OpenLLM-France/wikimedia huggingface dataset:
# World History/Major Figures

[Will order later...trying to get names down first. If you get here before I do (if anyone is still active), please try for chronological order subdivided by region of influence. Just quick snapshots of big achievements should do. The idea is for the reader to look at what the Wikibook includes and to get a short summary.]

## Philosophers / Intellectual Types / Artists / Inventors

Socrates, Plato, Aristotle, Eli Whitney, Johannes Gutenberg, Herodotus, Leonardo da Vinci, Niccolo Machiavelli, Galileo, Abu Rayhan Biruni, Avicenna, Muhammad ibn Mūsā al-Khwārizmī, Alhazen, Ibn al-Nafis, Rene Descartes, Isaac Newton, Jenner, Louis Pasteur, Thomas Hobbes, John Locke, Jean-Jacues Rousseau, Voltaire, Baron de Montesquieu, Emmanuel Kant, Mary Wollstonecraft, Thomas Malthus -- wrote "Essay on Population." Eventually, organisms will reproduce more than they have the means to support, Adam Smith -- wrote "Wealth of Nations." Pioneer of laissez faire, David Ricardo -- wrote "Iron Law of Wages", Karl Marx -- wrote "The Communist Manifesto, Confucius (Kung fuzi)

## Religious

Martin Luther, John Calvin, Saint Ignatius Loyola, Pope Urban II, Moses, Jesus of Nazareth, Paul of Tarsus, Augustine, Benedict, Muhammad, Thomas Aquinas

## Explorers

Prince Henry the Navigator, Christopher Columbus, Ferdinand Magellan, Commodore Perry, Ibn Battuta, Marco Polo

## Leaders

Hannibal, Julius Caesar, Augustus Caesar (Octavian), Attila the Hun, Cicero, Constantine, Justinian, Ivan the Terrible, Clovis, Charlemagne, Abu Bakr, Umar, William the Conqueror, Saladin, Henry VIII, Elizabeth I, Henry of Navarre, Charles V, Hernan Cortes, Francisco Pizarr, Oda Nobunaga, Tokugawa Ieyasu, Louis XIV, Peter the Great, Thomas Jefferson, Empress Dowager Cixi, Sun Yat-sen, Tsar Nicholas II, Emperor Wu Ti, Queen Victoria, Catherine The Great
# Umbraco/Samples and Articles/Dot Net Controls/Using Page Properties In UserControls

If you use usercontrols, wrapped in macros, in templates, this trick could save you a lot of time.
This is achieved by using squarebrackets and a "#". Remember, that in template you can also use # for dictionary item, but only when using GET_ITEM. Dictionary items will not work for Macro parameters.
Syntax: <?UMBRACO_MACRO macroAlias="myAlias" myProperty="[#propertyAlias]" />
Sample: <?UMBRACO_MACRO macroAlias="myAlias" myProperty="[#bodyText]" />
This also works with default properties like <?UMBRACO_MACRO macroAlias="myAlias" myProperty="[#pageID]" />
More:
# Umbraco/Samples and Articles/Multilingual sites
## Site structure

Usually best practice is to create separate branche for each language like:
Root
* en
** en-pages
* se
** se-pages
* dk
** dk-pages
By using dictionary items, templates and some xslt, most of the functionality built for one language is easily available for other languages.
Pros: - flexibility, sites can really be different and still share mutual resources - easy assignment of user rights for nodes
Cons: - sometimes sharing of resources, like media files, will become a mess. Consider case when you have 5 languages and documentation in 2, which you want to share depending on language.
One other option is to use multiple tabs in your text pages, one tab for each language and change body content depending on a language variable (might be the active sub-domain).
The draw-back (or advantage) with this second solution is that all editors will be able to change all languages since you can't restrict access to different tabs in umbraco. Another one might be that your menu XSLT's will become a bit complex if you have to check if the language exists for each node you want to display...

