nyuuzyou/wikis huggingface dataset:
Closed Balls Centered on P-adic Number is Countable

== Theorem ==
Let $p$ be a prime number.
Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers.
Let $a \in \Q_p$.
Then the set of all closed balls centered on $a$ is the countable set:
:$\BB^- = \set {\map {B^-_{p^{-n} } } a: n \in \Z}$
=== Corollary ===
== Proof ==
Let $\epsilon \in \R_{>0}$.
=== Lemma ===
From Closed Ball contains Smaller Closed Ball:
:$\map {B^-_{p^{-n}}} a \subseteq \map {B^-_\epsilon} a$
From Open Ball contains Strictly Smaller Closed Ball:
:$\map {B^-_\epsilon} a \subseteq \map {B_{p^{-\paren {n - 1} } } } a$
From Open Ball in P-adic Numbers is Closed Ball
:$\map {B_{p^{-\paren{n-1}}}} a = \map {B^-_{p^{-n} } } a$
Hence:
:$\map {B^-_\epsilon} a \subseteq \map {B^-_{p^{-n} } } a$
By definition of set equality:
:$\map {B^-_\epsilon} a = \map {B^{\, -}_{p^{-n} } } a$
Since $\epsilon \in \R_{>0}$ was arbitrary then:
:$\forall \epsilon \in \R_{>0} : \exists n \in \Z : \map {B^-_\epsilon} a = \map {B^-_{p^{-n} } } a$
Hence the set of all closed balls centered on $a$ is:
:$\BB^- = \set {\map {B^-_{p^{-n} } } a : n \in \Z}$
From Surjection from Countably Infinite Set iff Countable, it follows that $\BB^-$ is a countable set.
Category:Topology of P-adic Numbers
Category:Closed Balls Centered on P-adic Number is Countable
Closed Balls Centered on P-adic Number is Countable/Lemma

== Theorem ==
Let $p$ be a prime number.
Let $\epsilon \in \R_{\gt 0}$.
Then:
:$\exists n \in \Z : p^{-n} \le \epsilon < p^{-\paren {n - 1} }$
== Proof ==
From Sequence of Powers of Reciprocals is Null Sequence:
:$\exists n_1 \in \N: \forall k \ge n_1 : p^{-k} < \epsilon$
Similarly:
:$\exists n_2 \in \N: \forall k \ge n_2 : p^{-k} < \dfrac 1 \epsilon$
Hence: 
:$p^{-n_1} < \epsilon$ and $ p^{-n_2} < \dfrac 1 \epsilon$
That is:
:$p^{-n_1} < \epsilon < p^{n_2}$
From Power Function on Integer between Zero and One is Strictly Decreasing:
:$-n_2 < n_1$.
Let:
:$n = \min \set {k : -n_2 \mathop \le k \mathop \le n_1 \text{ and } p^{-k} \le \epsilon}$
By choice of $n$:
:$p^{-n} \le \epsilon < p^{n_2}$
From Power Function on Integer between Zero and One is Strictly Decreasing:
:$n > -n_2$
Also by the choice of $n$:
:$n \le n_1$
So:
:$n_1 > n - 1 \ge -n_2$
:
:$p^{-\paren {n - 1} } \le \epsilon$
Then:
:$n - 1 \in \set {k: -n_2 \le k \le n_1 \text{ and } p^{-k} \le \epsilon}$
 
So:
:$n = \min \set {k: -n_2 \le k \le n_1 \text { and } p^{-k} \le \epsilon} \le n - 1$
which is a contradiction.
Hence:
:$\epsilon < p^{-\paren {n - 1} }$
The result follows.
Category:Closed Balls Centered on P-adic Number is Countable
Closed Balls Centered on P-adic Number is Countable/Open Balls

