text:
\begin{document}

\title{Prime-universal diagonal quadratic forms}

\author[Jangwon Ju and et tal] {Jangwon Ju, Daejun Kim, Kyoungmin Kim, Mingyu Kim, and Byeong-Kweon Oh}

\address{Department of Mathematics, University of Ulsan, Ulsan, 44610, Republic of Korea}
\email{jangwonju@ulsan.ac.kr}
\thanks{This work of the first author was supported by the National Research Foundation of Korea(NRF) grant funded by the Korea government(MSIT) (NRF-2019R1F1A1064037).}

\address{Research Institute of Mathematics, Seoul National University, Seoul 08826, Korea}
\email{goodkdj@snu.ac.kr}

\address{Department of Mathematics, Sungkyunkwan University, Suwon 16419, korea}
\email{kiny30@skku.edu}
\thanks{This work of the third author was supported by the National Research Foundation of Korea(NRF) grant funded by the Korea government(MSIT) (NRF-2016R1A5A1008055  and NRF-2018R1C1B6007778)}

\address{Department of Mathematics, Sungkyunkwan University, Suwon 16419, korea}
\email{kmg2562@skku.edu}
\thanks{This work of the fourth author was supported by the National Research Foundation of Korea (NRF-2019R1A6A3A01096245).}

\address{Department of Mathematical Sciences and Research Institute of Mathematics, Seoul National University, Seoul 08826, Korea}
\email{bkoh@snu.ac.kr}
\thanks{This work of the second and fifth authors were supported by the National Research Foundation of Korea (NRF-2019R1A2C1086347).}

\subjclass[2010]{Primary 11E12, 11E20}

\keywords{Diagonal quaternary quadratic forms, Prime-universal}

\begin{abstract}  A (positive definite and integral) quadratic form  is said to be {\it prime-universal} if it represents all primes.
Recently, Doyle and Williams in \cite{dw} classified all prime-universal  diagonal  ternary quadratic forms, and all prime-universal  diagonal  quaternary quadratic forms under two conjectures proposed by themselves.
In this article, we classify all  prime-universal diagonal quadratic forms regardless of ranks. Furthermore, we prove, so called, $67$-Theorem for a diagonal quadratic form to be prime-universal.
\end{abstract}

\maketitle

\section{Introduction}

A positive definite integral quadratic form
$$
f(x_1,x_2,\dots,x_n)=\sum_{i,j=1}^n f_{ij}x_ix_j \quad (f_{ij}=f_{ji} \in \z)
$$
is said to be {\it prime-universal} if for any prime $p$, the diophantine equation
$$
f(x_1,x_2,\dots, x_n)=p
$$
has an integer solution $(x_1,x_2,\dots,x_n) \in \z^n$.  There are many examples of prime-universal quadratic forms. For example, any quadratic form that represents all positive integers, which we call  a {\it  universal} quadratic form, is also prime-universal.  Hence by 15-Theorem  given by Conway and Schneeberger (for this, see \cite{b}), any quadratic form that represents all integers less than or equal to $15$ is universal, which is also prime-universal.  On the other hand, there are many prime-universal quadratic forms that are not universal.

Recently,  Doyle and Williams \cite{dw}  classified all  prime-universal diagonal ternary quadratic forms, which are, in fact, not universal. Furthermore, they classified all prime-universal diagonal quaternary quadratic forms under the two conjectures proposed by themselves.

In this article, we classify all  prime-universal diagonal  quadratic forms without any assumption. Furthermore, we prove that  a diagonal quadratic form is prime-universal if and only if it represents
$$
  2, \ 3, \ 5, \ 7, \ 13, \ 17, \ 23, \ 41, \ 43, \ \text{and} \ \ 67.
$$

Let $f=\sum f_{ij} x_ix_j$ be a (positive definite integral) quadratic form.  The symmetric matrix corresponding to $f$ is defined by $M_f=(f_{ij})$. If $f$ is diagonal, then we simply write
$$
f=\df{f_{11},f_{22},\dots,f_{nn}}.
$$
Note that the symmetric matrix corresponding to a diagonal form is a diagonal matrix.  For a positive integer $a$,  we say $a$ is represented by $f$, denoted by $a \ra f$, if the diophantine equation $f(x_1,\dots,x_n)=a$ has an integer solution $(x_1,\dots,x_n) \in \z^n$.  For a prime $p$, if it has a solution $(x_1,\dots,x_n) \in \z_p$, where $\z_p$ is the $p$-adic integer ring, then we say $a$ is represented by $f$ over $\z_p$. Furthermore, if $a$ is represented by $f$ over $\z_p$  for any prime $p$, then we say $a$ is {\it locally} represented by $f$.
The set of all integers that are represented by $f$ will be denoted by $Q(f)$.
The genus of $f$, denoted by $\gen(f)$, is the set of quadratic forms that are isometric to $f$ over  $\z_p$ for any prime $p$. It is well known that an integer $a$ is locally represented by $f$ if and only if there is a quadratic form $g \in \gen(f)$ that represents $a$.
The set of integers that are locally represented by $f$ will be denoted by $Q(\gen(f))$.

Any unexplained notations and terminologies can be found in \cite{ki} or \cite{om}.

\section{Prime-universal diagonal quaternary quadratic forms}

In this section, we classify all prime-universal diagonal quaternary quadratic forms.   In \cite{dw}, the authors classified all prime-universal diagonal quaternary quadratic forms except for the following $27$ forms:

\begin{equation}\label{candi}
\begin{array}{llll}
\df{2,3,4,5}, &\df{2,3,4,11}, & \df{2,3,5,5}, & \df{2,3,5,11}, \\
\df{2,3,5,13},&\df{2,3,5,14}, & \df{2,3,5,16},& \df{2,3,5,17}, \\
\df{2,3,5,h} & \multicolumn{3}{l}{\text{for $h=20,\dots,27,29,\dots,33,35,36,38,40,41,43$.}}
\end{array}
\end{equation}
Hence it suffices to prove the prime-universalities  of these  $27$ diagonal forms.

In \cite{poly}, \cite{regular}, and \cite{pentagonal}, we developed a method that determines whether or not integers in an arithmetic progression are represented by some particular ternary quadratic form.
We briefly introduce this method for those who are unfamiliar with it.
Let $d$ be a positive integer and let $a$ be a nonnegative integer $(a\leq d)$.
We define
$$
S_{d,a}=\{dn+a \mid n \in \mathbb N \cup \{0\}\}.
$$
For two quadratic forms $f$ and $g$, we define
$$
R(g,d,a)=\{v \in (\mathbb{Z}/d\mathbb{Z})^3 \mid vM_gv^t\equiv a \ (\text{mod }d) \}
$$
and
$$
R(f,g,d)=\{T\in M_3(\mathbb{Z}) \mid  T^tM_fT=d^2M_g \}.
$$
A coset (or, a vector in the coset) $v \in R(g,d,a)$ is said to be {\it good} with respect to $f,g,d, \text{ and }a$ if there is a $T\in R(f,g,d)$ such that $\frac1d \cdot vT^t \in \mathbb{Z}^3$.
The set of all good vectors in $R(g,d,a)$ is denoted by $R_f(g,d,a)$.
If  $R(g,d,a)=R_f(g,d,a)$, we write
$$
g\prec_{d,a} f.
$$
If $g\prec_{d,a} f$, then by Lemma 2.2 of \cite{regular},  we have
\begin{equation}\label{good}
S_{d,a}\cap Q(g) \subset Q(f).
\end{equation}
In particular, if $g\prec_{d,a} f$ for any $g \in \gen(f)$, then we have $Q(\gen(f)) \cap S_{d,a} \subset Q(f)$.

 A MAPLE based  computer program for computing the above sets  is available upon request to the authors.

\begin{thm} \label{start} Both quaternary quadratic forms $f=\langle 2,3,4,5\rangle$ and $\langle2,3,4,11\rangle$  are prime-universal.
\end{thm}

\begin{proof}   Since the proofs are quite similar to each other, we only provide the proof of the former case.

Note that the class number of $g=\langle 2,3,4\rangle$ is $2$, and its genus mate is $\df{1,2,12}$. If an even integer is represented by $\df{1,2,12}$, then it is also represented by $\df{4,2,12}$, which is a subform of $g$. Therefore $Q(\gen(g)) \cap 2\z \subset Q(g)$, and consequently  $g$
represents all nonnegative even integers except for those integers of the form $2^{2s+1}(8t+5)$ for some nonnegative integers $s$ and $t$.

Since one may directly check that $p \ra f$ for any prime $p$ such that $p\le 113$, we assume that $p\ge 127$.
Then, one may easily check that
$$
p-5\cdot d^2 \ra g,
$$
where
$$
d=\begin{cases}  1 \qquad &\text{if $p \equiv 1,3 \Mod 8$,  $p \equiv 7\Mod {16}$, \text{or} \ $p \equiv 21\Mod {64}$}, \\
                           5 \qquad &\text{if $p \equiv 13, 45  \Mod {64}$,}\\
                           3 \qquad &\text{otherwise.}
                           \end{cases}
$$
This completes the proof.
\end{proof}

\begin{thm}\label{thm235alpha}
For an integer $\alpha$ that is not divisible by $7$ with $5\leq \alpha\leq 43$,  the quaternary quadratic form $f=\df{2,3,5,\alpha}$ listed in \eqref{candi} is prime-universal.
\end{thm}

\begin{proof}
Note that the class number of $g=\df{2,3,5}$ is $2$, and its genus mate is $g_1=\df{1,1,30}$.
One may easily check that
$$
g_1\prec_{7,0}g,\quad g_1\prec_{7,3}g,\quad g_1\prec_{7,5}g,\quad\text{and}\quad g_1\prec_{7,6}g.
$$
Therefore, for any $a\in\{0,3,5,6\}$, $Q(\text{gen}(g))\cap S_{7,a} \subset Q(g)$.
Consequently, $g$ represents all nonnegative integers that are not divisible by $3$, and are congruent to $0,3,5$, or $6$ modulo $7$.

