TITLE: Existence of whole number between two real numbers $x$ and $x +1$?
QUESTION [2 upvotes]: How to prove that there is a whole number, integer, between two real numbers $x$ and $x+1$ (in case $x$ is not whole)?
I need this for an exercise solution in my Topology class, so I can, probably, use more than just axioms from set theory.
Any ideas?

REPLY [1 votes]: Under the usual Lebesgue measure on $\Bbb R$, the quotient $\Bbb R/\Bbb Z$ has volume 1.
Also the interval $(x,x+1)$ has measure 1 and it is connected.
Therefore the image of $(x,x+1)$ under the quotient map $\pi$ is connected and has measure 1, so it is $\Bbb R/\Bbb Z\setminus\{\bar x\}$. Since $\{\bar x\}\neq\{\bar 0\}$, there must be a point $n\in(x,x+1)$ such that $\pi(n)=\bar 0$. Thus $n\in\Bbb Z$ by definition.


TITLE: show that $\int_{0}^{\infty}\frac{x\cos ax}{\sinh x}dx=\frac{\pi^2}{4} \operatorname{sech}^2 \left(\frac{a\pi}{2}\right) $
QUESTION [6 upvotes]: Show that $$\int_{0}^{\infty}\frac{x\cos ax}{\sinh x}dx=\frac{\pi^2}{4} \operatorname{sech}^2\left(\frac{a\pi}{2}\right) $$

I think we can solve it by contour integration, but how? and it would be better if there were two ways, the first with real analysis and the second with complex analysis.

REPLY [0 votes]: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{\bbox[5px,#ffd]{\int_{0}^{\infty}{x\cos\pars{ax}
\over \sinh\pars{x}}\,\dd x =
{\pi^{2} \over 4}\on{sech}^{2}\pars{\pi a \over 2}}:\
{\Large ?}.}$

\begin{align}
&\bbox[5px,#ffd]{\int_{0}^{\infty}{x\cos\pars{ax}
\over \sinh\pars{x}}\,\dd x} =
\totald{}{a}\int_{0}^{\infty}{\sin\pars{ax}
\over \sinh\pars{x}}\,\dd x
\\[5mm] = &\
2\,\totald{}{a}\Im\int_{0}^{\infty}
{\expo{\ic ax} - 1 \over \expo{x} - \expo{-x}}\,\dd x
\\[5mm] = &\
2\,\totald{}{a}\Im\int_{0}^{\infty}
{\expo{-\pars{1 - \ic a}x} - \expo{-x} \over 1 - \expo{-2x}}
\,\dd x
\\[5mm] \stackrel{2x\ \mapsto\ x}{=}\,\,\,&
\totald{}{a}\Im\int_{0}^{\infty}
{\expo{-\pars{1/2 - \ic a/2}x}\,\,\, - \expo{-x/2} \over 1 - \expo{-x}}\,\dd x
\\[5mm] = &\
\totald{}{a}\Im\left[\int_{0}^{\infty}
{\expo{-x} - \expo{-x/2} \over 1 - \expo{-x}}\,\dd x
\right.
\\[2mm] &\
\phantom{\totald{}{a}\Im\left[\right.}
\left.-\int_{0}^{\infty}
{\expo{-x} \expo{-\pars{1/2 - \ic a/2}x} \over 1 - \expo{-x}}
\,\dd x\right]
\\[5mm] = &\
\totald{}{a}\Im\bracks{\Psi\pars{1 \over 2} -
\Psi\pars{{1 \over 2} - {a \over 2}\,\ic}}
\\[5mm] = &\
\totald{}{a}\bracks{-\Psi\pars{1/2 - a\ic/2} +
\Psi\pars{1/2 + a\ic/2} \over 2\ic}
\\[5mm] = &\
-\,{\ic \over 2}
\totald{}{a}\bracks{\pi\cot\pars{\pi\bracks{{1 \over 2} - {a \over 2}\,\ic}}}
\\[5mm] = &\
-\,{\pi\,\ic \over 2}
\totald{}{a}\bracks{\tan\pars{{\pi a \over 2}\,\ic}}
=
{\pi \over 2}
\totald{}{a}\bracks{\tanh\pars{\pi a \over 2}}
\\[5mm] = &\
\bbx{{\pi^{2} \over 4}\on{sech}^{2}\pars{\pi a \over 2}} \\ &
\end{align}


TITLE: What should I do to tackle the following matrices calculation?
QUESTION [0 upvotes]: Through chapter 3 of Group Theory by Morton Hamermesh in part 3-6 (Equivalent representations; characters.) I stopped in some point.
It's told "If we change the basis in the n-dimensional space $L$, the matrices $D(R)$ will be replaced be their transforms by some matrix $C$.
The matrices
$$D^{'}(R) = C D(R) C^{-1}$$
also provide a representation of the group $G$, which is equivalent to the representation $D(R)$.
What we wish to find are quantities which are intrinsic properties of $D(R)$, i.e., are invariant under change of coordinate axes. One such invariant is easily found, for if we take the sum of the diagonal elements of the matrix, we obtain
$$\sum_{i}(CD(R)C^{-1}))_{ii} = \sum_{ikl} C_{ik}D_{kl}(R)C_{li}^{-1 }= \sum_{kl} \delta _{kl}D_{kl}(R)= \sum_{k} D_{kk}(R)$$
Thus the sum of the diagonal elements of matrix $D(R)$ is invariant under a transformation of coordinate axes."
Now, this is a stop point for me. I can't understand how $\delta_{kl}$ appears in last relation. Does matrix $C$ is diagonal?

REPLY [1 votes]: The equality $C^{-1}C=Id$ gives, by taking the $(l,k)$ component :$$\sum_{i} C^{-1}_{li}C_{ik}=\delta_{lk};$$now multiply this by $D_{kl}$ and sum over $k,l$ and you get the middle equality in your equation.


TITLE: Heat equation solution using Fourier transform
QUESTION [4 upvotes]: I want to solve the equation
$$x^2\frac{\partial^2 u(x,t)}{\partial x^2}+ax\frac{\partial u(x,t)}{\partial x}=\frac{\partial u(x,t)}{\partial t}$$
with $u(x,0)=f(x)$ for $0<x<\infty$ and $t>0$. Substituting $U(y,t)=u(e^{-y},t)$ and $F(y)=f(e^{-y})$ we get
$$ \frac{\partial^2 U(y,t)}{\partial y^2}+(1-a)\frac{\partial U(y,t)}{\partial y}=\frac{\partial U(y,t)}{\partial t},$$
with the solution
$$\hat U(\xi,t)=\hat F(\xi)e^{(-4\pi\xi^2+(1-a)2\pi i\xi)t},$$ since $\hat U(\xi,0)=\hat F(\xi)$. Taking the Fourier transform in the y variable (assuming that u satisfies the necessary conditions and $\hat {\frac{\partial U}{\partial t}}$=$\frac{\partial}{\partial t}\hat U$), and using $$\hat F(\xi)=\int_{-\infty}^{\infty} F(x)e^{-2\pi xi\xi} dx=\int_{0}^{\infty} \frac{f(y)}{y}e^{2\pi i\log(y)\xi} dy,$$ we are supposed to get
$$u(x,t)=\frac{1}{\sqrt{4\pi t}}\int_0^{\infty}e^{-(\log(v/x)+(1-a)t)^2/(4t)}f(v) \frac{dv}{v},$$
whereas I get
$$\int_{-\infty}^{\infty} \int_0^{\infty} e^{(-4\pi^2\xi^2+(1-a)2\pi i\xi)t}e^{2\pi\log(y/x)i\xi} \frac{f(y)}{y} dyd\xi$$
and don't know how to simplify it.

REPLY [3 votes]: Use $$-4\pi\xi^2t+(1-a)2\pi i\xi t+ 2\pi i\xi\log(y/x)=\frac{t}{\pi}\left(2\pi i\xi+\frac{(1-a)t+\log(y/x)}{2t} \pi\right)^2-\frac{((1-a)t+\log(y/x))^2}{4t}\pi$$ and $$\int_{-\infty}^{\infty}e^{(2\pi i \xi+a)^2} d\xi=\frac{1}{2\sqrt{\pi}}$$ for any $a\in\mathbb{R}$. The integral above you can calculate by transforming it to the case $a=0$ by substitution, and for the case $a=0$ you have a gaussian integral which is well known. By this and some further elementary calculation you should get your solution.


TITLE: the sum of 3 numbers on the second power to be sub 0
QUESTION [2 upvotes]: This is my exercise

$x^3 - 3 x^2 + 6 x - a$, the polynom is from $\mathbb{R}[X]$.

And my job is to calculate $x_1^2+x_2^2+x_3^2$. I solved it and I get $-3$ as a result.
My question is: can 3 numbers $a^2+b^2+c^2$ be lower than $0$? Because the power is supposted to make all 3 of them positive.

REPLY [2 votes]: your answer is correct
your polynomial is : $x^3 - 3 x^2 + 6 x - a  $
and its coefficients are : $a_0=a$, $a_1=6$, $a_2=-3$, $a_3=1$
$\sigma_k = (-1)^k \frac{a_{3-k}}{a3} =(-1)^ka_{3-k} $ for $k=1,2,3$
$$x_1+x_2+x_3=\sigma_1 = -a_2=3$$
$$x_1x_2+x_1x_3+x_2x_3 =\sigma_2 = a_1=6$$
$$(x_1+x_2+x_3)^2= x_1^2+x_2^2+x_3^2+2(x_1x_2+x_1x_3+x_2x_3) \implies x_1^2+x_2^2+x_3^2 = \sigma_1^2 - 2\sigma_2= 9-12 =-3 $$
and to answer your question yes it can be less than $0$ if $a,b,c \in \mathbb{C}$
for example : $a = i, b = 2i, c = -1$   $a^2+b^2+c^2= i^2+4i^2+(-1)^2=-1-4+1=-4$


TITLE: Meaning and Intuition behind product $\sigma$ algebras , Folland Definition
QUESTION [9 upvotes]: I am currently studying Real Analysis using Folland's book by the same name, but I have troubles in understanding the product $\sigma$-algebras definition which is as follows:
Let $\{ X_{\alpha} \}_{\alpha \in A}$ be an indexed collection of nonempty sets, $X = \prod_{\alpha \in A}X_{\alpha}$, and $\pi_{\alpha}: X \rightarrow X_{\alpha}$ the coordinate maps. If $\mathcal{M}_{\alpha}$ is a $\sigma$-algebra on $X_{\alpha}$ for each $\alpha$, the $\textbf{product $\sigma$-algebra}$ on $X$ is the $\sigma$-algebra generated by :
$$\{ \pi^{-1}_{\alpha}(E_{\alpha}):E_{\alpha} \in \mathcal{M}_{\alpha}, \alpha \in A \}.$$
First I don't know what is the definition for the coordinate maps and then what is the intuition or how can I understand more practically product $\sigma$-algebras ?
Thank you!

REPLY [0 votes]: How does the map look like when $X_1 = \{a,b\}$ and $X_2 = \{c,d\}$ and also assuming the sigma algebra on $X_1$ and $X_2$ are as follows: $$\sigma(X_1) = \mathcal P(X_1) = \{\phi,\{a\},\{b\},\{a,b\}\}$$ and $$\sigma(X_2) = \{\phi,X_2\}=\{\phi,\{c,d\}\}$$
In this case, $\pi_a^{-1}(\{a,b\})$ doesn't exist but there is still an element $\{a,b\}$ in the sigma algebra generated by $X_1$.


