nyuuzyou/wikis huggingface dataset:
Half Angle Formulas/Sine/Proof 2

== Theorem ==
== Proof ==
Define:
:$u = \dfrac \theta 2$
Then:
We also have that:
:In quadrant $\text I$, and quadrant $\text {II}$, $\sin \theta > 0$
:In quadrant $\text {III}$, and quadrant $\text {IV}$, $\sin \theta < 0$.
Category:Half Angle Formula for Sine
Half Angle Formulas/Tangent

== Theorem ==
where $\tan$ denotes tangent and $\cos$ denotes cosine.
When $\theta = \paren {2 k + 1} \pi$, $\tan \dfrac \theta 2$ is undefined.
=== Corollary 1 ===
=== Corollary 2 ===
=== Corollary 3 ===
== Proof ==
Since $\cos \theta \ge -1$, it follows that $\cos \theta + 1 \ge 0$.
When $\cos \theta = -1$ it follows that $\cos \theta + 1 = 0$ and then $\tan \dfrac \theta 2$ is undefined.
This happens when $\theta = \paren {2 k + 1} \pi$.
We have that:
:$\tan \dfrac \theta 2 = \dfrac {\sin \dfrac \theta 2} {\cos \dfrac \theta 2}$
=== Quadrant $\text I$ ===
In quadrant $\text I$, we have:
:Sine in First Quadrant: $\sin \dfrac \theta 2 > 0$
:Cosine in first Quadrant: $\cos \dfrac \theta 2 > 0$
and so in quadrant $\text I$:
=== Quadrant $\text {II}$ ===
In quadrant $\text {II}$, we have:
:Sine in Second Quadrant: $\sin \dfrac \theta 2 > 0$
:Cosine in Second Quadrant: $\cos \dfrac \theta 2 < 0$
and so in quadrant $\text {II}$:
=== Quadrant $\text {III}$ ===
In quadrant $\text {III}$, we have:
:Sine in Third Quadrant: $\sin \dfrac \theta 2 < 0$
:Cosine in Third Quadrant: $\cos \dfrac \theta 2 < 0$
and so in quadrant $\text {III}$:
=== Quadrant $\text {IV}$ ===
In quadrant $\text {IV}$, we have:
:Sine in Fourth Quadrant: $\sin \dfrac \theta 2 < 0$
:Cosine in Fourth Quadrant: $\cos \dfrac \theta 2 > 0$
and so in quadrant $\text {IV}$:
== Also see ==
* Half Angle Formula for Sine
* Half Angle Formula for Cosine
== Sources ==
* : $\S 5$: Trigonometric Functions: $5.43$
Category:Half Angle Formula for Tangent
Category:Tangent Function
Half Angle Formulas for Spherical Triangles

== Theorem ==
Let $\triangle ABC$ be a spherical triangle on the surface of a sphere whose center is $O$.
Let the sides $a, b, c$ of $\triangle ABC$ be measured by the angles subtended at $O$, where $a, b, c$ are opposite $A, B, C$ respectively.
=== Cosine of Half Angle for Spherical Triangles ===
=== Sine of Half Angle for Spherical Triangles ===
=== Tangent of Half Angle for Spherical Triangles ===
== Also known as ==
== Also see ==
* Half Side Formulas for Spherical Triangles
Category:Spherical Trigonometry
Category:Tangent Formulas
Category:Half Angle Formulas for Spherical Triangles
Half Angle Formulas for Spherical Triangles/Also known as

== Half Angle Formulas for Spherical Triangles: Also known as == 
Some sources render Half Angle Formulas for Spherical Triangles with a hyphen: Half-Angle Formulas for Spherical Triangles.
Some use the term Half-Angle Formulae for Spherical Triangles.
Category:Half Angle Formulas for Spherical Triangles
Half Side Formulas for Spherical Triangles

