COMMENT ⊗   VALID 00005 PAGES
C REC  PAGE   DESCRIPTION
C00001 00001
C00002 00002	\def\sample#1.{\yyskip\noindent
C00006 00003	\sample 6.
C00008 00004	\sample 11.
C00012 00005	\sample 16.
C00021 ENDMK
C⊗;
\def\sample#1.{\yyskip\noindent
{\:<Sample #1.\par\penalty1000\vskip 6pt}}
\def\Nbf{\hbox{\bf N}}
\def\starsub#1{*↓{\lower 1pt\hbox{\hskip-5pt
$\scriptstyle#1$}}}
\gdef\tpage{F}\gdef\rauth{}\gdef\rtitle{}\tenpoint

\sample 1.
\thbegin {10. Corollary}. For large $n$, $[k↓0(k↑{(n)}:\,k↓0]↓i=[\=k↓0]↓i
=[\=k↓0:\,k↓0]↓i$.

\proofbegin Proof. For $n≥r$, $\=k↓0(k↑{p↑n})=\=k↓0\oplus↓{\=k↓0↑s}k↓0(k↑{p↑n})$
since $\=k↓0/\=k↓0↑s$ is purely inseparable and $k↓0(k↑{p↑n})/k↓0$ is
separable [{\bf3}, Theorem 21, Part (1), p.\ 197]. Thus

\sample 2.
$$\twoline{\roman{gr} T↓1=\leftset (x↓i,x\starsub i) \relv x↓i\in X,
x\starsub i\in X*;\;\hbox{$∃$ a net $\in \roman{gr}↓i T$}}{\hbox{$x↓i$
bounded, $x↓i→x**$ weak** and $x\starsub i→x*$ in norm}\rightset}$$
(or something like that).

\sample 3.
\def\extralead{\noalign{\vskip 3pt}}
\noindent where $P↓1$, $P↓2$, $Q$ are integers such that $(P↓1,P↓2,Q)=1$.
Put $\delta=\rho↓2-\rho↓1$ and define $\Delta=\delta↑2=P↓{\!1}↑2-4P↓2$,\xskip
$E=(P↓2+4Q)↑2-4QP↓{\!1}↑2$,
$$V↓n={1\over\delta}\left|
\cpile{α↓1↑n+β↓1↑n\cr α↓2↑n+β↓2↑n\cr}\quad
\cpile{\rho↓1\cr \rho↓2\cr}\right|,\qquad
U↓n={1\over\delta}\left|\cpile{1\cr 1\cr}\quad
\cpile{α↓1↑n+β↓1↑n\cr α↓2↑n+β↓2↑n\cr}\right|.$$
The first few values for these functions are given in the table below.
$$\rpile{n\cr \extralead 0\cr 1\cr 2\cr 3\cr 4\cr}\qquad
\lpile{V↓n\cr \extralead 2\cr 0\cr -P↓2-2Q\cr -P↓1P↓2\cr
P↓{\!2}↑2-P↓{\!1}↑2P↓2+4P↓2Q+2Q↑2\cr}\qquad
\lpile{U↓n\cr \extralead 0\cr 1\cr P↓1\cr P↓{\!1}↑2-P↓2-3Q\cr
P↓{\!1}↑3-2P↓1P↓2-4P↓1Q\cr}$$

\sample 4.
\noindent and$$\eqalign{X↓{2m}⊗≡\left\{
\lpile{Q(X↓m↑2-P↓2W↓{\!m}↑2)-2S↑2\cr
P↓{\!2}↑2(X↓m↑2-P↓2W↓{\!m}↑2)-2S↑2\cr}\quad
\cpile{\hbox{($m$ odd)}\cr \hbox{($m$ even)}\cr}\right.
\mod N,\cr
\extralead
W↓{2m}⊗≡\left\{
\lpile{Q(2X↓mW↓m-P↓1W↓{\!m}↑2)\cr
P↓{\!2}↑2(2X↓mW↓m+P↓1W↓{\!m}↑2)\cr}\quad
\cpile{\hbox{($m$ odd)}\cr \hbox{($m$ even)}\cr}\right.
\mod N.\cr}$$