Since one may directly check that $p \ra f$ for any prime $p$ such that $p\le 36\alpha$, we assume that $p> 36\alpha$.
If $p$ is congruent to $3,5$, or $6$ modulo $7$, then $p$ is represented by $g$.
Assume that $p$ is congruent to $1,2$, or $4$ modulo $7$.
Then, one may easily check that
$$
p-\alpha\cdot3^2\cdot d^2\ra g,
$$
where
$$
d=\begin{cases} 1 \qquad &\text{if $p-2\alpha\equiv0,3,5$, or $6 \Mod 7$},\\
2 \qquad &\text{otherwise}.
\end{cases}
$$
This completes the proof.   \end{proof}

\begin{thm} \label{thm23514} The quaternary quadratic form $f=\df{2,3,5,14}$ is prime-universal.
\end{thm}

\begin{proof}
Note that the class number of $g=\df{2,3,14}$ is two and its genus mate is
$$
g_1=\begin{pmatrix} 1&0&0\\0&10&4\\0&4&10\end{pmatrix}.
$$
One may easily check that
$$
g_1\prec_{8,3}g,\quad g_1\prec_{32,12}g\quad \text{and}\quad g_1\prec_{32,16}g.
$$
Thus if we put
$$
A=\left\{ m\in \n : m\equiv 3\Mod 8 \ \ \text{or}\ \ m\equiv 12\Mod {32}\ \ \text{or}\ \ m\equiv 16\Mod {32} \right\},
$$
then we have $Q(\gen(g))\cap A\subset Q(g)$.
Consequently, $g$ represents all integers in $A$ except for those of the form $3^{2s+1}(3t+2)$ for some nonnegative integers $s$ and $t$.
One may check directly that $p\ra g$ for any prime $p<2207$. Thus we assume that $p$ is a prime greater than or equal to  $2207$.
If we define
$$
d=\begin{cases} 0&\ \text{if}\ p\equiv 3\Mod 8,\\
                           1&\ \text{if}\ p\equiv 25\ \text{or}\ 29\Mod {32},\\
                           2&\ \text{if}\ p\equiv 7\Mod 8,\\
                           3&\ \text{if}\ p\equiv 1\ \text{or}\ 5\Mod {32},\\
                           5&\ \text{if}\ p\equiv 17\ \text{or}\ 21\Mod {32},\\
                           7&\ \text{if}\ p\equiv 9\ \text{or}\ 13\Mod {32},
                           \end{cases}
$$
then one may easily check that
$$
n=p-5\cdot 3^2\cdot d^2\in A \quad \text{and} \quad n \not \equiv 0 \Mod 3.
$$
Thus we have $n\ra g$, which completes the proof.
\end{proof}

\begin{thm} \label{thm23521} The quaternary quadratic form $f=\df{2,3,5,21}$ is prime-universal.
\end{thm}

\begin{proof}
One may check directly that $f$ represents  all primes less than $88211$.
Thus we assume  that $p$ is a prime greater than or equal to $88211$.
Note that the class number of $g=\df{3,5,21}$ is three and its genus mates are
$$
g_1=\begin{pmatrix} 5&0&0\\0&6&3\\0&3&12\end{pmatrix} \ \ \text{and}\ \ g_2=\begin{pmatrix} 2&1&0\\1&8&0\\0&0&21\end{pmatrix}.
$$
One may easily show that any positive integer $m$ which is congruent to 2 modulo 3 and relatively prime to 35 is represented by the genus of $g$.
If we define
$$
d=\begin{cases} 2&\ \text{if}\ p\equiv 1\Mod {12},\\
                           6&\ \text{if}\ p\equiv 5\Mod {12},\\
                           1&\ \text{if}\ p\equiv 7\Mod {12},\\
                           3&\ \text{if}\ p\equiv 11\Mod {12}, \end{cases}
$$
and
$$
n=p-2\cdot 35^2\cdot d^2,
$$
then one may easily check that  $n$ is a positive integer congruent to $5$ modulo $12$ and is relatively prime to $35$.
Thus we have
$$
n\equiv 1\Mod 4\quad  \text{and}  \quad n\ra \gen(g).
$$
Hence the integer $n$ is represented by a quadratic form in the genus of $g$.
Suppose that $n\ra g_1$. Then there is a vector $(x,y,z)\in \z^3$ such that
$$
n=5x^2+6y^2+12z^2+6yz.
$$
Since $n\equiv 1\Mod 4$, we have either $y\equiv 0\Mod 2$ or $yz\equiv 1\Mod 2$.
If $y=2y'$ for some $y'\in \z$, then we have
$$
\begin{array}{rl}
n\!\!\!\!\!&=5x^2+24y'^2+12z^2+12y'z \\[0.4em]
&=3(y'+2z)^2+5x^2+21y'^2
\end{array}
$$
which means that $n\ra g$.
If $yz\equiv 1\Mod 2$, then we may write $y=z-2y'$.
Then we have
$$
\begin{array}{rl}
n\!\!\!\!\!&=5x^2+6(z-2y')^2+12z^2+6(z-2y')z \\[0.4em]
&=5x^2+24y'^2+24z^2-36y'z \\[0.4em]
&=3(y'+z)^2+5y^2+21(y'-z)^2
\end{array}
$$
and thus $n\ra g$.
Similarly, one may show that the supposition $n\ra g_2$ also implies that $n\ra g$.
This completes the proof.
\end{proof}

\begin{thm} \label{thm23535} The quaternary quadratic form $f=\df{2,3,5,35}$ is prime-universal.
\end{thm}

\begin{proof}
One may directly check that $f$ represents all primes less than $596243$.
Thus we assume that $p$ is a prime greater than or equal to $596243$.
Note that the class number of $g$ is three and its genus mates are
$$
g_1= \begin{pmatrix} 3&0&0\\0&10&5\\0&5&20\end{pmatrix}
\ \ \text{and}\ \ g_2=\begin{pmatrix} 2&1&0\\1&8&0\\0&0&35\end{pmatrix}.
$$
One may easily show that any positive integer $m$ which is congruent to 2 or 3 modulo 5, and is relatively prime to 21 is represented by the genus of $g$.
If we define
$$
d=\begin{cases} 13&\ \text{if}\ p\equiv 1\Mod {20},\\
                           10&\ \text{if}\ p\equiv 3\ \text{or}\ 7\Mod {20},\\
                           1&\ \text{if}\ p\equiv 9\Mod {20},\\
                           2&\ \text{if}\ p\equiv 11\Mod {20},\\
                           5&\ \text{if}\ p\equiv 13\ \text{or}\ 17\Mod {20},\\
                           26&\ \text{if}\ p\equiv 19\Mod {20},\\
                           \end{cases}  \quad \text{and} \quad n=p-2\cdot 21^2\cdot d^2,
$$
 then one may easily check that
$$
n\equiv 3\Mod 4,\ \ n\ra \gen(g).
$$
For each $i=1,2$, one may easily show that any positive integer $m$ represented by $g_i$ which is congruent to 3 modulo 4 is also represented by $g$ as in the proof of Theorem \ref{thm23521}.
Therefore we have $n\ra g$, which implies that $p\ra f$.
\end{proof}

\section{Higher dimensional prime-universal diagonal quadratic forms}

 A prime-universal diagonal quadratic form  is called {\it proper} if there does not exist a prime-universal diagonal subform  whose rank is less than the rank of it.  Note that to find all prime-universal diagonal quadratic forms, it suffices to find all proper ones.  In this section, we classify all proper prime-universal diagonal quadratic forms with ranks greater than $4$.

If a diagonal quadratic form $f$ fails to represent a prime, we call the smallest prime not represented by $f$ its {\em prime truant}.

According to the results in \cite{dw}, there are no prime-universal diagonal binary quadratic forms, and there are five proper prime-universal diagonal ternary quadratic forms, namely,
$$
\df{1,1,2}, \quad \df{1,1,3}, \quad \df{1,2,3}, \quad \df{1,2,4}, \quad \text{and} \quad \df{1,2,5}.
$$
Moreover, as proved in the previous section,  all prime-universal diagonal quaternary quadratic forms with minimum greater than $1$ are listed in Table 1 of \cite{dw}.
Therefore, it suffices to classify all proper prime-universal diagonal quadratic forms with rank greater than or equal to $5$, which are, in fact, listed in Table \ref{HigherCandidate}.

\begin{table}[ht]
\caption{Proper prime-universal forms $\langle a_1,a_2,\dots,a_k\rangle \ \ (5\le k\le6)$}
\label{HigherCandidate}
	\begin{center}
		\begin{tabular}{ccccccc}
			\Xhline{1pt}
			$a_1$&$a_2$&$a_3$&$a_4$&$a_5$&$a_6$& Conditions on $a_k  \ (5\le k\le6)$ \\
			\hline

			$2$&$2$&$2$&$2$&$3$&& \\
			$2$&$2$&$2$&$3$&$a_5$&& $a_5 = 8,11,17$\\

			\hline

			$2$&$2$&$3$&$8$&$17$&& \\
			$2$&$2$&$3$&$11$&$17$&& \\
			$2$&$2$&$3$&$16$&$17$&& \\
			$2$&$2$&$3$&$17$&$a_5$&& $17\le a_5 \le 41$, $a_5\neq 26,32,35,40$\\

			\hline

			$2$&$3$&$3$&$3$&$4$&& \\
			$2$&$3$&$3$&$4$&$a_5$&& $a_5 = 4,6,10,13$\\

			\hline
			$2$&$3$&$4$&$4$&$a_5$&& $4\le a_5 \le 17$, $a_5\neq 5,8,9,11,16$\\
			$2$&$3$&$4$&$6$&$a_5$&& $6\le a_5 \le 23$, $a_5\neq 8,9,11,22$\\
			$2$&$3$&$4$&$7$&$a_5$&& $7 \le a_5 \le 17$, $a_5\neq 8,9,11,16$ \\
			$2$&$3$&$4$&$10$&$a_5$&& $10\le a_5 \le 23$, $a_5\neq 11,22$\\
			$2$&$3$&$4$&$12$&$13$&& \\
			$2$&$3$&$4$&$13$&$a_5$&& $13 \le a_5 \le 23$, $a_5\neq 13,22$\\

			\hline

			$2$&$3$&$5$&$7$&$a_5$&& $a_5=7,19,28,34$ \\
			$2$&$3$&$5$&$19$&$19$&&  \\

			\hline
			$2$&$3$&$6$&$6$&$7$&&  \\
			$2$&$3$&$6$&$7$&$a_5$&& $7\le a_5 \le 23$, $a_5\neq 8,19,20,22$  \\
			$2$&$3$&$6$&$7$&$19$&$20$& \\
			$2$&$3$&$6$&$7$&$20$&$a_6$& $20 \le a_6 \le 67$, $a_6\neq 21,23,63,66$ \\

			\hline

			$2$&$3$&$7$&$7$&$a_5$&& $7\le a_5 \le 13$, $a_5\neq 7,8,9,10,12$ \\
			$2$&$3$&$7$&$7$&$7$&$10$&  \\
			$2$&$3$&$7$&$9$&$a_5$&& $9\le a_5 \le 13$, $a_5\neq 9,12$ \\
			$2$&$3$&$7$&$10$&$a_5$&& $10\le a_5 \le 23$, $a_5\neq 17,19,22$ \\
			$2$&$3$&$7$&$11$&$a_5$&& $11\le a_5 \le 17$, $a_5\neq 11,13,16$ \\
			$2$&$3$&$7$&$12$&$13$&& \\
			$2$&$3$&$7$&$13$&$a_5$&& $13\le a_5 \le 17$, $a_5\neq 13,16$ \\
			\Xhline{1pt}
		\end{tabular}
	\end{center}
\end{table}

\begin{thm}\label{HigherproperPU}
	Every diagonal quadratic form in Table \ref{HigherCandidate} is, in fact, proper prime-universal.
\end{thm}

We will prove Theorem \ref{HigherproperPU} at the end of this section.
Assuming that Theorem \ref{HigherproperPU} is proved, we first prove the following theorem, which guarantees that Table \ref{HigherCandidate} is the complete list of proper prime-universal diagonal quadratic forms of rank greater than or equal to $5$.