## Language specific char in page names

Umbraco can handle language specific characters so that you do not have them in page names. Like in Icelandic we have ð and þ among others. http://yourdomain.com/þaðan.aspx would become http://yourdomain.com/thadan.aspx if configured correctly. Umbraco handles spaces dots and danish char by default. To ad yours modify umbracosettings.xml from the config folder, find the <urlReplacing></urlReplacing> section and add your char to it. Like <char org="þ">th</char>
# Umbraco/Samples and Articles/Creating Newsletters
## IMPORTANT - The UltraSimpleMailer is *not* compatible with Umbraco 4 and DOESN'T WORK

* Create a membergroup named "Newsletter" - members in this group will receive the newsletter
* Create a new DataType in developer (call it "Newsletter")
* Choose Ntext as database datatype and choose "UltraSimpleMailer(tm)" as rendercontrol - hit the save icon
* A bunch of new options appear: Choose the Newsletter Member Group from the dropdown list, and type the name and email address of the sender
* Now, go to Settings and create a document type named Newsletter
* Add a tab named "Mail" and add a generic property with your Newsletter Type
* Allow the Newsletter to be created under the appropriate document type (in the structure)
* Create a template that has the design that you wish the newsletter should contain
* Go to content and start creating newsletters.
# Umbraco/Samples and Articles/Dot Net Controls
## .NET Controls

* Using .NET controls with umbraco
* Access Umbraco Data with .NET controls
* Using page properties in userControls
* Using Querystring Parameters in userControls
# Umbraco/Samples and Articles/Dot Net Controls/Using Querystring Parameters In UserControls

If you use usercontrols, wrapped in macros, in templates, this trick could save you a lot of time.
This is achieved by using squarebrackets and a "@".
Syntax: <?UMBRACO_MACRO macroAlias="myAlias" myProperty="[@parameter]" />
Sample: <?UMBRACO_MACRO macroAlias="myAlias" myProperty="[@id]" />
# Umbraco/Samples and Articles/Macros
## A macro for getting the url of an image

Go to Developer/XSLT Files, create a new xslt file GetImageURL with the following content:
Now go to Developer/Macros, create a new macro, Get Image URL, giving it the alias GetImageURL. In Use XSLT File, select the XSLT file you just created.
On Parameters, add a new parameter with the alias imageID and name Image ID. The type should be set to mediaCurrent and Show should be checked. Save the macro and your home!
# Umbraco/Samples and Articles/Creating umbraco pages programmatically
## Creating umbraco pages programmatically

This is based on an original article at http://www.umbraco.org/frontpage/documentation/implementingnetcontrols/creatingumbracopagesprogrammatically.aspx

### Getting Started

It's easy to create new pages programmatically using the umbraco api (businesslogic). Just make sure you reference the umbraco.dll (or if running 2.1: cms.dll and businesslogic.dll).
Here's a few lines of sample code that shows how to do it.
In umbraco version 2.1, you can use the DocumentType.GetByAlias(string Alias), so you're not dependent on identifiers. This is best practice and very important in order to make your code re-usable. If you're running 2.0, make sure that the id of the documenttype can be changed from umbraco by making a public property.
# Umbraco/Samples and Articles/Dot Net Controls/Access Umbraco Data With Dot NET Controls
## Access Umbraco Data with .NET controls

This is based on an original article at http://www.umbraco.org/frontpage/documentation/implementingnetcontrols/accessingumbracodata.aspx

### Getting Started

You can access all umbraco content by creating a reference to the umbraco.dll. Use the node factory located at umbraco.presentation.nodeFactory to manipulate the XML used to store all of Umbracos data.
First include:
To get the current node for the page the control is on use the Node.GetCurrent method:
Using your newly created node object, you can use work on the current node directly, or use it as a starting point to traverse the entire tree.
# Timeless Theorems of Mathematics/Law of Cosines

The law of cosines explains the relation between the sides and an angle of a triangle. The law states that for any triangle $\Delta ABC$, if $\angle ACB=\theta$, $AB=c$, $AC=b$ and $BC=a$, then $c^{2}=a^{2}+b^{2}-2ab\cdot \cos \theta$. For $\theta ={\frac {\pi }{2}},\Delta ABC$ is a right angle and $c^{2}=a^{2}+b^{2}$, we have proved it as the pythagorean theorem earlier.

