TITLE: Elementary Number Theory Divisibility Question
QUESTION [0 upvotes]: Let $a, b, c \in \mathbb Z$. I know that if $c|a$ and $c|b$, then $c|a \pm b$.  I was working on some elementary number theory and I began to wonder if the following were true:$$\text{if }c|a \text{ and } c|a+b \text{, then } c|b$$

I managed a very simple proof:
Suppose $c|a$ and $c|a+b$.  It follows that $a=c \cdot d$ and $a+b=c \cdot d'$, where $d, d' \in \mathbb Z$.  Note that if $a=c \cdot d$, then $c \cdot d +b=c \cdot d'$.  Hence $b=c \cdot d' -c \cdot d$, or $b=c(d-d')$. Letting $d''=d-d' \in \mathbb Z$, we have $b=c \cdot d''$, or $c|b$, as desired.

However, while trying to prove this the good ol' fashioned way (using propositional logic), I'm hitting a barrier that has me second-guessing if my above proof is correct or not.  For the life of me, I cannot fathom how $$(P \land Q) \Rightarrow R \equiv (P \land R) \Rightarrow Q$$
for $P:=c|a$, $Q:=c|b$ and $R:=c|a+b$.
Sorry for the silly question.  My propositional logic is a little rusty and I'm probably just overlooking something.  Any insight would be greatly appreciated!

REPLY [2 votes]: Your number theory proof is correct, but your proposed equivalence in propositioal logic is (in general) not correct, so you can't use it to proof what you want just with your initial knowledge and "propositional magic", you need actual number theory for that.
To see that the propositional formula is generally incorrect, set for example
$$P:=2|x, Q:=4|x, R=2|x$$
Of course, I may have misunderstood what you wanted to achieve with your propositional formula. If that's true, please explain.


TITLE: Deriving the cumulative density of a four parameter logistic function
QUESTION [0 upvotes]: I am using a four parameter logistic function $y = a + \dfrac{b-a}{1+e^{c(d-x)}}$. I would like to know how one finds the value $x$ that corresponds to $5\%$ (or $n\%$) cumulative density.
My intuition would be that I find the value between the lower and upper asymptotes that corresponds to an increase of $n\%$. Substituting this value for $y$ and solving the equation for x would give me the value $x$ at which there was a cumulative increase of $n\%$ from the lower asymptote. Is this correct?

REPLY [0 votes]: Yes, the procedure you described is correct (except for the misuse of the word "density", pointed out by Michael Hardy). As $x\to-\infty$, $y\to a$ (assuming $c>0$), and as $x\to+\infty$, $y\to b$. Therefore, you want to solve the equation   $y=a+0.05(b-a)$, or $y=a+(n/100)(b-a)$ in general. This amounts to
$$\frac{1}{1+e^{c(d-x)}} = \frac{n}{100}$$
$$  e^{c(d-x)}  = \frac{100}{n}-1$$
$$   c(d-x)   = \ln\left(\frac{100}{n}-1\right)$$
and you can take it from there.


TITLE: Can affine transformation of a matrix changes its rank?
QUESTION [1 upvotes]: I am exploring several ways to lower the rank of a matrix by preserving set of constraints. I am sure that performing positive affine transformation (f(x) = ax + b, where a is a non-zero value) preserves my set of constraints, but I want to determine if positive affine transformation can lower the rank of a matrix?

REPLY [2 votes]: I think I understand the problem now. The question is somewhat "artificial" in the context of matrices, so the answer is long, and probably not very fruitful, but I've gone through the logic. I'll present here a TL;DR, and you can read onwards if you want to see the path of the logic.
Suppose $M$ is an $m \times n$ matrix. If the vectors in $\mathbb{R}^m$ and $\mathbb{R}^n$, both containing only $1$ as entries, lie in the column/row spaces of $M$ respectively, then for all $b \neq 0$, there exists an $a \neq 0$ such that applying $f(x) = ax + b$ to the entries of $M$ will reduce the rank by exactly $1$.
Unless we allow $a = 0$, there are no other conditions under which $f(x) = ax + b$ will reduce the rank of $M$.