\sample 5.
\thbegin {1.2. Corollary}. The confluent image of
$$\left\{\,\lpile{\hbox{an arc}\cr\hbox{a circle}\cr\hbox{a fan}\cr}\,\right\}
\qquad\hbox{is}\qquad
\left\{\,\lpile{\hbox{an arc}\cr\hbox{an arc or circle}\cr
\hbox{a fan on an arc}\cr}\,\right\}.$$

\sample 6.
$$\twoline{\limsup↓{n→∞}\sum↓k|X↓{nk}|↑\gamma≤\lim↓{ε\down0}\limsup↓{n→∞}\sum↓k
|X↓{nk}|↑\gamma\,I↓{nk}(ε)}{\null+\lim↓{ε\down0}\limsup↓{n→∞}\sum↓k|X↓{nk}|↑\gamma
\,\biglp 1-I↓{nk}(ε)\bigrp,}$$

\sample 7.
$\|u\|≤\|u+λx\|-\delta$, for some $\delta=\delta(x,ε)>0$ for each $u\in U$,\xskip
$\|u\|=1$, and each $λ$,\xskip $|λ|≥ε$.

\sample 8.
\noindent Therefore
$${\int↓{-1}↑1 P↓{\!0}↑{\prime2}(x)\,dx\over\int↓{-1}↑1 P↓{\!0}↑2(x)\,dx}
={n\biglp n+{1\over2}\bigrp\over2(n-1)}={n\over 2}+{3\over4}+{3\over 4(n-1)},
\qquad n>1.$$

\sample 9.
{\sl$$\twoline{\lpile{\dispstyle{\max↓{-1≤x≤+1}|P↓{\!n}↑\prime(x)|\over
\max↓{-1≤x≤+1}|P↓n(x)|}≥{n\over 2}\qquad\hbox{if $n=2$, 3,}\cr
\dispstyle
\qquad≥{n\over\sqrt{n-1}}\bigglp 1-{1\over n-1}\biggrp↑{(n-2)/2},\qquad\hbox{$n$
even,\xskip $n≥4$,}\cr
\dispstyle
\qquad≥{n↑2\over(n-1)\sqrt{n+1}}\biggglp 1-{\sqrt{n+1}\over n-1}\bigggrp↑{(n-3)/2}
\biggglp 1+{1\over\sqrt{n+1}}\bigggrp↑{(n-1)/2},\cr}}{\hbox{if $n≥5$ and odd.}}$$}

\sample 10.
$$a\,\left(\cpile{x↓1\cr \vdots\cr x↓n\cr}\right)
+\left(\cpile{b↓1\cr \vdots\cr b↓n\cr}\right)
\,x↓{m+1}=
\left(\cpile{c↓1\cr \vdots\cr c↓m\cr}\right)\,,$$

\sample 11.
$$k↓0(k↑{(1)})=\leftset u\in K\relv u↑{p↑n}\in k↓0(k↑{1+p↑n})\rightset =
k↓0↑{p↑{-n}}(k↑p)∩k.$$
Let $V↓m=\{v↓{i,j}\}↓{i=m,j=1}↑{t-1,q↓i}$ and define $X↓m$, $Y↓m$, $Z↓m$ 
similarly.

\thbegin Contention. $k↓0↑{p↑{-n}}(k↑p)⊂k=k↓0(s↑p)(V↓1,w↑{p↑{-n}})$.