## Proof
### Statement

In any triangle, the square of one side's length is equal to the difference between the sum of the squares of the other two sides' lengths and twice the product of those two sides' lengths and the cosine of the included angle.
Assume, $\Delta ABC$ is a triangle where $\angle ACB=\theta ,$ $AB=c$ $AC=b$ and $BC=a.$ It is needed to be proved that, $c^{2}=a^{2}+b^{2}-2ab\cdot \cos \theta$.
Let us extend the line segment $BC$ to $BD$ (only for obtuse triangles), where $AD\perp BD$. Assume $AD=h$ and $CD=k$.

### Proof with the help of the Pythagorean Theorem
#### For obtuse triangles:

According to the Pythagorean theorem, we can say that, for $\Delta ACD,$ $b^{2}=k^{2}+h^{2}$
Similarly, for the triangle $\Delta ABD,$ $c^{2}=h^{2}+(a+k)^{2}$ $=h^{2}+a^{2}+k^{2}+2ak$ $=a^{2}+b^{2}+2ak$.
We will be using this value for further proof. But now, let's determine some trigonometric values for the triangles.
Here, $\angle ACD=\pi -\theta$.
Therefore, ${\frac {k}{b}}=\cos(\pi -\theta )$
Or, $k=b\cdot \cos(\pi -\theta )$ $=b[\cos(\pi )\cdot cos(\theta )+\sin(\pi )\cdot \sin(\theta )]$ $=-b\cdot \cos(\theta )$
Now, $c^{2}=a^{2}+b^{2}+2ak$.
$\therefore c^{2}=a^{2}+b^{2}-2ab\cdot \cos \theta$. [Proved]

#### For acute angles:

Like as the proof we have proved before, according to the Pythagorean theorem, we can say that, for $\Delta ACD,$ $b^{2}=k^{2}+h^{2}$
Similarly, for the triangle $\Delta ABD,$ $c^{2}=h^{2}+(a-k)^{2}$ $=h^{2}+a^{2}+k^{2}-2ak$ $=a^{2}+b^{2}-2ak$.
Here, $\angle ACD=\theta$.
Therefore, ${\frac {k}{b}}=\cos \theta$ $\Rightarrow k=b\cdot \cos \theta$
Now, $c^{2}=a^{2}+b^{2}-2ak$.
$\therefore c^{2}=a^{2}+b^{2}-2ab\cdot \cos \theta$. [Proved]
[Note: Whatever the triangle is, the formula, $c^{2}=a^{2}+b^{2}-2ab\cdot \cos \theta$ works.]
# Timeless Theorems of Mathematics/Rolle's Theorem

The Rolle's theorem says that, If a real-valued function $f$ is continuous on a closed interval $[a,b]$, differentiable on an open interval $(a,b)$ and $f(a)=f(b)$, then there exists at least a number $c$ such that $D_{x}f(c)=0$. It means that if a function satisfies the three conditions mentioned in the previous sentence, then there is at least a point in the graph of the function, where the slope of the tangent line at the point is $0$, or the tangent line is parallel to the $x$-axis.
# Timeless Theorems of Mathematics/Intermediate Value Theorem

The Intermediate Value Theorem is a fundamental theorem in calculus. The theorem states that if a function, $f(x)$ is continuous on a closed interval $[a,b],$ then for any value, $y$ defined between $(f(a)$ and $f(b),$ there exists at least one value $c\in (a,b)$ such that $f(c)=y$.

