nyuuzyou/wikis huggingface dataset:
Central Field is Field of Functional

== Theorem ==
Let $\mathbf y$ be an $N$-dimensional vector.
Let $J$ be a functional, such that:
:$\ds J \sqbrk {\mathbf y} = \int_a^b \map F {x, \mathbf y, \mathbf y'} \rd x$
Let the following be a central field:
:$\map {\mathbf y'} x = \map {\boldsymbol \psi} {x, \mathbf y}$
Then this central field is a field of functional $J$.
== Proof ==
Suppose:
:$\ds \map g {x, \mathbf y} = \int_c^{\paren {x, \mathbf y} } \map F {x, \hat {\mathbf y}, \hat {\mathbf y}'} \rd x$
where $\hat {\mathbf y}$ is an extremal of $J$ connecting points $c$ and $\tuple {x, \mathbf y}$.
Define a field of directions in $R$ by the following:
:$\map {\mathbf p} {x, \mathbf y, \mathbf y'} = \map {g_{\mathbf y} } {x, \mathbf y}$
where $\mathbf p$ stands for momentum.
Note, that it does not depend on the path defined by $\hat {\mathbf y}$, only on its endpoint at $\tuple {x, \mathbf y}$.
By definition, $ \map g {x, \mathbf y}$ is a geodetic distance $S$.
Hence, $g_\mathbf y$ does not explicitly depend on $\mathbf y'$.
Then:
:$\map {g_{\mathbf y} } {x, \mathbf y} = \map {\mathbf p} {x, \mathbf y, \mathbf z}$
where $\mathbf z = \map {\mathbf z} {x, \mathbf y}$ denotes slope of the curve joining $c$ and $\tuple {x, \mathbf y}$ at the point $\tuple {x, \mathbf y}$.
Furthermore, since $g$ is a geodetic distance, it satisfies Hamilton-Jacobi equation:
:$\ds \frac {\partial \map g {x, \mathbf y} } {\partial x} + \map H {x, \mathbf y, \map {g_{\mathbf y} } {x, \mathbf y} } = 0$
which together with the previous relation results in a system of equations:
$\quad \begin {cases}
\dfrac {\partial \map g {x, \mathbf y} } {\partial x} + \map H {x, \mathbf y, \map {\mathbf p} {x, \mathbf y, \mathbf z} } = 0 \\
\map {\mathbf z} {x, \mathbf y} = \map{\mathbf y'} x 
\end {cases}$
Differentiation $\mathbf y$ yields:
$\quad \begin {cases}
\dfrac {\partial \mathbf p} {\partial x} = -\dfrac {\partial H} {\partial \mathbf y} \\
\map {\mathbf z} {x,\mathbf y} = \map{\mathbf y'} x 
\end {cases}$
These are Euler's equations in canonical coordinates with a constraint on derivative.
Hence, the previously defined field of directions coincides with the field of functional.
== Sources ==
* : $\S 6.32$: The Field of a Functional
Category:Calculus of Variations
Central Limit Theorem

== Theorem ==
Let $X_1, X_2, \ldots$ be a sequence of independent and identically distributed real-valued random variables with:
:expectation $\expect {X_i} = \mu \in \R$
:variance $\var {X_i} = \sigma^2 > 0$
Let:
:$\ds S_n = \sum_{i \mathop = 1}^n X_i$
Then:
:$\dfrac {S_n - n \mu} {\sqrt {n \sigma^2} } \xrightarrow D \Gaussian 0 1$ as $n \to \infty$
that is, converges in distribution to a standard normal distribution.
=== Conditions ===
== Proof ==
Let $Y_i = \dfrac {X_i - \mu} \sigma$.
We have that:
:$\expect {Y_i} = 0$
and:
:$\expect {Y_i^2} = 1$
Then by Taylor's Theorem the characteristic function can be written:
:$\map {\phi_{Y_i} } t = 1 - \dfrac {t^2} 2 + \map o {t^2}$
Now let:
Then its characteristic function is given by:
Recall that the Characteristic Function of Normal Distribution is given by:
Indeed, the characteristic function of $S_n$ converges to the Characteristic Function of Normal Distribution:
:$\paren {1 - \dfrac {t^2} {2 n} + \map o {t^2} }^n \to e^{-\frac 1 2 t^2}$ as $n \to \infty$
Then Lévy's Continuity Theorem applies.
In particular, the convergence in distribution of the $U_n$ to some random variable with standard normal distribution is equivalent to continuity of the limiting characteristic equation at $t = 0$.
But $e^{-\frac 1 2 t^2}$ is clearly continuous at $0$.
So we have that $\dfrac {S_n - n \mu} {\sqrt {n \sigma^2} }$ converges in distribution to a standard normal random variable.
== Historical Note ==
== Sources ==
* 
* 
* 
* 
* 
Category:Central Limit Theorem
Category:Normal Distribution
Category:Probability Theory
Category:Named Theorems
Central Limit Theorem/Conditions

