nyuuzyou/wikis huggingface dataset:
Kronecker’s Theorem

== Theorem ==
Let $K$ be a field.
Let $f$ be a polynomial over $K$ of degree $n \ge 1$.
Then there exists a finite extension $L / K$ of $K$ such that $f$ has at least one root in $L$.
Moreover, we can choose $L$ such that the degree $\index L K$ of $L / K$ satisfies $\index L K \le n$.
== Proof ==
Let $K \sqbrk X$ be the ring of polynomial forms over $K$.
By Polynomial Forms over Field form Unique Factorization Domain, $K \sqbrk X$ is a unique factorisation domain.
Therefore, we can write $f = u g_1 \cdots g_r$, where $u$ a unit of $K \sqbrk X$ and $g_i$ is irreducible for $i = 1, \ldots, r$.
Clearly it is sufficient to find an extension of $K$ in which the irreducible factor $g_1$ of $f$ has a root.
Let $L = K \sqbrk X / \gen {g_1}$, where $\gen {g_1}$ is the ideal generated by $g_1$.
By Principal Ideal of Principal Ideal Domain is of Irreducible Element iff Maximal, $\gen {g_1}$ is maximal.
Therefore by Maximal Ideal iff Quotient Ring is Field, $L$ is a field.
Moreover, writing the quotient ring epimorphism $\overline {\map p X} = \map p X + \gen {g_1}$ for $\map p X \in K \sqbrk X$:
So $\overline X$ is a root of $g_1$ in $L$.
It remains to show that $\index L K \le n$.
By the Division Theorem for Polynomial Forms over Field, every polynomial $\map p X \in K \sqbrk X$ can be written as:
:$\map p X = \map q X \map {g_1} X + \map r X$
with $\map \deg {\map r X} < \map \deg {\map {g_1} X} \le \map \deg {\map f X} = n$.
Now we have by the definition of the quotient ring:
So if $\map r X = r_0 + r_1 X + \cdots + r_{n - 1} X^{n - 1}$, we have:
:$\overline {\map p X} = \overline {r_0 + r_1 X + \cdots + r_{n - 1} X^{n - 1} }$
Our choice of $p$ was arbitrary, so every polynomial can be written in this form.
In particular the set $\set {\overline 1, \overline X, \ldots, \overline {X^{n - 1} } }$ spans $L$ over $K$.
Thus by Generator of Vector Space Contains Basis, a basis of $L$ has at most $n$ elements.
That is, $\index L K \le n$.
Category:Field Extensions
Krull's Theorem

== Theorem ==
Let $R$ be a non-null ring with unity.
Then $R$ has a maximal ideal.
== Outline of Proof ==
We use Zorn's Lemma to construct a maximal ideal.
== Proof ==
Let $\struct {P, \subseteq}$ be the ordered set consisting of all proper ideals of $R$, ordered by inclusion.
The theorem is proved by applying Zorn's Lemma to $P$.
First, we check that the conditions for Zorn's Lemma are met: $P$ must be non-empty, and every non-empty chain in $P$ must have an upper bound.
=== $P$ is non-empty ===
Since $R$ is non-null, the zero ideal is a proper ideal of $R$, and thus an element of $P$.
=== Every non-empty chain in $P$ has an upper bound in $P$ ===
Let $\sequence {I_\alpha}_{\alpha \mathop \in A}$ be a non-empty chain of ideals in $P$.
Let $\ds I = \bigcup_{\alpha \mathop \in A} I_\alpha$.
We will show that $I$ is an upper bound in $P$ for the chain $\sequence {I_\alpha}_{\alpha \mathop \in A}$.
==== $I$ is a proper ideal of $R$ ====
By By Union of Chain of Proper Ideals is Proper Ideal, $I$ is a proper ideal of $R$.
==== $I$ is an upper bound for the chain $\sequence {I_\alpha}$ ====
Since $I$ is a proper ideal of $R$, it is an element of our ordered set $P$.
$I$ is the union of the $I_\alpha$, so $I_\alpha \subseteq I$ for all $\alpha \in A$.
This means that $I$ is an upper bound in $P$ for the chain $\sequence {I_\alpha}_{\alpha \mathop \in A}$.
=== Applying Zorn's Lemma ===
We have shown that the conditions for Zorn's Lemma are met:
:$(1): \quad P$ is non-empty
:$(2): \quad$ every non-empty chain in $P$ has an upper bound.
Applying Zorn's Lemma to $\struct {P, \subseteq}$ gives us a maximal element $M$.
This $M$ is a proper ideal of $R$ which is not contained in any other proper ideal.
So by definition, $M$ is a maximal ideal of $R$.
Category:Ideal Theory
Category:Maximal Ideals of Rings
Krull Dimension of Open Cover