## Proof

Statement: If a function, $f$ is continuous on $[a,b],$ then for every $y$ between $f(a)$ and $f(b),$ there exists at least one value $c\in (a,b)$ such that $f(c)=y$
Proof: Assume that $f(x)$ is a continuous function on $[a,b]$ and $f(a)<f(b).$
Consider a function $g(x)=f(x)-y.$ The purpose of defining $g(x)$ is to investigate the behavior of $f(x)$ concerning the value $y$.
Since $f$ is continuous on $[a,b]$ and $y$ is a constant, $g(x)=f(x)-y$ is also continuous on $[a,b],$ as the difference of two continuous functions is continuous.
Now, $f(a)<y$ [As $c\in (a,b)$ and $y=f(c)$]
Or, $f(a)-y<0$
$\therefore g(a)<0$
In the same way, $g(b)>0$
Since $g(x)$ is continuous and $g(a)$ is defined below the $x$-axis while $g(b)$ is defined above the $x$-axis, there must exist at least one point $c$ in the interval $[a,b]$ where $g(c)=0$.
Therefore, at the point $c$, $g(c)=f(c)-y=0\implies f(c)=y$
∴ There exists at least one point $c$ in the interval $[a,b]$ such that $f(c)=y.$ [Proved]
# Timeless Theorems of Mathematics/Napoleon's theorem

The Napoleon's theorem states that if equilateral triangles are constructed on the sides of a triangle, either all outward or all inward, the lines connecting the centers of those equilateral triangles themselves form an equilateral triangle. That means, for a triangle $\Delta ABC$, if three equilateral triangles are constructed on the sides of the triangle, such as $\Delta ACM$, $\Delta BCX$ and $\Delta ABZ$ either all outward or all inward, the three lines connecting the centers of the three triangles, $ML$, $LN$ and $MN$ construct an equilateral triangle $LMN$.

## Proof

Let, $\Delta ABC$ a triangle. Here, three equilateral triangles are constructed, $\Delta BCE$, $\Delta ABD$ and $\Delta ACF$ and the centroids of the triangles are $P$, $Q$ and $R$ respectively. Here, $AC=b$, $AB=c$, $BC=a$, $PQ=r$, $PR=q$ and $QR=p$. Therefore, the area of the triangle $\Delta ABC$, $T={\frac {1}{2}}bc\cdot \sin A$ $\Rightarrow bc\cdot \sin A=2T$
For our proof, we will be working with one equilateral triangle, as three of the triangles are similar (equilateral). A median of $\Delta ACF$ is $AG=(m+k)$, where $AR=m$ and $RG=k$. $AQ=n$ and, as $\Delta ACF$ is a equilateral triangle, $\angle AGC=90^{\circ }$.
Here, $m+k={\frac {b{\sqrt {3}}}{2}}$. As the centroid of a triangle divides a median of the triangle as $1:2$ ratio, then $m={\frac {2}{3}}\cdot {\frac {b{\sqrt {3}}}{2}}$ $={\frac {b}{\sqrt {3}}}$. Similarly, $n={\frac {c}{\sqrt {3}}}$.
According to the Law of Cosines, $a^{2}=b^{2}+c^{2}-2bc\cdot \cos A$ (for $\Delta ABC$) and for $\Delta AQR$, $p^{2}=m^{2}+n^{2}-2mn\cdot \cos(A+60^{\circ })$
$=({\frac {b}{\sqrt {3}}})^{2}+({\frac {c}{\sqrt {3}}})^{2}-2{\frac {b}{\sqrt {3}}}\cdot {\frac {c}{\sqrt {3}}}\cdot \cos(A+60^{\circ })$
$={\frac {b^{2}+c^{2}-2bc\cdot \cos(A+60^{\circ })}{3}}$
$={\frac {b^{2}+c^{2}-2bc(\cos A\cdot cos60^{\circ }-\sin A\cdot \sin 60^{\circ })}{3}}$
$={\frac {b^{2}+c^{2}-2bc(\cos A\cdot {\frac {1}{2}}-\sin A\cdot {\frac {\sqrt {3}}{2}})}{3}}$
$={\frac {b^{2}+c^{2}-bc(\cos A\cdot -\sin A\cdot {\sqrt {3}})}{3}}$
$={\frac {b^{2}+c^{2}-bc\cos A\cdot -2T{\sqrt {3}})}{3}}$
$={\frac {a^{2}+2bc\cos A-bc\cos A-2T{\sqrt {3}})}{3}}$ [According to the law of cosines for $\Delta ABC$]
$={\frac {a^{2}+bc\cos A-2T{\sqrt {3}})}{3}}$
$={\frac {a^{2}+{\frac {b^{2}+c^{2}-a^{2}}{2}}-2T{\sqrt {3}})}{3}}$
$={\frac {\frac {b^{2}+c^{2}+a^{2}-4T{\sqrt {3}}}{2}}{3}}$
Therefore, $p={\sqrt {{\frac {1}{6}}(a^{2}+b^{2}+c^{2}-4T{\sqrt {3}})}}$
In the same way, we can prove, $q={\sqrt {{\frac {1}{6}}(a^{2}+b^{2}+c^{2}-4T{\sqrt {3}})}}$ and $r={\sqrt {{\frac {1}{6}}(a^{2}+b^{2}+c^{2}-4T{\sqrt {3}})}}$. Thus, $p=q=r$.
$\therefore \Delta PQR$ is an equilateral triangle. [Proved]
# Timeless Theorems of Mathematics/Polynomial Remainder Theorem

