nyuuzyou/wikis huggingface dataset:
Left and Right Coset Spaces are Equivalent/Proof 2

== Theorem ==
== Proof ==
Let $G$ be a group and let $H \le G$.
Consider the mapping $\phi$ from the left coset space to the right coset space defined as:
:$\forall g \in G: \map \phi {g H} = H g^{-1}$
We need to show that $\phi$ is a bijection.
First we need to show that $\phi$ is well-defined.
That is, that $a H = b H \implies \map \phi {a H} = \map \phi {b H}$.
Suppose $a H = b H$.
But $a^{-1} \paren {b^{-1} }^{-1} = a^{-1} b \in H$ as $a H = b H$.
So $H a^{-1} = H b^{-1}$ and $\phi$ is well-defined.
Next we show that $\phi$ is injective:
Suppose $\exists x, y \in G: \map \phi {x H} = \map \phi {y H}$.
Then $H x^{-1} = H y^{-1}$, so $x^{-1} = e_G x^{-1} = h y^{-1}$ for some $h \in H$.
Thus $h = x^{-1} y \implies h^{-1} = y^{-1} x$.
As $H$ is a subgroup, $h^{-1} \in H$.
Thus:
: $y^{-1} x \in H$
So by Left Cosets are Equal iff Product with Inverse in Subgroup:
:$x H = y H$
Thus $\phi$ is injective.
Next we show that $\phi$ is surjective:
Let $H x$ be a right coset of $H$ in $G$.
Since $x = \paren {x^{-1} }^{-1}$, $H x = \map \phi {x^{-1} H}$ and so $\phi$ is surjective.
Thus $\phi$ constitutes a bijection from the left coset space to the right coset space, and the result follows.
== Sources ==
* : $\S 6.3$. Index. Transversals
* : $\S 1.9$: Lemma $9$
* : $\S 3.3$: Group actions and coset decompositions
* : Chapter $5$: Cosets and Lagrange's Theorem: Proposition $5.14$
Category:Left and Right Coset Spaces are Equivalent
Left and Right Identity are the Same

== Theorem ==
Let $\struct {S, \circ}$ be an algebraic structure.
Let $e_L \in S$ be a left identity, and $e_R \in S$ be a right identity.
Then:
:$e_L = e_R$
that is, both the left identity and right identity are the same, and are therefore an identity $e$.
Furthermore, $e$ is the only left identity and right identity for $\circ$.
== Proof ==
Let $\struct {S, \circ}$ be an algebraic structure such that:
:$\exists e_L \in S: \forall x \in S: e_L \circ x = x$
:$\exists e_R \in S: \forall x \in S: x \circ e_R = x$
Then $e_L = e_L \circ e_R = e_R$ by both the above, hence the result.
The uniqueness of the left and right identity is a direct result of Identity is Unique.
== Sources ==
* : Exercise $1.4: \ 13$
* : Chapter $4$. Groups: Exercise $5$
* : Chapter $\text I$: Algebraic Structures: $\S 4$: Neutral Elements and Inverses: Exercise $4.3 \ \text{(a)}$
Category:Identity Elements
Left and Right Inverse Mappings Implies Bijection

== Theorem ==
Let $f: S \to T$ be a mapping.
Let $f$ be such that:
: $\exists g_1: T \to S: g_1 \circ f = I_S$
: $\exists g_2: T \to S: f \circ g_2 = I_T$
where both $g_1$ and $g_2$ are mappings.
Then $f$ is a bijection.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Exercise $1.3: \ 11$
* : Chapter $4$: Mappings: Exercise $15$
Category:Bijections
Category:Inverse Mappings
Category:Left and Right Inverse Mappings Implies Bijection
Left and Right Inverse Mappings Implies Bijection/Proof 1

== Theorem ==
== Proof ==
Suppose:
:$\exists g_1: T \to S: g_1 \circ f = I_S$
:$\exists g_2: T \to S: f \circ g_2 = I_T$
We have that the Identity Mapping is Bijection, thus $I_S$ and $I_T$ are both bijections.
From Injection if Composite is Injection, if $I_S = g_1 \circ f$ is an injection, then so is $f$.
From Surjection if Composite is Surjection, if $I_T = f \circ g_2$ is a surjection, then so is $f$.
So $f$ is both an injection and a surjection and, by definition, therefore also a bijection.
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 5$: Composites and Inverses of Functions: Theorem $5.4$
Category:Left and Right Inverse Mappings Implies Bijection
Left and Right Inverse Mappings Implies Bijection/Proof 2