== Theorem ==
Let $p$ be a prime number.
Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers.
Let $a \in \Q_p$.
Then the set of all open balls centered on $a$ is the countable set:
:$\BB = \set {\map {B_{p^{-n} } } a : n \in \Z}$
== Proof ==
Let $\epsilon \in \R_{\ge 0}$.
=== Lemma ===
From Open Ball contains Smaller Open Ball:
:$\map {B_\epsilon} a \subseteq \map {B_{p^{-n} } } a$
From Open Ball contains Strictly Smaller Closed Ball:
:$\map {B^-_{p^{-\paren {n + 1} } } } a \subseteq \map {B_\epsilon} a$
From Open Ball in P-adic Numbers is Closed Ball
:$\map {B_{p^{-n} } } a = \map {B^-_{p^{-\paren {n + 1} } } } a$
Hence:
:$\map {B_{p^{-n} } } a \subseteq \map {B_\epsilon} a$
By definition of set equality:
:$\map {B_\epsilon} a = \map {B_{p^{-n} } } a$
Since $\epsilon \in \R_{>0}$ was arbitrary then:
:$\forall \epsilon \in \R_{>0} : \exists n \in \Z : \map {B_\epsilon} a = \map {B_{p^{-n} } } a$
Hence the set of all open balls centered on $a$ is:
:$\BB = \set {\map {B_{p^{-n} } } a : n \in \Z}$
From Surjection from Countably Infinite Set iff Countable, it follows that $\BB$ is a countable set.
Category:Topology of P-adic Numbers
Category:Closed Balls Centered on P-adic Number is Countable
Closed Balls Centered on P-adic Number is Countable/Open Balls/Lemma

== Theorem ==
Let $p$ be a prime number.
Let $\epsilon \in \R_{> 0}$.
Then:
:$\exists n \in \Z : p^{-\paren {n + 1} } < \epsilon \le p^{-n}$
== Proof ==
From Lemma for Closed Balls:
:$\exists m \in \Z : p^{-m} \le \epsilon < p^{-\paren {m - 1} }$
Suppose $\epsilon \ne p^{-m}$.
Then:
:$p^{-m} < \epsilon < p^{-\paren {m - 1} }$
and the theorem is proved with $n = m - 1$.
Now suppose $\epsilon = p^{-m}$.
 
From Power Function on Integer between Zero and One is Strictly Decreasing:
:$p^{-\paren{m + 1}} < p^{-m}$
So:
:$p^{-\paren{m + 1}} < \epsilon \le p^{-m}$
and the theorem is proved with $n = m$.
Category:Closed Balls Centered on P-adic Number is Countable
Closed Bounded Subset of Real Numbers is Compact

== Theorem ==
Let $\R$ be the real number line considered as an Euclidean space.
Let $S \subseteq \R$ be a closed and bounded subspace of $\R$.
Then $S$ is compact in $\R$.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $5$, Theorem $10$
Category:Analysis
Category:Euclidean Space
Category:Closed Bounded Subset of Real Numbers is Compact
Closed Bounded Subset of Real Numbers is Compact/Proof 2

== Theorem ==
== Proof ==
Let $S$ be closed and bounded.
As $S$ is bounded, there exist some $a, b \in \R$ such that:
:$S \subseteq \openint a b$
where $\openint a b$ is the open interval between $a$ and $b$.
It follows that $S \subseteq \closedint a b$.
Consider the set:
:$U = \relcomp \R S \cap \openint {a - 1} {b + 1}$
By inspection it can be seen that:
:$U = \openint {a - 1} a \cup \openint b {b + 1}$
By Union of Open Sets of Metric Space is Open, it follows that $U$ is open in $\R$.
Let $\CC$ be an open cover for $S$.
Then we can construct the open cover $\CC' = \CC \cup \set U$ for $\closedint a b$.
Let $\FF' \subseteq \CC'$ be a finite subcover of $\CC'$ for $\closedint a b$.
Then $\FF = \FF' \setminus \set U$ is the desired finite subcover for $S$.
Hence it is sufficient to prove that any open cover for $\closedint a b$, has a finite subcover.
So, suppose $S = \closedint a b$ and create the set $T \subseteq \closedint a b$ as follows:
Let $a \le x \le b$.
Then $x \in T$ $\closedint a x$ has a finite subcover of $\CC$.
We have that $a \in T$ and that $b$ is an upper bound for $T$.
Let $l = \map \sup T$ be the supremum of $T$.
Let $L \in \CC$ such that $l \in L$.
Since $L$ is open:
:$\exists \epsilon \in \R_{>0}: \openint {l - \epsilon} {l + \epsilon} \subseteq L$
Since $l = \map \sup T$ there exists $t \in T$ such that $t > l - \epsilon$.
We have that $\FF$ is a finite subset of $\CC$ such that $\ds \closedint a t \subseteq \bigcup \FF$.
Then $\FF \cup \set L$ is a finite subset of $\CC$ whose union contains $\closedint a {l + \delta}$ for every $\delta \in \openint 0 \epsilon$.
Since $l$ is an upper bound for $T$, it follows that $l + \delta \notin T$.
Thus $l + \delta > b$ for all $\delta > 0$.
That is, $l \ge b$.
But by definition, $l \le b$.
So $l = b$ and so $\CC$ has a finite subcover for $\closedint a b$.
== Sources ==
* : Chapter $\text{III}$: Metric Spaces: Compactness
Category:Closed Bounded Subset of Real Numbers is Compact
Closed Class under Progressing Mapping Lemma