The Polynomial Remainder Theorem is an application of Euclidean division of polynomials. It is one of the most fundamental and popular theorems of Algebra. It states that the remainder of the division of a polynomial $P(x)$ by a linear polynomial $(x-r)$ is equal to $P(r)$.

## Examples
### Example 1

Show that the remainder of the division of a polynomial $f(x)=x^{2}-2x+2$ by a linear polynomial $(x-1)$ is equal to $f(1)$. Solution : Divide $f(x)=x^{2}-2x+2$ by $(x-1)$ like the following one.
As, $f(1)=1^{2}-2(1)+2=1$, thus the remainder is equal to $f(1)$.

### Example 2

Show that the remainder of the division of a polynomial $F(x)=ax^{2}+bx+c$ by a linear polynomial $(x-m)$ is equal to $F(m)$. Solution : Divide $F(x)=ax^{2}+bx+c$ by $(x-m)$ like the following one.
As, $f(m)=am^{2}+bm+c$, thus the remainder $am^{2}+bm+c$ is equal to $f(m)$.

## Proof
### Proposition

If $P(x)$ is a polynomial of a positive degree and $a$ is any definite number, the remainder of the division of $P(x)$ by $(x-a)$ will be $P(a)$

### Proof

The remainder of the division of a polynomial of a positive degree $P(x)$ by $(x-a)$ is either 0 or a non-zero constant. Let the remainder is $R$ and the quotient is $Q(x)$. Then, for every value of $x$,
$P(x)=(x-a)$
$Q(x)+R....(i)$
Putting $x=a$ in the equation $(i)$, we get $P(a)=0$ · $Q(a)+R=R$. Thus, the remainder of $P(x)$÷$(x-a)$ is equal to $P(a)$.
# Timeless Theorems of Mathematics/Product, Quotient, Composition Rules

The product rule, the quotient rule and the composition (or chain) rule are the most fundamental rules or formulas of differential calculus. For the functions $u(x)$ and $v(x)$, the rules are,
* Product Rule: $D_{x}(uv)=v\cdot D_{x}u+u\cdot D_{x}v$
* Quotient Rule: $D_{x}({\frac {u}{v}})={\frac {1}{v^{2}}}(v\cdot D_{x}u-u\cdot D_{x}v)$
* Composition Rule: $D_{x}v=D_{y}v\cdot D_{x}y$

