TITLE: How to pick out numbers $0,2\pmod 6$ with a root of unity (possibly sixth)?
QUESTION [0 upvotes]: Let $f(n)=ab^{-n}:\Bbb N\to\Bbb C^\times$
Find some $a,b\in\Bbb C^\times$ such that $f$ has the following properties:

$f(n)$ as a sequence indexed by $n$ has precisely two positive integer subsequences $f(6n)$ and $f(6n+2)$ and every other sequence $f(6n+x\in\{1,3,4,5\})$ contains no positive integers.
$f(x\in\{0,2\})=f(x+6)$

[question ends]

Here's as much as I can do.  For the ordered pair $(a,b)$, I have that:
Setting $(a,b)$ to be the primitive 2nd root of unity $(-1,-1)$ gives a sequence that alternates between positive integers and not: $f(n+2)=f(n)$.
So I suspect the solution involves generalising this idea by setting $b$ to be a primitive sixth root of unity.  I think this because there are exactly two primitive sixth roots of unity and they're separated by exponents $2$ mod $6$, as the solution requires.
Let $\phi$ be the primitive sixth root of unity.  Then some function $f$ which was was integer if and only if $\phi^{n-1}$ was a primitive root of unity would give us the required result. One approach I tried is letting $a=\phi^{-1}$ and $b=\phi$, but that doesn't quite work

REPLY [1 votes]: I'm not quite sure what $\mathbb{C^\times}$ means, but I assume it's something like the set of complex numbers apart from $0$. Please correct me if this is inaccurate, especially if it affects the answer.
Assume that for some $a,b\in\Bbb C^\times$ your property $1$ holds for $n = 1$. Then you have, for some positive integers $m_1$ and $m_2$, that
$$f(6) = m_1 \tag{1}\label{eq1A}$$
$$\begin{equation}\begin{aligned}
f(8) & = f(6)(b^{-2}) \\
& = m_1(b^{-2}) \\
& = m_2
\end{aligned}\end{equation}\tag{2}\label{eq2A}$$
This gives that
$$b^{-2} = \frac{m_2}{m_1} \tag{3}\label{eq3A}$$
Next, you have
$$\begin{equation}\begin{aligned}
f(10) & = f(8)(b^{-2}) \\
& = m_2\left(\frac{m_2}{m_1}\right) \\
& = \frac{m_2^2}{m_1}
\end{aligned}\end{equation}\tag{4}\label{eq4A}$$
Since $f(10)$ is not a positive integer (as $10 = 6(1) + 4$), this means $\frac{m_2}{m_1}$ is a positive rational, non-integral number. Thus, there's at least one prime factor $p$ of $m_1$ where $m_1$ has more factors of $p$ than $m_2$. Let $i$ be the number of these factors of $p$ in $m_1$, and $j$ be the number of factors of $p$ in $m_2$. Then we have
$$j \lt i \tag{5}\label{eq5A}$$
Also, since \eqref{eq4A} is not a positive integer, we also have
$$2j \lt i \tag{6}\label{eq6A}$$
However, note that
$$\begin{equation}\begin{aligned}
f(12) & = f(10)(b^{-2}) \\
& = \left(\frac{m_2^2}{m_1}\right)\left(\frac{m_2}{m_1}\right) \\
& = \frac{m_2^3}{m_1^2}
\end{aligned}\end{equation}\tag{7}\label{eq7A}$$
Property $1$ for $n = 2$ says this is a positive integer. This means that, regarding the factors of $p$ in the numerator and denominator,
$$3j \ge 2i \tag{8}\label{eq8A}$$
However, adding \eqref{eq5A} and \eqref{eq6A} gives
$$3j \lt 2i \tag{9}\label{eq9A}$$
This contradicts \eqref{eq8A}, showing that property $1$ cannot hold.


TITLE: Phase of the standing wave constituents
QUESTION [0 upvotes]: Do 2 superimposed waves have to be 180/0 degree apart to form a standing wave? It seems to be the necessity for forming nodes of 0 displacement (taken that their amplitudes are the same) but can they form a standing wave with minimum displacement nodes if the phase is NOT 180 or 0 degrees?
As in a situation that is NOT the case of incident and it's reflected wave interfering but simply two waves of opposing directions, same frequencies and amplitudes but phases that are anything but 2 π or 1 π apart. I have trouble visualizing it and could not find a simulator that would let me change the phase of opposing waves.

