OpenLLM-France/wikimedia huggingface dataset:
# Vector space/Basis/Introduction/Section
## Definition

Let $K$ be a field, and let $V$ be a $K$-vector space. Then a linearly independent generating system $v_{i}\in V$, $i\in I$,
$V$
$V$

## Example

The standard vectors in $K^{n}$ form a basis. The linear independence was shown in example. To show that they also form a generating system, let
    $v={\begin{pmatrix}b_{1}\\b_{2}\\\vdots \\b_{n}\end{pmatrix}}\in K^{n}\,$
be an arbitrary vector. Then we have immediately
    $v=\sum _{i=1}^{n}b_{i}e_{i}\,.$
Hence, we have a basis, which is called the standard basis of $K^{n}$.

## Theorem

Let $K$ be a field, and let $V$ be a $K$-vector space. Let $v_{1},\ldots ,v_{n}\in V$
* The family is a basis of $V$.
* The family is a minimal generating system, that is, as soon as we remove one vector $v_{i}$, the remaining family is not a generating system any more.
* For every vector $u\in V$, there is exactly one representation
    $u=s_{1}v_{1}+\cdots +s_{n}v_{n}\,.$
* The family is maximal linearly independent, that is, as soon as some vector is added, the family is not linearly independent any more.

### Proof

$\Box$

## Remark

Let a basis $v_{1},\ldots ,v_{n}$ of a $K$-vector space $V$ be given. Due to fact, this means that for every vector $u\in V$, there exists a uniquely determined representation
    $u=s_{1}v_{1}+s_{2}v_{2}+\cdots +s_{n}v_{n}\,.$
The elements $s_{i}\in K$ (scalars) are called the coordinates of $u$ with respect to the given basis. Thus, for a fixed basis, we have a (bijective) correspondence between the vectors from $V$, and the coordinate tuples $(s_{1},s_{2},\ldots ,s_{n})\in K^{n}$. We express this by saying that a basis determines a linear coordinate system.

## Theorem

Let $K$ be a field, and let $V$ be a $K$-vector space with a finite generating system. Then $V$ has a finite

### Proof

Let $v_{i}$, $i\in I$, be a finite generating system of $V$, with a finite index set $I$. We argue with the characterization from fact. If the family is minimal, then we have a basis. If not, then there exists some $k\in I$, such that the remaining family where $v_{k}$ is removed, that is $v_{i}$, $i\in I\setminus \{k\}$, is also a generating system. In this case, we can go on with this smaller index set. With this method, we arrive at a subset $J\subseteq I$ such that $v_{i}$, $i\in J$, is a minimal generating set, hence a basis.
$\Box$
* ↑ Linear coordinates give a bijective relation between points and number tuples. Due to linearity, such a bijection respects the addition and the scalar multiplication. In many different contexts, also nonlinear (curvilinear) coordinates are important. These put points of a space and number tuples into a bijective relation. Examples are polar coordinates, cylindrical coordinates and spherical coordinates. By choosing suitable coordinates, mathematical problems, like the computation of volumes, can be simplified.
# Vector space/Basis/Subbasis/Orthogonal space/Example

Let $V$ be a finite-dimensional $K$-vector space with a basis $v_{i}$, $i\in I$, and the corresponding dual basis $v_{i}^{*}$, $i\in I$. Let
    $U=\langle v_{j},\,j\in J\rangle \,$
for a subset $J\subseteq I$. Then
    $U^{\perp }=\langle v_{i}^{*},\,i\not \in J\rangle \,.$
# Vector space/Bidual/Introduction/Section
## Definition

Let $K$ be a field, and let $V$ be a $K$-vector space. Then the dual space of the dual space ${V}^{*}$, that is,
    ${({V}^{*})}^{*},$
$V$

## Lemma

Let $K$ be a field, and let $V$ be a $K$-vector space. Then there exists a natural injective linear mapping
    $\Psi \colon V\longrightarrow {({V}^{*})}^{*},v\longmapsto {\left(f\mapsto f(v)\right)}.$
If $V$ has finite dimension, then $\Psi$ is an