TITLE: Is the identity true?
QUESTION [0 upvotes]: Is the following identity true:
$$\lim\limits_{\Delta x \to 0}  \dfrac{\displaystyle \int_0^{\Delta x} f(x)\ dx}{\displaystyle \int_0^{\Delta x} g(x)\ dx}
=\dfrac{\lim\limits_{\Delta x \to 0} \displaystyle \int_0^{\Delta x} f(x)\ dx}{\lim\limits_{\Delta x \to 0} \displaystyle \int_0^{\Delta x} g(x)\ dx}
=\dfrac{f(x) \lim\limits_{\Delta x \to 0} \displaystyle\int_0^{\Delta x}  dx}
       {g(x) \lim\limits_{\Delta x \to 0} \displaystyle\int_0^{\Delta x}  dx}
=\dfrac{f(x)}{g(x)}$$
If yes:
$(1)$ How is taking $f(x)$ and $g(x)$ out of the integral mathematically justifyable?
$(2)$ Can we extend this integral to higher integrals (surface and volume integrals)? For example:
$$\lim\limits_{\Delta V \to 0}  \dfrac{\displaystyle \int_{\Delta V} f(x,y,z)\ dV}{\displaystyle \int_{\Delta V} g(x,y,z)\ dV}$$

REPLY [1 votes]: Both are false in the manner you have stated them, for example see the comment about the RHS being a function of $x$, while the LHS is a constant. FOr your higher diemnsional geenralisation, when you write $\lim_{\Delta V \to 0},$ you're thinking of $\Delta V$ as a number, but when you write $\int_{\Delta V} (\cdots)$, you're thinking of $\Delta V$ as a subset of $\Bbb{R}^3$. However, your intuition seems to be correct.
I'll offer you two possible statements which can be proven easily. Note that I make no attempt at minimal hypothesis... so you could probably get away with weaker hypotheses, but then necessarily the proof will have to be more involved.

Suppose $f,g: \Bbb{R} \to \Bbb{R}$ are continuous functions, and that there is an interval containing $0$, where $g$ doesn't vanish. ($\exists \delta > 0$ such that $  \forall x\in (- \delta, \delta)$, $g(x) \neq 0$). Then, it is true that
\begin{equation}
\lim_{h \to 0} \dfrac{\displaystyle\int_0^hf}{\displaystyle\int_0^hg} = \dfrac{f(0)}{g(0)}.
\end{equation}

In higher dimensions:

Let $f,g: \Bbb{R}^n \to \Bbb{R}$ be continuous functions, and suppose that there is an open ball around the origin $0$, where $g$ doesn't vanish. For $r > 0$, let $B_r$ denote the closed ball of radius $r$ around the origin. Then,
\begin{equation}
\lim_{r \to 0} \dfrac{\displaystyle\int_{B_r}f}{\displaystyle\int_{B_r}g} = \dfrac{f(0)}{g(0)}
\end{equation}

In the single variable case, you can see this as an application of LHopital's rule and the fundamental theorem of calculus. But I think using the Mean-Value theorem for integrals is more easily generalised. So, I'll prove it using that instead.
Note that for any $h \in \Bbb{R}$, (since $f$ and $g$ are continuous) we can apply the mean value theorem of integrals to say
\begin{equation}
\int_0^h f = f(\xi_h) \cdot h \quad \text{and} \quad \int_0^h g = g(\eta_h) \cdot h
\end{equation}
for some $\xi_h, \eta_h$ between $0$ and $h$. So,
\begin{align}
\lim_{h \to 0} \dfrac{\displaystyle\int_0^hf}{\displaystyle\int_0^hg} &= \lim_{h \to 0}
\dfrac{f(\xi_h) \cdot h}{g(\eta_h) \cdot h} \\
&= \lim_{h \to 0} \dfrac{f(\xi_h)}{g(\eta_h)} \\
&= \dfrac{f(\lim_{h \to 0} \xi_h)}{g(\lim_{h \to 0} \eta_h)} \tag{$*$}\\
&= \dfrac{f(0)}{g(0)}.
\end{align}
By the way in $(*)$, we can bring the limit inside the functions because we assumed continuity of $f$ and $g$.
In the multidimensional case, the exact same proof works, with some minor notational changes: for any $r> 0$, by the mean-value theorem for integrals, we can say that
\begin{equation}
\int_{B_r} f = f(\xi_r) \cdot \text{vol}(B_r) \quad \text{and} \quad \int_{B_r} g = g(\eta_r) \cdot \text{vol}(B_r)
\end{equation}
for some $\xi_r, \eta_r \in B_r$. Hence,
\begin{align}
\lim_{r \to 0} \dfrac{\displaystyle \int_{B_r}f}{\displaystyle \int_{B_r}g} &=
\lim_{r \to 0} \dfrac{f(\xi_r) \cdot \text{vol}(B_r)}{g(\eta_r) \cdot \text{vol}(B_r)} \\
&= \lim_{r \to 0} \dfrac{f(\xi_r)}{g(\eta_r)} \\
&= \dfrac{f(\lim_{r \to 0} \xi_r)}{g(\lim_{r \to 0} \eta_r)} \tag{$**$} \\
&= \dfrac{f(0)}{g(0)}
\end{align}
Once again, we can bring the limits inside the functions in $(**)$ because we assumed continuity of the function $f$ and $g$.
(If you would like to see a proof of the mean-value theorem for integrals, let me know, I can add that into my post as well.)


TITLE: Semi positive definite Hessian Matrix
QUESTION [0 upvotes]: Given a function $f(x,y):\mathbb{R}^2\to \mathbb{R}$, $C^2$, and a point $x_{0}$ where:
1) The derivative is zero
2) The respective Hessian matrix is semi positive definite
Can I say $x_{0}$ is a local minimum point?
If not, what can I say about this point and how can I reach more information about it?

REPLY [0 votes]: Take $x_0=(0,0)$, $f(x,y)=x^4+y^4$, $g(x,y)=-x^4-y^4$, and $h(x,y)=x^3-y^3$. In each case:

$f'(x_0)$ is the null function;
the Hessian at $x_0$ is semi-definite positive (actually, it's $0$).

However, $x_0$ is a minimum of $f$, a maximum of $g$, and a saddle point of $h$. So, you have to examine how the function behaves near $x_0$ by other methods.


TITLE: Find point on circle's tangent based on point on circle, radius and angle
QUESTION [0 upvotes]: The circle is centered at (0,0)"P" with a radius of 5. I have a point on the circle at (4,-3)"A".
How would I find the points "B1" and "B2" on the tangent through point "A" given an arbitrary angle (A P Bx)?

REPLY [0 votes]: You can obtain the equation of the tangent at $A$ which is $3y=4x-25$
If the given angle $APB$ is $\theta$ then $PB=5\sec\theta$, so you can form and solve a quadratic equation like $$x^2+(\frac{4x-25}{3})^2=25\sec^2\theta$$


TITLE: Nonlinear ODE with integral
QUESTION [3 upvotes]: My question is to find the function $f(t)$ such that $$\frac{df}{dt} = -2f(t)\int_{0}^{t}f(s)\, ds$$ with $f(0) = 1$.
My idea is to divide both sides by $-2f$ and differentiate both sides, and then let $g = \frac{df}{dt}$ and consider $g$ as a function of $f$ which would reduce the order of the differential equation, but this doesn't seem to be working. Is this the right way? Is there another way to do this?

REPLY [4 votes]: I don't know if your way works or not, but here is how I would do it.  Define $F(t)=\int_0^t f(s)\,ds$.  The ODE becomes $F''(t)=-2F'(t)F(t)=-\frac{d}{dt}(F(t)^2)$.  Integrate both sides, get $F'(t)=-F(t)^2+c$.  $F(0) = 0$ and $F'(0) = f(0) = 1$, so $c=1$, and $\frac{dF}{dt}=1-F^2$.  This is a first-order separable ODE that can be solved by separation of variables.


TITLE: Where do the traces come from in Casimir's trick?
QUESTION [0 upvotes]: I am following the derivation for electron-muon scattering amplitude in Griffiths textbook and got to the section where they use Casimir's trick. I can't see where the traces come from. Equation 7.123 on pg. 251:
$$\require{cancel} \sum_{i,j =1}^{4} Q_{ij}(\cancel{p}_a + m_a)_{ji} = \sum_{i=1}^{4} [Q(\cancel{p}_a + m_a)]_{ii} = Tr[Q(\cancel{p}_a + m_a)] \tag{7.123}$$
$Q$ is a $4 \times4$ matrix. What I don't understand is the middle step - why is there suddenly only summation of $ii$?
When I write it out I get:
$$\sum_{i,j =1}^{4} Q_{ij}(\cancel{p}_a + m_a)_{ji} = \sum_{i=1}^{4}[Q_{i1}X_{1i} + Q_{i2}X_{2i} + Q_{i3}X_{3i} +  Q_{i4}X_{4i}] $$
Where I've replaced the $4 \times 4$ matrix $\cancel{p}_a - m_a$ with $X$.
Then when I carry the sum out over $i$ I get a load of cross-terms as well as the diagonals, so not sure how that results in a trace, unless Q and X are such that all the diagonal elements equal zero.
Am I doing sums wrong? I'd expect just to be left with the sum of the diagonals.

REPLY [1 votes]: The trace of any matrix is defined by the sum of its diagonal elements, as was pointed out in the comments.
You can basically view what you have inside your square brackets in Eq. (7.123) as one matrix, and then you sum over the diagonals because you only consider the $ii$-elements of the matrix, no mixed terms of the form $ij$ or $ji$, where $i\ne j$.


text:
\section{Parabolic Riemann Surface is Plane, Punctured Plane or Torus}
Tags: Riemann Surfaces

\begin{theorem}
A [[Definition:Parabolic Riemann Surface|parabolic Riemann surface]] is [[Definition:Conformal Isomorphism|conformally isomorphic]] to either:
:the [[Definition:Complex Plane|complex plane]]
:the [[Definition:Punctured Complex Plane|punctured complex plane]] $\C \setminus \set 0$
or:
:a [[Definition:Torus (Topology)|torus]].
\end{theorem}

\begin{proof}
{{ProofWanted}}
[[Category:Riemann Surfaces]]
a1rl0eigrhpgy6ha6rrzg3u3wsrszus
\end{proof}

config: wiki


text:
\newcounter{point}[section]
\newcommand{\point}{
\paragraph{(\roman{point})}
\stepcounter{point}
}

\newcommand{\describe}[1]{{\bf (#1)}}

\newcommand{\mynewline}{\mbox{}\\}

\newcommand{\separator}{\vspace{1em} \begin{center} $\bullet$ \end{center} \vspace{1em}}

\newcommand{\chapteroutline}{\P \hspace{1ex}}

\newcommand{\says}[1]{\vspace{1em}
{\sc #1} --}

\newcommand{\myheading}[2]{
\begin{center}
{\Large \bf Notes for #1 -- #2\\
 Snorre Christiansen, \today\\
}
\rule{\textwidth}{1pt}\\
\vspace{1cm}
\end{center}
}

\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{conjecture}[theorem]{Conjecture}

\theoremstyle{definition}
\newtheorem{definition}{Definition}[section]

\theoremstyle{remark}
\newtheorem{remark}{Remark}[section]
\newtheorem{exercise}{Exercise}[section]
\newtheorem{problem}{Problem}[section]
\newtheorem{example}{Example}[section]
\newtheorem{question}{Question}[section]