== Definition ==
Let $\struct {X, \tau}$ be a topological space.
Let $\CC \subseteq \tau$ be an open cover of $X$.
Then:
:$\map \dim X = \set { \map \dim U : U \in \CC }$
where $\dim$ denotes the Krull dimension.
== Proof ==
By Krull Dimension of Topological Subspace is Smaller:
:$\forall U \in \CC : \map \dim X \ge \map \dim U$
Thus:
:$\map \dim X \ge \set { \map \dim U : U \in \CC }$
Conversely, let:
:$A_0 \subsetneq A_1 \subsetneq \cdots \subsetneq A_n$
be a chain of closed irreducible sets of $X$.
There exists an $U_0 \in \CC$ such that:
:$U_0 \cap A_0 \ne \emptyset$
Let:
:$\forall i = 0, \ldots, n : \tilde A_i := U_0 \cap A_i$
By Open Set of Irreducible Space is Irreducible, each $\tilde A_i$ is irreducible.
Moreover:
:$\tilde A_0 \subsetneq \tilde A_1 \subsetneq \cdots \subsetneq \tilde A_n$
Indeed, $\tilde A_i = \tilde A_{i+1}$ would imply:
:$A_{i+1} = A_i \cup \paren {A_{i+1} \setminus U_0}$
which contradicts the irreduciblity of $A_{i+1}$.
Thus:
:$n \le \map \dim {U_0} \le \sup \set { \map \dim U : U \in \CC }$
As $\map \dim X$ is the spremum of such $n$, we have:
:$\map \dim X \le \sup \set { \map \dim U : U \in \CC }$
Category:Irreducible Spaces
Krull Dimension of Topological Subspace is Smaller

== Definition ==
Let $X$ be a topological space.
Let $Y \subseteq X$ be a subspace.
Then:
:$\map \dim Y \le \map \dim X$
where $\dim$ denotes the Krull dimension.
== Proof ==
Let:
:$A_0 \subsetneq A_1 \subsetneq \cdots \subsetneq A_n$
be a chain of closed irreducible sets of $Y$.
Let $\map \cl {A_i}$ be the closure of $A_i$ in $X$ for each $i = 0, \ldots, n$.
By Closure of Irreducible Subspace is Irreducible, each $\map \cl {A_i}$ is irreducible.
Furthermore:
:$\map \cl {A_0} \subsetneq \map \cl {A_1} \subsetneq \cdots \subsetneq \map \cl {A_n}$
since, by Closure of Subset in Subspace:
:$A_i = Y \cap \map \cl {A_i}$
Thus:
:$n \le \map \dim X$
As $\map \dim Y$ is the spremum of such $n$, we have:
:$\map \dim Y \le \map \dim X$
Category:Irreducible Spaces
Kruskal's Algorithm

