nyuuzyou/wikis huggingface dataset:
Binomial Theorem/Examples/6th Power of Difference

== Example of Use of Binomial Theorem ==
:$\paren {x - y}^6 = x^6 - 6 x^5 y + 15 x^4 y^2 - 20 x^3 y^3 + 15 x^2 y^4 - 6 x y^5 + y^6$
== Proof ==
Follows directly from the Binomial Theorem:
:$\ds \forall n \in \Z_{\ge 0}: \paren {x + \paren {-y} }^n = \sum_{k \mathop = 0}^n \binom n k x^{n - k} \paren {-y}^k$
putting $n = 6$.
== Sources ==
* : $\S 2$: Special Products and Factors: $2.10$
* : $\S 2$: Special Products and Factors: $2.10.$
Category:Examples of Use of Binomial Theorem
Category:Sixth Powers
Binomial Theorem/Examples/6th Power of Sum

== Example of Use of Binomial Theorem ==
:$\paren {x + y}^6 = x^6 + 6 x^5 y + 15 x^4 y^2 + 20 x^3 y^3 + 15 x^2 y^4 + 6 x y^5 + y^6$
== Proof ==
Follows directly from the Binomial Theorem:
:$\ds \forall n \in \Z_{\ge 0}: \paren {x + y}^n = \sum_{k \mathop = 0}^n \binom n k x^{n - k} y^k$
putting $n = 6$.
== Sources ==
* : $\S 2$: Special Products and Factors: $2.9$
* : $\S 2$: Special Products and Factors: $2.9.$
Category:Examples of Use of Binomial Theorem
Category:Sixth Powers
Binomial Theorem/Examples/Cube of Difference

== Example of Use of Binomial Theorem ==
:$\paren {x - y}^3 = x^3 - 3 x^2 y + 3 x y^2 - y^3$
== Proof ==
Follows directly from the Binomial Theorem:
:$\ds \forall n \in \Z_{\ge 0}: \paren {x + \paren {-y} }^n = \sum_{k \mathop = 0}^n \binom n k x^{n - k} \paren {-y}^k$
putting $n = 3$.
== Sources ==
* : $\S 2$: Special Products and Factors: $2.4$
* : $\S 20$: Binomial Series: $20.6$
* : $\S 2$: Special Products and Factors: $2.4.$
Category:Examples of Use of Binomial Theorem
Category:Third Powers
Binomial Theorem/Examples/Cube of Sum

== Example of Use of Binomial Theorem ==
:$\paren {x + y}^3 = x^3 + 3 x^2 y + 3 x y^2 + y^3$
=== Corollary ===
== Proof ==
Follows directly from the Binomial Theorem:
:$\ds \forall n \in \Z_{\ge 0}: \paren {x + y}^n = \sum_{k \mathop = 0}^n \binom n k x^{n - k} y^k$
putting $n = 3$.
== Sources ==
* : $\S 2$: Special Products and Factors: $2.3$
* : $\S 20$: Binomial Series: $20.6$
* : Chapter $1$: Some Preliminary Considerations: $1.2$ The Binomial Theorem
* : $\S 2$: Special Products and Factors: $2.3.$
* : $\S 22$: Taylor Series: Binomial Series: $22.6.$
Category:Examples of Use of Binomial Theorem
Category:Third Powers
Binomial Theorem/Examples/Cube of Sum/Corollary

== Corollary to Cube of Sum ==
:$\paren {x + 1}^3 = x^3 + 3 x^2 + 3 x + 1$
== Proof ==
Follows directly from the Cube of Sum:
setting $y = 1$.
== Sources ==
* : $24$
* : $24$
Category:Examples of Use of Binomial Theorem
Category:Third Powers
Binomial Theorem/Examples/Square Root of 2

== Example of Use of Binomial Theorem ==
:$\sqrt 2 = 2 \paren {1 - \dfrac 1 {2^2} - \dfrac 1 {2^5} - \dfrac 1 {2^7} - \dfrac 5 {2^{11} } - \cdots}$
== Proof ==
The first few terms of the real sequence are:
:$2, \dfrac 3 2, \dfrac {23} {16}, \dfrac {91} {64}, \dfrac {1451} {1024}, \dotsc$
Category:Examples of Use of Binomial Theorem
Category:Square Root of 2
Binomial Theorem/Extended

