OpenLLM-France/wikimedia huggingface dataset:
# UMD Analysis Qualifying Exam/Jan09 Real
## Problem 1

* (a) Let $f,g$ be real valued measurable functions on $[0,1]$ with the property that for every $x\in [0,1]$, $g$ is differentiable at $x$ and $g'(x)=(f(x))^{2}.$
* Prove that $f\in L^{1}[0,1]$
* (b) Suppose in addition that $f$ is bounded on $[0,1].$ Prove that
* $2\int _{0}^{1}g(x)f^{2}(x)\,dx=g^{2}(1)-g^{2}(0).$

### Solution 1
## Problem 3

* Let $f\in L^{1}(-\infty ,\infty )\!\,$ and suppose $\alpha >0\!\,$. Set $f_{n}(x)={\frac {f(nx)}{n^{\alpha }}}\!\,$ for $n=1,2,\ldots \!\,$. Prove that for almost every $x\in (-\infty ,\infty )\!\,$,
* $\lim _{n\rightarrow \infty }f_{n}(x)=0\!\,$

### Change of variable

By change of variable (setting u=nx), we have
$\int |f_{n}(x)|dx=\int {\frac {|f(u)|}{n^{\alpha +1}}}du\quad \quad (*)\!\,$

### Monotone Convergence Theorem

Define $u_{n}(x)=\sum _{i=1}^{n}|f_{i}(x)|\!\,$.
Then, $u_{n}\!\,$ is a nonnegative increasing function converging to $\sum _{i=1}^{\infty }|f_{i}(x)|\!\,$.
Hence, by Monotone Convergence Theorem and $(*)\!\,$
${\begin{aligned}\int \sum _{i=1}^{\infty }|f_{i}(x)|dx&=\sum _{i=1}^{\infty }\int |f_{i}(x)|dx\\&=\sum _{i=1}^{\infty }\int {\frac {|f(x)|}{i^{\alpha +1}}}dx\\&=\left(\int |f(x)|dx\right)\left(\sum _{i=1}^{\infty }{\frac {1}{i^{\alpha +1}}}\right)\\&<\infty \end{aligned}}\!\,$
where the last inequality follows because the series converges ($\alpha >0\!\,$ ) and $f\in L^{1}\!\,$

### Conclusion

Since
$\int \sum _{i=1}^{\infty }|f_{i}(x)|dx<\infty \!\,$,
we have almost everywhere
$\sum _{i=1}^{\infty }|f_{i}(x)|<\infty \!\,$
This implies our desired conclusion:
$\lim _{i\rightarrow \infty }f_{i}(x)=0\quad {\mbox{a.e.}}\!\,$
# UMD Analysis Qualifying Exam/Jan10 Complex
## Problem 2

* The function $\sec \pi z$ has a convergent Taylor expansion $\sum _{n=0}^{\infty }a_{n}(z+i)^{n}$. Find $\lim \sup _{n\to \infty }|a_{n}|^{1/n}$.

### Solution

$\sec \pi z={\frac {1}{\cos \pi z}}$. By using the definition $\cos z={\frac {e^{iz}+e^{-iz}}{2}}$, we get that $\cos(z)=0$ if and only if $e^{-y}e^{ix}=-e^{y}e^{-ix}$. It is not hard to show that this happens if and only if $y=0$ and $x={\frac {(2n+1)\pi }{2}}$. Therefore, the only zeros of $\cos \pi z$ all occur on the real axis at integer distances away from 1/2. Therefore, $\sec \pi z$ is analytic everywhere except at these points.
Our Taylor series $\sum _{n=0}^{\infty }a_{n}(z+i)^{n}$ is centered at $z=-i$. By simple geometry, the shortest distance from $z=-i$ to $z=1/2$ or $z=-1/2$ (the closest poles of $\sec \pi z$) is $R={\sqrt {1/2^{2}+1^{2}}}={\frac {\sqrt {5}}{2}}$. This is the radius of convergence of the Taylor series.
From calculus (root test), we know that $\lim \sup _{n\to \infty }|a_{n}|^{1/n}=1/R$. Therefore, $\lim \sup _{n\to \infty }|a_{n}|^{1/n}={\frac {2}{\sqrt {5}}}$.
# UMD Analysis Qualifying Exam/Jan09 Complex
## Problem 4

