nyuuzyou/wikis huggingface dataset:
Galois Connection Implies Order on Mappings

== Theorem ==
Let $\struct {S, \preceq}$, $\struct {T, \precsim}$ be ordered sets.
Let $g: S \to T$ and $d: T \to S$ be mappings such that
:$\tuple {g, d}$ is Galois connection.
Then $d \circ g \preceq I_S$ and $I_T \precsim g \circ d$
where
:$\preceq, \precsim$ denote the orderings on mappings,
:$I_S$ denotes the identity mapping of $S$
== Proof ==
Let $s \in S$.
By definition of reflexivity:
:$\map g s \precsim \map g s$
By definition of Galois connection:
:$\map d {\map g s} \preceq s$
By definition of composition:
:$\map {\paren {d \circ g} } s \preceq s$
By definition of identity mapping:
:$\map {\paren {d \circ g} } s \preceq \map {I_S} s$
Thus by definition of order on mappings:
:$d \circ g \preceq I_S$
Let $t \in T$.
By definition of reflexivity:
:$\map d t \preceq \map d t$
By definition of Galois connection:
:$t \precsim \map g {\map d t}$
By definition of composition:
:$t \precsim \map {\paren {g \circ d} } t$
By definition of identity mapping:
:$\map {I_T} t \precsim \map {\paren {g \circ d} } t$
Thus by definition of order on mappings:
:$I_T \precsim g \circ d$
== Sources ==
* 
* 
Category:Galois Connections
Galois Connection implies Upper Adjoint is Surjection iff Lower Adjoint is Injection

== Theorem ==
Let $L = \struct {S, \preceq}$ and $R = \struct {T, \precsim}$ be ordered sets.
Let $g: S \to T$ and $d: T \to S$ be mappings such that $\struct {g, d}$ is a Galois connection.
Then $g$ is a surjection $d$ is an injection.
== Proof ==
=== Sufficient Condition ===
Assume that
:$d$ is a surjection.
By Upper Adjoint of Galois Connection is Surjection implies Lower Adjoint at Element is Minimum of Preimage of Singleton of Element
:$\forall t \in T: \map d t = \min \set {g^{-1} \sqbrk {\set t} }$
By Lower Adjoint at Element is Minimum of Preimage of Singleton of Element implies Composition is Identity:
:$g \circ d = I_T$
Thus by Injection iff Left Inverse:
:$d$ is an injection.
=== Necessary Condition ===
Assume that
:$d$ is an injection.
By definition of Galois connection:
:$g$ and $f$ are increasing mappings.
By Galois Connection Implies Order on Mappings:
:$d \circ g \preceq I_S$ and $I_T \precsim g \circ d$
By Increasing and Ordering on Mappings implies Mapping is Composition:
:$d = \paren {d \circ g} \circ d$
By Composition of Mappings is Associative:
::$= d \circ \paren {g \circ d}$
By Injection iff Left Cancellable:
:$g \circ d = I_T$
By Identity Mapping is Right Identity:
:$d = d \circ I_T$
Thus by Surjection iff Right Inverse:
:$g$ is a surjection.
== Sources ==
* 
Category:Galois Connections
Galois Connection is Unique for Given Lower Adjoint

== Theorem ==
Let $\struct {S, \preceq}$ and $\struct {T, \precsim}$ be ordered sets.
Let $g: S \to T$, $d: T \to S$ be mappings such that $\tuple{g, d}$ is a Galois connection.
Then:
:$\tuple{g, d}$ is the unique Galois connection such that $d$ is the lower adjoint
== Proof ==
Let $g^\prime: S \to T$ be a mapping such that $\tuple{g^\prime, d}$ is a Galois connection.
We have:
where:
:$\max$ denotes the maximum
:$d^{-1} \sqbrk {x^\preceq}$ denotes the image of $x^\preceq$ under relation $d^{-1}$
:$x^\preceq$ denotes the lower closure of $x$
From Equality of Mappings:
:$g^\prime = g$
The result follows.
Category:Galois Connections
Galois Connection is Unique for Given Upper Adjoint