### Proof

Let $v\in V$ be fixed. First of all, we show that $\Psi (v)$ is a linear form on the dual space ${V}^{*}$. Obviously, $\Psi (v)$ is a mapping from ${V}^{*}$ to $K$. The additivity follows from
    $(\Psi (v))(f_{1}+f_{2})=(f_{1}+f_{2})(v)=f_{1}(v)+f_{2}(v)=(\Psi (v))(f_{1})+(\Psi (v))(f_{2})\,,$
where we have used the definition of the addition on the dual space. The compatibility with the scalar multiplication follows similarly from
    $(\Psi (v))(sf)=(sf)(v)=s(f(v))=s((\Psi (v))(f))\,.$
In order to prove the additivity of $\Psi$, let $v,w\in V$ be given. We have to show the equality
    $\Psi (v+w)=\Psi (v)+\Psi (w)\,.$
This is an equality inside of ${\left({V}^{*}\right)}^{*}$, in particular, it is an equality of mappings. So let $f\in {V}^{*}$ be given. Then, the additivity follows from
    $(\Psi (v+w))(f)=f(v+w)=f(v)+f(w)=(\Psi (v))(f)+(\Psi (w))(f)\,.$
The scalar compatibility follows from
    $(\Psi (sv))(f)=f(sv)=s(f(v))=s((\Psi (v))(f))\,.$
In order to prove injectivity, let $v\in V$ with $\Psi (v)=0$ be given. this means that for all linear forms $f\in {V}^{*}$, we have $f(v)=0$. But then, due to fact, we have
    $v=0\,.$
By the criterion for injectiviy, $\Psi$ is injective.
In the finite-dimensional case, the bijectivity follows from injectivity and from fact.
$\Box$
Thus, the mapping $\Psi$ sends a vector $v$ to the evaluation (or evaluation mapping) which evaluates a linear form $f$ an the point $v$.
# Vector space/Basis complement/Fact

Let $V$ denote a finite-dimensional vector space over a field $K$. Let
    $u_{1},\ldots ,u_{k}$
denote linearly independent vectors in $V$.
    $u_{k+1},\ldots ,u_{n}$
such that
    $u_{1},\ldots ,u_{k},u_{k+1},\ldots ,u_{n}$
form a basis
$V$
# QB/d cp2.9

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# QB/d cp2.6

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# QB/d cp2.8

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# Q^3/23-5/926/-14-1/Basis/Exercise

Establish if in $\mathbb {R} ^{3}$ the three vectors
    ${\begin{pmatrix}2\\3\\-5\end{pmatrix}}\,,{\begin{pmatrix}9\\2\\6\end{pmatrix}}\,,{\begin{pmatrix}-1\\4\\-1\end{pmatrix}}$
form a basis.
# QB/e HuntEdgeSpace

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# QB/d cp2.7

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# QB/d cp2.gaussC

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Wright State University Lake Campus/2018-9/Phy2410
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# Social Media/Photos and video online

* Photos: Yahoo's Flickr, Google's Picasa, Wikimedia Commons
* Video: Google's Youtube, Blip.tv, Wikimedia Commons, Internet Archive
# Social Media/Get ready

* Checking Internet connection
* Tips on using a web browser
* Overview of social media
# Social Media/Set up a blog

* Blogs in Plain English
* Blogger Help
* Sue Water's Tips for Better Blogging
# Social Media/Set up an RSS reader

* RSS in Plain English
* Looking at examples of RSS in action
* Set up a Google Reader and subscribe to RSS feeds
* Link RSS feeds to Facebook
# Social Media/It's the best of times, it's the worst of times
## In society generally

* Robin Sloan and Matt Thompson, with music by Aaron McLeran. EPIC2014. http://robinsloan.com/epic/ 2005.
* n+1 editors. Webism - The Internet as Social Movement. Change This, Issue 73 - 01, 2010-08-04.
* Jeffery Carino. The explosion of networking sites has opened a new can of worms in college and high school athletics. Daily Record. 2011-02-12.
* Jessica Halloran. AOC to use Monaghan, Rice as examples to avoid medal twits. The Sunday Telegraph 2011-02-20.
* Andrew Keen. The Cult of the Amateur: How Today's Internet Is Killing Our Culture. Currency 2007.
* Clay Shirky. Here Comes Everybody: The Power of Organizing Without Organizations. Penguin Press 2008.
* Rick Levine, Christopher Locke, Doc Searls, and David Weinberger. The Cluetrain Manifesto. Perseus Books 2000.
* Jonah Lehrer. We, Robots (A review of Sherry Turkle's new book, Alone Together. Basic Books 2011). New York Times, 2011-01-21.