* Suppose that $f,g\!\,$ are analytic on $\{|z|\leq 1\}\!\,$
* with $g\neq 0\!\,$ on $\{|z|<1\}\!\,$. Prove that $|f(z)|\leq |g(z)|\!\,$ for all $z\in \{|z|=1\}\!\,$ implies $|f(0)|\leq |g(0)|\!\,$

## Solution 4
### Define new function h(z)

Define $h(z)={\frac {f(z)}{g(z)}}\!\,$.

#### h is continuous on the closure of D

Since $g\neq 0\!\,$ on $D\!\,$, then by the Maximum Modulus Principle, $g\!\,$ is not zero in ${\overline {D}}\!\,$.
Hence, since $f\!\,$ and $g\!\,$ are analytic on ${\overline {D}}\!\,$ and $g\neq 0\!\,$ on ${\overline {D}}\!\,$, then $h\!\,$ is analytic on ${\overline {D}}\!\,$ which implies $h\!\,$ is continuous on ${\overline {D}}\!\,$

#### h is analytic on D

This follows from above

### Case 1: h(z) non-constant on D

If $h\!\,$ is not constant on ${\overline {D}}\!\,$, then by Maximum Modulus Principle, $|h|\!\,$ achieves its maximum value on the boundary of $D\!\,$.
But since $|h(z)|\leq 1\!\,$ on $\partial D\!\,$ (by the hypothesis), then
$|h(z)|\leq 1\!\,$ on ${\overline {D}}\!\,$.
In particular $|h(0)|\leq 1\!\,$, or equivalently
$|f(0)|\leq |g(0)|\!\,$

### Case 2: h(z) constant on D

Suppose that $h(z)\!\,$ is constant. Then
$|h(z)|=\left|{\frac {f(z)}{g(z)}}\right|=|\alpha |\!\,$ where $\alpha \in \mathbb {C} \!\,$
Then from hypothesis we have for all $z\in \{|z|=1\}\!\,$,
$|f(z)|=|\alpha ||g(z)|\leq |g(z)|\ \!\,$
which implies
$|\alpha |\leq 1\!\,$
Hence, by maximum modulus principle, for all $z\in D\!\,$
$\left|{\frac {f(z)}{g(z)}}\right|=|\alpha |\leq 1\!\,$
i.e.
$|f(z)|\leq |g(z)|\!\,$
Since $0\in D\!\,$, we also have
$|f(0)|\leq |g(0)|\!\,$
# Trigonometry/Derivative of Cosine

To find the derivative of $\cos(x)$ .
    ${\frac {d}{dx}}\cos(x)=\lim _{h\to 0}{\frac {\cos(x+h)-\cos(x)}{h}}=-\lim _{h\to 0}{\frac {2\sin \left(x+{\frac {h}{2}}\right)\sin \left({\frac {h}{2}}\right)}{h}}=-\lim _{h\to 0}\left[\sin \left(x+{\frac {h}{2}}\right){\frac {\sin \left({\frac {h}{2}}\right)}{\frac {h}{2}}}\right]=-\lim _{h\to 0}\left[\sin \left(x+{\frac {h}{2}}\right)\right]\cdot \lim _{h\to 0}\left[{\frac {\sin \left({\frac {h}{2}}\right)}{\frac {h}{2}}}\right]$ .
As in the proof of Derivative of Sine, the limit of the first term is $\sin(x)$ and the limit of the second term is 1. Thus
    ${\frac {d}{dx}}{\big\cos(x){\big ]}=-\sin(x)$ .
Thus
    ${\frac {d}{dx}}{\big\sin(x){\big ]}=\cos(x)$
    ${\frac {d}{dx}}{\big\cos(x){\big ]}=-\sin(x)$
    ${\frac {d}{dx}}{\big-\sin(x){\big ]}=-\cos(x)$
    ${\frac {d}{dx}}{\big-\cos(x){\big ]}=\sin(x)$
and so on for ever.
# Trigonometry/Derivative of Inverse Functions