== Theorem ==
== Proof ==
Suppose:
: $\exists g_1: T \to S: g_1 \circ f = I_S$
: $\exists g_2: T \to S: f \circ g_2 = I_T$
From Injection iff Left Inverse, it follows that $f$ is an injection.
From Surjection iff Right Inverse, it follows that $f$ is a surjection.
So $f$ is both an injection and a surjection and, by definition, therefore also a bijection.
Category:Left and Right Inverse Mappings Implies Bijection
Left and Right Inverse Relations Implies Bijection

== Theorem ==
Let $\RR \subseteq S \times T$ be a relation on a cartesian product $S \times T$.
Let:
:$I_S$ be the identity mapping on $S$
:$I_T$ be the identity mapping on $T$.
Let $\RR^{-1}$ be the inverse relation of $\RR$.
Let $\RR$ be such that:
:$\RR^{-1} \circ \RR = I_S$ and
:$\RR \circ \RR^{-1} = I_T$
where $\circ$ denotes composition of relations.
Then $\RR$ is a bijection.
== Proof ==
Let $\RR \subseteq S \times T$ be such that:
:$\RR^{-1} \circ \RR = I_S$
and:
:$\RR \circ \RR^{-1} = I_T$.
From Condition for Composite Relation with Inverse to be Identity, we have that:
:$\RR$ is many-to-one
:$\RR$ is right-total
:$\RR^{-1}$ is many-to-one
:$\RR^{-1}$ is right-total.
From Inverse of Many-to-One Relation is One-to-Many, it follows that both $\RR$ and $\RR^{-1}$ are by definition one-to-one.
From Inverses of Right-Total and Left-Total Relations, it also follows that both $\RR$ and $\RR^{-1}$ are left-total.
By definition, an injection is a relation which is:
:One-to-one
:left-total.
Also by definition, a surjection is a relation which is:
:Left-total
:Many-to-one
:Right-total.
It follows that $\RR$ is both an injection and a surjection, and so by definition a bijection.
By the same coin, the same applies to $\RR^{-1}$.
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 5$: Composites and Inverses of Functions: Exercise $5.8 \ \text{(f)}$
Category:Inverse Relations
Category:Bijections
Left and Right Inverses of Mapping are Inverse Mapping

== Theorem ==
Let $f: S \to T$ be a mapping such that:
:$(1): \quad \exists g_1: T \to S: g_1 \circ f = I_S$
:$(2): \quad \exists g_2: T \to S: f \circ g_2 = I_T$
Then:
: $g_1 = g_2 = f^{-1}$
where $f^{-1}$ is the inverse of $f$.
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
Category:Bijections
Category:Inverse Mappings
Category:Left and Right Inverses of Mapping are Inverse Mapping
Left and Right Inverses of Mapping are Inverse Mapping/Proof 1

== Theorem ==
== Proof ==
From Left and Right Inverse Mappings Implies Bijection it follows that $f$ is a bijection.
It follows from Composite of Bijection with Inverse is Identity Mapping that $g_1 = g_2 = f^{-1}$.
== Sources ==
* : $\S 5$. Induced mappings; composition; injections; surjections; bijections: Theorem $5.9$
* : Chapter $4$: Mappings: Exercise $15$
Category:Left and Right Inverses of Mapping are Inverse Mapping
Left and Right Inverses of Mapping are Inverse Mapping/Proof 2

== Theorem ==
== Proof ==
From Left and Right Inverse Mappings Implies Bijection, we have that $f$ is a bijection.
Let $y \in T$ and let $x = \map {f^{-1} } y$.
Such an $x$ exists because $f$ is surjective, and unique within $S$ as $f$ is injective.
Then $y = \map f x$ and so:
So $f^{-1} = g_1$.
Also:
So $f^{-1} = g_2$.
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 5$: Composites and Inverses of Functions: Theorem $5.4$
Category:Left and Right Inverses of Mapping are Inverse Mapping
Left and Right Inverses of Mapping are Inverse Mapping/Proof 3