## In sport

* Jennifer Van Grove. Sports and Social Media: Where Opportunity and Fear Collide. Mashable.com, 2009-09-05.
# Social Media/Radical transparency

* Yochai Benkler. Open-source economics - Video 17:53 - July 2005
* Nate Harrison Video explains the world's most important 6-sec drum loop Video 18min 2004
* Lawrence Lessig. 23C3 On Free, and the Differences between Culture and Code - Video 1:15:37 - Dec 2006
* Creative Commons licenses
* Crowd sourcing
* Participatory economics
# Python/pip (package manager)

pip is a de facto standard package-management used to install and manage software packages written in Python.

## Sub Commands

* pip install <package> This command is used to install packages from PyPI
* pip uninstall <package> This command is used to remove packages within your environment.
* pip list This command is used to list all installed packages
* pip show -f <package> Used to show general detail about a package
* pip check Checks that all installed packages have compatible dependencies.

## General Options

* -h Shows all the sub-commands of pip.
* --help Long version of -h, shows all the sub-commands of pip
* -r This enables the instillation of packages as defined in a 'requirements.txt' file, in order to specify a file, add the file's directory after the -r.

### Requirements File (-r)

This is a feature of pip, which allows the user to specify each package they wish to install in a text file. This is often done in repositories, and allows users to quickly and easily install packages necessary for a specific piece of code. Note that version numbers are optional.
# Student Projects/ICT in Indian school education
#### Introduction

In today’s era, knowledge and technology has become an integral part of the education. ICT has changed the scenario of Indian education in the last few years and is one of the most efficient tools in teaching-learning processes. It is a boon to both the teachers and the learners. It has become easier to look for the resources outside the textbook.
Modern technologies helped the teachers or educators to avail various resources and tools that can be used in the classrooms. Through ICT, teachers are able to prepare lessons, create interactive activities that cannot be done directly in the classrooms through Learning Management System (LMS). These resources can also be modified easily.
Technology is playing a vital role in assessing the students. Assessment has become easier and comprehensive. The assessment tools like assignment in Edmodo, quizzes in Kahoot, survey sheets in MS office or Google forms, Google classrooms have helped in assessing the students.
ICT has boosted the changes that are taking place in the school education. Teaching-learning process has changed from traditional lecture method to a constructivistic approach where students learn by doing by hands-on experience. By using various tools and techniques of ICT, the classroom experience can be made into a wonderful learning experience.
Students in today’s classroom are encouraged to create their own knowledge and build upon them. Teachers are the facilitators who provide the guidelines and skills to encourage the students to create their own knowledge.
ICT in schools was initially introduced in 2004 and is revised in 2010 and is currently subsumed under Rashtriya Madhyamik Shiksha Abhiyan. Various initiatives have been taken by the government for the implementation of ICT in secondary education in schools. E-platforms like e-pathashala, Saransh, Siddhi shala, Darpan, etc brought major changes in the classrooms.
Key challenges
Though ICT can change the face of Indian school education, its implementation still remains a challenge to an extent.
Training teachers with the latest developments, use of ever evolving technologies, improving their skills is a tremendous task.
Availability of technology and maintenance in schools requires a lot of financial investment. That’s a huge challenge.
# University of Florida/Egm6321/f09.team1.gzc/Mtg22

