text:
\section{\label{sec: Main results}Main results}

Now we formulate the main results of the  paper. The⋅first one deals
with approximation under minimal assumptions on the initial problem.
\begin{thm}
\label{thm: Convergence and Approximation with 1st corrector}If⋅\textup{(\ref{est: Original operator | Norm of (A=001D4B-=0003BC)=00207B=0000B9})},
\textup{(\ref{assu: N is Lipschitz})} and⋅\textup{(\ref{est: Norm of (A=002070-=0003BC)=00207B=0000B9})}
hold, then for any⋅$\varepsilon\in\set E_{\mu}$ and⋅$f\in L_{p}\rbr{\Omega}^{n}$
we⋅have
\begin{align}
\norm{\rbr{\op A_{\mu}^{\varepsilon}}^{-1}f-\rbr{\op A_{\mu}^{0}}^{-1}f}_{q,\Omega} & \lesssim\varepsilon^{s/p}\norm f_{p,\Omega},\label{est: Convergence}\\
\norm{D\rbr{\op A_{\mu}^{\varepsilon}}^{-1}f-D\rbr{\op A_{\mu}^{0}}^{-1}f-\varepsilon D\op K_{\mu}^{\varepsilon}f}_{p,\Omega} & \lesssim\varepsilon^{s/p}\norm f_{p,\Omega},\label{est: Approximation with 1st corrector}
\end{align}
where~$q\in\sbr{p,p^{*}}$. The~constants depend only on the parameters~$d$,
$s$, $p$, $q$, $n$, $\mu$, the domain~$\Omega$, the $C^{0,1}$\nobreakdash-norms
of⋅$A$ and⋅$N$ and the constants in the  bounds~\textup{(\ref{est: Original operator | Norm of (A=001D4B-=0003BC)=00207B=0000B9})}
and~\textup{(\ref{est: Norm of (A=002070-=0003BC)=00207B=0000B9})}.
\end{thm}
Notice that the inverse of⋅$\op A_{\mu}^{\varepsilon}$ actually does
converge in the operator norm from $L_{p}$ to $W_{p}^{r}$ with~$r<1$,
yet the rate may be not as good.
\begin{cor}
\label{cor: Corollary 1}Under⋅the⋅hypotheses of Theorem~\textup{\ref{thm: Convergence and Approximation with 1st corrector}},
for any⋅$r\in\rbr{0,1}$, $\varepsilon\in\set E_{\mu}$ and⋅$f\in L_{p}\rbr{\Omega}^{n}$
it holds⋅that
\begin{equation}
\Seminorm{\rbr{\op A_{\mu}^{\varepsilon}}^{-1}f-\rbr{\op A_{\mu}^{0}}^{-1}f}_{r,p,\Omega}\lesssim\varepsilon^{s/p\meet\rbr{1-r}}\norm f_{p,\Omega}.\label{est: Corollary 1}
\end{equation}
The⋅constant depends only on the parameters~$d$, $r$, $s$, $p$,
$n$, $\mu$, the domain~$\Omega$, the $C^{0,1}$\nobreakdash-norms
of⋅$A$ and⋅$N$ and the constants in the  bounds~\textup{(\ref{est: Original operator | Norm of (A=001D4B-=0003BC)=00207B=0000B9})}
and~\textup{(\ref{est: Norm of (A=002070-=0003BC)=00207B=0000B9})}.
\end{cor}
We can improve the estimate~(\ref{est: Convergence}) for⋅$q=p$
provided that the adjoint problem enjoys silimar regularity properties
as for the initial one.
\begin{thm}
\label{thm: Convergence with e}Suppose that \textup{(\ref{est: Original operator | Norm of (A=001D4B-=0003BC)=00207B=0000B9})},
\textup{(\ref{assu: N is Lipschitz})}, \textup{(\ref{est: Norm of (A=002070-=0003BC)=00207B=0000B9})}
and \textup{(\dualref{assu: N is Lipschitz})}, \textup{(\dualref{est: Norm of (A=002070-=0003BC)=00207B=0000B9})}
hold. Then for any⋅$\varepsilon\in\set E_{\mu}$ and⋅$f\in L_{p}\rbr{\Omega}^{n}$
we⋅have
\begin{equation}
\norm{\rbr{\op A_{\mu}^{\varepsilon}}^{-1}f-\rbr{\op A_{\mu}^{0}}^{-1}f}_{p,\Omega}\lesssim\varepsilon^{s/p+s^{\dual}{\mkern-4mu }/p^{\dual}}\norm f_{p,\Omega}.\label{est: Convergence with e}
\end{equation}
The⋅constant depends only on the parameters~$d$, $s$, $s^{\dual}$,
$p$, $n$, $\mu$, the domain~$\Omega$, the $C^{0,1}$\nobreakdash-norms
of⋅$A$, $N$ and⋅$N^{\dual}$ and the constants in the  bounds~\textup{(\ref{est: Original operator | Norm of (A=001D4B-=0003BC)=00207B=0000B9})},
\textup{(\ref{est: Norm of (A=002070-=0003BC)=00207B=0000B9})}
and⋅\textup{(\dualref{est: Norm of (A=002070-=0003BC)=00207B=0000B9})}.
\end{thm}
The⋅other estimate in Theorem~\ref{thm: Convergence and Approximation with 1st corrector}
can be improved as well, but only if restricted to an interior of~$\Omega$.