\newenvironment{proofof}[1]
{\noindent {\itshape -- Proof of #1:} }
{\hfill $\square$}

\newtheorem{theoreme}{Th\'eor\`eme}[section]

\newtheorem{lemme}[theoreme]{Lemme}
\newtheorem{corollaire}[theoreme]{Corollaire}

\theoremstyle{plain}
\newtheorem{teorem}{Teorem}[section]

\newtheorem{proposisjon}[teorem]{Proposisjon}
\newtheorem{korollar}[teorem]{Korollar}
\newtheorem{aksiom}{Aksiom}
\newtheorem{prinsipp}{Prinsipp}

\theoremstyle{definition}
\newtheorem{definisjon}{Definisjon}[section]

\theoremstyle{remark}
\newtheorem{bemerkning}{Bemerkning}[section]
\newtheorem{oppgave}{Oppgave}[section]
\newtheorem{eksempel}{Eksempel}[section]
\newtheorem{kommentar}{Kommentar}[section]

\newenvironment{bevis}
{\noindent{\itshape -- Bevis: }}
{\hfill $\square$}

\newenvironment{stikkord}
{$\bullet$}
{$\bullet$}

\newcommand{\myquote}[3]
{
\vspace{2mm}
\begin{flushright}
\begin{minipage}{9cm}
 #3
\end{minipage}\\
\vspace{2mm}
{\sc #1} \  {\it #2}
\end{flushright}
\vspace{2mm}
}

\newcommand{\bibauthor}[1]{{\scshape #1}: }
\newcommand{\bibtitle}[1]{{\itshape #1}; }
\newcommand{\bibspec}[1]{{\upshape #1}.}
\newcommand{\mybib}[3]{\bibauthor{#1}\bibtitle{#2}\bibspec{#3}}

config: arxiv


TITLE: Probability Of Two Boys Puzzle (Standard vs Tuesday Boy)
QUESTION [1 upvotes]: For those that aren't familiar with Gary Foshee's probability puzzle/paradox from 4-5 years ago, you can find an analysis here:
http://news.bbc.co.uk/2/hi/programmes/more_or_less/8735812.stm
While this puzzle has been discussed here on Stack Exchange:
The Tuesday Birthday Problem - why does the probability change when the father specifies the birthday of a son?
(two or three times in fact), the focus has been on the solution and not the LOGIC/MECHANICS behind why the solutions between the two statements below are different.
Statement 1: "I have two children. One of them is a boy. What is the probability I have two boys?" -- OK, the explanation makes sense, it's 1/3.
Statement 2: "I have two children. One is a boy born on a Tuesday. What is the probability I have two boys?" -- Again, the explanation makes sense, it's 13/27
What are the mechanics behind the change in probability? Is it because the more specific he is about one of his children the less chance it could be either child he is referring to as opposed to one in particular? So there's some kind of probability "overlap" that is removed (sorry, I don't know a better way to put it).
I've seen this explained similarly here (though still not quite as satisfactorily as I would like):
https://www.quora.com/What-is-the-reasoning-behind-Gary-Foshees-boy-born-on-a-Tuesday-problem-where-providing-additional-apparently-irrelevant-information-changes-the-resulting-probabilities
By adding specific information "You've effectively told people about one individual, not told them that one of your children is a member of a category."
Thank you!

REPLY [1 votes]: Well, let's try it a little differently.
Question #1: Mr. Smith, I hear you have two children. Is either of them a boy?
Answer : Why, yes. At least one is.
Solution 1: condider four Mr. Smiths. One would have said "no" because he has two girls. Three would say "yes"; one because he has two boys, and two because they have a boy and a girl. So the probabilty that a Mr. Smith who says "yes" has two boys is 1/3.
Question #2: Mr. Smith, I hear you have two children. Is either a boy who was born on a Tuesday?
Answer #2: What an odd question! But you guessed right, at least one is.
Solution #2: Consider 196 Mr. Smiths. 169 will say "no," 49 of them because they have two girls. Of the 49 who have two boys, 13 will say "yes." Of the 98 who have one boy, 14 will say "yes." So the answer is (13)/(13+14)=13/27.
Mechanism for the change: A father of two boys is nearly twice as likely to have one who was born on a Tuesday, than a father of one. That is, 13/49 to 14/98=7/49.
Question #3: Mr. Smith, I hear you have two children. Tell me an interesting fact that applies to at least one.
Answer #3: Well, at least one is a left-handed, red-haired boy who was born on Tuesday and grew up to be  chiropractor.
Solution #3: No matter how many additional facts he adds, or leaves out, the probability that the other child is a boy remains 1/2. Consider the same four Mr. Smiths as in Question #1. One had to tell you about a boy, and it is irrelevant whether the additional facts apply to both. One had to tell you about a girl. Two had a choice, and only one would choose to tell you about the boy.


TITLE: What does $:=$ mean?
QUESTION [7 upvotes]: What does $:=$ mean? For example:

Consider the subset $ \mathbb{S} = \{ p \in \mathbb {P_4} (
> \mathbb{R,R} )  \ | \ P(2)=0 \} $   Suppose $p$, $q$ are in
$\mathbb{S}$, so $p(2)=q(2)=0$. Then $r := p + q$ is also a polynomial
of degree at most $4$ and $r(2) = p(2) + q(2)=0+0=0$

Is it just another notation for the $=$ sign? Or is there any significance on having a : in front of it?

REPLY [0 votes]: In the mathematics world, the $ := $ generally refers to any function, variable, or constant that is "defined to be equal to" whichever is on the right hand side.


text:
\section{Holomorphic Chern-Simons theory}
As we have seen,   when we work in the BV formalism, the space of classical fields is a (typically infinite dimensional) differential graded manifold equipped with a symplectic form of cohomological degree $-1$.  The action functional is a secondary object in this approach.  The differential on the space of fields preserves the symplectic form, and thus, at least locally, is given by Poisson bracket with some Hamiltonian function $S$, of cohomological degree zero. This function is the classical action.

In the paper \cite{AleKonSch95}, Alexandrov, Kontsevich, Schwartz and Zabronovsky introduced a beautiful and general method for constructing classical field theories in the BV formalism.  Many quantum field theories studied in mathematics arise from the AKSZ construction.  For example, Chern-Simons theory,  Rozansky-Witten theory and the Poisson $\sigma$ model all fit very naturally into this framework.

For us, the relevance of the AKSZ construction is that the classical field theory related to the Witten genus arises most naturally from the AKSZ construction.

Before I introduce the AKSZ construction, we need some notation.
\begin{definition}
A \emph{differential graded manifold} is a smooth manifold $X$ equipped with a sheaf $\Oo_X$ of differential graded commutative algebras over $\C$, with the property that $\Oo_X$ is locally isomorphic as a graded algebra to $\cinfty_X [[x_1,\ldots, x_n]]$, where $x_i$ are formal variables of cohomological degree $d_i \in \Z$.
\end{definition}
In this definition, $\cinfty_X$ refers to the sheaf of complex-valued smooth functions on $X$.  One can talk about geometric structures -- such as Poisson or symplectic structures -- on a differential graded manifolds.

If $X$ is a smooth manifold, we will let $X_{dR}$ denote the dg manifold whose underlying smooth manifold is $X$, and whose sheaf of functions is the complexified de Rham complex $\Omega^\ast_X$ of $X$. If $X$ is a complex manifold, we will let $X_{\dbar}$ denote the dg manifold whose underlying smooth manifold is $X$, and whose sheaf of functions is the Dolbeault complex $\Omega^{0,\ast}_X$.

\numberedparagraph
Now we can explain the AKSZ construction.  Suppose we have a compact differential graded manifold $M$, equipped with volume element of cohomological degree $k$. Let $X$ be a differential graded manifold with a symplectic form of cohomological degree $l$.  Then, the infinite-dimensional differential graded manifold $\op{Maps}(M,X)$ acquires a symplectic form of cohomological degree $l - k$.

If $f : M \to X$ is a map, then the tangent space to $\op{Maps}(M,X)$ at $f$ is
$$
T_f \op{Maps}(M,X) = \Gamma(M, f^\ast T X).
$$
We define a pairing on $T_f \op{Maps}(M,X)$ by the formula
$$
\ip{\alpha,\beta} = \int_M \ip{\alpha,\beta}_X.
$$
Since the integration map $\int_M : \cinfty(M) \to \R$ is of cohomological degree $-k$, and the symplectic pairing on $T X$ is of cohomological degree $m$, the pairing on $T_f \op{Maps}(M,X)$ is of cohomological degree $m - k$.  The case of interest in the Batalin-Vilkovisky formalism is when $m-k = -1$.

There is a variation of this construction which applies when the source manifold $M$ is non-compact.  In this situation, the space $\op{Maps}(M,X)$ has a natural integrable distribution given by the subspace
$$
\Gamma_c(M, f^\ast T X) \subset  T_f^c \op{Maps}(M,X) \subset T_f \op{Maps}(M,X)
$$
consisting of compactly supported tangent vector fields.    In this situation,  instead of having a symplectic pairing on $T_f \op{Maps}(M,X)$, we only have one on the distribution $T_f^c\op{Maps}(M,X)$.    The action functional, instead of being a closed one-form on $\op{Maps}(M,X)$, is a closed one-form on the leaves of the foliation.

\numberedparagraph

There are two broad classes of AKSZ theories which are commonly considered.  These are the theories of Chern-Simons type, and the theories of holomorphic Chern-Simons type.

The two classes of theories are distinguished by the nature of the source dg manifold.  In theories of Chern-Simons type, the source differential graded ringed space is $M_{dR}$, where $M$ is an oriented manifold.   The orientation on $M$ gives rise to a volume element on $M_{dR}$ of cohomological degree $\op{dim}(M)$.

The target manifolds for Chern-Simons theories of dimension $k$ are dg symplectic manifolds of dimension $k-1$.   For example, perturbative Chern-Simons theory arises when we take the target to be the dg manifold whose underlying manifold is a point, and whose algebra of functions is the algebra $C^\ast(\mf g)$ of cochains on a semi-simple Lie algebra $\mf g$.  The Killing form endows this dg manifold with a symplectic form of cohomological degree $2$.   This theory is perturbative, because maps $M_{dR} \to C^\ast(\g)$ are the same as connections on the trivial principal $G$ bundle which are infinitesimally close to the trivial connection.

Non-perturbative Chern-Simons theory arises from a genearlized form of the AKSZ construction whcih takes the stack $BG$ as the target manifold.  Vector bundles on $B G$ are the same as $G$-modules; the tangent bundle of $B G$ is the adjoint module $\g[1]$.   The Killing form on $\g$ is $G$-equivariant, and so gives rise to a symplectic form on $B G$ of cohomological degree $2$.

Rozansky-Witten theory also arises from this framework.  Let $X_{\dbar}$ be a holomorphic symplectic manifold.  Let us work over the base ring $\C[q,q^{-1}]$ where $q$ is a parameter of degree $-2$.  Then the symplectic form $q^{-1} \omega$ on $X_{\dbar}$ is of cohomological degree $2$, and so we can define a $3$-dimensional Chern-Simons type theory.    The fields of this theory are maps $M_{dR} \to X_{\dbar}$, where $M$ is a $3$-manifold, and everything takes place over the base ring $\C[q,q^{-1}]$.

