nyuuzyou/wikis huggingface dataset:
Autocorrelation of Strictly Stationary Stochastic Process

== Example of Strictly Stationary Stochastic Process ==
Let $S$ be a strictly stationary stochastic process giving rise to a time series $T$.
It is necessary that:
:The autocorrelation between every two observations $z_t, z_{t + k}$ separated by a given lag $k$ 
is the same as:
:the autocorrelation between every other two observations $z_{t + m}, z_{t + m + k}$ separated by a given lag $k$
For such a strictly stationary stochastic process:
:$\rho_k = \dfrac {\gamma_k} {\gamma_0}$
where $\gamma_k$ denotes the autocovariance of $S$.
== Proof ==
The autocorrelation is defined as:
:$\rho_k := \dfrac {\expect {\paren {z_t - \mu} \paren {z_{t + k} - \mu} } } {\sqrt {\expect {\paren {z_t - \mu}^2} \expect {\paren {z_{t + k} - \mu}^2} } }$
The autocovariance is defined as:
:$\gamma_k := \expect {\paren {z_t - \mu} \paren {z_{t - k} - \mu} }$
Hence:
:$\rho_k := \dfrac {\gamma_k} {\sqrt {\expect {\paren {z_t - \mu}^2} \expect {\paren {z_{t + k} - \mu}^2} } }$
Then we have that for a strictly stationary stochastic process:
:$\expect {\paren {z_t - \mu}^2} = \sigma_z^2$
where $\sigma_z^2$ is the variance of $S$ and, for a strictly stationary stochastic process, is constant.
Thus:
:$\rho_k := \dfrac {\gamma_k} {\sigma_t^2}$
Then from Autocovariance at Zero Lag for Strictly Stationary Stochastic Process is Variance:
:$\sigma_z^2 = \gamma_0$
Hence:
:$\rho_k = \dfrac {\gamma_k} {\gamma_0}$
== Sources ==
* :
::Part $\text {I}$: Stochastic Models and their Forecasting:
:::$2$: Autocorrelation Function and Spectrum of Stationary Processes:
::::$2.1$ Autocorrelation Properties of Stationary Models:
:::::$2.1.2$ Stationary Stochastic Processes: Autocovariance and autocorrelation coefficients: $(2.1.6)$
Category:Autocorrelation
Category:Stationary Stochastic Processes
Autocovariance Matrix for Stationary Process is Variance by Autocorrelation Matrix

== Theorem ==
Let $S$ be a strictly stationary stochastic process giving rise to a time series $T$.
Let $\sequence {s_n}$ be a sequence of $n$ successive values of $T$:
:$\sequence {s_n} = \tuple {z_1, z_2, \dotsb, z_n}$
Let $\boldsymbol \Gamma_n$ denote the autocovariance matrix associated with $S$ for $\sequence {s_n}$.
Let $\mathbf P_n$ denote the autocorrelation matrix associated with $S$ for $\sequence {s_n}$.
Then:
:$\boldsymbol \Gamma_n = \sigma_z^2 \mathbf P_n$
where $\sigma_z^2$ denotes the variance of $S$.
== Proof ==
From Autocorrelation of Strictly Stationary Stochastic Process:
:$\rho_k = \dfrac {\gamma_k} {\gamma_0}$
Then from Autocovariance at Zero Lag for Strictly Stationary Stochastic Process is Variance:
:$\gamma_0 = \sigma_z^2$
Hence:
:$\gamma_k = \sigma_z^2 \rho_k$
and the result follows.
== Sources ==
* :
::Part $\text {I}$: Stochastic Models and their Forecasting:
:::$2$: Autocorrelation Function and Spectrum of Stationary Processes:
::::$2.1$ Autocorrelation Properties of Stationary Models:
:::::$2.1.3$ Positive Definiteness and the Autocovariance Matrix
Category:Autocovariance Matrices
Category:Autocorrelation Matrices
Autocovariance Matrix is Positive Definite