== Theorem ==
Let $\struct {S, \preceq}$ and $\struct {T, \precsim}$ be ordered sets.
Let $g: S \to T$, $d: T \to S$ be mappings such that $\tuple{g, d}$ is a Galois connection.
Then:
:$\tuple{g, d}$ is the unique Galois connection such that $g$ is the upper adjoint
== Proof ==
Let $d^\prime: T \to S$ be a mapping such that $\tuple{g, d^\prime}$ is a Galois connection.
We have:
where:
:$\min$ denotes the minimum
:$g^{-1} \sqbrk {y^\succsim}$ denotes the image of $y^\succsim$ under relation $g^{-1}$
:$y^\succsim$ denotes the upper closure of $y$
From Equality of Mappings:
:$d^\prime = d$
The result follows.
Category:Galois Connections
Galois Connection with Upper Adjoint Surjective implies Scond Ordered Set and Image of Lower Adjoint are Isomorphic

== Theorem ==
Let $L = \left({S, \preceq}\right), R = \left({T, \precsim}\right)$ be ordered sets.
Ley $g:S \to T, d:T \to S$ be mappings such that
:$\left({g, d}\right)$ is Galois connection
and
:$g$ is a surjection/
Let $N = \left({d\left[{T}\right], \preceq'}\right)$ be an ordered subset of $L$.
Then $R$ and $N$ are order isomorphic.
== Proof ==
By Galois Connection implies Upper Adjoint is Surjection iff Lower Adjoint is Injection:
:$d$ is an injection.
Define $d' = d:T \to g\left[{T}\right]$
By definition:
:$d'$ is an injection.
By definition of Galois connection:
:$d$ is an increasing mapping.
We will prove that
:$d$ is order embedding.
Let $x, y \in T$.
By definition of increasing mapping:
:$x \precsim y \implies d\left({x}\right) \preceq d\left({y}\right)$
Thus by definition of $d'$ and ordered subset:
:$x \precsim y \implies d'\left({x}\right) \preceq' d'\left({y}\right)$
Assume that
:$d'\left({x}\right) \preceq' d'\left({y}\right)$
By Upper Adjoint of Galois Connection is Surjection implies Lower Adjoint at Element is Minimum of Preimage of Singleton of Element
:$\forall t \in T: d\left({t}\right) = \min\left({g^{-1}\left[{\left\{ {t}\right\} }\right]}\right)$
By Lower Adjoint at Element is Minimum of Preimage of Singleton of Element implies Composition is Identity:
:$g \circ d = I_T$
By definition of Galois connection:
:$g$ is an increasing mapping.
Thus by definition of $d'$ and ordered subset:
:$d\left({x}\right) \preceq d\left({y}\right)$
By definition of increasing mapping:
:$g\left({d\left({x}\right)}\right) \precsim g\left({d\left({y}\right)}\right)$
Thus by definitions of composition of mappings and identity mapping:
:$x \precsim y$
By definition:
:$d'$ is a surjection.
Hence $d'$ is order isomorphism between $R$ and $N$.
== Sources ==
* 
Category:Galois Connections
Category:Order Isomorphisms
Galois Field/Examples/Order 4

== Example of Galois Field ==
The algebraic structure $\struct {\GF, +, \times}$ defined by the following Cayley tables is a Galois field:
:$\begin{array} {c|cccc}
+ & 0 & 1 & a & b \\
\hline
0 & 0 & 1 & a & b \\
1 & 1 & 0 & b & a \\
a & a & b & 0 & 1 \\
b & b & a & 1 & 0 \\
\end{array}
\qquad
\begin{array} {c|cccc}
\times & 0 & 1 & a & b \\
\hline
0 & 0 & 0 & 0 & 0 \\
1 & 0 & 1 & a & b \\
a & 0 & a & b & 1 \\
b & 0 & b & 1 & a \\
\end{array}$
== Proof ==
From Field with 4 Elements has only Order 2 Elements we have that a Galois field of order $4$, if it exists, must have this structure:
:$\struct {\GF, +}$ is the Klein $4$-group
:$\struct {\GF^*, \times}$ is the cyclic group of order $3$.
We have that $4 = 2^2$, and $2$ is prime.
From Galois Field of Order q Exists iff q is Prime Power, there exists at least one Galois field of order $4$.
As $\struct {\GF^*, +, \times}$ is the only such algebraic structure that can be a Galois field, it follows that $\struct {\GF^*, +, \times}$ must be a Galois field.
== Sources ==
* : Chapter $4$: Fields: $\S 17$. The Characteristic of a Field: Example $25$
* : Chapter $3$: Field Theory: Definition and Examples of Field Structure: $\S 87 \varepsilon$
Category:Examples of Galois Fields
Galois Field has Non-Zero Characteristic/Proof 2