REPLY [1 votes]: Consider two waves of equal amplitude, frequency and speed moving in opposite directions and overlapping.
There will be positions where the waves are exactly in phase, ie the displacements of the two waves are equal at all times thus at these position antinodes are formed whereas if the displacements are equal in magnitude but opposite in direction then nodes are formed.
One way to form a standing wave is by reflection and one can then think of waves from a source and waves from the virtual image of the source overlapping and forming a standing wave.
If there no phase change on reflection then standing waves are produced which have nodes and antimonides at different positions than those if there was a $\pi$ phase change on reflection.
This is equivalent to changing the phase of the virtual source relative to the actual source by $\pi$.
This simulation illustrates this point.
Changing the separation of two sources is equivalent to keep the separation of the two sources the same and changing the phase of one relative to the other.


TITLE: Question about point particle vs. wave equation location
QUESTION [0 upvotes]: Another uncertainty question, this came up in another forum.
As I understand it an electron, for example, is a point-like particle.  I take this to mean it exhibits dimensionless properties, but being only point-like, lacks a defined location.  Instead of a location it has a wave equation.
I asserted this means that all of the properties (eg. charge and mass), of an electron would factually act in a manner that directly matched the wave equation, whatever it was at that moment.  In other words, if the wave equation for the electron was an "electron cloud" in an atom, then the electron's properties would be that also.  It would generate an electric field exactly as if it factually was a cloud around the nucleus until such time as it's wave equation changed.
The other poster insisted I was wrong.  I am sure I can't explain their view exactly but they said electrons don't become "smeared out" (their term) like that and that the electron was just a point particle, we were just uncertain about where it was.
I just can't see it working that way, I believe that if the electron is in any wave equation state, all of it's properties and fields would have to be in the same wave equation state as well.  Of course the location can be confined to a smaller location, different wave equation, but it can never be confined to a point.  So as a practical matter, the properties of an point particle must always be spread over some finite area.
Is this wrong, and if so why?

REPLY [0 votes]: YES, allmost all of this is wrong.
1) The fact that the electron is (or more accurately can be in this model considered as) punctual has nothing to do with its wave behaviour.
2) The latter comes from its mass and (typical) velocity, which result in a de Broglie wavelength lamba_dB=h/mv in the range of 0.1 mm, that is to say the typical size of the atom. And thus imply that its position can not be known accurately at this length. This lead to use the wave equation and its probabilistic interpretation.
3) due to the attractive potential, the related wave is not the propagative wave you seem to have in mind, but a localised wave, that must fulfill a specific wave equation, named Schödinger equation. Typically, the most simple solution is psi[r]=A exp[-r/a_0] where a_0 is Bohr's radius, corresponding to the ground state of the atom. Its squares modulus defines the volumic  probability density of what is often described as the "electronic cloud" to account of the fuzziness of the electron's position.
4) this density is actually associated to an electric charge density which interacts with the positive nucleus. And the corresponding negatve potential energy is the opposite of the energy to provide to break the bounds between  them, that is to say the ionization energy of the atomic ground state.
Finally your last sentence must be reformulated into :
if the electron is in any state, (defined as a solution of the wave equation), all of it's properties ( including the electric field that it creates) are described by this solution.
Is it more clear like this ?


TITLE: Does there exist an uncountable separable metric space $X$ such that every subset of $X$ is a Borel set?
QUESTION [15 upvotes]: Is it consistent with ZFC that there exists an uncountable separable metric space $X$ such that every subset of $X$ is a Borel set?
If the continuum hypothesis holds, or more generally $2^{\aleph_{0}}<2^{\aleph_{1}}$
, then each uncountable separable metric space contains non-Borel sets since there are only $2^{\aleph_{0}}$ Borel sets. Therefore, in order for there to be an uncountable separable metric space $X$ where every subset of $X$ is Borel, we would need to at least have $2^{\aleph_{0}}=2^{\aleph_{1}}$. What other conditions are necessary in order for there to exist an uncountable separable metric space $X$ where every subset of $X$ is Borel?

REPLY [18 votes]: Under Martin's Axiom plus the negation of CH, every set $X$ of reals of size $<\mathfrak c$ is a Q-set, which means that every subset of $X$ is an $F_\sigma$-set with respect to the subspace topology that $X$ inherits from the real line.  For more about such sets, see Arnie Miller's chapter, "Special subsets of the real line" in the Handbook of Set-Theoretic Topology.


TITLE: Will frictional force work against the vertical direction of velocity of an object?
QUESTION [0 upvotes]: We assume that an object is placed on a horizontal plane. If this object is accelerated to vertical direction, will frictional force act against the vertical direction of velocity of object? For any kind of applied force, will frictional force work against the vertical direction of velocity of an object?

