nyuuzyou/wikis huggingface dataset:
Duality Principle (Order Theory)/Global Duality

== Theorem ==
Let $\Sigma$ be a statement about ordered sets (whether in natural or a formal language).
Let $\Sigma^*$ be the dual statement of $\Sigma$.
The following are equivalent:
:$(1): \quad \Sigma$ is true for all ordered sets
:$(2): \quad \Sigma^*$ is true for all ordered sets
== Proof ==
=== $(1)$ implies $(2)$ ===
Let $\struct {S, \preceq}$ be an ordered set.
Let $\struct {S, \succeq}$ be its dual.
By assumption, $\Sigma$ is true for $\struct {S, \succeq}$.
By Local Duality, this implies $\Sigma^*$ is true for $\struct {S, \preceq}$.
Since $\struct {S, \preceq}$ was arbitrary, the result follows.
=== $(2)$ implies $(1)$ ===
From Dual of Dual Statement (Order Theory):
:$\paren {\Sigma^*}^* = \Sigma$
The result thus follows from applying the other implication to $\Sigma^*$.
== Also see ==
* Definition:Dual Statement (Order Theory)
* Duality Principle (Category Theory), a more general duality principle.
Category:Order Theory
Category:Named Theorems
Duality Principle (Order Theory)/Local Duality

== Theorem ==
Let $\Sigma$ be a statement about ordered sets (whether in natural or a formal language).
Let $\Sigma^*$ be the dual statement of $\Sigma$.
Let $\struct {S, \preceq}$ be an ordered set, and let $\struct {S, \succeq}$ be its dual.
Then the following are equivalent:
:$(1): \quad \Sigma$ is true for $\struct {S, \preceq}$
:$(2): \quad \Sigma^*$ is true for $\struct {S, \succeq}$
== Proof ==
=== $(1)$ implies $(2)$ ===
By assumption, $\Sigma$ is true for $\struct {S, \preceq}$.
By Dual of Dual Ordering, the dual statement $\Sigma^*$ applied to $\struct {S, \succeq}$ is the same as $\Sigma$ applied to $\struct {S, \preceq}$.
Hence $\Sigma^*$ is true for $\struct {S, \succeq}$.
=== $(2)$ implies $(1)$ ===
From Dual of Dual Statement (Order Theory), $\paren {\Sigma^*}^* = \Sigma$.
From Dual of Dual Ordering, $\struct {S, \preceq}$ is the dual of $\struct {S, \succeq}$.
The result thus follows from applying the other implication to $\Sigma^*$ and $\struct {S, \succeq}$.
== Also see ==
* Definition:Dual Statement (Order Theory)
* Duality Principle (Category Theory), a more general duality principle.
Category:Order Theory
Category:Named Theorems
Duality Principle for Sets

== Theorem ==
Any identity in set theory which uses any or all of the operations:
:Set intersection $\cap$
:Set union $\cup$
:Empty set $\O$
:Universal set $\mathbb U$
and none other, remains valid if:
:$\cap$ and $\cup$ are exchanged throughout
:$\O$ and $\mathbb U$ are exchanged throughout.
== Proof ==
Follows from:
* Algebra of Sets is Huntington Algebra
* Principle of Duality of Huntington Algebras
== Sources ==
* : $\S 5$: Complements and Powers
* : $\S 1.5$. The algebra of sets
Category:Named Theorems
Category:Set Theory
Duals of Isomorphic Ordered Sets are Isomorphic

== Theorem ==
Let $\struct {S, \preccurlyeq_1}$ and $\struct {T, \preccurlyeq_2}$ be ordered sets.
Let $\struct {S, \succcurlyeq_1}$ and $\struct {T, \succcurlyeq_2}$ be the dual ordered sets of $\struct {S, \preccurlyeq_1}$ and $\struct {T, \preccurlyeq_2}$ respectively.
Let $f: \struct {S, \preccurlyeq_1} \to \struct {T, \preccurlyeq_2}$ be an order isomorphism.
Then $f: \struct {S, \succcurlyeq_1} \to \struct {T, \succcurlyeq_2} $ is also an order isomorphism.
== Proof ==
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Theorem $14.2$
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Exercise $14.8$
Category:Order Isomorphisms
Category:Dual Orderings
Dudeney's Property of 2592