\proofbegin Proof. The claim is that

\sample 12.
\eightpoint
$$\hbox par 200pt{{\nm Abstract}.\xskip Let $k⊃s⊃k↓0$ be fields of
characteristic $p≠0$,\xskip $k/k↓0$ finitely generated and $s$ a distinguished
subfield. The field $k↓0(k↑{(n)})=\leftset x\in k\relv x↑{p↑t}\in k↓0(k↑{p↑{n+t}})$
for some $t≥0\rightset$ has $k↓0(s↑{p↑n})$ as a distinguished subfield and is
}$$\tenpoint

\sample 13.
\proofbegin Proof. Consider a minimal normal subgroup $N\subset \Phi(G)≠1$ and
the factor group $G/N$. Then $\leftset NL↓j/N\relv j=1, \ldotss, n\rightset$ is a
lower nilpotent series for $G↓\Fscr/N$,\xskip$N\Nbf↓{G↓\Fscr}(\Sscr↓{n-j-1})/N$
is a relative s-stem normalizer for $NL↓{n-j-1}/N$ with respect to the Sylow
system $\Sscr*$ of $G↓\Fscr/N$ into which $\Sscr$ reduces, and the prefrattini
series for $(G/N)↓\Fscr$ is $N\subset\Fscr↓1/N=\=\Phi↓j⊂NL\starsub1/N=\=L\starsub1
\subset\cdots⊂NL\starsub{n-1}/N=\=L\starsub{n-1}⊂\Phi↓n/N=G↓\Fscr/N$. The result is
valid whenever $G↓\Fscr$ is nilpotent. By induction, $\inter↓{j=1}↑{n-1}\=\Phi↓j
(N\Nbf↓{G↓\Fscr}(\Sscr↓{n-j-1})/N$ is a prefrattini subgroup $W/N$ of $G/N$
with respect to the Sylow system $\Sscr N/N$ (see \ref7). Since
$$\eqalign{W/N⊗=\inter↓{j=1}↑{n-1}(\phi↓j/N)\biglp N\Nbf↓{G↓\Fscr}(\Sscr↓{n-j-1})/N
\bigrp\cr
⊗=\inter↓{j=1}↑{n-1}\biglp\Phi↓j \Nbf↓{G↓\Fscr}(\Sscr↓{n-j-1})/N\bigrp
=\bigglp\inter↓{j=1}↑{n-1}\biglp\Phi↓j\Nbf↓{G↓\Fscr}(\Sscr↓{n-j-1})\bigrp/N\biggrp,
\cr}$$

\sample 14.
$$\cpile{\null\cr\null\cr\null\cr\null\cr\null\cr\null\cr\null\cr\null\cr
\extralead\hbox{Total}\cr}\quad
\cpile{\hbox{Permutations}\cr\null\cr 123\cr 132\cr 231\cr 213\cr 312\cr 321\cr
\extralead 6=n!\cr}\quad
\cpile{\hbox{One peak}\cr\hbox{(underlined)}\cr 123\cr
\underline{123}\cr \underline{231}\cr 213\cr 312\cr 321\cr \extralead
2=P↓2\cr}\quad
\cpile{\hbox{Zero peak}\cr\hbox{(underlined)}\cr \underline{123}\cr
132\cr 231\cr \underline{231}\cr \underline{312}\cr \underline{321}\cr
\extralead 4=P↓1\cr}$$

\sample 15.
$$=1+\sum↑∞↓{n↓e=2;(n↓e\,\hbox{\:d even})}(-1)↑{n↓e/2}\;\left(
\sum↓{h↓1=0}↑{n↓e/2-1}b↓{h↓1h↓2h↓3}k↑{2h↓1}\right)\;{u↑{n↓e}\over n↓e!},
\qquad h↓3=n↓e/2$$

\sample 16.
\def\diagdots{\raise4pt\hbox{$\char'401$}\≥\char'401\≥\lower4pt\hbox{$\char'401$}}
$$A↓N=\left[\cpile{α↓0↑N+\gamma↓0 h\cr \null-1-β↓1↑N h\cr \diagdots\cr
\null\cr\null\cr}\quad
\cpile{-α↓0↑N\cr 2+(β↓1↑N+α↓1↑N)h+\gamma↓1↑N h↑2\cr \diagdots\cr
\null-1-β↓{N-1}↑N h\cr -β↓N↑N\cr}\quad
\cpile{\null\cr \null-1-α↓1↑N h\cr \diagdots\cr 2+(β↓{N-1}↑N+α↓{N-1}↑N)h+
\gamma↓{N-1}↑N h↑2\cr β↓N↑N+\gamma↓N\,h\cr}\right].$$