== Theorem ==
== Proof ==
Let $\GF$ be a Galois field.
Let $P$ be its prime subfield.
Suppose $\Char \GF = 0$.
Then from Field of Characteristic Zero has Unique Prime Subfield, $P$ is isomorphic to $\Q$ which is infinite.
But a Galois field can not have an infinite subfield.
== Sources ==
* : Chapter $\text {I}$: Elementary Definitions: $\S 3$. Homomorphisms: Theorem $3.2$: Corollary
Category:Galois Field has Non-Zero Characteristic
Galois Field is Perfect

== Theorem ==
Let $\GF$ be a Galois field.
Then $\GF$ is perfect.
== Proof ==
By Characteristic of Galois Field is Prime, $\Char \GF$ is a prime number, say $p$.
By Frobenius Endomorphism on Field is Injective, $\Frob$ is injective.
By Injection from Finite Set to Itself is Surjection, $\Frob$ is bijective.
By Bijective Ring Homomorphism is Isomorphism, $\Frob$ is an automorphism.
Category:Galois Fields
Category:Perfect Fields
Galois Group Acts Faithfully on Generating Set

== Theorem ==
Let $G$ be the Galois group of a finite field extension $E/ F$.
Let $\set {\alpha_1, \ldots , \alpha_n}$ be a generating set for field $E$ as an $F$-vector space.
Let $f_1, \ldots, f_n$ be their minimal polynomials.
Let $f = f_1 \dots f_n$ be the product of them.
Then the action of $G$ on roots of $f$ is faithful.
== Proof ==
Suppose that $\sigma \in G$ fixes each $\alpha_i$.
Then since $\sigma$ is an $F$-linear automorphism of $E$ and since $\set {\alpha_1, \ldots, \alpha_n}$ is a generating set for $E$, we see that $\sigma$ fixes every other element of $E$.
Then $\sigma$ must be the identity map $I_E: E \to E$.
Then the action of $G$ on roots of $f$ is faithful.
Category:Galois Groups of Field Extensions
Galois Group is Group

== Theorem ==
Let $L / K$ be a normal extension.
Let $\Gal {L / K}$ be the Galois group of $L / K$.
Then $\Gal {L / K}$ forms a group under the operation of composition of mappings.
== Proof ==
The Galois group $\Gal {L / K}$ of $L / K$ is defined as:
:$\Gal {L / K} = \leftset{\sigma: L \to L: \sigma}$ is an automorphism of $L$ such that $\sigma$ fixes $K$ point-wise$\rightset{}$
We have that $\Gal {L / K}$ is a subset of the automorphism group on $L / K$.
We initially note that the Identity Mapping is Automorphism which trivially fixes $K$.
Thus $\Gal {L / K}$ is not the empty set.
First, to show that $\Gal {L / K}$ is closed under composition of mappings.
Let $\sigma, \tau \in \Gal {L / K}$.
Let $\rho = \sigma \circ \tau$.
If $k \in K$, then:
:$\map \rho k = \map \sigma {\map \tau k} = \map \sigma k = k$
and so $\rho$ fixes $K$.
By Composite of Isomorphisms in Algebraic Structure is Isomorphism, $\rho$ is an automorphism.
Thus, $\Gal {L / K}$ is closed under composition.
Next, by Inverse of Algebraic Structure Isomorphism is Isomorphism, $\sigma^{-1}$ is an automorphism, and:
:$\forall k \in K: \map \sigma k = k \implies \map {\sigma^{-1} } k = k$
Thus, $\sigma^{-1} \in \Gal {L / K}$.
The result follows from the Two-Step Subgroup Test.
That is, $\Gal {L / K}$ is a group.
Category:Galois Theory
Gamma Difference Equation

== Theorem ==
The Gamma function satisfies:
:$\map \Gamma {z + 1} = z \, \map \Gamma z$
for any $z$ which is not a nonpositive integer.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $1$: The Laplace Transform: Some Special Functions: $\text {I}$. The Gamma function: $1$
* : $\S 16$: The Gamma Function: Recursion Formula: $16.2$
* : $0,88560 31944 \ldots$
* : $\S 1.2.5$: Permutations and Factorials: Exercise $10$
* 
* : $11.14 \ \text{(iii)}$
* 
* : $\S 25$: The Gamma Function: Recursion Formula: $25.2.$
Category:Gamma Function
Category:Gamma Difference Equation
Gamma Difference Equation/Proof 1