Let's suppose we have a matrix
$$M = \left(v_1 | v_2 | \ldots | v_n \right),$$
where $v_1, \ldots, v_n$ are column vectors of $M$. If we take an affine function
$$f : \mathbb{R} \to \mathbb{R} : x \mapsto ax + b,$$
then applying $f$ to the entries of $M$ corresponds to the matrix $aM + bJ$, where $J$ is the matrix of appropriate dimensions whose entries are all $1$. In particular, if we let $j = (b, b, \ldots, b)^T$, we get the matrix,
$$aM + bJ = \left(av_1 + j | av_2 + j | \ldots | av_n + j \right).$$
Let's first acknowledge that $b = 0$ is a waste of time; the matrix's rank will be perfectly preserved.
To find the rank of $aM + bJ$, we compute its nullity, comparing to the nullity of $M$. So, let's suppose we have $v = (c_1, c_2, \ldots, c_n)^T$ such that $(aM + bJ)v = 0$. Then we have,
\begin{align*}
c_1 (av_1 + j) + \ldots + c_n (av_n + j) &= 0 \\
c_1 v_1 + \ldots + c_n v_n &= -\frac{c_1 + \ldots + c_n}{a}j.
\end{align*}
Now, consider two possibilities: either $j$ is in the columnspace of $M$ or it isn't. If it isn't, then the above equation must have $c_1 + c_2 + \ldots + c_n = 0$. This may or may not increase the rank of $M$ by $1$, as it adds an extra constraint on the nullspace of $M$. This depends on whether or not the vector $(1, 1, \ldots, 1) \in \mathbb{R}^n$ is in the rowspace of $M$.
Note: there's no way, in the case of $j$ not being in the columnspace of $M$, that we can ever reduce the rank of $M$. Also, the (non-zero) value of $b$ is immaterial. We are really just testing whether $j/b = (1, 1, \ldots, 1)^T$ belongs to the columnspace of $M$. This already gives us a constraint as to when this technique could be useful to you.
The other possibility is that $j$ (or equivalently, $j/b$) is in the columnspace of $M$. In this case, we can write,
$$j = d_1 v_1 + \ldots + d_n v_n,$$
for some $d_1, \ldots, d_n$. Let $w = (d_1, \ldots, d_n)^T$, and $N = (w | w | \ldots | w) \in \mathbb{R}^{n \times n}$. Note that we have $j = Mw$ and $bJ = MN$. We therefore have,
$$aM + bJ = aM + MN = M(aI_n + N).$$
Let's consider the eigenvalues of $N$. Since $w \neq 0$, it is rank $1$. There are two possibilities: either $d_1 + \ldots + d_n = 0$ or not. If it is equal to $0$, it's not difficult to check that $N^2 = 0$, hence $0$ is the only eigenvalue of $N$. Therefore, $aI_n + N$ will have $a \neq 0$ as its only eigenvalue, hence is invertible, and so $aM + bJ$ will have the same rank as $M$.
If it's not equal to $0$, then $w$ is an eigenvector with non-zero eigenvalue $\lambda = d_1 + \ldots + d_n$, and it is the only eigenvalue that is non-zero. Thus, the eigenvalues of $aI_n + N$ are $a$ with multiplicity $n - 1$ and $a + \lambda$ with multiplicity $1$. If $a + \lambda \neq 0$, again, $N$ is invertible, hence $aM + bJ$ will have the same rank as $M$. Hence, we now assume that
$$a = -\lambda = -(d_1 + d_2 + \ldots + d_n).$$
Note that $N$ is diagonalisable, hence so is $aI_n + N$ and $(aI_n + N)^T = aI_n + N^T$. Using a classical result, the columnspace of $aI_n + N$ will be the perpendicular complement of the nullspace of $aI_n + N^T$, which simple computation reveals to be $V = (\operatorname{span} (1, 1, \ldots, 1))^\perp$.
Consider the map $x \to Mx$, when restricted to $V$. The nullspace of this map is the subspace of all vectors in $V$ that map to $0$, which is clearly $\operatorname{null} M \cap V$, a subspace of $\operatorname{null} M$. Thus, the nullity of the restricted map is less than or equal to the nullity of the unrestricted map. However, the nullity can only differ by at most $1$, as $\operatorname{dim} V = n - 1$. So the nullity of the restricted map is $\operatorname{nullity} M$ or $\operatorname{nullity} M - 1$.
Since $aI_n + N$ is diagonalisable, we can choose a basis of eigenvalues, starting with $w$: $w, w_1, \ldots, w_{n - 1}$. Since $w_1, \ldots, w_{n - 1}$ correspond to non-zero eigenvalue $a$, it follows that their span is the columnspace of $aI_n + N$, i.e. $V$. We can use this to show that $\operatorname{span} (w) \oplus (\operatorname{null} M \cap V)$ is the nullspace of $aM + bJ$. Hence, taking the dimension of each side, we see that we're only interested in the case where the nullity of the restricted map is equal to the nullity of $M$. In this case, we get
$$\operatorname{nullity}(aM + bJ) = \operatorname{nullity}(M) + 1 \implies \operatorname{rank}(aM + bJ) = \operatorname{rank}(M) - 1,$$
that is, a decrease in rank, and only by the value $1$.
These nullities are equal, i.e. we obtain a decrease in rank, if and only if
\begin{align*}
&\operatorname{null} M \cap V = \operatorname{null} M \\
\iff &\operatorname{null} M \subseteq V \\
\iff &(\operatorname{colspace} M^T)^\perp \subseteq V \\
\iff &(\operatorname{rowspace} M)^\perp \subseteq (\operatorname{span}(1,1, \ldots, 1))^\perp \\
\iff &\operatorname{rowspace} M \supseteq \operatorname{span}(1,1, \ldots, 1) \\
\iff &(1, 1, \ldots, 1) \in \operatorname{rowspace} M.
\end{align*}
This finally proves the TL;DR above.

To cement this, I thought I'd try a few examples. Suppose
$$M = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \end{pmatrix}.$$
Note that $M$ satisfies the conditions, and has rank $2$.
Choose $b = 2$. Our value of $a$ is forced to be $-4$, since we can form $2\begin{pmatrix} 1 \\ 1 \end{pmatrix}$ by $2$ lots of the first column vector, and a total of $2$ lots of the second and third vectors. So, we let $f(x) = 2 - 4x$. Applying $f$ to $M$ yields,
$$\begin{pmatrix} -2 & 2 & 2 \\ 2 & -2 & -2 \end{pmatrix},$$
which is rank $1$, as predicted.
If we take
$$M' = \begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \end{pmatrix},$$
then the conditions are no longer satisfied. The rank is still $2$. If we take $f(x) = 2 - 4x$ as above, and apply to the entries of $M$, then we get
$$\begin{pmatrix} -2 & 2 & -2 \\ 2 & -2 & -2 \end{pmatrix},$$
which is still rank $2$.


TITLE: Integral of function in $R^2$ - Sobolev space
QUESTION [0 upvotes]: I'm trying to solve this exercice:
Find all possible values of $\alpha\in\mathbb{R}$ such that $f(x)=\min(|x|^\alpha,1)$ is in $W^{1,2}(\mathbb{R}^2)$ with $|x|=\sqrt{x_1^2+x_2^2}$.
First I control if $f\in L^2(\mathbb{R}^2)$.
So I want to calculate or estimate ( I guess the answer is estimate)
$||f||_2^2 = \int_{\mathbb{R}^2}|f(x)| dx$
but I cannot do the integral because I don't know how to consider that $\min$.
Can you help me?

REPLY [0 votes]: If $\alpha \ge 0$ then $|x|^\alpha \le 1$ if and only if $|x| \le 1$, and you can write $f$ as
$$f(x) = \left\{ \begin{array}{ll} |x|^\alpha & |x| \le 1 \\ 1 & |x| > 1. \end{array} \right.$$
If $\alpha < 0$ then $|x|^\alpha \ge 1$ if and only if $|x| \le 1$, and you can write $f$ as
$$f(x) = \left\{ \begin{array}{ll} 1 & |x| \le 1 \\ |x|^\alpha & |x| > 1. \end{array} \right.$$
In both cases you have
$$\int_{\mathbb R^2} |f(x)|^2 \, dx = \int_{|x| \le 1} |f(x)|^2 \, dx + \int_{|x| > 1} |f(x)|^2 \, dx$$ and you can evaluate the integrals individually.


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 {\par(Subsubsubsection head:)#1\par }}{}
\@ifundefined{subparagraph}{\def\subparagraph#1
 {\par(Subsubsubsubsection head:)#1\par }}{}

\@ifundefined{therefore}{\def\therefore{}}{}
\@ifundefined{backepsilon}{\def\backepsilon{}}{}
\@ifundefined{yen}{\def\yen{\hbox{\rm\rlap=Y}}}{}
\@ifundefined{registered}{
   \def\registered{\relax\ifmmode{}\r@gistered
                    \else$\m@th\r@gistered$\fi}
 \def\r@gistered{^{\ooalign
  {\hfil\raise.07ex\hbox{$\scriptstyle\rm\text{R}$}\hfil\crcr
  \mathhexbox20D}}}}{}
\@ifundefined{Eth}{\def\Eth{}}{}
\@ifundefined{eth}{\def\eth{}}{}
\@ifundefined{Thorn}{\def\Thorn{}}{}
\@ifundefined{thorn}{\def\thorn{}}{}