\begin{thm}\label{ThmHigherescalation}
	Let $f$ be a proper prime-universal diagonal quadratic form with rank greater than or equal to $5$. Then $f$ is isometric to one of the quadratic forms listed in Table \ref{HigherCandidate}.
\end{thm}
\begin{proof}
	Let  $f=\langle a_1,a_2,\cdots,a_n \rangle$ be a proper prime-universal diagonal quadratic form, where $a_i$'s are all integers such that $1\le a_1\le \dots \le  a_n$.
	 For the sake of convenience, we define $f_i=\df{a_1,\ldots,a_i}$  for each $i$ with $1\le i\le n$.
Since we are assuming that $f$ is proper, $f_4$ should not be prime-universal.
	Following the argument used in the proof of Theorem 1.3 of \cite{dw}, one may easily obtain all candidates of $f_4$.
	After removing those forms that are already prime-universal among all possible candidates,  $f_4$ is isometric to one of the followings:
	$$
	\begin{array}{llll}
	\text{\bf (Case 1)}& \df{2,2,2,a_4} && \text{where  $a_4=2,3$,}\\
	\text{\bf (Case 2)}&\df{2,2,3,a_4} && \text{where  $a_4=8,11,16,17$,}\\
	\text{\bf (Case 3)}&\df{2,3,3,a_4} && \text{where  $a_4=3,4,6$,}\\
	\text{\bf (Case 4)}&\df{2,3,4,a_4} && \text{where  $a_4=4,6,7,10,12,13$,}\\
	\text{\bf (Case 5)}&\df{2,3,5,a_4} && \text{where  $a_4=7,19,28,34,37,39,42$,}
	\end{array}
	$$
	and
	$$
	\hspace{-0.65cm}\begin{array}{llll}
	\text{\bf (Case 6)}&\df{2,3,6,a_4} && \text{where  $a_4=6,7$,}\\
	\text{\bf (Case 7)}&\df{2,3,7,a_4} && \text{where  $a_4=7,9,10,11,12,13$.}
	\end{array}
	$$

Note that any diagonal form $\df{a_1,a_2,a_3,a_4}$ listed in the above is not prime-universal.
	Although what we have to check is different case by case, we repeat the following process, starting from $i=5$:

	Let $p$ be the prime truant of $f_{i-1}=\df{a_1,\ldots,a_{i-1}}$. Since $f$ is prime-universal, it should represent $p$. Hence we have $a_{i-1}\le a_i \le p$.
	For each integer $a$ with $a_{i-1}\le a\le p$,  it is possible that
	\begin{enumerate}[label={\rm(\arabic*)}]
		\item [(I)] The form $g_i=g_{i,a}:=\df{a_1,\ldots,a_{i-1},a}$ is prime-universal, and either
		\begin{enumerate}
			\item[(I-1)] $g_i$ is proper, or
			\item[(I-2)] $g_i$ is not proper.
		\end{enumerate}
		\item [(II)] The form $g_i$ is not prime-universal, and either
		\begin{enumerate}
			\item[(II-1)] $g_i$ fails to represent another prime $p'$ greater than $p$, or
			\item[(II-2)] $g_i$ still fails to represent the prime $p$.
		\end{enumerate}
	\end{enumerate}
	Note that any prime-universal form  contain a proper prime-universal form as its subform, one may use Table \ref{HigherCandidate} given above and Table 1 of \cite{dw} to check the prime-universality of $g_i$ as well as its properness.

	If (I-1) happens, then one may easily check that $g_i$ is listed in Table  \ref{HigherCandidate} and hence the theorem is proved. If (I-2) happens, then one may easily find a proper subform of $g_i$ which is listed in Table  \ref{HigherCandidate}.

	If (II) happens, we increase $i$ by $1$ and repeat the process.
	In particular, if (II-2) happens,  then one might come across the following situation:
	any diagonal quadratic form $\df{a_1,\ldots,a_{i-1},a,\ldots,a}$ fails to represent $p$, no matter how many times the integer $a$ is repeated.
	Even if it happens, the equation
	$$
	a_1x_1^2+\dots+a_{i-1}x_{i-1}^2+a_{i}x_{i}^2+\dots+a_nx_n^2=p
	$$
	has an integer solution, for $f$ represents the prime $p$. Therefore there is an integer $b$ with $a<b \le p$ such that $\df{a_1,\dots,a_i,a,b}$ represents $p$.  Now, we repeat the process from the diagonal quadratic form
	$$
	\df{a_1,\dots,a_i,\overbrace{a,\dots,a}^{\text{$s$-times}},b}
	$$
	 for each $s\ge 1$.  From (Case 1) to (Case 7) given above, one may check that this process terminates in a finite step. Since the proofs of all cases are quite similar to each other, we provide the proofs of (Case 2) and (Case 6).
\vskip 0.3cm

	\noindent {\bf (Case 2)} $f_4=\df{2,2,3,a_4}$ with $a_4=8,11,16,17$.

	First, we consider the case when $f_4=\df{2,2,3,8}$.
	Since the prime truant of $f_4$ is $17$, we have $8\le a_5 \le 17$.
	For each $a=9,10,12,13,14,15$, the form $\df{2,2,3,a}$ is already prime-universal, and hence $f$ is not proper if $a_i=a$ for some $i\ge 5$.
	Therefore, we may assume that
	$$
	a_i \in \{8,11,16,17\} \quad \text{or} \quad a_i \ge 18
	$$
	for any $i\ge 5$. If $a_i \ne 17$ for any $i \ge 5$, then the diagonal quadratic form $\df{2,2,3,8,a_5\dots,a_n}$ does not represent $17$. Therefore $a_i=17$ for some $i$.
	Since $\df{2,2,3,8,17}$ is prime-universal, the only proper prime-universal diagonal quadratic form $f$ with $f_4=\df{2,2,3,8}$ is $\df{2,2,3,8,17}$.
	One may similarly prove the case when $a_4=11,16$.

	Next, we consider the case when $f_4=\df{2,2,3,17}$. Note that the prime truant of $\df{2,2,3,17}$ is $41$, and hence we have $17\le a_5 \le 41$.
	Since $\df{2,2,3,17}$ does not represent $1,6,9,15,41$, $f$ represents $41$ if and only if
$$
a_i\in \mathcal A= \{17\le a \le 41 : a\neq 26,32,35,40\}
$$
	 for some $i\ge 5$. Furthermore, one may easily check by using Table \ref{HigherCandidate} that any quadratic form $\df{2,2,3,17,a_5}$ with $a_5 \in \mathcal A$ is proper prime-universal.
\vskip 0.3cm

	\noindent {\bf (Case 6)} $f_4=\df{2,3,6,a_4}$ with $a_4=6,7$.

	First, we consider the case when $f_4=\df{2,3,6,6}$.
	Note that the prime truant of $f_4$ is $7$, and $f$ can represent $7$ if and only if $a_i=7$ for some $i\ge5$. Also, note that $\df{2,3,6,6,7}$ is prime-universal, whereas $\df{2,3,6,7}$ does not represent $23$.

	Next, we consider the case when $f_4=\df{2,3,6,7}$.
	Since the prime truant of $f_4$ is $23$, we have $7\le a_5\le 23$.
	Note that every form $\df{2,3,6,7,a_5}$ with $7\le a_5\le 23$ and $a_5\neq 19,20,22$ is proper prime-universal.
	On the other hand, the prime truants of $\df{2,3,6,7,19}$, $\df{2,3,6,7,20}$, and $\df{2,3,6,7,22}$ are $23$, $67$, and $23$, respectively.

	Now, assume that $a_5=19$. If $a_i\in \{21,23\}$ for some $i\ge 6$, then $f$ is prime-universal but not proper. Since $\df{2,3,6,7,19}$ does not represent integers $1,4,23$, $f$ can represent $23$ if and only if $a_i=20$ for some $i\ge 6$. Note that $\df{2,3,6,7,19,20}$ is proper prime-universal.
	Therefore, the only proper prime-universal diagonal quadratic form $f$ with $f_5=\df{2,3,6,7,19}$ is $\df{2,3,6,7,19,20}$.

	Next, assume that $a_5=20$. Since the prime truant of $\df{2,3,6,7,20}$ is $67$, we have $20\le a_6 \le 67$. Note that $\df{2,3,6,7,20,a_6}$ with  $20\le a_6 \le 67$ and $a_6\neq 63,66$ is proper  prime-universal, and $f=\df{2,3,6,7,20,\{63 \text{ or } 66\},a_7,\cdots}$ can represent $67$ if and only if $a_i=64,65,67$ or $a_i=67$, respectively, for some $i\ge 7$. However, since $\df{2,3,6,7,20,a}$ with $a=64,65,$ or $67$ is already prime-universal, any prime-universal form $\df{2,3,6,7,20,\{63 \text{ or }66\},a_7,\cdots}$ is not proper.

	Finally, assume that $a_5=22$.  Since $f_5=\df{2,3,6,7,22}$ does not represent $23$, $a_i=23$ for some $i\ge 6$. Note that $\df{2,3,6,7,23}$ is proper prime-universal.
\end{proof}

\begin{thm}\label{thm222}
The quadratic form $f=\df{2,2,2,3}$ represents all primes but not $17$.
Hence, the quadratic forms $\df{2,2,2,2,3}$, and $\df{2,2,2,3,a_5}$ with $3 \le a_5 \le 17$ and $a_5\neq 16$ are prime-universal.
\end{thm}

\begin{proof}
Note that the class number of $g=\df{2,2,2}$ is $1$.
Therefore $Q(\gen(g))=Q(g)$, and consequently $g$ represents all nonnegative even integers except for those of the form $4^s(16t+14)$ for some nonnegative integers $s$ and $t$.

Since one may directly check that $p\ra f$ for any prime $p$ such that $p\le 73$ except for $17$, we assume that $p\ge 79$.
Then, one may easily check that
$$
p-3\cdot d^2 \ra g,
$$
where
$$
d=
\begin{cases}
5&\ \text{if}\ p\equiv 27 \Mod {32},\\
3&\ \text{if}\ p\equiv 1,3,17 \Mod {12},\\
1&\ \text{otherwise}.
\end{cases}
$$
Therefore, $p \ra f$.
Now, the proof of the last assertion is straight forward.
\end{proof}

\begin{thm}\label{thm223}
The quadratic form $f=\df{2,2,3,17}$ represents all primes but not $41$.
Hence, the quadratic forms $\df{2,2,3,a_4,17}$ with $a_4=8,11,16$, and $\df{2,2,3,17,a_5}$ with $17 \le a_5 \le 41$ and $a_5\neq 26,32,35,40$ are prime-universal.
\end{thm}

\begin{proof}
Note that the class number of $g=\df{2,2,3}$ is $1$.
Therefore $Q(\gen(g))=Q(g)$, and consequently $g$ represents all nonnegative integers except for those of the forms $8t+1$ and $9^s(9t+6)$ for some nonnegative integers $s$ and $t$.