\sample 17.
\ctrline{\nm Table 1}
\penalty1000\vskip 6pt
\ctrline{\sl Interpolation Formulas for $(x↓\ast,t↓\ast)=(.2,.4)$}
\penalty1000\vskip 6pt
\vbox{\tabskip0pt plus 100pt\halign to size{$\ctr{#}$⊗\!
$\rt{#}$\tabskip 0pt⊗$#\hfill$\tabskip0pt plus 100pt⊗\!
$\rt{#}$\tabskip 0pt⊗$#\hfill$\tabskip0pt plus 100pt⊗\!
$\rt{#}$\tabskip 0pt⊗$#\hfill$\tabskip0pt plus 100pt\cr
N⊗⊗\hfill A↓i⊗⊗\hfill x↓i⊗⊗\hfill t↓i\cr
\extralead
2⊗⊗.4588\ 3147⊗-⊗.7435\ 5958⊗0⊗.0\cr
⊗⊗.5411\ 6853⊗1⊗.0⊗⊗.4171\ 7446(-1)\cr
3⊗⊗.2281\ 3315⊗-1⊗.0⊗⊗.1148\ 8151\cr
⊗⊗.3726\ 6462⊗⊗.7763\ 2581(-1)⊗0⊗.0\cr
4⊗⊗.3992\ 0223⊗1⊗.0⊗⊗.1978\ 9729\cr}}

\sample 18.
\ctrline{\nm Table 2}
\penalty1000\vskip 6pt
\ctrline{\sl Approximate Solution of an Initial-Boundary Value Problem}
\penalty1000\vskip 6pt
\ctrline{\vbox{\halign{$\ctr{#}$⊗\qquad
$\ctr{#}$⊗\qquad
$\ctr{#}$⊗\qquad
$\ctr{#}$⊗\qquad
$\ctr{#}$⊗\qquad
$\ctr{#}$⊗\qquad
$\ctr{#}$\cr
(x↓\ast,t↓\ast):⊗(.1,.1)⊗(.2,.1)⊗(.3,.1)⊗(.1,.2)⊗(.2,.2)⊗(.3,.2)\cr
\extralead
\hbox{Exact solution:}⊗7.7172⊗7.4310⊗6.9610⊗6.0298⊗5.8062⊗5.4396\cr
N\cr
2⊗7.5884⊗7.2589⊗6.8005⊗5.3908⊗5.1908⊗4.8631\cr
3⊗7.7261⊗7.4395⊗6.9691⊗6.0806⊗5.8527⊗5.4799\cr
4⊗7.7171⊗7.4308⊗6.9616⊗6.0276⊗5.8040⊗5.4376\cr
5⊗7.7172⊗7.4310⊗6.9618⊗6.0299⊗5.8063⊗5.4397\cr}}}