\def\TEXTsymbol#1{\mbox{$#1$}}
\@ifundefined{degree}{\def\degree{{}^{\circ}}}{}

\newdimen\theight
\@ifundefined{Column}{\def\Column{
 \vadjust{\setbox\z@=\hbox{\scriptsize\quad\quad tcol}
  \theight=\ht\z@\advance\theight by \dp\z@\advance\theight by \lineskip
  \kern -\theight \vbox to \theight{
   \rightline{\rlap{\box\z@}}
   \vss
   }
  }
 }}{}

\@ifundefined{qed}{\def\qed{
 \ifhmode\unskip\nobreak\fi\ifmmode\ifinner\else\hskip5\p@\fi\fi
 \hbox{\hskip5\p@\vrule width4\p@ height6\p@ depth1.5\p@\hskip\p@}
 }}{}

\@ifundefined{cents}{\def\cents{\hbox{\rm\rlap c/}}}{}
\@ifundefined{tciLaplace}{\def\tciLaplace{L}}{}
\@ifundefined{tciFourier}{\def\tciFourier{F}}{}
\@ifundefined{textcurrency}{\def\textcurrency{\hbox{\rm\rlap xo}}}{}
\@ifundefined{texteuro}{\def\texteuro{\hbox{\rm\rlap C=}}}{}
\@ifundefined{textfranc}{\def\textfranc{\hbox{\rm\rlap-F}}}{}
\@ifundefined{textlira}{\def\textlira{\hbox{\rm\rlap L=}}}{}
\@ifundefined{textpeseta}{\def\textpeseta{\hbox{\rm P\negthinspace s}}}{}

\@ifundefined{miss}{\def\miss{\hbox{\vrule height2\p@ width 2\p@ depth\z@}}}{}

\@ifundefined{vvert}{\def\vvert{\Vert}}{}

\@ifundefined{tcol}{\def\tcol#1{{\baselineskip=6\p@ \vcenter{#1}} \Column}}{}

\@ifundefined{dB}{\def\dB{\hbox{{}}}}{}
\@ifundefined{mB}{\def\mB#1{\hbox{$#1$}}}{}
\@ifundefined{nB}{\def\nB#1{\hbox{#1}}}{}

\@ifundefined{note}{\def\note{$^{\dag}}}{}

\def\newfmtname{LaTeX2e}

\ifx\fmtname\newfmtname
  \DeclareOldFontCommand{\rm}{\normalfont\rmfamily}{\mathrm}
  \DeclareOldFontCommand{\sf}{\normalfont\sffamily}{\mathsf}
  \DeclareOldFontCommand{\tt}{\normalfont\ttfamily}{\mathtt}
  \DeclareOldFontCommand{\bf}{\normalfont\bfseries}{\mathbf}
  \DeclareOldFontCommand{\it}{\normalfont\itshape}{\mathit}
  \DeclareOldFontCommand{\sl}{\normalfont\slshape}{\@nomath\sl}
  \DeclareOldFontCommand{\sc}{\normalfont\scshape}{\@nomath\sc}
\fi

\def\alpha{{\Greekmath 010B}}
\def\beta{{\Greekmath 010C}}
\def\gamma{{\Greekmath 010D}}
\def\delta{{\Greekmath 010E}}
\def\epsilon{{\Greekmath 010F}}
\def\zeta{{\Greekmath 0110}}
\def\eta{{\Greekmath 0111}}
\def\theta{{\Greekmath 0112}}
\def\iota{{\Greekmath 0113}}
\def\kappa{{\Greekmath 0114}}
\def\lambda{{\Greekmath 0115}}
\def\mu{{\Greekmath 0116}}
\def\nu{{\Greekmath 0117}}
\def\xi{{\Greekmath 0118}}
\def\pi{{\Greekmath 0119}}
\def\rho{{\Greekmath 011A}}
\def\sigma{{\Greekmath 011B}}
\def\tau{{\Greekmath 011C}}
\def\upsilon{{\Greekmath 011D}}
\def\phi{{\Greekmath 011E}}
\def\chi{{\Greekmath 011F}}
\def\psi{{\Greekmath 0120}}
\def\omega{{\Greekmath 0121}}
\def\varepsilon{{\Greekmath 0122}}
\def\vartheta{{\Greekmath 0123}}
\def\varpi{{\Greekmath 0124}}
\def\varrho{{\Greekmath 0125}}
\def\varsigma{{\Greekmath 0126}}
\def\varphi{{\Greekmath 0127}}

\def\nabla{{\Greekmath 0272}}
\def\FindBoldGroup{
   {\setbox0=\hbox{$\mathbf{x\global\edef\theboldgroup{\the\mathgroup}}$}}
}

\def\Greekmath#1#2#3#4{
    \if@compatibility
        \ifnum\mathgroup=\symbold
           \mathchoice{\mbox{\boldmath$\displaystyle\mathchar"#1#2#3#4$}}
                      {\mbox{\boldmath$\textstyle\mathchar"#1#2#3#4$}}
                      {\mbox{\boldmath$\scriptstyle\mathchar"#1#2#3#4$}}
                      {\mbox{\boldmath$\scriptscriptstyle\mathchar"#1#2#3#4$}}
        \else
           \mathchar"#1#2#3#4
        \fi
    \else
        \FindBoldGroup
        \ifnum\mathgroup=\theboldgroup
           \mathchoice{\mbox{\boldmath$\displaystyle\mathchar"#1#2#3#4$}}
                      {\mbox{\boldmath$\textstyle\mathchar"#1#2#3#4$}}
                      {\mbox{\boldmath$\scriptstyle\mathchar"#1#2#3#4$}}
                      {\mbox{\boldmath$\scriptscriptstyle\mathchar"#1#2#3#4$}}
        \else
           \mathchar"#1#2#3#4
        \fi
	  \fi}

\newif\ifGreekBold  \GreekBoldfalse
\let\SAVEPBF=\pbf
\def\pbf{\GreekBoldtrue\SAVEPBF}

\@ifundefined{theorem}{\newtheorem{theorem}{Theorem}}{}
\@ifundefined{lemma}{\newtheorem{lemma}[theorem]{Lemma}}{}
\@ifundefined{corollary}{\newtheorem{corollary}[theorem]{Corollary}}{}
\@ifundefined{conjecture}{\newtheorem{conjecture}[theorem]{Conjecture}}{}
\@ifundefined{proposition}{\newtheorem{proposition}[theorem]{Proposition}}{}
\@ifundefined{axiom}{\newtheorem{axiom}{Axiom}}{}
\@ifundefined{remark}{\newtheorem{remark}{Remark}}{}
\@ifundefined{example}{\newtheorem{example}{Example}}{}
\@ifundefined{exercise}{\newtheorem{exercise}{Exercise}}{}
\@ifundefined{definition}{\newtheorem{definition}{Definition}}{}