== Theorem ==
Let $N$ be a class which is closed under a progressing mapping $g$.
Let $g$ be such that:
:$\forall x, y \in N: \map g x \subseteq y \lor y \subseteq x$
:if $\map g x = x$, then $x$ is the greatest element of $N$.
Let the following hold:
:$A \subseteq N$ is a subclass of $N$
:$x \in N$ is an element of $N$
Let $x$ be:
:a proper subset of all elements of $A$
and:
:the greatest element of $A$ with that property.
Then $\map g x \in A$ and is the smallest element of $A$.
== Proof ==
Let the hypothesis be assumed.
Let $A$ be an arbitrary non-empty subclass of $N$.
Let $L$ be the class of all elements $y$ of $N$ such that $y$ is a proper subset of all elements of $A$.
Let $x$ be the greatest element of $L$.
It is to be shown that $\map g x$ is the smallest element of $A$.
We have that $x$ is a proper subset of all elements of $A$.
Then by Corollary 1 to the Sandwich Principle:
:$\map g x$ is a subset of all elements of $A$.
It remains to be shown that $\map g x$ is itself an element of $A$.
We have that $x$ is a proper subset of all elements of $A$.
We also have that $x$ contains at least one element.
Therefore $x$ cannot be the greatest element of $N$.
Therefore, by hypothesis:
:$x \ne \map g x$
Thus $x$ is a proper subset of $\map g x$.
Because $x$ is the greatest element of $L$, $\map g x$ is not therefore an element of $L$.
Therefore there exists at least $1$ element $y$ of $A$ such that $\map g x$ is not a proper subset of $y$.
But we have shown that $\map g x \subseteq y$.
Hence:
:$\map g x = y$
and so:
:$\map g x \in A$
Hence the result.
== Sources ==
* : Chapter $3$: The Natural Numbers: $\S 4$ A double induction principle and its applications: Lemma $4.15$
Category:Progressing Mappings
Category:Closedness under Mappings
Closed Continuous Injection is Embedding

== Theorem ==
Let $T_1 = \struct{S_1, \tau_1}$ and $T_2 = \struct{S_2, \tau_2}$ be topological spaces.
Let $f: S_1 \to S_2$ be a closed and continuous injection.
Then $f$ is an embedding of $T_1$ into $T_2$.
== Proof ==
Let $g: S_1 \to f \sqbrk {S_1}$ be the restriction of $f$ to $S_1 \times f \sqbrk {S_1}$.
It must be shown that $g$ is a homeomorphism.
=== $g$ is a bijection ===
From Restriction of Mapping to Image is Surjection, $g$ is a surjection.
From Restriction of Injection is Injection, $g$ is a injection.
Thus, $g$ is a bijection by definition.
=== $g$ is continuous ===
Follows from Restriction of Continuous Mapping is Continuous: Topological Spaces
=== $g$ is closed ===
Let $A \subseteq S_1$ be closed in $T_1$.
Since $f$ is an closed mapping, $f \sqbrk {A}$ is closed in $T_2$.
Because $f$ is an closed mapping, $f \sqbrk {S_1}$ is closed in $T_2$, as $S_1$ is open in $T_1$ by Definition of a topology.
Thus Closed Set in Closed Subspace applies: $f \sqbrk {A}$ is closed in $T_2$ it is closed in the subspace topology of $f \sqbrk {S_1}$.
Therefore, $f \sqbrk {A}$ is closed in the subspace topology of $f \sqbrk {S_1}$.
Since $f$ and $g$ agree on $S_1$, $f \sqbrk {A} = g \sqbrk {A}$.
Therefore, $g$ is a closed mapping.
== Also See ==
* Open Continuous Injection is Embedding
== Sources ==
* : $\S 3$: New Spaces From Old: Subspaces. Topological Embeddings
Category:Embeddings (Topology)
Closed Convex Hull in Normed Vector Space is Convex

