nyuuzyou/wikis huggingface dataset:
Chiu Chang Suann Jing/Examples/Example 2

== Example of Problem from ==
:Suppose that there are a number of rabbits and pheasants confined in a cage,
:in all thirty-five heads and ninety-four feet;
:required the number of each?
== Solution ==
There are $12$ rabbits and $23$ pheasants.
== Proof ==
Let $r$ be the number of rabbits.
Let $p$ be the number of pheasants.
It is assumed that the rabbits all have $4$ feet and the pheasants $2$.
Then:
== Sources ==
* 
* 
* : The Nine Chapters: $60$
Category:Chiu Chang Suann Jing
Chiu Chang Suann Jing/Examples/Example 3

== Example of Problem from ==
:A number of men bought a number of articles, neither of which are known;
:it is only known that if each man paid $8$ cash, there would be a surplus of $3$ cash,
:and if each man paid $7$ cash, there would be a deficiency of $4$ cash.
:Required the respective numbers?
== Solution ==
There are $7$ men and $53$ objects.
== Proof ==
Let $m$ be the number of men.
Let $p$ be the number of objects.
It is assumed that the objects each cost $1$ cash each.
Then:
== Sources ==
* 
* 
* 
* : The Nine Chapters: $61$
Category:Chiu Chang Suann Jing
Chiu Chang Suann Jing/Examples/Example 4

== Example of Problem from ==
:If $5$ oxen and $2$ sheep together cost $10$ taels of gold,
:and $2$ oxen and $5$ sheep together cost $8$ taels,
:what are the prices of oxen and sheep respectively?
== Solution ==
An ox costs $1 \frac {11} {21}$ tael.
A sheep costs $\frac {20} {21}$ tael.
It would appear that the numbers for the problem were chosen with no thought to the convenience or plausibility of the answer.
== Proof ==
Let $x$ be the price per ox.
Let $y$ be the price per sheep.
Then:
== Sources ==
* 
* 
* : The Nine Chapters: $62$
Category:Chiu Chang Suann Jing
Chiu Chang Suann Jing/Examples/Example 5

== Example of Problem from ==
:There are $3$ classes of corn, of which
::$3$ bundles of the first class,
::$2$ of the second class, and
::$1$ of the third class
:make $39$ measures.
::$2$ of the first,
::$3$ of the second, and
::$1$ of the third
:make $34$ measures.
:And:
::$1$ of the first,
::$2$ of the second, and
::$3$ of the third 
:make $26$ measures.
:How many measures of grain are contained in $1$ bundle of each class?
== Solution ==
The first class bundle contains $9 \frac 1 4$ measures.
The second class bundle contains $4 \frac 1 4$ measures.
The third class bundle contains $2 \frac 3 4$ measures.
== Proof ==
Let $x$, $y$ and $z$ denote the measures of grain contained in one bundle of each of the $1$st, $2$nd and $3$rd class respectively.
We have:
== Sources ==
* 
* 
* : The Nine Chapters: $63$
Category:Chiu Chang Suann Jing
Chiu Chang Suann Jing/Examples/Example 6

== Example of Problem from ==
:There is a pool $10$ feet square, with a reed growing vertically in the centre,
:its roots at the bottom of the pool, which rises $1$ foot above the surface;
:when drawn towards the shore it reaches exactly to the brink of the pool;
:what is the depth of the water?
== Solution ==
The water is $12$ feet deep.
== Proof ==
Let the depth of the water be $d$.
The length of the reed is $d + 1$.
When drawn to the edge of the pool, the reed forms the hypotenuse of a right triangle.
One of the legs of that right triangle is the depth of the pool, which is $d$.
The other leg is the distance from the centre of the pool, which is $5$ feet.
Hence:
The right triangle in question here is the $\text{5-12-13}$ triangle.
== Sources ==
* 
* 
* : The Nine Chapters: $64$
Category:Chiu Chang Suann Jing
Chiu Chang Suann Jing/Examples/Example 7