== Theorem ==
== Proof ==
Let $z \in \C$, with $\map \Re z > 0$.
Then:
If $z \in \C \setminus \set {0, -1, -2, \ldots}$ such that $\map \Re z \le 0$, then the statement holds by the definition of $\Gamma$ in this region.
== Sources ==
* : Chapter $1$: The Laplace Transform: Solved Problems: The Gamma Function: $28 \ \text{(a)}$
* : $\S 17.4 \ (4)$
Category:Gamma Difference Equation
Gamma Distribution with Parameter 1 is Exponential Distribution

== Theorem ==
Let $X$ be a continuous random variable with a gamma distribution
:$\map f x = \dfrac {x^{\alpha - 1} e^{-\alpha / \beta} } {\beta^\alpha \map \Gamma \alpha}$
Let $X$ be such that the value of $\alpha$ is $1$.
Then $X$ has the exponential distribution.
== Proof ==
== Sources ==
* 
* 
Category:Gamma Distribution
Gamma Function/Examples/One Quarter

== Theorem ==
Let $\Gamma$ denote the Gamma function.
Then:
:$\map \Gamma {\dfrac 1 4} = 3 \cdotp 62560 \, 99082 \, 21908 \ldots$
== Sources ==
* : Table $1.1$. Mathematical Constants
* : $\S 1$: Special Constants: $1.25$
* : $\S 1$: Special Constants: $1.8.$
Category:Examples of Gamma Function Values
Gamma Function/Examples/One Third

== Theorem ==
Let $\Gamma$ denote the Gamma function.
Then:
:$\map \Gamma {\dfrac 1 3} = 2 \cdotp 67893 \, 85347 \, 07747 \, 63 \ldots$
== Sources ==
* : Table $1.1$. Mathematical Constants
* : $\S 1$: Special Constants: $1.24$
* : $\S 1$: Special Constants: $1.7.$
Category:Examples of Gamma Function Values
Gamma Function Extends Factorial

== Theorem ==
:$\forall n \in \N: \map \Gamma {n + 1} = n!$
== Proof ==
For $n = 0$:
Then by Gamma Difference Equation:
:$\forall z \in \Z_{> 0}: \map \Gamma {z + 1} = z \, \map \Gamma z$
Hence the result.
== Sources ==
* : Chapter $1$: The Laplace Transform: Some Special Functions: $\text {I}$. The Gamma function: $1$
* : Chapter $1$: The Laplace Transform: Solved Problems: The Gamma Function: $28 \ \text{(b)}$
* : $\S 16$: The Gamma Function: Recursion Formula: $16.3$
* : Back endpapers: A Brief Table of Integrals: $139$.
* : $\S 17.4 \ (4)$
* : $0,88560 31944 \ldots$
* : $1 \cdotp 772 \, 453 \, 850 \, 905 \, 516 \, 027 \, 298 \, 167 \, 483 \, 341 \, 145 \, 182 \, 797 \ldots$
* : $\S 1.2.5$: Permutations and Factorials
* : $1 \cdotp 77245 \, 38509 \, 05516 \, 02729 \, 81674 \, 83341 \, 14518 \, 27975 \ldots$
* 
* 
* : $\S 25$: The Gamma Function: Recursion Formula: $25.3.$
Category:Gamma Function
Category:Factorials
Gamma Function as Integral of Natural Logarithm

== Theorem ==
Let $x \in \R_{>0}$ be a strictly positive real number.
Then:
:$\ds \map \Gamma x = \int_{\to 0}^1 \paren {\ln \frac 1 t}^{x - 1} \rd t$
where $\Gamma$ denotes the Gamma function.
== Proof ==
By definition of the Gamma function:
:$\ds \map \Gamma x = \int_0^{\to \infty} t^{x - 1} e^{-t} \rd t$
In order to allow the limits to be evaluated, this is to be expressed as:
:$\ds \map \Gamma x = \lim_{\delta \mathop \to 0^+, \ \Delta \mathop \to +\infty} \int_\delta^\Delta t^{x - 1} e^{-t} \rd t$
where $0 < \delta < \Delta$.
Let $-t = \ln u$.
Then by Derivative of Logarithm Function and the Chain Rule for Derivatives:
:$-1 = \dfrac 1 u \dfrac {\rd u} {\rd t}$
Also:
Hence:
Next:
Thus:
The result follows by renaming the bound variable $u$ to $t$.
== Sources ==
* : $\S 17.7 \ (1)$
Category:Gamma Function
Gamma Function for Non-Negative Integer Argument

