in the rod.
Would it be correct to say that there are 2 sources of induced emf in the rod: one from the separation of positive and negative charges and another one from induced emf via Faraday's law (which is doing nothing).
Does that mean that when a rod first enters a magnetic field, the magnetic flux through the rod will increase and there will be an emf induced? Would it mean there will be 2 contributing emfs in the rod, one from the separation of charges and one from the increase in flux linkage by Faraday's Law?
I can't find answers for this online. For a bit more about my knowledge level on this subject, I do not know anything about Maxwell's equations and what they might imply.

REPLY [2 votes]: The emf in a moving wire arises from the magnetic Lorentz forces acting on the free electrons, as they are carried along by the wire...
$$\vec F = (-e)\ \vec w \times \vec B$$
Here, $\vec w$ is the velocity of the free electron. This is the vector sum of the wire's velocity, $\vec v,$ and the drift velocity, $\vec {v_{dr}},$ of the electron through the wire. So
$$\vec F = (-e)\ \vec v \times \vec B\ +\ (-e)\ \vec {v_{dr}} \times \vec B$$
The second term on the right acts at right angles to the wire and implies that we have to do work as we push the wire. The first term on the right gives rise to the emf. If the wire has length $\ell$, and is being moved in a direction at right angles to itself, with $\vec B$ at right angles to plane containing the wire and $\vec B$, then the emf is $$\mathscr E =\frac{\text{work done on charge}}{\text{charge}}=\frac{-evB \ell}{-e}=vB \ell$$
Now let's deal with the flux approach. In my opinion, for a moving conductor this approach is not as fundamental as the Lorentz force approach just given.
For the flux approach we need to include the moving conductor in a circuit. But it's important that the rest of the circuit isn't moving with the conductor, or emf's in the conductor and the rest of the circuit will cancel out. This is why elementary textbooks have the conductor sliding on fixed rails. The circuit might be completed by a fixed voltmeter connected between the rails. As the conductor slides, the area of the circuit changes, so the rate of change of flux through the circuit is
$$\frac{d \Phi}{dt}=\frac{\ell v dt\ B}{dt}=\ell v\ B$$
So you see that the flux approach leads to the same emf as the Lorentz force approach. Why use the flux approach? (a) Because it automatically sums the emfs in any parts of the circuit that are moving, (b) It works both for a moving conductor (provided it is thought of as part of a circuit) and for changing flux density in a fixed circuit (the province of the Faraday-Maxwell law): rather neat.


TITLE: Lipschitz function $\mathrm{dist}(x,\partial U)$
QUESTION [0 upvotes]: The following function ($U\subset\mathbb{R}^n$ open set) $$\delta:\mathbb{R}^n\longrightarrow\mathbb{R}$$ $$\delta(x) = \begin{cases}\!-\mathrm{dist}(x,\partial U)\!\!\!\!&,\;x\in {U}\\\! \;\;\,\mathrm{dist}(x,\partial U)  \!\!\!\!&,\;x\in \mathbb{R}^n-U\end{cases}$$ is Lipschitz-continuous?
Any hint would be appreciated.

REPLY [3 votes]: For both points in the interior or exterior, we follow this answer https://math.stackexchange.com/a/370238/119285, and see that, using the triangle inequality
$$d(x,\partial U)\leq |x-y|+d(y,\partial U),$$
so that if both points are in the interior this is equivalent to
$$\delta(y)-\delta(x)\leq |x-y|$$
and if both points are in the exterior this is equivalent to
$$\delta(x)-\delta(y)\leq |x-y|.$$
Switching the roles of $x,y$ gives the estimate for the absolute value, and thus the Lipschitz constant obtained is $1$.
When one point is in the interior and the other is in the exterior it is easier: as
$$|\delta(x)|\leq |x-y|$$ and simmetrically for $y$, the result follows from the triangle inequality
$$|\delta(x)-\delta(y)|\leq |\delta(x)|+|\delta(y)|\leq 2|x-y|,$$
so the Lipschitz constant is at most $2$.


TITLE: Why is this not correct (fermat number)
QUESTION [0 upvotes]: If $$F_{n}=2^{2^{n}}+1$$
then compute $F_{10}$ mod $14$,
I thought since $2^{10}=2$ mod $14$
that it would be equal to $5$ mod $14$
but apparently it is $3$.
Anyone can help explain?

REPLY [2 votes]: Look at what happens to powers of $2$ modulo $14$.  They are:
\begin{align*}
2^1&=2\\
2^2&=4\\
2^3&=8\\
2^4&=16\equiv 2\\
2^5&=32\equiv 4\\
2^6&=64\equiv 8
\end{align*}
Observe that the pattern repeats every $3$ terms.  Therefore, the power should be computed modulo $3$, not $14$.  You need to compute $2^{10}\pmod 3$ and use that as the exponent on $2$, add $1$, and take that result modulo $14$.


TITLE: Why do I see discrete images when moving an object in front of a screen?
QUESTION [2 upvotes]: I observed this while doing my homework. I have a habit of shaking my pen between my fingers vigorously while thinking something. My PC's screen is right in front of me (on my study table itself).
There was a white background on the screen. Whenever I shook the pen with the screen behind it, I could see 8-10 blurred distinct images in the region  instead of a continuous one which I generally see when the background is dark. Could anyone explain the reason for this?

REPLY [2 votes]: The picture you observe when you are shaking a pen in front of a computer or TV screen is a result of the averaging action of the eye.
When you are shaking a pen in front of a wall illuminated by natural light, you see a continuous blur. Both the wall and the pen contribute to the brightness of the blur. The brighter the wall in comparison with the pen, the greater its relative contribution - so you'll see more of the wall than the pen. If the pen has a bright band in the middle, the blur will have a corresponding bright band as well, i.e., you'll see more of the pen there.
When you are shaking a pen in front of a TV screen, what you see, similarly, depends on the brightness of the part of the screen behind the pen, the background.
If the background is black, you'll see a lot of the pen in the blur. If the background is bright, you'll see mostly the background itself, except at the moments, when the screen goes dark as part of the refresh cycle and, correspondingly, at those moments, you'll see mostly the pen and, overall, a darker image due to a lower average brightness.

In summary, the blur is caused by the inertia (averaging action) of the eye, but the discrete nature of the image of the shaking pen in front of a TV screen is due to the refresh cycle and would be apparent even if the eye had no inertia.