## Proof
### Product Rule

$D_{x}(uv)$ $=\lim _{\Delta x\to 0}{\frac {\Delta (uv)}{\Delta x}}$ $=\lim _{\Delta x\to 0}{\frac {u(x+\Delta x)\cdot v(x+\Delta x)-u(x)\cdot v(x)}{\Delta x}}$ $=\lim _{\Delta x\to 0}{\frac {u(x+\Delta x)\cdot v(x+\Delta x)-u(x)\cdot v(x+\Delta x)+u(x)\cdot v(x+\Delta x)-u(x)\cdot v(x)}{\Delta x}}$ $=\lim _{\Delta x\to 0}{\frac {[u(x+\Delta x)-u(x)]\cdot v(x+\Delta x)+u(x)\cdot [v(x+\Delta x)-v(x)]}{\Delta x}}$ $=\lim _{\Delta x\to 0}{\frac {(\Delta u)\cdot v(x+\Delta x)+u(x)\cdot (\Delta v)}{\Delta x}}$ $=\lim _{\Delta x\to 0}[{\frac {\Delta u}{\Delta x}}\cdot v(x+\Delta x)+{\frac {\Delta v}{\Delta x}}\cdot u(x)]$ $=\lim _{\Delta x\to 0}[D_{x}u\cdot v(x+\Delta x)+u\cdot D_{x}v]$ $=v\cdot D_{x}u+u\cdot D_{x}v$
$\therefore D_{x}(uv)=v\cdot D_{x}u+u\cdot D_{x}v$ [Proved]

### Quotient Rule

$D_{x}({\frac {u}{v}})$ $=\lim _{\Delta x\to 0}{\frac {\Delta ({\frac {u}{v}})}{\Delta x}}$ $=\lim _{\Delta x\to 0}{\frac {1}{\Delta x}}\cdot [{\frac {u(x+\Delta x)}{v(x+\Delta x)}}-{\frac {u(x)}{v(x)}}]$ $=\lim _{\Delta x\to 0}{\frac {1}{\Delta x}}\cdot [{\frac {u(x+\Delta x)\cdot v(x)-u(x)\cdot v(x+\Delta x)}{v(x)\cdot v(x+\Delta x)}}]$ $=\lim _{\Delta x\to 0}{\frac {1}{\Delta x}}\cdot [{\frac {u(x+\Delta x)\cdot v(x)-v(x)\cdot u(x)+v(x)\cdot u(x)-u(x)\cdot v(x+\Delta x)}{v(x)\cdot v(x+\Delta x)}}]$ $=\lim _{\Delta x\to 0}{\frac {1}{\Delta x}}\cdot [{\frac {v(x)\cdot (u(x+\Delta x)-u(x))+u(x)\cdot (v(x)-v(x+\Delta x)}{v(x)\cdot v(x+\Delta x)}}]$ $=\lim _{\Delta x\to 0}{\frac {1}{\Delta x}}\cdot [{\frac {v(x)\cdot (\Delta u)+u(x)\cdot (\Delta v)}{v(x)\cdot v(x+\Delta x)}}]$ $=\lim _{\Delta x\to 0}{\frac {{\frac {v(x)\cdot (\Delta u)}{\Delta x}}+{\frac {u(x)\cdot (\Delta v)}{\Delta x}}}{v(x)\cdot v(x+\Delta x)}}$ $=\lim _{\Delta x\to 0}{\frac {v(x)\cdot {\frac {\Delta u}{\Delta x}}+u(x)\cdot {\frac {\Delta v}{\Delta x}}}{v(x)\cdot v(x+\Delta x)}}$ $=\lim _{\Delta x\to 0}{\frac {v(x)\cdot D_{x}u+u(x)\cdot D_{x}v}{v(x)\cdot v(x+\Delta x)}}$ $={\frac {v(x)\cdot D_{x}u+u(x)\cdot D_{x}v}{v(x)\cdot v(x)}}$ $={\frac {v\cdot D_{x}u+u\cdot D_{x}v}{v^{2}}}$
$\therefore D_{x}({\frac {u}{v}})={\frac {v\cdot D_{x}u-u\cdot D_{x}v}{v^{2}}}$