== Theorem ==
== Proof ==
Because Composition of Mappings is Associative, brackets do not need to be used.
Thus $g_1 = g_2$.
Now suppose there exists $g_3: T \to S: g_3 \circ f = I_S$.
By the same argument as above, $g_3 = g_2$.
This means that $g_1 (= g_3)$ is the only left inverse of $f$.
Similarly, suppose there exists $g_4: T \to S: f \circ g_4 = I_T$.
By the same argument as above, $g_4 = g_1$.
This means that $g_2 (= g_4)$ is the only right inverse of $f$.
So $g_1 = g_2 = g_3 = g_4$ are all the same.
By Composite of Bijection with Inverse is Identity Mapping, it follows that this unique mapping is the inverse $f^{-1}$.
== Sources ==
* : Appendix $\text{A}.7$: Inverses: Proposition $\text{A}.7.3$
Category:Left and Right Inverses of Mapping are Inverse Mapping
Left and Right Inverses of Product

== Theorem ==
Let $\struct {S, \circ}$ be a monoid whose identity is $e_S$.
Let $x, y \in S$.
Let:
:$(1): \quad x \circ y$ have a left inverse for $\circ$
:$(2): \quad y \circ x$ have a right inverse for $\circ$.
Then both $x$ and $y$ are invertible for $\circ$.
== Proof ==
Let $z_L$ be the left inverse of $x \circ y$ and $z_R$ be the right inverse of $y \circ x$. Then:
Thus $y$ has both a left inverse $z_L \circ x$ and a right inverse $x \circ z_R$.
From Left Inverse and Right Inverse is Inverse:
:$z_L \circ x = x \circ z_R$
and $y$ has an inverse, that is, is invertible.
From the above, we have:
and:
Thus $x$ has both a left inverse $y \circ z_L$ and a right inverse $z_R \circ y$.
From Left Inverse and Right Inverse is Inverse:
:$y \circ z_L = z_R \circ y$
and $x$ has an inverse, that is, is invertible.
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 4$: Neutral Elements and Inverses: Exercise $4.9 \ \text{(b)}$
Category:Monoids
Left and Right Inverses of Square Matrix over Field are Equal

== Theorem ==
Let $\Bbb F$ be a field, usually one of the standard number fields $\Q$, $\R$ or $\C$.
Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Let $\map \MM n$ denote the matrix space of order $n$ square matrices over $\Bbb F$.
Let $\mathbf B$ be a left inverse matrix of $\mathbf A$.
Then $\mathbf B$ is also a right inverse matrix of $\mathbf A$.
Similarly, let $\mathbf B$ be a right inverse matrix of $\mathbf A$.
Then $\mathbf B$ is also a right inverse matrix of $\mathbf A$.
== Proof ==
Consider the algebraic structure $\struct {\map \MM {m, n}, +, \circ}$, where:
:$+$ denotes matrix entrywise addition
:$\circ$ denotes (conventional) matrix multiplication.
From Ring of Square Matrices over Field is Ring with Unity, $\struct {\map \MM {m, n}, +, \circ}$ is a ring with unity.
Hence a fortiori $\struct {\map \MM {m, n}, +, \circ}$ is a monoid.
The result follows directly from Left Inverse and Right Inverse is Inverse.
== Sources ==
* : Part $\text I$: Matrices and vector spaces: $1$ Matrices: $1.3$ The inverse of a matrix: Fact $1.1$
Category:Inverse Matrices
Left and Right Operation are Closed for All Subsets

== Theorem ==
Let $S$ be a set.
Let:
:$\leftarrow$ be the left operation on $S$
:$\rightarrow$ be the right operation on $S$.
That is:
:$\forall x, y \in S: x \leftarrow y = x$
:$\forall x, y \in S: x \rightarrow y = y$
Let $\powerset S$ be the power set of $S$.
Then for all $T \in \powerset S$, both $\leftarrow$ and $\rightarrow$ are closed on $T$.
That is, for all $T \in \powerset S$:
=== Left Operation is Closed for All Subsets ===
=== Right Operation is Closed for All Subsets ===
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 8$: Compositions Induced on Subsets: Exercise $8.1$
Category:Left and Right Operations
Left and Right Zero are the Same