Since one may directly check that $p\ra f$ for any prime $p$ such that $p\le 607$ except for $41$, we assume that $p\ge 613$.
Then, one may easily check that
$$
p-17\cdot 6^2 \ra g.
$$
Therefore, $p \ra f$.
Now, the proof of the last assertion is straight forward.
\end{proof}

\begin{thm}\label{thm233}
The quadratic form $f=\df{2,3,3,4}$ represents all primes but not $13$.
Hence, the quadratic forms $\df{2,3,3,3,4}$ and $\df{2,3,3,4,a_5}$ with $4 \le a_5 \le 13$ and $a_5\neq 12$ are prime-universal.
\end{thm}

\begin{proof}
Note that the class number of $g=\df{2,3,3}$ is $1$.
Therefore $Q(\gen(g))=Q(g)$, and consequently $g$ represents all nonnegative integers except for those of the form $9^s(3t+1)$ for some nonnegative integers $s$ and $t$.

Since one may directly check that $p\ra f$ for any prime $p$ such that $p\le 13$ except $13$, we assume that $p\ge 17$.
Then, one may easily check that
$$
p-4\cdot d^2 \ra g,
$$
where
$$
d=
\begin{cases}
1&\ \text{if}\ p\equiv 1,7 \Mod {9},\\
2&\ \text{if}\ p\equiv 4 \Mod {9}.\\
\end{cases}
$$
Therefore, $p \ra f$.
Now, the proof of the last assertion is straight forward.
\end{proof}

\begin{thm}\label{thm234}
The quadratic forms
$$
\begin{array}{llll}
\df{2,3,4,4},& \df{2,3,4,7} &&\text{represents all primes but not } 17,\\
\df{2,3,4,6},& \df{2,3,4,10},& \df{2,3,4,13}&\text{represents all primes but not } 23,\\
\df{2,3,4,12}&&&\text{represents all primes but not } 13.\\
\end{array}
$$
Hence, all quadratic forms of the form $\df{2,3,4,a_4,a_5}$ listed in Table \ref{HigherCandidate} are prime-universal.
\end{thm}

\begin{proof}
Let $f$ be one of the six quaternary forms given above.
Let
$$
\alpha=
\begin{cases}
7&\ \text{if}\ f=\df{2,3,4,7},\\
13&\ \text{if}\ f=\df{2,3,4,13},\\
3&\ \text{otherwise},\\
\end{cases}
$$
and let $g$ be the ternary sub-form such that $f=g \perp \df{\alpha}$. Note that
$g$ represents all even integers except for those of the form
$$
\begin{cases}{}
4^s(16t+10)						&\text{if}\ g=\df{2,3,4} \text{ or } \df{2,4,12}, \\
4^s(16t+14)						&\text{if}\ g=\df{2,4,4},\\
4^s(32t+20) 					&\text{if}\ g=\df{2,4,6},\\
25^s(50t+\{20 \text{ or } 30\}) &\text{if}\ g=\df{2,4,10}.
\end{cases}
$$

For the proof of the case when $g=\df{2,3,4}$, see the proof of Theorem \ref{start}, and for all the other cases, note that the class number of $g$ is $1$.

Since one may directly check that $p\ra f$ for any prime $p$ such that $p\le 317$ except for the only one prime given above, we assume that $p\ge 331$.
Then one may easily check that
$$
p-\alpha\cdot d^2 \ra g,
$$
where
$$
d=\begin{cases}
5 & \text{if } \ell=\df{2,4,10},\\
1,3, \text{ or }5 & \text{otherwise, defined according as the residue of $p$ modulo $32$},
\end{cases}
$$
Therefore, $p \ra f$.
Now, the proof of the last assertion is straight forward.
\end{proof}

\begin{thm}\label{thm236}
The quadratic form $f=\langle 2,3,6,7\rangle$ represents all primes but not $23, 47$, and $67$.
Hence, the quadratic forms
$$
\df{2,3,6,6,7}, \ \df{2,3,6,7,a_5},  \quad \text{and} \quad  \df{2,3,6,7,a_5,a_6}
$$
 listed in Table \ref{HigherCandidate} are prime-universal.
Moreover, the quadratic forms
$$
\begin{array}{lll}
\df{2,3,6,7,19}, & \df{2,3,6,7,22} &\text{ represents all primes but not } 23,\\
\df{2,3,6,7,20} && \text{ represents all primes but not } 67.
\end{array}
$$
\end{thm}
\begin{proof}
Note that the class number of $g=\df{2,3,6}$ is $1$.
Thus $Q(\gen(g))=Q(g)$, and consequently $g$ represents all nonnegative integers except for those of the forms $4^s(8t+7)$ and $9^s(3t+1)$ for some nonnegative integers $s$ and $t$.

Since one may directly check that $p\ra f$ for any prime $p$ such that $p\le 251$ except for $23,47,$ and $67$, we assume that $p\ge 257$.
Then, one may easily check that
$$
p-7\cdot d^2 \ra g,
$$
where
$$
d=\begin{cases}
0&\ \text{if}\ p\equiv 5,11,17 \Mod {24},\\
2&\ \text{if}\ p\equiv 1, 7, 13 \Mod {24},\\
4&\ \text{if}\ p\equiv 19 \Mod {24},\\
6&\ \text{if}\ p\equiv 23 \Mod {24}.\\
\end{cases}
$$
Therefore, $p \ra f$.
Now, the proof of the last assertion is straight forward.

\end{proof}

\begin{thm}\label{thm235-237}
The quadratic form $\df{2,3,5,19}$ represents all primes but not $43$. Hence, the quadratic form $\df{2,3,5,19,19}$ is prime-universal.
Moreover, the quadratic forms
$\df{2,3,5,7,a_5}$, $\df{2,3,7,a_4,a_5}$, and $\df{2,3,7,7,7,10}$ listed in Table \ref{HigherCandidate} are prime-universal.
\end{thm}

\begin{proof}
Since $19$ is not divisible by $7$, $\df{2,3,5,19}$ represents every prime $p\ge36\times 19=684$ (see the proof of Theorem \ref{thm235alpha}). One may easily check that $\df{2,3,5,19}$ represents all primes less than $684$ except  for $43$. Also, since $\df{2,3,5,19,19}$ represents $43$, it is prime-universal.

Now, we prove ``Moreover part" of the theorem. Let $f=\df{2,3,7,\alpha,\beta}$ be one of the quadratic forms that we should prove it is prime-universal. (for the quadratic form $\df{2,3,5,7,a_5}$, we let $\alpha=5$ and $\beta=a_5$, and we keep the senary quadratic form $\df{2,3,7,7,7,10}$ in mind for a moment).
Since one may easily check that $p\ra f$ for any prime $p$ such that $p\le 13121$, we assume that $p\ge 13127$.
Note that the class number of $g=\df{2,3,7}$ is $3$ and all isometric classes in the genus are
$$
g=\df{2,3,7},\ \ g_1=\df{1,3,14} \ \ \text{and}\ \
g_2= \begin{pmatrix} 2&1&1\\1&3&1\\ 1&1&9\end{pmatrix}.
$$
Now, one may easily check that
$$
g_1\prec_{3,0}g, \ \ \text{and} \ \ g_2\prec_{3,0}g.
$$
Therefore, $Q(\gen (g)) \cap 3\z \subset Q(g)$, and consequently $g$ represents all nonnegative odd integers that are multiple of $3$.
In the case when both $\alpha$ and $\beta$ are not divisible by $3$, then one may easily check that
$$
p-\alpha\cdot d^2-\beta \cdot e^2 \ra g
$$
for some integers $d,e\in\{0,1,2\}$.
Therefore, $p \ra f=\df{2,3,7,\alpha,\beta}$ in this case, and we may also have $p\ra \df{2,3,7,7,7,10}$.

Now, the remaining quadratic forms $f$ under consideration are
$$
\begin{array}{llll}
\df{2,3,7,9,\beta} &\text{with } \beta=11,13, & \df{2,3,7,10,\beta} &\text{with } \beta=12,15,18,21,\\
\df{2,3,7,11,\beta} &\text{with } \beta=12,15,&\df{2,3,7,12,13}, &  \df{2,3,7,13,15}.\\
\end{array}
$$
One may check by a computer program that
$$
g_1\prec_{30,r}g, \ \ \text{and} \ \ g_2\prec_{30,r}g
$$
for any integer $r$ in $R_{30}=\{0,2,3,6,8,9,10,12,15,18,20,21,22,24,27,28\}$.
Therefore, $Q(\gen(g))\cap \{m : m\equiv r \Mod{30} \text{ for some }r\in R_{30}\}\subset g$, and consequently $g$ represents all nonnegative integers in the set
$$
M_{30}=\{m \mid m \not \equiv 0,6,22,24 \Mod{32}, \ m \equiv r \Mod{30} \text{ for some } r\in R_{30}\}.
$$
Using this, for each $\gamma=9,11,13,21$, one may check that
$$
p-\gamma\cdot d^2 \ra g
$$
for some $0\le d \le 25$, defined according as the residue of $p$ modulo $32\cdot 15$.
Therefore, $p \ra \df{2,3,7,\gamma}$.
This proves the prime-universality of $f$ except for those with $\alpha=10$ and $\beta=12,15,18$, namely,
$$
\df{2,3,7,10,12},\ \ \df{2,3,7,10,15},\ \ \text{and} \ \ \df{2,3,7,10,18}.
$$
To prove in these cases, one may check by a computer program that
$$
g_1\prec_{42,r}g, \ \ \text{and} \ \ g_2\prec_{42,r}g
$$
for any $r$ in $R_{42}=\{0, 3, 6, 7, 9, 12, 14, 15, 18, 21, 24, 27, 28, 30, 33, 35, 36, 39\}$. Therefore, $Q(\gen(g))\cap \{m : m\equiv r \Mod{42} \text{ for some }r\in R_{42}\}\subset g$, and consequently $g$ represents all nonnegative integers in the set
$$
M_{42}=\{m \mid m \not \equiv 0,6,22,24 \Mod{32}, \ m \equiv r \Mod{42} \text{ for some } r\in R_{42}\}.
$$
Using this, for any $\alpha=10$ and $\beta=12,15,18$,
one may check that
$$
n=p-\alpha\cdot d^2 -\beta\cdot e^2 \ra g
$$
for some integers $d\in\{0,1,2\}$ and $0\le e \le 6$, defined according as the residue of $p$ modulo $32\cdot 21$.
Therefore, $n\ra f$ in this case.
This completes the proof.
\end{proof}

\begin{rmk}\label{rmk235-237}
From the proof of Theorem \ref{thm235-237}, one may verify that
$$
\begin{array}{lll}
\df{2,3,7,9} 		&\text{represents all primes but not } 13,97,\\
\df{2,3,7,11}	 	&\text{represents all primes but not } 17,\\
\df{2,3,7,13} 	&\text{represents all primes but not } 17,\\
\df{2,3,10,21} 	&\text{represents all primes but not } 13,17,43,47.\\
\end{array}
$$
\end{rmk}

 Now, we are ready to prove Theorem \ref{HigherproperPU}.

