nyuuzyou/wikis huggingface dataset:
Isomorphism between Symmetry Group of Regular Pentagon and Subgroup of Symmetric Group

== Theorem ==
Let $\PP = ABCDE$ denote a regular pentagon.
Let $\struct {\PP, \circ}$ be the symmetry group of $\PP$, where the various symmetries are identified as:
:the identity mapping $e$
:the rotations $r, r^2, r^3, r^4$ of $72^\circ, 144^\circ, 216^\circ, 288^\circ$ around the center of $\PP$ anticlockwise respectively
:the reflections $t_A, t_B, t_C, t_D, t_E$ about the lines through the center of $\PP$ and the vertices $A$ to $E$ respectively.
:
Let $\struct {S_5, \circ}$ denote the symmetry group on $5$ letters.
Then $\PP$ is isomorphic to a subgroup of $\struct {S_5, \circ}$ of order $10$.
== Proof ==
Let the $A$, $B$, $C$, $D$ and $E$ of $\PP$ be identified with the integers $1$, $2$, $3$, $4$ and $5$ of $S_5$.
Let each of the symmetries of $\PP$ be identified with permutations of $S_n$ according to where the symmetry moves the vertices of $\PP$.
We express these in two-row notation, which we construct by inspection:
These are now identified with permutations of $S_5$ and hence elements of $\struct {S_5, \circ}$, which we express in cycle notation:
The isomorphism is implicit in the construction.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 8$: Compositions Induced on Subsets: Exercise $8.5$
Category:Symmetry Group of Regular Pentagon
Category:Symmetric Group on 5 Letters
Isomorphism by Cayley Table

== Theorem ==
Let $\struct {S, \circ}$ and $\struct {T, *}$ be algebraic structures whose underlying sets are both finite.
Then $\struct {S, \circ}$ and $\struct {T, *}$ are isomorphic :
:a bijection $f: S \to T$ can be found such that:
::the Cayley table of $\struct {T, *}$ can be generated from the Cayley table of $\struct {S, \circ}$ by replacing each entry of $S$ with its image under $f$.
== Proof ==
=== Necessary Condition ===
Let $S$ and $T$ be isomorphic.
Then by definition there exists an isomorphism $f: S \to T$.
An isomorphism is a bijection by definition.
Thus the existence of the posited bijection has been demonstrated.
By the definition of set equivalence, $S$ and $T$ have the same cardinality.
Let $\card S = \card T = n$.
Let:
:$S = \set {s_1, s_2, \ldots, s_n}$
and:
:$T = \set {t_1, t_2, \ldots, t_n}$
be such that:
:$\forall i \in \set {1, 2, \ldots, n}: \map f {s_i} = t_i$
Let $S \sqbrk {j, k}$ denote the entry of $S$ in row $j$ and column $k$.
Thus:
:$S \sqbrk {j, k} = s_j \circ s_k$
for some $s_j, s_k \in S$.
We have that $f$ is an isomorphism.
Thus:
:$\map f {S \sqbrk {j, k} } = \map f {s_j} * \map f {s_k}$
But:
:$\map f {S \sqbrk {j, k} } = T \sqbrk {j, k}$
:$\map f {s_j} = t_j$
:$\map f {s_k} = t_k$
Thus the Cayley table of $\struct {T, *}$ has been generated from the Cayley table of $\struct {S, \circ}$ by replacing each entry of $S$ with its image under $f$.
=== Sufficient Condition ===
Let there exist a bijection $f: S \to T$ such that the Cayley table of $\struct {T, *}$ can be generated from the Cayley table of $\struct {S, \circ}$ by replacing each entry of $S$ with its image under $f$.
By the definition of set equivalence, $S$ and $T$ have the same cardinality.
Let $\card S = \card T = n$.
Let the rows and columns of the Cayley table of $\left({S, \circ}\right)$ both be numbered from $1$ to $n$.
Let the entries heading the rows and columns be denoted $s_1, s_2, \ldots, s_n$.
Let $S \sqbrk {j, k}$ denote the entry of $S$ in row $j$ and column $k$.
By the method of construction:
:$S \sqbrk {j, k} = s_j \circ s_k$
for some $s_j, s_k \in S$.
Let $f$ be used to generate the Cayley table of $\struct {T, *}$ from the Cayley table of $\struct {S, \circ}$.
Then for all $i, j \in \set {1, 2, \ldots, n}$:
:$\map f {S \sqbrk {j, k} } = T \sqbrk {j, k}$
:$\map f {s_j} = t_j$
:$\map f {s_k} = t_k$
But by definition of the Cayley table of $\struct {T, *}$:
:$T \sqbrk {j, k} = t_j * t_k$
Thus:
:$\forall j, k \in \set {1, 2, \ldots, n}: \map f {s_j \circ s_k} = \map f {s_j} * \map f {s_k}$
demonstrating that $f$ is an isomorphism.
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 6$: Isomorphisms of Algebraic Structures
Category:Cayley Tables
Category:Isomorphisms (Abstract Algebra)
Isomorphism from R^n via n-Term Sequence