== Theorem ==
Let $\struct {S, \circ}$ be an algebraic structure.
Let $z_L \in S$ be a left zero, and $z_R \in S$ be a right zero.
Then $z_L = z_R$, that is, both the left and right zero are the same, and are therefore a zero $z$.
Furthermore, $z$ is the only left and right zero for $\circ$.
== Proof ==
Let $\struct {S, \circ}$ be an algebraic structure such that:
:$\exists z_L \in S: \forall x \in S: z_L \circ x = z_L$
:$\exists z_R \in S: \forall x \in S: x \circ z_R = z_R$
Then $z_L = z_L \circ z_R = z_R$ by both the above, hence the result.
The uniqueness of the left and right zero is a direct result of Zero Element is Unique.
== Sources ==
* : Chapter $4$. Groups: Exercise $6$
Category:Zero Elements
Left or Right Inverse of Matrix is Inverse

== Theorem ==
Let $\mathbf A, \mathbf B$ be square matrices of order $n$ over a commutative ring with unity $\left({R, +, \circ}\right)$.
Suppose that:
: $\mathbf A \mathbf B = \mathbf I_n$
where $\mathbf I_n$ is the unit matrix of order $n$.
Then $\mathbf A$ and $\mathbf B$ are nonsingular matrices, and furthermore:
:$\mathbf B = \mathbf A^{-1}$
where $\mathbf A^{-1}$ is the inverse of $\mathbf A$.
== Proof ==
When $1_R$ denotes the unity of $R$, we have:
 
From Matrix is Nonsingular iff Determinant has Multiplicative Inverse, it follows that $\mathbf A$ and $\mathbf B$ are nonsingular.
Then:
== Sources ==
* : $\S 5.2$
Category:Inverse Matrices
Legendre's Condition

== Theorem ==
Let $y =\map y x$ be a real function, such that:
:$\map y a = A,\quad \map y b = B$
Let $J \sqbrk y$ be a functional, such that:
:$\ds J \sqbrk y = \int_a^b \map F {x, y, y'} \rd x$
where
:$F \in C^2 \closedint a b$ 
 all its variables, and $C$ stands for differentiability class.
Then a necessary condition for $J \sqbrk y$ to have a minimum at $y = \hat y$ is:
:$\bigintlimits {F_{y'y'} } {y \mathop = \hat y} {} \ge 0 \quad \forall x \in \closedint a b$
== Proof ==
=== Lemma 1 ===
Let $y = \map y x$ be a real function, such that:
:$\map y a = A, \quad \map y b = B$
Let $J \sqbrk y$ be a functional, such that:
:$\ds J \sqbrk y = \int_a^b \map F {x, y, y'} \rd x$
where:
:$F \in C^2 \closedint a b$ 
 all its variables.
Then:
:$\ds \delta^2 J \sqbrk {y; h} = \int_a^b \paren {\map P {x, \map y x } h'^2 + \map Q {x, \map y x} h^2} \rd x$
where:
:$\map P {x, \map y x} = \dfrac 1 2 F_{y' y'}, \quad \map Q {x, \map y x} = \dfrac 1 2 \paren {F_{yy} - \dfrac \d {\d x} F_{yy'} }$
=== Lemma 2 ===
Let $h$ be a real function such that:
:$h \in C^1 \openint a b, \quad \map h a = 0, \quad \map h b = 0$
Let:
:$\ds \delta^2 J \sqbrk {y; h} = \int_a^b \paren {\map P {x, \map y x} h'^2 + \map Q {x, \map y x} h^2} \rd x$
where $P \in C^0 \closedint a b$. 
Then a necessary condition for:
:$\delta^2 J \sqbrk {y; h} \ge 0$ 
is:
:$\map P {x, \map y x} \ge 0 \quad \forall x \in \closedint a b$
== Sources ==
* : $\S 5.25$: The Formula for the Second Variation. Legendre's Condition 
Category:Calculus of Variations
Legendre's Condition/Lemma 1