== Definition ==
Let $\struct {X, \norm \cdot}$ be a normed vector space over $\R$. 
Let $U \subseteq X$. 
Let $C$ be the closed convex hull of $U$. 
Then: 
:$C$ is convex.
== Proof ==
From the definition of closed convex hull, we have: 
:$C$ is the closure of the convex hull $\map {\operatorname {conv} } U$ of $U$.
From Convex Hull is Smallest Convex Set containing Set, we have: 
:$\map {\operatorname {conv} } U$ is convex.
So, from Closure of Convex Subset in Normed Vector Space is Convex, we have: 
:$C$ is convex.
Category:Closed Convex Hulls
Category:Normed Vector Spaces
Closed Convex Set in terms of Bounded Linear Functionals

== Theorem ==
Let $X$ be a normed vector space over $\R$.
Let $X^\ast$ be the normed dual of $X$. 
Let $C$ be a closed convex subset of $X$.
Then: 
:$\ds C = \bigcap_{f \in X^\ast} \set {x \in X : \map f x \le \sup_{c \in C} \map f c}$
== Proof ==
Clearly if $x \in C$ then: 
:$\ds \map f x \le \sup_{c \in C} \map f c$
for all $f \in X^\ast$, by the definition of supremum.
That is: 
:$\ds C \subseteq \bigcap_{f \in X^\ast} \set {x \in X : \map f x \le \sup_{c \in C} \map f c}$
To show that: 
:$\ds \bigcap_{f \in X^\ast} \set {x \in X : \map f x \le \sup_{c \in C} \map f c} \subseteq C$
we prove that if $x_0 \not \in C$ then: 
:$\ds x_0 \not \in \bigcap_{f \in X^\ast} \set {x \in X : \map f x \le \sup_{c \in C} \map f c}$
The Rule of Transposition will then give us the result. 
Let $x_0 \not \in C$.
From Finite Topological Space is Compact, we have that $\set {x_0}$ is compact. 
Clearly $\set {x_0}$ is disjoint from $C$. 
We can therefore apply Hahn-Banach Separation Theorem: Compact Convex Set and Closed Convex Set, which gives the existence of $f \in X^\ast$, $c \in \R$ and $\epsilon > 0$ such that: 
:$\map f {x_0} \le c - \epsilon < c + \epsilon \le \map f x$ for each $x \in C$.
From the definition of infimum, we therefore have: 
:$\ds c + \epsilon \le \inf_{x \in C} \map f x$
so that: 
:$\ds \map f {x_0} \le c - \epsilon < c + \epsilon \le \inf_{x \in C} \map f x$
Then, from Negative of Supremum is Infimum of Negatives, we have:
:$\ds -\map f {x_0} \ge \epsilon - c > -\paren {c + \epsilon} \ge \sup_{x \in C} \paren {-\map f x}$
Setting $g = -f \in X^\ast$ we have: 
:$\ds \map g {x_0} > \sup_{x \in C} \map g x$
So:
:$\ds x_0 \not \in \bigcap_{f \in X^\ast} \set {x \in X : \map f x \le \sup_{c \in C} \map f c}$
as required.
Hence we obtain: 
:$\ds \bigcap_{f \in X^\ast} \set {x \in X : \map f x \le \sup_{c \in C} \map f c} \subseteq C$
and:
:$\ds C = \bigcap_{f \in X^\ast} \set {x \in X : \map f x \le \sup_{c \in C} \map f c}$
Category:Normed Dual Spaces
Category:Convex Sets (Vector Spaces)
Closed Element of Composite Closure Operator

== Theorem ==
Let $\struct {S, \preceq}$ be an ordered set.
Let $f, g: S \to S$ be closure operators.
Let $h = f \circ g$, where $\circ$ represents composition.
Suppose that $h$ is also a closure operator.
Then an element $x \in S$ is closed $h$ it is closed $f$ and $g$.
== Proof ==
An element is closed with respect to a closure operator it is a fixed point of that operator.
Since $f$ and $g$ are closure operators, they are inflationary.
Thus the result follows from Fixed Point of Composition of Inflationary Mappings.
Category:Closure Operators
Category:Closed Elements
Closed Elements Uniquely Determine Closure Operator