== Theorem ==
Let $r, \alpha \in \C$ be complex numbers.
Let $z \in \C$ be a complex number such that $\cmod z < 1$.
Then:
:$\ds \paren {1 + z}^r = \sum_{k \mathop \in \Z} \dbinom r {\alpha + k} z^{\alpha + k}$
where $\dbinom r {\alpha + k}$ denotes a binomial coefficient.
== Proof ==
== Sources ==
* 
* : $\S 1.2.6$: Binomial Coefficients: Exercise $42$ (Solution)
Category:Binomial Theorem
Binomial Theorem/General Binomial Theorem

== Theorem ==
Let $\alpha \in \R$ be a real number.
Let $x \in \R$ be a real number such that $\size x < 1$.
Then:
where:
:$\alpha^{\underline n}$ denotes the falling factorial
:$\dbinom \alpha n$ denotes a binomial coefficient.
=== Convergence ===
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Examples ==
== Historical Note ==
== Sources ==
* : Chapter $\text{VI}$: On the Seashore
* : $\text I$. Algebra: The Binomial Theorem: Negative and fractional indices
* : $\S 20$: Binomial Series: $20.4$
* : $4.$ Mathematics: $4.2$ Series
* : $8$. Taylor Series and Power Series: Appendix: Table $8.2$: Power Series of Important Functions
* : $\S 1.2.6$: Binomial Coefficients: $\text{F} \ (15)$
* : $\S 1.2.9$: Generating Functions: $(19)$
* 
* 
* 
* 
* : $\S 22$: Taylor Series: Binomial Series: $22.4.$
Category:Binomial Theorem
Binomial Theorem/General Binomial Theorem/Historical Note

== Theorem ==
The General Binomial Theorem was first conceived by during the years $1665$ to $1667$ when he was living in his home in Woolsthorpe.
He announced the result formally, in letters to Henry Oldenburg on $13$th June $1676$ and $24$th October $1676$ but did not provide a proper proof (at that time the need for the appropriate level of rigor had not been recognised).
 made an incomplete attempt in $1774$, but the full proof had to wait for to provide it in $1812$.
This was, in fact, the first time anything about infinite summations was proved adequately.
== Sources ==
* : Chapter $\text{VI}$: On the Seashore
* : $\S 3$: Appendix $\text B$: Newton
* : Chapter $\text {A}.18$: Newton ()
* : $\S 1.2.6$: Binomial Coefficients
Category:Historical Notes
Binomial Theorem/General Binomial Theorem/Proof 1

== Theorem ==
== Proof ==
Let $R$ be the radius of convergence of the power series:
:$\ds \map f x = \sum_{n \mathop = 0}^\infty \frac {\prod \limits_{k \mathop = 0}^{n - 1} \paren {\alpha - k} } {n!} x^n$
Then:
Thus for $\size x < 1$, Power Series is Differentiable on Interval of Convergence applies:
:$\ds D_x \map f x = \sum_{n \mathop = 1}^\infty \frac {\prod \limits_{k \mathop = 0}^{n - 1} \paren {\alpha - k} } {n!} n x^{n - 1}$
This leads to:
Gathering up:
:$\paren {1 + x} D_x \map f x = \alpha \map f x$
Thus:
:$\map {D_x} {\dfrac {\map f x} {\paren {1 + x}^\alpha} } = -\alpha \paren {1 + x}^{-\alpha - 1} \map f x + \paren {1 + x}^{-\alpha} D_x \map f x = 0$
So $\map f x = c \paren {1 + x}^\alpha$ when $\size x < 1$ for some constant $c$.
But $\map f 0 = 1$ and hence $c = 1$.
== Historical Note ==
== Sources ==
* : $\S 15.10 \ (1)$
Category:Binomial Theorem
Binomial Theorem/General Binomial Theorem/Proof 2

== Theorem ==
== Proof ==
From Sum over k of r-kt choose k by r over r-kt by z^k:
:$\ds \sum_n \dbinom {\alpha - n t} k \dfrac \alpha {\alpha - n t} z^n = x^\alpha$
where:
:$z = x^{t + 1} - x^t$
:$x = 1$ for $z = 0$.
Setting $t = 0$:
== Historical Note ==
== Sources ==
* : $\S 1.2.9$: Generating Functions: Exercise $4$
Category:Binomial Theorem
Binomial Theorem/General Binomial Theorem/Proof 3