== Lemma ==
Let $y = \map y x$ be a real function, such that:
:$\map y a = A$
:$\map y b = B$
Let $J \sqbrk y$ be a functional, such that:
:$\ds J \sqbrk y = \int_a^b \map F {x, y, y'} \rd x$
where:
:$F \in C^2 \closedint a b$ 
 all its variables.
Then:
:$\ds \delta^2 J \sqbrk {y; h} = \int_a^b \paren {\map P {x, \map y x} h'^2 + \map Q {x, \map y x} h^2} \rd x$
where:
== Proof ==
The minimising function $y$ has fixed end-points.
Therefore, consider an increment of a functional with $h$ such that:
:$h \in C^1 \closedint a b: \paren {\map h a = 0} \land \paren {\map h b = 0}$
Then:
where omitted variables are $\paren {x, y, y'}$, and the overbar indicates derivatives being taken along some intermediate curves:
with $0 < \theta < 1$.
If $\overline F_{yy}$, $\overline F_{yy'}$, $\overline F_{y'y'} $ are to be replaced by $F_{yy}$, $F_{yy}$, $F_{y'y'}$ evaluated at the point $\tuple {x, \map y x, \map {y'} x}$, then:
:$\ds \Delta J \sqbrk {y; h} = \int_a^b \paren {\map {F_y} {x, y, y'} h + \map {F_{y'} } {x, y, y'} h'} \rd x + \frac 1 2 \int_a^b \paren {\map {F_{yy} } {x, y, y'} h^2 + 2 \map {F_{yy'} } {x, y, y'} h h'+ \map {F_{y'y'} } {x, y, y'} h'^2} \rd x + \epsilon$
where:
:$\ds \epsilon = \int_a^b \paren {\epsilon_1 h^2 + \epsilon_2 h h' + \epsilon_3 h'^2}$
By continuity of $F_{yy}$, $F_{yy}$, $F_{y'y'}$:
:$\size h_1 \to 0 \implies \epsilon_1, \epsilon_2, \epsilon_3 \to 0$
Thus, $\epsilon$ is an infinitesimal of the order higher than 2 $\size h$.
The first and second term on the of $\Delta J \sqbrk {y; h}$ are $\delta J \sqbrk {y; h}$ and $\delta^2 J \sqbrk {y; h}$ respectively.
Integrate the second term of $\delta^2 J \sqbrk {y; h}$ by parts:
Therefore:
:$\ds \delta^2 J \sqbrk {y; h} = \int_a^b \paren {\frac 1 2 F_{y'y'} h'^2 + \frac 1 2 \paren {F_{yy} - \frac \d {\d x} F_{yy'} } h^2} \rd x$
== Mistake ==
: $\S 5.25$: The Formula for the Second Variation. Legendre's Condition p. 102
states that
:$P = \dfrac 1 2 F_{y'y'} \quad Q = \dfrac 1 2 \paren {F_{yy'} - \dfrac \d {\d x} F_{yy'} }$
This is a mistake, since the second variation should contain both pure and mixed partial derivatives of the order 2.
However, $F_{yy} $ is missing and could not have been lost during derivation of the proof.
It should be:
:$Q = \dfrac 1 2 \paren {F_{yy} - \dfrac \d {\d x} F_{yy'} }$
== Sources ==
: $\S 5.25$: The Formula for the Second Variation. Legendre's Condition 
Category:Calculus of Variations
Legendre's Condition/Lemma 1/Dependent on N Functions

== Lemma ==
Let $\mathbf y = \paren {\sequence {\map {y_i} x}_{1 \mathop \le i \mathop \le N} }$ be a vector real function, such that:
:$\map {\mathbf y} a = A$
:$\map {\mathbf y} b = B$
Let $J \sqbrk {\mathbf y}$ be a functional, such that:
:$\ds J \sqbrk {\mathbf y} = \int_a^b \map F {x, \mathbf y, \mathbf y'} \rd x$
where:
:$F \in C^2 \closedint a b$ 
 all its variables.
Then:
:$\ds \delta^2 J \sqbrk {\mathbf y; \mathbf h} = \int_a^b \paren {\mathbf h' \mathbf P \mathbf h' + \mathbf h \mathbf Q \mathbf h} \rd x$
where:
:$\mathbf P = \dfrac 1 2 F_{y_i'y_j'}$
:$\mathbf Q = \dfrac 1 2 \paren {F_{ y_i y_j} - \dfrac \d {\d x} F_{y_i y_j'} }$
== Proof ==
== Sources ==
: $\S 5.29$: Generalization to n Unknown Functions
Category:Calculus of Variations
Legendre's Condition/Lemma 2