== Theorem ==
Let $\triangle ABC$ be a spherical triangle on the surface of a sphere whose center is $O$.
Let the sides $a, b, c$ of $\triangle ABC$ be measured by the angles subtended at $O$, where $a, b, c$ are opposite $A, B, C$ respectively.
=== Cosine of Half Side for Spherical Triangles ===
=== Sine of Half Side for Spherical Triangles ===
=== Tangent of Half Side for Spherical Triangles ===
== Also known as ==
== Also see ==
* Half Angle Formulas for Spherical Triangles
Category:Half Side Formulas for Spherical Triangles
Category:Tangent Formulas
Category:Spherical Trigonometry
Half Side Formulas for Spherical Triangles/Also known as

== Half Side Formulas for Spherical Triangles: Also known as == 
Some sources render Half Side Formulas for Spherical Triangles with a hyphen: Half-Side Formulas for Spherical Triangles.
Some use the term Half-Side Formulae for Spherical Triangles.
== Sources ==
* 
* 
Category:Half Side Formulas for Spherical Triangles
Hall's Marriage Theorem/Explanation

== Hall's Marriage Theorem: Explanation ==
This Hall's Marriage Theorem is so called for the following reason:
Let $I$ be a set of women.
Suppose that each woman $k$ is romantically interested in a finite set $S_k$ of men.
Suppose also that:
:each woman would like to marry exactly one of these men
and:
:each man in $\ds \bigcup_{k \mathop \in I} S_k$ would like to marry at most one woman in $I$.
Then this theorem gives a condition under which it is possible to match every woman to a man.
Category:Hall's Marriage Theorem
Hall's Marriage Theorem/Finite Set

== Theorem ==
Let $\SS = \family {S_k}_{k \mathop \in I}$ be a finite indexed family of finite sets.
For each $F \subseteq I$, let $\ds Y_F = \bigcup_{k \mathop \in F} S_k$.
Let $Y = Y_I$.
Then the following are equivalent:
:$(1): \quad \SS$ satisfies the marriage condition: 
:$(2): \quad$ There exists an injection $f: I \to Y$ such that $\forall k \in I: \map f k \in S_k$.
== Proof ==
=== $(2)$ implies $(1)$ ===
Let:
:$\exists P \subseteq I: \card P > \card {Y_P}$
Then:
:$\card P \not \le \card {Y_P}$
By contrapositive of Injection implies Cardinal Inequality, there can be no injection from $P$ to $Y_P$.
Thus there can be no injection from $I$ to $Y$ satisfying the requirements.
=== $(1)$ implies $(2)$ ===
The proof proceeds by induction on $n$, the cardinality of $I$.
If $n = 0$, then the empty set is the necessary injection.
Suppose the theorem holds for all $m < n$.
Let $I$ be a set with $n$ elements, where $n \ge 1$.
Let $\SS = \family {S_k}_{k \mathop \in I}$ be an indexed family of finite sets that satisfies the marriage condition.
Let $e \in I$.
Let $y \in S_e$.
Let $\SS_y = \family {S_k \setminus \set y}_{k \mathop \in I \setminus \set e}$.
By Law of Excluded Middle, one of the following must hold:
:$(a): \quad \SS_y$ satisfies the marriage condition.
:$(b): \quad \SS_y$ violates the marriage condition.
=== Case $(a)$ ===
Suppose $\SS_y$ satisfies the marriage condition.
By Cardinality Less One:
:$\card {I \setminus \set e} = n - 1 < n$
Then by the inductive hypothesis, there is an injection $\ds g: I \setminus \set e \to \bigcup \SS_y$ such that for each $k \in I \setminus \set e$, $\map g k \in S_k \setminus \set y$.
Let $f: I \to Y$ be defined thus:
:$\map f k = \begin{cases}
\map g k & : k \ne e \\
y & : k = e \end{cases}$
Then $f$ is an injection satisfying the requirements.
=== Case $(b)$ ===
Suppose $\SS_y$ does not satisfy the marriage condition.
Then there exists $P \subseteq I \setminus \set e \subsetneqq I$ such that:
:$\ds \card P > \card {\bigcup_{i \mathop \in P} \paren {S_i \setminus \set y} }$
Note that $P$ cannot be empty.
Since $\SS$ satisfies the marriage condition:
:$\card P \le \card {Y_P}$
so $y \in Y_P$.
By Cardinality Less One:
:$\card {Y_P \setminus \set y} = \card {Y_P} - 1$
Thus:
:$\card P > \card {Y_P} - 1$
Thus:
:$\card P = \card {Y_P}$
We have for $\card P < n$ that $P \subsetneqq I$.
So by the inductive hypothesis there is an injection $g: P \to Y_P$ such that:
:$\forall k \in P: \map g k \in S_k$
It is to be shown that $\TT = \family {S_k \setminus Y_P}_{k \mathop \in I \setminus P}$ satisfies the marriage condition.
Let $Q \subseteq I \setminus P$.
We have that:
:$P$ and $Q$ are disjoint sets
:$\SS$ satisfies the marriage condition.
Thus:
As this holds for all such $Q$, $\TT$ satisfies the marriage condition.
As $\card P \ne 0$, $\card {I \setminus P} < n$.
Thus there is an injection $h: I \setminus P \to Y_{I \setminus P} \setminus Y_P$ such that:
:$\forall k \in I \setminus P: \map h k \in S_k \setminus Y_P$
Then $f = g \cup h$ is the injection we seek.
== Explanation ==
Category:Hall's Marriage Theorem
Hall's Marriage Theorem/General Set