\begin{thm}
\label{thm: Interior approximation with 1st corrector}Suppose that
\textup{(\ref{est: Original operator | Norm of (A=001D4B-=0003BC)=00207B=0000B9})},
\textup{(\ref{assu: N is Lipschitz})}, \textup{(\ref{est: Norm of (A=002070-=0003BC)=00207B=0000B9})}
and \textup{(\dualref{assu: N is Lipschitz})}, \textup{(\dualref{est: Norm of (A=002070-=0003BC)=00207B=0000B9})}
hold. Suppose further that for a given⋅$\chi\in C^{0,1}\rbr{\wb{\Omega}}$
with⋅$\supp\chi\subset\Omega$ there is $\chi^{\prime}\in C^{0,1}\rbr{\wb{\Omega}}$
with⋅$\supp\chi\subset\supp\chi^{\prime}\subset\Omega$ such that
for all⋅$\varepsilon\in\set E_{\mu}$  the interior energy estimate
\begin{equation}
\norm{D\chi u}_{p,\Omega}\lesssim\norm u_{p,\Omega}+\norm{\chi^{\prime}\op A_{\mu}^{\varepsilon}u}_{-1,p,\Omega}^{*},\qquad u\in\set W_{p}^{1}\rbr{\Omega;\C^{n}},\label{est: Local enery estimate}
\end{equation}
holds. Then for any⋅$\varepsilon\in\set E_{\mu}$ and~$f\in L_{p}\rbr{\Omega}^{n}$
\begin{equation}
\norm{D\chi\rbr{\rbr{\op A_{\mu}^{\varepsilon}}^{-1}f-\rbr{\op A_{\mu}^{0}}^{-1}f-\varepsilon\op K_{\mu}^{\varepsilon}f}}_{p,\Omega}\lesssim\varepsilon^{s/p+s^{\dual}{\mkern-4mu }/p^{\dual}}\norm f_{p,\Omega}.\label{est: Interior approximation with 1st corrector}
\end{equation}
The⋅constant depends only on the parameters~$d$, $s$, $s^{\dual}$,
$p$, $n$, $\mu$, the domain~$\Omega$,  the $C^{0,1}$\nobreakdash-norms
of⋅$A$, $N$, $N^{\dual}$ and⋅$\chi^{\prime}$ and the constants
in the  bounds~\textup{(\ref{est: Original operator | Norm of (A=001D4B-=0003BC)=00207B=0000B9})},
\textup{(\ref{est: Norm of (A=002070-=0003BC)=00207B=0000B9})},
\textup{(\dualref{est: Norm of (A=002070-=0003BC)=00207B=0000B9})}
and⋅\textup{(\ref{est: Local enery estimate})}.
\end{thm}
As⋅a corollary we have:
\begin{cor}
\label{cor: Interior approximation with 1st corrector}Let⋅$\Sigma$
be a subdomain of⋅$\Omega$ with~$\dist\rbr{\Sigma,\partial\Omega}>0$.
Suppose that the hypotheses of Theorem~\textup{\ref{thm: Interior approximation with 1st corrector}}
hold for $\chi$ such that⋅$\chi\restrict{\Sigma}=1$. Then, for any⋅$\varepsilon\in\set E_{\mu}$
and⋅$f\in L_{p}\rbr{\Omega}^{n}$,
\begin{equation}
\norm{D\rbr{\op A_{\mu}^{\varepsilon}}^{-1}f-D\rbr{\op A_{\mu}^{0}}^{-1}f-\varepsilon D\op K_{\mu}^{\varepsilon}f}_{p,\Sigma}\lesssim\varepsilon^{s/p+s^{\dual}{\mkern-4mu }/p^{\dual}}\norm f_{p,\Omega}.\label{est: Interior approximation with 1st corrector-2}
\end{equation}
The⋅constant depends only on the parameters~$d$, $s$, $s^{\dual}$,
$p$, $n$, $\mu$, the domain~$\Omega$, the $C^{0,1}$\nobreakdash-norms
of⋅$A$, $N$, $N^{\dual}$ and⋅$\chi^{\prime}$ and the constants
in the  bounds~\textup{(\ref{est: Original operator | Norm of (A=001D4B-=0003BC)=00207B=0000B9})},
\textup{(\ref{est: Norm of (A=002070-=0003BC)=00207B=0000B9})},
\textup{(\dualref{est: Norm of (A=002070-=0003BC)=00207B=0000B9})}
and⋅\textup{(\ref{est: Local enery estimate})}.
\end{cor}
The⋅next result follows from Corollary~\ref{cor: Interior approximation with 1st corrector}
in the same manner as Corollary~\ref{cor: Corollary 1} comes from
Theorem~\ref{thm: Convergence and Approximation with 1st corrector}.
\begin{cor}
\label{cor: Interior convergence for fractional derivative}Let⋅hypotheses
be as in Corollary~\textup{\ref{cor: Interior approximation with 1st corrector}}.
Then, for any⋅$r\in\rbr{0,1}$, $\varepsilon\in\set E_{\mu}$ and⋅$f\in L_{p}\rbr{\Omega}^{n}$,
\begin{equation}
\Seminorm{\rbr{\op A_{\mu}^{\varepsilon}}^{-1}f-\rbr{\op A_{\mu}^{0}}^{-1}f}_{r,p,\Sigma}\lesssim\varepsilon^{\rbr{s/p+s^{\dual}{\mkern-4mu }/p^{\dual}}\meet\rbr{1-r}}\norm f_{p,\Omega}.\label{est: Interior convergence for fractional derivative}
\end{equation}
The⋅constant depends only on the parameters~$d$, $r$, $s$, $s^{\dual}$,
$p$, $n$, $\mu$, the domain~$\Omega$, the $C^{0,1}$\nobreakdash-norms
of⋅$A$, $N$, $N^{\dual}$ and⋅$\chi^{\prime}$⋅and the constants
in the  bounds~\textup{(\ref{est: Original operator | Norm of (A=001D4B-=0003BC)=00207B=0000B9})},
\textup{(\ref{est: Norm of (A=002070-=0003BC)=00207B=0000B9})},
\textup{(\dualref{est: Norm of (A=002070-=0003BC)=00207B=0000B9})}
and⋅\textup{(\ref{est: Local enery estimate})}.
\end{cor}

\begin{rem}
The⋅corrector~$\varepsilon\op K_{\mu}^{\varepsilon}$ is usually
involved in an approximation for⋅$\rbr{\op A_{\mu}^{\varepsilon}}^{-1}$
in the ``energy'' norm. If⋅we want to approximate⋅$D\rbr{\op A_{\mu}^{\varepsilon}}^{-1}$
only, we may use the operator~$\tau^{\varepsilon}\op S^{\varepsilon}D_{2}\op K_{\mu}\rbr{\varepsilon}$
instead, because
\[
\varepsilon D\op K_{\mu}^{\varepsilon}=\varepsilon\tau^{\varepsilon}\op S^{\varepsilon}D_{1}\op K_{\mu}\rbr{\varepsilon}+\tau^{\varepsilon}\op S^{\varepsilon}D_{2}\op K_{\mu}\rbr{\varepsilon},
\]
where
\[
\norm{\tau^{\varepsilon}\op S^{\varepsilon}D_{1}\op K_{\mu}\rbr{\varepsilon}\hairspace f}_{p,\Omega}\lesssim\norm f_{p,\Omega}
\]
by⋅Lemma~\ref{lem: =0003C4=001D4BS=001D4B} and the estimate~(\ref{est: K(delta) is bounded}),
so this term can be absorbed into the error.
\end{rem}

\begin{rem}
The⋅results of Theorem~\ref{thm: Interior approximation with 1st corrector}
and Corollaries~\ref{cor: Interior approximation with 1st corrector}
and~\ref{cor: Interior convergence for fractional derivative} rely
on an a~priory bound~(\ref{est: Local enery estimate}). In⋅view
of Lemma~\ref{lem: Original operator | Interior energy estimate},
for a compactly supported function~$\chi$ this can be reduced to
a similar bound with a smaller exponent~$q\ge1$, provided that
 the resolvent of⋅$\op A^{\varepsilon}$ is bounded from⋅$\rbr{W_{q^{\dual}}^{1}\rbr{\Omega}^{n}}^{*}$
to~$W_{q}^{1}\rbr{\Omega}^{n}$.
\end{rem}