== Algorithm ==
The purpose Kruskal's Algorithm is to produce a minimum spanning tree for any given weighted graph $G$.
:Step 1: Start with the edgeless graph $T$ whose vertices correspond with those of $G$.
:Step 2: Choose an edge $e$ of $G$ such that:
::a): Adding $e$ to $T$ would not make a cycle in $T$;
::b): $e$ has the minimum weight of all the edges remaining in $G$ that fulfil the condition in a).
:Step 3: Add $e$ to $T$.
:Step 4: If $T$ spans $G$, stop. Otherwise, go to Step 2.
The above constitutes an algorithm, for the following reasons:
=== Finiteness ===
For each iteration through the algorithm, step 3 is executed, which increases the number of edges in $T$ by 1.
As a tree with $n$ nodes has $n-1$ edges, the algorithm will terminate after $n-1$ iterations.
=== Definiteness ===
:Step 1: Trivially definite.
:Step 2: As the edges of a graph can be arranged in order of weight, there is a definite edge (or set of edges) with minimal weight. It is straightforward to select an edge $e$ which does not make a cycle in $T$, by ensuring that at least one end of $e$ is incident to a vertex which has not so far been connected into $T$.
:Step 3: Trivially definite.
:Step 4: It is straightforward to determine whether all the vertices are connected.
=== Inputs ===
The input to this algorithm is the weighted graph $G$.
=== Outputs ===
The output to this algorithm is the minimum spanning tree $T$.
=== Effective ===
Each step of the algorithm is basic enough to be done exactly and in a finite length of time.
== Also known as ==
== Also see ==
* Kruskal's Algorithm produces Minimum Spanning Tree
* Prim's Algorithm
== Historical Note ==
== Sources ==
* : $\S 4.1$: The Minimal Connector Problem: An Introduction to Trees
* 
* 
* 
* 
* 
Category:Kruskal's Algorithm
Category:Graph Theory
Category:Greedy Algorithms
Kruskal's Algorithm/Also known as

== Kruskal's Algorithm: Also known as == 
It is clear that Kruskal's Algorithm is a greedy algorithm: at each stage the minimum possible weight is chosen, without any analysis as to whether there may be a combination of larger weights which may produce a smaller-weight spanning tree.
For this reason, it is sometimes called Kruskal's Greedy Algorithm.
Category:Kruskal's Algorithm
Kruskal's Algorithm produces Minimum Spanning Tree

== Theorem ==
Kruskal's Algorithm produces a minimum spanning tree.
== Proof ==
Let $N = \struct {V, E, f}$ be an undirected network.
Let $H$ be a subgraph of the underlying graph $G = \struct {V, E}$ of $N$.
Let $\ds \map f H = \sum_{e \mathop \in H} \map f e$.
Let $N$ be of order $p$.
Let $T$ be a spanning tree of $N$, created using Kruskal's Algorithm.
Then from Finite Connected Simple Graph is Tree iff Size is One Less than Order, $T$ has $p - 1$ edges.
Let the edges of $T$ be ordered $e_1, e_2, \ldots, e_{p - 1}$ according to the construction of Kruskal's Algorithm.
Then:
:$\ds \map f t = \sum_{i \mathop = 1}^{p - 1} \map f {e_i}$
Let $T_0$ be a minimum spanning tree of $N$.
Suppose that $T_0$ and $T$ are not equal.
Then $T$ has one or more edges which are not in $T_0$.
Using the ordering on the edges of $T$ created above, let $e_i$ be the first edge of $T$ not in $T_0$.
By construction, $1 \le i \le p-1$.
Let $e_i$ be added to $T_0$, to obtain the graph $G_0$.
Let $e_i = u v$ where $u, v \in V$ are vertices of $N$
Then there exists a $u-v$ path $P$ in $T_0$.
Together with $e_i$ this makes a cycle $C$ in $G_0$.
Since $T$ contains no cycles, there must be an edge $e_0$ of $C$ which is not in $T$.
Thus the edge deletion $T'_0 = G_0 - e_0$ is also a spanning tree of $N$, and:
:$\map f {T'_0} = \map f {T_0} + \map f {e_i} - \map f {e_0}$
But as $T_0$ is a minimum spanning tree:
:$\map f {T_0} \le \map f {T'_0}$
So:
:$\map f {T_0} - \map f {T'_0} \le 0$
But since:
:$\map f {T_0} - \map f {T'_0} = \map f {e_0} - \map f {e_i}$
it follows that:
$\map f {e_0} \le \map f {e_i}$
However, the manner in which $T$ was constructed, $e_i$ is an edge of smallest value which can be added to $\set {e_1, e_2, \ldots, e_{i - 1} }$ without producing a cycle.
Also, if $e_0$ is added to $\set {e_1, e_2, \ldots, e_{i - 1} }$, again no cycle is produced.
Therefore $\map f {e_i} = \map f {e_0}$ and so $\map f {T'_0} = \map f {T_0}$.
Thus $T'_0$ is a minimum spanning tree which has one more edge in common with $T$ than $T_0$, that is, $e_i$.
The procedure can be repeated as many times as there are edges in $T_0$ which are not in $T$.
Finally the spanning tree $T$ is arrived at, which by the above is a minimum spanning tree.
Hence the result.
== Sources ==
* : $\S 4.1$: The Minimal Connector Problem: An Introduction to Trees: Theorem $4.3$
Category:Tree Theory
Kummer's Hypergeometric Theorem