== Theorem ==
Let $\SS = \family {S_k}_{k \mathop \in I}$ be an indexed family of finite sets.
For each $F \subseteq I$, let $\ds Y_F = \bigcup_{k \mathop \in F} S_k$.
Let $Y = Y_I$.
Then the following are equivalent:
:$(1): \quad \SS$ satisfies the marriage condition: 
:$(2): \quad$ There exists an injection $f: I \to Y$ such that $\forall k \in I: \map f k \in S_k$.
== Proof ==
=== $(2)$ implies $(1)$ ===
Suppose that for some finite $F \subseteq I$, $\card F > \card {Y_F}$.
Then $\card F \not \le \card {Y_F}$.
By contrapositive of Injection implies Cardinal Inequality, there can be no injection from $F$ to $Y_F$.
Thus there can be no injection from $I$ to $Y$ satisfying the requirements.
=== $(1)$ implies $(2)$ ===
Let $\map \Phi x$ be the set of all elements of $y$ in the expression of $x$.
Let $\FF$ be the set of all partial functions $g$ from $I$ to $Y$ such that:
:$g$ is one-to-one.
:If $\tuple {k, y} \in g$, then $y \in S_k$.
Then for any $k \in I$, $\set {\map g x: g \in \FF \land k \in \Dom g}$ is finite.
$\FF$ has finite character:
Let $g \in \FF$.
Let $F$ be a finite subset of $I$.
Then $g \restriction F$ is obviously in $\FF$.
Suppose instead that $g$ is a partial function from $I$ to $Y$.
Let the restriction of $g$ to each finite subset of $\Dom g$ be in $\FF$.
Then $g$ is one-to-one, as follows:
Let $p, q \in \Dom g$ with $p \ne q$.
Then $\set {p, q}$ is a finite subset of $\Dom g$.
Thus:
:$g \restriction \set {p, q} \in \FF$
Thus:
:$\map g p \ne \map g q$
Let $\tuple {k, y} \in g$.
Then:
:$\tuple {k, y} \in g \restriction \set k$
So:
:$y \in S_k$
Thus:
:$g \in \FF$
So we see that $\FF$ has finite character.
Let $F$ be a finite subset of $I$.
By the finite case, there is an element of $\FF$ with domain $F$.
By the Cowen-Engeler Lemma, $\FF$ has an element $\phi$ whose domain is $I$.
That is, $\phi$ is a mapping from $I$ to $Y$.
But a one-to-one mapping is an injection, so $\phi$ is an injection from $X$ to $Y$.
By the definition of $\FF$:
:$\forall k \in I: \map \phi k \in S_k$
== Explanation ==
Category:Hall's Marriage Theorem
Halley's Method