\begin{rem}
\label{rem: Boundary layer term}A⋅glance at⋅(\ref{est: Approximation with 1st corrector})
and⋅(\ref{est: Interior approximation with 1st corrector-2})  suggests
that the rate of approximation for⋅$D\rbr{\op A_{\mu}^{\varepsilon}}^{-1}$
becomes worse  only near the boundary of~$\Omega$. In⋅fact, one
can introduce a boundary-layer correction term~$\op B_{\mu}^{\varepsilon}$
so that for any⋅$\varepsilon\in\set E_{\mu}$ and⋅$f\in L_{p}\rbr{\Omega}^{n}$
\begin{equation}
\norm{\rbr{\op A_{\mu}^{\varepsilon}}^{-1}f-\rbr{\op A_{\mu}^{0}}^{-1}f-\varepsilon\op K_{\mu}^{\varepsilon}f-\op B_{\mu}^{\varepsilon}f}_{1,p,\Omega}\lesssim\varepsilon^{s}\norm f_{p,\Omega},\label{est: Boundary-layer correction term}
\end{equation}
see⋅(\ref{def: Boundary part}) and Lemmas~\ref{lem: Ie} and~\ref{lem: De}.
For⋅$s=1$ and⋅$p=2$, such a result was the starting point of the
approach suggested in~\cite{ZhPas:2005} (see⋅also~\cite{PasT:2007},
\cite{PSu:2012}, \cite{Su:2013-1} and~\cite{Su:2013-2}). However,
the construction of⋅$\op B_{\mu}^{\varepsilon}$ is no simpler than
the original problem and actually amounts to finding the inverse of~$\op A_{\mu}^{\varepsilon}$.
Thus, that approach required  further analysis of the boundary-layer
correction term to obtain bounds on its norms.  We also note that
if $\Omega=\R^{d}$ (or, more generally, $\Omega$ is a flat manifold
without boundary, such as, e.g.,~$\T^{d}$), then⋅$\op B_{\mu}^{\varepsilon}=0$.
This enables one to improve the rates in⋅(\ref{est: Convergence})–(\ref{est: Approximation with 1st corrector})
to⋅$\varepsilon^{s}$, which, at least for⋅$s=1$, is known to be
sharp.
\end{rem}

\begin{rem}
\label{rem: Main results | Regularity of the effective operator}By⋅inspection
of the proofs, one can see that the estimates in  Theorem~\ref{thm: Convergence and Approximation with 1st corrector}–Corollary~\ref{cor: Interior convergence for fractional derivative}
follow from  inequalities with⋅$\abs{\rbr{\op A_{\mu}^{0}}^{-1}f}_{1+s,p,\Omega}$
in place of⋅$\norm f_{p,\Omega}$ on the right. Thus, if⋅(\ref{est: Norm of (A=002070-=0003BC)=00207B=0000B9})
fails to hold, but $\rbr{\op A_{\mu}^{0}}^{-1}f\in\set W_{p}^{1}\rbr{\Omega;\C^{n}}\cap\Lambda_{p}^{1+s}\rbr{\Omega}^{n}$
for some⋅$f\in L_{p}\rbr{\Omega}^{n}$, then for fixed such~$f$
we still have, e.g., results similar to Theorem~\ref{thm: Convergence and Approximation with 1st corrector}
and⋅Corollary~\ref{cor: Corollary 1}.
\end{rem}


TITLE: What is meant by “combinatorial background” in experimental high energy physics
QUESTION [10 upvotes]: My guess is that they find a certain tracks coming from a certain source by "combinatorially" selecting all track pairs and finding their invariant mass. If this is true, of which I am not sure, how is the background defined, i.e how far away from the invariant mass peaks? Any references will help too.

REPLY [3 votes]: In a collision lots of particle are produced . Very often one looks for a particle not directly but through its decay products (look at particle data book for the known particles and their decays). Asan example consider jpsi decay in a muon pair. In each collision one look for a pair of muons. Depending on the collision energy  there will be less or more than one muon per event but anywayone will construct all the possible pairs of muons in each event with the muons measured in the event.  If the muons come from a jpsi, like fragments of a bomb explosion they will carry the energy of the jpsi, so one calculate the total energy of each pair (its mass).  Pions are copiously produced in the collision and they have a probability of decay into muons. This makes another way to produce a pair by random combination of muons produced independantly by pions or other sources.  This is a typical combinatorial background. If it leads to the same mass there is no way to separate it from a psi decay. But globally these combinations will lead to a distributions of masses so one will be able to separate the psi peak from the much broader combinatorial contribution. this combiatorial problem is more general: each time one look for a specific pair one will have a certain probability a random combination of other products will look similar


TITLE: A better way to prime factorize a set of numbers?
QUESTION [3 upvotes]: Let's say I have a range of numbers starting from 1 to 10^9 and I need to prime factorize each one of them.My basic algorithm is:
1.Use prime-sieve algorithms(Atkins or Eratosthenes(segmented approach more preferable to avoid Buffer overflows)) to generate primes less than 10^9.
2.Divide each n in the list with these primes in ascending order till the numerator becomes one and use of count array to store the powers of each prime needed to factorize the given number.
Even though this basic algorithm works fine I need to optimize it in any way possible to increase it's speed.I also want to know is there anyway to guess the minimum range of primes required to factorize a given number.For ex:The naive algorithm to check if a given number n is prime or not is to check for possible divisors from 2 to sqrt(n)....I want to know if it's enough to generate primes to a certain limit to prime-factorize a given n.To be more precise:
  999,824 = 2^4 × 7 × 79 × 113

My algorithm would generate primes in the range of 1 to 999824 to prime factorize the above given n and it's a solid waste of resources and degrades the performance and hence I want to know if it's possible to know in beforehand that it would be enough to find primes less than say 115 to factorize above given n.

REPLY [6 votes]: All of them?  The eratosthenes seive, or rather something similar, will probably win: most non-seiving systems need to duplicate their work for each prime, as opposed to getting some economy of scale.  You don't need to generate any numbers higher than sqrt(highest number you care about).
Also instead of doing this in two passes, simply give each number a pair of data points when you hit it: the first prime you hit that divides it evenly, and what that number divides it into.  So 522 would have the data 2, 261.  Any number that doesn't have one of these is prime; all others can be factored by appending its prime factor to the factor list of its prime factor's counterpart.  So factors(522) = 2, factors(261) = 2, 3, factors(87) = 2,3,3,factors(29) = 2,3,3,29.


text:
{\bf Problem.} Find the least positive integer such that when its leftmost digit is deleted, the resulting integer is 1/29 of the original integer.
{\bf Level.} Level 5
{\bf Type.} Number Theory
{\bf Solution.} The desired integer has at least two digits.  Let $d$ be its leftmost digit, and let $n$ be the integer that results when $d$ is deleted.  Then for some positive integer $p$, $10^p\cdot
d+n=29n$, and so $10^p\cdot d=28n$. Therefore 7 is a divisor of $d$, and because $1\le d\le9$, it follows that $d=7$. Hence $10^p=4n$, so $\displaystyle n={{10^p}\over4}=
{{100\cdot10^{p-2}}\over4}=25\cdot10^{p-2}$.  Thus every positive integer with the desired property must be of the form $7\cdot10^p+25\cdot10^{p-2}=10^{p-2}(7\cdot10^2+25)=725\cdot10^{p-2}$ for some $p\ge2$. The smallest such integer is $\boxed{725}$.

set_name: MATH


TITLE: Self-linkage of the orthogonal group $O_n({\mathbb R})$.
QUESTION [5 upvotes]: In Exercise 153 of my list, it is proved that the connected components $SO_2({\mathbb R})$ and $O_2^-({\mathbb R})$ of the orthogonal group are linked as curves in the three-dimensional sphere defined by $\|M\|=1$ in the operator norm.
Question. Consider the case of $n\times n$ matrices. The real orthogonal group has two connected components $SO_n({\mathbb R})$ and $O_n^-({\mathbb R})$, each one a submanifold of dimension $n(n-1)/2$. Let $s_1(M)\ge\cdots\ge s_n(M)(\ge0)$ be the singular values of a matrix $M$. We have $s_1(M)=\|M\|$. Define the set $V_n$ by the identities $s_1(M)=\cdots=s_{n-1}(M)=1$. Is it a submanifold of dimension $n^2-n+1$ ? In $V_n$, are $SO_n({\mathbb R})$ and $O_n^-({\mathbb R})$ linked ?
Nota. The exercise in $M_2({\mathbb R})$ is included in the forthcoming second edition of my book ``Matrices'' as Exercise 21 in chapter 10.