== Theorem ==
:$2592 = 2^5 \times 9^2$
It is the only number $n$ that has the property that:
:$n = \sqbrk {abcd} = a^b \times c^d$
where $\sqbrk {abcd}$ denotes the decimal representation of $n$.
== Proof ==
First we verify that $2592$ does indeed satisfy the given property.
It remains to be shown that this is the only such number.
Because $\sqbrk {abcd} = a^b \times c^d$, neither $a^b$ nor $c^d$ can have more than $4$ digits.
Hence, for each digit, the highest power is:
Neither $a$ or $c$ can be zero, or that would make $\sqbrk {abcd} = 0$.
Suppose $a = 1$ or $b = 0$.
Then:
:$\sqbrk {abcd} = c^d$
Apart from the above powers which have $4$ digits, we also have:
Hence, by inspection, it is seen that none of these fit the pattern $\sqbrk {1bcd}$ or $\sqbrk {a0cd}$.
Similarly, suppose $c = 1$ or $d = 0$.
Then:
:$\sqbrk {abcd} = a^b$
Again, by inspection, it is seen that none of the above $4$-digit powers fit the pattern $\sqbrk {ab1d}$ or $\sqbrk {abc0}$.
Suppose either $a^b$ or $c^d$ has $4$ digits.
Then the other is less than $10$, giving:
We try multiplying these by all the above $4$-digit powers such that their product is $4$ digits (there are not many).
First note that $1^1$ can be ruled out as none of these $4$-digit powers either begins or ends with $11$.
So none of the above $4$-digit powers multiplied by any of these single-digit powers satisfies the condition.
We have that $\sqbrk {abcd}$ cannot end in $0$, as that can also be ruled out by inspection of the $4$-digit powers.
Similarly, because $a \ne 1$, $\sqbrk {abcd} > 2000$.
Also, it cannot be the case that both $a^b$ and $c^d$ are smaller than $44$, as $44^2 < 2000$.
== Historical Note ==
== Sources ==
* : Various Arithmetical and Algebraical Problems: $115$. A Printer's Error
* 
* 
* : $2592$
* : $2592$
Category:Recreational Mathematics
Category:2592
Dudeney's Property of 2592/Historical Note

== Historical Note on Dudeney's Property of 2592 ==
This result is generally attributed to , who stated it in his $1917$ book .
It continues to crop up occasionally in puzzle pages of journals.
== Sources ==
* : Various Arithmetical and Algebraical Problems: $115$. A Printer's Error
* 
* 
* : $2592$
* : $2592$
Category:Historical Notes
Dummy Variable/Examples/Regression Equation

== Example of Dummy Variable ==
Consider the regression equation with the two explanatory variables $x_1$ and $x_2$:
:$y = \beta_0 + \beta_1 x_1 + \beta_2 x_2$
It is sometimes more convenient to write it as:
:$y = \beta_0 x_0 + \beta_1 x_1 + \beta_2 x_2$
where in this context $x_0$ is a dummy variable which always takes the value $1$.
Hence, if we also want to consider the regression model where $\beta_0 = 0$, this can be accommodated by setting the dummy variable $x_0$ equal to $0$.
== Sources ==
* 
* 
Category:Examples of Dummy Variables
Duodecimal Notation/Examples/Arbitrary Example 1

== Example of Duodecimal Notation ==
The number written in decimal notation as:
:$1139_{10}$
is expressed in duodecimal notation as:
:$7 \mathrm A \mathrm B_{12}$
where:
:$\mathrm A$ denotes $10$
:$\mathrm B$ denotes $11$.
== Proof ==
== Sources ==
* 
* 
Category:Examples of Duodecimal Notation
Duplicated Triangle Forms a Kite or a Parallelogram