== Proof Technique ==
Halley's Method is a method of solving an equation $\map f x = 0$ involving the real function $f$ for which there may be no convenient closed form solution.
The derivative and the second derivative of $f$ have to be known in order to use Halley's Method.
Let $x_0$ be an initial guess.
Halley's Method uses the iteration:
:$x_{n + 1} = x_n - \dfrac {\map f {x_n} } {\map {f'} {x_n} - \dfrac {\map f {x_n} \map {f' '} {x_n} } {2 \map {f'} {x_n} } }$
where:
:$\map {f'} {x_n}$ is the derivative of $f$ $x$ evaluated at $x_n$
:$\map {f' '} {x_n}$ is the second derivative of $f$ $x$ evaluated at $x_n$.
== Proof ==
The function $\map f x$ can be expanded about $x_n$ using Taylor's Theorem:
:$\map f x = \map f {x_n} + \map {f'} {x_n} \paren {x - x_n} + \dfrac 1 2 \map {f' '} {x_n} \paren {x - x_n}^2 + \dotsb$
This series can be rewritten with an error term as follows:
== Historical Note ==
== Also see ==
* Newton's Method, which is slower but uses less computation per iteration
== Sources ==
* 
* 
Category:Halley's Method
Category:Calculus
Category:Numerical Analysis
Halting Problem is Not Algorithmically Decidable

== Theorem ==
Let $H: \N^2 \to \N$ be the function given by:
:$\map H {m, n} = 1$ if $m$ codes a URM program which halts with input $n$
:$\map H {m, n} = 0$ otherwise.
Then $H$ is not recursive.
== Proof ==
We perform a proof by Cantor's Diagonal Argument.
 $H$ is recursive.
Consider the universal URM computable function $\Phi_1: \N^2 \to \N$.
Let $f: \N \to \N$ be the function given by:
:$\map f n = \begin{cases}
\map {\Phi_1} {n, n} & : \map H {n, n} = 1 \\
0 & : \text{otherwise}
\end{cases}$
As $H$ is recursive, the relation $\map H {n, n} = 1$ is also recursive.
The universal function is recursive as all URM computable functions are recursive.
Also, $\map {\Phi_1} {n, n}$ is defined when $\map H {n, n} = 1$.
The constant $0$ is (primitive) recursive and always defined.
So $f$ is total, and by Combination of Recursive Functions it is also recursive.
It follows immediately that the function $g: \N \to \N$ given by:
:$\map g n = \map f n + 1$
is also recursive.
So by Universal URM Computable Functions there exists some $e \in \N$ such that:
:$\forall n: \map g n = \map {\Phi_1} {e, n}$
that is, the URM program coded by $e$ computes $g$.
Hence:
:$\map g e = \map {\Phi_1} {e, e}$
But since $e$ codes a URM program which halts with input $e$, we have:
:$\map H {e, e} = 1$
and so:
:$\map f e = \map {\Phi_1} {e, e}$
Therefore, by definition of $g$, we have:
:$\map g e = \map {\Phi_1} {e, e} + 1$
This contradiction arose because we assumed that $H$ is recursive.
Hence the result.
Category:URM Programs
Category:Recursion Theory
Hamilton's Principle

== Physical Law ==
In a conservative field, the motion of a mechanical system can be characterized by requiring that:
:$\ds \int_{t_1}^{t_2} \paren {T - V} \rd t$
where:
:$T$ denotes the kinetic energy
:$V$ denotes the potential energy
be stationary in an actual motion during the time interval from $t_1$ to $t_2$.
== Proof ==
== Also see ==
* Definition:Calculus of Variations
== Sources ==
* 
* 
Category:Hamiltonian Mechanics
Hamiltonian Graph is not necessarily Ore Graph