== Theorem ==
== Proof ==
The series is the Maclaurin series expansion of the function $\map f x = \paren {1 + x}^\alpha$.
The derivatives of $f$ are:
Evaluated at $x = 0$, we have:
The Maclaurin series of $f$ is:
Category:Binomial Theorem
Binomial Theorem/Hurwitz's Generalisation

== Theorem ==
:$\ds \paren {x + y}^n = \sum x \paren {x + \epsilon_1 z_1 + \cdots + \epsilon_n z_n}^{\epsilon_1 + \cdots + \epsilon_n - 1} \paren {y - \epsilon_1 z_1 - \cdots - \epsilon_n z_n}^{n - \epsilon_1 - \cdots - \epsilon_n}$
where the summation ranges over all $2^n$ choices of $\epsilon_1, \ldots, \epsilon_n = 0$ or $1$ independently.
== Proof ==
Follows from this formula:
:$(1): \quad \ds \sum x \paren {x + \epsilon_1 z_1 + \cdots + \epsilon_n z_n}^{\epsilon_1 + \cdots + \epsilon_n - 1} \paren {y + \paren {1 - \epsilon_1} z_1 - \cdots + \paren {1 - \epsilon_n} z_n}^{n - \epsilon_1 - \cdots - \epsilon_n} = \paren {x + y} \paren {x + y + z_1 + \cdots + z_n}^{n - 1}$
== Sources ==
* 
* : $\S 1.2.6$: Binomial Coefficients: Exercise $51$
Category:Binomial Theorem
Binomial Theorem/Hurwitz's Generalisation/Mistake

== Source Work ==
:
:Chapter Two: Information Structures
::$2.3.$ Trees:
:::$2.3.4.$ Basic Mathematical Properties of Trees
::::$2.3.4.4.$ Enumeration of Trees
:::::Exercise $30$: Solution
== Mistake ==
:Use this to prove Hurwitz' generalization of the binomial theorem:
::$\ds \sum x \paren {x + \epsilon_1 z_1 + \cdots + \epsilon_n z_n}^{\epsilon_1 + \cdots + \epsilon_n - 1} y \paren {y + \paren {1 - \epsilon_1} z_1 - \cdots + \paren {1 - \epsilon_n} z_n}^{n - \epsilon_1 - \cdots - \epsilon_n} = \paren {x + y} \paren {x + y + z_1 + \cdots + z_n}^{n - 1}$
== Correction ==
There is an extraneous $y$ here. It should read:
::$\ds \sum x \paren {x + \epsilon_1 z_1 + \cdots + \epsilon_n z_n}^{\epsilon_1 + \cdots + \epsilon_n - 1} \paren {y + \paren {1 - \epsilon_1} z_1 - \cdots + \paren {1 - \epsilon_n} z_n}^{n - \epsilon_1 - \cdots - \epsilon_n} = \paren {x + y} \paren {x + y + z_1 + \cdots + z_n}^{n - 1}$
== Sources ==
* 
Category:Mistakes/The Art of Computer Programming
Binomial Theorem/Integral Index

== Theorem ==
Let $X$ be one of the standard number systems $\N$, $\Z$, $\Q$, $\R$ or $\C$.
Let $x, y \in X$.
Then:
where $\dbinom n k$ is $n$ choose $k$.
== Proof ==
=== Basis for the Induction ===
For $n = 0$ we have:
:$\ds \paren {x + y}^0 = 1 = \binom 0 0 x^{0 - 0} y^0 = \sum_{k \mathop = 0}^0 \binom 0 k x^{0 - k} y^k$
This is the basis for the induction.
=== Induction Hypothesis ===
This is our induction hypothesis:
:$\ds \paren {x + y}^n = \sum_{k \mathop = 0}^n \binom n k x^{n - k} y^k$
=== Induction Step ===
This is our induction step:
The result follows by the Principle of Mathematical Induction.
== Also presented as ==
== Examples ==
== Also known as ==
== Also see ==
* Definition:Binomial Coefficient
== Sources ==
* : Chapter $\text{VI}$: On the Seashore
* : $\text I$. Algebra: The Method of Induction: Example $3$
* : $\text I$. Algebra: The Binomial Theorem
* : $3$: Elementary Analytic Methods: $3.1$ Binomial Theorem etc.: Binomial Theorem: $3.1.1$
* : $\S 2.6$. Algebra of congruences: Example $42 \ (1)$
* : $\S 0.6$: Theorem $8: \ 5$
* : $\S 3$: The Binomial Formula and Binomial Coefficients: Binomial Coefficients: $3.4$
* : $\text {3-1}$ Permutations and Combinations: Exercise $10$
* : Chapter $1$: Properties of the Natural Numbers: $\S 20 \gamma$
* : $\S 1.18$: Sequences Defined Inductively: Exercise $4$
* : $\S 3$: Natural Numbers: Exercise $\S 3.11 \ (4)$
* : Chapter $1$: Some Preliminary Considerations: $1.2$ The Binomial Theorem
* : $1$: Complex Numbers: Solved Problems: De Moivre's Theorem: $21$
* : Chapter $2$: Integers and natural numbers: $\S 2.1$: The integers: Exercise $12$
* : $35$
* : $\S 1.2.4$: Factorials and binomial coefficients
* : $\S 1.2.6$: Binomial Coefficients: $\text{F} \ (13)$
* : $\S 1.2.6$: Binomial Coefficients: Exercise $15$
* : $35$
* 
* 
* 
* 
* : $\S 3$: The Binomial Formula and Binomial Coefficients: Binomial Coefficients: $3.4$
Category:Binomial Theorem
Category:Proofs by Induction
Binomial Theorem/Multiindex