REPLY [2 votes]: The dimension of $V_n$ is at most $\binom{n}{2} +n$. For $n>2$, this is less than $2 \binom{n}{2} +1$.
It is convenient to solve a more general problem. Fix $s_1 \geq s_2 \geq \cdots \geq s_n \geq 0$. Let $M(s_1, \ldots, s_n)$ be the space of matrices with singular values $s_i$. Let $d_1$, $d_2$, ..., $d_k$ be the list of multiplicites with which the $s$'s occur, so $\sum d_i = n$. For example, if the $s$'s are $(1,1,1,\ldots, 1, r)$ then $(d_1, d_2) = (n-1, 1)$.
I claim that $\dim M(s) = 2 \binom{n}{2} - \sum \binom{d_i}{2}$. Proof: The group $O(n) \times O(n)$ acts transitively on $M(s)$, so the dimension of $M(s)$ is $\dim ( O(n) \times O(n)) = 2 \binom{n}{2}$ minus the dimensional of the stabilizer of the diagonal matrix $d(s) := \mathrm{diag}(s_1, s_2, \ldots, s_{n-1}, s_n)$. We compute the Lie algebra of the stabilizer. Let $g$ and $h$ be skew-symmetric matrices. Then, to first order, $e^g d(s) e^h = d(s)$ if and only if $g d(s) + d(s) h=0$.
For every $(i,j)$, this gives the equations $g_{ij} s_j + s_i h_{ij}$ and $s_i g_{ij} + s_j h_{ij} =0$. If $s_i \neq \pm s_j$, this forces $g_{ij} = h_{ij}=0$. If $s_i = \pm s_j$ then the sign must be $+$, as the $s_i$ are nonnegative, and we get $g_{ij} = - h_{ij}$. So the Lie algebra of the stabilizer has a basis element for every $(i,j)$ with $s_i = s_j$, and we see that the dimension of the stabilizer is $\sum \binom{d_i}{2}$.
You are interested in $\bigcup_{0 \leq r \leq 1} M(1,1,\ldots,1,r)$. Each space in the union has dimension $2\binom{n}{2} - \binom{n-1}{2} = \binom{n}{2} + n-1$, except for the $r=1$ term which is even smaller. Thus, the union has dimension at most $\binom{n}{2} +n$. I have not been able to work out whether the union is a manifold near the boundary points $r=0$ and $r=1$.
One way to salvage your question would be to look at matrices whose singular values are $(1,1-r/(n-1), 1-2r/(n-1), \ldots, 1-(n-2)r/(n-1), 1-r)$, for $r \in [0,1]$.


TITLE: The error of the midpoint rule for quadrature
QUESTION [3 upvotes]: Wikipedia says the midpoint formula for numerical integration has error of order $h^3 f''(\xi)$. I am trying to replicate this result.
I'm guessing that I want to use Lagrange's formulation of the remainder for Taylor series. Let $x_0=\frac{a+b}{2}$ (i.e. the midpoint).
The midpoint method says $\int_a^b f(x)dx \approx (b-a)f(\frac{a+b}{2})$, so to get the error I find $(b-a) f(\frac{a+b}{2}) - \int_a^bf(x)dx$. If I expand this using Taylor's theorem I get:
$ \begin{aligned}
error & =(b-a) f(x_0) - \int_a^bf(x_0)+\frac{f'(\xi)(x-x_0)}{2}dx \\
& =\frac{f'(\xi)}{2}\int_a^b(x-x_0)dx \\
& =\frac{f'(\xi)}{2}\int_a^b(x-\frac{a+b}{2})dx \\
& = 0
\end{aligned}$
So apparently I have just proven that it has zero error? Any hints as to what I did wrong? (I realize that since wikipedia gives it in terms of $f''$ I probably want to take the expansion one level further to match them, but I don't understand why this doesn't work.)

REPLY [1 votes]: I guess you are more concerned with why you cannot take the $f'(\xi)$ out in this case. The other answers in right in saying that $\xi$ depends on $x$, but I think you can show that $g(x) = f(\xi(x))$ depends on $x$ continuously, and that's not the reason why you cannot take $f'(\xi)$ outside the integral. In fact, if you want to derive the remainder for the rectangular rule, you have to do this at some point.
The main reason why your argument does not work is because the function $x - \frac{b + a}{2}$ is neither nonpositive nor nonnegative on the interval $[a, b]$, which makes it invalid to apply the mean value theorem for integrals. That's exactly why you can apply M.V.T for integrals for the quadratic term in the Taylor series. Hope this helps!


TITLE: Congruence modulo $n$ forms an equivalence class
QUESTION [2 upvotes]: I am learning about equivalence classes, and I learned that congruence modulo $n$ forms an equivalence relation. In the proof below, I didn't understand how at the very end of the second picture they said "we assumed a result known as the division algorithm" because no where in their proof did it look like they used it.

REPLY [1 votes]: The Division algorithm is the theorem that states when you divide $n$ by $m$ (both integers, and $m>0$), you get a unique quotient and remainder, and the remainder is between $0$ and $m-1$.
This was used in determining that there were three equivalence classes modulo $3$.  More generally, there are $m$ equivalence classes modulo $m$.


TITLE: Random graph is simple cycle or tree
QUESTION [1 upvotes]: Consider an undirected random graph $G$ of $n$ vertices. The probability that there is an edge between a pair of vertices is $p$.
What is the probability that $G$ is simple cycle of $n$ vertices or tree of $n$ vertices?
My solution:
Number of simple cycles is $\frac{(n - 1)!}{2}$ and number of trees is $n^{n - 2}$.
For every simple cycle prolability is $p^{n - 1}$ $(1 - p)^{(\frac{n}{2}) - n + 1}$ and for every tree probability is $p^{n - 1}$ $(1 - p)^{(\frac{n}{2}) - n + 1}$.
Answer is $(\frac{(n - 1)!}{2} + n^{n - 2}) p^{n - 1} (1 - p)^{(\frac{n}{2}) - n + 1}$.
Is my answer true?
Thank you for any help!