\@ifundefined{mathletters}{

  \newcounter{equationnumber}
  \def\mathletters{
     \addtocounter{equation}{1}
     \edef\@currentlabel{\theequation}
     \setcounter{equationnumber}{\c@equation}
     \setcounter{equation}{0}
     \edef\theequation{\@currentlabel\noexpand\alph{equation}}
  }
  \def\endmathletters{
     \setcounter{equation}{\value{equationnumber}}
  }
}{}

\@ifundefined{BibTeX}{
    \def\BibTeX{{\rm B\kern-.05em{\sc i\kern-.025em b}\kern-.08em
                 T\kern-.1667em\lower.7ex\hbox{E}\kern-.125emX}}}{}
\@ifundefined{AmS}
    {\def\AmS{{\protect\usefont{OMS}{cmsy}{m}{n}
                A\kern-.1667em\lower.5ex\hbox{M}\kern-.125emS}}}{}
\@ifundefined{AmSTeX}{\def\AmSTeX{\protect\AmS-\protect\TeX\@}}{}

\def\@@eqncr{\let\@tempa\relax
    \ifcase\@eqcnt \def\@tempa{& & &}\or \def\@tempa{& &}
      \else \def\@tempa{&}\fi
     \@tempa
     \if@eqnsw
        \iftag@
           \@taggnum
        \else
           \@eqnnum\stepcounter{equation}
        \fi
     \fi
     \global\tag@false
     \global\@eqnswtrue
     \global\@eqcnt\z@\cr}

\def\TCItag{\@ifnextchar*{\@TCItagstar}{\@TCItag}}
\def\@TCItag#1{
    \global\tag@true
    \global\def\@taggnum{(#1)}}
\def\@TCItagstar*#1{
    \global\tag@true
    \global\def\@taggnum{#1}}

\def\QATOP#1#2{{#1 \atop #2}}
\def\QTATOP#1#2{{\textstyle {#1 \atop #2}}}
\def\QDATOP#1#2{{\displaystyle {#1 \atop #2}}}
\def\QABOVE#1#2#3{{#2 \above#1 #3}}
\def\QTABOVE#1#2#3{{\textstyle {#2 \above#1 #3}}}
\def\QDABOVE#1#2#3{{\displaystyle {#2 \above#1 #3}}}
\def\QOVERD#1#2#3#4{{#3 \overwithdelims#1#2 #4}}
\def\QTOVERD#1#2#3#4{{\textstyle {#3 \overwithdelims#1#2 #4}}}
\def\QDOVERD#1#2#3#4{{\displaystyle {#3 \overwithdelims#1#2 #4}}}
\def\QATOPD#1#2#3#4{{#3 \atopwithdelims#1#2 #4}}
\def\QTATOPD#1#2#3#4{{\textstyle {#3 \atopwithdelims#1#2 #4}}}
\def\QDATOPD#1#2#3#4{{\displaystyle {#3 \atopwithdelims#1#2 #4}}}
\def\QABOVED#1#2#3#4#5{{#4 \abovewithdelims#1#2#3 #5}}
\def\QTABOVED#1#2#3#4#5{{\textstyle
   {#4 \abovewithdelims#1#2#3 #5}}}
\def\QDABOVED#1#2#3#4#5{{\displaystyle
   {#4 \abovewithdelims#1#2#3 #5}}}

\def\tint{\mathop{\textstyle \int}}
\def\tiint{\mathop{\textstyle \iint }}
\def\tiiint{\mathop{\textstyle \iiint }}
\def\tiiiint{\mathop{\textstyle \iiiint }}
\def\tidotsint{\mathop{\textstyle \idotsint }}
\def\toint{\mathop{\textstyle \oint}}
\def\tsum{\mathop{\textstyle \sum }}
\def\tprod{\mathop{\textstyle \prod }}
\def\tbigcap{\mathop{\textstyle \bigcap }}
\def\tbigwedge{\mathop{\textstyle \bigwedge }}
\def\tbigoplus{\mathop{\textstyle \bigoplus }}
\def\tbigodot{\mathop{\textstyle \bigodot }}
\def\tbigsqcup{\mathop{\textstyle \bigsqcup }}
\def\tcoprod{\mathop{\textstyle \coprod }}
\def\tbigcup{\mathop{\textstyle \bigcup }}
\def\tbigvee{\mathop{\textstyle \bigvee }}
\def\tbigotimes{\mathop{\textstyle \bigotimes }}
\def\tbiguplus{\mathop{\textstyle \biguplus }}

\def\dint{\mathop{\displaystyle \int}}
\def\diint{\mathop{\displaystyle \iint}}
\def\diiint{\mathop{\displaystyle \iiint}}
\def\diiiint{\mathop{\displaystyle \iiiint }}
\def\didotsint{\mathop{\displaystyle \idotsint }}
\def\doint{\mathop{\displaystyle \oint}}
\def\dsum{\mathop{\displaystyle \sum }}
\def\dprod{\mathop{\displaystyle \prod }}
\def\dbigcap{\mathop{\displaystyle \bigcap }}
\def\dbigwedge{\mathop{\displaystyle \bigwedge }}
\def\dbigoplus{\mathop{\displaystyle \bigoplus }}
\def\dbigodot{\mathop{\displaystyle \bigodot }}
\def\dbigsqcup{\mathop{\displaystyle \bigsqcup }}
\def\dcoprod{\mathop{\displaystyle \coprod }}
\def\dbigcup{\mathop{\displaystyle \bigcup }}
\def\dbigvee{\mathop{\displaystyle \bigvee }}
\def\dbigotimes{\mathop{\displaystyle \bigotimes }}
\def\dbiguplus{\mathop{\displaystyle \biguplus }}

\if@compatibility\else
  \RequirePackage{amsmath}
  \makeatother
  \endinput
\fi

\typeout{TCILATEX defining AMS-like constructs in LaTeX 2.09 COMPATIBILITY MODE}
\def\ExitTCILatex{\makeatother\endinput}

\bgroup
\ifx\ds@amstex\relax
   \message{amstex already loaded}\aftergroup\ExitTCILatex
\else
   \@ifpackageloaded{amsmath}
      {\message{amsmath already loaded}\aftergroup\ExitTCILatex}
      {}
   \@ifpackageloaded{amstex}
      {\message{amstex already loaded}\aftergroup\ExitTCILatex}
      {}
   \@ifpackageloaded{amsgen}
      {\message{amsgen already loaded}\aftergroup\ExitTCILatex}
      {}
\fi
\egroup