== Conditions under which Central Limit Theorem holds ==
The Central Limit Theorem holds under the following conditions:
* The variance of any one of the contributory random variables does not dominate.
* The samples are not from the Cauchy distribution, as from Cauchy Distribution has no Finite Moments, the Cauchy distribution has no expectation.
== Sources ==
* 
* 
Category:Central Limit Theorem
Central Moment of Exponential Distribution

== Theorem ==
Let $X$ be a continuous random variable of the exponential distribution with parameter $\beta$ for some $\beta \in \R_{> 0}$
Let $n$ be a strictly positive integer. 
Then the $n$th central moment $\mu_n$ of $X$ is given by: 
:$\ds \mu_n = n! \beta^n \sum_{k \mathop = 0}^n \frac {\paren {-1}^k} {k!}$
== Proof == 
From definition of central moment we have:
:$\mu_n = \expect {\paren {x - \mu}^n}$
By Expectation of Exponential Distribution we have:
:$\mu = \beta$
So: 
Category:Exponential Distribution
Category:Central Moments
Central Product/Examples/D4 with D4

== Example of Central Product ==
Let $G$ be the dihedral group $D_4$ whose group presentation is:
:$G = \gen {a, b: a^4 = b^2 = e_G, a b = b a^{-1} }$
From Center of Dihedral Group $D_4$, the center of $G$ is:
:$\map Z G = \set {e_G, a^2}$
Let:
:$Z = W = \set {e_G, a^2}$
Let $\theta: Z \to W$ be the mapping defined as:
:$\map \theta g = \begin{cases} e_G & : g = e_G \\ a^2 & : g = a^2 \end{cases}$
Let $X$ be the set defined as:
:$X = \set {\tuple {z, \map \theta z^{-1} }: z \in Z}$
The central product of $G$ and $H$ via $\theta$ has $19$ elements of order $2$.
== Proof ==
== Sources ==
* : Chapter $13$: Direct products: Example $13.10$
Category:Dihedral Group D4
Category:Examples of Central Products
Central Product/Examples/D4 with Q

== Example of Central Product ==
Let $G$ be the dihedral group $D_4$ whose group presentation is:
:$G = \gen {a, b: a^4 = b^2 = e_G, a b = b a^{-1} }$
From Center of Dihedral Group $D_4$, the center of $G$ is:
:$\map Z G = \set {e_G, a^2}$
Let $H$ be the quaternion group $Q$ whose group presentation is:
:$Q = \gen {x, y: x^4 = e_H, y^2 = x^2, x y = y x^{-1} }$
From Center of Quaternion Group, the center of $H$ is:
:$\map Z H = \set {e_H, x^2}$
Let:
:$Z = \set {e_G, a^2}$
:$W = \set {e_H, x^2}$
Let $\theta: Z \to W$ be the mapping defined as:
:$\map \theta g = \begin{cases} e_H & : g = e_G \\ x^2 & : g = a^2 \end{cases}$
Let $X$ be the set defined as:
:$X = \set {\tuple {z, \map \theta z^{-1} }: z \in Z}$
The central product of $G$ and $H$ via $\theta$ has $32$ elements.
== Proof ==
We have that every element $h$ of $H$ which is not in its center:
: is of order $4$
: is such that $h^2 = x^2$.
The set $X$ consists of:
:$X = \set {\tuple {e_G, e_H}, \tuple {a^2, x^2} }$
The central product of $G$ and $H$ via $\theta$ is:
::$\dfrac {G \times H} X$
Thus:
In the direct product $G \times H$:
:$\tuple {e_G, e_H}$ has order $1$ and is in $X$
For all $h \in H$ such that $h$ is of order $4$:
For all $h \in H$ such that $h$ is of order $4$:
For all $g \in G$ such that $g$ is of order $2$:
For all $g \in G$ such that $g$ is of order $2$:
All remaining elements of $G \times H$ are of order $4$, and none of their squares is in $X$.
== Sources ==
* : Chapter $13$: Direct products: Example $13.10$
Category:Dihedral Group D4
Category:Quaternion Group
Category:Examples of Central Products
Central Product/Examples/Q with Q