REPLY [1 votes]: Let $T$ be the event that the random graph is a tree, and let $C$ be the event that it is a simple cycle. We are interested in
$$P(T\mbox{ or }C) = P(T) + P(C) - P(T\mbox{ and }C) = P(T) + P(C).$$
In a complete graph there are ${n\choose 2}$ edges, and therefore for a fixed graph $G$ with $m$ edges on $n$ vertices, a random graph will be equal to $G$ with probability
$$p^m(1-p)^{{n\choose 2}-m}.$$
If $t$ and $c$ are the number of trees and cycles on $n$ vertices, respectively, and $m_t$ and $m_c$ the number of edges in a tree and cycle, respectively, you have
$$P(T) = t\cdot p^{m_t}(1-p)^{{n\choose 2}-m_t},\qquad P(C) = c\cdot p^{m_c}(1-p)^{{n\choose 2}-m_c}.$$
I think you can pick it up from here...


TITLE: Find what values of $n$ give $\varphi(n) = 10$
QUESTION [7 upvotes]: For what values of $n$ do we get $\varphi(n) = 10$?

Here, $\varphi$ is the Euler Totient Function. Now, just by looking at it, I can see that this happens when $n = 11$. Also, my friend told me that it happens when $n = 22$, but both of these are lucky guesses, or educated guesses. I haven't actually worked it out, and I don't know if there are any more.
How would I go about answering this question?

REPLY [1 votes]: the general method will be demonstrated for 12.
The factors of 12 are 2*2*3.  this is the product of totients of the prime powers (p-1)p^q.  So we divide this up into eg products to 12 that are numbers of this form.
 1*12    1,2 * 13      13, 26
 1*2*6   1,2 *3* 7     21, 42
 2*6     4 * 7, 9      28, 36

Since 3 can not be the totient of a prime power, the factorisation of 12=3*4 is not considered.
The case for 10involves 11 and 2 having totient products of 10 and 1, which leads directly to 11 and 22.


TITLE: Does $H^i(E) = 0$ for $i \geq 0$ imply $\operatorname{Hom}_{\mathbf{D}(\mathcal{A})}(E, A) = 0$ for any $A \in \mathcal{A}$?
QUESTION [1 upvotes]: Let $\mathcal{A}$ be an abelian category, $\mathbf{D}(\mathcal{A})$ be the derived category, and consider an object $E \in \mathbf{D}(\mathcal{A})$ with $H^i(E) = 0$ for all $i \geq 0$. Is it true that
$$\operatorname{Hom}_{\mathbf{D}(\mathcal{A})}(E, A) = 0 \tag{$*$}$$
for all objects $A \in \mathcal{A}$ (considered as a cochain complex in degree $0$)?
I realized, that $E$ is quasi-isomorphic to the subcomplex $E' \subset E$, given by
$$\dots \to E^{-2} \to E^{-1} \to \operatorname{im}(d^{-1}) \to 0 \to \cdots,$$
and that any chain map $E' \to A$ is zero-homotopic, but apparently that is not enough to conclude $(*)$.
This argument seems to appear in Bridgeland's Stability conditions on K3 surfaces, in the proof of Lemma 10.1.

REPLY [0 votes]: $E$ is quasi-isomorphic to a complex $P$ of projectives so $$Hom_{D}(E,A) = Hom_{K}(P,A) = Hom(H^0P,A) = 0$$
See 1.49-1.51 of https://www.math.univ-toulouse.fr/~dmattei/MemoireM2_Dominique_Mattei.pdf


TITLE: Lipschitz type estimate for the Green function for fractional Laplacian
QUESTION [0 upvotes]: Let  $G(x, y)$ be the Green function of the  fractional Laplacian $(-\Delta)^s$ in a bounded interval $I$ of $\mathbb{R}$ with $G(x, y)=0$ on $\mathbb{R}\setminus I$ and $s\in (0, 1).$ Is it possible to obtain an estimate of the form
$|G(x, y)- G(x, z)|\leq C|y-z|$ where $C>0$ might depend on $x.$

REPLY [2 votes]: No: the estimate fails when $y, z$ are close to each other and also close to either $x$ or one of the endpoints of $I$.
Observe that we have $$G(x, y) = C(s) |x - y|^{2 s - 1} + h(x, y) ,$$ where $C(s)$ is some constant and $h(x, y)$ is a smooth function in the interior of $I$ (for $h(x, \cdot)$ is the $(-\Delta)^s$-harmonic reduction of $C(s) |x - \cdot|^{2 s - 1}$ in $I$). Thus your estimate cannot hold when $y, z \approx x$. (The case $s = \tfrac{1}{2}$ is different, you have $\log |x - y|$ instead of $|x - y|^{2 s - 1}$).
Similarly, it is known that $G(x, y) \sim \tilde{C}(s) (\operatorname{dist}(y, \partial I))^{s/2}$ as $y$ converges to one of the endpoints of $I$, so $G(x, \cdot)$ is not Lipshitz continuous there.
By the way, $G(x, y)$ is known explicitly (in terms of $_2F_1$ or in an integral form), the expression was given by M. Riesz, Kac and Blumenthal–Getoor–Ray. Any standard reference on the fractional Laplacian and its potential theory (e.g. Landkoff, Bliedtner–Hansen) discusses these subjects.


TITLE: Find the limit of $\lim_{n\to \infty} \sum_{x=n}^\infty \frac{1}{x^3}$
QUESTION [2 upvotes]: Find the limit of $\lim_{n\to \infty} \sum_{x=n}^\infty \frac{1}{x^3}$. Im having issues understanding how to prove this since I have never done something like this. I want to say it goes to zero since I know the series itself converges meaning it also tends to zero. Since x scales with n, once n is sufficiently large the sum would also essentially be 0 is my thinking. I am just unsure of how to prove this rigorously.

REPLY [1 votes]: If you already know that a series $\displaystyle\sum_{n=m}^\infty a_n$ converges, then you can write
$$\sum_{n=m}^\infty a_n = \sum_{n=1}^\infty a_n - \sum_{n=1}^{m-1} a_n$$
Taking limits on both sides we have $$\lim_{m\to\infty}\sum_{n=m}^\infty a_n = \sum_{n=1}^\infty a_n - \lim_{m\to\infty}\sum_{n=1}^{m-1} a_n =  \sum_{n=1}^\infty a_n - \sum_{n=1}^\infty a_n = 0$$


TITLE: The sum of the roots
QUESTION [5 upvotes]: The function $f(x)$ has exactly six different roots, and is such that $f(11+x)=f(11-x)$. How do you find the sum of the roots ?

REPLY [6 votes]: The condition implies that if $r$ is a root so is its reflection with respect to $11$. Thus the six roots can be organized into three pairs, each summing 22, and the sum of all six roots is 66.