The inverse functions $\arcsin(x)$ , etc. have derivatives that are purely algebraic functions.
If $y=\arcsin(x)$ then $x=\sin(y)$ and
    ${\frac {dx}{dy}}=\cos(y)={\sqrt {1-\sin ^{2}(y)}}={\sqrt {1-x^{2}}}$ .
So
    ${\frac {dy}{dx}}={\frac {1}{\frac {dx}{dy}}}={\frac {1}{\sqrt {1-x^{2}}}}$
Similarly,
    ${\frac {d}{dx}}{\bigl\arccos(x){\bigr ]}=-{\frac {1}{\sqrt {1-x^{2}}}}$ .
If $y=\arctan(x)$ then $x=\tan(y)$ and
    ${\frac {dx}{dy}}=\sec ^{2}(y)=1+\tan ^{2}(y)=1+x^{2}$ .
So
    ${\frac {dy}{dx}}={\frac {1}{\frac {dx}{dy}}}={\frac {1}{1+x^{2}}}$
If $y=\operatorname {arcsec}(x)$ then $x=\sec(y)$ and
    ${\frac {dx}{dy}}=\sec(y)\tan(y)=x{\sqrt {x^{2}-1}}$ .
So
    ${\frac {dy}{dx}}={\frac {1}{\frac {dx}{dy}}}={\frac {1}{x{\sqrt {x^{2}-1}}}}$

## Power series

The above results provide an easy way to find the power series expansions of these functions.
    ${\frac {1}{\sqrt {1-x^{2}}}}=1+{\frac {x^{2}}{2}}+{\frac {3x^{4}}{8}}+{\frac {5x^{6}}{16}}+{\frac {35x^{8}}{128}}+\cdots$
This is uniformly convergent if $|x|<1$ so can be integrated term by term. The constant of integration is zero since $\arcsin(0)=0$ , so
    $\arcsin(x)=x+{\frac {x^{3}}{6}}+{\frac {3x^{5}}{40}}+{\frac {5x^{7}}{112}}+{\frac {35x^{9}}{1152}}+\cdots$
    ${\frac {1}{1+x^{2}}}=1-x^{2}+x^{4}-x^{6}+\cdots$
This is uniformly convergent if $|x|<1$ so can be integrated term by term. The constant of integration is zero since $\arctan(0)=0$ , so
    $\arctan(x)=x-{\frac {x^{3}}{3}}+{\frac {x^{5}}{5}}-{\frac {x^{7}}{7}}+\cdots$
Note that $\operatorname {arcsec}(x)$ has no power series expansion about $x=0$ , as it is not defined for $x<1$ and has an infinite derivative when $x=1$ . An expansion about any point $x=a>1$ in powers of $x-a$ can be found using Taylor's theorem; it will converge for $1<x<2a-1$ .
# Trigonometry/Cyclic Quadrilaterals and Ptolemy's Theorem

A cyclic quadrilateral is one where all four vertices lie on the same circle.
A quadrilateral is cyclic if and only if the two pairs of opposite angles each sum to 180º.
Outline proof
Draw the radii from two opposite vertices to the centre; they make two angles that must add to 360º. Since the angle at the circumference is half the angle at the centre, the sum of the opposite angles must be 180º. To prove the converse, consider the circumcircle of any three vertices; if the fourth vertex does not lie on this circle, opposite angles will not add to 180º.