== Example of Problem from ==
:A chain suspended from an upright post has a length of $2$ feet lying on the ground,
:and on being drawn out to its full length, so as just to touch the ground,
:the end is found to be $8$ feet from the post.
:What is the length of the chain?
== Solution ==
The chain is $17$ feet long.
== Proof ==
Let the height of the post be $h$.
The length of the chain is then $h + 2$.
When drawn out to its full length, the chain forms the hypotenuse of a right triangle.
One of the legs of that right triangle is the post, which is $h$ feet long.
The other leg is the distance of the end of the chain from the post, which is $8$ feet.
Hence:
The right triangle in question here is the $\text{8-15-17}$ triangle.
== Sources ==
* 
* 
* : The Nine Chapters: $65$
Category:Chiu Chang Suann Jing
Chiu Chang Suann Jing/Examples/Example 8

== Example of Problem from ==
:There is a bamboo $10$ feet high,
:the upper end of which being broken down on reaching the ground,
:the tip is just $3$ feet from the stem;
:what is the height of the break?
== Solution ==
The height of the break is $4 \tfrac {11} {20}$ feet.
== Proof ==
Let the height of the break be $x$ feet.
The length of the broken section is then $10 - x$ feet.
The broken section forms the hypotenuse of a right triangle.
One of the legs of that right triangle is the remaining stalk of bamboo, which is $x$ feet long.
The other leg is the distance of the end of the tip from the stalk, which is $3$ feet.
Hence:
== Historical Note ==
== Sources ==
* 
* 
* : The Nine Chapters: $66$
Category:Chiu Chang Suann Jing
Chiu Chang Suann Jing/Examples/Example 9

== Example of Problem from ==
:What is the largest circle that can be inscribed within a right-angled triangle,
:the two short sides of which are respectively $8$ and $15$?
== Solution ==
The diameter of the circle is $6$.
== Proof ==
Let the lengths of the short sides be denoted $a$ and $b$.
Let the length of the hypotenuse be denoted $c$.
 let $a \le b$.
Let the right-angled triangle be half of a rectangle whose sides are of length $a$ and $b$.
Let the rectangle be dissected along the straight lines shown.
Let $2$ copies of this rectangle be assembled into one long rectangle whose length is $a + b + c$ and whose width is $D$.
Note that $D$ is the diameter of the circle that we are required to find.
:
Thus we have that:
:$D \paren {a + b + c} = 2 a b$
Setting $a = 8$, $b = 15$, from Pythagoras's Theorem we have that $c = 17$.
Thus we have that:
== Sources ==
* 
* 
* 
* : The Nine Chapters: $67$
Category:Chiu Chang Suann Jing
Choice Function/Examples/Doubleton

== Examples of Choice Functions ==
Let $S$ be a doubleton.
Then there exist $2$ choice functions for $S$.
== Proof ==
Let $S = \set {1, 2}$.
The non-empty subsets of $S$ are $\set 1$, $\set 2$ and $\set {1, 2}$.
A choice function for $S$ must include $1$ element from each of these subsets.
There is $1$ way to select $1$ element from a singleton.
There are $2$ ways to select $1$ element from a soubleton.
Hence by the Product Rule for Counting, there are $1 \times 1 \times 2$ choice functions for $S$.
These choice functions can be defined as:
:$\map {f_1} {\set 1} = 1, \map {f_1} {\set 2} = 2, \map {f_1} {\set {1, 2} } = 1$
:$\map {f_2} {\set 1} = 1, \map {f_2} {\set 2} = 2, \map {f_2} {\set {1, 2} } = 2$
== Sources ==
* 
* 
* 
Category:Examples of Choice Functions
Choice Function/Examples/Doubletons of Real Numbers

== Example of Choice Function ==
Let $\FF$ be a set of sets of the form $\set {a, b}$ where $a$ and $b$ are real numbers.
Then there exists a choice function on $\FF$.
== Proof ==
Let $f: \FF \to \bigcup \FF$ be the mapping defined as:
:$\forall \set {a, b} \in \FF: \map f {\set {a, b} } = \map \min {a, b}$
where $\min$ denotes the minimum operation.
Then $f$ is a choice function on $\FF$.
== Sources ==
* : $1.1$ The Axiom of Choice
Category:Examples of Choice Functions
Category:Doubletons
Category:Real Numbers
Choice Function/Examples/Singletons