== Theorem ==
Let $G$ be a unitary $R$-module.
Let $\sequence {a_k}_{1 \mathop \le k \mathop \le n}$ be an ordered basis of $G$.
Let $R^n$ be the $R$-module $R^n$.
Let $\psi: R^n \to G$ be defined as:
:$\ds \map \psi {\sequence {\lambda_k}_{1 \mathop \le k \mathop \le n} } = \sum_{k \mathop = 1}^n \lambda_k a_k$
Then $\psi$ is an isomorphism.
== Proof ==
By Unique Representation by Ordered Basis, $\psi$ is a bijection.
We have:
and we have:
thus proving that $\psi$ is also a homomorphism.
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 27$. Subspaces and Bases: Theorem $27.5$
Category:Module Theory
Category:Unitary Modules
Category:R-Algebraic Structure Isomorphisms
Isomorphism is Equivalence Relation

== Theorem ==
Isomorphism is an equivalence on a set of magmas.
This result applies to all magmas: rings, groups, R-algebraic structures etc.
== Proof ==
To prove a relation is an equivalence, we need to prove it is reflexive, symmetric and transitive.
So, checking in turn each of the criteria for equivalence:
=== Reflexive ===
From Identity Mapping is Automorphism, the identity mapping $I_S: S \to S$ is an automorphism, which is an isomorphism from a magma onto itself.
So $S \cong S$, and isomorphism is seen to be reflexive.
=== Symmetric ===
If $S \cong T$, then $\exists \phi: S \to T$ such that $\phi$ is an isomorphism.
Thus, from Inverse of Algebraic Structure Isomorphism is Isomorphism, $\phi^{-1}: T \to S$ is also an isomorphism, and $T \cong S$.
Thus we have shown that $S \cong T \implies T \cong S$, so $\cong$ is symmetric.
=== Transitive ===
Let $S_1 \cong S_2$, and $S_2 \cong S_3$.
We can define these two isomorphisms:
: $\phi: S_1 \to S_2$
: $\psi: S_2 \to S_3$
From Composite of Isomorphisms in Algebraic Structure is Isomorphism, we have that $\psi \circ \phi$ is an isomorphism between $S_1$ and $S_3$.
Thus we have shown that $\cong$ is transitive.
Thus isomorphism is reflexive, symmetric and transitive, and therefore an equivalence.
== Sources ==
* : $\S 7.3$. Isomorphism
* : Chapter $\text {V}$: Vector Spaces: $\S 26$. Vector Spaces and Modules: Theorem $26.1$
* : Chapter $\text{II}$: Groups: Exercise $\text{H}$
* : $\S 2.2$: Homomorphisms
* : Chapter $2$: Group Homomorphism and Isomorphism: $\S 62$
* : $0$: Some Conventions and some Basic Facts
* : $\S 46.1$ Isomorphic groups
* : Chapter $8$: The Homomorphism Theorem: Proposition $8.9$
Category:Isomorphisms (Abstract Algebra)‎
Category:Examples of Equivalence Relations
Isomorphism of Abelian Groups