## Ptolemy's Theorem

With the above notation, if AB means the length of the line AB, etc., then
    AB.CD + BC.DA = AC.BD
Proof
Let AC and BD be drawn. Find a point X on AC such that ∠ABX = ∠CBD. Since ∠ABX + ∠CBX = ∠ABC = ∠CBD + ∠ABD, it follows that ∠CBX = ∠ABD.
△ABX is similar to △DBC, and △ABD is similar to △XBC. Thus AX/AB = CD/BD, and CX/BC = DA/BD, so
    AX·BD = AB·CD, and CX·BD = BC·DA
Adding these two equalities,
    AX·BD + CX·BD = AB·CD + BC·DA
hence
    (AX+CX)·BD = AB·CD + BC·DA;
Since AX+CX = AC, the result follows.

### Special case

Suppose ABCD is cyclic and is such that ABC is an equilateral triangle. Then regardless of where D is on the arc AC, DB = DA + DC.

## Maltitudes

A maltitude of a quadrilateral is a perpendicular to a side passing through the midpoint of the opposite side.
For a cyclic quadrilateral, the four maltitudes are concurrent. If this point is X and the centre of the circumcircle of the quadrilateral is O, then the centroid of the vertices bisects OX.

## Other theorems

If the vertices in clockwise order are A, B, C and D, this means that the triangles ABC, BCD, CDA and DAB all have the same circumcircle and hence the same circumradius. The incentres of these four triangles always lie on the four vertices of a rectangle; these four points plus the twelve excentres form a rectangular 4x4 grid. If P, Q, R and S are the centres of arcs AB, BC, CD and DA respectively on the circumcircle, then the lines PR and QS are the perpendicular bisectors of the sides of the rectangle.
The four orthocentres form a quadrilateral congruent to the original; the centroids and the centres of the nine-point circles each form a quadrilateral similar to the original.
# VirtualBox/Installation

VirtualBox can be downloaded from the official website at virtualbox.org. The main page has a big blue download button that takes you to the download page. From there, you can select your system.
    Windows
    Linux
    macOS
VirtualBox software requires a 64-bit host operating system. VirtualBox supports running both 64-bit and 32-bit guest operating systems in its virtual machines.
# VirtualBox/Installation/Linux

Specific Instruction Vary depending on your distribution. The first step is to download the version that is correct for your distro. However, Debian and Red Hat based systems now have an automatic install/update option for the OSE.

## Officially supported Linux distributions

* Ubuntu 18.04 LTS, 19.03 and 19.10
* Debian GNU/Linux 9 ("Stretch") and 10 ("Buster")
* Oracle Linux 6, 7 and 8
* Red Hat Enterprise Linux 6, 7 and 8
* Fedora 30 and 31
* Gentoo Linux
* SUSE Linux Enterprise server 12 and 15
* openSUSE Leap 15.1
Note that Linux 2.4-based host OSes are no longer supported.

## Installation for Debian Systems

These include Debian and Ubuntu

### Building a Kernel Mod

See Below

## Installation for Ubuntu

Open a terminal and type
and follow the prompts.

## Installation for Red Hat based Systems

These Include Red Hat Enterprise Linux (RHEL), CentOS, and Fedora.

## PUEL Edition

After downloading the rpm file for your distro, there are a number of methods of installation.
1. Double Click on the file, it should install VirtualBox 2. In a Terminal using YUM

### Using YUM in a Terminal

Load up a terminal and change directory to the one where you saved the rpm. To continue beyond this point you will need root privileges. You will either need to be logged on as root (not recommended), su to root, or use sudo (if you have that configured). The following assumes that you are using sudo. If you are logged on as root or you used su then the instructions will work as shown. However, you may also exclude the "sudo" in front of the commands.
In the terminal type
Replace *name* with the name of the rpm file.
Yum will load up and check for dependencies. After doing that it should ask if you want to install VirtualBox. Type 'Y' (without quotations) to continue. VirtualBox will now be installed on your machine.