Mtg 30: Mon, 14 Mar 11
    page22-1 Comp. Trap. rule error: $h={\color {red}{\frac {\color {black}(b-a)}{n}}}{\color {blue}\leftarrow }{\color {red}(3)}{\color {blue}p.7-4}$ ${E}_{n}^{\color {blue}T(Trap.)}{\underset {{\color {red}(1)}{\color {blue}p.7-4}}{:=}}\ I-{I}_{n}^{T}=\int _{a}^{b}f(x)dx-h[{\color {red}{\frac {1}{2}}}f(x_{\color {red}0})+f(x_{1})+\cdots +f(x_{n-1}){\color {red}{\frac {1}{2}}}f(x_{\color {red}n})]\ {\color {red}(1)}$ $=\sum _{i=1}^{n}{\color {blue}\left\{{\color {black}\int _{x_{i-1}}^{x_{i}}f(x)dx-{\frac {h}{2}}[f(x_{i-1})+f(x_{i})]}\right\}}$ (1)p.17-2: $\left|{E}_{n}^{T}\right|\leqslant {\frac {h^{3}}{12}}{\underset {\color {blue}{\underset {\overset {\color {red}(3)}{=:{\overline {M_{2}}}}}{\color {blue}\underbrace {\color {black}\xi \in ]x_{i-1},x_{i} }}}{\sum _{i=1}^{n}(max{\color {red}\left|{\color {black}f^{\color {blue}(2)}(\xi )}\right|})}}{\color {red}\ (2)}$ (4)p.17-1: ${\overline {M_{2}}}\leqslant M_{2}\ {\color {red}(4)}$ $\left|{\color {blue}{\color {black}E}_{n}^{T}}\right|\leqslant {\frac {(b-a)^{3}}{12n^{3}}}nM_{2}={\frac {(b-a)^{3}}{12{\color {blue}n^{2}}}}M_{2}={\frac {(b-a)h^{\color {red}2}}{12}}M_{2}$ ${\color {blue}A.P.253}$ ${\color {red}(5)}$
page22-2
HW$^*$4.8: Comp. Simpson error
$\left|{E}_{\color {blue}n}^{\color {blue}S}\right|\leqslant {\frac {(b-a)^{5}}{2880n^{4}}}{\color {blue}{\underset {{\color {red}(4)}p.17-1}{\underbrace {\color {black}M_{4}} }}}={\frac {(b-a)h^{\color {red}(4)}}{2880}}M_{4}\ \color {red}(1)$
HW$^*$4.8: See HW$^*$2.4 p.7-3
1) Use error estimate for Taylor Series, Compare Trap., Compare Simpson, to findn
Q(10$^{-6}$), and compare to number results.
2)Numerically find the power ofhin error : Plot
logerror vslogh , and meas. slop with least square
(lin. regression).
    $E=ah^{\color {red}k}+Q(h^{k+1}),\$ $\ logE=loga+{\color {red}k}logh$
page22-3
HW$^*$4.9: pf. of SSET, G(.) in (1) p.19-1
A)Redo the pf for 2 cases
${\color {blue}1)}\ G(t):=\ e(t)-t^{\color {red}4}e(1)\ {\color {red}(1)}$
${\color {blue}2)}\ G(t):=\ e(t)-t^{\color {red}6}e(1)\ {\color {red}(2)}$
Print out where pf breaks down.
B)For G(t) as in(1)p.19-1(w/t$^5$), find
G$^{(3)}$(0) and follow same steps in pf to
see what happen.
HW$^*$4.10:1)Don'tUse matlabtrapz forcompare Trap.in
your code to produce Table 5.1 n A.p.255(p.22-4)
${\color {red}(3)}I=\int _{0}^{\pi }e^{x}{\color {blue}sinx}dx$(use WAto find I)
# University of Florida/Egm6321/f09.team1.gzc/Mtg20