== Theorem ==
Let a triangle be copied either by rotation or reflection.
Then if you mung them together by corresponding sides what you get is either a parallelogram or a kite.
== Proof ==
Let $\triangle ABC$ be any triangle. We can form a congruent triangle by reflection or translation.
=== Case $1$: Reflection ===
Reflect $\triangle ABC$ across one of its sides.
 draw the two triangles with the longest side shared. 
The resulting $\Box ABCD$ has two pairs of equal sides adjacent (it forms a kite).
=== Case $2$: Translation ===
Translate a second congruent copy of $\triangle ABC$, rotate it.
 draw the two triangles together with the longest side shared.
The resulting $\Box ABCD$ has two pairs of opposite sides equal.
By Quadrilateral is Parallelogram iff Both Pairs of Opposite Sides are Equal or Parallel:
:$\Box ABCD$ is a parallelogram
By definition of parallelogram:
:Both pairs of opposite angles in $\Box ABCD$ are equal
=== Special case: Right Triangle ===
Consider the special case where $\triangle ABC$ is a right triangle with $\angle ABC$ and $\angle ABD$ both right angles.
Then $\Box ABCD$ has opposite sides parallel and both pairs of opposite angles equal.
Since one pair of angles is a pair of right angles:
:the other pair consists of two right angles
Therefore by the definition of rectangle:
:$\Box ABCD$ is a rectangle.
Category:Triangles
Category:Quadrilaterals
Category:Parallelograms
Category:Kites
Duplication Formula for Weierstrass's Elliptic Function

== Theorem ==
Let $\omega_1$, $\omega_2$ be non-zero complex constants with $\dfrac {\omega_1} {\omega_2}$ having a positive imaginary part.
Let $z$ be a complex number where $z \notin \set {2 m \omega_1 + 2 n \omega_2: \tuple {n, m} \in \Z^2}$.
Then:
:$\map \wp {2 z; \omega_1, \omega_2} = \dfrac 1 4 \paren {\dfrac {\map {\wp} {z; \omega_1, \omega_2} } {\map {\wp'} {z; \omega_1, \omega_2} } }^2 - 2 \map \wp {z; \omega_1, \omega_2}$
where:
:$\wp$ is Weierstrass's elliptic function 
:$\wp'$ and $\wp$ denote its first and second derivative with respect to $z$.
== Proof ==
== Sources == 
* : $20.311$: The duplication formula for $\wp \left({z}\right)$
Category:Weierstrass's Elliptic Function
Dx - k is Hypoelliptic

== Theorem ==
Let $f \in \map {C^\infty} \R$ be a smooth real function.
Let $T \in \map {\DD'} \R$ be a distribution.
Let $T_f$ be the distribution associated with $f$.
Let $k \in \R$.
Suppose in the distributional sense it holds that:
:$\paren {\dfrac \d {\d x} - k} T = T_f \quad (1)$
Then there is a $c \in \R$ such that:
:$T = T_{F + c \map \exp {k x} }$
where $F \in \map {C^\infty} \R$ is a solution to:
:$\paren {\dfrac \d {\d x} - k} F = f$
In other words, $\dfrac \d {\d x} - k$ is a hypoelliptic operator.
== Proof ==
By assumption $f \in \map {C^\infty} \R$.
By Solution to Linear First Order ODE with Constant Coefficients:
:$\exists F \in \map {C^\infty} \R : \paren {\dfrac \d {\d x} - k} F = f$
By Differentiable Function as Distribution and multiplication of distribution by a smooth function:
:$\exists F \in \map {C^\infty} \R : \paren {\dfrac \d {\d x} - k} T_F = T_f \quad (2)$
Subtract $(2)$ from $(1)$:
where $\mathbf 0$ is the zero distribution.
This is solved by:
:$T - T_F = T_{c \map \exp {k x} }$
or:
:$T = T_{F + c \map \exp {k x} }$
where $c \in \C$.
Finally:
:$\ds \paren {\dfrac \d {\d x} - k} \paren {F + c \map \exp {k x} } = f$
== Sources ==
* : Chapter $\S 6.4$: A glimpse of distribution theory. Multiplication by $C^\infty$ functions
Category:Examples of Hypoelliptic Operators
Category:Examples of Distributional Solutions
Dynamic Programming/Examples/Arbitrary Example 1