== Theorem ==
Let $G = \struct {V, E}$ be a simple graph of order $n \ge 3$.
Let $G$ be a Hamiltonian graph.
Then $G$ is not necessarily an Ore graph.
== Proof ==
Proof by Counterexample:
Recall the definition of an Ore graph:
:For each pair of non-adjacent vertices $u, v \in V$:
::$\deg u + \deg v \ge n$
Let $n \in \N$ such that $n \ge 5$.
Consider the cycle graph $C_n$.
We have from Cycle Graph is Hamiltonian that $C_n$ is a Hamiltonian graph.
We also have from Cycle Graph is 2-Regular that all the vertices of $C_n$ are of degree $2$.
Let $u, v \in C_n$ such that $u$ and $v$ are non-adjacent.
Such a pair is always possible to find in $C_n$.
Then:
:$\deg u + \deg v = 4 < n$
so $C_n$ is not an Ore graph.
 
== Also see ==
* Dirac's Theorem
* Ore's Theorem
Category:Hamiltonian Graphs
Category:Ore Graphs
Hamiltonian formalism

Prior to the dawn of Quantum Mechanics, there was Classical Mechanics. This was a theory initiated by the Italian polymath Galileo Galilei with his studies of kinematics and inertial frames of reference and later formalized by Sir Isaac Newton with his three well-known Laws. This Force-oriented theory of movement soon proved to be as useful as it was complex, since it requires to solve a system of up to three second order ODEs which are probably coupled. The need for a more modern, elegant and general approach was evident, and it was Joseph-Louis Lagrange who brought it to life (with the noteworthy mention of D'Alembert), and with it the advent of Theoretical Mechanics. Despite this great achievement, it would be William Rowan Hamilton who would accomplish the feat of reducing Classical Mechanics to a single first principle ("the path followed by an elementary particle is such that the action is stationary"), and introduced the Hamiltonian formalism.
In Quantum Physics, one works with probabilities rather than with certainties. Therefore, the formalism must be different than the Classical one: Erwin Schrödinger introduced the concept of "wavefunction" one can assign to a particle, which encodes all of the available information about the particle (this is a postulate). For a wavefunction to be a candidate to describe a quantum system, it must fulfill two conditions: (1) to be a solution to Schrödinger's equation and (2) to be normalizable (that is to say, to be square-integrable). Schrödinger's work was based on the Classical Hamilton-Jacobi theory, so it is not surprising that the operator we find in his time-independent equation is called, precisely, Hamiltonian.
$H\Psi = E\Psi$
where $H$ is the Hamiltonian operator, $\Psi$ is the spatial wavefunction and $E$ is the energy of the particle in state $\Psi$. As one can see, this is an eigenvalue problem for some differential operator $H$, which is defined as:
$H \equiv \frac{p²}{2m}+V$In total resemblance to the Hamiltonian used in Classical Mechanics, but with the essential difference that the so-called observables ($x$ for position and $p$ for momentum) are now operators acting on functions of a Hilbert space.
Hamiltonian of Standard Lagrangian is Total Energy

== Theorem ==
Let $P$ be a physical system of classical particles.
Let $L$ be a standard Lagrangian associated with $P$.
Then the Hamiltonian of $P$ is the total energy of $P$.
== Proof ==
By definition, the last expression is the total energy of $P$.
== Sources ==
* : $\S 4.21$: The Principle of Least Action
* 
* 
Category:Hamiltonians
Category:Lagrangian Mechanics
Category:Mathematical Physics
Category:Physics
Hamming Distance/Examples/Codewords in V(4,2)

== Example of Hamming Distance ==
The Hamming distance between the codewords $0101$ and $1110$ is $3$.
== Proof ==
This is because they are different in $3$ digits: the $1$st, $3$rd and $4$th.
== Sources ==
* 
Category:Examples of Hamming Distance
Hamming Distance/Examples/Codewords in V(4,3)