Mtg 20: Wed, 16 Feb 11
    page20-1 Pf of SSET p.19-3 cont'd ${\color {red}(1)}{\color {blue}p.19-3\ Rolle's\ thm}\ \Rightarrow \ \exists \ \varsigma _{1}\in \ ]\ 0,1\ [\ st\ G^{\color {blue}(1)(\varsigma _{1})}{\overset {\color {red}(1)}{=}}0$ $Now\ G^{(1)}(0)\ {\overset {\color {red}(2)}{=}}0\ {\color {red}why?}$ ${\color {red}(1)}\ {\color {blue}p.19-1:}\ G^{(1)}(t){\overset {\color {red}(3)}{=}}e^{(1)}(t)-5t^{4}e(1)$ ${\color {red}(5)}{\color {blue}p.18-3:}\ e(t)=A(t)-{A}_{2}^{L}(t)\ {\color {red}(4)}$ $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ e^{(1)}(t)=A^{(1)}-{A}_{2}^{L{\color {red}(1)}}(t)\ {\color {red}(5)}$ page20-2 $A(t)=\int _{+t}^{-t}-=\int _{k}^{-t}-+\int _{t}^{k}-$ $A^{(1)}{\overset {\color {red}(1)}{=}}F(-t)+F(t)$ $k\in {\color {red}]}-t,t{\color {red}\ ({\color {red}kisconstant})$ (4)p.20-1& (5)p.18-3: $A_{2}^{L{\color {red}(1)}}(t){\overset {\color {red}(2)}{=}}{\frac {1}{3}}[F(-t)+4F(0)+F(t)]+{\frac {t}{3}}[F^{(1)}(-t)+F^{(1)}(t)$ (5) p.20-1: $e^{(1)}(0)=A^{(1)}(0)-A_{2}^{L(1)}(0)=2F(0)-[2F(0)+0]=0\ {\color {red}(3)}$ ${\color {red}(3)\ and\ (3)}{\color {blue}p.20-1}\Rightarrow {\color {red}(2)}{\color {blue}p.20-1}$ Recall: $G^{(1)}(\xi _{1})=0\ {\color {red}(1)}{\color {blue}p.20-1}$ $G^{(1)}(0)=0\ {\color {red}(2)}{\color {blue}p.20-1}$
page20-3
$Rolle'stheorem\Rightarrow \ \forall \xi _{2}\in ]0,\xi _{1}[st\ \ G^{(2)}(\xi _{2})=0\ {\color {red}(1)}$
$Again,\ G^{(2)}(0){\overset {\color {red}(2)}{=}}0\ {\color {blue}HW^{*}4.3}$
${\color {red}(1)\ and\ (2)}\ Rolle'stheorem\ \Rightarrow \ \forall \xi _{3}\in ]0,\xi _{2}[\ G^{(2)}(\xi _{2})=0\ st\ \ G^{(3)}(\xi _{3}){\overset {\color {red}(2)}{=}}0$
${\color {red}(1)}{\color {blue}p.19-1:}\ G^{(3)}(t){\overset {\color {red}(4)}{e}}^{(3)}(t)-{\color {red}{\underset {(\xi )(4)(3)}{\underbrace {\color {black}t^{2}} }}}e(1)$
$e^{(3)}(t){\underset {\color {blue}HW^{*}4.4}{=}}-{\frac {t}{3}}[F^{(3)}(t)-F^{(3)}(-t)]\ {\color {red}(5)}$
$G^{(3)}(\xi _{3})=-{\frac {\xi _{3}}{3}}{\color {green}{\overset {Apply\ DMVT}{\overbrace {\color {black}[F^{(3)}(\xi _{3})-F^{(3)}(-\xi _{3})]} }}}-60(\xi _{3})^{2}e(1){\underset {{\color {blue}from}{\color {red}(3)}}{\overset {\color {red}(6)}{=}}}0$
${\underset {\color {red}(f)}{\overset {\color {blue}DMVT}{=}}}-{\frac {\xi _{3}}{3}}[{\color {blue}{\underset {\xi _{3}-(-\xi _{3})}{\underbrace {\color {black}2\xi _{3}} }}}F^{(4)}(\xi _{4})]-60(\xi _{3})^{2}e(1)\ \xi _{4}\in \ ]-\xi _{3},\xi _{3}[$
# University of Florida/Egm6321/f10.team03/Hwk1

Egm6321.f10.team03

## Problem 3
### Given
### Find

Show that the following first order ODE is non-linear

### Solution
## Contributing Team Members

Egm6321.f10.team3.franklin
"Add your names and pages in similar fashion."
# University of Florida/Egm6321/f09.team1.gzc/Mtg8