== Example of Choice Function ==
Let $\FF$ be a set of singletons.
Then there exists a choice function on $\FF$.
== Proof ==
Let $f: \FF \to \bigcup \FF$ be the mapping defined as:
:$\forall \set a \in \FF: \map f {\set a} = a$
Then $f$ is trivially a choice function on $\FF$.
== Sources ==
* : $1.1$ The Axiom of Choice
Category:Examples of Choice Functions
Category:Singletons
Choice Function Exists for Set of Well-Ordered Sets

== Theorem ==
Let $\mathbb S$ be a set of sets such that:
:$\forall S \in \mathbb S: S \ne \O$
that is, none of the sets in $\mathbb S$ may be empty.
Let every element of $\mathbb S$ be well-ordered.
Then there exists a choice function $f: \mathbb S \to \bigcup \mathbb S$ satisfying:
:$\forall S \in \mathbb S: \exists x \in S: \map f S = x$
== Proof ==
Every member of $\mathbb S$ is a well-ordered set. 
Thus, for $S \in \mathbb S$, there is a minimal element $s$ for $S$ (with respect to the ordering of $S$). 
By Well-Ordering Minimal Elements are Unique, $s$ is unique.
Therefore, we can define $f$ by:
:$\forall S \in \mathbb S: \map f S = s$
== Warning ==
Note that this only applies if we are given a well-ordering for each $S \in \mathbb S$.
More formally, this means: if there is a mapping that maps $S \in \mathbb S$ to a well-ordering of $S$.
If we just know that each $S \in \mathbb S$ is well-orderable, we generally do need AoC to get a choice function (to apply the proof above, we have to pick a well-order for each $S\in \mathbb S$, which requires AoC. This is related to the fact that generally we need AoC to show that, for example, the countable union of countable sets is countable.)
Category:Well-Orderings
Category:Choice Functions
Choice Function Exists for Well-Orderable Union of Sets

== Theorem ==
Let $\mathbb S$ be a set of sets such that:
:$\forall S \in \mathbb S: S \ne \O$
that is, none of the sets in $\mathbb S$ may be empty.
Let the union $\bigcup \mathbb S$ be well-orderable.
Then there exists a choice function $f: \mathbb S \to \bigcup \mathbb S$ defined as:
:$\forall S \in \mathbb S: \exists x \in S: \map f S = x$
== Proof ==
Suppose $T = \bigcup \mathbb S$ is well-orderable.
Then we can create a well-ordering $\preceq$ on $T$ so as to make $\struct {T, \preceq}$ a well-ordered set.
From the definition of well-ordered set, every subset of $T$ is itself well-ordered.
From Set is Subset of Union we have that:
:$\forall S \in \mathbb S: S \subseteq T$
So every $S \in \mathbb S$ is well-ordered and Choice Function Exists for Set of Well-Ordered Sets applies.
Indeed, if every element of $\mathbb S$ is well-ordered, then we can create a choice function $f$ defined as:
:$\forall S \in \mathbb S: \map f S = \inf S$
It may be the case that we are making infinitely many choices, but we have a rule for doing so.
Category:Set Union
Category:Well-Orderings
Category:Choice Functions
Choice Function for Power Set implies Choice Function for Set

== Theorem ==
Let $S$ be a set.
Let $\powerset S$ denote the power set of $S$.
Let there exists a choice function for $\powerset S$.
Then there exists a choice function for $S$.
== Proof ==
== Sources ==
* : Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text I$ -- Superinduction and Well Ordering: $\S 4$ Well ordering and choice: Exercise $4.2$
Category:Choice Functions
Category:Power Set
Choice Function for Set does not imply Choice Function for Union of Set