TITLE: Convexity of variance
QUESTION [4 upvotes]: How can I prove the following function is convex? Is variance  concave or convex?
$$\psi (x)=\left(x_1-\frac{x_1+x_2+x_3}{3}\right)^2+ \left(x_2-\frac{x_1+x_2+x_3}{3}\right)^2+ \left(x_3-\frac{x_1+x_2+x_3}{3}\right)^2$$

REPLY [5 votes]: The simplest way to see that this is convex is to note that it is the composition of a trivial quadratic function and an affine function (actually a linear one); that is, $\psi(x)=g(h(x))$, where
$$g:\mathbb{R}^3\rightarrow\mathbb{R}, \quad g(y) = y_1^2+y_2^2+y_3^3$$
$$h:\mathbb{R}^3\rightarrow\mathbb{R}^3, \quad h(x)=(x_1-(x_1+x_2+x_3)/3,x_2-(x_1+x_2+x_3)/3,x_3-(x_1+x_2+x_3)/3)$$
It is of course easy to verify that $g$ is convex; its Hessian is twice the identity matrix. It truly is the prototypical multivariate convex function! And the composition of a convex function and an affine function is always convex.
EDIT: This is readily extended to all data sizes. That is, the function
$$\psi(x) = \sum_{i=1}^n \left( x_i - \left( \textstyle \tfrac{1}{n} \sum_{j=1}^n x_j \right)\right)^2$$
is convex for any $n$. It is the composition of the trivially convex function
$$g:\mathbb{R}^n\rightarrow\mathbb{R}, \quad g(y) = \sum_{i=1}^n y_i^2$$
and the affine function
$$h:\mathbb{R}^n\rightarrow\mathbb{R}^n, \quad h(x) = (h_1(x),h_2(x),\dots,h_n(x)), \quad h_i(x) = x_i - \tfrac{1}{n}\textstyle\sum_{j=1}^n x_j$$
As the composition of a convex function and an affine function, it is convex. If you are getting an indefinite Hessian for $n>3$, as you suggest you are below, then your derivation is incorrect.


TITLE: Piece-wise Function with Variable
QUESTION [0 upvotes]: I have this piecewise function:

$g(x)=\begin{cases}
\frac{x^2-a^2}{x-a},  &  x\ne a \\
8,        &  x=a
\end{cases}$

From there I broke it down to:

$g(x)=\begin{cases}
\frac{(x-a)(x+a)}{x-a}, & x\ne a \\
8,                      & x=a
\end{cases}$

I have to find the $a$ that makes the function continuous, hence I took the limit of the piecewise function at $a$. I used the definition of limit definition of continuity, and got this:

$$\lim_\limits{x\to a}(g(x))=g(a)$$
$$\lim_\limits{x \to a}(\frac{(x-a)(x+a)}{(x-a)})=8$$
$$\lim_\limits{x \to a}(x+a)=8$$
$$((a)+a)=8$$
$$2a=8$$
$$a=4$$

Usually if I have the right answer but the wrong steps the problem is generally wrong: is my solution, and method to the problem correct?

REPLY [2 votes]: Your solution is mostly correct. You proved that the function is continuous at $a$. However, to make it complete, I would also mention that the function is continuous on $\mathbb{R}-\{a\}.$


TITLE: Can a proposition be false and its negation be false in Intuitionistic logic?
QUESTION [0 upvotes]: The principle of non-contradiction say that p and $\neg$ p can not be both true.

If you consider the law of excluded middle as relevant, then either
p or $\neg$ p is true, and thus the other is false by the principle
of non-contradiction.
But if you consider the excluded middle as not relevant, just like
intuitionistic logic, what would avoid some p and $\neg$ p to be
both false ?

Actually, if true is the same than provable, then such a proposition is just a Gödel one : p is not provable and neither $\neg$ p is.
I think this is a very good interpretation and a big demystification of Gödel theorem : it is not about some true that we can not reach, it is about some absurdity that is still an absurdity when you negate it.

REPLY [4 votes]: The statement "$p$ and $\neg p$ are both false" is formally expressed as
$$\neg p \land \neg \neg p,$$
which is false by the law of non-contradiction, which states that for any $q$ the proposition $q \land \neg q$ is false (take $q = \neg p$).
More importantly, these consideration do not "demistyfy Gödel", as they refer to proposition with free propositional variables in them. Gödel constructed a sentence, i.e., a logical formula without free parameters, which is neither provable nor falsifiable (in a given formal system).
To put it another way, it is very easy to find a formula which is neither provable nor falsifiable, if we allow free variables, for example
$$x < 42$$
where $x$ ranges over the natural numbers (say, in Peano arithmetic) is not provable, but neither is
$$\neg (x < 42).$$


TITLE: Representation space for the Alternating groups $A_n$
QUESTION [1 upvotes]: Is there a systematic way to construct the representation spaces for an irreducible representation $\phi:A_n\rightarrow {GL}(V)$, where $V$ is called as representation space. For example doubly transitive actions give rise to an irreducible representation spaces. See here. But not all can be constructed this way.
Over $\mathbb{C}$, the alternating group $A_7$ has an irreducible representation of degree $15$. But how do I find the corresponding representation space? This can not be constructed by doubly transitive actions on $16$ points, since $A_7$  doesn't have any transitive action on $16$ points, which can be found here.
I would like to get some reference for this.

REPLY [2 votes]: One technique is to ask GAP:
gap> LoadPackage("repsn");;
gap> grp := AlternatingGroup(7);
Alt( [ 1 .. 7 ] )
gap> chi := First(Irr(grp), chi -> Degree(chi)=15);
Character( CharacterTable( Alt( [ 1 .. 7 ] ) ), [ 15, -1, 3, -1, 0, -1, 0, 1, 1 ] )
gap> rep := IrreducibleAffordingRepresentation(chi);;

If you want to know the matrix for a particular permutation, you use:
gap> Image(rep, (1,2)(3,4));
[ [ 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0 ],
  [ 1, -1, 0, -1, 0, 0, 0, 0, 0, 0, 0, -1, 0, 0, 0 ],
  [ -1, 0, 0, 1, 1, 1, 1, 0, 1, 1, 0, 0, 1, 0, -1 ],
  [ -1, 0, -1, 1, 1, 0, 0, 0, -1, -1, 0, 1, 0, -1, 0 ],
  [ 0, 0, 0, 0, 0, 0, 1, 0, 1, 1, 1, -1, 1, 0, -1 ],
  [ 1, 0, 0, -1, -1, -1, -1, 0, -1, -1, -1, 0, -1, 0, 1 ],
  [ 0, 0, 1, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 0, 0 ],
  [ 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0 ],
  [ 0, 0, 1, 1, 0, 1, 1, 0, 0, 1, 0, 1, 0, 0, 0 ],
  [ 0, 0, -1, -1, 0, -1, 0, 0, 0, -1, 1, -1, 0, 0, 0 ],
  [ 0, 0, -1, 0, 0, -1, 0, 0, 0, 0, 0, 0, 0, 0, 0 ],
  [ 1, 0, 1, 0, -1, 0, 0, 0, 1, 1, 0, 0, 0, 1, 0 ],
  [ 0, 0, 1, 1, 1, 0, 0, -1, 1, 1, 0, 0, 1, 1, -2 ],
  [ 0, 0, 0, -1, -1, -1, -1, 0, 0, 0, 0, 0, 0, 0, 0 ],
  [ -1, 0, 0, 1, 1, 0, 1, -1, 1, 1, 1, 0, 1, 0, -2 ] ]

A similar technique to permutation representations is induced representations, but this representation is not induced (the only subgroups of index dividing 15 are the PGL(3,2)s, and they are index 15 but with no nontrivial linear characters).
You could restrict a representation of the symmetric group on 7 points. The Specht module with partition [3,1,1,1,1] or [5,1,1] will work. This works for many of the irreducible representations of alternating groups. If you like that technique, then Sagan's book on the representations of The Symmetric Group is a popular choice.