Mtg 8: Wed, 19 Jan 11
page8-1
Legendre poly. P$_n$(x):
    ${P}_{n}(x)=\sum _{i=0}^{\color {red}[{\color {black}n/2}]}{(-1)}^{i}{\frac {(2n-2i)!{x}^{n-2i}}{{2}^{n}i!(n-i)!(n-2i)!}}$ (1)
    ${\color {red}[{\color {black}n/2}]}=integer\ part\ of\ n/2$ (2)
    e.g., m = 5 , n/2 = 2.5 , [2.5] = 2 ${\color {red}(3)}\ {P}_{0}=1\color {blue}\in {P}_{0}$ ${\color {red}(4)}\ {P}_{1}=x\color {blue}\in {P}_{1}$ ${\color {red}(5)}\ {P}_{2}={\frac {1}{2}}(3{x}_{2}-1)\color {blue}\in {P}_{1}$ ${\color {red}(6)}\ {P}_{3}={\frac {1}{2}}(5{x}_{3}-3x)\color {blue}\in {P}_{1}$ ${\color {red}(7)}\ {P}_{4}={\frac {35}{8}}{x}^{4}-{\frac {15}{4}}{x}^{2}+{\frac {3}{8}}\color {blue}\in {P}_{1}$ ${\color {blue}{P}_{n}}\ =\ set\ of\ poly.\ of\ degree\ \leqslant \ n$ HW 2.6: Verify(3)-(7) using(1)-(2) ${\color {red}(5)}\Rightarrow \ {P}^{2}=0\Rightarrow {x}_{1,2}={\color {red}-}{\frac {1}{\sqrt {3}}}\ or{\color {red}+}{\frac {1}{\sqrt {3}}}$ page8-2 Weights w$_i$ , i = 1,...,n ((1) p.7-5) Thm: $I(f)={I}_{n}(f)+{E}_{n}(f)\ {\color {red}(1)}$ ${w}_{i}={\frac {-2}{(n+1){P}_{n}^{'}({x}_{i}){P}_{n+1}({x}_{i})}}\ {\color {red}(2)}$ ${E}_{n}(f)={\frac {{2}^{2n+1}{(n)!}^{4}}{(2n+1){[(2n)!]}^{2}}}{\frac {f^{\color {red}(2n)}(\xi )}{(2n)!}}\ {\color {red}(3)}$ $\xi \in [-1,1]$
${\underline {\color {blue}{NOTE:}}}\ {E}_{n}(f)=0\ \forall f\in {P}_{2n-1}\ since\ f^{\color {red}(2n)}(x)=0\ .\ Only\ need\ to\ use\ n\ int.\ pts\ \left\{x_{i},\ i=1,...n\right\}\ to\ int.\ exactly\ any\ poly\ in\ P_{2n-1}\ i.e.,\ of\ degree\ \leqq \ 2n-1$
${\color {blue}{Newton-cotes\ method:}}\ I(f)$
$History\ Newton\ cotes\ \color {blue}\rightarrow \ Lecture\ plan\ suli+Meyers(2003)$
page8-3
${\color {blue}1)}\ Approx.\ f(.)\ using\ {\color {blue}{\underline {\color {black}Lagrange}}}\ interp.\ funcs\ \Rightarrow \ {f}_{n}^{\color {red}L}(x)$
$Int.\ exactly\ {f}_{n}^{\color {red}L}(x)\ \Rightarrow \ \color {blue}\underbrace {\color {black}I({f}_{n}^{\color {red}L})}$
$I_{n}(f)\ :=I({f}_{n}^{\color {red}L})=\int {f}_{n}^{\color {red}L}(x)dx\ \color {red}(1)$
${f}_{n}^{\color {red}L}(x)=P_{n}(x)=\sum _{i=0}^{\infty }{\color {blue}{\underset {Lagrange\ interp.\ func}{\color {black}\underbrace {l_{i,n}(x)} }}}\ f(x_{i})\ \color {red}(2)$
NOTE: Demonstrated Wolfram Alpha(WA) e.g., (debt usa)/(gdp usa) integrate x from 0 to 1 link WA comp. results in HW reportsAvoid plagiarism