== Theorem ==
Let $S$ be a strictly stationary stochastic process giving rise to a time series $T$.
Let $\sequence {s_n}$ be a sequence of $n$ successive values of $T$:
:$\sequence {s_n} = \tuple {z_1, z_2, \dotsb, z_n}$
Let $\boldsymbol \Gamma_n$ denote the autocovariance matrix associated with $S$ for $\sequence {s_n}$.
Then $\boldsymbol \Gamma_n$ is a positive definite matrix.
== Proof ==
== Sources ==
* :
::Part $\text {I}$: Stochastic Models and their Forecasting:
:::$2$: Autocorrelation Function and Spectrum of Stationary Processes:
::::$2.1$ Autocorrelation Properties of Stationary Models:
:::::$2.1.3$ Positive Definiteness and the Autocovariance Matrix
Category:Autocovariance Matrices
Autocovariance at Zero Lag for Strictly Stationary Stochastic Process is Variance

== Example of Strictly Stationary Stochastic Process ==
Let $S$ be a strictly stationary stochastic process giving rise to a time series $T$.
Then the autocovariance at zero lag is given by:
:$\gamma_0 = \sigma_z^2$
where $\sigma_z^2$ is the variance of $S$.
== Proof ==
By definition, the autocovariance of $S$ at lag $k$ is defined as:
:$\gamma_k := \cov {z_t, z_{t + k} } = \expect {\paren {z_t - \mu} \paren {z_{t - k} - \mu} }$
where:
:$z_t$ is the observation at time $t$
:$\mu$ is the mean of $S$
:$\expect \cdot$ is the expectation.
For a strictly stationary stochastic process:
:$\expect {\paren {z_t - \mu}^2} = \sigma_z^2$
where:
:$\mu$ is the constant mean level of $S$
:$\expect {\paren {z_t - \mu}^2}$ is the expectation of $\paren {z_t - \mu}^2$
:$\sigma_z^2$ is the variance of $S$ and, for a strictly stationary stochastic process, is constant.
Hence we have that:
== Sources ==
* :
::Part $\text {I}$: Stochastic Models and their Forecasting:
:::$2$: Autocorrelation Function and Spectrum of Stationary Processes:
::::$2.1$ Autocorrelation Properties of Stationary Models:
:::::$2.1.2$ Stationary Stochastic Processes: Autocovariance and autocorrelation coefficients
Category:Stationary Stochastic Processes
Category:Autocovariance
Autocovariance is Autocorrelation by Variance

== Theorem ==
Let $\map S z$ be a stochastic process giving rise to a time series $T$.
Let $S$ be weakly stationary of order $2$ or greater.
Let $\gamma_k$ denote the autocovariance coefficient of $S$ at lag $k$.
Let $\rho_k$ denote the autocorrelation coefficient of $S$ at lag $k$.
Then:
:$\gamma_k = \rho_k \sigma_z^2$
where $\sigma_z^2$ denotes the variance of $S$.
== Proof ==
Hence the result.
== Sources ==
* :
::Part $\text {I}$: Stochastic Models and their Forecasting:
:::$2$: Autocorrelation Function and Spectrum of Stationary Processes:
::::$2.1$ Autocorrelation Properties of Stationary Models:
:::::$2.1.4$ Autocovariance and Autocorrelation Functions
Category:Autocovariance
Category:Autocorrelation
Automorphic Numbers in Base 10

== Theorem ==
If leading zeroes are allowed, there are exactly $4$ $n$-digit automorphic numbers in base $10$:
:$00 \dots 00$
:$00 \dots 01$
:$5^{2^{n - 1} } \pmod {10^n}$
:$6^{5^{n - 1} } \pmod {10^n}$
== Proof ==
The proof proceeds by induction on $n$.
For all $n \in \Z_{> 0}$, let $\map P n$ be the proposition:
:There are exactly $4$ $n$-digit automorphic numbers of the forms above.
=== Basis for the Induction ===
For $n = 1$:
$0, 1, 5, 6$ are the only $1$-digit automorphic numbers, and we have:
:$5^{2^0} = 5$
:$6^{5^0} = 6$
Thus $\map P 1$ is seen to hold.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
We assume that for some $k \ge 1$, there are exactly $4$ $k$-digit automorphic numbers of the forms above.
=== Induction Step ===
This is the induction step:
We aim to construct $x$, a $\paren {k + 1}$-digit automorphic number.
By Left-Truncated Automorphic Number is Automorphic, after removing the leftmost digit, what remains is a $k$-digit automorphic number.
Write $x = 10^k a + b$, where $a$ is the leftmost digit of $x$.
We have:
By Left-Truncated Automorphic Number is Automorphic, $b$ must end in $0, 1, 5, 6$.
We see that:
:$a \equiv N \pmod {10}$ for $b$ ending in $0$ or $5$;
:$a \equiv -N \pmod {10}$ for $b$ ending in $1$ or $6$;
So the choice of $x$ is completely determined by $b$.
This shows that there are exactly $4$ $\paren {k + 1}$-digit automorphic numbers.
Now we show that the $4$ numbers are indeed of the forms given above.
For any $n \in \Z_{>0}$, we have:
Since $2^{n - 1} \ge n$ and the each factor on the right is divisible by $2$, the above product is divisible by $5^n 2^n = 10^n$.
Since $5^{n - 1} \ge n$, the above product is divisible by $2^n 5^n = 10^n$.
Hence for each number above:
:$x^2 \equiv x \pmod {10^n}$
showing they are indeed automorphic.
So $\map P k \implies \map P {k + 1}$ and thus it follows by the Principle of Mathematical Induction that these are all the automorphic numbers.
Category:Automorphic Numbers
Automorphic Numbers with 10 Digits