== Example of Central Product ==
Let $G$ be the quaternion group $Q$ whose group presentation is:
:$Q = \gen {x, y: x^4 = e_H, y^2 = x^2, x y = y x^{-1} }$
From Center of Quaternion Group, the center of $H$ is:
:$\map Z H = \set {e_G, x^2}$
Let:
:$Z = W = \set {e_G, x^2}$
Let $\theta: Z \to W$ be the mapping defined as:
:$\map \theta g = \begin{cases} e_G & : g = e_G \\ x^2 & : g = x^2 \end{cases}$
Let $X$ be the set defined as:
:$X = \set {\tuple {z, \map \theta z^{-1} }: z \in Z}$
The central product of $G$ and $H$ via $\theta$ has $19$ elements.
== Proof ==
== Sources ==
* : Chapter $13$: Direct products: Exercise $4$
Category:Quaternion Group
Category:Examples of Central Products
Central Subgroup is Normal

== Theorem ==
Let $G$ be a group.
Let $H$ be a central subgroup of $G$.
Then $H$ is a normal subgroup of $G$.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 11$: Quotient Structures
Category:Normal Subgroups
Category:Central Subgroups
Category:Central Subgroup is Normal
Central Subgroup is Normal/Proof 1

== Theorem ==
== Proof ==
Let $H$ be a central subgroup of $G$.
By definition of central subgroup:
:$H \subseteq \map Z G$
where $\map Z G$ is the center of $G$.
Thus we have that $H$ is a group which is a subset of $\map Z G$.
Therefore by definition $H$ is a subgroup of $\map Z G$.
We also have from Center of Group is Abelian Subgroup that $\map Z G$ is an abelian group.
It follows from Subgroup of Abelian Group is Normal that $Z$ is a normal subgroup of $G$.
Category:Central Subgroup is Normal
Central Subgroup is Normal/Proof 2

== Theorem ==
== Proof ==
Let $H$ be a central subgroup of $G$.
By definition of central subgroup:
:$H \subseteq \map Z G$
where $\map Z G$ is the center of $G$.
Then:
== Sources ==
* : $\S 49.5$ Normal subgroups
Category:Central Subgroup is Normal
Centralizer in Subgroup is Intersection

== Theorem ==
Let $G$ be a group.
Let $H$ be a subgroup of $G$.
Then:
:$\forall x \in G: \map {C_H} x = \map {C_G} x \cap H$
That is, the centralizer of an element in a subgroup is the intersection of that subgroup with the centralizer of the element in the group.
== Proof ==
It is clear that:
:$g \in \map {C_H} x \iff g \in \map {C_G} x \land g \in H$
The result follows by definition of set intersection.
== Sources ==
* : Chapter $10$: The Orbit-Stabiliser Theorem: Proposition $10.19$
* : Chapter $10$: The Orbit-Stabiliser Theorem: Proposition $10.25$
Category:Subgroups
Category:Centralizers
Category:Set Intersection
Centralizer is Normal Subgroup of Normalizer