== Example of Dynamic Programming ==
The following diagram represents the problem of minimizing the fare for a $3$-stage bus journey from $A$ to $D$:
:
$B_1$, $B_2$ and $B_3$ are the possible end points of the first stage of the journey.
$C_1$ and $C_2$ are the possible end points of the second stage of the journey.
The number on each line represents the bus fare for that route.
The problem is to find the route from $A$ to $D$ which has the lowest fare.
The dynamic programming approach first finds the optimum one-stage strategy for the final (third) stage, from either $C_1$ or $C_2$ to $D$.
This is trivially found to be $C_2$, as there is only one route from each.
The next step is to find the optimum two-stage strategy from $B_1$, $B_2$ and $B_3$ to $D$.
Here it can be found by inspection that:
:from $B_1$ to $D$, $C_1$ is the optimal route, which costs $7$, as opposed to $10$ via $C_2$
:from $B_2$ to $D$, $C_1$ is the optimal route, which costs $11$, as opposed to $12$ via $C_2$
:from $B_3$ to $D$, $C_2$ is the optimal route, which costs $13$, as opposed to $19$ via $C_1$.
Now we note that no matter how many additional earlier stages we add, the above routes are always optimal once we have arrived at one of $B_1$, $B_2$ and $B_3$.
Hence we see that from $A$, the three-stage total fare is:
:via $B_1$: $10 + 7 = 17$
:via $B_2$: $5 + 11 = 16$
:via $B_3$: $5 + 13 = 18$.
Hence the optimum route is via $B_2$ and $C_1$.
== Sources ==
* 
* 
Category:Examples of Dynamic Programming
Dynamical System/Examples/Arbitrary Example 1

== Example of Dynamical System ==
Let $T: \C \to \C$ be the mapping on the complex plane defined as:
:$\forall z \in \C: \map T z = z^2 - 1$
Hence $T$ and the iteration $z_{n + 1} = {z_n}^2 - 1$ together define a dynamical system in the complex plane.
== Sources ==
* 
* 
Category:Examples of Dynamical Systems
Dynkin System Closed under Disjoint Union

== Theorem ==
Let $X$ be a set, and let $\DD$ be a Dynkin system on $X$.
Let $D, E \in \DD$ be disjoint.
Then the union $D \cup E$ is also an element of $\DD$.
== Proof ==
Define $D_1 = D, D_2 = E$, and for $n \ge 2$, $D_n = \O$.
Then by Dynkin System Contains Empty Set:
:$\forall n \in \N: D_n \in \DD$
Also, by Intersection with Empty Set, it follows that $\sequence {D_n}_{n \mathop \in \N}$ is a pairwise disjoint sequence.
Hence, by property $(3)$ of a Dynkin system:
:$\ds D \cup E = \bigcup_{n \mathop \in \N} D_n \in \DD$
== Sources ==
* : $5.2$, $\S 5$: Problem $1$
Category:Dynkin Systems
Dynkin System Closed under Intersections is Sigma-Algebra

== Theorem ==
Let $X$ be a set, and let $\DD$ be a Dynkin system on $X$.
Suppose that $\DD$ satisfies the following condition:
:$(1):\quad \forall D, E \in \DD: D \cap E \in \DD$
That is, $\DD$ is closed under intersection.
Then $\DD$ is a $\sigma$-algebra.
== Proof ==
The first two conditions for a Dynkin system are identical to those for a $\sigma$-algebra.
Hence it is only required to verify that $(1)$ implies that $\DD$ is closed under arbitrary countable unions.
So let $\sequence {D_n}_{n \mathop \in \N}$ be a sequence in $\DD$.
Now define the sequence $\sequence {E_n}_{n \mathop \in \N}$ by:
:$\ds E_n := D_n \cap \paren {X \setminus \bigcup_{m \mathop < n} D_m}$
By Dynkin System Closed under Union and $(1)$, it follows that $E_n \in \DD$ for all $n \in \N$.
=== Lemma ===
For all $n \in \N$, it holds that:
:$\ds \bigcup_{k \mathop = 0}^n E_k = \bigcup_{k \mathop = 0}^n D_k$
=== Proof of Lemma ===
The lemma, combined with the definition of the $E_n$, gives immediately that for all $n \in \N$:
:$\ds E_n \in X \setminus \bigcup_{m \mathop < n} D_m = X \setminus \bigcup_{m \mathop < n} E_m$
whence the $E_n$ are pairwise disjoint.
Another consequence is that:
:$\ds \bigcup_{n \mathop \in \N} D_n = \bigcup_{n \mathop \in \N} E_n$
Now since the $E_n$ are pairwise disjoint, it follows that:
:$\ds \bigcup_{n \mathop \in \N} E_n \in \DD$
which, combined with above equality, concludes in:
:$\ds \bigcup_{n \mathop \in \N} D_n \in \DD$
Therefore, $\DD$ is closed under countable unions, making it a $\sigma$-algebra.
== Sources ==
* : $5.4$
Category:Dynkin Systems
Category:Sigma-Algebras
Dynkin System Closed under Set Difference