REPLY [1 votes]: In physics, the normal force is always perpendicular to the surface of contact between two bodies. And when there is friction, the force of friction is always, by definition, perpendicular to the normal force. The magnitude of the friction force is also proportional to the magnitude of the normal force, and the constant of proportionality is denoted as $\mu$.
Note that the normal force and friction force are actually two perpendicular components of a more general, single "contact" force, but we split them up to make solving problems easier.

In a simple scenario involving an upright table with a flat surface, the normal force the table exerts on an object placed upon it will be vertical, at 90-degree angles with the tabletop. Friction will consequently be parallel to the tabletop. Regardless of the acceleration of the body on the table, the direction of the normal force on it from the table will always be vertical, and the direction of the force of friction will be always be some horizontal direction, depending on the motion of the body.
To answer your title: No, friction will not affect the vertical motion of the body if the body is on a horizontal plane.


TITLE: Time dependent perturbation, particle on a segment
QUESTION [0 upvotes]: Consider a particle of mass $M$ moving alongside a segment of length $a$.
Wavefunction:
$$\psi_n(x)=\sqrt\frac{2}{a} \sin(\frac{n\pi}{a}x)$$
Energy:
$$E_n=\frac{\hbar^2\pi^2 n^2}{2Ma^2}$$
Time-dependent perturbation:
$$\varepsilon(t)=\varepsilon_0e^{-\frac{t}{\tau}}.$$
Particle is in a ground state $n = 1$. What is the probability to find the system  in state $n = 4$?
This formula:
$$p_{m\leftarrow n}=\frac{1}{\hbar^2} \left| \int_0^t e^{i\omega_{mn}t}V_{mn}(t)\,dt \right|^2, m\neq n$$
$$V_{mn}=\int_0^a \psi_m^*\varepsilon (t) \psi_n \, dx.$$
Okay, I try to compute $V_{m1}$:
$$V_{m1}=\int_0^a \frac{2}{a}\varepsilon_0 e^{-\frac{t}{\tau}} \sin(\frac{m\pi}{a}x) \sin(\frac{\pi}{a}x) \,dx = \frac{2}{a}\varepsilon_0 e^{-\frac{t}{\tau}} \int_0^a  \sin(\frac{m\pi}{a}x) \sin(\frac{\pi}{a}x) \,dx = 0$$
Right off I find this integral to be equal to zero. Am I doing anything wrong? Or is the answer zero?

REPLY [0 votes]: Your perturbation is constant in space, so that any integral of the type
$$
V_{nm}=\epsilon \int_0^a dx\, \psi_m(x)\psi_n(x)=0
$$
by orthogonality of the wavefunctions.
There are two possibilities.

there’s a typo and $\epsilon(t)$ depends on $x$ somehow,
it’s a trick question.


TITLE: An estimator for the c.d.f $F$ at a point $x_0$?
QUESTION [6 upvotes]: Problem: Let $X_1,X_2,\ldots,X_n$ be independent identically distributed random variables (i.i.d's) with common CDF $F$. Fix $x_0\in\mathbb{R}$ and find an unbiased estimator for $F(x_0)$. Show that your estimator is an UMVUE for $F(x_0)$ (i.e. a uniformly minimum variance unbiased estimator).

So the problem is to find an unbiased estimator $\hat{F}(x_0)$ for $F(x_0)$ such that if $T$ is any unbiased estimator for $F(x_0)$, then $\operatorname{Var}(T)\geq \operatorname{Var}(\hat{F}(x_0))$.

Attempt: I think I found what should be our unbiased estimator for $F(x_0)$. But after many, many attempts, I have no idea on how to show that it is an UMVUE. My estimator is
$$\hat{F}(x_0)=\frac{1}{n}\sum_{i=1}^nI\{X_i\leq x_0\}$$
where $I\{X_i\leq x_0\}$ is the indicator function that equals $1$ when $X_i\leq x_0$ and $0$ otherwise. We have
$$E[I\{X_i\leq x_0\}]=\int_{-\infty}^{x_0}f(x)dx=F(x_0),$$
so it is clear that $\hat{F}(x_0)$ is unbiased.
But to show that it is UMVUE, the only tool that I have is the Cramer-Rao Inequality:
$$\operatorname{Var}(T)\geq\frac{[g'(\theta)]^2}{n E[(\frac{\partial}{\partial \theta}\log f(X;\theta))^2]}\tag{1}$$
for any unbiased estimator $T$ of $g(\theta)$. But here if $\theta=F(x_0)$ how do we differentiate $f$ with respect to $\theta$?
I got
$$\operatorname{Var}(\hat{F}(x_0))=\frac{F(x_0)(1-F(x_0))}{n},\tag{2}$$
so I am trying to show that the r.h.s of $(1)$ is $(2)$ if $\theta=F(x_0)$ and $g(\theta)=\theta$.