\begin{proof}[Proof of Theorem \ref{HigherproperPU}]
The properness can be checked through a straight forward calculation.
The prime-universality of each diagonal quadratic form in Table \ref{HigherCandidate} follows from Theorems \ref{thm222}-\ref{thm235-237}.
\end{proof}

\begin{cor}
Let $f$ be a diagonal quadratic form and let
$$
\mathcal{S}:=\{2,3,5,7,13,17,23,41,43,67\}.
$$
The diagonal quadratic form $f$ represents every prime in $\mathcal{S}$  if and only if $f$ is prime-universal.
Moreover, the set $S$ is minimal in the sense that any proper subset of $\mathcal{S}$ does not guarantee the prime-universality.
\end{cor}

\begin{proof}
Note that if we repeat the proof of Theorem \ref{ThmHigherescalation} under the assumption that $f$ represents all  primes in $S$, we arrive at the same conclusion.
Also, note that for each prime $p \in S$, some quadratic forms representing all primes except $p$ are given in Table \ref{exceptOne}.
The representabilities of primes for each of the first four quadratic forms $\df{1,3,4}$, $\df{1,1,6}$, $\df{1,2,6,10}$, and $\df{1,1,1,9}$ in Table \ref{exceptOne} can be proved as usual, since the class number of each of the forms $\df{1,3,4}$, $\df{1,1,6}$, $\df{1,2,6}$, and $\df{1,1,1}$ is $1$.
Representabilities of primes for each of the other quadratic forms are proved in Theorems \ref{thm222}-\ref{thm235-237} and Remark \ref{rmk235-237}. Therefore, the set $S$ is minimal.
\end{proof}

\begin{table}[ht]
	\caption{Diagonal forms representing all primes but only one} \label{exceptOne}
	\begin{center}
		\begin{tabular}{c|c}
			\Xhline{1pt}
			$f$ & Exception of $f$\\
			\hline
$\df{1,3,4}$	&	 $2$\\

$\df{1,1,6}$	&	 $3$\\
\hline
$\df{1,2,6,10}$	&	 $5$\\

$\df{1,1,1,9}$	&	 $7$\\

$\df{2,3,3,4}$, $\df{2,3,4,12}$ &	$13$\\
\hline
$\df{2,2,2,3}$, $\df{2,3,4,4}$, $\df{2,3,4,7}$,	&	 \multirow{2}{*}{$17$}\\
$\df{2,3,7,11}$, $\df{2,3,7,13}$&\\
\hline
$\df{2,3,4,6}$, $\df{2,3,4,10}$, $\df{2,3,4,13}$, &  \multirow{2}{*}{$23$} \\
$\df{2,3,6,7,19}$, $\df{2,3,6,7,22}$	&  \\
\hline
$\df{2,2,3,17}$			&$41$\\

$\df{2,3,5,19}$			&$43$\\

$\df{2,3,6,7,20}$		&$67$\\
\Xhline{1pt}
		\end{tabular}
	\end{center}
\end{table}

\begin{rmk} {\rm As pointed out by \cite{dw},  Bhargava asserted that if any quadratic form, which is not necessarily diagonal, represents
$$
2, \ 3, \ 5, \ 7, \ 11, \ 13, \ 17, \ 23, \ 29, \ 31, \ 37, \ 41, \ 43, \ 67, \ 73,
$$
then it is prime-universal.  However,  no proof of this has appeared in the literature to the author's knowledge.}
\end{rmk}


TITLE: Walks of odd Lengths in a Matrix
QUESTION [1 upvotes]: Consider the following matrix
$$
 A=\left[
\begin {array}{cccc}
 1&1&0&0\\ 0&0&1&0\\ 0&0&1&1\\ 1&0&0&0
\end {array}
\right].
$$
Assume that $B=A^k$ for some positive integer $k$.
My Question:
How to prove there is no $k$ such that
 all entries of $B$ are odd numbers.
In terminology of graph theory, we should prove there is no positive integer $k$ such that the numbers of walks of length $k$ from any vertex $v_i$ to $v_j$ with $1\leq i,j \leq 4$ are  odd numbers.
Thanks for any suggestions.

REPLY [8 votes]: This matrix is invertible modulo 2, thus so is each its power, but all-ones matrix is singular.


TITLE: If $A$ and $B$ are linear operators such that $A(x)=B(x)$ then $x=0$?
QUESTION [0 upvotes]: Let $V$ be a vector space. If $A,B:V \longrightarrow V$ are linear operators such that $A(x)=B(x)$ for some $x \in V$, then $x=0$?
Is that true or not? And I couldn't think of a counterexample.

REPLY [0 votes]: Considering the matrix of the transformations A and B as M and M' respectively, A(x)=B(x) means that Mx=M'x -> Mx-M'x=0 -> (M-M')x=0, considering that we can write x in a matrix representation, and as matrices form a Ring with divisors of zero, so there is nothing we can conclude from this. If you want a counter example try thinking about matrices


TITLE: Prove that if $\int_{a}^{b} f(x)$ exists, $\delta >0 $ such that $|\sigma_1 -\sigma_2|<\epsilon$
QUESTION [1 upvotes]: Prove if $\int_{a}^{b} f(x)$ exists, then for every $\epsilon>0$, there is a $\delta >0 $ such that $|\sigma_1 -\sigma_2|<\epsilon$,  if $\sigma_1$ and $\sigma_2$ are Reimann sums of $f$ over partitions P$_1$ and P$_2$ of $[a,b] $ with norms less than $\delta$.

First Recalling the primary definition: As $\int_{a}^{b} f(x)$ exists, there is a unique $L$ s.t. $L = \int_{a}^{b} f(x)$ and for every $\epsilon>0$, there is a $\delta >0 $ s.t.
$$|\sigma - L| < \epsilon$$
Second, I want to show that because $f$ is integrable on $[a,b]$, the function has to be bounded on $[a,b]$:
If $f$ is unbounded on $[a,b]$, for any partition P$_1$,P$_2$$\in P$
we have with $M>0$
$$|\sigma_1 - \sigma_2|>M$$
Third, Because $\int_{a}^{b} f(x)$ exists $f$ has to be bounded
I am new to this. I have to admit, I am quite lost. I just do not see how i can prove part 2 . Any help is much appreciated.

REPLY [2 votes]: Fix $\epsilon>0$ and let $L = \int_a^b f(x) \ dx$.
The normal definition states that there is a $\delta>0$ such that if $\sigma$ is a Riemann sum such that its partition $P_{\sigma}$ satisfies $\text{mesh}\left(P_{\sigma}\right) < \delta$, then
$$|\sigma - L| < \epsilon$$
Now consider $\frac{\epsilon}{2}$. The above definition implies there is a $\delta_1>0$  such that if $\sigma_1, \sigma_2$ are arbitrary Riemann sums such that $\max\left\{\text{mesh}\left(P_{\sigma_1}\right), \text{mesh}\left(P_{\sigma_2}\right)\right\} < \delta_1$, then we have
$$|\sigma_1 - L| < \frac{\epsilon}{2}$$
and
$$|\sigma_2 - L|  < \frac{\epsilon}{2}$$
and hence, by the triangle inequality,
$$|\sigma_2 - L| + |\sigma_1 - L| = |L - \sigma_2| + |\sigma_1 - L| \leq |\sigma_1 - \sigma_2| < \epsilon$$


TITLE: Asking About Best Upper Bound And lowerBound
QUESTION [2 upvotes]: Find the best lower and upper bounds for
$$\left(\cos A-\sin A\right)\left(\cos B-\sin B\right)\left(\cos C-\sin C\right),$$
$1)~~$ overall acute-angled $\Delta ABC.$
first of all we know that $\cos(A)-\sin(A) = -\sqrt{2}\sin(A - \frac{\pi}{4})$
$\Pi { -\sqrt{2}\sin(A - \frac{\pi}{4})}=-2\sqrt{2}*\Pi{(\sin(A-\frac{\pi}{4}))}$
$-\frac{1}{2}*\sqrt{2}<\sin(A-\frac{\pi}{4})<\frac{1}{2}*\sqrt{2}$
$-\frac{1}{2}*\sqrt{2}<\sin(B-\frac{\pi}{4})<\frac{1}{2}*\sqrt{2}$
$-\frac{1}{2}*\sqrt{2}<\sin(C-\frac{\pi}{4})<\frac{1}{2}*\sqrt{2}$
Then Lagrange? or there is any other method you all could suggest
Edit:
For Everyone who wanna post or finish using every possible method just write it down here

REPLY [2 votes]: One of my (who happens to be the problem's author) solutions:
We'll prove the following equivalent statement:
If
$$x+y+z=\frac\pi4;~~~-\frac\pi4\leq x,y,z\leq\frac\pi4,$$
then
$$-1\leq2\sqrt2\sin x\sin y\sin z\leq\frac{\left(\sqrt3-1\right)^3}8.$$
Study the equality cases.
WLOG assume that $x\leq y\leq z.$ If $xyz=0,$ then we have nothing to prove.
Otherwise, if $y<0$ then $z>\frac\pi4,$ which is false. So $y>0.$
Let's prove the right inequality. If $x<0$ then we are done. So let $x>0.$
In this case the right inequality is proved, due to Jensen's inequality. Equality iff $x=y=z,$ or $ABC$ is equilateral.
We prove now the left one. If $x>0,$ then we are done. So let $x<0.$
We need to prove
$$2\sqrt2\sin (-x)\sin y\sin z\leq1.$$
But this is obviously true, since $0<\sin (-x),\sin y,\sin z\leq\frac1{\sqrt2}.$
The equality can be achieved for $x=-\frac\pi4;y=z=\frac\pi4$ for example. The proof is complete.
References: https://artofproblemsolving.com/community/c6h2815170p24872115
The general problem is:If $\Delta ABC$ is acute angled and $\alpha\in\left(0,\frac\pi3\right]$ is fixed,
then
$$\max\left\{\sin^2\alpha\cos\alpha;\cos^3\left(\frac\pi3+\alpha\right)\right\}\geq\prod_{\text{cyc}}{\cos\left(A+\alpha\right)}\geq\min\left\{-\sin\alpha\cos^2\left(\frac\pi4+\alpha\right);\cos^3\left(\frac\pi3+\alpha\right)\right\}.$$
(C. Mateescu and L. Giugiuc, 2022)
References: https://artofproblemsolving.com/community/c6t243f6h2826343_cosines_product_with_parameter


TITLE: How to find the diagonizable matrix of the linear map $T:M^{\mathbb R}_{2 \times 2} \to M^{\mathbb R}_{2 \times 2}$ defined as $T(A)=A-A^t$?
QUESTION [0 upvotes]: Let linear map $T:M^{\mathbb R}_{2 \times 2} \to M^{\mathbb R}_{2 \times 2}$ defined as $T(A)=A-A^t$ for all $A \in M^{\mathbb R}_{2 \times 2}$. Find the diagonizable matrix $D$ and the matrix $P$ which makes $A$ diagonizable.