== Theorem ==
Let $x, n \in \C$.
Let $n \notin \Z_{< 0}$.
Let $\map \Re {x + 1} > 0$.
Then:
:$\map F {n, -x; x + n + 1; -1} = \dfrac {\map \Gamma {x + n + 1} \map \Gamma {\dfrac n 2 + 1} } {\map \Gamma {x + \dfrac n 2 + 1} \map \Gamma {n + 1} }$
where:
:$\map F {n, -x; x + n + 1; -1}$ is the Gaussian hypergeometric function of $-1$
:$\map \Gamma {n + 1} = n!$ is the Gamma function.
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Examples ==
== Also see ==
* Dixon's Hypergeometric Theorem
* Gauss's Hypergeometric Theorem
* Properties of Generalized Hypergeometric Function
== Sources ==
* : Chapter $\text {10}$. Hypergeometric Series: $\text I$
* 
Category:Kummer's Hypergeometric Theorem
Category:Hypergeometric Functions
Category:Gamma Function
Kummer's Hypergeometric Theorem/Examples

== Examples of Use of Kummer's Hypergeometric Theorem==
=== Example: $\map F {\dfrac 1 2, \dfrac 1 2; 1; -1}$ ===
=== Example: $\map F {\dfrac 1 3, \dfrac 1 3; 1; -1}$ ===
=== Example: $\map F {\dfrac 2 5, \dfrac 1 {10}; \dfrac {13} {10}; -1}$ ===
Category:Examples of Use of Kummer's Hypergeometric Theorem
Kummer's Hypergeometric Theorem/Examples/2F1(0.33,0.33;1;-1)

== Example of Use of Kummer's Hypergeometric Theorem ==
:$1 - \paren {\dfrac 1 3}^2 + \paren {\dfrac {1 \times 4} {3 \times 6} }^2 - \paren {\dfrac {1 \times 4 \times 7} {3 \times 6 \times 9} }^2 + \cdots = \dfrac {\pi} {\paren {\map \Gamma {\dfrac 5 6} }^2 \map \Gamma {\dfrac 1 3} }$
== Proof ==
From Kummer's Hypergeometric Theorem:
:$\ds \map F {n, -x; x + n + 1; -1} = \dfrac {\map \Gamma {x + n + 1} \map \Gamma {\dfrac n 2 + 1} } { \map \Gamma {x + \dfrac n 2 + 1} \map \Gamma {n + 1} } $
where: 
:$\ds \map F {n, -x; x + n + 1; -1}$ is the Gaussian hypergeometric function of $-1$: $\ds \sum_{k \mathop = 0}^\infty \dfrac { n^{\overline k} \paren {-x}^{\overline k} } { \paren {x + n + 1}^{\overline k} } \dfrac {\paren {-1}^k} {k!}$
:$x^{\overline k}$ denotes the $k$th rising factorial power of $x$
:$\map \Gamma {n + 1} = n!$ is the Gamma function.
We have:
and:
Recall from the Euler Reflection Formula: $\map \Gamma z \map \Gamma {1 - z} = \dfrac \pi {\map \sin {\pi z} }$
Therefore:
Substituting this result back into our equation above:
Therefore:
:$1 - \paren {\dfrac 1 3}^2 + \paren {\dfrac {1 \times 4} {3 \times 6} }^2 - \paren {\dfrac {1 \times 4 \times 7} {3 \times 6 \times 9} }^2 + \cdots = \dfrac {\pi} {\paren {\map \Gamma {\dfrac 5 6} }^2 \map \Gamma {\dfrac 1 3} }$
== Sources ==
* : Chapter $\text {10}$. Hypergeometric Series: $\text I$
Category:Examples of Use of Kummer's Hypergeometric Theorem‎
Kummer's Hypergeometric Theorem/Examples/2F1(0.4,0.1;1.3;-1)