== Theorem ==
Let $\phi: \struct {G, \circ} \to \struct {H, *}$ be a group isomorphism.
Then $\struct {G, \circ}$ is abelian $\struct {H, *}$ is abelian.
== Proof ==
We have that Isomorphism Preserves Commutativity.
Thus:
:$\forall x, y \in G: x \circ y = y \circ x \implies \map \phi x * \map \phi y = \map \phi y * \map \phi x$
Thus if $G$ is abelian, so is $H$.
As $\phi^{-1}: H \to G$ is also an isomorphism, it is clear that if $H$ is abelian, then so is $G$.
== Sources ==
* : $\S 7.3$. Isomorphism: Example $137$
Category:Abelian Groups
Category:Group Isomorphisms
Isomorphism of External Direct Products

== Theorem ==
Let:
:$\struct {S_1 \times S_2, \circ}$ be the external direct product of two algebraic structures $\struct {S_1, \circ_1}$ and $\struct {S_2, \circ_2}$
:$\struct {T_1 \times T_2, *}$ be the external direct product of two algebraic structures $\struct {T_1, *_1}$ and $\struct {T_2, *_2}$
:$\phi_1$ be an isomorphism from $\struct {S_1, \circ_1}$ onto $\struct {T_1, *_1}$
:$\phi_2$ be an isomorphism from $\struct {S_2, \circ_2}$ onto $\struct {T_2, *_2}$.
Then the mapping $\phi_1 \times \phi_2: \struct {S_1 \times S_2, \circ} \to \struct {T_1 \times T_2, *}$ defined as:
:$\map {\paren {\phi_1 \times \phi_2} } {x, y} = \tuple {\map {\phi_1} x, \map {\phi_2} y}$
is an isomorphism from $\struct {S_1 \times S_2, \circ}$ to $\struct {T_1 \times T_2, *}$.
=== General Result ===
== Proof ==
From Homomorphism of External Direct Products, we have that $\phi_1 \times \phi_2: \struct {S_1 \times S_2, \circ} \to \struct {T_1 \times T_2, *}$ is a homomorphism.
From Cartesian Product of Bijections is Bijection, we have that $\phi_1 \times \phi_2$ is a bijection.
Thus $\phi_1 \times \phi_2$ is a bijective homomorphism, and so an isomorphism.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 13$: Compositions Induced on Cartesian Products and Function Spaces: Theorem $13.2$
* : Chapter $\text {II}$: New Structures from Old: $\S 13$: Compositions Induced on Cartesian Products and Function Spaces: Exercise $13.2$
Category:Isomorphism of External Direct Products
Category:Isomorphisms (Abstract Algebra)
Category:External Direct Products
Isomorphism of External Direct Products/General Result

== Theorem ==
Let:
:$(1): \quad \ds \struct {S, \circ} = \prod_{k \mathop = 1}^n S_k = \struct {S_1, \circ_1} \times \struct {S_2, \circ_2} \times \cdots \times \struct {S_n, \circ_n}$
:$(2): \quad \ds \struct {T, \ast} = \prod_{k \mathop = 1}^n T_k = \struct {T_1, \ast_1} \times \struct {T_2, \ast_2} \times \cdots \times \struct {T_n, \ast_n}$
be external direct products of algebraic structures.
Let $\phi_k: \struct {S_k, \circ_k} \to \struct {T_k, \ast_k}$ be an isomorphism for each $k \in \closedint 1 n$.
Then:
:$\phi: \struct {s_1, \ldots, s_n} \to \struct {\map {\phi_1} {s_1}, \ldots, \map {\phi_n} {s_n} }$
is an isomorphism from $\struct {S, \circ}$ to $\struct {T, \ast}$.
== Proof ==
By definition of isomorphism, each $\phi_k$ is a homomorphism which is a bijection.
From Cartesian Product of Bijections is Bijection: General Result, $\phi$ is a bijection.
From Homomorphism of External Direct Products: General Result, $\phi$ is a homomorphism.
Hence the result.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 18$: Induced $N$-ary Operations: Theorem $18.11$
Category:Isomorphism of External Direct Products
Isomorphism of Finite Group with Permutations of Quotient with Subgroup