== Theorem ==
Let $S$ be a set of sets.
Let $\bigcup S$ denote the union of $S$.
Let there exist a choice function for $S$.
Then there does not necessarily exist a choice function for $\bigcup S$.
== Proof ==
== Sources ==
* : Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text I$ -- Superinduction and Well Ordering: $\S 4$ Well ordering and choice: Exercise $4.3$
Category:Choice Functions
Category:Set Union
Choice Function for Set does not imply Choice Function for Union of Set/Mistake

== Source Work ==
:
:Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text I$ -- Superinduction and Well Ordering:
::$\S 4$ Well ordering and choice:
:::Exercise $4.3$
== Mistake ==
:Show that if there exists a choice function for $S$ then there exists a choice function for $\bigcup S$.
== Analysis ==
The statement is not provable from the axioms of Zermelo-Fraenkel Set Theory because the statement implies the Axiom of Choice but Axiom of Choice is Independent of ZF.
To show this, let $A$ be an arbitrary set.
Then, by the Axiom of Pairing, there exists a set
:$S = \set A$
with $A$ as its only element.
There exists a choice function:
:$f: \powerset S \setminus \set \O \to S$
for $S$ that maps the single element:
:$\set A \in \powerset S \setminus \set \O$
in its domain to $A$.
The union of $S$ is
:$\bigcup S = A$
If the given statement is true, then there exists a choice function for every set $A$.
== Sources ==
* : Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text I$ -- Superinduction and Well Ordering: $\S 4$ Well ordering and choice: Exercise $4.3$
Category:Mistakes/Set Theory and the Continuum Problem
Cholesky Factorization/Examples/Arbitrary Example 1

== Example of Cholesky Factorization ==
Arbitrary simple example of a positive definite matrix with its Cholesky factorization:
:$\begin {pmatrix} 1 & -1 \\ -1 & 5 \end {pmatrix} = \begin {pmatrix} 1 & 0 \\ -1 & 2 \end {pmatrix} \begin {pmatrix} 1 & -1 \\ 0 & 2 \end {pmatrix}$
== Sources ==
* 
* 
Category:Examples of Cholesky Factorizations
Choquet's Theorem

== Theorem ==
Let $X$ be a locally convex vector space over $\R$.
Let $K$ be a non-empty metrizable compact convex subspace of $X$.
Let $K_e$ be the set of extreme points of $K$.
Then each $x \in K$ is a barycenter of a Borel probability measure $m_x$ on $K$ such that:
:$\map {m_x} {K_e} = 1$
That is:
:$\ds \forall \ell \in X^\ast : \map \ell x = \int_{K_e} \map \ell u \rd \map {m_x} u$
where $X^\ast$ is the dual of $X$.
== Proof ==
== Sources ==
* : $13.4$: Choquet's Theorem
* $8.6.2$: Choquet's Theorem
Category:Topological Vector Spaces
Chord Length for Regular Polygon

== Theorem ==
Let $P$ be a regular polygon of $n$ sides.
Let $P$ be inscribed into a circumcircle with radius $r$.
:
Let $CF$ be a chord of $P$, also a chord of the circumcircle.
$CF$ divides $P$ into two polygons containing $k$ and $n - k$ sides of $P$.
The length of the chord is $2 r \map \sin {\dfrac {k \pi} n}$.
== Proof ==
Let $\theta = \angle BAC$ be an inscribed angle, subtending one of the sides of $P$, with its vertex coinciding in one of the other vertices of $P$.
Because $P$ is a regular $n$-gon, its sides are all the same length.
We have that $P$ is inscribed into a circle.
Therefore, the central angles subtending a single side all have the same angular measure: $\dfrac {2 \pi} n$.
By the Inscribed Angle Theorem, the angles formed at the vertices of $P$ subtending a single side all have equal angular measure.
That is:
:$\theta = \dfrac \pi n$
Let $\angle CAF = \phi$ be an inscribed angle subtending the chord $CF$.
By addition:
:$\angle CAF = \angle CAD + \angle DAE + \angle EAF$
Thus, the angle $\phi$ is the sum of $k$ smaller angles each with measure $\theta$.
Hence:
:$\phi = k \theta = \dfrac {k \pi} n$
By Length of Chord of Circle:
:$CF = 2 r \map \sin {\dfrac {k \pi} n}$
== Examples ==
For the diagonal of a square inscribed in a circle of radius $r$, we have $k=2$. The central angle subtending one side has measure $\frac {pi} 2$, so the inscribed angle subtending one side has measure $\frac {\pi} 4$. Thus, the length of the diagonal is $2r \map \sin 2 \frac \pi 4 = 2r$.
For the regular hexagon, there are chords enclosing $k=2$ and $k=3$ sides. The latter are diameters of the circumcircle. The central angle subtending one side has measure $\frac \pi 3$, so the corresponding inscribed angle has measure $\frac \pi 6$.
For the case $k=3$ we have $2r \sin 3 \frac \pi 6 = 2r$ and for the case $k=2$ we have $2r \sin 2 \frac \pi 6 = \sqrt{3} r$
For the general case with $n$ sides, the inscribed angle subtending one side has measure $\frac \pi n$ and the length of the chord cutting off $k$ sides is $2r \sin k \frac \pi n$.
Category:Regular Polygons
Chord Lies Inside its Circle