== Theorem ==
The only $10$-digit automorphic numbers are:
:$1 \, 787 \, 109 \, 376$
:$8 \, 212 \, 890 \, 625$
== Proof ==
We have:
thus demonstrating they are automorphic.
By Automorphic Numbers in Base 10, there are no others.
== Sources ==
* : $1,787,109,376$
* : $1,787,109,376$
Category:Automorphic Numbers
Automorphic Numbers with 5 Digits

== Theorem ==
The only $5$-digit automorphic number which does not begin with a zero is $90 \, 625$.
== Proof ==
We have:
:$90 \, 625^2 = 8 \, 212 \, 8 \mathbf {90 \, 625}$
thus demonstrating it is automorphic.
By Automorphic Numbers in Base 10, the only other possible candidate is $6^{5^4}$.
However:
:$6^{5^4} \equiv 09 \, 376 \pmod {10^5}$
begins with a zero.
Hence there are no others.
== Sources ==
* : $90,625$
* : $90,625$
Category:Automorphic Numbers
Automorphism Group/Examples/Cyclic Group C3

== Example of Automorphism Group ==
Consider the cyclic group $C_3$, which can be presented as its Cayley table:
The automorphism group of $C_3$ is given by:
:$\Aut {C_3} = \set {\phi, \theta}$
where $\phi$ and $\theta$ are defined as:
The Cayley table of $\Aut {C_3}$ is then:
:$\begin{array}{r|rr}
 & \phi & \theta \\
\hline
\phi & \phi & \theta
\\
\theta & \theta & \phi
\\
\end{array}$
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $8$: The Homomorphism Theorem: Exercise $5$
Category:Examples of Automorphism Groups
Category:Cyclic Group of Order 3
Automorphism Group/Examples/Cyclic Group C3/Proof 1

== Example of Automorphism Group ==
== Proof ==
Let $\xi$ be a general automorphism on $C_3$.
Then by Group Homomorphism Preserves Identity we immediately have that:
:$\map \xi {\eqclass 0 3} = \eqclass 0 3$
Investigating $\map \xi {\eqclass 1 3}$, we find $2$ options:
:$\map \xi {\eqclass 1 3} = \eqclass 1 3$
:$\map \xi {\eqclass 1 3} = \eqclass 2 3$
Each leads to one and only one bijection from $C_3$ to $C_3$, that is, $\phi$ and $\theta$ as defined.
It is determined by inspection that both $\phi$ and $\theta$ are automorphisms.
Hence Automorphism Group is Subgroup of Symmetric Group is applied to confirm that $\set {\phi, \theta}$ forms a group.
The Cayley table follows.
Category:Examples of Automorphism Groups
Category:Cyclic Group of Order 3
Automorphism Group/Examples/Cyclic Group C3/Proof 2

== Example of Automorphism Group ==
== Proof ==
This is an example of Order of Automorphism Group of Prime Group:
:$\order {\Aut G} = p - 1$
for a group of prime order $p$.
The only group of order $2$ is the cyclic group of order $2$.
The result follows.
Category:Examples of Automorphism Groups
Category:Cyclic Group of Order 3
Automorphism Group/Examples/Cyclic Group C8