== Example of Hamming Distance ==
In the master code $\map V {4, 3}$, the Hamming distance between the codewords $1201$ and $2211$ is $2$.
== Proof ==
This is because they are different in $2$ digits: the $1$st and the $3$rd.
== Sources ==
* : Chapter $6$: Error-correcting codes: Example $6.6$
Category:Examples of Hamming Distance
Hamming Distance between Codewords of Perfect Code

== Theorem ==
Let $\KK$ be a $k$-perfect code.
The Hamming distance between codewords of $\KK$ may well be less than the maximum Hamming distance of $2 k + 1$.
== Proof ==
== Also see ==
* Upper Bound of Hamming Distance for Length of K-Error-Correcting Code
== Sources ==
* 
Category:Perfect Codes
Category:Hamming Distance
Hamming Distance is Distance Function

== Theorem ==
Let $\map V {n, p}$ be a master code.
Let $d: V \times V \to \Z$ be the mapping defined as:
:$\forall u, v \in V: \map d {u, v} =$ the Hamming distance between $u$ and $v$
that is, the number of corresponding terms at which $u$ and $v$ are different.
Then $d$ defines a distance function in the sense of a metric space.
== Proof ==
It is to be demonstrated that $d$ satisfies all the metric space axioms.
Let $u, v, w \in \map V {n, p}$ be arbitrary.
=== Proof of ===
By definition of Hamming distance:
:$\map d {u, u} = 0$
So holds for $d$.
=== Proof of ===
Consider $\map d {u, v} + \map d {v, w}$.
Let $\map d {u, w} \ne 0$.
Then at each term at which $u$ and $w$ are different, those corresponding terms in either $u$ and $v$ or $v$ and $w$ must be different.
So every contribution to the value of $\map d {u, w}$ is present in either $\map d {u, v}$ or $\map d {v, w}$.
It follows that $\map d {u, v} + \map d {v, w} \ge \map d {u, w}$.
So holds for $d$.
=== Proof of ===
$\map d {u, v} = \map d {v, u}$ by definition of Hamming distance.
So holds for $d$.
=== Proof of ===
So holds for $d$.
Thus $d$ satisfies all the metric space axioms and so is a metric.
== Sources ==
* : Chapter $6$: Error-correcting codes: Proposition $6.7$
Category:Hamming Distance
Handshake Lemma

== Theorem ==
Let $G$ be a $\tuple {p, q}$-undirected graph, which may be a multigraph or a loop-graph, or both.
Let $V = \set {v_1, v_2, \ldots, v_p}$ be the vertex set of $G$.
Then:
:$\ds \sum_{i \mathop = 1}^p \map {\deg_G} {v_i} = 2 q$
where $\map {\deg_G} {v_i}$ is the degree of vertex $v_i$.
That is, the sum of all the degrees of all the vertices of an graph is equal to twice its size.
This result is known as the Handshake Lemma or Handshaking Lemma.
=== Corollary ===
== Proof ==
In the notation $\tuple {p, q}$-graph, $p$ is its order and $q$ its size.
That is, $p$ is the number of vertices in $G$, and $q$ is the number of edges in $G$.
Each edge is incident to exactly two vertices.
The degree of each vertex is defined as the number of edges to which it is incident.
So when we add up the degrees of all the vertices, we are counting all the edges of the graph twice.
== Examples ==
== Historical Note ==
== Sources ==
* : Chapter $2$: Elementary Concepts of Graph Theory: $\S 2.1$: The Degree of a Vertex: Theorem $2.1$
Category:Graph Theory
Category:Vertices of Graphs
Category:Direct Proofs
Category:Named Theorems
Category:Handshake Lemma
Handshake Lemma/Corollary