== Theorem ==
Let $\struct {S, \preceq}$ be an ordered set.
Let $f, g: S \to S$ be closure operators on $S$.
Suppose that $f$ and $g$ have the same closed elements.
Then $f = g$.
== Proof ==
Let $x \in S$.
Let $C$ be the set of closed elements of $S$ ( either $f$ or $g$) that succeed $x$.
By Closure is Smallest Closed Successor, $\map f x$ and $\map g x$ are smallest closed successors of $x$.
That is, $\map f x$ and $\map g x$ are smallest elements of $C \cap x^\succeq$, where $x^\succeq$ denotes the upper closure of $x$.
By Smallest Element is Unique:
:$\map f x = \map g x$
As $x \in S$ is arbitrary it follows that:
:$\forall x \in S: \map f x = \map g x$
Hence by Equality of Mappings:
:$f = g$
Category:Closure Operators
Closed Euclidean Ball/Examples/Dimension 1

== Example of Closed Euclidean Ball ==
Let $\map { { {\overline B}_\epsilon}^1} a$ be the closed Euclidean ball of center $a$ and radius $\epsilon$ in the real number line $\R$.
Then $\map { { {\overline B}_\epsilon}^1} a$ is the closed real interval $\closedint {a - \epsilon} {a + \epsilon}$.
Category:Examples of Closed Euclidean Balls
Closed Euclidean Ball/Examples/Dimension 2

== Example of Closed Euclidean Ball ==
Let $\map { { {\overline B}_\epsilon}^2} a$ be the closed Euclidean ball of center $a$ and radius $\epsilon$ in the Euclidean plane $\R^2$.
Then $\map { { {\overline B}_\epsilon}^2} a$ is of the form of a disc of radius $\epsilon$ and center $a$.
Category:Examples of Closed Euclidean Balls
Closed Extension Space is Irreducible

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $T^*_p = \struct {S^*_p, \tau^*_p}$ be the closed extension space of $T$.
Then $T^*_p$ is irreducible.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Part $\text {II}$: Counterexamples: $12$. Closed Extension Topology: $21$
Category:Irreducible Spaces
Category:Closed Extension Topology
Category:Closed Extension Space is Irreducible
Closed Extension Space is Irreducible/Proof 1

== Theorem ==
== Proof ==
Trivially, by definition, every open set in $T^*_p$ contains $p$.
So:
:$\forall U_1, U_2 \in \tau^*_p: p \in U_1 \cap U_2$
for $U_1, U_2 \ne \O$.
Category:Closed Extension Space is Irreducible
Closed Extension Space is Irreducible/Proof 2

== Theorem ==
== Proof ==
From Closure of Open Set of Closed Extension Space we have that:
:$\forall U \in \tau^*_p: U \ne \O \implies U^- = S$
where $U^-$ is the closure of $U$.
The result then follows by definition of irreducible space.
Category:Closed Extension Space is Irreducible
Closed Extension Topology is Topology

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $\tau^*_p$ be the closed extension topology of $\tau$.
Then $\tau^*_p$ is a topology on $S^*_p = S \cup \set p$.
== Proof ==
By definition:
:$\tau^*_p = \set {U \cup \set p: U \in \tau} \cup \set \O$
We have that $\O \in \tau^*_p$ by definition.
We also have that $S \in \tau$ so $S \cup \set p \in \tau^*_p$.
Now let $U_1, U_2 \in \tau^*_p$.
Then $U_1^* = U_1 \setminus \set p \in \tau, U_2^* = U_2 \setminus \set p \in \tau$.
Then $p \in U_1$ and $p \in U_2$ and so $p \in U_1 \cap U_2$.
So $U_1 \cap U_2 = \paren {U_1^* \cap U_2^*} \cup \set p \in \tau^*_p$. 
Finally consider $\UU \subseteq \tau^*_p$.
:$\ds \bigcup \UU = \bigcup_{U \mathop \in \UU} \paren {U \setminus \set p} \cup \set p$
So $\ds \bigcup \UU \in \tau^*_p$.
So $\tau^*_p$ is a topology on $S \cup \set p$.
== Sources ==
* : Part $\text {II}$: Counterexamples: $12$. Closed Extension Topology: $20$
Category:Closed Extension Topology
Closed Extension Topology is not Hausdorff