### Composition Rule

$D_{x}v$ $=\lim _{\Delta x\to 0}{\frac {\Delta v}{\Delta x}}$ $=\lim _{\Delta x\to 0}{\frac {\Delta v\cdot \Delta y}{\Delta x\cdot \Delta y}}$ $=\lim _{\Delta x\to 0}{\frac {\Delta v}{\Delta y}}\cdot \lim _{\Delta x\to 0}{\frac {\Delta y}{\Delta x}}$ $=\lim _{\Delta y\to 0}{\frac {\Delta v}{\Delta y}}\cdot \lim _{\Delta x\to 0}{\frac {\Delta y}{\Delta x}}$ $=D_{y}v\cdot D_{x}y$
$\therefore D_{x}v=D_{y}v\cdot D_{x}y$
# Timeless Theorems of Mathematics/Rational Root Theorem

The rational root theorem states that, if a rational number ${\frac {p}{q}}$ (where $p$ and $q$ are relatively prime) is a root of a polynomial with integer coefficients, then $p$ is a factor of the constant term and $q$ is a factor of the leading coefficient. In other words, for the polynomial, $P(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+...+a_{0}$, if $P\left({\frac {p}{q}}\right)=0$, (where $a_{i}\in \mathbb {Z}$ and $a_{0},a_{n}\neq 0$) then ${\frac {a_{0}}{p}}\in \mathbb {Z}$ and ${\frac {a_{n}}{q}}\in \mathbb {Z}$

## Proof

Let $P(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+...+a_{0}$, where $a_{i}\in \mathbb {Z}$.
Assume $P({\frac {p}{q}})=0$ for coprime $p,q\in \mathbb {Z}$. Therefore, $P({\frac {p}{q}})=a_{n}({\frac {p}{q}})^{n}+a_{n-1}({\frac {p}{q}})^{n-1}+a_{n-2}({\frac {p}{q}})^{n-2}+...+a_{1}({\frac {p}{q}})+a_{0}=0$ $\Rightarrow a_{n}p^{n}+a_{n-1}p^{n-1}q+a_{n-2}p^{n-2}q^{2}+...+a_{1}pq^{n-1}+a_{0}q^{n}=0$ $\Rightarrow p(a_{n}p^{n}-1+a_{n-1}p^{n-2}q+a_{n-3}p^{n-2}q^{2}+...+a_{1}q^{n-1})=-a_{0}q^{n}$
Let $w=a_{n}p^{n}-1+a_{n-1}p^{n-2}q+a_{n-3}p^{n-2}q^{2}+...+a_{1}q^{n-1}\in \mathbb {Z}$
Thus, $w=-{\frac {a_{0}q^{n}}{p}}$
As $p$ is coprime to $q$ and $w\in \mathbb {Z} .$, thus ${\frac {a_{0}}{p}}\in \mathbb {Z}$.
Again, $a_{n}p^{n}+a_{n-1}p^{n-1}q+a_{n-2}p^{n-2}q^{2}+...+a_{1}pq^{n-1}+a_{0}q^{n}=0$ $\Rightarrow q(a_{n-1}p^{n-1}+a_{n-2}p^{n-2}q+...+a_{1}pq^{n-2}+a_{0}q^{n-1})=-a_{n}p^{n}$
Let $(a_{n-1}p^{n-1}+a_{n-2}p^{n-2}q+...+a_{1}pq^{n-2}+a_{0}q^{n-1})=v\in \mathbb {Z}$
Thus, $qv=-a_{n}p^{n}$ $\Rightarrow v=-{\frac {a_{n}p^{n}}{q}}$
As $q$ is coprime to $p$ and $v\in \mathbb {Z} .$, thus ${\frac {a_{n}}{p}}\in \mathbb {Z}$.
$\therefore$ For $P(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+...+a_{0}$, if $P({\frac {p}{q}})=0$, (where $a_{i}\in \mathbb {Z}$ and $a_{0},a_{n}\neq 0$) then ${\frac {(a_{0})}{p}}\in \mathbb {Z}$ and ${\frac {a_{n}}{q}}\in \mathbb {Z}$. [Proved]
# Timeless Theorems of Mathematics/Polynomial Factor Theorem

The Polynomial Factor Theorem is a theorem linking factors and zeros of a polynomial. It is an application of the Polynomial Remainder Theorem. It states that a polynomial $f(x)$ has a factor $(x-a)$ if and only if $f(a)=0$. Here, $a$ is also called the root of the polynomial.