== Lemma ==
Let $ h$ be a real function such that:
:$h \in C^1 \openint a b$
:$\map h a = 0$
:$\map h b = 0$
Let:
:$\ds \delta^2 J \sqbrk {y; h} = \int_a^b \paren {\map P {x, \map y x} h'^2 + \map Q {x, \map y x} h^2} \rd x$
where $P \in C^0 \closedint a b$. 
Then a necessary condition for:
:$\delta^2 J \sqbrk {y; h} \ge 0$ 
is:
:$\forall x \in \closedint a b: \map P {x, \map y x} \ge 0$
== Proof ==
Assume that above is not true.
Then:
:$\paren {\exists x_0 \in \closedint a b} \land \paren {\exists \beta \in \R_{<0} }: \map P {x_0} = -2 \beta$
$P$ is continuous.
Thus:
:$\exists \alpha \in \R_{>0}: \paren {a \ge x_0 - \alpha} \land \paren {x_0 + \alpha \ge b}$
and:
:$\forall x \in \openint {x_0 - \alpha} {x_0 + \alpha}: \map P x < -\beta$
In other words:
$\map P x \begin {cases}
= 0 & : x \in \closedint a {x_0 - \alpha} \lor \closedint {x_0 + \alpha} b \\
< 0 & : x \in \closedint {x_0 - \alpha} {x_0 + \alpha}
\end {cases}$
Let
:$h = \begin {cases}
\sin^2 \paren {\dfrac {\map \pi {x - x_0} } \alpha} & : x_0 - \alpha \ge x \ge x_0 + \alpha \\
0 & : \text {otherwise}
\end {cases}$
It belongs to $C^1 \openint a b$ because:
In other words, only $h$ and $h'$ are continuous in $\closedint a b$
Then:
where:
:$\ds M = \max_{a \mathop \le x \mathop \le b} \size {\map Q x}$
For sufficiently small $\alpha$ the is negative.
Hence, $\delta^2 J$ is negative for the corresponding $h$.
To conclude, it has been shown that
:$P \ge 0 \quad \neg \forall x \in \closedint a b \implies \delta^2 J<0$
Then, by contrapositive statement this is equivalent to:
:$\forall x \in \closedint a b: \delta^2 J \ge 0 \implies P \ge 0$
== Sources ==
: $\S 5.25$: The Formula for the Second Variation. Legendre's Condition 
Category:Calculus of Variations
Legendre's Condition/Lemma 2/Dependent on N Functions

== Lemma ==
Let $J \sqbrk {\mathbf y}$ be a functional, such that:
:$\ds J \sqbrk {\mathbf y} = \int_a^b \map F {x, \mathbf y, \mathbf y'} \rd x$
where $\mathbf y = \paren {\sequence {y_i}_{1 \mathop \le i \mathop \le N} }$ is an N-dimensional vector.
Let $F \in C^2 \closedint a b$ all its variables.
Let $\mathbf P$, $\mathbf Q$ be $N \times N$ real matrices, such that:
:$\mathbf P = \dfrac 1 2 \dfrac {\partial^2 F} {\partial y_i'\partial y_j'}$
:$\mathbf Q = \dfrac 1 2 \paren {\dfrac {\partial^2 F} {\partial y_i \partial y_j} - \dfrac \d {\d x} \dfrac {\partial^2 F} {\partial y_i \partial y_j'} }$
Let:
:$\ds \int_a^b \paren {\mathbf h' \mathbf P \mathbf h' + \mathbf h \mathbf Q \mathbf h} \rd x \ge 0$
Then $\mathbf P$ is nonnegative.
== Proof ==
== Sources ==
: $\S 5.29$: Generalization to n Unknown Functions
Category:Calculus of Variations
Legendre's Conjecture

== Open Question ==
It is not known whether:
:$\exists n \in \N_{>1}: \map \pi {n^2 + 2 n + 1} = \map \pi {n^2}$
where $\pi$ denotes the prime-counting function.
That is:
:Is there always a prime number between consecutive squares?
== Landau's Problems ==
This is the $3$rd of Landau's problems.
== Sources ==
* : Chapter $2$: 'And you do addition?': $\S 2.1$: Functions: Example $2.1.1$
Category:Open Questions
Category:Prime-Counting Function
Category:Landau's Problems
Legendre's Differential Equation/(1 - x^2) y'' - 2 x y' + 2 y = 0