Another example is the Poisson $\sigma$-model of \cite{Kon97, CatFel01}.  Here, the source is $\Sigma_{dR}$ where $\Sigma$ is a smooth surface. The target is the differential graded manifold $T^\ast[1] X$, whose underlying smooth manifold is $X$, and whose algebra of functions is $\Gamma(X, \wedge^\ast T X)$.    The Schouten-Nijenhuis bracket $\{-,-\}$ endows this dg manifold with a symplectic form of cohomological degree $1$.  The differential on $\Gamma(X,\wedge^\ast TX)$ is given by bracketing with the Poisson tensor $\pi$.

\numberedparagraph
Let us now discuss holomorphic Chern-Simons theory, which is the only quantum field theory we will be concerned with in this paper.     In holomorphic Chern-Simons theory, the source dg manifold is $M_{\dbar}$, where $M$ is a complex manifold equipped with a never-vanishing holomorphic volume element $\omega$ (thus, $M$ is a Calabi-Yau manifold).  This volume form can be thought of as a volume element on $M_{\dbar}$ of cohomological degree $\op{dim}_{\C}(M)$.    Integration against this volume element is simply the map
\begin{align*}
\Omega^{0,\op{dim}_{\C}(M)} (M) &\to \C \\
\alpha &\mapsto \int_M \omega \wedge \alpha.
\end{align*}

Theories of holomorphic Chern-Simons type on Calabi-Yau manifolds of complex dimension $k$ can thus be constructed from dg symplectic manifolds with a symplectic form of cohomological degree $k-1$.

In this paper, we are only interested in one-dimensional holomorphic Chern-Simons theories.  In these theories, the source dg manifold is $\Sigma_{\dbar}$, where $\Sigma$ is a Riemann surface equipped with a never-vanishing holomorphic volume form. The target is $X_{\dbar}$, where $X$ is a holomorphic symplectic manifold.  (The holomorphic symplectic form can be thought of as a dg symplectic form on $X_{\dbar}$).

\numberedparagraph
We can now give a more precise statement of the theorem relating elliptic cohomology and the Witten genus.

\begin{theorem}
Let $X$ be a complex manifold.  Then,
\begin{enumerate}
\item The obstruction to quantizing the holomorphic Chern-Simons theory whose fields are maps $\C_{\dbar} \to (T^\ast X)_{\dbar}$ is
$$\op{ch}_2 ( T X ) \in H^2(X, \Omega^2_{cl}(X))$$
where $\Omega^2_{cl}(X)$ is the sheaf of closed holomorphic $2$-forms on $X$.
\item If this obstruction vanishes (or, more precisely, is trivialized), then we can quantize holomorphic Chern-Simons theory to yield a factorization algebra on $\C$ with values in quasi-coherent sheaves on $X_{dR}\times \op{Spec} \C[\hbar]$.  We will call this factorization $D^{\hbar}_{X,ch}$.
\item If $E$ is an elliptic curve, then there is a quasi-isomorphism of $BD$ algebras in quasi-coherent sheaves on $X_{dR} \times \C[\hbar]$
$$
FH (  E,  D^{\hbar}_{X,ch } ) \simeq \left(  \Omega^{-\ast} ( T^\ast X ) [\hbar] , \hbar L_\pi + \hbar \{ \log \op{Wit}(X,E) - \} \right) .
$$
\end{enumerate}
\end{theorem}

\numberedparagraph
Recall that the factorization homology complex $FH ( E, D^{\hbar}_{X,ch} )$ is defined  by first constructing a factorization algebra $D^{E,\hbar}_{X,ch}$ on $E$, using the descent property of factorization algebras; and then taking global sections.

Quantum field theories in the sense of \cite{Cos10} have a descent property similar to that satisfied by factorization algebras, and the construction of a factorization algebra from a quantum field theory is compatible with descent.    The quantum field theory on an elliptic curve $E$ which arises by descent from holomorphic Chern-Simons theory on $\C$ is simply holomorphic Chern-Simons theory on $E$.

Thus, one can interpret the factorization homology group $FH ( E, D^{\hbar}_{X,ch} )$ in terms of holomorphic Chern-Simons theory on the elliptic curve $E$.   From this point of view, $FH( E, D^{\hbar} _{X,ch} ) $ is the cochain complex of global observables for the holomorphic Chern-Simons theory of maps $E \to T^\ast X$.

This theorem is proved using the Wilsonian approach to quantum field theory developed in \cite{Cos10}.  The result is then translated into the language of factorization algebras.    The proof appears in \cite{Cos10b}.

\numberedparagraph
As I mentioned earlier, this factorization algebra is an analytic avatar of the chiral algebra of chiral differential operators constructed by Gorbounov, Malikov and Schechtman \cite{GorMalSch00}.  A detailed description of the factorization algebra constructed from holomorphic Chern-Simons theory, and its relationship to chiral differential operators, will appear elsewhere.

In the physics literature, Kapustin\cite{Kap05} and Witten \cite{Wit05} have argued that chiral differential operators arise from what is called, in section 3.3 of \cite{Wit05}, a non-linear $\beta$-$\gamma$ system.     This non-linear $\beta$-$\gamma$ system is the same as what I have called holomorphic Chern-Simons theory.

\begin{comment}

\numberedparagraph

One can also ask how this construction relates to the original suggestion of Witten \cite{Wit87, Wit88}. Witten argued that the Witten genus arises as the partition function of a certain supersymmetric $\sigma$-model.   In this last section I will explain briefly how the  the $N = 1$ supersymmetric $\sigma$ model, when rewritten in th

Let us first consider the ordinary $\sigma$-model of maps $f : \Sigma \to X$, where $X$ is a Kahler manifold.  The action of this theory is
$$
S(f) = \int_{\Sigma}\ip{\d f, \d f} .
$$
One can argue \cite{} that the infinite-volume limit of this theory is decouples as a tensor product of the holomorphic and anti-holomorphic Chern-Simons theory.  It seems that introducing the

\end{comment}


TITLE: How to find the expectation of joint non-linear SDEs?
QUESTION [0 upvotes]: I'm working with the following pair of joint stochastic differential equations
\begin{align}
X_t &= (\alpha X_t - Y_t)dt + \sigma\left(X_t - \frac{Y_t}{a}\right)dW_t^{(1)} \\
Y_t &= (b(X_t)X_t + \beta Y_t)dt + \nu \sqrt{Y_t\left(X_t - \frac{Y_t}{a}\right)}dW_t^{(2)} \\
\end{align}
In the case that $b(X_t) \equiv b \in \mathbb{R}$ is a constant, we can explicitly recover the expectations of both $X$ and $Y$ by solving a system of ordinary differential equations. My question is this, if the function $b(X_t)$ is non-constant (and hence the term $b(X_t)X_t$ becomes non-linear), are there any other situations in which we can still find the expectations of the system?

REPLY [0 votes]: Setting $f_t:=\mathbb{E}\left[X_t\right],g_t:=\mathbb{E}\left[Y_t\right],h_t:=\mathbb{E}\left[b(X_t)X_t\right]$, we still have that
$$\dot{f}_t=af_t-g_t$$
such that
$$f_t=-e^{at}\int_0^te^{-as}g(s)ds.$$ Similarly,
$$g_t=e^{\beta t}\int_0^te^{-\beta s}h(s)ds$$ so it depends what $h$ is. If it is $x\mapsto \frac{1}{x}$, then this would be an example for a case, where one obtains an explicit solution.


text:
\def\gps{s}

\def\gps{s}
\def\GK{\cc{G}}
\def\Excgr{\diamondsuit}

\def\dimh{m}
\def\LCS{C}
\def\Bc{B_{0}}
\def\AF{A}
\def\CF{C}
\def\Ab{A_{\rdb}}
\def\Am{A_{\rdm}}

\def\DPrp{\DPr_{\dimh}}
\def\DPrb{\DPr_{\rdb}}
\def\DPrm{\DPr_{\rdm}}
\def\Cb{\cc{C}_{\rdb}}
\def\Cm{\cc{C}_{\rdm}}
\def\Ub{\cc{U}_{\rdb}}
\def\xivrb{\breve{\xiv}_{\rd}}
\def\VPrb{\breve{\VP}_{\rdb}}
\def\Larb{\breve{\La}_{\rdb}}
\def\Lar{\breve{\La}}
\def\Larm{\breve{\La}_{\rdm}}
\def\VH{Q}
\def\VHc{\VH_{0}}
\def\kappavrm{\kappavr_{\rdm}}

\def \Xv{\bb{X}}

\def\fvh{\bb{\dimh}}
\def\N{\mathbb{N}}
\def\Z{\mathbb{Z}}

\def\iic{\IF}
\def\iif{\breve{\iic}}
\def\DP{\text{D}}
\def\HH{\text{H}}
\def\A{\text{A}}

\def\ifc{\breve{\iic}}

\def\HF{\mathcal H}

\def\Thetathetav#1{\substack{\\[0.1pt] \upsilonv\in\Theta \\[1pt] \Proj \upsilonv = #1}}
\def\Span{\operatorname{span}}
\def\Exc{{\square}}
\def\UUs{U_{\circ}}
\def\errbm{\errb^{*}}
\def\corrDF{\rho}
\def\BBr{\breve{\BB}}
\def\taua{\tau}
\def\AssId{\mathcal{I}}
\def\AFD{\cc{A}}

\def\BanX{\cc{X}}
\def\basX{\ev}
\def\apprX{\alpha}
\def\fvs{\fv^{*}}
\def\lkh{\ell}
\def\Bc{B_{0}}
\def\h{\frac{1}{2}}
\def\basis{\ev}
\def\Proj{\Pi_{0}}
\def\Projh{\Pi_{\dimtotal}}

\def\Ij{\mathcal{I}}

\def\Mn{M_{\nsize}}
\def\bA{\breve{A}}
\def\cA{\bA_{\dimh}}

\def\Sdr{\cc{S}}
\def\xxn{\xx_{\nsize}}

\def\CONST{\mathtt{C}}
\def\Ij{\mathcal{I}}
\def\etas{\eta^{*}}
\def\kappavs{\kappav^{*}}
\def\kappavc{\kappav'}

\def\kappav{\boldsymbol{\kappa}}
\def\kappavs{\kappav^{*}}

\def\omegav{\bb{\phi}}
\def\omegavs{\omegav^{*}}
\def\omegavc{\omegav'}

\def\dimn{\dimp_{\nsize}}
\def\betan{\beta_{\nsize}}

\def\bA{\breve{A}}
\def\cA{\bA_{\dimh}}

\def\corrDF{\rho}
\def\upsilonv{\boldsymbol{\upsilon}}
\def\upsilonvs{\boldsymbol{\upsilon}^*}
\def\upsilonvd{\boldsymbol{\upsilon}^\circ}