== Corollary to the Handshake Lemma ==
Let $G$ be a $\tuple {p, q}$-graph, which may be a multigraph or a loop-graph, or both.
The number of odd vertices in $G$ is even.
== Proof ==
Let $G$ be a $\tuple {p, q}$-graph.
Consider the sum of the degrees of its vertices:
:$\ds K = \sum_{v \mathop \in V} \map {\deg_G} v$
From the Handshake Lemma:
:$K = 2 \card E$
which is an even integer.
Subtracting from $K$ the degrees of all even vertices, we are left with the sum of all degrees of odd vertices in $V$.
That is:
:$\ds \paren {\sum_{v \mathop \in V} \map {\deg_G} v} - \paren {\sum_{v \mathop \in V : \map {\deg_G} v \mathop = 2 k} \map {\deg_G} v} = \paren {\sum_{v \mathop \in V : \map {\deg_G} v \mathop = 2 k + 1} \map {\deg_G} v}$
This must still be an even number, as it is equal to the difference of two even numbers.
Because this is a sum of exclusively odd terms, there must be an even number of such terms for the sum on the to be even.
Hence the number of odd vertices in $G$ must be even.
== Sources ==
* : Chapter $2$: Elementary Concepts of Graph Theory: $\S 2.1$: The Degree of a Vertex: Theorem $2.2$
* : Chapter $\text {A}.21$: Euler ()
* : The Bridges of Königsberg: $134$
Category:Handshake Lemma
Handshake Lemma/Examples/Arbitrary Order 8 Graph

== Examples of Use of Handshake Lemma ==
:
The above simple graph has $8$ vertices and $10$ edges (which can be ascertained by counting).
== Sources ==
* : Chapter $2$: Elementary Concepts of Graph Theory: $\S 2.1$: The Degree of a Vertex: Problem $1$
Category:Handshake Lemma
Handshake Lemma/Examples/Impossible Order 6 Graph

== Examples of Use of Handshake Lemma ==
There exists no undirected graph whose vertices have degrees $2, 3, 3, 4, 4, 5$.
== Proof ==
If such a graph were to exist, it would have exactly $3$ odd vertices.
This contradicts the Handshake Lemma, which states that the number of odd vertices is even.
== Sources ==
* : Chapter $2$: Elementary Concepts of Graph Theory: $\S 2.1$: The Degree of a Vertex: Problem $2$
Category:Handshake Lemma
Handshake Lemma/Examples/No Graph with One Odd Vertex

== Examples of Use of Handshake Lemma ==
There exists no undirected graph with exactly one odd vertex.
== Proof ==
The Handshake Lemma which states that the number of odd vertices in any undirected graph is even.
== Sources ==
* : Chapter $2$: Elementary Concepts of Graph Theory: $\S 2.1$: The Degree of a Vertex: Problem $7 \ \text{(c)}$
Category:Handshake Lemma
Hankel Matrix is Symmetric

== Theorem ==
Let $\mathbf H$ be a Hankel matrix.
Then $\mathbf H$ is a symmetric matrix.
== Proof ==
Recall the definition of symmetric matrix:
$\mathbf A$ is symmetric :
:$\mathbf A = \mathbf A^\intercal$
where $\mathbf A^\intercal$ is the transpose of $\mathbf A$.
By definition of transpose of $\mathbf A$:
:$\mathbf A^\intercal_{i j}: = \mathbf A_{j i}$
Recall the definition of Hankel matrix:
That is, by definition of antidiagonal:
:$\forall i, j \in \closedint 1 n: \mathbf A_{i j}: = \mathbf A_{j i}$
Hence $\mathbf A^\intercal = \mathbf A$ and the result follows.
== Sources ==
* 
Category:Hankel Matrices
Category:Symmetric Matrices
Hankel Representation of Riemann Zeta Function

== Theorem ==
Let $C$ be the Hankel contour.
Then for $s \in \C \setminus \Z_{>0}$: 
:$\ds \map \zeta s = \frac {i \Gamma \paren {1 - s} } {2 \pi} \oint_C \frac {\paren {-z}^{s - 1} } {e^z - 1} \rd z$
where:
:$\zeta$ is the Riemann Zeta function
:$\Gamma$ is the Gamma function.
== Proof ==
== Sources == 
* 
Category:Riemann Zeta Function