== Theorem ==
Let $G$ be a group.
Let $H \le G$ be a subgroup of $G$. 
Let $\map {C_G} H$ be the centralizer of $H$ in $G$.
Let $\map {N_G} H$ be the normalizer of $H$ in $G$.
Let $\Aut H$ be the automorphism group of $H$.
Then:
:$(1): \quad \map {C_G} H \lhd \map {N_G} H$
:$(2): \quad \map {N_G} H / \map {C_G} H \cong K$
where:
:$\map {N_G} H / \map {C_G} H$ is the quotient group of $\map {N_G} H$ by $\map {C_G} H$
:$K$ is a subgroup of $\Aut H$.
== Proof ==
In order to invoke the First Isomorphism Theorem for Groups, we must construct a group homomorphism $\phi: \map {N_G} H \to \Aut H$.
Consider the mapping $\phi: x \mapsto \paren {g \mapsto x g x^{-1}}$.
From Inner Automorphism is Automorphism, $g \mapsto x g x^{-1}$ is an automorphism of $G$, so $\phi$ is well-defined.
As $\phi$ is an group automorphism, $\phi$ is a group homomorphism.
Now we prove that $\ker \phi = \map {C_G} H$.
Note that for $x \in \map {N_G} H$:
Hence $\ker \phi = \map {C_G} H$.
By Kernel is Normal Subgroup of Domain:
:$\map {C_G} H \lhd \map {N_G} H$
By First Isomorphism Theorem for Groups:
:$\map {N_G} H / \map {C_G} H \cong \Img \phi$
By Image of Group Homomorphism is Subgroup:
:$\Img \phi \le \Aut H$
Hence the result.
== Sources ==
* : Chapter $10$: The Orbit-Stabiliser Theorem: Proposition $10.26$
Category:Normal Subgroups
Category:Normalizers
Category:Quotient Groups
Centralizer of Group Element is Subgroup

== Theorem ==
Let $\struct {G, \circ}$ be a group and let $a \in G$.
Then $\map {C_G} a$, the centralizer of $a$ in $G$, is a subgroup of $G$.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 8$: Compositions Induced on Subsets: Exercise $8.9$
* : $1$: Introduction to Finite Group Theory: $1.13$
* : Chapter $10$: The Orbit-Stabiliser Theorem: Exercise $4$
Category:Subgroups
Category:Centralizers
Category:Centralizer of Group Element is Subgroup
Centralizer of Group Element is Subgroup/Proof 1

== Theorem ==
== Proof ==
Let $\struct {G, \circ}$ be a group.
We have that:
:$\forall a \in G: e \circ a = a \circ e \implies e \in C_G \paren a$
Thus $C_G \paren a \ne \O$.
Let $x, y \in C_G \paren a$.
Then:
Thus $C_G \paren a$ is closed under $\circ$.
Let $x \in C_G \paren a$.
Then:
So:
:$x \in C_G \paren a \implies x^{-1} \in C_G \paren a$
Thus, by the Two-Step Subgroup Test, the result follows.
== Sources ==
* : $\S 37.1$. Some important general examples of subgroups
Category:Centralizer of Group Element is Subgroup
Centralizer of Group Element is Subgroup/Proof 2

== Theorem ==
== Proof ==
Let $\struct {G, \circ}$ be a group.
We have that:
:$\forall a \in G: e \circ a = a \circ e \implies e \in \map {C_G} a$
Thus $\map {C_G} a \ne \O$.
Let $x, y \in \map {C_G} a$.
Then from Commutation with Group Elements implies Commuation with Product with Inverse:
:$a \circ x \circ y^{-1} = x \circ y^{-1} \circ a$
so:
:$x \circ y^{-1} \in\map {C_G} a$
The result follows by the One-Step Subgroup Test, the result follows.
== Sources ==
* : $\S 3.2$: Groups; the axioms: Exercise $(11)$
Category:Centralizer of Group Element is Subgroup
Centralizer of Ring Subset is Subring

== Theorem ==
Let $S$ be a subset of a ring $\struct {R, +, \circ}$
Then $\map {C_R} S$, the centralizer of $S$ in $R$, is a subring of $R$.
== Proof ==
Certainly $0_R \in \map {C_R} S$ as $0_R$ commutes (trivially) with all elements of $R$.
Suppose $x, y \in \map {C_R} S$.
Then:
So:
:$x + \paren {-y} \in \map {C_R} s$
Suppose $x, y \in \map {C_R} S$ again.
Then from Element Commutes with Product of Commuting Elements:
:$x \circ y \in \map {C_R} S$
Thus all the conditions are fulfilled for Subring Test, and $\map {C_R} S$ is a subring of $R$.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $21$. Rings and Integral Domains: Theorem $21.5$
Category:Ring Theory
Category:Subrings
Centralizer of Self-Inverse Element of Non-Abelian Finite Simple Group is not That Group