\section{Introduction}
This paper presents a way to control the bias in a sieve profile contrast estimation problem, which we elaborate for simplicity for parameters in \(l^2\eqdef \{(x_1,x_2,\ldots)\subset \R:\, \sum_{k=1}^\infty x^2<\infty\}\). More precisely consider a contrast functional \(\E\LL:\R^{\dimp}\times l^2\to \R\). Assume that the goal is to calculate the target parameter \( \thetavs \) defined as
\begin{EQA}[c]
    \thetavs
    \eqdef
    \argmax_{\thetav} \sup_{\etav: (\thetav,\etav) \in \Ups \subset l^2} \E \LL(\thetav,\etav) =\Pi_{\thetav}\upsilonvs\eqdef\Pi_{\thetav} \argmax_{\upsilonv=(\thetav,\etav) \in \Ups\subset  l^2} \E \LL(\upsilonv),
\label{thetavsBanX}
\end{EQA}
with a set \(\Ups\subset \R^{\dimp}\times l^2\).
To circumvent the problem of maximizing over an infinite dimensional set \(\Ups\subset\R^{\dimp}\times  l^2\) define for some \(\dimh\in\N\) the following approximation contrast functional
\begin{EQA}
\E\LL_{\dimh}:\R^{\dimp+\dimh}&\to& \R,\\
(\thetav,\etav)&\mapsto& \E\LL(\thetav,E_{l^2}\etav),
\end{EQA}
where \(E_{l^2}: \R^{\dimh} \to l^2\) is the natural embedding operator. Further define the biased target
\begin{EQA}[c]
    \thetavs_{\dimh}
    \eqdef
    \argmax_{\thetav} \sup_{\etav \in \R^{\dimh}} \E \LL_{\dimh}(\thetav,\etav)
    =\Pi_{\thetav}\upsilonvs_{\dimh} \eqdef\Pi_{\thetav} \argmax_{\upsilonv=(\thetav,\etav) \in \R^{\dimtotal}} \E \LL_{\dimh}(\thetav,\etav).
\label{thetavsBanX}
\end{EQA}
We are interested in a bound for the euclidean distance \(\|\thetavs-\thetavs_{\dimh}\|\).
Further consider

the following block representations of the hessian operator
\( \DF^{2}(\upsilonv) = - \nabla^{2} \E \LL(\thetav,\fv) \):
\begin{EQA}
    \label{eq: introduction of operator representations upsilonv}
    \DF^{2}&=&
     \left(
      \begin{array}{ccc}
        \DF_{\dimh}^2 & \AF_{\dimh} \\
        \AF_{\dimh}^\T & \HF^2_{\dimh}
      \end{array}
    \right)\in L(\R^{\dimtotal}\times l^2,\R^{\dimtotal}\times l^2),
\label{DFc012se}
\end{EQA}
where for a vectorspace \(V\) the symbol \(L(V,V)\) denotes the set of linear operators from \(V\)  to \(V\), \(\dimtotal=\dimp+\dimh\in\N\) and where
\begin{EQA}[c]
\DF^2_{\dimh}(\upsilonv)\eqdef\nabla_{\dimh}^2\E\LL(\upsilonv)\in\R^{\dimtotal\times\dimtotal},
\end{EQA}
i.e. the derivatives of \(\E\LL\) are only taken with respect to the first \(\dimp+\dimh\in\N\) coordinates of \(\upsilonv=(\thetav,\etav)\in \R^{\dimp}\times l^2\)

Define the following two matrices
\begin{EQA}
\DPr^{2}_{\dimh}(\upsilonv)\eqdef\left(\Pi_{\thetav}\DF_{\dimh}^{-2}(\upsilonv)\Pi_{\thetav}^\T\right)^{-1}\in\R^{\dimp\times\dimp},
&\text{
and }&
\DPr^{2}(\upsilonv)\eqdef \left(\Pi_{\thetav}\DF^{-2}(\upsilonv)\Pi_{\thetav}^\T\right)^{-1}\in\R^{\dimp\times\dimp}.
\end{EQA}

The second result we want to derive is a bound for the difference between \(\DPrp(\upsilonvs_{\dimh})\in\R^{\dimp\times\dimp}\) and \(\DPr(\upsilonvs)\in\R^{\dimp\times\dimp}\) in spectral norm, where
\begin{EQA}[c]
\upsilonvs_{\dimh}\eqdef \argmax_{\upsilonv\in\R^{\dimp+\dimh}}\E\LL.
\end{EQA}

This kind of problem arises for instance when a sieve profile estimator is analyzed as in \cite{AASP2013}. Given a (random) contrast functional \(\LL:\R^{\dimp}\times l^2\to \R\) one defines \(\LL_{\dimh}\) analogously to \(\E\LL_{\dimh}\) above and the sieve profile estimator
\begin{EQA}[c]
\label{ttdhttdhs}
\tilde \thetav_{\dimh}\eqdef \argmax_{\thetav\in\R^{\dimp}}\max_{\etav\in\R^{\dimh}}\LL_{\dimh}(\thetav,\etav)\eqdef \Pi_{\thetav}\upsilonvs_{\dimh}\eqdef \argmax_{\upsilonv\in\R^{\dimp+\dimh}}\E\LL.
\end{EQA}
The parametric results obtained in \cite{AASP2013} claim that the profile estimator
\( \tilde{\thetav}_{\dimh} \) estimates well \( \thetavs_{\dimh} \) if the spread \(\Excgr(\rr,\xx)>0\) is small. More precisely we have for fixed \(\xx \) with Theorem 2.1 of \cite{AASP2013} applied to
\( \tilde{\thetav}_{\dimh} \) from \eqref{ttdhttdhs} that with probability greater \(1-22.8 \ex^{-\xx}\)
\begin{EQA}[c]
\label{eq: convergence of estimator for dimh}
    \|\DPrp \bigl( \tilde{\thetav}_{\dimh} - \thetavs_{\dimh} \bigr)
    - \xivr_{\dimh}(\upsilonvs_{\dimh})\|\le \Excgr\big(\rr,\xx\big),
\label{DPrctthsh}
\end{EQA}
where \( \upsilonvs_{\dimh} = (\thetavs_{\dimh},\etavs_{\dimh})
= \argmax_{\upsilonv} \E \LL_{\dimh}(\upsilonv) \).

This result involves exactly the two kinds of bias from above, i.e. one that concerns the difference \(\thetavs_{\dimh} - \thetavs\) and the other the difference between \(\DPrp\in\R^{\dimp\times\dimp}\) and \(\DPr\in\R^{\dimp\times\dimp}\).

\cite{AASP2013} use two assumptions to address this bias. The first one reads:

\begin{description}
\item[\(\bb{(bias)}\)] There exists a  decreasing function \(\alpha:\N\to \R_+\) such that
\begin{EQA}[c]
\|\DPrp(\thetavs_{\dimh} - \thetavs) \|\le \hat\alpha(\dimh).
\end{EQA}
\end{description}

The second one:

\begin{description}
\item[\(\bb{(bias')}\)] As \(\dimh\to \infty\)
\begin{EQA}
\| I-\DPrp(\upsilonvs)^{-1}\DPr(\upsilonvs)^2\DPrp(\upsilonvs)^{-1}\|&=& o(1),\\
\| I-\DPrp(\upsilonvs_{\dimh})^{-1}\DPrp(\upsilonvs)^2\DPrp(\upsilonvs_{\dimh})^{-1}\|&=& o(1).

 \end{EQA}
\end{description}
In this paper we want to present a particular way to obtain such a function \(\hat\alpha(\dimh)\) and to derive \(\bb{(bias')}\), which relies on less high level conditions on the smoothness and structure of \(\E\LL\).

\section{Main result}
Denote by \(\Pi_{\dimtotal}:l^2\to \R^{\dimtotal}\) the projection to the first \(\dimtotal\in\N\) coordinates of an element of \(l^2\).
To bound the bias \(\|\DPrp(\thetavs_{\dimh} - \thetavs) \|>0\) we present the following condition:

\begin{description}
  \item[\( \bb{(\kappav)} \)]
  The vector \( \kappavs\eqdef (Id_{l^2}-\Pi_{\dimtotal})\upsilonvs\) satisfies \(\|\HF_{\dimh}\kappavs\|^2\le \CONST_{\kappavs}\dimh\) for some \(\CONST_{\kappavs}>0\) and with \( \alpha(\dimh)\to 0 \)
\begin{EQA}
    \|\DF_{\dimh}^{-1} \AF_{\dimh} \kappavs\|
    &\le&
    \alpha(\dimh).
\label{apprXdimhledimh}
\end{EQA}
Further for any \(\lambda\in[0,1]\) with some \(\tau(\dimh)\to 0\)
\begin{EQA}
\label{eq: condition on smoothness of DF}
    \|\DF_{\dimh}^{-1} \left(\nabla_{\upsilonv\kappav}\E\LL(\upsilonvs,\lambda\kappavs) -\AF_{\dimh}\right) \kappavs\|
    &\le&
    \tau(\dimh),\\
\left|{\kappavs}^\T(\DF_{\kappav\kappav}-\nabla_{\kappav\kappav}\E\LL(\upsilonvs,\lambda\kappavs))\kappavs\right| &\le& \CONST_{\kappavs}^2\dimh.
\label{eq: condition on smoothness of DF-zeta}
\label{eq: new condition for bias}
\end{EQA}
\end{description}

To ensure that \( \DPr_{\dimh} \) is close to \( \DPr \) we impose the following second condition.

\begin{description}
    \item[\( \bb{(\upsilonv\kappav)} \)]
Assume that with some \(\beta(\dimh)\to 0\)
\begin{EQA}[c]
\label{eq: convergence of breve matrix}
\|\HF^{-1}\AF^{\T}\DF_{\dimh}^{-1}\|\le \beta(\dimh).
\end{EQA}
\end{description}

\begin{theorem}
\label{theo: bias cond}
Let the conditions \(\bb{ (\kappav)} \) and \( \bb{(\upsilonv\kappav)} \) be
fulfilled. Further let the condition \( \bb{(\cc{L}{\rr}_{\infty})}\) from Section \ref{sec: conditions} be satisfied for \( \E\LL: l^2\to \R \) with \(\gmi(\rr)\equiv \gmi>0\). Set \( {\rr^*}^{2}= 4\CONST_{\kappavs}^2\dimh/\gmi\) and let for some \(\dimh_0\in\N\) and all \(\dimh\ge \dimh_0\) the condition \( \bb{(\LL_{0})} \) be fulfilled for \(\DFc=\DF_{\dimh}\) and for any \(\rr\le \rr^*\). Then \(\bb{(bias)}\) and \(\bb{(bias')}\) are satisfied with
 \begin{EQA}
   \hat\alpha(\dimh)&=&  \sqrt{\frac{1+\corrDF^2}{1-\corrDF^2}}\Bigg(\alpha(\dimh) +\tau(\dimh)+2\delta(2\rr^*)\rr^*\Bigg).

\end{EQA}
\end{theorem}

\section{Application to single-index model}
\label{sec: application to single index model}
We present an example to illustrate how these results can be derived for single-index modeling.
Consider the following model
\begin{EQA}[c]
\label{eq: single index model introduced}
    \Yv_{i}
    =
    \fs(\Xv_{i}^{\T} \thetavs) + \varepsilon_{i},
    \qquad
    i=1,...,\nsize,
\end{EQA}
or some \(\fs:\R\to \R\) and \(\thetavs\in S_{1}^{\dimp,+}\subset\R^{\dimp}\), i.i.d errors \(\varepsilon_i\in\R\) and \(\Var(\varepsilon_i)=\sigma^2\) and i.i.d random variables \(\Xv_{i}\in \R^{\dimp}\) with distribution denoted by \(\P^{\Xv}\).