== Example of Use of Kummer's Hypergeometric Theorem ==
:$1 - \paren {\dfrac {\paren {2} \paren {1} } {\paren {5} \paren {13} } } + \paren {\dfrac {\paren {2 \times 7} \paren {1 \times 11} } {\paren {5 \times 10} \paren {13 \times 23} } } - \paren {\dfrac {\paren {2 \times 7 \times 12} \paren {1 \times 11 \times 21} } {\paren {5 \times 10 \times 15} \paren {13 \times 23 \times 33} } } + \cdots = \dfrac {1944^{\frac 1 5} \pi^{\frac 3 2} } {\phi \map \Gamma {\dfrac 1 {10} } \paren {\map \Gamma {\dfrac 7 {10} } }^2 }$
== Proof ==
From Kummer's Hypergeometric Theorem:
:$\ds \map F {n, -x; x + n + 1; -1} = \dfrac {\map \Gamma {x + n + 1} \map \Gamma {\dfrac n 2 + 1} } { \map \Gamma {x + \dfrac n 2 + 1} \map \Gamma {n + 1} } $
where: 
:$\ds \map F {n, -x; x + n + 1; -1}$ is the Gaussian hypergeometric function of $-1$
:$x^{\overline k}$ denotes the $k$th rising factorial power of $x$
:$\map \Gamma {n + 1} = n!$ is the Gamma function.
We have:
and:
Recall from the Euler Reflection Formula: $\map \Gamma z \map \Gamma {1 - z} = \dfrac \pi {\map \sin {\pi z} }$
Therefore:
Recall from the Legendre Duplication Formula: $\map \Gamma z \map \Gamma {z + \dfrac 1 2} = 2^{1 - 2 z} \sqrt \pi \, \map \Gamma {2 z}$
Therefore:
Substituting these results back into our equation above:
Therefore:
:$1 - \paren {\dfrac {\paren {2} \paren {1} } {\paren {5} \paren {13} } } + \paren {\dfrac {\paren {2 \times 7} \paren {1 \times 11} } {\paren {5 \times 10} \paren {13 \times 23} } } - \paren {\dfrac {\paren {2 \times 7 \times 12} \paren {1 \times 11 \times 21} } {\paren {5 \times 10 \times 15} \paren {13 \times 23 \times 33} } } + \cdots = \dfrac {1944^{\frac 1 5} \pi^{\frac 3 2} } {\phi \map \Gamma {\dfrac 1 {10} } \paren {\map \Gamma {\dfrac 7 {10} } }^2 }$
== Sources ==
* : Chapter $\text {10}$. Hypergeometric Series: $\text I$
Category:Examples of Use of Kummer's Hypergeometric Theorem‎
Kummer's Hypergeometric Theorem/Examples/2F1(0.5,0.5;1;-1)