== Theorem ==
== Proof ==
:
Let $ABC$ be a circle where $A, B$ are two points chosen arbitrarily on its circumference.
 the straight line $AB$ does not lie entirely within the circle $ABC$.
Then it will lie wholly outside (otherwise it will cross the circumference somewhere, and so we can take the part of $AB$ which does lie outside).
Find the center $D$ of circle $ABC$.
Draw $DA$, $DB$ and $DE$, and let $F$ be the point where $DE$ crosses the circumference.
As $DA = DB$ then from Isosceles Triangle has Two Equal Angles it follows that $\angle DAE = \angle DBE$.
Since one side of $AEB$ of $\triangle DAE$ is produced, it follows from External Angle of Triangle is Greater than Internal Opposite that $\angle DEB$ is greater than $\angle DAE$.
But $\angle DAE = \angle DBE$, so $\angle DEB$ is greater than $\angle DBE$.
But from Greater Angle of Triangle Subtended by Greater Side, $DB$ is greater than $DE$.
But $DB = DF$, so $DF$ is greater than $DE$.
But $DF$ is less than $DE$.
Thus we have a contradiction.
So the straight line $AB$ does not fall outside the circle.
Similarly, we can show that $AB$ does not lie on the circumference itself.
Hence the result.
== Sources ==
* : Book $\text{III}$. Propositions
Category:Circles
Chords do not Bisect Each Other

== Theorem ==
If in a circle two chords (which are not diameters) cut one another, then they do not bisect one another.
== Proof ==
:
Let $ABCD$ be a circle, in which $AC$ and $BD$ are chords which are not diameters (i.e. they do not pass through the center).
Let $AC$ and $BD$ intersect at $E$.
 they were able to bisect one another, such that $AE = EC$ and $BE = ED$.
Find the center $F$ of the circle, and join $FE$.
From Conditions for Diameter to be Perpendicular Bisector, as $FE$ bisects $AC$, then it cuts it at right angles.
So $\angle FEA$ is a right angle.
Similarly, $\angle FEB$ is a right angle.
So $\angle FEA = \angle FEB$, and they are clearly unequal.
From this contradiction, it follows that $AC$ and $BD$ can not bisect each other.
== Sources ==
* : Book $\text{III}$. Propositions
Category:Circles
Christoffel Symbols vanish at Origin of Normal Neighborhood

== Theorem ==
Let $\struct {M, g}$ be an $n$-dimensional Riemannian or pseudo-Riemannian manifold.
Let $U_p$ be the normal neighborhood for $p \in M$.
Let $\struct {U_p, \tuple {x^i}}$ be a normal coordinate chart.
Let $\set {\Gamma^i_{jk}}$ be Christoffel symbols.
Then:
:$\map {\Gamma^i_{jk} } {\map {x^r}p} = 0$
== Proof ==
== Sources ==
* : $\S 5$: The Levi-Civita Connection. Normal Neighborhoods and Normal Coordinates
Category:Normal Neighborhoods
Category:Christoffel Symbols
Chu-Vandermonde Identity