To ensure identifyability of \(\thetavs\in\R^{\dimp}\) we assume that it lies in the half sphere \(S_{1}^{\dimp,+}:=\{\thetav\in\R^{\dimp}:\, \|\thetav\|=1,\, \theta_1> 0\}\subset\R^{\dimp}\). We assume that the support of the \(\Xv_{i}\in \R^{\dimp}\) is contained in the ball of radius \(s_{\Xv}>0\).
Further we assume that \(\fs\in \{f:[-s_{\Xv},s_{\Xv}]\mapsto \R\} \) can be well approximated by a orthonormal \(C^2\)-Daubechies-wavelet basis, i.e. for a suitable function \(\basX_{0}:=\psi:[-s_{\Xv},s_{\Xv}]\mapsto \R\) we set for \(k=2^{r_k}+j_k\)
\begin{EQA}[c]
\basX_{k}(t)=2^{k/2}\psi\left(2^{k}(t-2j_ks_{\Xv})\right),\, k\in\N.
\end{EQA}

Our aim is to analyze the properties of the profile MLE
\begin{EQA}[c]
    \tilde{\thetav}
    \eqdef
    \argmax_{\thetav} \max_{\etav \in \R^{\dimh}} \LL(\thetav,\etav) ,
\label{ttSI}
\end{EQA}
where
\begin{EQA}[c]
    \LL(\thetav,\etav)
    =
    - \frac{1}{2}
    \sum_{i=1}^{\nsize} | \Yv_{i} - \sum_{k=0}^{m} \etav_{k} \basX_{k}(\Xv_{i}^{\T} \thetav) |^{2}.
\end{EQA}

Consider the following assumptions.

\begin{description}
 \item[\((\mathbf{Cond}_{\Xv})\)]
  The measure \(\P^{\Xv}\) is absolutely continuous with respect to the Lebesgue measure. The Lebeque density \(d_{\Xv}\) of \(\P^{\Xv}\) is only positive on the ball \(B_{s_{\xv}}(0)\subset \R^{\dimp}\) and Lipshitz continuous with Lipshitz constant \(L_{d_{\Xv}}>0\).

\end{description}

Of course we need some regularity of the link function \(\fs\in \{f:[-s_{\Xv},s_{\Xv}]\mapsto \R\} \):
\begin{description}

  \item[\( (\mathbf{Cond}_{\fvs}) \)]
   For some \(\fvs\in\R^{\N}\)
   \begin{EQA}[c]
   \label{eq: expansion of indexfunciton}
    \fs=\fs_{\fvs}=\sum_{k=1}^\infty f^*_k\basX_{k},
   \end{EQA}
   where with some \(\alpha>2\) and a constant \(C_{\|\fvs\|}>0\)
   \begin{EQA}[c]\label{eq: smoothness of fs}
    \sum_{l=0}^\infty l^{2\alpha}{f^*_{l}}^2 \le C_{\|\fvs\|}^2< \infty.
   \end{EQA}
\end{description}

\begin{lemma}
\label{lem: conditions theta eta}
Assume \( (\mathbf{Cond}_{\fvs}) \) and \((\mathbf{Cond}_{\Xv})\). Using our orthogonal and sufficiently smooth wavelet basis we get for any \(\lambda\in[0,1]\)
\begin{EQA}
 \|\DF_{\kappav\kappav}^{1/2}\kappavs\|^2<\CONST_1 n\dimh^{-2\alpha}, &&
\alpha(\dimh)\le \CONST_2\dimh^{-\alpha-1/2}\sqrt n,\\
\beta(\dimh)\le \CONST_3 \dimh^{-1/2},&&
 \tau(\dimh)\le \CONST_2\dimh^{-2\alpha+1}\sqrt{n},
\end{EQA}
and \(\left|{\kappavs}^\T(\DF_{\kappav\kappav}-\nabla_{\kappav\kappav}\E\LL(\upsilonvs,\lambda\kappavs))\kappavs\right|=0\).
\end{lemma}

For details see \cite{Andresensingleindex}.

\appendix
\section{Appendix}

\subsection{The conditions}
\label{sec: conditions}
We adopt the conditions from Section~3 of \cite{SP2011} with some
minor changes. First we present the parametric conditions that apply to parametric models with finite dimensional parameter. Then explain two new conditions that arise in the infinite dimensional setting.

For some finite dimension \(\dimtotal\in\N\) the parametric conditions involve a matrix \( \DF_{0}^{2} \) and a central point \(\upsilonvd\in\R^{\dimtotal}\) that have to be specified before the conditions can be checked.

\begin{remark}
For Theorem \ref{theo: bias cond} the matrix equals
\begin{EQA}[c]
\DFc^{2}
    =
    - \nabla_{\dimh}^{2} \E \LL(\upsilonvs),
\end{EQA}
and \(\upsilonvd=\upsilonvs_{\dimh}\), i.e. the central point does not coincide with the element that defines the matrix \(\DFc^{2}\). It is important to note that condition \( \bb{(\LL_{0})} \) thus becomes another constraint on the bias.
\end{remark}

The matrix \( \DFc^{2} \) has to satisfy certain regularity conditions.
We begin by representing the information matrix in block form:
\begin{EQA}
    \DF_{0}^2
    &=&
    \left(
      \begin{array}{cc}
        \DP_{0}^{2} & A_{0} \\
        A_{0}^{\T} & \HH_{0}^{2} \\
      \end{array}
    \right).
\label{DFcseg0}
\end{EQA}
Here we restate \emph{identifiability conditions}:

\begin{description}
  \item[\( (\bb{\AssId}) \)]
   It holds
\begin{EQA}[c]
    \| \HH_{0}^{-1} A_{0}^{\T} \DP_{0}^{-1} \|^2_{\infty}
    =:\corrDF< 1 .
\label{regularity2}
\end{EQA}
\end{description}

Using the matrix \(\DFc\in\R^{\dimtotal\times\dimtotal}\) and the central point \(\upsilonvd\in\R^{\dimtotal}\) we define the local set \(\Theta(\rr)\) with some
\(\rr\ge 0\)
\begin{EQA}[c]
\Theta(\rr)\eqdef \{\upsilonv=(\thetav,\etav)\in\Theta,\, \|\DFc(\upsilonv-\upsilonvd)\|\le \rr\}.
\end{EQA}
The local conditions only describe the properties of the process \( \LL(\upsilonv) \)
for \( \upsilonv \in \Theta(\rr) \) with some fixed value \( \rr>0 \). The global
conditions have to be fulfilled on the whole \( \Theta \).
We start with the local conditions.

\begin{description}
    \item[\( \bb{(\LL_{0})} \)]
    \textit{

    For each \( \rr \le \rups \),
    there is a constant \( \rddelta(\rr) \) such that
    it holds on the set \( \Upsilon(\rr) \) and with spectral norm \(\|\cdot\|\):
    }
\begin{EQA}[c]
\label{LmgfquadELGP}
    \bigl\|
       \DF_{0}^{-1} \nabla^{2}\E\LL(\upsilonv) \DF_{0}^{-1} - \Id_{\dimtotal}
    \bigr\|
    \le
    \rddelta(\rr).
\end{EQA}

\end{description}

We also need:
\begin{description}
  \item[\(\bb{(\cc{L}{\rr}_{\infty})}\)]   For any \( \rr > \rups\) there exists a value \( \gmi(\rr) > 0 \),
     such that
\begin{EQA}[c]
    \frac{-\E \LL(\upsilonv,\upsilonvs)}{\|\DF(\upsilonv-\upsilonvs)\|^{2}}
    \ge
    \gmi\left(\|\DF(\upsilonv-\upsilonvs)\|\right).
\end{EQA}
\end{description}

\subsection{Proof of Theorem~\ref{theo: bias cond}}

\begin{lemma}
\label{lem: apriori bias in sieve approach}
Assume that \( \bb{(\cc{L}{\rr}_{\infty})} \) is satisfied with \( \gmi(\rr) \equiv \gmi \) and that the condition \( \bb{(\kappav)} \) is satisfied. Then we get \(\|\DF(\upsilonvs_{\dimh}-\upsilonvs)\|\le \rr^*\) where \( {\rr^*}^{2}= 4\CONST_{\kappavs}\dimh/ \gmi \).
\end{lemma}

\begin{proof}
Note that
\begin{EQA}[c]
 \|\DF(\upsilonvs-\Pi_{\dimtotal}\upsilonvs)\|=\|\HF_{\dimh} \kappavs\|,
\end{EQA}
such that \(\upsilonvs\in\Theta(\rr^*)\). Further we have \(\nabla\E\LL(\upsilonvs)=0\) such that by the Taylor expansion with some \(\lambda\in[0,1]\)
\begin{EQA}
\E\LL(\Pi_{\dimtotal}\upsilonvs,\upsilonvs)&=&\|\HF_{\dimh} \kappavs\|^2+{\kappavs}^\T(\HF_{\dimh}-\nabla_{\kappa\kappa}\E\LL(\upsilonvs,\lambda\kappavs))\kappavs.
\end{EQA}
which gives with \eqref{eq: condition on smoothness of DF-zeta} and \(\bb{(\kappav)}\) on \(\Theta(\rr^*)\) that
\begin{EQA}
\label{eq: bound for logratio projected oracle}
|\E\LL(\Pi_{\dimtotal}\upsilonvs,\upsilonvs)|&\le& \|\DF(\upsilonvs-\upsilonvs)\|^2 +\CONST_{\kappavs}\dimh\le 2\CONST_{\kappavs}\dimh.
\end{EQA}
Now we show that \( \upsilonvs_{\dimh} \) also belongs to \( \Theta(\rr^*) \) for \( {\rr^*}^{2} \ge 4\CONST_{\kappavs}\dimh/\gmi \). Suppose for the moment that
\( \bigl\| \DF (\upsilonvs_{\dimh} - \upsilonvs) \bigr\| > \rr^* \).
By \( \bb{(\cc{L}{\rr}_{\infty})} \), it holds
\begin{EQA}[c]
    2 \bigl| \E \LL(\upsilonvs_{\dimh},\upsilonvs) \bigr|
    \ge
    \gmi \bigl\| \DF (\upsilonvs_{\dimh} - \upsilonvs) \bigr\|^{2}
    >
    \gmi {\rr^*}^{2} .
\label{2ELL1s2}
\end{EQA}
This contradicts \( \bigl| \E \LL(\upsilonvs_{\dimh},\upsilonvs) \bigr|
\le \bigl| \E \LL(\Pi_{\dimtotal}\upsilonvs,\upsilonvs) \bigr| \)
in view of \( {\rr^*}^{2} \ge 4\CONST_{\kappavs}\dimh/ \gmi  \) and \eqref{eq: bound for logratio projected oracle}, so \( \upsilonvs_{\dimh} \in \Theta(\rr^*) \).
\end{proof}

\begin{lemma}
\label{lem: bias in sieve approach}
Assume that \( \bb{(\cc{L}{\rr}_{\infty})} \) is satisfied with \( \gmi(\rr) \equiv \gmi \). Further assume \( \bb{(\kappav)} \) and \( \bb{(\cc{L}_{0})} \) with central point \(\upsilonvs_{\dimh}\in\R^{\dimtotal}\) and operator \(\DF_{\dimh}\).