== Example of Use of Kummer's Hypergeometric Theorem ==
:$1 - \paren {\dfrac 1 2}^2 + \paren {\dfrac {1 \times 3} {2 \times 4} }^2 - \paren {\dfrac {1 \times 3 \times 5} {2 \times 4 \times 6} }^2 + \cdots = \dfrac {\sqrt \pi} {\sqrt 2 \paren {\map \Gamma {\dfrac 3 4} }^2 }$
== Proof ==
From Kummer's Hypergeometric Theorem:
:$\ds \map F {n, -x; x + n + 1; -1} = \dfrac {\map \Gamma {x + n + 1} \map \Gamma {\dfrac n 2 + 1} } { \map \Gamma {x + \dfrac n 2 + 1} \map \Gamma {n + 1} } $
where: 
:$\ds \map F {n, -x; x + n + 1; -1}$ is the Gaussian hypergeometric function of $-1$: $\ds \sum_{k \mathop = 0}^\infty \dfrac { n^{\overline k} \paren {-x}^{\overline k} } { \paren {x + n + 1}^{\overline k} } \dfrac {\paren {-1}^k} {k!}$
:$x^{\overline k}$ denotes the $k$th rising factorial power of $x$
:$\map \Gamma {n + 1} = n!$ is the Gamma function.
We have:
and:
Recall from the Euler Reflection Formula: $\map \Gamma z \map \Gamma {1 - z} = \dfrac \pi {\map \sin {\pi z} }$
Therefore:
Substituting this result back into our equation above:
Therefore:
:$1 - \paren {\dfrac 1 2}^2 + \paren {\dfrac {1 \times 3} {2 \times 4} }^2 - \paren {\dfrac {1 \times 3 \times 5} {2 \times 4 \times 6} }^2 + \cdots = \dfrac {\sqrt \pi} {\sqrt 2 \paren {\map \Gamma {\dfrac 3 4} }^2 }$
== Sources ==
* : Chapter $\text {10}$. Hypergeometric Series: $\text I$
Category:Examples of Use of Kummer's Hypergeometric Theorem‎
Kummer's Hypergeometric Theorem/Lemma 1

== Lemma for Kummer's Hypergeometric Theorem ==
:$\ds \lim_{y \mathop \to \infty} \dfrac {y^{\underline k} } {\paren {y + n + 1}^{\overline k} } = 1$
where $y^{\underline k}$ denotes the $k$th falling factorial of $y$.
== Proof ==
From Properties of Limit at Infinity of Real Function: Product Rule, we have:
From L'Hôpital's Rule:Corollary 2, we have:
:$\ds \lim_{y \mathop \to a^+} \frac {\map f y} {\map g y} = \lim_{y \mathop \to a^+} \frac {\map {f'} y} {\map {g'} y}$
In the present example, for the $k$th limit, we have:
:$\map {f_k} y = \paren {y + 1 - k}$
:$\map {g_k} y = \paren {y + n + k}$
Therefore taking the derivative of the numerator $\map {f_k} y$ and denominator $\map {g_k} y$ $y$, we proceed:
Therefore:
Category:Kummer's Hypergeometric Theorem
Kummer's Hypergeometric Theorem/Lemma 2

== Lemma for Kummer's Hypergeometric Theorem ==
:$\ds \lim_{y \mathop \to \infty} \dfrac {\paren {y + \dfrac n 2 + 1}^{\overline x} } {\paren {y + n + 1}^{\overline x} } = 1$
where $y^{\overline x}$ denotes the $x$th rising factorial of $y$.
== Proof ==
From Properties of Limit at Infinity of Real Function: Product Rule, we have:
From L'Hôpital's Rule:Corollary 2, we have:
:$\ds \lim_{y \mathop \to a^+} \frac {\map f y} {\map g y} = \lim_{y \mathop \to a^+} \frac {\map {f'} y} {\map {g'} y}$
In the present example, for the kth limit, we have:
:$\ds \map {f_k} y = \paren {y + \dfrac n 2 + x}$ and 
:$\ds \map {g_k} y = \paren {y + n + x}$
Therefore taking the derivative of the numerator $\map {f_k} y$ and denominator $\map {g_k} y$ $y$, we proceed:
Therefore:
Category:Kummer's Hypergeometric Theorem
Kummer's Hypergeometric Theorem/Proof 1