== Theorem ==
Let $\alpha$ be a multiindex, indexed by $\set {1, \ldots, n}$ such that $\alpha_j \ge 0$ for $j = 1, \ldots, n$.
Let $x = \tuple {x_1, \ldots, x_n}$ and $y = \tuple {y_1, \ldots, y_n}$ be ordered tuples of real numbers.
Then:
:$\ds \paren {x + y}^\alpha = \sum_{0 \mathop \le \beta \mathop \le \alpha} \dbinom \alpha \beta x^\beta y^{\alpha - \beta}$
where $\dbinom n k$ is a binomial coefficient.
== Proof ==
First of all, by definition of multiindexed powers:
:$\ds \paren {x + y}^\alpha = \prod_{k \mathop = 1}^n \paren {x_k + y_k}^{\alpha_k}$
Then:
On the other hand:
This shows that:
:$\ds \paren {x + y}^\alpha = \sum_{0 \mathop \le \beta \mathop \le \alpha} \dbinom \alpha \beta x^\beta y^{\alpha - \beta}$
as required.
Category:Binomial Theorem
Binomial Theorem/Ring Theory

== Theorem ==
Let $\struct {R, +, \odot}$ be a ringoid such that $\struct {R, \odot}$ is a commutative semigroup.
Let $n \in \Z: n \ge 2$.
Then:
:$\ds \forall x, y \in R: \odot^n \paren {x + y} = \odot^n x + \sum_{k \mathop = 1}^{n - 1} \binom n k \paren {\odot^{n - k} x} \odot \paren {\odot^k y} + \odot^n y$
where $\dbinom n k = \dfrac {n!} {k! \ \paren {n - k}!}$ (see Binomial Coefficient).
If $\struct {R, \odot}$ has an identity element $e$, then:
:$\ds \forall x, y \in R: \odot^n \paren {x + y} = \sum_{k \mathop = 0}^n \binom n k \paren {\odot^{n - k} x} \odot \paren {\odot^k y}$
== Proof ==
First we establish the result for when $\struct {R, \odot}$ has an identity element $e$.
For $n = 0$ we have:
:$\ds \odot^0 \paren {x + y} = e = {0 \choose 0} \paren {\odot^{0 - 0} x} \odot \paren {\odot^0 y} = \sum_{k \mathop = 0}^0 {0 \choose k} x^{0 - k} \odot y^k$
For $n = 1$ we have:
:$\ds \odot^1 \paren {x + y} = \paren {x + y} = {0 \choose 1} \paren {\odot^{1 - 0} x} \odot \paren {\odot^0 y} + {1 \choose 1} \paren {\odot^{1 - 1} x} \odot \paren {\odot^1 y} = \sum_{k \mathop = 0}^1 {1 \choose k} x^{1 - k} \odot y^k$
=== Basis for the Induction ===
For $n = 2$ we have:
This is the basis for the induction.
=== Induction Hypothesis ===
This is our inductive hypothesis:
:$\ds \forall n \ge 2: \odot^n \paren {x + y} = \odot^n x + \sum_{k \mathop = 1}^{n - 1} \dbinom n k \paren {\odot^{n - k} x} \odot \paren {\odot^k y} + \odot^n y$
=== Induction Step ===
This is the induction step:
The result follows by the Principle of Mathematical Induction.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $21$. Rings and Integral Domains: Theorem $21.1$
Category:Binomial Theorem
Category:Proofs by Induction
Category:Ring Theory
Binomial Theorem Approximations/Examples/Arbitrary Example 1