== Theorem ==
The Gamma function satisfies:
:$\map \Gamma z = \dfrac {\map \Gamma {z + 1} } z$
for any $z$ which is not a nonpositive integer.
== Proof ==
From Gamma Difference Equation:
:$\map \Gamma {z + 1} = z \, \map \Gamma z$
which is valid for all $z \notin \Z_{\le 0}$.
The result follows by dividing by $z$.
== Sources ==
* : Chapter $1$: The Laplace Transform: Some Special Functions: $\text {I}$. The Gamma function: $5$
* : $\S 16$: The Gamma Function for $n < 0$: $16.4$
* : $\S 25$: The Gamma Function: The Gamma Function for $n < 0$: $25.4.$
Category:Gamma Function
Gamma Function is Continuous on Positive Reals

== Theorem ==
Let $\Gamma: \R_{>0} \to \R$ be the Gamma function, restricted to the strictly positive real numbers.
Then $\Gamma$ is continuous.
== Proof ==
Let $0 < \alpha < a \le x \le y \le b < \beta$.
Let $0 < \delta < \Delta$.
Then:
:$(1): \quad \ds \size {\int_\delta^\Delta t^{x - 1} e^{-t} \rd t - \int_\delta^\Delta t^{y - 1} e^{-t} } \le \int_\delta^\Delta \paren {t^{x - 1} - t^{y - 1} }e^{-t} \rd t$
From the Mean Value Theorem:
:$(2): \quad \dfrac {t^{x - 1} - t^{y - 1} }{x - y} = \paren {\ln t} t^{\xi - 1}$
for some $\xi \in \R$ such that $x \le \xi \le y$.
Let $r \in \R_{>0}$.
Then:
:$t^{-r} \ln t \to 0$ as $t \to \infty$
:$t^r \ln t \to 0$ as $t \to 0^+$
From $(2)$ it follows that:
:$\exists H \in \R: \size {t^{x - 1} - t^{y - 1} } \le H \paren {t^{\alpha - 1} + t^{\beta - 1} } \size {x - y}$
From $(1)$ it follows that:
:$\size {\map \Gamma x - \map \Gamma y} \le H \size {x - y} \paren {\map \Gamma \alpha - \map \Gamma \beta}$
The result follows from the Squeeze Theorem.
== Sources ==
* : $\S 17.4 \ (5)$
* : $11.14 \ \text{(i)}$
Category:Gamma Function
Gamma Function is Unique Extension of Factorial

== Theorem ==
Let $f: \R_{>0} \to \R$ be a real function which is positive and continuous.
Let $\ln \mathop \circ f$ be convex on $\R_{>0}$.
Let $f$ satisfy the conditions:
:$\map f {x + 1} = \begin{cases}
1 & : x = 0 \\
x \map f x & : x > 0 \end{cases}$
Then:
:$\forall x \in \R_{>0}: \map f x = \map \Gamma x$
where $\map \Gamma x$ is the Gamma function.
== Proof ==
From the Gamma Function Extends Factorial:
:$\map f {x + 1} = \begin{cases}
1 & : x = 0 \\
x \map f x & : x > 0 \end{cases}$
From Gamma Function is Continuous on Positive Reals, $\Gamma$ is positive and continuous on $\R_{>0}$.
From Log of Gamma Function is Convex on Positive Reals, $\ln \mathop \circ \Gamma$ is convex on $\R_{>0}$.
It remains to be shown that $\Gamma$ is the only real function which satisfies these conditions.
Let $s, t \in \R_{>0}$ such that $s \le t \le s + 1$.
Let $t = \alpha s + \beta \paren {s + 1}$ where $\alpha, \beta \in \R_{>0}, \alpha + \beta = 1$.
Then:
:$t = \paren {\alpha + \beta} s + \beta = s + \beta$
and so:
:$\beta = t - s$
Thus:
Because $s \le t \le s + 1$, it follows that $t - 1 \le s \le t$.
Substituting as appropriate in $(1)$:
Combining $(1)$ and $(2)$:
Let $x \in \R$ such that $0 < x \le 1$.
Let $n \in \N$.
Let:
:$s = n + 1$
:$t = x + n + 1$
Then substituting in $(3)$:
Hence:
Thus when $0 < x \le 1$:
:$\ds \map f x = \lim_{n \mathop \to \infty} \frac {x \paren {x + 1} \cdots \paren {x + n} } {n^x n!}$
which is the Euler form of the Gamma function.
The general case is deduced by the use of:
:$\map f {x + 1} = x \map f x$
== Sources ==
* : $\S 17.6$
Category:Gamma Function
Gamma Function of Minus 1