### Building a Kernel Mod

See Below

## Other
### Building a Kernel Mod

See Below

## Building a Kernel Mod

In order to run VirtualBox requires a Kernel Mod. VirtualBox installed a script that can build the mod for you. In order to do this you will require the Kernel Development Tools and GCC to be installed on your machine. Assuming you have those you may continue by running the following command in a terminal:
If all is successful the script to successfully compile and initialise a VirtualBox Kernel Mod. This mod should load automatically each time you reboot your host computer.
NOTE: You will need to rebuild the mod (using the script mentioned above) every time you update the Kernel.

## Further reading

* Using Ubuntu Linux/Virtualization
# VirtualBox/Installation/Windows
## Supported Windows host operating systems (64 bit)

Virtualbox 7 supports the following Windows hosts:
* Windows 8.1
* Windows 10 RTM (1507 / 2015 LTSB) build 10240
* Windows 10 Anniversary Update (1607 / 2016 LTSB) build 14393
* Windows 10 Fall Creators Update (1709) build 16299
* Windows 10 April 2018 Update (1803) build 17134
* Windows 10 October 2018 Update (1809 / 2019 LTSC) build 17763
* Windows 10 May 2019 Update (19H1 / 1903) build 18362
* Windows 10 November 2019 Update (19H2 / 1909) build 18363
* Windows 11 (21H2)
* Windows Server 2012
* Windows Server 2012 R2
* Windows Server 2016
* Windows Server 2019
* Windows Server 2022

## Download and Install with Chocolatey

If you have already installed Chocolatey, the easiest way to download and and install VirtualBox is (in PowerShell: Run as administrator):
That's it. Then you're ready for Setting up a Virtual Machine.

## Download

The VirtualBox installer can be downloaded from the official VirtualBox website. You might want to compare the checksums to verify the integrity of downloaded packages. The SHA256 checksums should be favored as the MD5 algorithm must be treated as insecure!

## Installation

Run the downloaded installer and follow the on-screen instruction to install Oracle VM VirtualBox. You can opt to add a desktop and start menu shortcut.
NOTE: Your network connections (including the Internet connection) will be interrupted for a short time while the installer installs VirtualBox network drivers.
* ↑ "1.4. Supported Host Operating Systems".
* ↑ https://chocolatey.org/
* ↑ "Grid Singularity: Step 2: Install Virtualbox"
* ↑ "Installing Virtualbox via choco "
* ↑ "Install oracle virtual box using PowerShell"
# VirtualBox/Guest Additions

The VirtualBox Guest Additions are a plugin which improves the interoperability between the host and the hosted systems.
* Windows
* Ubuntu
* Mac OS X
# VirtualBox/Guest Additions/Windows

Here we install the Guest Additions.
* Go to the host operating system. At the top of the VM window, choose Devices >> "Insert Guest Additions CD image...". Then, VirtualBox simulates inserting a CD into the simulated optical drive of the VM.
* If you need to eject this, you can either restart the VM or click on Devices >> optical drives >> eject.
If you didn't add a Windows shared folder into the "configuration" menu of the VirtualBox graphical interface, you can do it in Command Prompt with: VBoxManage sharedfolder add "VMName" --name "SharedName" --hostpath "C:\Users\Public\Documents\SharedFolderPath".

## Linux guest

The Guest Additions CD should be read from the guest OS like this:
* sudo mount /dev/cdrom /media/cdrom
* cd /media/cdrom
* sudo ./VBoxLinuxAdditions.run
Then, just mount the Windows folder from Linux: mkdir ~/vboxshare sudo mount -t vboxsf SharedName ~/vboxshare
That's it, you can browse your Windows files: ls ~/vboxshare

## References

* ↑ VirtualBox manual, "Guest Additions"
* ↑ http://www.htpcbeginner.com/mount-virtualbox-shared-folder-on-windows/
* ↑ http://superuser.com/questions/950431/how-to-install-virtual-box-guest-additions-on-debian
* ↑ https://help.ubuntu.com/community/VirtualBox/SharedFolders
# VirtualBox/Guest Additions/Mac OS X

Sorry, no Guest Additions are provided for Mac OS X.
# User-Generated Content in Education/User-Generated Music
## User-Generated Music

Music is an art form whose medium is sound and silence.
User-generated music is a term that is used for music that is created by an individual user or group of users and then distributed over the Internet. This music is often created for the purpose of entertainment. There has recently been a major shift the distribution of music with the advent of Web 2.0 technologies. With the current technology creators of music are able to get their music heard by millions of people simply by placing it on Youtube.