== Example of Binomial Theorem Approximation ==
:$\paren {1 \cdotp 0 6}^{1/3} \approx 1 \cdotp 019613$
to $6$ decimal places.
== Proof ==
== Sources ==
* : $\text I$. Algebra: The Binomial Theorem: Approximations: Example $1$.
Category:Examples of Binomial Theorem Approximations
Binomial Theorem Approximations/Examples/Arbitrary Example 2

== Example of Binomial Theorem Approximation ==
:$\sqrt {25 \cdotp 1} \approx 5 \cdotp 0100$
to $4$ decimal places.
== Proof ==
== Sources ==
* : $\text I$. Algebra: The Binomial Theorem: Approximations: Example $2$.
Category:Examples of Binomial Theorem Approximations
Binomial Theorem for Integral Index/Also presented as

== Binomial Theorem: Also known as == 
The binomial theorem (for integral index) can also be presented in the form:
which is seen to be the same as the given form by definition of binomial coefficient.
== Sources ==
* : $\S 3$: The Binomial Formula and Binomial Coefficients: Binomial Formula for Positive Integral $n$: $3.3$
* : $\S 3$: The Binomial Formula and Binomial Coefficients: Binomial Formula for Positive Integral $n$: $3.3.$
Category:Binomial Theorem
Binomial Theorem for Negative Index and Negative Parameter

== Theorem ==
Let $n \in \Z_{\ge 0}$ be a positive integer.
Let $z \in \R$ be a real number such that $\size z < 1$.
Then:
where $\dbinom {n + k} n$ denotes a binomial coefficient.
== Proof ==
== Sources ==
* : $\S 1.2.9$: Generating Functions: $(20)$
Category:Binomial Theorem
Binomial is Irrational

== Theorem ==
== Proof ==
Let $AB$ and $BC$ be two rational straight lines which are commensurable in square only.
By definition, $AB$ and $BC$ are incommensurable in length.
We have:
:$AB : BC = AB \cdot BC : BC^2$
Therefore from :
: $AB \cdot BC$ is incommensurable with $BC^2$.
But by :
:$2 AB \cdot BC$ is commensurable with $AB \cdot BC$.
We have that $AB$ and $BC$ are commensurable in square.
So from :
:$AB^2 + BC^2$ is commensurable with $BC^2$.
So from :
:$2 AB \cdot BC$ is incommensurable with $AB^2 + BC^2$.
Thus from :
:$AC^2 = \left({AB + BC}\right)^2 = 2 AB \cdot BC + AB^2 + BC^2$ is incommensurable with $AB^2 + BC^2$.
But $AB^2 + BC^2$ is rational.
Therefore $AC^2$ is irrational.
Thus by :
:$AC$ is irrational.
Such a straight line is called binomial.
== Sources ==
* : Book $\text{X}$. Propositions
Category:Euclidean Number Theory
Biot-Savart Law

== Theorem ==
Let $s$ be a wire carrying a steady current $I$.
Let $\map {\mathbf B} {\mathbf r}$ be the total magnetic field due to $I$ flowing through $s$ at a point $P$ whose position vector is $\mathbf r$.
Then:
:$\ds \map {\mathbf B} {\mathbf r} = \dfrac {\mu_0 I} {4 \pi} \oint_s \dfrac {\d \mathbf l \times \paren {\mathbf r - \mathbf r'} } {\size {\mathbf r - \mathbf r'}^3}$
where:
:$\d \mathbf l$ is an infinitesimal vector associated with $s$
:$\mathbf r$ is the position vector of $P$
:$\mathbf r'$ is the position vector of $\d \mathbf l$
:$\mu_0$ denotes the vacuum permeability.
== Proof ==
== Sources ==
* : Chapter $4$: Steady currents and magnetic fields: $4.4$ Ampère's Law: $4.4.2$ The Biot-Savart law: $(4.32)$
Category:Biot-Savart Law
Category:Electromagnetism
Birch and Swinnerton-Dyer Conjecture