== Theorem ==
The special case of Legendre's differential equation:
:$(1): \quad \paren {1 - x^2} y - 2 x y' + 2 y = 0$
has the general solution:
:$y = C_1 x + C_2 \paren {\dfrac x 2 \, \map \ln {\dfrac {1 + x} {1 - x} } - 1}$
== Proof ==
Note that:
and so by inspection:
:$y_1 = x$
is a particular solution of $(1)$.
$(1)$ can be expressed as:
:$(2): \quad y - \dfrac {2 x} {1 - x^2} y' + \dfrac 2 {1 - x^2} y = 0$
which is in the form:
:$y + \map P x y' + \map Q x y = 0$
where:
:$\map P x = - \dfrac {2 x} {1 - x^2}$
:$\map Q x = \dfrac 2 {1 - x^2}$
From Particular Solution to Homogeneous Linear Second Order ODE gives rise to Another:
:$\map {y_2} x = \map v x \, \map {y_1} x$
where:
:$\ds v = \int \dfrac 1 { {y_1}^2} e^{-\int P \rd x} \rd x$
is also a particular solution of $(1)$.
We have that:
Hence:
and so:
From Two Linearly Independent Solutions of Homogeneous Linear Second Order ODE generate General Solution:
:$y = C_1 x + C_2 \paren {\dfrac x 2 \, \map \ln {\dfrac {1 + x} {1 - x} } - 1}$
== Sources ==
* : $\S 3.16$: Problem $5$
Category:Examples of Legendre's Differential Equation
Legendre's Duplication Formula

== Theorem ==
Let $\Gamma$ denote the gamma function.
Then:
:$\forall z \notin \set {-\dfrac n 2: n \in \N}: \map \Gamma z \map \Gamma {z + \dfrac 1 2} = 2^{1 - 2 z} \sqrt \pi \, \map \Gamma {2 z}$
where $\N$ denotes the natural numbers.
== Proof 1 ==
== Proof 2 ==
== Also presented as ==
== Also known as ==
== Also see ==
* Gamma Function of Positive Half-Integer
== Sources ==
* : $\S 16$: Relationships Among Gamma Functions: $16.9$
* : $\S 25$: The Gamma Function: Relationships Among Gamma Functions: $25.9.$
Category:Legendre's Duplication Formula
Category:Gamma Function
Category:Functional Equations
Legendre's Duplication Formula/Also known as

== Legendre's Duplication Formula: Also known as == 
Some sources refer to Legendre's duplication formula as just the duplication formula.
== Sources ==
* : $\S 16$: Relationships Among Gamma Functions: $16.9$
* : $\S 25$: The Gamma Function: Relationships Among Gamma Functions: $25.9.$
Category:Legendre's Duplication Formula
Legendre's Duplication Formula/Also presented as

== Legendre's Duplication Formula: Also presented as == 
Some sources report Legendre's Duplication Formula in the form:
:$\forall z \notin \set{-\dfrac n 2: n \in \N}: 2^{2 z - 1} \map \Gamma z \, \map \Gamma {z + \dfrac 1 2} = \sqrt \pi \, \map \Gamma {2 z}$
== Sources ==
* : $\S 16$: Relationships Among Gamma Functions: $16.9$
* : $\S 25$: The Gamma Function: Relationships Among Gamma Functions: $25.9.$
Category:Legendre's Duplication Formula
Legendre's Duplication Formula/Proof 1

== Theorem ==
== Proof ==
From the definition of the Beta function:
Letting $z_1 = z_2 = z$ gives:
Now substituting $u = x^2$ into the Beta function:
:$\ds \map \Beta {z_1, z_2} = \int_0^1 x^{2 z_1 - 2} \paren {1 - x^2}^{z_2 - 1} 2 x \rd x$
Letting $z_1 = \dfrac 1 2$ and $z_2 = z$ gives:
:$(2): \quad \ds \map \Beta {\frac 1 2, z} = 2 \int_0^1 \paren {1 - x^2}^{z - 1} \rd x$
Combining results $(1)$ and $(2)$:
From Gamma Function of One Half:
:$\map \Gamma {\dfrac 1 2} = \sqrt \pi$
It follows that:
:$\map \Gamma z \map \Gamma {z + \dfrac 1 2} = 2^{1 - 2 z} \sqrt \pi \, \map \Gamma {2 z}$
Category:Legendre's Duplication Formula
Legendre's Duplication Formula/Proof 2

== Theorem ==
== Proof ==
From Gauss Multiplication Formula:
:$\ds \prod_{k \mathop = 0}^{n - 1} \map \Gamma {z + \frac k n} = \paren {2 \pi}^{\paren {n - 1} / 2} n^{1/2 - n z} \map \Gamma {n z}$
Substituting $n = 2$ yields:
Category:Legendre's Duplication Formula