\let\DOTSI\relax
\def\RIfM@{\relax\ifmmode}
\def\FN@{\futurelet\next}
\newcount\intno@
\def\iint{\DOTSI\intno@\tw@\FN@\ints@}
\def\iiint{\DOTSI\intno@\thr@@\FN@\ints@}
\def\iiiint{\DOTSI\intno@4 \FN@\ints@}
\def\idotsint{\DOTSI\intno@\z@\FN@\ints@}
\def\ints@{\findlimits@\ints@@}
\newif\iflimtoken@
\newif\iflimits@
\def\findlimits@{\limtoken@true\ifx\next\limits\limits@true
 \else\ifx\next\nolimits\limits@false\else
 \limtoken@false\ifx\ilimits@\nolimits\limits@false\else
 \ifinner\limits@false\else\limits@true\fi\fi\fi\fi}
\def\multint@{\int\ifnum\intno@=\z@\intdots@
 \else\intkern@\fi
 \ifnum\intno@>\tw@\int\intkern@\fi
 \ifnum\intno@>\thr@@\int\intkern@\fi
 \int}
\def\multintlimits@{\intop\ifnum\intno@=\z@\intdots@\else\intkern@\fi
 \ifnum\intno@>\tw@\intop\intkern@\fi
 \ifnum\intno@>\thr@@\intop\intkern@\fi\intop}
\def\intic@{
    \mathchoice{\hskip.5em}{\hskip.4em}{\hskip.4em}{\hskip.4em}}
\def\negintic@{\mathchoice
 {\hskip-.5em}{\hskip-.4em}{\hskip-.4em}{\hskip-.4em}}
\def\ints@@{\iflimtoken@
 \def\ints@@@{\iflimits@\negintic@
   \mathop{\intic@\multintlimits@}\limits
  \else\multint@\nolimits\fi
  \eat@}
 \else
 \def\ints@@@{\iflimits@\negintic@
  \mathop{\intic@\multintlimits@}\limits\else
  \multint@\nolimits\fi}\fi\ints@@@}
\def\intkern@{\mathchoice{\!\!\!}{\!\!}{\!\!}{\!\!}}
\def\plaincdots@{\mathinner{\cdotp\cdotp\cdotp}}
\def\intdots@{\mathchoice{\plaincdots@}
 {{\cdotp}\mkern1.5mu{\cdotp}\mkern1.5mu{\cdotp}}
 {{\cdotp}\mkern1mu{\cdotp}\mkern1mu{\cdotp}}
 {{\cdotp}\mkern1mu{\cdotp}\mkern1mu{\cdotp}}}

\def\RIfM@{\relax\protect\ifmmode}
\def\text{\RIfM@\expandafter\text@\else\expandafter\mbox\fi}
\let\nfss@text\text
\def\text@#1{\mathchoice
   {\textdef@\displaystyle\f@size{#1}}
   {\textdef@\textstyle\tf@size{\firstchoice@false #1}}
   {\textdef@\textstyle\sf@size{\firstchoice@false #1}}
   {\textdef@\textstyle \ssf@size{\firstchoice@false #1}}
   \glb@settings}

\def\textdef@#1#2#3{\hbox{{
                    \everymath{#1}
                    \let\f@size#2\selectfont
                    #3}}}
\newif\iffirstchoice@
\firstchoice@true

\def\Let@{\relax\iffalse{\fi\let\\=\cr\iffalse}\fi}
\def\vspace@{\def\vspace##1{\crcr\noalign{\vskip##1\relax}}}
\def\multilimits@{\bgroup\vspace@\Let@
 \baselineskip\fontdimen10 \scriptfont\tw@
 \advance\baselineskip\fontdimen12 \scriptfont\tw@
 \lineskip\thr@@\fontdimen8 \scriptfont\thr@@
 \lineskiplimit\lineskip
 \vbox\bgroup\ialign\bgroup\hfil$\m@th\scriptstyle{##}$\hfil\crcr}
\def\Sb{_\multilimits@}
\def\endSb{\crcr\egroup\egroup\egroup}
\def\Sp{^\multilimits@}
\let\endSp\endSb

\newdimen\ex@
\ex@.2326ex
\def\rightarrowfill@#1{$#1\m@th\mathord-\mkern-6mu\cleaders
 \hbox{$#1\mkern-2mu\mathord-\mkern-2mu$}\hfill
 \mkern-6mu\mathord\rightarrow$}
\def\leftarrowfill@#1{$#1\m@th\mathord\leftarrow\mkern-6mu\cleaders
 \hbox{$#1\mkern-2mu\mathord-\mkern-2mu$}\hfill\mkern-6mu\mathord-$}
\def\leftrightarrowfill@#1{$#1\m@th\mathord\leftarrow
\mkern-6mu\cleaders
 \hbox{$#1\mkern-2mu\mathord-\mkern-2mu$}\hfill
 \mkern-6mu\mathord\rightarrow$}
\def\overrightarrow{\mathpalette\overrightarrow@}
\def\overrightarrow@#1#2{\vbox{\ialign{##\crcr\rightarrowfill@#1\crcr
 \noalign{\kern-\ex@\nointerlineskip}$\m@th\hfil#1#2\hfil$\crcr}}}
\let\overarrow\overrightarrow
\def\overleftarrow{\mathpalette\overleftarrow@}
\def\overleftarrow@#1#2{\vbox{\ialign{##\crcr\leftarrowfill@#1\crcr
 \noalign{\kern-\ex@\nointerlineskip}$\m@th\hfil#1#2\hfil$\crcr}}}
\def\overleftrightarrow{\mathpalette\overleftrightarrow@}
\def\overleftrightarrow@#1#2{\vbox{\ialign{##\crcr
   \leftrightarrowfill@#1\crcr
 \noalign{\kern-\ex@\nointerlineskip}$\m@th\hfil#1#2\hfil$\crcr}}}
\def\underrightarrow{\mathpalette\underrightarrow@}
\def\underrightarrow@#1#2{\vtop{\ialign{##\crcr$\m@th\hfil#1#2\hfil
  $\crcr\noalign{\nointerlineskip}\rightarrowfill@#1\crcr}}}
\let\underarrow\underrightarrow
\def\underleftarrow{\mathpalette\underleftarrow@}
\def\underleftarrow@#1#2{\vtop{\ialign{##\crcr$\m@th\hfil#1#2\hfil
  $\crcr\noalign{\nointerlineskip}\leftarrowfill@#1\crcr}}}
\def\underleftrightarrow{\mathpalette\underleftrightarrow@}
\def\underleftrightarrow@#1#2{\vtop{\ialign{##\crcr$\m@th
  \hfil#1#2\hfil$\crcr
 \noalign{\nointerlineskip}\leftrightarrowfill@#1\crcr}}}

\def\qopnamewl@#1{\mathop{\operator@font#1}\nlimits@}
\let\nlimits@\displaylimits
\def\setboxz@h{\setbox\z@\hbox}

\def\varlim@#1#2{\mathop{\vtop{\ialign{##\crcr
 \hfil$#1\m@th\operator@font lim$\hfil\crcr
 \noalign{\nointerlineskip}#2#1\crcr
 \noalign{\nointerlineskip\kern-\ex@}\crcr}}}}