Then we get with \( {\rr^*}^{2}= 4\CONST_{\kappavs}\dimh/ \gmi\)
\begin{EQA}[c]
\label{eq: bounds for bias}
 \|\breve\DP_{\dimh}(\thetavs_{\dimh}-\thetavs)\|\vee\|\DF_{\dimh}(\upsilonvs_{\dimh}-\upsilonvs)\|\le\sqrt{\frac{1+\corrDF^2}{1-\corrDF^2}} \Big(\alpha(\dimh) +\tau(\dimh)+2\delta(2\rr^*)\rr^*\Big).
\end{EQA}

\end{lemma}

\begin{proof}
Using condition \( \bb{(\cc{L}_{0})} \) and Taylor expansion we have on \({\Theta }_{\dimh}(\rr)=\{\|\DF_{\dimh}(\upsilonv-\upsilonvs_{\dimh})\|\le\rr\}\subset\R^{\dimp+\dimh}\)
\begin{EQA}
\label{eq: bound for expected value of approx error}
&&\nquad\sup_{\upsilonv\in\Theta(\rr)}\|\DF_{\dimh}^{-1}\nabla_{\dimh} \E\LL_{\dimh}(\upsilonv) - \DF_{\dimh}^{-1}\nabla_{\dimh} \E\LL_{\dimh}(\upsilonvs)-\DF_{\dimh} \, (\upsilonv - \upsilonvs)\|\\
  &\le& \sup_{\upsilonv\in\Theta(\rr)}\|\DF_{\dimh}^{-1}\nabla_{\dimh}^2 \E\LL_{\dimh}(\upsilonv)^2\DF_{\dimh}^{-1}-I_{\dimtotal}\|_{\DF\Upss^-(\rr),\DF\Upss^-(\rr)}\rr\\
  &\le& \delta(\rr)\rr.
\end{EQA}
Because of Lemma \ref{lem: apriori bias in sieve approach} we know that
\begin{EQA}
\|\DF_{\dimh}(\Pi_{\dimtotal}\upsilonvs-\upsilonvs_{\dimh})\|&=&\|\DF E_{l^2}(\Pi_{\dimtotal}\upsilonvs-\upsilonvs_{\dimh})\|\\
	&\le&\|\DF(\Pi_{\dimtotal}\upsilonvs-\omegavs)\|+\|\DF(\upsilonvs-E_{l^2} \upsilonvs_{\dimh})\|
	\le 2\rr^*,
\end{EQA}
such that \(\Pi_{\dimtotal}\upsilonvs_{\dimh}\in{\Theta}_{0,\dimh}(2\rr^*)\), which gives
\begin{EQA}[c]
\big\|\DF_{\dimh}\, (\upsilonvs_{\dimh} -\Pi_{\dimtotal}\upsilonvs)- \DF_{\dimh}^{-1} \nabla_{\dimp+\dimh} \E\Big(\LL(\Pi_{\dimtotal}\upsilonvs)-\LL(\upsilonvs_{\dimh})\Big)
	\big\|\le 2\delta(2\rr^*)\rr^*,
\end{EQA}
from which we derive with the triangle inequality
\begin{EQA}
\big\|\DF_{\dimh}\, (\upsilonvs_{\dimh} - \upsilonvs)\big\|
	 &\le& 2\delta(2\rr^*)\rr^*+\Big\|\DF_{\dimh}^{-1} \nabla_{\dimp+\dimh} \E\big(\LL(\Pi_{\dimtotal}\upsilonvs)-\LL(\upsilonvs_{\dimh})\big)\Big\|.
\end{EQA}
Because \(\nabla_{\dimp+\dimh} \E\LL(\upsilonvs_{\dimh})=0\) and \(\nabla \E\LL(\upsilonvs)=0\) we find
\begin{EQA}
&&\nquad\Big\|\DF_{\dimh}^{-1} \nabla_{\dimp+\dimh} \E\big(\LL(\Pi_{\dimtotal}\upsilonvs)-\LL(\upsilonvs_{\dimh})\big)\Big\|=\Big\|\DF_{\dimh}^{-1} \Pi_{\dimp+\dimh} \nabla\E\big(\LL(\Pi_{\dimtotal}\upsilonvs)-\LL(\upsilonvs)\big)\Big\|
\end{EQA}
Using that \(\|\DF(\Pi\upsilonvs-\upsilonvs)\|\le\rr^*\) and condition \(\bb{(\kappav)} \) we may infer by the Taylor expansion that with some \(\lambda\in[0,1]\)
\begin{EQA}
&&\nquad\Big\|\DF_{\dimh}^{-1} \Pi_{\dimp+\dimh} \nabla\E\big(\LL(\Pi_{\dimtotal}\upsilonvs)-\LL(\upsilonvs)\big)\Big\|\\
	&\le&\Big\|\DF_{\dimh}^{-1}\AF_{\dimh}\big(E_{l^2}\Pi_{\dimtotal}\upsilonvs-\upsilonvs\big)\Big\|+\Big\|\DF_{\dimh}^{-1} (\nabla_{\upsilonv\kappav}\E[\LL(\Pi_{\dimtotal}\upsilonvs))]- \AF_{\dimh}) \kappavs\Big\| \\
	&=& \Big\|\DF_{\dimh}^{-1} \AF_{\dimh}\kappavs\Big\|+\Big\|\DF_{\dimh}^{-1} \left(\nabla_{\upsilonv\kappav}\E[\LL\big((\Pi_{\dimtotal}\upsilonvs,\lambda\kappavs)\big)] -\AF_{\dimh}\right) \kappavs\Big\|.
\end{EQA}
Due to assumption  \( \bb{(\kappav)} \) the last sum is bounded by \(\alpha(\dimh)+\tau(\dimh) \). Together this gives that
\begin{EQA}[c]
\big\|\DF_{\dimh}\, (\upsilonvs_{\dimh} - \upsilonvs)\big\|=\alpha(\dimh) + \tau(\dimh)+2\delta(2\rr^*)\rr^*.
\end{EQA}
Finally we can represent
\begin{EQA}
\DF_{\dimh}^2= \left(
      \begin{array}{ccc}
        \DP^2 & \A\\
        \A^\T & \HH^2
      \end{array}
    \right), && \breve\DP_{\dimh}^2=\DP^2-\A^\T\HH^{-2}\A.
\end{EQA}
and due to \( (\bb{\AssId}) \) this gives
\begin{EQA}
 \|\breve\DP_{\dimh}(\thetavs_{\dimh}-\thetavs)\|^2&\le&\frac{1+\corrDF}{1-\corrDF}\|\DF_{\dimh}(\upsilonvs_{\dimh} - \upsilonvs)\|^2.
\end{EQA}

\begin{comment}
It remains to estimate
\begin{EQA}[c]
\Big\|\DF_{\dimh}^{-1} \Pi_{\dimp+\dimh}\DF\Big\|^2\eqdef \sup_{\mathbf w\in\R^{\dimtotal}}\frac{\langle \mathbf w,\DF_{\dimh}^{-1} \Pi_{\dimp+\dimh}\DF^2 \Pi_{\dimp+\dimh}^\T\DF_{\dimh}^{-1}\mathbf w \rangle }{\|\mathbf w\|^2} =1,
\end{EQA}
which completes the proof.
\end{comment}
\end{proof}

\begin{lemma}
 \label{lem: convergence of breve matrix}
Assume \( \bb{(\upsilonv\kappav)} \) then
\begin{EQA}[c]
\|I-\DPr_{\dimh}^{-1} \DPr^2\DPr_{\dimh}^{-1} \|\le\frac{1+\corrDF^2+\beta^2(\dimh)}{1-\corrDF^2}\frac{\beta^2(\dimh)}{1-\beta^2(\dimh)}.
\end{EQA}
\end{lemma}

\begin{proof}
Take any \(\mathbf v\in \R^{\dimp}\) with \(\|\mathbf v\|\le 1\) and note that with \(\upsilonv=(\thetav,\etav,\kappav)\in  l^2\)
\begin{EQA}
&&\nquad\DPr^{-2}\DPr_{\dimh}\mathbf v\\
&=&\Pi_{\thetav}\argmax_{\upsilonv\in l^2}\left\{\thetav^\T \DPr_{\dimh}\mathbf v-\| \DF \upsilonv\|^2/2\right\}\\
&=&\Pi_{\thetav}\argmax_{\upsilonv\in \R^{\dimp+\dimh}}\left\{\thetav^\T \DPr_{\dimh}\mathbf v-\| \DF_{\dimh} \upsilonv\|^2/2-\inf_{\kappav}(\kappav^\T\AF_{\dimh}^{\T}\upsilonv+\| \HF_{\dimh} \kappav\|^2/2)\right\}\\
&=&\Pi_{\thetav}\argmax_{\upsilonv\in \R^{\dimp+\dimh}}\left\{\thetav^\T \DPr_{\dimh}\mathbf v-\| \DF_{\upsilonv\upsilonv} \upsilonv\|^2/2-\|\DF_{\kappav\kappav}^{-1/2}\AF_{\dimh}^{\T}\upsilonv \|^2/2\right\}\\
&\eqdef&\Pi_{\thetav}\argmax_{\upsilonv\in \R^{\dimp+\dimh}}g(\upsilonv)\eqdef \Pi_{\thetav}\upsilonvd.
\end{EQA}
Setting the gradient of \(g(\cdot)\) equal to zero gives that the maximizer \(\upsilonvd\in \R^{\dimp+\dimh}\) satisfies
\begin{EQA}[c]
\upsilonvd=\DF_{\dimh}^{-1}(I_{\dimtotal}-\DF_{\dimh}^{-1}\AF_{\dimh}\HF_{\dimh}^{-2}\AF_{\dimh}^{\T}
\DF_{\dimh}^{-1})^{-1}\DF_{\dimh}^{-1}\Pi_{\thetav}^\T\DPr_{\dimh}\mathbf v,
\end{EQA}
where \(\Pi_{\thetav}^\T:\,\R^{\dimp}\to \R^{\dimp+\dimh}\) denotes the canonical embedding of \(\R^{\dimp}\) into \(\R^{\dimp+\dimh}\). By assumption we have
\begin{EQA}[c]
\|(I_{\dimtotal}-\DF_{\dimh}^{-1}\AF_{\dimh}\HF_{\dimh}^{-2}\AF_{\dimh}^{\T}\DF_{\dimh}^{-1})^{-1} -I_{\dimtotal}\|\le \frac{\beta^2(\dimh)}{1-\beta^2(\dimh)}.
\end{EQA}
Note that \(\DPr_{\dimh}\Pi_{\thetav}\DF_{\dimh}^{-2}\Pi_{\thetav}^\T\DPr_{\dimh}\mathbf v= \mathbf v\) which gives
\begin{EQA}
&&\nquad\|(I-\DPr_{\dimh} \DPr^{-2}\DPr_{\dimh})\mathbf v \|=\|\mathbf v-\DPr_{\dimh}\Pi_{\thetav}\upsilonvd\|=\|\DPr_{\dimh}\Pi_{\thetav}\DF_{\dimh}^{-2}\Pi_{\thetav}^\T\DPr_{\dimh}\mathbf v-\DPr_{\dimh}\Pi_{\thetav}\upsilonvd\|\\
&=&\|\DPr_{\dimh}\Pi_{\thetav}\DF_{\dimh}^{-1}\bigg((I_{\dimtotal}-\DF_{\dimh}^{-1}
\AF_{\dimh}\HF_{\dimh}^{-2}\AF_{\dimh}^{\T}\DF_{\dimh}^{-1})^{-1}-I_{\dimtotal}\bigg)\DF_{\dimh}^{-2}\Pi_{\thetav}^\T\DPr_{\dimh}\mathbf v\|\\
&\le&  \frac{\beta^2(\dimh)}{1-\beta^2(\dimh)}\|\DPr_{\dimh}\Pi_{\thetav}\DF_{\dimh}^{-2}\Pi_{\thetav}^\T\DPr_{\dimh}\mathbf v\|\\
  &=& \frac{\beta^2(\dimh)}{1-\beta^2(\dimh)}\| \mathbf v\|= \frac{\beta^2(\dimh)}{1-\beta^2(\dimh)}.
\end{EQA}
This implies
\begin{EQA}[c]
\|I-\DPr_{\dimh}^{-1} \DPr^2\DPr_{\dimh}^{-1} \|\le \|I-\DPr_{\dimh} \DPr^{-2}\DPr_{\dimh}\|\|\DPr_{\dimh}^{-1} \DPr^2\DPr_{\dimh}^{-1}\|\le \frac{1+\corrDF^2+\beta^2(\dimh)}{1-\corrDF^2}\frac{\beta^2(\dimh)}{1-\beta^2(\dimh)}.
\end{EQA}