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $T^*_p = \struct {S^*_p, \tau^*_p}$ be the closed extension space of $T$.
Then $T^*_p$ is not a $T_2$ (Hausdorf) space.
== Proof ==
 $T^*_p$ is not a $T_2$ (Hausdorf) space.
From $T_2$ Space is $T_1$ Space, $T^*_p$ is a $T_1$ (Fréchet) space.
But this contradicts Closed Extension Topology is not $T_1$
Hence by Proof by Contradiction $T^*_p$ can not be a $T_2$ (Hausdorf) space.
== Sources ==
* : Part $\text {II}$: Counterexamples: $12$. Closed Extension Topology: $21$
Category:Hausdorff Spaces
Category:Closed Extension Topology
Closed Extension Topology is not T1

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $T^*_p = \struct {S^*_p, \tau^*_p}$ be the closed extension space of $T$.
Then $T^*_p$ is not a $T_1$ (Fréchet) space.
== Proof ==
By definition:
:$\tau^*_p = \set {U \cup \set p: U \in \tau} \cup \set \O$
Let $x \in S^*_p, x \ne p$.
Let $U = \set p$.
Then $U \in \tau^*_p$ such that $p \in U, x \notin U$.
But there is no $V \in \tau^*_p$ such that $x \in V, p \notin V$, by definition of the closed extension topology.
Hence $T^*_p$ can not be a $T_1$ (Fréchet) space.
== Sources ==
* : Part $\text {II}$: Counterexamples: $12$. Closed Extension Topology: $21$
Category:T1 Spaces
Category:Closed Extension Topology
Closed Extension Topology is not T3

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $T^*_p = \struct {S^*_p, \tau^*_p}$ be the closed extension space of $T$.
Then $T^*_p$ is not a $T_3$ space.
== Proof ==
By Underlying Set of Topological Space is Closed, $S$ is closed in $T$.
By Closed Sets of Closed Extension Topology, $S$ is closed in $T^*_p$.
 gives:
:$p \notin S$
:Every open set in $T^*_p$ is either $\O$ or it contains $p$.
Thus no open set containing $S$ is disjoint from $\set p$.
This shows that $T^*_p$ is not a $T_3$ space.
== Sources ==
* : Part $\text {II}$: Counterexamples: $12$. Closed Extension Topology: $21$
Category:T3 Spaces
Category:Closed Extension Topology
Closed Form for Heptagonal Numbers

== Theorem ==
The closed-form expression for the $n$th heptagonal number is:
:$H_n = \dfrac {n \paren {5 n - 3} } 2$
== Proof ==
Heptagonal numbers are $k$-gonal numbers where $k = 7$.
From Closed Form for Polygonal Numbers we have that:
:$\map P {k, n} = \dfrac n 2 \paren {\paren {k - 2} n - k + 4}$
Hence:
Hence the result.
== Sources ==
* : $81$
Category:Heptagonal Numbers
Category:Closed Forms
Closed Form for Heptagonal Pyramidal Numbers

== Theorem ==
The closed-form expression for the $n$th heptagonal pyramidal number is:
:$Q_n = \dfrac {n \paren {n + 1} \paren {5 n - 2} } 6$
== Proof ==
== Sources ==
* : $196$
Category:Closed Forms
Category:Pyramidal Numbers
Closed Form for Hexagonal Numbers

== Theorem ==
The closed-form expression for the $n$th hexagonal number is:
:$H_n = n \paren {2 n - 1}$
== Proof ==
Hexagonal numbers are $k$-gonal numbers where $k = 6$.
From Closed Form for Polygonal Numbers we have that:
:$\map P {k, n} = \dfrac n 2 \paren {\paren {k - 2} n - k + 4}$
Hence:
Hence the result.
== Sources ==
* : $45$
* : $45$
Category:Hexagonal Numbers
Category:Closed Forms
Closed Form for Lucas Numbers