 \def\rightarrowfill@#1{\m@th\setboxz@h{$#1-$}\ht\z@\z@
  $#1\copy\z@\mkern-6mu\cleaders
  \hbox{$#1\mkern-2mu\box\z@\mkern-2mu$}\hfill
  \mkern-6mu\mathord\rightarrow$}
\def\leftarrowfill@#1{\m@th\setboxz@h{$#1-$}\ht\z@\z@
  $#1\mathord\leftarrow\mkern-6mu\cleaders
  \hbox{$#1\mkern-2mu\copy\z@\mkern-2mu$}\hfill
  \mkern-6mu\box\z@$}

\def\projlim{\qopnamewl@{proj\,lim}}
\def\injlim{\qopnamewl@{inj\,lim}}
\def\varinjlim{\mathpalette\varlim@\rightarrowfill@}
\def\varprojlim{\mathpalette\varlim@\leftarrowfill@}
\def\varliminf{\mathpalette\varliminf@{}}
\def\varliminf@#1{\mathop{\underline{\vrule\@depth.2\ex@\@width\z@
   \hbox{$#1\m@th\operator@font lim$}}}}
\def\varlimsup{\mathpalette\varlimsup@{}}
\def\varlimsup@#1{\mathop{\overline
  {\hbox{$#1\m@th\operator@font lim$}}}}

\def\stackunder#1#2{\mathrel{\mathop{#2}\limits_{#1}}}

\begingroup \catcode `|=0 \catcode `[= 1
\catcode`]=2 \catcode `\{=12 \catcode `\}=12
\catcode`\\=12
|gdef|@alignverbatim#1\end{align}[#1|end[align]]
|gdef|@salignverbatim#1\end{align*}[#1|end[align*]]

|gdef|@alignatverbatim#1\end{alignat}[#1|end[alignat]]
|gdef|@salignatverbatim#1\end{alignat*}[#1|end[alignat*]]

|gdef|@xalignatverbatim#1\end{xalignat}[#1|end[xalignat]]
|gdef|@sxalignatverbatim#1\end{xalignat*}[#1|end[xalignat*]]

|gdef|@gatherverbatim#1\end{gather}[#1|end[gather]]
|gdef|@sgatherverbatim#1\end{gather*}[#1|end[gather*]]

|gdef|@gatherverbatim#1\end{gather}[#1|end[gather]]
|gdef|@sgatherverbatim#1\end{gather*}[#1|end[gather*]]

|gdef|@multilineverbatim#1\end{multiline}[#1|end[multiline]]
|gdef|@smultilineverbatim#1\end{multiline*}[#1|end[multiline*]]

|gdef|@arraxverbatim#1\end{arrax}[#1|end[arrax]]
|gdef|@sarraxverbatim#1\end{arrax*}[#1|end[arrax*]]

|gdef|@tabulaxverbatim#1\end{tabulax}[#1|end[tabulax]]
|gdef|@stabulaxverbatim#1\end{tabulax*}[#1|end[tabulax*]]

|endgroup

\def\align{\@verbatim \frenchspacing\@vobeyspaces \@alignverbatim
You are using the "align" environment in a style in which it is not defined.}
\let\endalign=\endtrivlist

\@namedef{align*}{\@verbatim\@salignverbatim
You are using the "align*" environment in a style in which it is not defined.}
\expandafter\let\csname endalign*\endcsname =\endtrivlist

\def\alignat{\@verbatim \frenchspacing\@vobeyspaces \@alignatverbatim
You are using the "alignat" environment in a style in which it is not defined.}
\let\endalignat=\endtrivlist

\@namedef{alignat*}{\@verbatim\@salignatverbatim
You are using the "alignat*" environment in a style in which it is not defined.}
\expandafter\let\csname endalignat*\endcsname =\endtrivlist

\def\xalignat{\@verbatim \frenchspacing\@vobeyspaces \@xalignatverbatim
You are using the "xalignat" environment in a style in which it is not defined.}
\let\endxalignat=\endtrivlist

\@namedef{xalignat*}{\@verbatim\@sxalignatverbatim
You are using the "xalignat*" environment in a style in which it is not defined.}
\expandafter\let\csname endxalignat*\endcsname =\endtrivlist

\def\gather{\@verbatim \frenchspacing\@vobeyspaces \@gatherverbatim
You are using the "gather" environment in a style in which it is not defined.}
\let\endgather=\endtrivlist

\@namedef{gather*}{\@verbatim\@sgatherverbatim
You are using the "gather*" environment in a style in which it is not defined.}
\expandafter\let\csname endgather*\endcsname =\endtrivlist

\def\multiline{\@verbatim \frenchspacing\@vobeyspaces \@multilineverbatim
You are using the "multiline" environment in a style in which it is not defined.}
\let\endmultiline=\endtrivlist

\@namedef{multiline*}{\@verbatim\@smultilineverbatim
You are using the "multiline*" environment in a style in which it is not defined.}
\expandafter\let\csname endmultiline*\endcsname =\endtrivlist

\def\arrax{\@verbatim \frenchspacing\@vobeyspaces \@arraxverbatim
You are using a type of "array" construct that is only allowed in AmS-LaTeX.}
\let\endarrax=\endtrivlist

\def\tabulax{\@verbatim \frenchspacing\@vobeyspaces \@tabulaxverbatim
You are using a type of "tabular" construct that is only allowed in AmS-LaTeX.}
\let\endtabulax=\endtrivlist

\@namedef{arrax*}{\@verbatim\@sarraxverbatim
You are using a type of "array*" construct that is only allowed in AmS-LaTeX.}
\expandafter\let\csname endarrax*\endcsname =\endtrivlist

\@namedef{tabulax*}{\@verbatim\@stabulaxverbatim
You are using a type of "tabular*" construct that is only allowed in AmS-LaTeX.}
\expandafter\let\csname endtabulax*\endcsname =\endtrivlist

 \def\endequation{
     \ifmmode\ifinner
      \iftag@
        \addtocounter{equation}{-1}
        $\hfil
           \displaywidth\linewidth\@taggnum\egroup \endtrivlist
        \global\tag@false
        \global\@ignoretrue
      \else
        $\hfil
           \displaywidth\linewidth\@eqnnum\egroup \endtrivlist
        \global\tag@false
        \global\@ignoretrue
      \fi
     \else
      \iftag@
        \addtocounter{equation}{-1}
        \eqno \hbox{\@taggnum}
        \global\tag@false
        $$\global\@ignoretrue
      \else
        \eqno \hbox{\@eqnnum}
        $$\global\@ignoretrue
      \fi
     \fi\fi
 }