\end{proof}

\begin{lemma}
 \label{lem: convergence of breve matrix}
Assume \( \bb{(\upsilonv\kappav)} \) then we get with \( {\rr^*}^{2}= 4\CONST_{\kappavs}\dimh/ \gmi\)
\begin{EQA}[c]
\| I-\DPrp(\upsilonvs_{\dimh})^{-1}\DPrp(\upsilonvs)^2\DPrp(\upsilonvs_{\dimh})^{-1}\|\le \delta(\rr^*)/(1-2\delta(\rr^*)).
 \end{EQA}
\end{lemma}

\begin{proof}
Denote \(\DPr_{\dimh\dimh}\eqdef\DPrp(\upsilonvs_{\dimh})\), \(\DF_{\dimh\dimh}\eqdef\DF_{\dimh}(\upsilonvs_{\dimh})\) and \(\DPrp\eqdef\DPrp(\upsilonvs)\), \(\DF_{\dimh}\eqdef\DF_{\dimh}(\upsilonvs)\).
We simply calculate
\begin{EQA}
\|I-\DPr_{\dimh\dimh}^{-1} \DPrp^2\DPr_{\dimh\dimh}^{-1} \|&\le& \|\DPr_{\dimh\dimh}\DPrp^{-2}\DPr_{\dimh\dimh}-I_{\dimp}\|\|\DPr_{\dimh\dimh}^{-1} \DPrp^2\DPr_{\dimh\dimh}^{-1}\|.
\end{EQA}
Now we get with condition \( \bb{(\LL_{0})} \) and Lemma \ref{lem: apriori bias in sieve approach}
\begin{EQA}
 \|\DPr_{\dimh\dimh}\DPrp^{-2}\DPr_{\dimh\dimh}-I_{\dimp}\|& =&\|\DPr_{\dimh\dimh}\left(\Pi_{\thetav} \DF_{\dimh}^{-2}\Pi_{\thetav}^\T-\Pi_{\thetav}\DF_{\dimh\dimh}^{-2}\Pi_{\thetav}^\T\right)\DPr_{\dimh\dimh}\|\\
 &=&\|\DPr_{\dimh\dimh}\left(\Pi_{\thetav}\DF_{\dimh\dimh}^{-1}\left\{I-\DF_{\dimh\dimh}\DF_{\dimh}^{-2}\DF_{\dimh\dimh} \right)\DF_{\dimh\dimh}^{-1}\Pi_{\thetav}^\T\right)\DPr_{\dimh\dimh}\|\\
 &\le&\|I-\DF_{\dimh\dimh}\DF_{\dimh}^{-2}\DF_{\dimh\dimh}\|\\
 &\le&\|I-\DF_{\dimh\dimh}^{-1}\DF_{\dimh}^{2}\DF_{\dimh\dimh}^{-1}\|\|\DF_{\dimh\dimh}\DF_{\dimh}^{-2}\DF_{\dimh\dimh}\|\\
 &\le& \delta(\rr^*)\|\DF_{\dimh\dimh}\DF_{\dimh}^{-2}\DF_{\dimh\dimh}\|.
\end{EQA}
This implies
\begin{EQA}
\|\DF_{\dimh\dimh}\DF_{\dimh}^{-2}\DF_{\dimh\dimh}\|\le1/(1-\delta(\rr^*)),&& \|\DPr_{\dimh\dimh}^{-1} \DPrp^2\DPr_{\dimh\dimh}^{-1}\|\le (1-\delta(\rr^*))/(1-2\delta(\rr^*)).
\end{EQA}
Together this gives the claim.
\end{proof}


TITLE: Deriving the law of reflection for a moving mirror in relativity
QUESTION [10 upvotes]: I am following a training course and came across this proof, from my colleague, that the ordinary law of reflection $\theta_i = \theta_r$ does not hold in relativity:

Let $S$ be a perfectly reflecting mirrror.
Obviously if it is at rest the canonical law of reflection $\theta_i=\theta_r$ holds. Now suppose the mirror is moving with velocity $\overline v$ in the frame $\Sigma$, and let the frame of the mirror be $\Sigma'$. Then in $\Sigma'$,
$$\theta'_i\equiv\theta'_r$$
which means that
$$
\cos \theta =\frac{\cos \theta'_i+\beta}{1+\beta \cos \theta'_i}.
$$
For the reflected ray changes $\theta$ before was $\theta_i $ is now $\theta =\pi-\theta_r $, similarly $\theta'= \pi- \theta'_r $ waves $ \cos \theta = - \cos (\theta_r)$ which is replaced provides:
$$
-\cos \theta_r =\frac{-\cos \theta'_r+\beta}{1-\beta \cos \theta'_r}, \quad
\cos \theta_r =\frac{\cos \theta'_r-\beta}{1-\beta \cos \theta'_r}
$$
$$\tan \left(\frac{\theta_i}{2}\right) =\sqrt{\frac{1-\beta}{1+\beta}}\tan \left(\frac{\theta'_i}{2}\right), \qquad \tan  \left(\frac{\theta_r}{2}\right) =\sqrt{\frac{1+\beta}{1-\beta}}\tan \left(\frac{\theta'_r}{2}\right)
$$
hence
$$
\tan\left( \frac{\theta_i}{2}\right)=\left(\frac{1-\beta}{1+\beta}\right)\tan \left( \frac{\theta_r}{2} \right)
$$
  and the law of reflection is no longer valid.

I don't think this derivation is very clear. Is there anyone who can help me to understand the steps in this unclear derivation, or recommend a book where I can find a proof of the final result?
If more information is needed, a picture of the derivation can be found here and here.

REPLY [0 votes]: I would post this video as an answer:
https://www.youtube.com/watch?v=v8lzsYh3JRY&list=PLj6DWzIvBi4PFDXCCV1bNhVUgDLTwVbFc&index=13
A variation would be to shine a horizontal light beam at a 45 degree mounted mirror in a moving ship. Either way (light clock or mirror), the resultant beam would be vertical. The video shows how to calculate the angle from vertical seen from a stationary perspective of a moving light clock or mirror. Instead of using time dilation, the same problem can be solved by length contraction of the 45 degree mirror's base although this is a far more complex way to tackle the problem. That solution can be seen here:
https://www.physicsforums.com/threads/45-dergree-mirrors-in-special-relativity.352427/


TITLE: Condition of acceleration when using $x=vt$
QUESTION [0 upvotes]: So my teacher told me that since $v = \delta x/ \delta t$, $\delta x = v • \delta t$ (naturally), and that is equal to the "area under the velocity-time graph", or displacement. This all makes sense to me, except that he told me that I could only use $\delta x = v • \delta t$ when the acceleration is $0$. Shouldn't it be rather only when the acceleration is constant (so velocity is linear), not necessarily $0$?

REPLY [1 votes]: You have the expression for displacement as a function of time, $t$ -
$$
\mathbf{x}(t) = \mathbf{u}t + \dfrac{1}2 \mathbf{a}t^2
$$
Where $\mathbf{u}$ is the initial velocity, $\mathbf{a}$ is the acceleration of the particle.
Set $\mathbf{a} = 0$, and you have $\mathbf{x}(t) = \mathbf{u}t $. This is the case of uniform velocity, not acceleration. You are confusing the terms, as only when velocity, $\mathbf{v} = \text{Constant/Uniform} \implies \mathbf{a} = \dfrac{d\mathbf{v}}{dt} = 0$. So this condition wouldn't correspond to uniform acceleration.


TITLE: Rewrite predicate formulas in propositional calculus
QUESTION [0 upvotes]: Suppose that the universe of discourse of the atomic formula P(x,y) is the set {0,1,2,3,4,5}. Write each of the following propositions using dis-junctions, conjunctions and only one negation:
1) ∃x P(x,0)

2) ∀y P(1,y)

3) ∀y¬P(2,y)

REPLY [1 votes]: HINT. Existential quantification over finite domains (say, of size $n$) is equivalent to a $n$ term disjunction over all elements. So from your example, if the universe of discourse was $\{0,1\}$, then the solution to question 1) would be $P(0,0)\lor P(1,0)$.
Similarly, universal quantification over finite domains (say, of size $n$) is equivalent to a $n$ term conjunction over all elements, so the solution to question 2) would be $P(1,0)\land P(1,1)$.
For the negation in 3) use De Morgan laws, i.e. $\neg(A\lor B)$ is equivalent to $(\neg A\land \neg B)$. This way you can convert an $n$ conjunction of negated atomic formulae to a negated $n$ disjunction.
Thanks Mauro ALLEGRANZA for pointing out an obvious typo.


TITLE: Name of a particular matrix close to projection
QUESTION [1 upvotes]: I am wondering if there is a special name for an $m\times n$ matrix $A=(a_{i,j})$,  with $a_{i,j}\in\{0,1\}$ that will pick $m$ unique components from a vector $v\in\mathbb{R}^n$ ($m\le n$), it is similar to a projection matrix, except that the image of such matrix is the subspace $\mathbb{R}^m$ instead of $\mathbb{R}^n$. Is decimation matrix correct?
For example, given a vector $v=(v_i)_{1\le i\le n}\in\mathbb{R}^n$, such matrix may return $w=(v_i)_{i=1,3,\ldots,2\lfloor \frac{n-1}{2}\rfloor+1}$ or $(v_i)_{1\le i\le 4}$ or any subset of the vector components.
Closely related question (without the coordinate position advancing requirement): Matrix with exactly one 1 in each row

REPLY [0 votes]: I suppose you are talking about matrices with a single $1$ in each row, and their positions advancing to the right for successive rows. I would call this a "coordinate-selection matrix", but that is in no means standard terminology. If you drop the advancing requirement, you can also permute or repeat any selected coordinates. If your matrix randomly skips one out of every ten columns, then "decimation matrix" would indeed be an appropriate term.


TITLE: Solving an equation for X
QUESTION [1 upvotes]: I am really stuck on a very simple question! I am sorry, it's been 3 years since I did this.
The equation is: $$\frac{5}{x+1}-\frac{1}{2x-1}=0$$
The problem I'm having is I have been taught to get rid of the fractions first by multiplying out, and then clearing the subsequent brackets, but doing this introduces lots of terms of $x$ on both sides which I then seem to get lost on, but I don't know if I can find a common denominator for this since they are different (+/-) signs?
Any help in how to proceed would be much appreciated, even if no answer.
KR,
Q

REPLY [0 votes]: Both of the other answers are completely right, but I think they seem to leave out the explanation of how they work.
We have
$$\frac{5}{x+1}-\frac{1}{2x-1}=0$$
What we want to do is combine these two fractions into one fraction. So we multiply
$\frac{5}{x+1}$ by $\frac{2x-1}{2x-1}$ and $\frac{1}{2x-1}$ by $\frac{x+1}{x+1}$. As we can see, we're multiplying both of these fractions by 1. This leaves us with
$$\frac{(10x-5)-(x+1)}{(x+1)(2x-1)} = 0$$
Now we multiply both sides by $(x+1)(2x-1)$ and simplify, which gives us
$$9x-6=0$$
I'm sure at this point you can solve for $x$.