== Theorem ==
Let $G$ be a non-abelian finite simple group.
Let $t \in G$ be a self-inverse element of $G$.
Then:
:$\map {C_G} t \ne G$
where $\map {C_G} t$ denotes the centralizer of $t$ in $G$.
== Proof ==
Let $G$ be a non-abelian finite simple group.
Let $t \in G$ which is not the identity.
By definition of a simple group and Center of Group is Normal Subgroup:
:either $\map Z G = G$ or $\map Z G$ is the trivial group.
By definition of an abelian group:
:$\map Z G = G$ $G$ is abelian
Hence we must have $\map Z G$ is the trivial group.
Thus $t \notin \map Z G$.
From definition of center:
:$\exists x \in G: t x \ne x t$
For this $x$, $x \notin \map {C_G} t$.
Hence $\map {C_G} t \ne G$.
== Sources ==
* 
* : $1$: Introduction to Finite Group Theory: $1.14$
Category:Centralizers
Category:Simple Groups
Category:Self-Inverse Elements
Category:Finite Groups
Centralizer of Subset is Intersection of Centralizers of Elements

== Theorem ==
Let $\struct {G, \circ}$ be a group.
Let $S \subseteq G$.
Let $\map {C_G} S$ be the centralizer of $S$ in $G$.
Then:
:$\ds \map {C_G} S = \bigcap_{x \mathop \in S} \map {C_G} x$
where $\map {C_G} z$ is the centralizer of $x$ in $G$.
== Proof ==
== Sources ==
* : $\S 3.2$: Groups; the axioms: Exercise $(11)$
Category:Group Theory
Centrifugal Force on Particle

== Theorem ==
Let $P$ be a particle which is stationary in a rotating frame of reference which itself is rotating with constant angular velocity $\omega$.
Let the mass of $P$ be $m$.
Let $P$ be a distance $r$ from the axis of rotation of the frame of reference.
Then $P$ can be treated as a particle in an inertial frame of reference experiencing a radial centrifugal force of magnitude $m r \omega^2$.
== Proof ==
== Sources ==
* 
* 
Category:Centrifugal Force
Centripetal Force on Body in Circular Path

== Theorem ==
Let $B$ be a body of mass $m$ constrained to move at constant speed $v$ in a circular path $C$.
Let $\mathbf F$ denote the centripetal force on $B$.
:$(1): \quad$ The direction of $\mathbf F$ is towards the center of $C$
:$(2): \quad$ The magnitude of $\mathbf F$ is given by:
::::$\size {\mathbf F} = \dfrac {m v^2} r$
:::or:
::::$\size {\mathbf F} = m r \omega^2$
:::where $\omega$ denotes the angular speed of $B$.
== Proof ==
== Sources ==
* 
* 
* 
* 
Category:Centripetal Force
Centroid of Combined Systems of Weighted Points

== Theorem ==
Let $O'$ be a point whose position vector from $O$ is $\mathbf l$.
Let $S = \set {A_1, A_2, \ldots, A_n}$ be a set of $n$ points in Euclidean space.
Let $W_S: S \to \R$ be a weight function on $S$.
Let $T = \set {B_1, B_2, \ldots, B_m}$ be a set of $m$ points in Euclidean space.
Let $W_T: T \to \R$ be a weight function on $T$.
Let $H$ be the centroid of $S$ with weight function $W_S$.
Let $H'$ be the centroid of $T$ with weight function $W_T$.
Then the centroid of $S \cup T$ with weight function $W_S \cup W_T$ is the centroid of the two points $H$ and $H'$ with weights $\ds \sum_{i \mathop = 1}^n \map {W_S} {A_i}$ and $\ds \sum_{i \mathop = 1}^m \map {W_T} {B_m}$ respectively.
== Proof ==
Let the position vectors of the points in $S = \set {A_1, A_2, \ldots, A_n}$ be given by $\mathbf a_1, \mathbf a_2, \dotsc, \mathbf a_n$ repectively.
Let the position vectors of the points in $T = \set {B_1, B_2, \ldots, B_m}$ be given by $\mathbf b_1, \mathbf b_2, \dotsc, \mathbf b_n$ repectively.
We have that:
:$\vec OH = \dfrac {\ds \sum_{i \mathop = 1}^n \map {W_S} {A_i} \mathbf a_i} {\ds \sum_{i \mathop = 1}^n \map {W_S} {A_i} }$
and:
:$\vec OH' = \dfrac {\ds \sum_{i \mathop = 1}^m \map {W_T} {B_i} \mathbf b_i} {\ds \sum_{i \mathop = 1}^m \map {W_T} {B_i} }$
Let $\vec OG$ denote the centroid of the two points $H$ and $H'$ with weights $\ds \sum_{i \mathop = 1}^n \map {W_S} {A_i}$ and $\ds \sum_{i \mathop = 1}^m \map {W_T} {B_m}$ respectively.
Then:
Hence the result.
== Sources ==
* : Chapter $\text I$. Addition and Subtraction of Vectors. Centroids: Centroids: $9$. Centroid, or centre of mean position
Category:Centroids
Centroid of Triangle is Centroid of Medial