== Theorem ==
Let $G$ be a finite group.
Let $H$ be a subgroup of $G$.
Let $H$ contain no non-trivial normal subgroup of $G$.
Let $G / H$ denote the left coset space of $G$ by $H$.
Then $G$ is isomorphic to a subgroup of the group of permutations $\map \Gamma {G / H}$ of $G / H$.
== Proof ==
Let a homomorphism $\phi: G \to \map \Gamma {G / H}$ be defined as:
:$\forall x \in G: \map {\map \phi g} {x H} = \paren {g x} H$
By Kernel is Normal Subgroup of Domain, $\ker \phi$ is a normal subgroup of $G$.
So:
That is, $\ker \phi$ is the intersection of $H$ and all its conjugates.
Thus:
:$\ker \phi \subseteq H$
But $H$ contains no non-trivial normal subgroup of $G$.
Thus $\ker \phi$ is the trivial subgroup of $G$
Thus by Kernel is Trivial iff Group Monomorphism:
:$G$ is isomorphic to $\Img \phi$.
== Sources ==
* : Chapter $2$: The Symmetric Groups: $\S 85$
Category:Group Isomorphisms
Category:Symmetric Groups
Isomorphism to Closed Interval

== Theorem ==
Let $m, n \in \N$ such that $m < n$.
Let $\closedint {m + 1} n$ denote the integer interval from $m + 1$ to $n$.
Let $h: \N_{n - m} \to \closedint {m + 1} n$ be the mapping defined as:
:$\forall x \in \N_{n - m}: \map h x = x + m + 1$
Let the orderings on $\closedint {m + 1} n$ and $\N_{n - m}$ be those induced by the ordering of $\N$.
Then $h$ a unique order isomorphism.
== Proof ==
First note that the cardinality of $\closedint {m + 1} n$ is given by:
:$\card {\closedint {m + 1} n} = n - m$
From Unique Isomorphism between Equivalent Finite Totally Ordered Sets, it suffices to show that $h$ is an order isomorphism.
To this end, remark that, for all $x, y \in \N_{n - m}$:
proving $h$ is an injection.
Hence from Equivalence of Mappings between Finite Sets of Same Cardinality, $h$ is also a bijection.
By Ordering on Natural Numbers is Compatible with Addition and Natural Number Addition is Cancellable for Ordering, it follows that:
:$x \le y \iff \map h x \le \map h y$
so $h$ is an order isomorphism.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 17$: Finite Sets: Theorem $17.11$
Category:Set Theory
Category:Order Theory
Category:Mapping Theory
Isomorphisms between Additive Group of Integers Modulo 4 and Reduced Residue System Modulo 5 under Multiplication

== Theorem ==
Let $\struct {\Z_4, +_4}$ denote the additive group of integers modulo $4$.
Let $\struct {\Z'_5, \times_5}$ denote the multiplicative group of reduced residues modulo $5$.
There are $2$ (group) isomorphisms from $\struct {\Z_4, +_4}$ onto $\struct {\Z'_5, \times_5}$.
== Proof ==
Let us recall the Cayley table of $\struct {\Z_4, +_4}$:
and the Cayley Table of $\struct {\Z'_5, \times_5}$:
Each of these is the cyclic group of order $4$.
Each has $2$ generators, each of $1$ element.
Hence you can get an isomorphism from $\struct {\Z_4, +_4}$ to $\struct {\Z'_5, \times_5}$ by setting up the mappings:
:$\phi: \struct {\Z_4, +_4} \to \struct {\Z'_5, \times_5}: \forall x \in \Z_4: \map \phi x = \begin {cases} \eqclass 4 0 & : x = \eqclass 5 1 \\ \eqclass 4 1 & : x = \eqclass 5 2 \\ \eqclass 4 2 & : x = \eqclass 5 4 \\ \eqclass 4 3 & : x = \eqclass 5 3 \end {cases}$
:$\psi: \struct {\Z_4, +_4} \to \struct {\Z'_5, \times_5}: \forall x \in \Z_4: \map \psi x = \begin {cases} \eqclass 4 0 & : x = \eqclass 5 1 \\ \eqclass 4 1 & : x = \eqclass 5 3 \\ \eqclass 4 2 & : x = \eqclass 5 4 \\ \eqclass 4 3 & : x = \eqclass 5 2 \end {cases}$
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 6$: Isomorphisms of Algebraic Structures: Exercise $6.5$
Category:Examples of Group Isomorphisms
Category:Multiplicative Group of Reduced Residues Modulo 5
Category:Cyclic Group of Order 4
Isomorphisms between Lower Sections of Well-Ordered Classes are Nested