== Theorem ==
Let $r, s \in \R, n \in \Z_{\ge 0}$.
Then:
where $\dbinom r k$ is a binomial coefficient.
=== Rising Factorial Variant ===
=== Falling Factorial Variant ===
=== Extended Chu-Vandermonde Identity ===
== Proof 1 ==
== Proof 2 ==
== Proof 3 (Informal) ==
== Proof 4 ==
== Examples ==
== Also known as ==
== Historical Note ==
== Sources ==
* 
* 
* : $\S 3$: The Binomial Formula and Binomial Coefficients: Properties of Binomial Coefficients: $3.13$
* : $\S 1.2.6$: Binomial Coefficients: $\text {I} \ (21)$
* 
* : $\S 3$: The Binomial Formula and Binomial Coefficients: Properties of Binomial Coefficients: $3.13.$
* 
Category:Chu-Vandermonde Identity
Category:Binomial Coefficients
Chu-Vandermonde Identity/Also known as

== Chu-Vandermonde Identity: Also known as == 
When $r$ and $s$ are integers, the Chu-Vandermonde identity is more commonly known as Vandermonde's identity, Vandermonde's convolution (formula) or Vandermonde's convolution.
Sometimes it is seen referred to as the Chu-Vandermonde formula, or Vandermonde's theorem.
== Sources ==
* : $\S 1.2.6$: Binomial Coefficients: $\text {I}$
* 
* 
Category:Chu-Vandermonde Identity
Chu-Vandermonde Identity/Examples/2 from e + pi

== Example of Use of Chu-Vandermonde Identity ==
Let $r = e$, $s = \pi$ and $n = 2$
:$\ds \binom {e + \pi} 2 = \sum_{k \mathop = 0}^2 \binom e k \binom \pi {2 - k}$
== Proof 1 ==
== Proof 2 ==
Category:Chu-Vandermonde Identity/Examples/2 from e + pi
Category:Examples of Use of Chu-Vandermonde Identity
Chu-Vandermonde Identity/Examples/3 from 4 + 5

== Example of Use of Chu-Vandermonde Identity ==
:$\ds \binom 9 3 = \binom {4 + 5} 3 = \sum_{k \mathop = 0}^3 \binom 4 k \binom 5 {3 - k}$
== Proof 1 ==
== Proof 2 ==
Category:Chu-Vandermonde Identity/Examples/3 from 4 + 5
Category:Examples of Use of Chu-Vandermonde Identity
Chu-Vandermonde Identity/Examples/3 from 4 + 5/Proof 1

== Example of Use of Chu-Vandermonde Identity ==
== Proof ==
From the Chu-Vandermonde Identity:
The result follows on setting $r = 4$, $s = 5$ and $n = 3$.
Category:Chu-Vandermonde Identity/Examples/3 from 4 + 5
Chu-Vandermonde Identity/Extended

== Theorem ==
Let $r, s, \alpha, \beta \in \C$ be complex numbers.
Then:
:$\ds \sum_{k \mathop \in \Z} \dbinom r {\alpha + k} \dbinom s {\beta - k} = \dbinom {r + s} {\alpha + \beta}$
where $\dbinom r {\alpha + k}$ denotes a binomial coefficient.
== Proof ==
From the Chu-Vandermonde Identity, we have:
:$\ds \sum_{k \mathop \in \Z} \binom r k \binom s {n - k} = \binom {r + s} n$
Let $n = \alpha + \beta$
Let $k = \alpha + k$
Then:
:$\ds \sum_{k \mathop \in \Z} \binom r {\alpha + k} \binom s {\alpha + \beta - \paren {\alpha + k} } = \binom {r + s} {\alpha + \beta}$
:$\ds \sum_{k \mathop \in \Z} \binom r {\alpha + k} \binom s {\beta - k } = \binom {r + s} {\alpha + \beta}$
== Sources ==
* 
* : $\S 1.2.6$: Binomial Coefficients: Exercise $42$ (Solution)
Category:Chu-Vandermonde Identity
Chu-Vandermonde Identity/Falling Factorial Variant