== Theorem ==
Let $\triangle ABC$ be a triangle.
Let $\triangle DEF$ be the medial triangle of $\triangle ABC$.
Let $G$ be the centroid of $\triangle ABC$.
Then $G$ is also the centroid of $\triangle DEF$.
== Proof ==
:
By definition of centroid and medial triangle, the lines $AE$, $BF$ and $CD$ intersect at $G$.
It remains to be shown that $AE$, $BF$ and $CD$ bisect the sides of $DF$, $DE$ and $EF$ respectively.
, let $AE$ intersect $DF$ at $H$.
From the working of Triangle is Medial Triangle of Larger Triangle, we have that:
:$DE \parallel AF$
:$EF \parallel AD$
and so $\Box ADEF$ is a parallelogram whose diagonals are $DF$ and $AE$.
From Diameters of Parallelogram Bisect each other, $AE$ bisects $DF$ at $H$.
Similarly:
:$BF$ bisects $DE$
:$DC$ bisects $EF$.
Hence the result.
Category:Medial Triangles
Category:Centroids of Triangles
Centroid of Weighted Pair of Points

== Theorem ==
Let $A$ and $B$ be two points in Euclidean space.
Let $A$ and $B$ have weights $p$ and $q$ respectively.
Let $G$ be the centroid of $A$ and $B$.
Then $G$ divides the line $AB$ in the ratio $q : p$.
That is:
:$AG = \dfrac q {p + q} AB$
:$BG = \dfrac p {p + q} AB$
== Proof ==
Let the position vectors of $A$ and $B$ be given by $\mathbf a$ and $\mathbf b$ repectively.
By definition of centroid:
:$\vec {O G} = \dfrac {p \mathbf a + q \mathbf b} {p + q}$
The result follows from Point dividing Line Segment between Two Points in Given Ratio.
== Sources ==
* : Chapter $\text I$. Addition and Subtraction of Vectors. Centroids: Centroids: $9$. Centroid, or centre of mean position
Category:Centroids
Centroid of Weighted Set of Points is Independent of Origin

== Theorem ==
Let $O'$ be a point whose position vector from $O$ is $\mathbf l$.
Let $S = \set {A_1, A_2, \ldots, A_n}$ be a set of $n$ points in Euclidean space whose position vectors are given by $\mathbf a_1, \mathbf a_2, \dotsc, \mathbf a_n$ repectively.
Let $W: S \to \R$ be a weight function on $S$.
Let $G$ be the centroid of $S$ with weight function $W$:
:$\vec {OG} = \dfrac {w_1 \mathbf a_1 + w_2 \mathbf a_2 + \dotsb + w_n \mathbf a_n} {w_1 + w_2 + \dotsb + w_n}$
where $w_i = \map W {A_i}$ for each $i$.
Let $\vec {OG'}$ be the centroid of $S$ with weight function $W$ with respect to $O'$.
Then:
:$\vec {OG} - \mathbf l = \vec {OG'}$
That is, the centroid of $S$ with weight function $W$ is independent of the position of the origin.
== Proof ==
The position vectors of the elements of $S$ are given by:
:$\mathbf a_1 - \mathbf l, \mathbf a_2 - \mathbf l, \dotsc, \mathbf a_n - \mathbf l$
Hence the centroid of $S$ with weight function $W$ with respect to $O'$ ias:
:$\vec {OG'} = \dfrac {w_1 \paren {\mathbf a_1 - \mathbf l} + w_2 \paren {\mathbf a_2 - \mathbf l} + \dotsb + w_n \paren {\mathbf a_n - \mathbf l} } {w_1 + w_2 + \dotsb + w_n}$
== Sources ==
* : Chapter $\text I$. Addition and Subtraction of Vectors. Centroids: Centroids: $9$. Centroid, or centre of mean position
Category:Centroids
Cesàro Mean