== Example of Automorphism Group ==
The automorphism group of the cyclic group $C_8$ is the Klein $4$-group.
== Proof ==
The cyclic group $C_8$ is isomorphic to the underlying group of the ring $\Z_8$.
So the elements of $C_8$ can be denoted by $\eqclass 0 8, \eqclass 1 8, \ldots, \eqclass 7 8$.
By Power of Generator of Cyclic Group is Generator iff Power is Coprime with Order, the generators of $C_8$ are exactly $\eqclass 1 8, \eqclass 3 8, \eqclass 5 8, \eqclass 7 8$.
By Automorphism Maps Generator to Generator and Homomorphism of Generated Group, there are exactly $4$ automorphisms for the group $C_8$, namely:
:$\phi_1: \eqclass 1 8 \mapsto \eqclass 1 8$
:$\phi_3: \eqclass 1 8 \mapsto \eqclass 3 8$
:$\phi_5: \eqclass 1 8 \mapsto \eqclass 5 8$
:$\phi_7: \eqclass 1 8 \mapsto \eqclass 7 8$
By direct computation, the Cayley table for these $4$ elements can be presented as:
:$\begin{array}{c|cccc}
 & \phi_1 & \phi_3 & \phi_5 & \phi_7 \\
\hline
\phi_1 & \phi_1 & \phi_3 & \phi_5 & \phi_7 \\
\phi_3 & \phi_3 & \phi_1 & \phi_7 & \phi_5 \\
\phi_5 & \phi_5 & \phi_7 & \phi_1 & \phi_3 \\
\phi_7 & \phi_7 & \phi_5 & \phi_3 & \phi_1 \\
\end{array}$
which is the same as the Cayley table of the Klein $4$-group.
This shows that the automorphism group of $C_8$ is isomorphic to the Klein $4$-group.
== Sources ==
* : Chapter $8$: Homomorphisms, Normal Subgroups and Quotient Groups: Exercise $27 \ \text {(ii)}$
Category:Examples of Automorphism Groups
Category:Cyclic Group of Order 8
Automorphism Group/Examples/Infinite Cyclic Group

== Example of Automorphism Group ==
The automorphism group of the infinite cyclic group $\Z$ is the cyclic group of order $2$.
== Proof ==
Let $g$ be a generator of $\Z$.
Let $\varphi$ be an automorphism on $\Z$.
By Homomorphic Image of Cyclic Group is Cyclic Group, $\map \varphi g$ is a generator of $\Z$.
By Homomorphism of Generated Group, $\varphi$ is uniquely determined by $\map \varphi g$.
By Generators of Infinite Cyclic Group, there are $2$ possible values for $\map \varphi g$.
Therefore there are $2$ automorphisms on $\Z$:
:$\order {\Aut \Z} = 2$
By Prime Group is Cyclic, the automorphism group of the infinite cyclic group $\Z$ is the cyclic group of order $2$.
== Sources ==
* : Chapter $8$: Homomorphisms, Normal Subgroups and Quotient Groups: Exercise $27 \ \text {(iii)}$
Category:Examples of Automorphism Groups
Category:Infinite Cyclic Group
Automorphism Group/Examples/Klein Four-Group

== Example of Automorphism Group ==
The automorphism group of the Klein $4$-group is the Symmetric Group on 3 Letters $S_3$.
== Proof ==
Let $f:V\rightarrow V$ be a permutation that fixes the identity element.
By definition, this map is bijective.
$V$ has the property that the product of any 2 distinct non-identity elements is the 3rd non-identity element.
This implies:
:$\map f {vv'} = \map f v \cdot \map f {v'}$
for all $v, v' \in V$ with $v, v' \ne e$.
For products involving the identity element:
:$\map f {e v} = \map f e \cdot \map f v = e \cdot \map f v = \map f v$
for all $v \in V$.
Also, since all non-identity elements of $V$ have order $2$:
:$\map f {v^2} = \map f {v v} = \map f v \cdot \map f v = e$
for all $v \in V$.
Hence, any permutation of the elements of $V$ that fixes the identity element is an automorphism of $V$.
Since these permutations exhaust all possible automorphisms of $V$, they are the elements of $\Aut V$.
Since the $3! = 6$ elements of $\Aut V$ represent all permutations of $3$ objects, $\Aut V$ is isomorphic to $S_3$.
== Sources ==
* : Chapter $2$: Group Homomorphism and Isomorphism: $\S 64 \zeta$
* 
Category:Klein Four-Group
Category:Examples of Automorphism Groups
Automorphism Group Acts Faithfully on Generating Set