## Types of licenses

Music can be protected by several different licenses. The traditional licensing of music falls under the copyrighted music. This traditional form of licensing is in place to ensure that the creator of the music gets paid for their work. When you purchase music, you are buying the right to listen to that music, not to distribute it in any other fashion.
Music without any type of license is said to be in the public domain. Music is also said to be in the public domain if the intellectual property rights have expired. When music is in the public domain then it may be used freely.
Creative Commons is another type of license that can be sought by a creator of music. By giving a user-created piece of music a Creative Commons license, the creator of the musical piece is defining how a piece of music may be used. This type of protection may allow for commercial use of the work or modifications of the work.
No matter what type of license a piece of music has, you should always be sure to give credit to the creator of the piece.

## Where to find user-generated music

User-generated music can be found all over the web.
The most popular website for user-created music is YouTube. On YouTube the author of music can post their piece for free. By doing so the author can get their work out to millions of viewers.
Jamendo is another website that is a music community for free and legal music downloads. Music on Jamendo is licensed by under Creative Commons licensing. In addition to downloading, Jamendo also allows for music to be streamed over the Internet.

## References

* ↑ en.wikipedia.org. Music. Retrieved October 16, 2011 from: http://en.wikipedia.org/wiki/Music
* ↑ en.wikipedia.org. Music Licensing. Retrieved October 17, 2011 from: http://en.wikipedia.org/wiki/Music_licensing
* ↑ en.wikipedia.org. Public Domain. Retrieved October 17, 2011 from: http://en.wikipedia.org/wiki/Public_domain
* ↑ en.wikipedia.org. Jamendo. Retrieved October 16, 2011 from: http://en.wikipedia.org/wiki/Jamendo
# Web 2.0 and Emerging Learning Technologies/Google Groups

What the resource is – Its features, gadgets Google Groups is a free service from Google where groups of people have discussions about common interests. Internet users can find discussion groups related to their interests and participate in threaded conversations, either through the Google Groups web interface, or by e-mail.
This kind of tools make students learn by doing, think, and generate individual knowledge by increasing the amount of shared information.Also, students learn or find the way of organizing the information they got to avoid chaos.
Students can also upload, open and edit their own files, activities, videos or images to the group.What is more, users may create public profiles which are linked from all of their posts.
Google introduced group pages which can be edited by group members or group managers.
Group members can also discuss pages.
Teaching and learning possibilities. What a teacher/student can do How this resource can facilitate teaching and learning. Students and teachers may create a meeting point in the web to share comments and files on a specific topic. By including this tool in our syllabuses students profit from searching the web as they have to look for information, files like videos or images and exchanging and uploading information in a collaborative way. It is a good way of learning how to differentiate what information may be useful or not for a specific purpose .It does not mean that the google group helps find this difference, but the students are the ones who profit from it by deciding what information is worth considering.
It lets communication flow between teachers and students. Moreover, it is a negociation of knowledge and information which enriches deeply any learning process. It may serve any kind of pedagogical objective that the teacher can think of depending on the activity designed for that specific purpose.
Example links
Students or ordinary people can collaborate in the finding of information along the web about an specific topic. They make a deep investigation about history, art and geography closely tied to education as the topic is World War.
Google group about pedagogical implications for elementary students.
Parents and citizens interested in an equitable alternative to school scripture classes for students/families who choose non-scripture. Currently schools are not allowed to offer an alternative activity for these students. This group seeks to change this government policy.
# Nontraditional, Alternative, and Informal Learning with the Web 2.0
## Sub chapter

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