This question builds on this post: How to find basis of $\ker T$ and $\mathrm{Im} T$ for the linear map $T$?.
We can build the matrix representation for $T$ according to the standard basis $E$:
$$
[T]_E=\begin{pmatrix} 0&0&0&0\\0&1&-1&0\\0&-1&1&0\\0&0&0&0 \end{pmatrix}
$$
Hence in order to find the eigenvalues we need the calculate the determinant of the following matrix:
$$
\det(tI-[T]_E)=\det \begin{pmatrix} t&0&0&0\\0&t-1&1&0\\0&1&t-1&0\\0&0&0&t \end{pmatrix}=t^3(t-2)
$$
Therefore we have 2 eigenvalues $t_1=0$ and $t_2=2$. Because the algebraic and geometric (this can be seen in the question I referenced above) multiplicities are equal there's a diagonizable matrix.
What I'm not sure is how to write $D$ and $P$.
Should $D$ be like this:
$$
D=\begin{pmatrix} 0&0&0&0\\0&0&0&0\\0&0&2&0\\0&0&0&0 \end{pmatrix} \quad \text{or in the}\quad M^{\mathbb R}_{2 \times 2} \quad \text{format} \quad D=\begin{pmatrix} 0&0\\0&2 \end{pmatrix}
$$
After plugging in the eigenvalues instead of $t$'s above I got that the subspace for $t_1=0$ is $span\{(1,1)\}$ while the subspace for $t_2=2$ is $span\{(-1,1)\}$.
So should $P$ be:
$$
P=\begin{pmatrix} 1&1\\1&-1 \end{pmatrix}
$$
?

REPLY [0 votes]: This solution handles the $\mathbb{K}$-linear operator $T:\text{Mat}_{n\times n}(\mathbb{K})\to\text{Mat}_{n\times n}(\mathbb{K})$ over a field $\mathbb{K}$ of characteristic not equal to $2$.  Suppose that $\lambda\in\mathbb{K}$ is an eigenvalue of $T$ with an eigenvector $A\in\text{Mat}_{n\times n}(\mathbb{K})$ with $A\neq 0$.  Then, $T(A)=\lambda A$ implies that $(1-\lambda)A=A^\top$, so $$A=\left(A^\top\right)^\top=\big((1-\lambda)A\big)^\top=(1-\lambda)\,A^\top=(1-\lambda)^2\,A\,.$$
Therefore, $\lambda\,\left(\lambda-2\right)=0$, or $\lambda\in\{0,2\}$.
If $\lambda=0$, then $A$ is a symmetric matrix.  If $\lambda=2$, then $A$ is an skew-symmetric matrix.  Since the subspace consisting of symmetric matrices is $\frac{n(n+1)}{2}$-dimensional and the subspace consisting of skew-symmetric matrices is $\frac{n(n-1)}{2}$-dimensional, $T$ is diagonalizable.  In fact, every $X\in\text{Mat}_{n \times n}(\mathbb{K})$ can be written as
$$X=X_\text{sym}+X_{\text{skew}}\,,$$
with $X_\text{sym}:=\frac{A+A^\top}{2}$ and $X_\text{skew}:=\frac{A-A^\top}{2}$.  Then,
$$T(X)=2\,X_{\text{skew}}\,.$$
It is an interesting exercise to find a Jordan canonical form of $T$ in the case where $\text{char}(\mathbb{K})=2$.  The eigenspace with the eigenvalue $0=2$ is still the subspace consisting of symmetric matrices.


TITLE: A Regular Space that isn't Tychonoff
QUESTION [1 upvotes]: A known example of a regular space that isn't Tychonoff is the "Mysior Plane" $X=X_0\cup\{p\}$, where
$$X_0=\{(x,y)\in\mathbb{R}^2 \ |\ y\geq 0\}$$
and $p\notin X_0$. The topology here is defined by declaring every point $(x,y)$ with $y>0$ to be isolated, an open set around a point $(x,0)$ be on the form $U_x\setminus F$, where
$$U_x=\{(x,y)\ |\ 0\leq y\leq 2\}\cup\{(x+y,y)\ |\ 0\leq y\leq 2\}$$
and $F$ is a finite set, and a neighborhood of $p$ to be on the form $U_n=\{p\}\cup\{(x,0)\ |\ x\geq n\}$ for $n\in\mathbb{N}$.
However, I'm having some trouble showing this space in not Tychonoff:
Let $H_j=\{(x,0)\ |\ j-1\leq x \leq j\}$, $H$ be the $x$-axis and let $f:X\rightarrow\mathbb{R}$ be a continuous function such that $f(H_1)=0$. It can be shown that the sets $K_j=f^{-1}(0)\cap H_j$ are infinite for each $j\in\mathbb{R}$. Why does that implies that $f(p)=0$?

REPLY [3 votes]: You have a typo in the definition of $K_j$: $K_j=f^{-1}[\{0\}]\cap H_j$. The proof that $K_j$ is infinite for each $j\in\Bbb Z^+$ is by induction on $j$. Clearly $K_1=H_1$ to get the induction started. For the induction step it will be convenient to let
$$U_x^-=\{\langle x,y\rangle:0\le y\le 2\}$$
and
$$U_x^+=\{\langle x+y,y\rangle:0\le y\le 2\}$$
for each $x\in\Bbb R$.
Now assume that $K_n$ is infinite. Then there is a countably infinite $C\subseteq K_n$. Let $\langle c,0\rangle\in C$. Suppose that $G$ is a $G_\delta$-set containing $\langle c,0\rangle$, i.e., $G=\bigcap_{n\in\Bbb Z^+}G_n$ for some open nbhds $G_n$ of $\langle c,0\rangle$. Then each $G_n$ contains all but finitely many points of $U_c^+$, so $G$ contains all but countably many points of $U_c^+$. Now
$$\begin{align*}
U_c^+\setminus f^{-1}[\{0\}]&=\bigcup_{n\in\Bbb Z^+}\left(U_c^+\cap f^{-1}\left[\left[\frac1n,\to\right)\right]\right)\\
&\qquad\cup\bigcup_{n\in\Bbb Z^+}\left(U_c^+\cap f^{-1}\left[\left(\leftarrow,-\frac1n\right]\right]\right)
\end{align*}$$
is a subset of $U_c^+$ that is a countable union of closed sets and does not contain $\langle c,0\rangle$, so its complement is a $G_\delta$-set containing $\langle c,0\rangle$. We’ve just seen that this complement contains all but countably many points of $U_c^+$, so $U_c^+\setminus f^{-1}[\{0\}]$ must be countable. $C$ is countable, so $B=\bigcup\limits_{\langle c,0\rangle\in C}\big(U_c^+\setminus f^{-1}[\{0\}]\big)$ is countable, and so is
$$P=\{\langle x,0\rangle:\langle x,y\rangle\in B\}\,.$$
Let $H=H_{n+1}\setminus P$; clearly $H$ is infinite. (In fact $|H|=|\Bbb R|$.)
Now let $\langle x,0\rangle\in H$. Then
$$U_x\cap U_c^+\cap f^{-1}[\{0\}]\supseteq U_x^-\cap U_c^+\cap f^{-1}[\{0\}]\ne\varnothing$$
for each $\langle c,0\rangle\in C$. That is, every open nbhd of $\langle x,0\rangle$ meets $f^{-1}[\{0\}]$, and $f^{-1}[\{0\}]$ is closed, so $\langle x,0\rangle\in f^{-1}[\{0\}]$. Thus, $H\subseteq f^{-1}[\{0\}]$, and therefore $K_{n+1}=H_{n+1}\cap f^{-1}[\{0\}]$ is infinite. This completes the induction step.
We now know that $K_n$ is infinite for each $n\in\Bbb Z^+$. In particular,
$$U_n\cap f^{-1}[\{0\}]\ne\varnothing$$
for each $n\in\Bbb Z^+$, so $p\in\operatorname{cl}f^{-1}[\{0\}]=f^{-1}[\{0\}]$, i.e., $f(p)=0$.
You may also find the answers to this question of some interest.


TITLE: Why can't I use vector addition in this way here?
QUESTION [1 upvotes]: I know that this exact question has been asked here a number of times, but none of the answers sit right with me. The question says that the ends of the strings are pulled with a velocity of "u" units. We are to find the velocity of the block in terms of u. The answer is u/cos(θ) units. However, my approach included adding two vectors of magnitude "u" seperated by an angle of 2θ. Evaluating that gives an answer of 2ucos(θ), which is wrong. Where am I going wrong?

REPLY [2 votes]: The model you are assuming here is not correct for the following reason: Imagine you have a horse carriage that is pulled by one horse with a speed of $v$. Then (if the horse is strong enough) the carriage will travel at a speed of $v$. Now, if you put 10 horses in front of the carriage, the carriage will not travel at the speed of $10v$, but still at a speed of $v$. Otherwise it would overtake the horses and this is clearly not physical.
Remember that these are velocities and not forces!
This might be confusing since sometimes velocities are broken up in their coordinate components, one horizontal and one vertical component. However, this is a different setup here. The velocities are not components of a larger velocity, but the velocity of some points of the ropes. You can also think of the points of the ropes as the horses in front of the carriage!
Also, a good check is always taking the limiting case. Here this would correspond to $\theta\rightarrow0$. The physical situation is then that the mass is very far below from the setup which is outlined here. Then, if you pull by some distance $x$, you will lift the package by the same distance $x$. This is also exactly what you find with the formula since $\cos(0)=1$!


text:
\section{Consecutive Subsets of N}
Tags: Natural Numbers

\begin{theorem}
Let $\N_k$ denote the [[Definition:Initial Segment of Natural Numbers|initial segment of the natural numbers]] determined by $k$:
:$\N_k = \left\{{0, 1, 2, 3, \ldots, k - 1}\right\}$
Then:
:$\N_k = \N_{k + 1} \setminus \left\{{k}\right\}$
In particular:
:$\N_{k - 1} = \N_k \setminus \left\{{k - 1}\right\}$
\end{theorem}

\begin{proof}
The result follows as a direct application of [[Strict Lower Closure of Sum with One]].
{{qed}}
[[Category:Natural Numbers]]
ebnlj9h3zjmkiqhic6nj6grj7lm6oxj
\end{proof}

config: wiki


TITLE: Every Borel map into a separable metric space has a sequence of functions, taking countable values, which converges to the Borel map uniformly.
QUESTION [0 upvotes]: For a Borel map $f$ from a Borel space $(\Omega, S)$ into a separable metric space $(X, d)$, we need to construct a sequence of Borel maps $\{f_n\}$ such that for every $n, f_n$ can only take countable many values and $\{f_n\}$ converges to $f$ uniformly.
I think I have a valid strategy to prove this, but I can't figure the last step:
To do this, we start by writing $X$ as a countable union of open sets $X = \cup_iB_i$. This is possible since $(X, d)$ is a separable metric space and hence is second countable. Since the Borel $\sigma-$algebra on $X$ is generated by open sets, we can write $X = \cup_iB_i = \cup_iA_i$, where $A_i$'s are disjoint Borel sets. Then, $\Omega = \cup_if^{-1}(A_i)$, where $f^{-1}(A_i)$'s are disjoint Borel subsets of $\Omega$.
So, we can construct a sequence $\{f_n\}$ such that for every $n$, $f_n(\omega) = x_{i_n}$ is same for  every $\omega \in f^{-1}(A_i)$. Thus, each $f_n$ can only take countably many values.
However, I cannot figure out how to define $f_n(\omega)$ so that $f_n$ converges to $f$ uniformly.