Edit: I looked it up and this estimator for $F(x_0)$ appears to be known as the empirical distribution function. However, there is no clue that this is an UMVUE...

REPLY [1 votes]: Based on the problem statement, we can't assume that a pdf $f(x)$ necessarily exists, so we cannot apply the Cramér-Rao bound. But given that $F$ ranges over all possible CDFs (i.e., we're not restricted to any particular parametric family), $(X_1,\dots,X_n)$ is a complete, sufficient statistic for $F(x_0)$ (for a proof, see Jun Shao, Mathematical Statistics, 2nd ed., Example 2.17, pp. 111-2.). Therefore, by the Lehmann-Scheffé theorem, any unbiased estimator for $F(x_0)$ is a UMVUE, in fact the unique UMVUE; since the empirical distribution function $\hat F(x_0)$ is unbiased, it is therefore the UMVUE.


TITLE: marginal pdf of exponential functions
QUESTION [1 upvotes]: Hi I have a question about marginal PDFs of exponentials. Given is :
$$f(x,y)=y \text{ for } y\leqslant \ e^{-x/4} \text{ and } 0 \text{ otherwise}.$$
I have sketched the graph and got an exponential curve. Question is what parameters is the marginal PDF of $X$ and $Y$ supposed to take? My assumptions:
$$f_X(x)= \int_0^{e^{-x/4}}y \ dy$$
and $f_Y(y)= \int_0^{e^{-x/4}}y \ dx$ which doesn't make sense to me for $f(y)$ (according to the solutions given) why is the parameter in terms of $x$ and not $y$?
moreover, what would be the parameters given to the expected values?
for $\operatorname{E}(X)=\int_0^\infty x \ f(x) \ dx$ because of the asymptote.
what would it be for $\operatorname{E}(Y)= \int_0^\text{?} y \ f(y) \, dy$?
Apart from that, how would the independence be determined from the graph? I know the equation for indepedence is $f(x,y) = f(x)f(y)$ but is the correct way  of using it by plugging in x=0 and y=0 and see if they are equal to each other?

REPLY [1 votes]: It looks as if your pdf must be
$$
f(x,y) = \begin{cases} y & \text{if } x\ge0 \text{ and } 0\le y\le e^{-x/4}, \\[5pt] 0 & \text{if } x<0 \text{ or } y<0 \text{ or } y>e^{-x/4}. \end{cases}
$$
The marginal pdf for $Y$ is
$$
f_Y(y) = \int_0^\infty f_{X,Y} (x,y) \, dx.
$$
You "integrate out" $x,$ getting a function of $y.$ This is similar to a situation with a function of two discrete variables, $i$ and $j,$ both taking values in the set $\{1,2,3\}:$
$$
p_I(i) = \sum_{j=1}^3 p_{I,J}(i,j) = \overbrace{p(1,j)+p(2,j)+p(3,j)}
$$
The part under the $\overbrace{\text{overbrace}}$ is a function only  of $j$ and not of $i;$ the variable $i$ has been "summed out."
The density is $0$ when $y> e^{-x/4},$ and that is the same as $\log y > -x/4,$ or $-4\log y > x.$ And not that $\log y$ is negative since $y<1,$ so $-4\log y$ is positive.
You have
\begin{align}
f_Y(y) & = \int_0^\infty f_{X,Y}(x,y)\, dx = \int_0^\infty \left. \begin{cases} y & \text{if } x< -4\log y \\ 0 & \text{if } x>-4\log y \end{cases}  \right\} \, dx \\[10pt]
& = \int_0^{-4\log y} y \, dx = -4y\log y \quad \text{provided } 0<y<1.
\end{align}
This density is $0$ at the two endpoints $0$ and $1,$ and is not symmetric.
As for "parameters," you can't talk about those unless you choose some parametrized family of distributions that contains this one.
You also have
$$
f_X(x) = \int_0^1 f_{X,Y}(x,y)\,dy = \int_0^{e^{-x/4}} y \, dy = \frac{e^{-x/2}} 2 \quad \text{provided } x>0.
$$
This is an exponential distribution with expected value $2.$