== Theorem ==
:$\map \Gamma {-1}$ is not defined.
where $\Gamma$ denotes the Gamma function.
== Proof ==
But from Gamma Function of Zero, $\map \Gamma 0$ is not defined.
Hence the result.
== Sources ==
* : Chapter $1$: The Laplace Transform: Solved Problems: The Gamma Function: $33 \ \text{(e)}$
Category:Examples of Gamma Function Values
Gamma Function of Minus 2

== Theorem ==
:$\map \Gamma {-2}$ is not defined.
where $\Gamma$ denotes the Gamma function.
== Proof ==
But from Gamma Function of Minus 1, $\map \Gamma {-1}$ is not defined.
Hence the result.
== Sources ==
* : Chapter $1$: The Laplace Transform: Solved Problems: The Gamma Function: $33 \ \text{(f)}$
Category:Examples of Gamma Function Values
Gamma Function of Minus 3 over 2

== Theorem ==
:$\map \Gamma {-\dfrac 3 2} = \dfrac {4 \sqrt \pi} 3$
where $\Gamma$ denotes the Gamma function.
== Proof ==
== Sources ==
* : Chapter $1$: The Laplace Transform: Solved Problems: The Gamma Function: $33 \ \text{(b)}$
Category:Examples of Gamma Function Values
Gamma Function of Minus 5 over 2

== Theorem ==
:$\map \Gamma {-\dfrac 5 2} = -\dfrac {8 \sqrt \pi} {15}$
where $\Gamma$ denotes the Gamma function.
== Proof ==
== Sources ==
* : Chapter $1$: The Laplace Transform: Solved Problems: The Gamma Function: $33 \ \text{(c)}$
Category:Examples of Gamma Function Values
Gamma Function of Minus One Half

== Theorem ==
:$\map \Gamma {-\dfrac 1 2} = -2 \sqrt \pi$
where $\Gamma$ denotes the Gamma function.
== Proof ==
== Sources ==
* : Chapter $1$: The Laplace Transform: Solved Problems: The Gamma Function: $33 \ \text{(a)}$
* : $\S 1.2.5$: Permutations and Factorials: Exercise $9$
Category:Examples of Gamma Function Values
Gamma Function of Negative Half-Integer

== Theorem ==
where:
:$-m + \dfrac 1 2$ is a half-integer such that $m > 0$
:$\Gamma$ denotes the Gamma function.
== Proof ==
Proof by induction:
For all $m \in \Z_{> 0}$, let $\map P m$ be the proposition:
:$\map \Gamma {-m + \dfrac 1 2} = \dfrac {\paren {-1}^m 2^{2 m} m!} {\paren {2 m}!} \sqrt \pi$
=== Basis for the Induction ===
$\map P 1$ is the case:
and so $\map P 1$ holds.
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
So this is our induction hypothesis:
:$\map \Gamma {-k + \dfrac 1 2} = \dfrac {\paren {-1}^k 2^{2 k} k!} {\paren {2 k}!} \sqrt \pi$
Then we need to show:
:$\map \Gamma {-\paren {k + 1} + \dfrac 1 2} = \dfrac {\paren {-1}^{k + 1} 2^{2 \paren {k + 1} } \paren {k + 1}!} {\paren {2 \paren {k + 1} } !} \sqrt \pi$
=== Induction Step ===
This is our induction step:
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Finally:
Therefore:
== Sources ==
* : Chapter $1$: The Laplace Transform: Solved Problems: The Gamma Function: $33$
* : $\S 16$: Special Values for the Gamma Function: $16.7$
* : $\S 25$: The Gamma Function: Special Values for the Gamma Function: $25.7.$
Category:Gamma Function
Gamma Function of One Half