== Theorem ==
The Lucas numbers have a closed-form solution:
:$L_n = \phi^n + \paren {1 - \phi}^n = \paren {\dfrac {1 + \sqrt 5} 2}^n + \paren {\dfrac {1 - \sqrt 5} 2}^n$
where $\phi$ is the golden mean.
Putting $\hat \phi = 1 - \phi = -\dfrac 1 \phi$ this can be written:
:$L_n = \phi^n + \hat \phi^n$
== Proof ==
Proof by induction:
For all $n \in \N$, let $\map P n$ be the proposition:
:$L_n = \phi^n + \paren {1 - \phi}^n = \paren {\dfrac {1 + \sqrt 5} 2})^n + \paren {\dfrac {1 - \sqrt 5} 2}^n = \phi^n + \hat \phi^n$
=== Basis for the Induction ===
$\map P 0$ is true, as this just says:
:$\phi^0 + \hat \phi^0 = 1 + 1 = 2 = L_0$
$\map P 1$ is the case:
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $\map P j$ is true for all $0 \le j \le k + 1$, then it logically follows that $\map P {k + 2}$ is true.
So this is our induction hypothesis:
:$\forall 0 \le j \le k + 1: L_j = \phi^j + \hat \phi^j$
Then we need to show:
:$L_{k + 2} = \phi^{k + 2} + \hat \phi^{k + 2}$
=== Induction Step ===
This is our induction step:
We observe that we have the following two identities:
:$\phi^2 = \paren {\dfrac {1 + \sqrt 5} 2}^2 = \dfrac 1 4 \paren {6 + 2 \sqrt 5} = \dfrac {3 + \sqrt 5} 2 = 1 + \phi$
:$\hat \phi^2 = \paren {\dfrac {1 - \sqrt 5} 2}^2 = \dfrac 1 4 \paren {6 - 2 \sqrt 5} = \dfrac {3 - \sqrt 5} 2 = 1 + \hat \phi$
This can also be deduced from the definition of the golden mean: the fact that $\phi$ and $\hat \phi$ are the solutions to the quadratic equation $x^2 = x + 1$.
Thus:
The result follows by the Second Principle of Mathematical Induction.
Therefore:
:$\forall n \in \N: L_n = \phi^n + \hat \phi^n$
== Sources ==
* : $11$
* : $11$
Category:Lucas Numbers
Category:Golden Mean
Category:Closed Forms
Closed Form for Millin Series

== Theorem ==
The Millin series has the closed form expression:
:$\ds \sum_{n \mathop = 0}^\infty \frac 1 {F_{2^n} } = \frac {7 - \sqrt 5} 2$
where $F_k$ is the $k$th Fibonacci number.
== Proof 1 ==
== Proof 2 ==
== Historical Note ==
== Sources ==
* 
* 
Category:Fibonacci Numbers
Category:Closed Form for Millin Series
Closed Form for Millin Series/Proof 1

== Theorem ==
== Proof ==
First we will prove that:
:$\ds \sum_{r \mathop = 0}^n \frac 1 {F_{2^r} } = 3 - \frac {F_{2^n - 1} } {F_{2^n} }$
for $n \ge 1$.
We see that:
:$\dfrac 1 {F_1} + \dfrac 1 {F_2} = 2 = 3 - \dfrac {F_1} {F_2}$ 
so the proposition holds for $n = 1$.
Suppose the proposition is true for $n = k$.
Then:
Thus by Principle of Mathematical Induction, the proof is complete. 
Now taking the limit, we have:
as required.
== Sources ==
* (as an outline only)
Category:Closed Form for Millin Series
Closed Form for Millin Series/Proof 2

== Theorem ==
== Proof ==
Let:
:$\ds \map F x := \sum_{k \mathop = 1}^\infty \frac {x^{2^{k - 1} } } {F_{2^k} }$
:$\alpha := \dfrac {1 + \sqrt 5} 2$
:$\beta := \dfrac {1 - \sqrt 5} 2$
Note that for $\size x \le 1$ we have:
:$\size {\dfrac {x^{2^{k - 1} } } {F_{2^k} } } \le \dfrac 1 {F_{2^k} }$
Given that the Millin Series converges, by Comparison Test:
:$\ds \map F x = \sum_{k \mathop = 1}^\infty \frac {x^{2^{k - 1} } } {F_{2^k} }$ converges for $\size x \le 1$
We have:
:$\ds \map F {\alpha x} := \sum_{k \mathop = 1}^\infty \frac {\alpha^{2^{k - 1} } x^{2^{k - 1} } } {F_{2^k} }$
Hence:
Since $\beta^2 = \dfrac {3 - \sqrt 5} 2 \le 1$, we can substitute $x = -\beta$:
== Sources ==
* 
Category:Closed Form for Millin Series
Closed Form for Number of Derangements on Finite Set

== Theorem ==
The number of derangements $D_n$ on a finite set $S$ of cardinality $n$ is:
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Definition:Subfactorial
== Historical Note ==
== Sources ==
* 
* : Chapter $2$: The Symmetric Groups: $\S 78 \alpha$
* : The misaddressed letters: $130$
Category:Closed Form for Number of Derangements on Finite Set
Category:Number of Derangements on Finite Set
Category:Derangements
Category:Counting Arguments
Category:Closed Forms