\sample 19.
$$\left(\cpile{\mu↓l\cr -1\cr \null\cr \null\cr \null\cr}\quad
\cpile{-1\cr λ↓l\cr \cdot\cr \null\cr \null\cr}\quad
\cpile{0\cr -1\cr \cdot\cr \cdot\cr \null\cr}\quad
\cpile{0\cr \null\cr \cdot\cr \cdot\cr \cdot\cr}\quad
\cpile{\null\cr \null\cr \null\cr \cdot\cr \cdot\cr}\quad
\cpile{\null\cr \null\cr \null\cr -1\cr \cdot\cr}\quad
\cpile{\null\cr \null\cr \null\cr λ↓l\cr -1\cr}\quad
\cpile{\null\cr \null\cr \null\cr -1\cr \mu↓1\cr}\right)\left(
\cpile{\A u↓{N↓-}↑{(l)}\cr \A u↓{N↓-+1}↑{(l)}\cr \vdots\cr
\A u↓{N↓+-1}↑{(l)}\cr \A u↓{N↓+}↑{(l)}\cr}\right)=\left(
\cpile{\A f↓{N↓-}↑{(l)}\cr \A f↓{N↓-+1}↑{(l)}\cr \vdots\cr
\A f↓{N↓+-1}↑{(l)}\cr \A f↓{N↓+}↑{(l)}\cr}\right).$$
This system can be solved very easily by an $LU$-decomposition because, as is
easily verified
$$\cpile{\left(
\cpile{\mu↓l\cr -1\cr \null\cr \null\cr \null\cr \null\cr}\quad
\cpile{\-1\cr λ↓l\cr \cdot\cr \null\cr \null\cr \null\cr}\quad
\cpile{\null\cr -1\cr \cdot\cr \cdot\cr \null\cr \null\cr}\quad
\cpile{\null\cr \null\cr \cdot\cr \cdot\cr \null\cr \null\cr}\quad
\cpile{\null\cr \null\cr \null\cr \cdot\cr -1\cr \null\cr}\quad
\cpile{\null\cr \null\cr \null\cr \null\cr λ↓l\cr -1\cr}\quad
\cpile{\null\cr \null\cr \null\cr \null\cr -1\cr \mu↓l\cr}\right)\cr
=\left(
\cpile{1\cr -\mu↓l↑{-1}\cr \null\cr \null\cr \null\cr \null\cr \null\cr}\quad
\cpile{\null\cr 1\cr -\mu↓l↑{-1}\cr \null\cr \null\cr \null\cr \null\cr}\quad
\cpile{\null\cr \null\cr 1\cr \cdot\cr \null\cr \null\cr \null\cr}\quad
\cpile{\null\cr \null\cr \null\cr \cdot\cr \cdot\cr \null\cr \null\cr}\quad
\cpile{\null\cr \null\cr \null\cr \null\cr \cdot\cr -\mu↓l↑{-1}\cr \null\cr}\quad
\cpile{\null\cr \null\cr \null\cr \null\cr \null\cr 1\cr -\mu↓l↑{-1}\cr}\quad
\cpile{\null\cr \null\cr \null\cr \null\cr \null\cr \null\cr 1\cr}\right)\left(
\cpile{\mu↓l\cr \null\cr \null\cr \null\cr \null\cr \null\cr \null\cr}\quad
\cpile{-1\cr \mu↓l\cr \null\cr \null\cr \null\cr \null\cr \null\cr}\quad
\cpile{\null\cr -1\cr \mu↓l\cr \null\cr \null\cr \null\cr \null\cr}\quad
\cpile{\null\cr \null\cr  -1\cr \cdot\cr \null\cr \null\cr \null\cr}\quad
\cpile{\null\cr \null\cr \null\cr \cdot\cr \cdot\cr \null\cr \null\cr}\quad
\cpile{\null\cr \null\cr \null\cr \null\cr \cdot\cr \mu↓l\cr \null\cr}\quad
\cpile{\null\cr \null\cr \null\cr \null\cr \null\cr -1\cr \sigma↓l\cr}\right).\cr}$$

\sample 20.
In order to determine the behavior of the Ben-Israel iteration, we consider its
application to the following linear least squares problem: $\min↓x\|Ax-b\|$, where
$$A=\left(\cpile{a\cr 0\cr 0\cr}\quad\cpile{0\cr 0\cr 0\cr}
\right),$$
$a≠0$, and $b$ is arbitrary. Clearly, the rank of $A$ is one and the minimum
norm solution is given by $A↑+b$ which is
$$\left(\cpile{1/a\cr 0\cr}\quad\cpile{0\cr 0\cr}\quad\cpile{0\cr 0\cr}\right)
\left(\cpile{b↓1\cr b↓2\cr b↓3\cr}\right)=\left(\cpile{b↓1/a\cr 0\cr}\right)\,.$$
However, application of the Ben-Israel iteration from the initial guess $x↓0=
{1\choose1}$ yields
$$x↓1=x↓0-A↑+(Ax↓0-b)={1\choose 1}-{1\choose 0}+{b↓1/a\choose 0}=
{b↓1/a\choose 1}.$$