 \newif\iftag@ \tag@false

 \def\TCItag{\@ifnextchar*{\@TCItagstar}{\@TCItag}}
 \def\@TCItag#1{
     \global\tag@true
     \global\def\@taggnum{(#1)}}
 \def\@TCItagstar*#1{
     \global\tag@true
     \global\def\@taggnum{#1}}

  \@ifundefined{tag}{
     \def\tag{\@ifnextchar*{\@tagstar}{\@tag}}
     \def\@tag#1{
         \global\tag@true
         \global\def\@taggnum{(#1)}}
     \def\@tagstar*#1{
         \global\tag@true
         \global\def\@taggnum{#1}}
  }{}

\def\tfrac#1#2{{\textstyle {#1 \over #2}}}
\def\dfrac#1#2{{\displaystyle {#1 \over #2}}}
\def\binom#1#2{{#1 \choose #2}}
\def\tbinom#1#2{{\textstyle {#1 \choose #2}}}
\def\dbinom#1#2{{\displaystyle {#1 \choose #2}}}

\makeatother
\endinput

config: arxiv


TITLE: Board games and dice, landing on a specific number with dice rolls?
QUESTION [2 upvotes]: I ended up with a complicated problem.
I want to know what the chance is if you have a board game that you will land on a specific tile assuming that you will continue until you pass that tile.
the chance of landing on tiles is not equal (although I assume the further you get in the game the more this should even out).

landing on the first tile is a $1/6$ chance.
landing on the second
tile, throwing either $2$ or $1,1$ will give you a $1/6+1/36=7/36$ chance.
3rd gives, $3;1,2;2,1;1,1,1$ (roll a $3$, a $2$ and $1$ or three times a $1$),
which equals to $1/6+2\times1/36+1/216=49/216$ chance.
4th tile,
$4;1,3;3,1;2,2;2,1,1;1,2,1;1,1,2;1,1,1,1$ which gives
$1/6+2\times1/36+1/36+3\times1/216+1/6^4=343/1296$ chance.
5th tile,
$5;2\times(2,3);2\times(1,4);3\times(2,2,1);3\times(3,1,1);4\times(2,1,1,1);(1,1,1,1,1)$, which
gives
$1/6+2\times1/36+2\times1/36+3\times1/6^3+3\times1/6^3+4\times1/6^4+1/6^5 =
   2401/7776$ chance.

As you can see it becomes more complicated to do this by hand the higher the number is. So I want to make a script (in matlab) to calculate this for up to a 100 or so tiles, where I know the chances of landing on each of those 100.
I have an idea how to get the 6th and 7th tiles
but after that I'm not that sure anymore.
I'm sure that up to the sixth tile you could multiply the previous chances by $1/6$, add them up and add $1/6$.
for the 7th tile I have a good idea (multiply the previous 6 by $1/6$ and add them up) but I need to verify this.
Any help and verification is greatly appreciated.
Also I might possibly look into situations where you have two dice and thus an uneven chance distribution.
Ps. Thanks AMD for the edit.
Ps.Ps the $771/2497$ is a numerical error from matlab since apparently when using rational format it only takes into consideration up to 8 digits. This means that $771/2497$ is so close to $2401/7776$ that it can't see the difference.
Also for those interested a crude matlab code is below
format rat
nd=6; %Has to be bigger than 1, highest number on die
n=100; %tiles on board
A=zeros(n,1);
A(1,1)=1/nd;

for ind=2:n
    if (ind <= nd)
       k=ind-1;
       A(ind,1)=1/nd;
       for t=1:k
           A(ind,1)=A(ind,1)+A(t,1)/nd;
       end
    else
       k=ind-1;
       p=ind-nd;
       for t=p:k
           A(ind,1)=A(ind,1)+A(t,1)/nd;
       end
    end
end

hold on
plot(1:50,A(1:50,1))

%display(A)
% format short
% display(A)

REPLY [4 votes]: You have the right idea.
The recursion of $P_n$ (the probability that the tile $n$ will be landed on) is
$$
P_n = \frac16(P_{n-1}+P_{n-2}+P_{n-3}+P_{n-4}+P_{n-5}+P_{n-6})
$$
with $P_0 = 1$ and for all $k>0$, $P_{-k}=0$.
This recursion can almost be solved in closed form.  The key equation related to it, obtained by assuming a solution of the form $x^n$, is
$$
x^6 = \frac16(x^5+x^4+x^3+x^2+x+1)
$$
which has a root of $1$, two pairs of complex conjugate roots whose absolute values are about $0.73$ and $.68$, and a real root of about $-0.670$.
Using the values of $P_0$ and $P_{-k}$ for small $k>0$ to determine the coefficients in front of each of those $x^n$, one obtains
$$ \begin{array}{lllll}P_n = \frac27 + \frac17 \left[ (-0.670332)^n + (-0.375695-0.570175i)^n \\ + (-0.375695+0.570175i)^n  + (0.294195-.668367i)^n + (0.294195+.668367i)^n \right] \end{array}
$$
From this we can see that as $n$ grows very large, the probability of landing on any particular tile approaches $\frac27$.  One can "justify" this as follows:  The average step size is $3.5$ and the reciprocal of that is $\frac27$.
Notice that the real parts of those five roots (other than the root of value $1$) add up to $-\frac56$ and if you plug in $n=1$ you find that $P_1 = \frac27 - \frac17 \cdot \frac56 = \frac16$, as expected.


TITLE: Nonlinear optimization question
QUESTION [2 upvotes]: For (x,y) in $\mathbb R^2$, consider f(x,y) = $x^2 -2xy + \frac{4}{3}y^2 - 4y$ Find the local minimum of f. Is it a strict local minimum? Compute the $\lim\limits_{|(x,y)|\to \infty}$ f(x,y) to decide if the local minimum is a global minimum.
My work: I found the critical point (6,6) and for any v (since it is in the interior of $\mathbb R^2$) is feasible. Therefore $\vec\nabla$F.v = 0 since (6,6) is in $\mathbb R^2$.
I'm having trouble how taking the shown limit will prove whether the local min is a global min.
Thanks,

REPLY [1 votes]: Hint: Evaluate the determinant of the Hessian matrix at your point. (Second Derivative test). I.E.
$f_{xx}f_{yy}-f_{xy}^2|_{(6,6)}$
This PDF: http://www.math.osu.edu/~kwa.1/254au10/14.7.pdf will give you a nice in-depth explanation of the second derivative test.
This is for your first question.
As per your second question:
$\lim_{|(x,y)|→∞} f(x,y) = \infty$
Evaluating the limit we see $\lim_{|(x,y)|→∞} f(x,y)= \lim_{|(x,y)|→∞} x^2 +\frac{4}{3}y^2 - (2xy + 4y)$ ~ $\lim_{|(x,y)|\to \infty} x^2 + \frac{4}{3}y^2 = \infty$
We get to do this because $x^2 + \frac{4}{3}y^2 $ > $ (2xy + 4y)$
for all values outside of and on the boundary of the ellipse. (See picture for more info)