REPLY [0 votes]: Let $(x_i)$ be a counatble dense set. Then $X=\bigcup_n B(x_i,\frac 1 N)$ for each $n$. Considering the sets $ B(x_i,\frac 1 N) \setminus \bigcup_{j \leq i} B(x_j,\frac 1 N)$ we see that $X$ is a countable union of disjoint sets with diameter at most $\frac 2 N$. Now take these set as $A_i$'s in your argument. You see that $|f_n(\omega)-f(\omega)| \leq \frac 2 N$.


TITLE: One spot and distance known, Second spot unknown
QUESTION [0 upvotes]: I know the coordinates of points E and Q, so I know their euclidean distance L.
I'm looking for the point W with coordinates (a,b) related to other known values?

REPLY [1 votes]: I'll assume you know $\theta$, the coordinates $(x,y)$ of $Q$, and $r$.  The angle of rotation from a vertical line through $Q$ to $\overline{QW}$ is also $\theta$, so if we take the point that is $r$ above $Q$ and rotate it by $\theta$ about $Q$, we'll get $W$.  That means $$W=(a,b)=(x-r\sin\theta,y+r\cos\theta).$$


text:
Although the presented primal-dual algorithms can be applied virtually  to any area
where optimization problems have to be solved, we now mention a few common applications of these
techniques.

\subsection{Inverse problems}
For a long time, convex optimization approaches have been successfully used for solving
inverse problems such as signal restoration, signal reconstruction, or interpolation of missing data.
Most of the time, these problems are ill-posed and, in order to recover the signal
of interest in a satisfactory manner, some prior information needs to be introduced. To do this, an objective function
can be minimized which includes a data fidelity term modelling knowledge about the noise statistics and
possibly involves a linear observation matrix (e.g. a convolutive blur), and a regularization (or penalization) term which corresponds to the additional
prior information.  This formulation can also often be justified statistically as the determination of a Maximum A Posteriori (MAP) estimate.
In early developed methods, in particular in Tikhonov regularization, a quadratic penalty function is employed.
Alternatively, hard constraints can be imposed on the solution (for example, bounds on the signal values), leading to
signal feasibility problems. Nowadays, a hybrid regularization \cite{Pustelnik_N_2011_j-ieee-tip_par_pai}
may be prefered so as to combine various kinds of regularity measures, possibly computed for different representations
of the signal (Fourier, wavelets,...), some of them like total variation \cite{Rudin_L_1992_tv_atvmaopiip} and its nonlocal extensions \cite{Zhang_X_2010_j-siam-is_Bregmanized_nrfd} being taylored for
preserving discontinuities such as image edges. In this context, constraint sets can be translated into penalization terms
being equal to the indicator functions of these sets (see \eqref{e:defindicator}). Altogether, these lead to global cost functions which can be quite involved,
often with many variables, for which the splitting techniques described in Section \ref{se:extsplit} are very useful. An extensive literature
exists on the use of ADMM methods for solving inverse problems
(e.g., see \cite{Giovanelli_JF_2005-astoastro-positive-dse,Goldstein_T_2009_j-siam-is_split_bml,Figueiredo_M_2009_ssp_Deconvolution_opiuvsaalo,Figueiredo_M_2010_t-ip_restoration_piado,AfonsoBF11}).
With the advent of more recent primal-dual algorithms, many works have been mainly focused on image recovery applications
\cite{Chambolle_A_2010_first_opdacpai,Esser_E_2010_j-siam-is_gen_fcf,He_B_2012_j-siam-is_conv_apd,Pock_T_2008_p-iccv_diagonal_pffo,Goldstein_T_2013_adaptive_pdh,Bot_R_2014_jmiv_conv_primal_apd,Loris_I_2011_generalization_ist,Jezierska_A_2012_p-icassp_primal_dps,Chen_P_2013_j-inv-prob_prim_dfp,Bec_S_2014_j-nonlinear-conv-anal_alg_sps,Bonettini_S_2012_j-math-imaging-vis_convergence_pdhgatvir,Repetti_A_2012_p-eusipco_penalized_wlsardcsdn,Harizanov_S_2013_Epigraphical_psls,Teuber_T_2014_j-inv-prob_Minimization_pes,Burger_M_2014_First_oavip}. Two illustrations are now provided.

In \cite{Couprie_C_j-siam-is_dual_ctvb}, a generalization of the total variation is defined for an arbitrary graph in order to address a variety of inverse problems. {For denoising applications,
the optimization problem to be solved is of the form \eqref{e:primalvar}
where
\begin{equation}
f = 0,\qquad g = \sigma_C,\qquad h\colon x \mapsto \frac12 \|x-y\|^2,

\end{equation}

$x$ is a vector of variables associated with each vertex of a weighted graph, and $y\in \HH$ is a vector of data observed at each vertex.}
The matrix $L\in \RR^{K\times N}$ is equal to $\operatorname{Diag}(\sqrt{\omega}_1,\ldots,\sqrt{\omega}_K)\,A$ where $(\omega_1,\ldots\omega_K) \in \RP^K$ is the vector of edge weights and $A\in \RR^{K\times N}$ is the graph incidence matrix playing a role similar to a gradient operator on the graph. The set $C$ is defined as an intersection of closed semi-balls in such a way that
its support function $\sigma_C$ (see \eqref{e:defsupfonc}) allows us to define a class of functions extending the total variation semi-norm (see \cite{Couprie_C_j-siam-is_dual_ctvb} for more details).
Good image denoising results can be obtained by building the graph in a nonlocal manner following the strategy in \cite{Zhang_X_2010_j-siam-is_Bregmanized_nrfd}. Results obtained for Barbara image are
displayed in Fig. \ref{fig:nonlocal}. Interestingly, the ability of methods such as those presented in Section \ref{se:FBFPF} to circumvent matrix inversions leads to a significant
decrease of the convergence time for irregular graphs in comparison with algorithms based on the Douglas-Rachford iteration or ADMM (see Fig. \ref{fig:Compa_FBF_PPXA}).

\begin{figure}[ht]
\centering
\subfigure[Original image]{\includegraphics[scale=0.29]{images/barbara.eps}}
\subfigure[Noisy~{\scriptsize  SNR~=~$14.47$ dB}]{\includegraphics[scale=0.29]{images/barbara_n20.eps}}
\subfigure[{\scriptsize Nonlocal TV~{\scriptsize SNR~=~$20.78$~dB}}]{\includegraphics[scale=0.29]{images/patchPPXAbarbara_n20.eps}}

\caption{{\small Nonlocal denoising (additive white zero-mean Gaussian noise with variance $\sigma^2=20$).}}
\label{fig:nonlocal}
\end{figure}

\begin{figure}[h]
\begin{center}

\includegraphics[scale=0.48]{images/logtime.eps}

\end{center}

\caption{\small Comparison of the convergence speed of a Douglas-Rachford based algorithm (PPXA \cite{Combettes_PL_2008_j-ip_proximal_apdmfscvip}) (blue) and
an FBF-based primal-dual algorithm (red) for image denoising using a non-regular graph, Matlab implementation on an
Intel Xeon 2.5GHz 8-core system.}
\label{fig:Compa_FBF_PPXA}
\end{figure}

Another application example of primal-dual proximal algorithms is Parallel Magnetic Resonance Imaging (PMRI) reconstruction. A set of measurement vectors $(z_j)_{1 \leq j\leq J}$ is acquired
from $J$ coils. These observations are related to the original full FOV (Field Of View) image $\overline{x}\in\CC^N$ corresponding to a spin density. An estimate of $\overline{x}$
is obtained by solving the following problem:
{
\begin{equation}\label{e:probMRI}
\minimize{x\in \CC^N}{f(x)+g(L x)+\underbrace{\sum_{j=1}^J \| \Sigma F S_j x  - z_j\|^2_{\Lambda_j^{-1}}}}_{\mbox{$h(x)$}}
\end{equation}
}
where $(\forall j \in \{1,\ldots,J\})$ $\|\cdot\|^2_{\Lambda_j^{-1}} = (\cdot)^{\rm H}\Lambda_j^{-1} (\cdot)$,
$\Lambda_j$ is the noise covariance matrix for the $j$-the channel,
$S_{j}\in \CC^{N\times N}$ is a diagonal matrix modelling the sensitivity of the coil, $F \in \CC^{N\times N}$  is a 2D discrete Fourier transform,
$\Sigma \in \{0,1\}^{\lfloor\frac{N}{R}\rfloor\times N}$ is a subsampling matrix,
$g \in \Gamma_0(\CC^K)$ is a sparsity measure (e.g. a weighted $\ell_1$-norm), $L\in \CC^{K\times N}$ is a (possibly redundant) frame analysis operator, and {$f$ is the indicator function
of a vector} subspace of $\CC^N$ serving to set to zero the
image areas corresponding to the background.\footnote{$(\cdot)^{\rm H}$ denotes the transconjugate operation and $\lfloor \cdot \rfloor$ designates the
lower rounding operation.} Combining suitable subsampling strategies in the k-space with the use of an array of coils allows us to reduce the acquisition time
while maintaining a good image quality. The subsampling factor $R > 1$ thus corresponds to an \emph{acceleration factor}. For a more detailed account on the considered approach,
the reader is refered to \cite{Chaari_L_2011_j-media_wav_brr,Florescu_A_2014_j-sp_majorize_mmg} and the references therein. Reconstruction results are shown in Fig.~\ref{Fig:Reconst}. Fig.~\ref{fig:convergencePMRI} also allows us to evaluate the convergence time for various algorithms. It can be observed that smaller differences between the implemented primal-dual strategies are apparent in this example.
Due to the form of the subsampling matrix, the matrix inversion involved at each iteration of ADMM however requires to make use of a few subiterations of a linear conjugate gradient method.

\begin{figure}
\subfigure[]{
\includegraphics[width=4cm]{images/S12.eps}

\includegraphics[width=4cm]{images/S20.eps}}

\hspace{10pt}
\subfigure[]{
\includegraphics[width=4cm]{images/Orig_PMRI.eps}
\includegraphics[width=4cm]{images/im_poly1_cvx_wav_82.eps}}
\caption{\small \textbf{(a)} Effects of the sensitivity matrices in the spatial domain in the absence of subsampling: the moduli of the images corresponding
to $(S_j \overline{x})_{2\leq j \leq 3}$ are displayed for 2 channels out of 32.
\textbf{(b)} Reconstruction quality:  moduli of the original slice $\overline{x}$ and the reconstructed one  with SNR~=~$20.03$ dB (from left to right)
 using polynomial sampling of order 1 with $R=5$, a wavelet frame, and an $\ell_1$ regularization.}
\label{Fig:Reconst}
\end{figure}

\begin{figure}
\begin{center}
\includegraphics[height=6cm]{images/conv_PMRI}
\caption{\small Signal-to-Noise Ratio as a function of computation time using ADMM, and FB or FBF-based primal-dual methods for a given slice,
Matlab implementation on an Intel i7-3520M CPU@2.9 GHz system.}
\label{fig:convergencePMRI}
\end{center}
\vspace{-15pt}
\end{figure}