== Unsolved Problem ==
When the solution to a Diophantine equation in polynomials are the points of an Abelian variety, the order of the group of rational points is related to the behavior of an associated $\zeta$ (zeta) function $\map \zeta s$ near $s = 1$.
In particular:
:if $\map \zeta 1 = 0$ then there is an infinite set of rational points
:if $\map \zeta 1 \ne 0$ then there is a finite set of rational points.
== Progress ==
As of now, the Birch and Swinnerton-Dyer Conjecture has been confirmed only for special cases.
== Also known as ==
== Sources ==
* 
* 
* 
* 
* : Appendix $18$: Millennium Prize problems: $6$.
* : Appendix $23$: Millennium Prize problems: $6$.
Category:Birch and Swinnerton-Dyer Conjecture
Category:Millennium Problems
Category:Unsolved Problems
Birch and Swinnerton-Dyer Conjecture/Also known as

== Birch and Swinnerton-Dyer Conjecture: Also known as == 
The Birch and Swinnerton-Dyer Conjecture is also presented as the Birch–Swinnerton-Dyer Conjecture, where the first hyphen is longer than the second.
This style of presentation is not endorsed on , as the long dash is not simple to implement.
Category:Birch and Swinnerton-Dyer Conjecture
Birkhoff's Ergodic Theorem

== Theorem ==
Let $\struct {X, \BB, \mu, T}$ be a measure-preserving dynamical system.
Let $f: X \to \overline \R$ be a $\mu$-integrable function.
Then a $\mu$-integrable function $f^\ast$ exists such that:
:$\forall x \in X : \map {f^\ast} {\map T x} = \map {f^\ast} x$
and:
:$\ds \lim_{n \mathop \to \infty} \dfrac 1 n \sum_{n \mathop = 0}^{n - 1} f \circ T^n = f^\ast$
converges $\mu$-almost everywhere and in $L^1$-norm.
Furthermore, we have:
:$f^\ast = \expect {f \mid \II_T}$
where:
:$\expect {f \mid \II_T}$ denotes the conditional expectation of $f$ given $\II_T$
:$\II_T := \set { A \in \BB : T^{-1} \sqbrk A = A }$
== Proof ==
== Also known as ==
Birkhoff's Ergodic Theorem is also known as 
* The Birkhoff ergodic theorem
* The strong ergodic theorem
* The pointwise ergodic theorem.
== Also see ==
* Mean Ergodic Theorem (also known as the Weak Ergodic Theorem) of 
== Sources ==
* 
* $2.6$: Pointwise Ergodic Theorem
Category:Dynamical Systems Theory
Category:Ergodic Theory
Birkhoff's Theorem

* Theorems named for the American mathematician :
** Birkhoff's Theorem (Relativity)
** Birkhoff's Theorem (Electromagnetism)
** Birkhoff's Ergodic Theorem
* Theorems named for his son, :
** Birkhoff's HSP Theorem, concerning the closure operations of homomorphism, subalgebra and product
** Birkhoff's Representation Theorem for distributive lattices
** Birkhoff-von Neumann Theorem (with )
Birkhoff-Kakutani Theorem/Topological Vector Space/Corollary

== Corollary ==
Let $\GF \in \set {\R, \C}$.
Let $\struct {X, \tau}$ be a topological vector space over $\GF$.
Then $\struct {X, \tau}$ is metrizable $\struct {X, \tau}$ is first-countable and Hausdorff.
Further, if $\struct {X, \tau}$ is metrizable then there exists an invariant metric $d$ on $X$ such that:
:$(1): \quad$ $d$ induces $\tau$
:$(2): \quad$ the open balls in $\struct {X, d}$ are balanced.
== Proof ==
=== Necessary Condition ===
Suppose that $\struct {X, \tau}$ is metrizable.
From Metric Space is First-Countable, $\struct {X, \tau}$ is first-countable.
From Metric Space is Hausdorff, $\struct {X, \tau}$ is Hausdorff.
=== Sufficient Condition ===
Suppose that $\struct {X, \tau}$ is first-countable and Hausdorff.
By Birkhoff-Kakutani Theorem: Topological Vector Space, there exists an invariant pseduometric $d$ on $X$ such that:
:$(1) \quad$ $d$ induces $\tau$
:$(2) \quad$ the open balls in $\struct {X, d}$ are balanced.
From Topological Vector Space is Hausdorff iff T1 and T1 Space is T0 Space, we have that $\struct {X, \tau}$ is a Kolmogorov space.
Hence from Pseudometric Space is Metric Space iff Kolmogorov, $\struct {X, d}$ is a metric space.
That is, $d$ is in fact an invariant metric.
So $\struct {X, \tau}$ is metrizable and there exists an invariant metric $d$ on $X$ such that:
:$(1): \quad$ $d$ induces $\tau$
:$(2): \quad$ the open balls in $\struct {X, d}$ are balanced.
Birthday Paradox