== Theorem ==
Let $\struct {A, \preccurlyeq_A}$ and $\struct {B, \preccurlyeq_B}$ be well-ordered classes.
Let $\phi_1$ and $\phi_2$ be order isomorphisms from a lower section of $\struct {A, \preccurlyeq_A}$ to a lower section of $\struct {B, \preccurlyeq_B}$.
Then either:
:$\phi_1 \subseteq \phi_2$
or:
:$\phi_2 \subseteq \phi_1$
where $\subseteq$ denotes the subset relation on mappings:
:$f \subseteq g \iff \forall x \in \Dom f: \map f x = \map g x$
== Proof ==
Let us label the domains of $\phi_1$ and $\phi_2$:
:$D_1 = \Dom {\phi_1}$
:$D_2 = \Dom {\phi_2}$
Because $D_1$ and $D_2$ are both lower sections of $\struct {A, \preccurlyeq_A}$, they are comparable under the subset relation.
, suppose $D_1 \subseteq D_2$.
Then the restriction $\phi_2 {\restriction_{D_1} }$ is an order isomorphisms from $D_1$ to a lower section of $\struct {B, \preccurlyeq_B}$.
From Lower Sections of Well-Ordered Classes are Order Isomorphic at most Uniquely:
:$\phi_2 {\restriction_{D_1} } = \phi_1$
That is:
:$\phi_1 \subseteq \phi_2$
Similarly if $D_2 \subseteq D_1$ then:
:$\phi_2 \subseteq \phi_1$
== Sources ==
* : Chapter $6$: Order Isomorphism and Transfinite Recursion: $\S 2$ Isomorphisms of well orderings: Corollary $2.8$
Category:Well-Orderings
Category:Lower Sections
Category:Order Isomorphisms
Category:Nests
Isomorphisms between Symmetry Groups of Isosceles Triangle and Equilateral Triangle

== Theorem ==
Let $\TT = ABC$ be an isosceles triangle whose apex is $A$.
Let $\struct {\TT, \circ}$ be the symmetry group of $\TT$, where the symmetries are identified as:
:the identity mapping $e$
:the reflection $d$ in the line through $A$ and the midpoint of $BC$.
:
Let $\SS = A'B'C'$ be an equilateral triangle.
We define in cycle notation the following symmetries on $\triangle A'B'C'$:
Then $\struct {\TT, \circ}$ is isomorphic to the $3$ subgroups of $S_3$:
== Proof ==
We have that $\struct {\TT, \circ}$ is of order $2$.
We also have that:
are also groups of order $2$.
From Parity Group is Only Group with 2 Elements, all these groups are isomorphic.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 8$: Compositions Induced on Subsets: Exercise $8.6$
Category:Symmetry Group of Isosceles Triangle
Category:Symmetry Group of Equilateral Triangle
Isoperimetric Inequality

== Theorem ==
Let $F$ be a plane figure.
Let $P$ be the length of the perimeter of $F$.
Let $A$ be the area of $F$.
Then:
:$P^2 \ge 4 \pi A$
where the equality happens when $F$ is a circle.
That is, the circle is the largest of all the isoperimetric figures whose perimeters are of length $P$.
== Proof ==
== Also see ==
* Dido's Problem
== Historical Note ==
== Sources ==
* : Chapter $\text{VIII}$: Nature or Nurture?
* 
* 
Category:Isoperimetrical Problems
Category:Inequalities
Category:Named Theorems
Isoperimetrical Problem