== Source Work ==
:
: $5$: Uniqueness of measures
:: Problem $5.3$
This mistake can be seen in the first edition (2005), reprinted in 2007: ISBN 0-521-85015-5 (hardback) and ISBN 0-521-61525-9 (paperback).
== Mistake ==
:Let $\DD$ be a Dynkin system. Show that for all $A, B \in \DD$ the difference $B \setminus A \in D$.
This is, however, in general, only true if it is also stipulated that $A \subseteq B$, which case is proved on:
* Dynkin System Closed under Set Difference with Subset
The falsehood of the general statement is demonstrated by the following example:
Let $X = \set {1, 2, 3, 4}$.
Define $\DD$ to be the collection of subsets of $X$ having an even number of elements:
:$\DD := \set {S \subseteq X: \text {$\map {\#} S$ is even}}$
As for example $\set {1, 2}$ and $\set {2, 3}$ are in $\DD$, it is seen that $\DD$ is not closed under set difference.
It remains to show that $\DD$ constitutes a Dynkin system:
* It is clear that $X \in \DD$ as $\map {\#} X = 4$.
* Suppose that $\map {\#} S = 2 k$ for $k \in \N$. Then $\map {\#} {X \setminus S} = 4 - 2 k$, which is also even. Hence $X \setminus S \in \DD$.
* Under disjoint union, the cardinalities are added, and Sum of Even Integers is Even guarantees that the disjoint union is again in $\DD$.
Hence $\DD$ is a Dynkin system, and it provides a counterexample to the claim.
== Acknowledgements ==
This error has been corrected in the third (2010) and later printings.
It appears in the online list of misprints maintained by Schilling.
Category:Mistakes/Measures, Integrals and Martingales
Dynkin System Closed under Set Difference with Subset

== Theorem ==
Let $X$ be a set.
Let $\DD$ be a Dynkin system on $X$.
Let $D, E \in \DD$ and suppose that $E \subseteq D$.
Then the set difference $D \setminus E$ is also an element of $\DD$.
== Proof ==
For brevity, write for example $E^\complement$ for $\relcomp X E = X \setminus E$.
We reason as follows:
Now this implies that $D \setminus E \in \DD$ $D^\complement \cup E \in \DD$.
It is already known that $D^\complement$ and $E$ are in $\DD$ by axiom $(2)$ for a Dynkin system.
Since $E \subseteq D$, it follows that $D^\complement \cap E = \O$, and thus Dynkin System Closed under Disjoint Union applies to give:
:$D^\complement \cup E \in \DD$
which, combined with above reasoning, yields $D \setminus E \in \DD$.
== Sources ==
* : $\S 5$: Problem $3$
Category:Dynkin Systems
Dynkin System Contains Empty Set

== Theorem ==
Let $X$ be a set, and let $\DD$ be a Dynkin system on $X$.
Then the empty set $\O$ is an element of $\DD$.
== Proof ==
As $\DD$ is a Dynkin system, $X \in \DD$.
By Set Difference with Self is Empty Set, $X \setminus X = \O$.
Hence, by property $(2)$ of a Dynkin system, $\O = X \setminus X \in \DD$.
== Sources ==
* : $5.2$, $\S 5$: Problem $1$
Category:Dynkin Systems
Category:Empty Set
Dynkin System with Generator Closed under Intersection is Sigma-Algebra