text:
\section{Integral over 2 pi of Cosine of m x by Cosine of n x}
Tags: Definite Integrals involving Cosine Function

\begin{theorem}
Let $m, n \in \Z$ be [[Definition:Integer|integers]].
Let $\alpha \in \R$ be a [[Definition:Real Number|real number]].
Then:
:$\ds \int_\alpha^{\alpha + 2 \pi} \cos m x \cos n x \rd x = \begin {cases}
0 & : m \ne n \\
\pi & : m = n
\end {cases}$
That is:
:$\ds \int_\alpha^{\alpha + 2 \pi} \cos m x \cos n x \rd x = \pi \delta_{m n}$
where $\delta_{m n}$ is the [[Definition:Kronecker Delta|Kronecker delta]].
\end{theorem}

\begin{proof}
Let $m \ne n$.
{{begin-eqn}}
{{eqn | l = \int \cos m x \cos n x \rd x
      | r = \frac {\sin \paren {m - n} x} {2 \paren {m - n} } + \frac {\sin \paren {m + n} x} {2 \paren {m + n} } + C
      | c = [[Primitive of Cosine of a x by Cosine of p x|Primitive of $\cos m x \cos n x$]]
}}
{{eqn | ll= \leadsto
      | l = \int_\alpha^{\alpha + 2 \pi} \cos m x \cos n x \rd x
      | r = \intlimits {\frac {\sin \paren {m - n} x} {2 \paren {m - n} } + \frac {\sin \paren {m + n} x} {2 \paren {m + n} } } \alpha {\alpha + 2 \pi}
      | c =
}}
{{eqn | r = \paren {\frac {\sin \paren {\paren {m - n} \paren {\alpha + 2 \pi} } } {2 \paren {m - n} } + \frac {\sin \paren {\paren {m + n} \paren {\alpha + 2 \pi} } } {2 \paren {m + n} } }
      | c =
}}
{{eqn | o =
      | ro= -
      | r = \paren {\frac {\sin \paren {m - n} \alpha} {2 \paren {m - n} } + \frac {\sin \paren {m + n} \alpha} {2 \paren {m + n} } }
      | c =
}}
{{eqn | r = \paren {\frac {\sin \paren {m - n} \alpha} {2 \paren {m - n} } + \frac {\sin \paren {m + n} \alpha} {2 \paren {m + n} } }
      | c = [[Sine of Angle plus Full Angle/Corollary|Corollary to Sine of Angle plus Full Angle]]
}}
{{eqn | o =
      | ro= -
      | r = \paren {\frac {\sin \paren {m - n} \alpha} {2 \paren {m - n} } + \frac {\sin \paren {m + n} \alpha} {2 \paren {m + n} } }
      | c =
}}
{{eqn | r = 0
      | c =
}}
{{end-eqn}}
{{qed|lemma}}
When $m = n$ we have:
{{begin-eqn}}
{{eqn | l = \int \cos m x \cos m x \rd x
      | r = \int \cos^2 m x \rd x
      | c =
}}
{{eqn | r = \frac x 2 + \frac {\sin 2 m x} {4 m} + C
      | c = [[Primitive of Square of Cosine of a x|Primitive of $\cos^2 m x$]]
}}
{{eqn | ll= \leadsto
      | l = \int_\alpha^{\alpha + 2 \pi} \cos m x \cos m x \rd x
      | r = \intlimits {\frac x 2 + \frac {\sin 2 m x} {4 m} } \alpha {\alpha + 2 \pi}
      | c =
}}
{{eqn | r = \paren {\frac {\alpha + 2 \pi} 2 + \frac {\map \sin {2 m \paren {\alpha + 2 \pi} } } {4 m} } - \paren {\frac \alpha 2 + \frac {\sin 2 m \alpha} {4 m} }
      | c =
}}
{{eqn | r = \paren {\frac {\alpha + 2 \pi} 2 + \frac {\sin 2 m \alpha} {4 m} } - \paren {\frac \alpha 2 + \frac {\sin 2 m \alpha} {4 m} }
      | c = [[Sine of Angle plus Full Angle/Corollary|Corollary to Sine of Angle plus Full Angle]]
}}
{{eqn | r = \frac {\alpha + 2 \pi - \alpha} 2
      | c =
}}
{{eqn | r = \pi
      | c =
}}
{{end-eqn}}
{{qed}}
\end{proof}

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