== Theorem ==
Let $r, s \in \R, n \in \Z_{\ge 0}$.
Then:
:$\ds \sum_{k \mathop = 0}^n \dbinom n k r^{\underline k} s^{\underline {n-k} } = \paren {r + s}^{\underline n}$
== Proof ==
== Also known as ==
== Sources ==
* 
Category:Chu-Vandermonde Identity
Category:Binomial Coefficients
Category:Binomial Theorem
Chu-Vandermonde Identity/Proof 1

== Theorem ==
== Proof ==
Therefore:
:$\ds \binom {r + s} n = \sum_{k \mathop = 0}^n \binom r k \binom s {n - k}$
== Sources ==
* : $\text {3-4}$ Generating Functions: Exercise $3$
* : $\S 1.2.4$: Factorials and binomial coefficients: $1.35$
* : $\S 1.2.6$: Binomial Coefficients: Exercise $17$
Category:Chu-Vandermonde Identity
Chu-Vandermonde Identity/Proof 2

== Theorem ==
== Proof ==
The Chu-Vandermonde Identity is a special case of Gauss's Hypergeometric Theorem:
:$\map { {}_2F_1} {a, b; c; 1} = \dfrac {\map \Gamma c \map \Gamma {c - a - b} } {\map \Gamma {c - a} \map \Gamma {c - b} }$
when:
:$\map \Re {c - a - b} \gt 0$
where:
:$\map { {}_2F_1} {a, b; c; 1}$ is the hypergeometric series: $\ds \sum_{k \mathop = 0}^\infty \dfrac { a^{\overline k} b^{\overline k} } { c^{\overline k} } \dfrac {1^k} {k!}$
:$x^{\overline k}$ denotes the $k$th rising factorial power of $x$
:$\map \Gamma {n + 1} = n!$ is the Gamma function.
Starting on the :
Moving to the , we let $a = -n$:
Finally, setting the equal to the , we see the Chu-Vandermonde Identity:
:$\ds \dfrac 1 {c^{\overline n} } \sum_{k \mathop = 0}^n \dbinom n k \paren {1 - b - k}^{\overline k} \paren {c - 1 + k}^{\overline {n - k} } = \dfrac {\paren {c - b}^{\overline n} } {c^{\overline n} }$
Category:Chu-Vandermonde Identity
Chu-Vandermonde Identity/Proof 3

== Theorem ==
== Informal Proof ==
The can be thought of as the number of ways to select $n$ people from among $r$ men and $s$ women.
Each term in the is the number of ways to choose $k$ of the men and $n - k$ of the women.
== Sources ==
* : $\S 1.2.6$: Binomial Coefficients: $\text {I}$
Category:Chu-Vandermonde Identity
Chu-Vandermonde Identity/Proof 4

== Theorem ==
== Proof ==
From Sum over $k$ of $\dbinom {r - t k} k \dbinom {s - t \paren{n - k}} {n - k} \dfrac r {r - t k}$:
where $r, s, t \in \R, n \in \Z$.
Setting $t = 0$:
:$\ds \sum_{k \mathop \ge 0} \binom r k \binom s {n - k} = \binom {r + s} n$
which is the result required.
== Sources ==
* : $\S 1.2.6$: Binomial Coefficients: $\text {I}$
Category:Chu-Vandermonde Identity
Chu-Vandermonde Identity/Rising Factorial Variant

== Theorem ==
Let $r, s \in \R, n \in \Z_{\ge 0}$.
Then:
:$\ds \sum_{k \mathop = 0}^n \dbinom n k r^{\overline k} s^{\overline {n-k} } = \paren {r + s}^{\overline n}$
== Proof ==
From Rising Factorial as Factorial by Binomial Coefficient, we have:
Therefore:
== Also known as ==
== Sources ==
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Category:Chu-Vandermonde Identity
Category:Binomial Coefficients
Category:Binomial Theorem