== Theorem ==
Let $E/F$ be a field extension.
Let $\operatorname{Aut}(E/F)$ be its automorphism group.
Let $S\subset E$ be a generating set of the extension.
Let $S$ be stable under the group action of $\operatorname{Aut}(E/F)$.
Then the induced group action on $S$ is faithful.
== Proof ==
Let $\sigma \in \operatorname{Aut}(E/F)$ stabilize $S$.
Then $S$ is contained in the fixed field of $\sigma$.
By definition of generating set, $\sigma$ fixes $E$.
Thus $\operatorname{Aut}(E/F)$ acts faithfully.
Category:Field Extensions
Automorphism Group is Subgroup of Symmetric Group

== Theorem ==
Let $\struct {S, *}$ be an algebraic structure.
Let $\Aut S$ be the automorphism group of $\struct {S, *}$.
Then $\Aut S$ is a subgroup of the symmetric group $\struct {\Gamma \paren S, \circ}$ on $S$.
== Proof ==
An automorphism is an isomorphism $\phi: S \to S$ from an algebraic structure $S$ to itself.
The Identity Mapping is Automorphism, so $\Aut S$ is not empty.
The composite of isomorphisms is itself an isomorphism, as demonstrated on Isomorphism is Equivalence Relation.
So:
:$\phi_1, \phi_2 \in \Aut S \implies \phi_1 \circ \phi_2 \in \Aut S$
demonstrating closure.
If $\phi \in \Aut S$, then $\phi$ is bijective and an isomorphism.
Hence from Inverse of Algebraic Structure Isomorphism is Isomorphism, $\phi^{-1}$ is also bijective and an isomorphism.
So $\phi^{-1} \in \Aut S$.
The result follows by the Two-Step Subgroup Test.
== Sources ==
* : $\S 7.2$. Some lemmas on homomorphisms: Example $134$
* : Chapter $\text {II}$: New Structures from Old: $\S 8$: Compositions Induced on Subsets: Theorem $8.6$
* : Chapter $\text{II}$: Groups: Problem $\text{AA}$
* : Chapter $2$: Group Homomorphism and Isomorphism: $\S 64 \alpha$
* : $\S 1.2$
* : Chapter $8$: Homomorphisms, Normal Subgroups and Quotient Groups: Exercise $24$
* : Chapter $8$: The Homomorphism Theorem: Proposition $8.11$
* 
* 
Category:Examples of Groups
Category:Automorphism Groups
Category:Group Automorphisms
Automorphism Group of Complex Numbers over Real Numbers

== Theorem ==
The field extension $\C / \R$ of complex numbers $\C$ over real numbers $\R$ has automorphism group $\operatorname{Aut}$:
:$\operatorname{Aut} \paren {\C / \R} = \set {\operatorname{id}, \sigma}$
where:
:$\operatorname{id}$ denotes the identity mapping
:$\sigma$ denotes complex conjugation
== Proof ==
Category:Field Extensions
Automorphism Group of Cyclic Group is Abelian

== Theorem ==
Let $G$ be a cyclic group.
Let $\Aut G$ denote the automorphism group of $G$.
Then $\Aut G$ is abelian.
== Proof ==
Let $G = \gen g$
Let $\phi, \psi \in \Aut G$.
As $G$ is cyclic:
Thus:
Thus in particular, $\phi \circ \psi$ and $\psi \circ \phi$ are equal on the generator $g$.
Since $g$ generates $G$, they must be equal as automorphisms.
== Sources ==
* : Chapter $8$: Homomorphisms, Normal Subgroups and Quotient Groups: Exercise $27$
Category:Cyclic Groups
Category:Automorphism Groups
Automorphism Maps Generator to Generator

== Theorem ==
Let $G$ be a cyclic group.
Let $g$ be a generator of $G$.
Let $\phi$ be an automorphism on $G$.
Then $\map \phi g$ is also a generator of $G$.
== Proof ==
By definition of automorphism, $\phi$ is a homomorphism
It follows that this result is a specific instance of Homomorphic Image of Cyclic Group is Cyclic Group.
Category:Cyclic Groups
Category:Group Automorphisms
Autoregressive Model is Special Case of Linear Filter Model