Note that convex primal-dual proximal optimization algorithms have been applied to other fields than image recovery, in particular to
machine learning \cite{Bach_F_2012_j-ftml_optimization_sip,Mahadevan_S_2014_Proximal_rlntsdmpds}, system identification \cite{Ono_S_2013_p-icassp_sparse_siu}, audio processing \cite{Bayram_I_2014_sivp_prmal_daad}, optimal transport \cite{Papadkis_N_2014_j-siam-is_optimal_tps}, empirical mode decomposition \cite{Pustelnik_N_2014_j-sp_empirical_mdr}, seimics~\cite{Pham_M_2014_sparse_tbaf}, database management \cite{Moerkotte_G_2014_Proximal_oqf}, and data streaming over networks
\cite{Towfic_Z_2014_Stability_plapdn}.

config: arxiv


TITLE: Why is argon a noble gas but not, say, beryllium or palladium?
QUESTION [34 upvotes]: Why is argon a noble gas given that the 3d subshell is still empty?
More generally, why is it that the filling of a p sub-shell makes an element noble rather than s, d, or f sub-shells, or completed n-levels?
Let's start with the prior that in a hydrogen atom the principle quantum number determines the energy - so maybe full shells correspond to noble gases.
Now, I understand that in helium the n=1 level is full, so it is a noble gas.
Then, beryllium has a full 2s subshell, but empty 2p, so it is a metal.
Neon has the n=2 level full, and is a noble gas.
Magnesium has a full 3s subshell, but not a full n=3 level and is a metal - so far so good.
But, oops, argon has full 3s and 3p subshells but empty 3d, yet it is a noble gas.
At this point we look at the aufbau principle and say "ah, your prior is no good, the 4s is lower energy than 3d, and this is the first time it has kicked in - so there you go".
But, as we continue down the periodic table there are more noble gasses and they happen only with the filling of a p subshell.  Why is krypton noble, but not zinc?  Even palladium cleanly fills n=4 up through 4d (with only 4f empty) but it is still not a noble gas.  Ytterbium fills an f subshell for the first time and it isn't noble.  What makes the filling of a p subshell so specially inert compared to s, d, and f subshells?
The aufbau principle does not explain this.  Draw the classic "diagonal aufbau chart" and circle where the noble gasses are.  It doesn't add any insight.  Noble gasses only happen when you fill a p subshell.  Why?
Edited to add:  When an electron is promoted out of a two electron s-shell into an empty p-shell the spin will flip and energy is saved because of the Pauli exchange (two same spin electrons can avoid each other).  The one remaining s electron, and the one new p electron benefit from each having a sub-shell to themselves - kinda a push vs. the s to p promotion.  This, to me, intuitively explains why full s shells  aren't noble.
User4552 points out that brute force calculation shows that energy gaps after a p shell is filled are large.  But it doesn't help my intuition to just hear "the computer calculations reveal that ...".  So can anyone help with intuition on why full p-shells, rather that full d-shells, are noble?

REPLY [8 votes]: As a proxy for an atom's reluctance to form chemical bonds, we can use the energy required to lift one of its outer electrons to the next higher orbital, pretending as usual that the "orbital" concept is still at least approximately valid in a multi-electron atom. Then the question is why this energy gap tends to be significantly larger after filling a p-shell ($\ell=1$) than it is after filling s- or d- or f-shells ($\ell=0,2,3$).
As usual, let $n$ and $\ell$ denote the radial and angular quantum numbers, respectively. Let's accept that orbitals are filled in order of increasing $n+\ell$, and then in order of increasing $n$ whenever the first rule is neutral. (Reality is a little more complicated, but these rules mostly work pretty well.) The following list shows the filling order defined by these two rules, along with the number of nodes in the orbital wavefunction's radial factor:
\begin{align}
n+\ell &\hskip1cm (n,\ell) &\hskip1cm \text{radial nodes} & \\
\hline
1 &\hskip1cm (1,0) &\hskip1cm 0 &\hskip1cm  \text{helium}\\
2 &\hskip1cm (2,0) &\hskip1cm 1 \\
3 &\hskip1cm (2,1) &\hskip1cm 0 &\hskip1cm  \text{neon}\\
3 &\hskip1cm (3,0) &\hskip1cm 2 \\
4 &\hskip1cm (3,1) &\hskip1cm 1 &\hskip1cm  \text{argon} \\
4 &\hskip1cm (4,0) &\hskip1cm 3 \\
5 &\hskip1cm (3,2) &\hskip1cm 0 \\
5 &\hskip1cm (4,1) &\hskip1cm 2 &\hskip1cm  \text{krypton} \\
5 &\hskip1cm (5,0) &\hskip1cm 4 \\
6 &\hskip1cm (4,2) &\hskip1cm 1 \\
6 &\hskip1cm (5,1) &\hskip1cm 3 &\hskip1cm  \text{xenon} \\
6 &\hskip1cm (6,0) &\hskip1cm 5 \\
7 &\hskip1cm (4,3) &\hskip1cm 0 \\
7 &\hskip1cm (5,2) &\hskip1cm 2 \\
7 &\hskip1cm (6,1) &\hskip1cm 4 &\hskip1cm  \text{radon} \\
7 &\hskip1cm (7,0) &\hskip1cm 6 \\
8 &\hskip1cm (5,3) &\hskip1cm 1 \\
\end{align}
In every noble-gas case, and only in these cases, the next available level is an s-orbital ($\ell=0$), which has more radial nodes than the levels immediately before or after it in the sequence. (This pattern works for helium, too, even though helium doesn't have a filled p-shell.) If we could understand intuitively why energy gap between the ground state and the first excited state is relatively large when the first excited state is an s-orbital, then we would have at least a partial answer to the question.
The order in which the shells are filled (the sequence shown above) indicates that radial nodes are more costly than "azimuthal" nodes, because for a given $n+\ell$, the cases with smaller $\ell$ (fewer "azimuthal" nodes) have higher energy. Accepting this trend as an axiom, we can focus our intuition on the radial part.
Intuitively, if we think of the radial nodes as "no-fly zones" for that electron, then having a larger number of radial nodes may correspond to having less freedom to re-arrange the multi-electron system to minimize the energy in the presence of electron-electron interactions. Here's the key idea: by analogy with a construction-induced traffic jam, the impact of each additional node (each additional no-fly zone) could be an increasing function of the number of nodes already present. This picture suggests that lifting an electron from a filled p-shell up to the next available s-shell should be more costly than, say, lifting an electron from a filled d-shell up to the next available p-shell, because the former lift requires increasing an already-larger number of radial nodes. Using the magnitude of the energy-gap as a proxy for an atom's reluctance to form bonds, this suggests that the noble gases should be less reactive (relatively), at least among atoms with the same value of $n+\ell$.
The key idea is that the energy-cost of each additional radial node is an increasing function of the number of radial nodes already present. This seems to be consistent with the information I've seen, but I have no real justification for anticipating it except for the dubious traffic-jam analogy. Even if the intuition is correct, it isn't quantitative enough to predict just how noble the noble gases are. It is suggestive at best, but at least it doesn't rely entirely on a computer. That's why I thought it was worth posting.


TITLE: Fourier Transform of a modulated time shifted function
QUESTION [0 upvotes]: How would one go about calculating the fourier transform of the signal?
$$ y(t) =  x(t-t_0)cos(\omega_0t) $$

REPLY [0 votes]: In summary, you would:

Apply the Fourier Transform convolution theorem
$$\mathcal{F}\left\{x(t-t_0)\cos(\omega_0 t)\right\} = \dfrac{1}{2\pi}\left[\mathcal{F}\left\{x(t-t_0)\right\}*\mathcal{F}\left\{\cos(\omega_0 t)\right\}\right]$$
Apply the Fourier Transform shift/modulation theorem for $x(t-t_0)$
$$\mathcal{F}\left\{x(t-t_0)\right\} = e^{-it_0 \omega}\mathcal{F}\left\{x(t)\right\}= e^{-it_0 \omega}X(\omega)$$
Look up the Fourier Transform of $\cos(\omega_0 t)$ in a table
$$\mathcal{F}\left\{\cos(\omega_0 t)\right\} = \pi\left[\delta(\omega-\omega_0) + \delta(\omega+\omega_0)\right] $$
Apply the convolution properties of the Dirac Delta function
$$ \dfrac{1}{2\pi} e^{-it_0 \omega}X(\omega) * \pi\left[\delta(\omega-\omega_0) + \delta(\omega+\omega_0)\right] = \dfrac{1}{2}e^{-it_0 (\omega-\omega_0)}X(\omega-\omega_0)+\dfrac{1}{2}e^{-it_0 (\omega+\omega_0)}X(\omega+\omega_0) $$

You end up with a Fourier transform that is the sum of two shifted, scaled, and modulated copies of $\mathcal{F}\{x(t)\}$


TITLE: Is The Set Of Subsequential Limits Always Closed/Compact?
QUESTION [4 upvotes]: I've been working my way through Rudin's PoMA, and I was thinking about subsequential limits.
If you take a sequence with finitely many subsequential limits, then the set of subsequential limits is clearly closed and compact.
Theorem 3.17 in Rudin states that the upper limit of a sequence is the limit of some subsequence. This gives me the feeling that perhaps the set of subsequential limits for a divergent sequence with infinitely many subsequnetial limits is also closed or compact. Admittedly, however, I don't think this follows directly from theorem 3.17.
I've toyed with this idea for a bit, but I haven't been able to come up with a proof or a counter-example. Is this true or false, and what's the proof? Does this have any other interesting extensions/implications?

REPLY [5 votes]: If $x$ is not a subsequential limit, then (i) $x$ occurs in the sequence finitely often or not at all, and (ii) $x$ has an open neighbourhood, $N,$ whose intersection with the set of points in the sequence is $\varnothing$ or $\{x\}.$ Every point of $N \setminus \{x\},$ therefore, has a neighbourhood disjoint from the sequence, therefore it is not a subsequential limit. It follows that the complement of the set of subsequential limits is open, i.e. the set of subsequential limits is closed. As has already been pointed out, it need not be bounded, therefore it need not be compact.


TITLE: Will a consumer freezer survive air travel with a pressurization/depressurization cycle?
QUESTION [0 upvotes]: Is it possible to ship a consumer freezer by air cargo (such as this one: https://www.danby.com/products/freezers/dcf072a2wp/ )? Or will the refrigerant gas expand too much and break the freezer? We all know a bag of chips cannot withstand air travel.
What are the risks to the freezer itself?
I am not talking about the hazmat implications (I know it is classified as a "dangerous goods": UN3358) but even if it is allowed on board, is it a good idea? or will the freezer be damaged by the pressurization/depressurization cycle?
Thank you!
transferred from here: https://aviation.stackexchange.com/questions/77810/will-a-consumer-freezer-survive-air-travel-with-a-pressurization-depressurizatio?noredirect=1#comment204081_77810

REPLY [2 votes]: The coils in any refrigeration system are designed to withstand a substantial pressure difference beyond ambient pressure.  It is common to see pressures approaching 200 psig on the high pressure side of these systems while in operation.  When your freezer is not in operation, the highest pressure inside the coils should be in the range of 50 psig or so.  Since a high altitude aircraft will only add approximately 1/2 of an atmosphere to that (e.g., 7 psi), the answer is "no", the pressurization/depressurization cycle will have no effect on the refrigeration coils.