Credits: http://www.wolframalpha.com
Explicitly: $x^2 + \frac{4}{3}y^2 $ is growing a lot faster than the other term: $2xy + 4y$,  (This does not prove it is an absolute minimum, I just was showing you the limit.)
Explicitly, for all $(x,y)$, we have:
$$
f_{xx}\mid_{(x,y)} = 2, \;\;\; f_{yy}\mid_{(x,y)} = \frac{8}{3},\;\;\; f_{xy}\mid_{(x,y)}= -2
$$
This implies:
$$
D(H)\mid_{(x,y)} = f_{xx}f_{yy} - f_{xy}^2 = \frac{16}{3} - 4 = \frac{4}{3}
$$
Where $D(H)$ is the determinant of the hessian matrix. Since $D(H)\mid_{(6,6)}>0$ and $f_{xx}\mid_{(6,6)} > 0$ we have a local minimum.
Moreover, since both $f_{xx}$ and $f_{yy}$ are positive at the critical point and the discriminant is positive, the
surface must be concave upward in every direction.
Thus, this must be an elliptic paraboloid with a global minimum at $(6,6)$.


TITLE: Question about Euclid's infinite prime proof
QUESTION [0 upvotes]: Suppose that $p_1=2 < p_2 = 3 < \cdots < p_r$ are all of the primes. Let $P = p_1p_2...p_r+1$ and let $p_s$ be a prime dividing $P$ where $p_s$ is not in our original list $p_1, p_2, \cdots, p_r$. If $p_s$ can divide $p$, wouldn't it be able to divide 1?
i.e.
$7 = 2*3 + 1$ meaning $p_s | 2*3$ and $p_s | 1$?

REPLY [2 votes]: "Let $p_s$ be a prime dividing $P$ where $p_s$ is not in our original list." - This doesn't make sense. You assumed you have a list of all primes $\{p_i\}$. So $p_s$ must belong to $\{p_i\}$. This is what results in the contradiction.


TITLE: Visualization of subspaces
QUESTION [1 upvotes]: Are there any ways in MATLAB or MuPAD to visualize the nullspace and range for the given vector space?

REPLY [2 votes]: Here I just say how to do it with Maple for other softwares maybe one other will tell you, also for Mathematica you can ask the Mathematica section of stackexchange.
As an example I'll compute the NullSpace of the following matrix;
$$\begin{bmatrix}1 & 1 & 1\\ 0 & 2 & -1\\ 2 & 0 & 3\end{bmatrix}$$
with(LinearAlgebra):
M := Matrix([[1, 1, 1], [0, 2, -1], [2, 0, 3]]):
L:= NullSpace(M);

You should call the LinearAlgebra package for using matrixes and the command NullSpace, and for calling a package we do it with with(...) and in ... you should write name of the package. The ":" sign is for when you don't want to see the output and ";" sign for when you want to see the output. So when you finished writing a command, you use one of these two signs and then press Enter from keyboard. Then you should give the matrix to Maple. You can define a matrix as I did, writing its rows. You can do it using with columns in another way as following.
M:=<<1,0,2>|<1,2,0>|<1,-1,3>>:

And finally you using NullSpace(...) you will see a basis for the Null space of your matrix. You could do it in one step if you are not going to use your matrix again and again in the sequel.
NullSpace(Matrix([[1, 1, 1], [0, 2, -1], [2, 0, 3]]));

For plotting you can use the two following ways.
plot3d([-(3/2)*t, (1/2)*t, t], t = -5 .. 0, s = 0 .. 5);

X := <x, y, z>:
f := M.X:
with(plots):
implicitplot3d([f[1] = 0, f[2] = 0, f[3] = 0], x = -5 .. 5, y = -5 .. 5, z = -5 .. 5);

The null space of $M$ is the line in the first figure which is the intersection of the three planes in the second figure.


TITLE: Fixing up a homotopy to respect boundaries
QUESTION [1 upvotes]: Suppose that $M$and $N$ are manifolds with nonempty boundary and that $f,g : M \to N$ with $f(\partial M), g(\partial M) \subseteq \partial N$.  Also, suppose that $f \simeq g$ and that as maps $f|_{\partial M}, g|_{\partial M} : \partial M \to \partial N$, we have $f|_{\partial M} \simeq g|_{\partial M}$.
Is it then true that $f$ can be homotoped to $g$ with a homotopy $H : M \times I \to M$ with $H_t(\partial M) \subseteq \partial N$ for all $t \in I$?  In particular, I am interested in the case where $M$ and $N$ are handlebodies.

REPLY [1 votes]: No. Consider a manifold $M$ with connected boundary $\partial M$ so that $\pi_1(M,\partial M) $ has more than one element. Let f,g:I\rightarrow M represent distinct elements of $\pi_1(M,\partial M) $ . They satisfy the hypotheses of your question but cannot be homotoped to each other keeping the boundary of $I$ in $\partial M$.


TITLE: Finding a cubic equation from the relation between the roots
QUESTION [1 upvotes]: I'm trying to solve this problem:
$ x^3 - x^2 - 3x -10 = 0$ has roots α,β,γ.
Let u = −α+β+γ. Show that u+2α=1, and hence find a cubic equation having roots −α+β+γ, α−β+γ, α+β−γ.
I was able to complete the first part easily using Vieta's theorem, although I am having trouble with the second part.

REPLY [1 votes]: For the second part, let $y=\alpha + \beta + \gamma -2x$, where x is a root of the original polynomial.
Then $y=1-2x \implies x=\frac{1}{2}-\frac{y}{2}$.
Substitute this into $x^3-x^2-3x-10=0$ to find the required equation.


TITLE: Show why the proof for the following is false $u\mid mn$ and $\gcd(u,m) = 1$, then $m = \pm1$
QUESTION [0 upvotes]: Let $u,m,n$ be three integers.
If $u\mid mn$ and $\gcd(u,m) = 1$, then $m = \pm 1.$
If $\gcd(u,m) = 1$, then $1 = us + mt$ for some integers $s, t$.
If $u \mid mn$, then
$us = mn$ for some integer s.
Hence, $1 = mn + mt = m(n + t)$, which
implies that $m|1$, and therefore $m = 1$.
First dips into discrete math and having trouble with proofs
I've been playing around with various values and found $u=2, m=3, n=4,$ to be a counterexample to the proof but have yet to find the flash of inspiration needed to understand why.
Any help is appreciated.

REPLY [0 votes]: There exist integers $s ,t$ with $1=us+mt.$ There exists integer $s'$ with $us'=mn.$ IF $s=s'$ THEN $m=\pm 1.$ But if $s\ne s'$ then ... ?