== Theorem ==
Let $\Gamma$ denote the Gamma function.
Then:
:$\map \Gamma {\dfrac 1 2} = \sqrt \pi$
=== Decimal Expansion ===
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Proof 4 ==
== Proof 5 ==
== Sources ==
* : Chapter $1$: The Laplace Transform: Some Special Functions: $\text {I}$. The Gamma function: $2$
* : $\S 1$: Special Constants: $1.23$
* : $\S 16$: Special Values for the Gamma Function: $16.5$
* : $1 \cdotp 772 \, 453 \, 850 \, 905 \, 516 \, 027 \, 298 \, 167 \, 483 \, 341 \, 145 \, 182 \, 797 \ldots$
* : $\S 1.2.5$: Permutations and Factorials: Exercise $9$
* : $1 \cdotp 77245 \, 38509 \, 05516 \, 02729 \, 81674 \, 83341 \, 14518 \, 27975 \ldots$
* 
* 
* : $\S 1$: Special Constants: $1.6.$
* : $\S 25$: The Gamma Function: Special Values for the Gamma Function: $25.5.$
Category:Gamma Function of One Half
Category:Examples of Gamma Function Values
Gamma Function of One Half/Decimal Expansion

== Gamma Function of One Half ==
Let $\Gamma$ denote the Gamma function.
The decimal expansion of $\map \Gamma {\dfrac 1 2}$ starts:
:$\map \Gamma {\dfrac 1 2} = 1 \cdotp 77245 \, 38509 \, 05516 \, 02729 \, 81674 \, 83341 \, 14518 \, 27975 \ldots$
== Sources ==
* : Table $1.1$. Mathematical Constants
* : $\S 1$: Special Constants: $1.23$
* : $1 \cdotp 772 \, 453 \, 850 \, 905 \, 516 \, 027 \, 298 \, 167 \, 483 \, 341 \, 145 \, 182 \, 797 \ldots$
* : $1 \cdotp 77245 \, 38509 \, 05516 \, 02729 \, 81674 \, 83341 \, 14518 \, 27975 \ldots$
* : $\S 1$: Special Constants: $1.6.$
Category:Gamma Function of One Half
Gamma Function of One Half/Proof 1

== Theorem ==
== Proof ==
From the definition of the Beta function:
:$\map \Beta {x, y} := \dfrac {\map \Gamma x \map \Gamma y} {\map \Gamma {x + y} }$
Setting $x = y = \dfrac 1 2$:
Then from Beta Function of Half with Half:
:$\map \Beta {\dfrac 1 2, \dfrac 1 2} = \pi$
Hence the result.
== Sources ==
* : $\S 17.7 \ (6)$
* : $\S 1.2.6$: Binomial Coefficients: Exercise $43$
Category:Gamma Function of One Half
Gamma Function of One Half/Proof 2

== Theorem ==
== Proof ==
From Euler's Reflection Formula:
:$\forall z \notin \Z: \map \Gamma z \map \Gamma {1 - z} = \dfrac \pi {\map \sin {\pi z} }$
Setting $z = \dfrac 1 2$:
By definition of the gamma function:
:$\forall z \in \R_{\ge 0}: \map \Gamma z > 0$
and so the negative square root can be discarded.
Hence:
:$\map \Gamma {\dfrac 1 2} = \sqrt \pi$
as required.
Category:Gamma Function of One Half
Gamma Function of Positive Half-Integer

== Theorem ==
where:
:$m + \dfrac 1 2$ is a half-integer such that $m > 0$
:$\Gamma$ denotes the Gamma function.
== Proof ==
Proof by induction:
For all $m \in \Z_{> 0}$, let $\map P m$ be the proposition:
:$\map \Gamma {m + \dfrac 1 2} = \dfrac {\paren {2 m}!} {2^{2 m} m!} \sqrt \pi$
=== Basis for the Induction ===
$\map P 1$ is the case:
and so $\map P 1$ holds.
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
So this is our induction hypothesis:
:$\map \Gamma {k + \dfrac 1 2} = \dfrac {\paren {2 k}!} {2^{2 k} k!} \sqrt \pi$
Then we need to show:
:$\map \Gamma {k + 1 + \dfrac 1 2} = \dfrac {\paren {2 \paren {k + 1} }!} {2^{2 \paren {k + 1} } \paren {k + 1}!} \sqrt \pi$
=== Induction Step ===
This is our induction step:
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Finally:
Therefore:
== Also see ==
* Legendre's Duplication Formula
== Sources ==
* : $\S 16$: Special Values for the Gamma Function: $16.6$
* 
* 
* : $\S 25$: The Gamma Function: Special Values for the Gamma Function: $25.6.$
Category:Gamma Function