== Theorem ==
Let $X$ be a set.
Let $\GG \subseteq \powerset X$ be a collection of subsets of $X$.
Suppose that $\GG$ satisfies the following condition:
:$(1):\quad \forall G, H \in \GG: G \cap H \in \GG$
That is, $\GG$ is closed under intersection.
Then:
:$\map \delta \GG = \map \sigma \GG$
where $\delta$ denotes generated Dynkin system, and $\sigma$ denotes generated $\sigma$-algebra.
== Proof ==
From Sigma-Algebra is Dynkin System and the definition of generated Dynkin system, it follows that:
:$\map \delta \GG \subseteq \map \sigma \GG$
Let $D \in \map \delta \GG$, and define:
:$\delta_D := \set {E \subseteq X: E \cap D \in \map \delta \GG}$
Let us verify that these $\delta_D$ form Dynkin systems.
First of all, note $X \cap D = D$, hence $X \in \delta_D$.
Next, compute, for any $E \in \delta_D$:
Now from Intersection is Associative, Set Difference Intersection with Second Set is Empty Set and Intersection with Empty Set:
:$\paren {E \cap D} \cap \paren {X \setminus D} = E \cap \paren {D \cap \paren {X \setminus D} } = E \cap \O = \O$
Thus, since $E \cap D, X \setminus D \in \map \delta \GG$, it follows that their disjoint union is as well.
Finally, combining the above, it follows that:
:$\paren {X \setminus E} \cap D \in \map \delta \GG$
Thus:
:$E \in \delta_D \implies X \setminus E \in \delta_D$
Finally, let $\sequence {E_n}_{n \mathop \in \N}$ be a pairwise disjoint sequence of sets in $\delta_D$.
Then it is immediate that $\sequence {E_n \cap D}_{n \mathop \in \N}$ is also pairwise disjoint.
Hence:
and since the $E_n \cap D$ are in $\map \delta \GG$ by assumption, this disjoint union is also in $\map \delta \GG$.
Therefore, $\delta_D$ is a Dynkin system.
Now by definition of generated Dynkin system, $\GG \subseteq \map \delta \GG$.
By this observation and $(1)$, we immediately obtain, for any $G \in \GG$:
:$\GG \subseteq \delta_G$
Therefore, by definition of generated Dynkin system, for all $G \in \GG$:
:$\map \delta \GG \subseteq \delta_G$
Hence, for any $D \in \map \delta \GG$ and $G \in \GG$:
:$D \cap G \in \map \delta \GG$
Thus we have established that, for all $D \in \map \delta \GG$:
:$\GG \subseteq \delta_D$
whence, by the definition of $\delta_D$ and generated Dynkin system:
:$\map \delta \GG \subseteq \delta_D$
That is to say:
:$\forall D, E \in \map \delta \GG: D \cap E \in \map \delta \GG$
Hence, by Dynkin System Closed under Intersections is Sigma-Algebra, $\map \delta \GG$ is a $\sigma$-algebra.
Thus, it follows that $\map \sigma \GG \subseteq \map \delta \GG$.
Hence the result, by definition of set equality.
== Sources ==
* : $5.5$
Category:Dynkin Systems
Category:Sigma-Algebras
EPORN/Examples

== Examples of EPORNs ==
=== Example: $2520$ ===
=== Example: $63 \, 504$ ===
=== Example: $144 \, 648$ ===
=== Example: $7 \, 683 \, 984$ ===
== Also see ==
* Definition:EPORN
== Sources ==
* 
* : $63,504$
* http://www.shyamsundergupta.com/eporns.htm
Category:EPORNs
EPORN/Examples/63,504

== Example of EPORN ==
The number $63 \, 504$ is the smallest EPORN which is not a multiple of $10$:
== Also see ==
* Definition:EPORN
== Sources ==
* 
* : $63,504$
* http://www.shyamsundergupta.com/eporns.htm
Category:EPORNs
Category:63,504
EPORN/Examples/7,683,984

== Example of EPORN ==
The EPORN $7 \, 683 \, 984$ includes a palindrome and is therefore square:
== Also see ==
* Definition:EPORN
== Sources ==
* 
* : $63,504$
* http://www.shyamsundergupta.com/eporns.htm
Category:EPORNs
Category:7,683,984
Earthworm Sequence/Examples