== Paradox ==
Let there be $23$ or more people in a room.
The probability that at least $2$ of them have the same birthday is greater than $50 \%$.
== Proof ==
Let there be $n$ people in the room.
Let $\map p n$ be the probability that no two people in the room have the same birthday.
For simplicity, let us ignore leap years and assume there are $365$ days in the year.
Let the birthday of person $1$ be established.
The probability that person $2$ shares person $1$'s birthday is $\dfrac 1 {365}$.
Thus, the probability that person $2$ does not share person $1$'s birthday is $\dfrac {364} {365}$.
Similarly, the probability that person $3$ does not share the birthday of either person $1$ or person $2$ is $\dfrac {363} {365}$.
And further, the probability that person $n$ does not share the birthday of any of the people indexed $1$ to $n - 1$ is $\dfrac {365 - \paren {n - 1} } {365}$.
Hence the total probability that none of the $n$ people share a birthday is given by:
Setting $n = 23$ and evaluating the above gives:
:$\map p {23} \approx 0.493$
Hence the probability that at least $2$ people share a birthday is $1.000 - 0.493 = 0.507 = 50.7 \%$
== Conclusion ==
This is a veridical paradox.
Counter-intuitively, the probability of a shared birthday amongst such a small group of people is surprisingly high.
== General Birthday Paradox ==
== Sources ==
* : $23$
* : $23$
Category:Probability Theory
Category:Combinatorics
Category:Veridical Paradoxes
Category:Birthday Paradox
Birthday Paradox/General/3

== Theorem ==
Let $n$ be a set of people.
Let the probability that at least $3$ of them have the same birthday be greater than $50 \%$.
Then $n \ge 88$.
== Proof ==
Let $\map F {r, n}$ be the number of ways to distribute $r$ objects into $n$ cells such that there are no more than $2$ objects in each cell.
Let there be $d$ cells which are each occupied by $2$ objects.
These can be chosen in $\dbinom n d$ ways.
There remain $s = r - 2 d$ objects which can then be distributed among $n - d$ cells in $\dbinom {n - d} s$ ways.
In each such arrangement, the $r$ objects may be permuted in:
:$\dbinom r 2 \dbinom {r - 2} 2 \cdots \dbinom {r - 2 d + 2} 2 \paren {r - 2 d}! = \dfrac {r!} {2^d}$
different ways.
Hence:
:$\map F {r, n} = \dbinom n d \dbinom {n - d} s \dfrac {r!} {2^d}$
So the probability of exactly $d$ pairs and $s$ singletons, where $d - s \le n$, is given by:
:$\dfrac {\map F {r, n} } {n^r}$
If we assume a $365$-day year, we have that the probability that at least $3$ of them have the same birthday is given by:
:$\map \Pr r = 1 - \ds \sum_{d \mathop = 0}^{\floor {r / 2} } \dfrac {n! \, r!} {n^r 2^d d! \paren {r - 2 d}! \paren {n + d - r}!}$
where $n = 365$.
We require the smallest $r$ for which $\map \Pr r > \dfrac 1 2$.
The result yields to calculation.
== Sources ==
* 
* : $23$
Category:Birthday Paradox
Bisection Method/Examples/Arbitrary Example 1

== Example of Use of Bisection Method ==
Suppose that, using the bisection method, a solution to $\map f x = 0$ has been found in the interval $\closedint {1.717} {1.724}$.
Then the solution is $x = 1.72$ to $3$ significant figures.
== Sources ==
* 
* 
Category:Bisection Method
Bisection of Angle

== Theorem ==
It is possible to bisect any given rectilineal angle.
== Construction ==
== Proof ==
We have:
:$AD = AE$
:$AF$ is common
:$DF = EF$
Thus from :
:$\triangle ADF \sim \triangle AEF$
Thus:
:$\angle DAF = \angle EAF$
Hence $\angle BAC$ has been bisected by $AF$.
== Sources ==
* : Book $\text{I}$. Propositions
Category:Bisection of Angle
Category:Angles
Category:Angle Bisectors