== Theorem ==
== Proof ==
First we note the definition of Gaussian hypergeometric function:
:$\map F {n, -x; x + n + 1; -1} = \ds \sum_{k \mathop = 0}^\infty \dfrac { n^{\overline k} \paren {-x}^{\overline k} } {\paren {x + n + 1}^{\overline k} } \dfrac {\paren {-1}^k} {k!}$
where $x^{\overline k}$ denotes the $k$th rising factorial power of $x$.
Two lemmata:
=== Lemma 1 ===
=== Lemma 2 ===
We use Dixon's Hypergeometric Theorem:
:$\ds \map { {}_3 \operatorname F_2} { { {n, -x, -y} \atop {x + n + 1, y + n + 1} } \, \middle \vert \, 1} = \dfrac {\map \Gamma {x + n + 1} \map \Gamma {y + n + 1} \map \Gamma {\dfrac n 2 + 1} \map \Gamma {x + y + \dfrac n 2 + 1} } { \map \Gamma {n + 1} \map \Gamma {x + y + n + 1} \map \Gamma {x + \dfrac n 2 + 1} \map \Gamma {y + \dfrac n 2 + 1} }$
where $\ds \map { {}_3 \operatorname F_2} { { {n, -x, -y} \atop {x + n + 1, y + n + 1} } \, \middle \vert \, 1}$ is the generalized hypergeometric function of $1$.
So:
== Sources ==
* : Chapter $\text {10}$. Hypergeometric Series: $\text I$
Category:Kummer's Hypergeometric Theorem
Kummer's Hypergeometric Theorem/Proof 2

== Theorem ==
== Proof ==
From Euler's Integral Representation of Hypergeometric Function, we have:
:$\ds \map F {a, b; c; x} = \dfrac {\map \Gamma c } {\map \Gamma b \map \Gamma {c - b} } \int_0^1 t^{b - 1} \paren {1 - t}^{c - b - 1} \paren {1 - x t}^{- a} \rd t$
Where $a, b, c \in \C$.
and $\size x < 1$
and $\map \Re c > \map \Re b > 0$.
Since Euler's Integral Representation only applies where $\size x < 1$, we will determine the limit of the integral as $x \to -1$.
By symmetry, we have:
:$\ds \map F {n, -x; x + n + 1; -1} = \ds \map F {-x, n; x + n + 1; -1}$
Therefore:
We now apply a u-substitution:
Let $u = t^2$
Substituting back into our equation, we have:
Category:Kummer's Hypergeometric Theorem
Kummer's Hypergeometric Theorem/Proof 3

== Theorem ==
== Proof ==
From Kummer's Quadratic Transformation, we have:
:$\ds \map F {a, b; 1 + a - b; z} = \paren {1 - z}^{-a} \map F {\dfrac a 2, \dfrac {1 + a} 2 - b; 1 + a - b; \dfrac {-4 z} {\paren {1 - z }^2} }$
Let $z \to -1$ and we have:
:$\ds \map F {a, b; 1 + a - b; -1} = 2^{-a} \map F {\dfrac a 2, \dfrac {1 + a} 2 - b; 1 + a - b; 1 }$
From Gauss's Hypergeometric Theorem, we have:
:$\map F {a, b; c; 1} = \dfrac {\map \Gamma c \map \Gamma {c - a - b} } {\map \Gamma {c - a} \map \Gamma {c - b} }$
Therefore, the becomes:
Substituting $a = n$ and $b = -x$, we obtain:
== Sources ==
* Chapter $\text {2}$. The hypergeometric series
Category:Kummer's Hypergeometric Theorem
Kummer's Quadratic Transformation