TITLE: How to calculate the area of a 3D triangle?
QUESTION [55 upvotes]: I have coordinates of 3d triangle and I need to calculate its area. I know how to do it in 2D, but don't know how to calculate area in 3d. I have developed data as follows.
(119.91227722167969, 122.7717056274414, 39.3568115234375),
(119.8951187133789, 122.7717056274414, 39.38057327270508),
(121.11941528320312, 123.2818832397461, 38.41301345825195)

REPLY [0 votes]: If a,b,c are the position vectors, use
1/2abs(a×b+b×c+c×a).
Take the magnitude of the area vector to get the solution.


TITLE: Calculus based or calculus-based?
QUESTION [0 upvotes]: I've noticed something strange in the literature. Regardless of context, there seems to be a trend where some authors will say 'calculus-based solution' or 'calculus-based physics' whereas others will not use the hyphen and say something like 'calculus based bound'. In my own writing, I've usually stuck to avoiding the hyphen, however I'm starting to have my doubts. Is one of these choices always correct? Or is there a relevant set of rules of grammar?

REPLY [0 votes]: Solved by caverac's link in the comments. https://www.grammarbook.com/punctuation/hyphens.asp suffix rule 1. Use the hyphen always.


TITLE: Why everything wants to come into its lowest state of potential energy and why there manybe more than one position of stability?
QUESTION [0 upvotes]: Suppose we have a solid hemisphere , one can observe that from both curve and flat surface is kept on ground it acquires a stable position on its own , why its does so ? And if its true then why not the flat or the curve one rotates/moves to transform into other by seeing which among $\frac{4 R}{3 \pi}$ and $R - \frac{4 R}{3 \pi}$ is minimum using the fact of minimum PE(Potential energy(. So only one stable position ? R is the radius of hemisphere and 4R/3pi is the centre of mass location of the hemisphere from flat surface from centre . By move/rotate i mean to be able to go to the lowest energy configuration out of Mg(4R/3pi) or Mg(3pi-4)R/3pi by itself

REPLY [2 votes]: It is very common to have a situation where the potential energy as a function of some parameter has more than one minimum. In this case there are various local minima and there is usually just one global minimum (the lowest of all the local minima), but it can happen that there are several minima which have the same energy and are lower than all the others.
First all of, we have energy conservation so a system of given energy cannot move just anywhere on its potential energy surface unless it is able to exchange energy with other systems. In the example of the hemisphere, in order to role from the state with the curved surface down to the state with the flat surface down, the system first needs to go to the half-way rolled-over state where it has higher potential energy. It will only be able to do that if another system pushes it or if it had enough kinetic energy in the first place. If it did have enough kinetic energy (e.g. in a rotational motion) then it could indeed flip over. It would flip over and what happens next depends on the interaction with the ground. A real hemisphere would then impact on the ground in such a way as to deliver some energy to the ground in the form of sound and heat, so the hemisphere loses energy and will eventually settle with the flat side down in this case.
More generally, the fact that systems tend to move towards minimum potential energy is related not just to energy but also to entropy. If a system is to settle at a place of lower total energy then, as I already said, it must be in interaction with other systems. Such an interaction will move energy between the systems, and in principle the energy could go either way, and oscillate between the systems. But in practice there is always some mechanism whereby entropy can increase, and as a result the joint system changes state irreversibly towards the higher entropy configuration. In the example of the hemisphere this could be an overall state where there is more entropy in the floor, for example. Or it could be that the floor is unchanged but the sphere itself absorbs some heat and thus has higher entropy and less rotational kinetic energy. In other examples it could be owing to giving off thermal radiation or something like that.


TITLE: Easiest way to verify that $4x^2+y^2=1$ is an ellipse?
QUESTION [1 upvotes]: Normally I would just divide both sides by the number $4$ because it's not good in there, but I can't do it for
$$4x^2+y^2=1$$
I must have $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$
So what's the easiest way?

REPLY [0 votes]: For general case, a quadratic equation of two variables $x$ & $y$ is given as
$$ax^2+2hxy+by^2+2gx+2fy+c=0$$
The above equation will represent an ellipse iff the following conditions are satisfied
$$\Delta=(abc+2fgh-af^2-bg^2-ch^2)\neq0$$
$h^2-ab<0$ & $a\neq b$
As per your question, we have
$a=4$, $h=0$, $b=1$, $g=0$, $f=0$ & $c=-1$
$$\Delta=(4)(1)(-1)=-4\neq0$$
& $ a\neq b$
Hence, the equation represents an ellipse.


TITLE: What is the justification and intuition behind Muller's method's quadratic equation?
QUESTION [4 upvotes]: Usually we write the quadratic formula like this:

$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$

But Muller's is written like this: Muller's method.
Why is that?

REPLY [8 votes]: Here the author is  using the Citardauq Formula $x=\frac{2c}{-b\mp \sqrt{b^2-4ac}}$. It can be useful in avoiding loss of significance in the computation of one of the roots.
We can obtain the Citardauq Formula by multiplying top and bottom of the Quadratic Formula by $-b\mp\sqrt{b^2-4ac}$.
Suppose for example that $b$ is fairly large positive, and $4ac$ is quite a bit smaller. Then when we compute the root $x=\frac{-b+\sqrt{b^2-4ac}}{2a}$, there may be a serious loss of significant digits in the computation of the numerator. There is no such loss if we use the equivalent $x=\frac{2c}{-b-\sqrt{b^2-4ac}}$.
Another way of looking at the formula is that if $ax^2+bx+c=0$, where $ac\ne 0$, then $\frac{1}{x}$ is a solution of $cu^2+bu+a=0$. Solve for $u$ using the Quadratic Formula, and take the reciprocal.


TITLE: Cauchy problem with specific value of starting point, find limit
QUESTION [0 upvotes]: I have the following Cauchy problem:
$$ y' = \vert y \vert - \arctan{e^x} $$
$$ y(0) = y_0 $$
I want to prove that there exists a specific value of $y_0$ such that:
$$ \lim_{x\to\infty} y(x) = \frac{\pi}{2}.$$
I am not too sure how to even start on this, beside saying that f(x,y) is Lipschizt and therefore I have local uniqueness and existence of solution which can be extended globally to $\mathbb{R}$. Any help?

REPLY [1 votes]: Let $y($ be a solution and suppose that $y(x)=0$ for some $x\ge0$. Then $y'(x)<0$. This implies that once a solution takes a negative value, it remains negative, Thus, the only way to solve the problem is to look for positive solutions. This means solving the linear equation $y'=y-\arctan(e^x)$. Its solution is
$$
y(x)=e^x\Bigl(y_0-\int_0^xe^{-t}\arctan(e^t)\,dt\Bigr).
$$
This will be positive if and only if
$$
y_0\ge \int_0^\infty e^{-t}\arctan(e^t)\,dt=a.
$$
If $y_0>a$, then $\lim_{x\to\infty}y(x)=\infty$. If $y_0=a$, then
$$
\lim_{x\to\infty}y(x)=\lim_{x\to\infty}e^x\int_x^\infty e^{-t}\arctan(e^t)\,dt=\frac\pi2.
$$
Note. Although it is not needed for the argument, the actual value of $a$ is
$$
\frac{\pi+\log4}{4}.
$$


TITLE: Finding the basis of a lattice
QUESTION [1 upvotes]: Let $B=(b_1,b_2)$ be the linearly independent vectors and generate a lattice $ L(B)=\{xb_1+yb_2:x,y \in \mathbb{Z}\}$. If any two linearly independent vectors, $b_1^\prime,b_2^\prime$ are taken from the lattice $L(B)$, then $L(B^\prime)$ need not be equal to $L(B^\prime)$. For example,
$b_1=[1,2],b_2=[1,-1]$ and generate the lattice $L(b_1,b_2)$
Take two lattice vectors $b_1^\prime=b_1+b_2,b_2^\prime=b_1-b_2$.
Clearly, these are linearly independent but it doesn't form a basis of lattice $L(b_1,b_2)$($b_1$ cannot be generated using $b_1^\prime,b_2^\prime$). This clearly indicates that any set of $n$($n=2$ in the example) independent lattice vectors may not be the basis of the lattice.
Given a lattice vectors, how can we generate a basis of the lattice?