== Theorem ==
Let $S$ be a stochastic process based on an equispaced time series.
Let the values of $S$ at timestamps $t, t - 1, t - 2, \dotsc$ be $z_t, z_{t - 1}, z_{t - 2}, \dotsc$
Let $\tilde z_t, \tilde z_{t - 1}, \tilde z_{t - 2}, \dotsc$ be deviations from a constant mean level $\mu$:
:$\tilde z_t = z_t - \mu$
Let $a_t, a_{t - 1}, a_{t - 2}, \dotsc$ be a sequence of independent shocks at timestamps $t, t - 1, t - 2, \dotsc$
Let $M$ be an autoregessive model on $S$ of order $p$:
:$(1): \quad \tilde z_t = \phi_1 \tilde z_{t - 1} + \phi_2 \tilde z_{t - 2} + \dotsb + \phi_p \tilde z_{t - p} + a_t$
Then $M$ is a special case of a linear filter model.
</onlyinclude>
== Proof ==
We can eliminate $\tilde z_{t - 1}$ from the of $(1)$ by substituting:
:$\tilde z_{t - 1} = \phi_1 \tilde z_{t - 2} + \phi_2 \tilde z_{t - 3} + \dotsb + \phi_p \tilde z_{t - p - 1} + a_{t - 1}$
Similarly we can substitute for $\tilde z_{t - 2}$, and so on.
Eventually we get an infinite series in $a_{t - j}$.
Hence:
:$\map \phi B \tilde z_t = a_t$
is equivalent to:
:$\tilde z_t = \map \psi B a_t$
such that:
:$\map \psi B = \map {\phi^{-1} } B$
Hence the result by definition of linear filter model.
== Sources ==
* :
::$1$: Introduction:
:::$1.2$ Stochastic and Deterministic Dynamic Mathematical Models
::::$1.2.1$ Stationary and Nonstationary Stochastic Models for Forecasting and Control: Autoregressive models
Category:Autoregressive Models
Auxiliary Angle/Examples/3 cos x minus 2 sin x

== Example of Auxiliary Angle ==
:$3 \cos x - 2 \sin x = \sqrt {13} \map \cos {x + \arctan \dfrac 2 3}$
Hence the greatest value of $3 \cos x - 2 \sin x$ is $\sqrt {13}$ which happens when $x = -\arctan \dfrac 2 3$.
== Proof ==
From Multiple of Sine plus Multiple of Cosine: Cosine Form:
:
From the diagram:
where:
:$\tan \alpha = \dfrac 2 3$
The result follows.
== Sources ==
* : $\text V$. Trigonometry: The auxiliary angle: Example
Category:Auxiliary Angles
Auxiliary Angle/Examples/3 cos x minus 2 sin x equals 1

== Example of Auxiliary Angle ==
Consider the equation:
:$(1): \quad 3 \cos x - 2 \sin x = 1$
The solutions to $(1)$ between $0 \degrees$ and $360 \degrees$ are:
== Proof ==
From Multiple of Sine plus Multiple of Cosine: Cosine Form:
Hence the auxiliary angle form of $(1)$ is given by:
== Sources ==
* : $\text V$. Trigonometry: Solution of equations: Example
Category:Auxiliary Angles
Auxiliary Approximating Relation has Quasi Interpolation Property

== Theorem ==
Let $L = \struct {S, \vee, \wedge, \preceq}$ be a complete lattice.
Let $x, z \in S$.
Let $\RR$ be an auxiliary approximating relation on $S$ such that
:$\tuple {x, z} \in \RR \land x \ne z$
Then:
:$\exists y \in S: x \preceq y \land \tuple {y, z} \in \RR \land x \ne y$
== Proof ==
By definition of auxiliary relation:
:$x \preceq z$
By definition of $\prec$:
:$x \prec z$
By definition of antisymmetry:
:$z \nprec x$
Then:
:$z \npreceq x$
By Not Preceding implies Approximating Relation and not Preceding:
:$\exists u \in S: \tuple {u, z} \land u \npreceq x$
Define $y = x \vee u$.
Thus by Join Succeeds Operands:
:$x \preceq y$
Thus by definition of auxiliary relation:
:$\tuple {y, z} \in \RR$
Thus by Preceding iff Join equals Larger Operand:
:$x \ne y$
== Sources ==
* 
* 
Category:Approximating Relations
Category:Complete Lattices
Category:Auxiliary Relations
Auxiliary Relation Image of Element is Upper Section