== Classic Problem ==
What shape is the largest area that can be surrounded by a plane curve of a given length?
== Solution ==
The solution is: a circle.
== Proof ==
== Also see ==
* Dido's Problem, which is a variant of this.
== Sources ==
* : Chapter $\text I$. The First Variation: $1.1$. Introduction
Category:Euclidean Geometry
Isosceles Triangle has Two Equal Angles

== Theorem ==
In isosceles triangles, the angles at the base are equal to each other.
Also, if the equal straight lines are extended, the angles under the base will also be equal to each other.
== Proof ==
Let $\triangle ABC$ be an isosceles triangle whose side $AB$ equals side $AC$.
We extend the straight lines $AB$ and $AC$ to $D$ and $E$ respectively.
Let $F$ be a point on $BD$.
We cut off from $AE$ a length $AG$ equal to $AF$.
We draw line segments $FC$ and $GB$.
Since $AF = AG$ and $AB = AC$, the two sides $FA$ and $AC$ are equal to $GA$ and $AB$ respectively.
They contain a common angle, that is, $\angle FAG$.
So by Triangle Side-Angle-Side Congruence, $\triangle AFC = \triangle AGB$.
Thus $FC = GB$, $\angle ACF = \angle ABG$ and $\angle AFC = \angle AGB$.
Since $AF = AG$ and $AB = AC$, then $BF = CG$.
But $FC = GB$, so the two sides $BF, FC$ are equal to the two sides $CG, GB$ respectively.
Then $\angle BFC = \angle CGB$ while $CB$ is common to both.
Therefore by Triangle Side-Angle-Side Congruence:
:$\triangle BFC = \triangle CGB$
Therefore $\angle FBC = \angle GCB$ and $\angle BCF = \angle CBG$.
So since $\angle ACF = \angle ABG$, and in these $\angle BCF = \angle CBG$, then $\angle ABC = \angle ACB$.
But $\angle ABC$ and $\angle ACB$ are at the base of $\triangle ABC$.
Also, we have already proved that $\angle FBC = \angle GCB$, and these are the angles under the base of $\triangle ABC$.
Hence the result.
== Also known as ==
It appears to have originally been created by .
== Sources ==
* : $\text I$: Introductory: The Greeks and Mathematics
* : Book $\text{I}$. Propositions
* : Chapter $1$: Triangles and Polygons: Theorems and Corollaries $1.12$
* : Chapter $\text {A}.1$: Thales (ca. B.C.)
* 
* 
* : $(2)$
* 
* 
* : $(2)$
* : Chapter $2$: The Logic of Shape: Euclid
* 
* 
Category:Isosceles Triangle has Two Equal Angles
Category:Isosceles Triangles
Isosceles Triangle has Two Equal Angles/Also known as

== Isosceles Triangle has Two Equal Angles: Also known as ==
This proposition is the famous pons asinorum, that is, the bridge of donkeys, supposedly for two reasons:
:$(1): \quad$ The diagram accompanying it is supposed to look a bit like a bridge
:$(2): \quad$ If you can't cross this bridge (that is, understand this theorem), you're supposedly a bit of a donkey.
Don't blame the students -- this is one of the more tortuous ways of proving this theorem.
Commentators have speculated that may not have known what he was doing when he wrote this.
== Sources ==
* 
* 
* : Chapter $2$: The Logic of Shape: Euclid
* 
* 
Category:Isosceles Triangle has Two Equal Angles
Isotropic at Point and Homogeneous Riemannian Manifold is Isotropic

== Theorem ==
Let $\struct {M, g}$ be a Riemannian manifold.
Let $M$ be isotropic at any $p \in M$.
Let $M$ be homogeneous.
Then $M$ is isotropic.
== Proof ==
== Sources ==
* : $\S 3$: Model Riemannian Manifolds. Symmetries of Riemannian Manifolds
Category:Riemannian Geometry
Iteration of Ruelle-Perron-Frobenius Operator/One-Sided Shift Space of Finite Type