## Proof
### Statement

If $P(x)$ is a polynomial of a positive degree and if $P(a)=0$ so $(x-a)$ is a factor of $P(x)$.

### Proof

According to the Polynomial Remainder Theorem, the remainder of the division of $P(x)$ by $(x-a)$ is equal to $P(a)$. As $P(a)=0$, so the polynomial $P(a)$ is divisible by $(x-a)$
∴ $(x-a)$ is a factor of $P(x)$. [Proved]

### Converse of Factor Theorem

Proposition : If $(x-a)$ is a factor of the polynomial $P(x),$ then $P(a)=0$

## Factorization
### Example 1

Problem : Resolve the polynomial $P(x)=18x^{3}+15x^{2}-x-2$ into factors.
Solution : Here, the constant term of $P(x)$ is $-2$ and the set of the factors of $-2$ is $F$$_1$$=${±1, ±2}
Here, the leading coefficient of $P(x)$ is $18$ and the set of the factors of $18$ is $F$$_2$$=${±1, ±2, ±3, ±6, ±9, ±18}
Now consider $P(a)$, where $a={\frac {r}{s}},r\in F$$_1$$,s\in F$$_2$
When,
$a=1,P(1)=18+15-1-2\neq 0$
$a=-1,P(-1)=-18+15+1-2\neq 0$
$a=-{\frac {1}{2}},P(-{\frac {1}{2}})=18(-{\frac {1}{8}})+15(-{\frac {1}{4}})+{\frac {1}{2}}-2=0$
Therefore, $(x+{\frac {1}{2}})={\frac {1}{2}}(2x+1)$ is a factor of $P(x)$
Now, $P(x)=18x^{3}+15x^{2}-x-2$ $=18x^{3}+9x^{2}+6x^{2}+3x-4x-2$ $=9x^{2}(2x+1)+3x(2x+1)-2(2x+1)$ $=(2x+1)(9x^{2}+3x-2)$ $=(2x+1)(9x^{2}+6x-3x-2)$ $=(2x+1)(3x(3x+2)-1(3x+2))$ $=(2x+1)(3x+2)(3x-1)$
∴$P(x)=(2x+1)(3x+2)(3x-1)$

### Example 2

Problem : Resolve the polynomial $P(x)=-3x^{2}-2xy+8y^{2}+11x-8y-6$ into factors.
Solution : Considering only the terms of $x$ and constant, we get $-3x^{2}+11x-6$.
$-3x^{2}+11x-6\equiv (-3x+2)(x-3)..(i)$
In the same way, considering only the terms of $y$ and constant, we get $8y^{2}-8y-6$.
$8y^{2}-8y-6\equiv (4y+2)(2y-3)..(ii)$
Combining factors of above (i) and (ii), the factors of the given polynomial can be found. But the constants $+2,-3$ must remain same in both equations just like the coefficients of $x$ and $y$.
∴$P(x)=(-3x+4y+2)(x+2y-3)$

## References

* ↑ Byjus.com, Maths, Factor Theorem
* ↑ Byjus.com, Maths, Roots of Polynomials
# The Voynich Manuscript/Nymphs

Just a note, nothing final; a few comments.
* Sometimes the nymphs seem to me as spirits that operate on various levels of reality. In a few images they seem to be creating weather: rain, wind, snow, rainbows.
* A shot in the dark: If the author was a madman, maybe he saw his mother naked in the bath once, and her image populated his visions.
* A strange detail that hasn't been noted in commentary that I've read on the VMs: many of the nymphs hold their arms behind their bodies in positions that would be uncomfortable or impossible if they were human.
* When they're not bathing in tubs of green liquid, some of the nymphs stand in vehicles that look like hovering Segways.