== Theorem ==
Let $L = \struct {S, \preceq}$ be an ordered set.
Let $R$ be an auxiliary relation on $S$.
Let $x \in S$.
Then $\map R x$ is an upper section
where $\map R x$ denotes the image of $x$ under $R$.
== Proof ==
Let $a \in \map R x, b \in S$ such that
:$a \preceq b$
By definition of $R$-image of element:
:$\tuple {x, a} \in R$
By definition of reflexivity:
:$x \preceq x$
By definition of auxiliary relation:
:$\tuple {x, b} \in R$
Thus by definition of $R$-image of element:
:$b \in \map R x$
== Sources ==
* 
Category:Auxiliary Relations
Auxiliary Relation is Congruent

== Theorem ==
Let $\struct {S, \vee, \preceq}$ be a bounded below join semilattice.
Let $\RR$ be relation on $S$ satisfying conditions $(2)$ and $(3)$ of auxiliary relation.
Then:
:$\forall x, y, z, u \in S: \tuple {x, z} \in \RR \land \tuple {y, u} \in \RR \implies \tuple {x \vee y, z \vee u} \in \RR$
== Proof ==
Let $x, y, z, u \in S$ such that:
:$\tuple {x, z} \in \RR \land \tuple {y, u} \in \RR$
By definition of reflexivity:
:$x \preceq x$ and $y \preceq y$
By Join Succeeds Operands:
:$z \preceq z \vee u$ and $u \preceq z \vee u$
By condition $(2)$ of auxiliary relation:
:$\tuple {x, z \vee u} \in \RR$ and $\tuple {y, z \vee u} \in \RR$
Thus by condition $(3)$ of auxiliary relation:
:$\tuple {x \vee y, z \vee u} \in \RR$
== Sources ==
* 
* 
Category:Auxiliary Relations
Auxiliary Relation is Transitive

== Theorem ==
Let $\struct {S, \vee, \preceq}$ be a bounded below join semilattice.
Let $\RR$ be relation on $S$ satisfying conditions $(1)$ and $(2)$ of auxiliary relation.
Then
:$\RR$ is a transitive relation.
== Proof ==
Let $x, y, z \in S$ such that
:$\tuple {x, y} \in \RR \land \tuple {y, z} \in \RR$
By definition of reflexivity:
:$z \preceq z$ 
By condition $(1)$ of auxiliary relation:
:$x \preceq y$
Thus by condition $(2)$ of auxiliary relation:
:$\tuple {x, z} \in \RR$
Thus by definition
:$\RR$ is a transitive relation.
== Sources ==
* 
* 
* 
Category:Auxiliary Relations
Axes of Symmetry for Ellipse

== Theorem ==
Let $K$ be an ellipse.
Then $K$ has exactly $2$ axis of bilateral symmetry:
:the major axis of $K$
:the minor axis of $K$.
== Proof ==
From:
:Major Axis of Ellipse is Axis of Symmetry
:Minor Axis of Ellipse is Axis of Symmetry
we have that these diameters of $K$ are in fact axis of bilateral symmetry.
It remains to be shown that there are no more.
== Sources ==
* 
* 
Category:Axes of Symmetry for Ellipse
Category:Ellipses
Axes of Symmetry for Ellipsoid

== Theorem ==
Let $\EE$ be an ellipsoid.
Then $\EE$ has exactly $3$ axes of symmetry:
:the major axis of $\EE$
:the mean axis of $\EE$
:the minor axis of $\EE$.
== Proof ==
From:
:Major Axis of Ellipsoid is Axis of Symmetry
:Mean Axis of Ellipsoid is Axis of Symmetry
:Minor Axis of Ellipsoid is Axis of Symmetry
we have that these diameters of $\EE$ are in fact axes of symmetry.
It remains to be shown that there are no more.
== Sources ==
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Category:Axes of Symmetry for Ellipsoid
Category:Ellipsoids