== Theorem ==
Let $\struct {X_\mathbf A ^+, \sigma}$ be a one-sided shift of finite type.
Let $\map C {X _\mathbf A ^+}$ be the complex-valued continuous mapping space.
Let $f \in \map C {X _\mathbf A ^+}$.
Let $\LL_f$ be the Ruelle-Perron-Frobenius operator.
For all $w \in \map C {X _\mathbf A ^+}$, $n \in \N$ and $x \in X _\mathbf A ^+$, we have:
:$\ds \map {\LL_f^n w} x = \sum_{y \mathop \in \map {\sigma^{-n} } x} e^{\map {f_n} y} \map w y $
where
:$f_n := f + f \circ \sigma + \cdots + f \circ \sigma ^{n-1}$
:$\map {\sigma ^{-n} } x$ denotes the preimage of $x$ under $\sigma^n$, i.e.:
::$\map {\sigma^{-n} } x = \set {\sequence {i_1, i_2, \ldots , i_n, x_0, x_1, \ldots} : \map {\mathbf A} {i_1, i_2} = \cdots = \map {\mathbf A} {i_{n-1}, i_n} = \map {\mathbf A} {i_n, x_0} = 1}$
== Proof ==
Category:Ergodic Theory
Category:Functional Analysis
Jack-Knife Resampling/Examples/Arbitrary Example 1

== Example of Jack-Knife Resampling ==
Let us attempt to determine the bias in the correlation coefficient $r$ of a sample of $n$ pairs used as an estimator of the correlation coefficient $\rho$ of the population.
The procedure is to compute successively the values $r_{\paren i}$ for $i = 1, 2, \ldots, n$ for sample identical to the original but with the $i$th sample value omitted.
Let $r_{\paren .}$ be the mean of the $r_{\paren i}$ values.
Then the jack-knife estimator of bias is given by:
:$B = \paren {n - 1} \paren {r_{\paren .} - r}$
In general, this is not an exact measure of bias, but it is usually a good approximation.
== Sources ==
* 
Category:Examples of Jack-Knife Resampling
Jack-Knife Resampling/Examples/Standard Errors

== Example of Jack-Knife Resampling ==
Jack-knifing can be used to obtain estimates of standard errors.
While the jack-knife involves only $n$ samplings of the original data, each of a specified form, the bootstrapping method often uses $N$ samplings, where $N$ is considerably greater than $n$.
== Sources ==
* 
Category:Examples of Jack-Knife Resampling
Jacobi's Equation is Variational Equation of Euler's Equation

== Theorem ==
The Variational equation of Euler's equation is Jacobi's equation.
== Proof ==
Let Euler's equation be
:$\map {F_y} {x, \hat y, \hat y'} - \dfrac \d {\d x} \map {F_{y'} } {x, \hat y, \hat y'} = 0$
which is derived from:
:$\ds \int_a^b \paren {\map {F_y} {x, \hat y, \hat y'} - \frac \d {\d x} \map {F_{y'} } {x, \hat y, \hat y'} } \rd x = 0$
Let $\map {\hat y} x = \map y x$ and $\map {\hat y} x = \map y x + \map h x$ be solutions of Euler's equation.
By Taylor's Theorem:
where the omitted ordered tuple of variables is $\tuple {x, y, y'}$, and $\map {\hat y} x = \map y x$ has been used as a solution to $F_y - \dfrac \d {\d x} F_{y'} = 0$.
Therefore, Euler's equation is to be derived from
:$\ds \int_a^b \paren {\paren {F_{yy} - \frac \d {\d x} F_{y'y} } h - \map {\frac \d {\d x} } {F_{y'y'} h'} + \map \OO {h^2, h h', h'^2} } \rd x = 0$
By integration by parts, 
:$\ds \int_a^b \map \OO {h^2, h h', h'^2} \rd x = \int_a^b \map \OO {h^2} \rd x$
Thus, the equivalent differential equation is:
:$\paren {F_{yy} - \dfrac \d {\d x} F_{y'y} } h - \map {\dfrac \d {\d x} } {F_{y'y'} h'} + \map \OO {h^2} = 0$
Omission of $\map \OO {h^2}$ and multiplication of equation by $\frac 1 2$ yields Jacobi's equation.
== Sources ==
* : $\S 5.27$: Jacobi's Necessary Condition. More on Conjugate Points
Category:Calculus of Variations