== Examples of Earthworm Sequence ==
The earthworm sequence is analysed for all $n \in \set {00, 01, 02, \dotsc, 99}$, not only for the range $\set {10, 11, 12, \dotsc, 99}$.
The first table lists the integers that end up in a loop of $20$ terms:
:
$n$ 	$\map h n$	$\map h {\map h n}$	$\map h {\map h {\map h n} }$
$01$, $51$	$02$ 	$04$ 	$08$
$27$, $77$	$54$ 	$08$ 	$16$
$29$, $79$	$58$ 	$16$ 	$32$
$33$, $83$	$66$ 	$32$ 	$64$
$41$, $91$	$82$ 	$64$ 	$28$
$07$, $57$	$14$ 	$28$ 	$56$
$39$, $89$	$78$ 	$56$ 	$12$
$03$, $53$	$06$ 	$12$ 	$24$
$31$, $81$	$62$ 	$24$ 	$48$
$37$, $87$	$74$ 	$48$ 	$96$
$49$, $99$	$98$ 	$96$ 	$92$
$23$, $73$	$46$ 	$92$ 	$84$
$21$, $71$	$42$ 	$84$ 	$68$
$17$, $67$	$34$ 	$68$ 	$36$
$09$, $59$	$18$ 	$36$ 	$72$
$43$, $93$	$86$ 	$72$ 	$44$
$11$, $61$	$22$ 	$44$ 	$88$
$47$, $97$	$94$ 	$88$ 	$76$
$19$, $69$	$38$ 	$76$ 	$52$
$13$, $63$	$26$ 	$52$ 	$04$
The second table lists the integers that end up in a loop of $4$ terms:
:
$n$ 	$\map h n$	$\map h {\map h n}$	$\map h {\map h {\map h n} }$
$05$, $55$	$10$ 	$20$ 	$40$
$35$, $85$	$70$ 	$40$ 	$80$
$45$, $95$	$90$ 	$80$ 	$60$
$15$, $65$	$30$ 	$60$ 	$20$
The second table lists the integers that end up in a loop of $1$ term, that is, the degenerate case that ends in $0$:
:
$25$, $75$	$50$	$00$	$00$
Category:Earthworm Sequence
Eccentricity of Ellipse in terms of Semi-Major and Semi-Minor Axes

== Theorem ==
Let $K$ be a ellipse such that:
:$a$ denotes the length of the semi-major axis of $K$
:$b$ denotes the length of the semi-minor axis of $K$
:$e$ denotes the eccentricity of $K$.
Then:
:$e^2 = 1 - \dfrac {b^2} {a^2}$
== Proof ==
Let $c$ be the distance of the foci of $K$ from the center.
== Sources ==
* 
* 
Category:Eccentricity of Ellipse
Category:Major Axis of Ellipse
Category:Ellipses
Eccentricity of Ellipse in terms of Semi-Major and Semi-Minor Axes/Mistake

== Source Work ==
:
:ellipse
== Mistake ==
Alternatively [the eccentricity] is given by
:$e^2 - 1 - \dfrac {b^2} {a^2}$
== Correction ==
This should of course read:
:$e^2 = 1 - \dfrac {b^2} {a^2}$
This glitch does not appear in the second edition.
== Sources ==
* 
Category:Mistakes/The Penguin Dictionary of Mathematics
Eccentricity of Ellipse is Interfocal Distance over Major Axis

== Theorem ==
Let $K$ be a ellipse such that:
:$a$ denotes the length of semi-major axis of $K$
:$c$ denotes the distance between the foci of $K$
:$e$ denotes the eccentricity of $K$.
Then:
:$e = \dfrac c {2 a}$
== Proof ==
== Sources ==
* 
* 
Category:Eccentricity of Ellipse
Category:Major Axis of Ellipse
Category:Ellipses
Eccentricity of Hyperbola in terms of Semitransverse and Semiconjugate Axes

== Theorem ==
Let $\KK$ be a hyperbola such that:
:$a$ denotes the length of the semitransverse axis of $\KK$
:$b$ denotes the length of the semiconjugate axis of $\KK$
:$e$ denotes the eccentricity of $\KK$.
Then:
:$e^2 = 1 + \dfrac {b^2} {a^2}$
== Proof ==
== Sources ==
* 
* 
Category:Hyperbolas
Eccentricity of Hyperbola is Interfocal Distance over Transverse Axis

== Theorem ==
Let $\KK$ be a hyperbola such that:
:$a$ denotes the length of the semitransverse axis of $\KK$
:$c$ denotes the distance between the foci of $\KK$
:$e$ denotes the eccentricity of $\KK$.
Then:
:$e = \dfrac c {2 a}$
== Proof ==
== Sources ==
* 
* 
Category:Hyperbolas