== Theorem ==
Let $\sequence {a_n}$ be a sequence of complex numbers.
Suppose that $\sequence {a_n}$ converges to $\ell$ in $\C$:
:$\ds \lim_{n \mathop \to \infty} a_n = \ell$
Then also:
:$\ds \lim_{n \mathop \to \infty} \frac {a_1 + \dotsb + a_n} n = \ell$
== Proof ==
For every fixed integer $n_0$, we write:
:$\ds \cmod {\frac {a_1 + \dotsb + a_n} n - \ell} \le \frac {\cmod {a_1 - \ell} + \dotsb + \cmod {a_n - \ell} } n \le \frac {n_0 \ds \sup_{k \mathop \le n_0} \cmod {a_k - \ell} } n + \sup_{n_0 \mathop < k \mathop \le n} \cmod {a_k - \ell}$
As $n$ tends to $+\infty$, we get:
:$\ds \limsup_{n \mathop \to \infty} \cmod {\frac {a_1 + \dotsb + a_n} n - \ell} \le \sup_{k \mathop > n_0} \cmod {a_k - \ell}$
As $n_0$ tends to $+\infty$, we finally conclude:
:$\ds \limsup_{n \mathop \to \infty} \cmod {\frac {a_1 + \dotsb + a_n} n - \ell} = 0$
== Remarks ==
* The theorem and its proof hold in any normed vector space.
* When working with an arbitrary sequence $\sequence {a_n}$ of real numbers, the same truncation trick leads to:
:$\ds \liminf_{n \mathop \to \infty} a_n \le \liminf_{n \mathop \to \infty} \frac {a_1 + \dotsb + a_n} n \le \limsup_{n \mathop \to \infty} \frac {a_1 + \dotsb + a_n} n \le \limsup_{n \mathop \to \infty} a_n$
As a corollary, the conclusion of the theorem holds in the real case when $\ell = \pm \infty$.
Category:Analysis
Category:Topology
Category:Limits of Sequences
Cesàro Summation Operator is Continuous Linear Transformation

== Theorem ==
Let $\ell^\infty$ be the space of bounded sequences.
Let $A : \ell^\infty \to \ell^\infty$ be the Cesàro summation operator.
Then $A$ is a continuous linear transformation.
== Proof ==
=== Well-Definedness ===
Let $\mathbf x = \sequence {x_n}_{n \mathop \in \N} \in \ell^\infty$.
Then:
Thus:
Hence:
:$A \mathbf x \in \ell^\infty$
=== Linearity ===
Let $\mathbf x = \sequence {x_n}_{n \mathop \in \N}, \mathbf y = \sequence {y_n}_{n \mathop \in \N} \in \ell^\infty$.
Let $\lambda \in \C$.
By Space of Bounded Sequences with Pointwise Addition and Pointwise Scalar Multiplication on Ring of Sequences forms Vector Space:
:$\mathbf x + \lambda \mathbf y \in \ell^\infty$
Then:
By definition, $A$ is a linear transformation.
=== Continuity ===
Let $\mathbf x = \sequence {x_n}_{n \mathop \in \N} \in \ell^\infty$.
Then:
By Continuity of Linear Transformation between Normed Vector Spaces, $A$ is continuous. 
All together:
:$A \in \map {CL} {\ell^\infty}$
== Sources ==
* : Chapter $\S 2.3$: The normed space $\map {CL} {X, Y}$. Operator norm and the normed space $\map {CL} {X, Y}$
Category:Cesàro Summation Operator
Category:Operator Theory
Category:Linear Transformations
Category:Continuous Transformations
Ceva's Theorem

== Theorem ==
Let $\triangle ABC$ be a triangle.
Let $L$, $M$ and $N$ be points on the sides $BC$, $AC$ and $AB$ respectively.
Then the lines $AL$, $BM$ and $CN$ are concurrent :
:$\dfrac {BL} {LC} \times \dfrac {CM} {MA} \times \dfrac {AN} {NB} = 1$
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Menelaus's Theorem
== Historical Note ==
== Sources ==
* 
* 
* 
* 
Category:Ceva's Theorem
Category:Triangles