REPLY [0 votes]: There is a general answer for modules over any commutative ring:

Let $L$ be a finitely generated free $R$-module with basis $\mathcal B=(b_1,\dots b_n)$, and $b'_1, \dots, b'_n$ be $ n$ vectors in $L$. Then  $b'_1, \dots, b'_n$ are a basis of $L$ if and only if
  $\;\det_\mathcal{B}(b'_1, \dots, b'_n)$ is a unit in $R$.

In the specific case, as a lattice is just a free $\mathbf Z$-module of rank $2$, this means
$$\det\nolimits_{\{b_1,b_2\}}(b'_1,b'_2)=\pm 1.$$


TITLE: Compute area of a sphere through a Dirac delta
QUESTION [6 upvotes]: I've been having issues with integrating with a Dirac delta. To compute the area of a sphere centered at $(0,0,0)$ it seems to work just fine:
$$\int_{0}^{2\pi}{\int_{0}^{\pi}{\int_{0}^{\infty}{\delta(r-\rho)r^2\sin\theta\, dr}\,d\theta}\,d\phi} = 4\pi\rho^2$$
Now I will take the same sphere but offset by $(0,0,\rho)$, that is: $x^2 + y^2 + (z-\rho)^2 = \rho^2$. Going to spherical coordinates yields: $r^2\cos^2\phi\sin^2\theta + r^2\sin^2\phi\sin^2\theta + (r\cos\theta-\rho)^2 = \rho^2$, which yields: $r(r-2\rho\cos\theta)=0$, and we can express the sphere in spherical coordinates as: $r(\theta) = 2\rho\cos\theta, \theta \in [0,\pi/2], \phi\in[0,2\pi]$.
Integrating yields:
$$\int_{0}^{2\pi}{\int_{0}^{\frac{\pi}{2}}{\int_{0}^{\infty}{\delta(r-2\rho\cos\theta)r^2\sin\theta\, dr}\,d\theta}\,d\phi} = \frac{8\pi\rho^2}{3}$$
Now this is not right clearly. The only reason I can think of has to do something with properties of the Dirac  delta I am unaware of. Note that I have not studied measure theory. I need the Dirac delta and not a surface integral  because I will be using this to compute transformations of probability density functions which I will need to write through a Dirac delta.
Edit: References covering this for engineers/computer science students are welcome.
Edit 2:
Taking into account David Holden's answer I came up with the following fact which must hold (I hope it's correct):
$$\int_{V}{\delta(f(x)) \,dx} = \int_{S = \{x|f(x) =0\}}{\,dA}$$
Edit 3:
I found some more information on the subject:
Impulse functions over curves and surfaces
Properties_in_n_dimensions
Surface area from indicator function
Property of Dirac delta function in $\mathbb{R}^n$
Does the coarea formula hold for delta-function?
I believe the issue was that whenever I offset the sphere the Dirac delta changed such that $\delta(f(r)) \rightarrow \delta(g(r,\theta))$ and $g$ was then a non-trivial mapping (so it's not the one-dimensional dirac delta I am used to anymore). Based on the first article I believe that I can rewrite it as a surface Dirac delta $\delta(g(r,\theta)) = \delta_S(r,\theta)$ which yields the surface integral giving a correct result. The other threads and wikipedia state that I should have a normalization by the magnitude of the gradient. I think I am missing an important piece since for the result to be correct this normalization factor should cancel out with something. More precisely:
$$\int_V{\delta(r-2\rho\cos\theta)r^2\sin\theta \,dr\,d\theta\,d\phi} = \int_S{\frac{\,d\sigma}{\sqrt{r^2+\rho^2-2r\rho\cos\theta}}}$$
The only idea I have is that somehow the normalization factor will pop out of the $\,d\sigma$. No idea though since it's supposed to be a 'Minkowski content measure' which is way over my head as a computer science student.
To add to this I would also like to be able to solve the same problem with a heaviside function (for integrating the volume of the offset ball). I am unsure whether similar considerations apply there, however if I integrate it, the result seems correct. I still want to make sure this is valid for other volumes also (maybe it's just a coincidence like the sphere centered at $(0,0,0)$). So I would be grateful if somebody with more knowledge on geometric measure theory could clarify all of the points.

REPLY [2 votes]: To summarize the discussion in the comments, the definition of $\delta(f)$ is derived from postulating two basic properties: the substitution rule
$$\int_{\mathbb R^n} \delta(f(\mathbf x)) \,\phi(\mathbf x) \,d\mathbf x =
\int_U \delta(f(\mathbf x(\mathbf u))) \,\phi(\mathbf x(\mathbf u))
 \left| \det D \mathbf x(\mathbf u) \right| d\mathbf u$$
and
$$\int_{\mathbb R^n} \delta(x_1) \,\phi(\mathbf x) \,d\mathbf x =
\int_{\mathbb R^{n - 1}} \phi(\mathbf x) \rvert_{x_1 = 0} \,dx_2 \cdots dx_n.$$
If you try to set
$$\small \int \delta(f(\mathbf x)) d\mathbf x =
\int_{f(\mathbf x) = 0} dS =
\int_{2 f(\mathbf x) = 0} dS =
\int \delta(2 f(\mathbf x)) d\mathbf x,$$
you violate the first rule. If you try to set
$$\small \iint \delta(r - f(\theta)) \phi(r, \theta) dr d\theta \neq
\int \phi(f(\theta), \theta) d\theta,$$
you violate the second rule. If you adopt the definition that the rest of the world is using derived from the two stated properties, you get the identity
$$\int_{\mathbb R^n} \delta(f(\mathbf x))
 \left| \nabla f(\mathbf x) \right| \phi(\mathbf x) \,d\mathbf x =
\int_{f(\mathbf x) = 0} \phi(\mathbf x) \,dS(\mathbf x),$$
which is formally the same as the coarea formula because both are essentially the same change of variables formula. The first two formulas in your question will in fact be correct, while the last two will be incorrect.


text:
\section{Biconditional Equivalent to Biconditional of Negations/Formulation 2}
Tags: Biconditional Equivalent to Biconditional of Negations

\begin{theorem}
: $\vdash \left({p \iff q}\right) \iff \left({\neg p \iff \neg q}\right)$
\end{theorem}

config: wiki