== Theorem ==
Let $a, b \in \C$.
Let $z < 3 - 2 \sqrt 2$
Let $\size z < 1$
Then:
:$\ds \map F {a, b; 1 + a - b; z} = \paren {1 - z}^{-a} \map F {\dfrac a 2, \dfrac {1 + a} 2 - b; 1 + a - b; \dfrac {-4 z} {\paren {1 - z}^2} }$
where:
:$\map F {a, b; 1 + a - b; z}$ is the Gaussian hypergeometric function of $z$: $\ds \sum_{k \mathop = 0}^\infty \dfrac { a^{\overline k} b^{\overline k} } {\paren {1 + a - b}^{\overline k} } \dfrac {z^k} {k!}$
:$x^{\overline k}$ denotes the $k$th rising factorial power of $x$
:$\map \Gamma {n + 1} = n!$ is the Gamma function.
== Proof ==
On the , our $z$ variable transforms to $\dfrac {-4 z} {\paren {1 - z}^2}$, therefore:
By the Quadratic Formula, either $z > 3 + 2 \sqrt 2$ or $z < 3 - 2 \sqrt 2$.
Because $\size z < 1$, we have:
:$\size z < 3 - 2 \sqrt 2$
This means our sum is analytic and can be expanded in powers in $z$ for $\size z < 3 - 2 \sqrt 2$
From the definition of Gaussian hypergeometric function, the can be written as:
The coefficient of $z^n$ on the is:
We now observe that the coefficients of $z^n$ on the are identical to the coefficients of $z^n$ on the .
Therefore:
:$\ds \map F {a, b; 1 + a - b; z} = \paren {1 - z}^{-a} \map F {\dfrac a 2, \dfrac {1 + a} 2 - b; 1 + a - b; \dfrac {-4 z} {\paren {1 - z}^2} }$
== Also see ==
* Euler's Transformation
* Pfaff's Transformation
== Sources ==
* : Chapter $\text {2}$. The Hypergeometric Functions
* Chapter $\text {2}$. The hypergeometric series
* 
Category:Kummer's Quadratic Transformation
Category:Gaussian Hypergeometric Function
Category:Gamma Function
Kummer's Theorem

== Theorem ==
Let $p$ be a prime number.
Let $a, b \in \Z_{\ge 0}$.
Let:
:$p^n \divides \dbinom {a + b} b$
but
:$p^{n + 1} \nmid \dbinom {a + b} b$
where:
:$\divides$ denotes divisibility
:$\nmid$ denotes non-divisibility
:$\dbinom {a + b} b$ denotes a binomial coefficient.
Then $n$ equals the number of carries that occur when $a$ is added to $b$ using the classical addition algorithm in base $p$.
== Proof ==
== Sources ==
* 
* : $\S 1.2.6$: Binomial Coefficients: Exercise $11$
Category:Binomial Coefficients
Category:Number Bases
Category:Prime Numbers
Kuratowski's Closure-Complement Problem

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $A \subseteq S$ be a subset of $T$.
By successive applications of the operations of complement relative to $S$ and the closure, there can be as many as $14$ distinct subsets of $S$ (including $A$ itself).
=== Example ===
:
== Proof ==
That there can be as many as $14$ will be demonstrated by example.
=== Complement ===
=== Interior ===
=== Closure ===
=== Exterior ===
=== Closure of Complement ===
=== Closure of Interior ===
=== Interior of Closure ===
=== Interior of Closure of Interior ===
=== Interior of Complement of Interior ===
=== Closure of Interior of Complement ===
=== Closure of Interior of Closure ===
=== Interior of Complement of Interior of Closure ===
=== Complement of Interior of Closure of Interior ===
=== Proof of Maximum ===
It remains to be shown that there can be no more than $14$.
== Also known as ==
This result is also known as Kuratowski's Closure-Complement Theorem.
== Sources ==
* 
* : Part $\text I$: Basic Definitions: Section $1$: General Introduction: Closures and Interiors
* : Part $\text {II}$: Counterexamples: $32$. Special Subsets of the Real Line: $9$
Category:Set Closures
Category:Relative Complement
Category:Kuratowski's Closure-Complement Problem
Category:14