nyuuzyou/wikis huggingface dataset:
Banach Isomorphism Theorem

== Theorem ==
Let $\struct {X, \norm \cdot_X}$ and $\struct {Y, \norm \cdot_Y}$ be Banach spaces.
Let $T : X \to Y$ be a bijective bounded linear transformation.
Then the inverse of $T$ is a bounded linear transformation.
That is, $T$ is a normed vector space isomorphism.
== Proof ==
Let $B_X^-$ be the closed unit ball of $X$.
Let $B_Y^-$ be the closed unit ball of $Y$.
Let $T^{-1} : Y \to X$ be the inverse of $T$.
From Inverse of Linear Transformation is Linear Transformation, $T^{-1} : Y \to X$ is a linear transformation.
It remains to show that $T^{-1}$ is bounded.
Since $T$ is bijective, it is surjective.
So, by the Banach-Schauder Theorem:
:$T$ is an open mapping.
From Characterization of Open Linear Transformation between Normed Vector Spaces, there exists $\delta > 0$ such that: 
:$\delta B_Y^- \subseteq T \sqbrk {B_X^-}$
Then we have from Image of Subset under Mapping is Subset of Image and Image of Preimage under Mapping: 
:$T^{-1} \sqbrk {\delta B_Y^-} \subseteq B_X^-$
From Image of Dilation of Set under Linear Transformation is Dilation of Image, we have: 
:$T^{-1} \sqbrk {B_Y^-} \subseteq \delta^{-1} B_X^-$
From Norm of Bounded Linear Transformation in terms of Closed Unit Ball, $T^{-1}$ is therefore bounded.
== Also known as ==
This theorem is also known as the inverse mapping theorem.
== Sources ==
* $23.1$: The Open Mapping and Inverse Mapping Theorems
Category:Linear Transformations on Banach Spaces
Banach Space Valued Function is Analytic iff Weakly Analytic

== Theorem ==
Let $U$ be an open subset of $\C$.
Let $\struct {X, \norm {\, \cdot \,} }$ be a Banach space over $\C$.
Let $f : U \to X$ be a function.
Then $f$ is analytic it is weakly analytic.
== Proof ==
=== Necessary Condition ===
Suppose that $f$ is analytic. 
Define $f' : U \to X$ by: 
:$\ds \map {f'} z = \lim_{w \mathop \to z} \frac {\map f w - \map f z} {w - z}$
for each $z \in U$. 
Let $\struct {X^\ast, \norm {\, \cdot \,}_{X^\ast} }$ be the normed dual space of $\struct {X, \norm {\, \cdot \,} }$.
Let $\phi \in X^\ast$. 
Then for each $z, w \in U$ with $z \ne w$ we have: 
:$\ds \map \phi {\frac {\map f w - \map f z} {w - z} } = \frac {\map {\paren {\phi \circ f} } w - \map {\paren {\phi \circ f} } z} {w - z}$
since $\phi$ is linear. 
From Continuity of Linear Functionals, $\phi$ is continuous. 
So, we have: 
:$\ds \lim_{w \mathop \to z} \map \phi {\frac {\map f w - \map f z} {w - z} } = \map \phi {\map {f'} z}$
So:
:$\ds \frac {\map {\paren {\phi \circ f} } w - \map {\paren {\phi \circ f} } z} {w - z}$ exists for each $z \in U$. 
So $\phi \circ f$ is analytic for each $\phi \in X^\ast$. 
So $f$ is weakly analytic.
=== Sufficient Condition ===
Category:Analytic Functions (Banach Spaces)
Banach Space is F-Space

== Theorem ==
Let $\GF \in \set {\R, \C}$. 
Let $\struct {X, \norm {\, \cdot \,} }$ be a Banach space over $\GF$. 
From Normed Vector Space is Hausdorff Topological Vector Space, we may consider $\struct {X, \norm {\, \cdot \,} }$ as a topological vector space.
With this identification, $\struct {X, \norm {\, \cdot \,} }$ is an $F$-Space.
== Proof ==
Let $d$ be the metric induced by $\norm {\, \cdot \,}$.
From Norm Topology Induced by Metric Induced by Norm, $d$ induces the topology on $\struct {X, \norm {\, \cdot \,} }$.
Since $\struct {X, \norm {\, \cdot \,} }$ is a Banach space, $\struct {X, d}$ is a complete metric space. 
From Metric Induced by Norm is Invariant Metric, $d$ is invariant.
Hence $\struct {X, \norm {\, \cdot \,} }$ is an $F$-Space.
Category:F-Spaces
Category:Banach Spaces
Banach Space is Reflexive iff Normed Dual is Reflexive

== Theorem ==
Let $\Bbb F \in \set {\R, \C}$. 
Let $X$ be a Banach space over $\Bbb F$.
Let $X^\ast$ be the normed dual space of $X$. 
Then: 
:$X$ is reflexive $X^\ast$ is reflexive.
== Proof ==
Let $X^{\ast \ast}$ be the second normed dual of $X$.
Let $X^{\ast \ast \ast}$ be the normed dual of $X^{\ast \ast}$. 
=== Necessary Condition ===
Suppose that $X$ is reflexive.
We want to show that $X^\ast$ is reflexive.
From Normed Vector Space is Reflexive iff Surjective Evaluation Linear Transformation, for each $\Phi \in X^{\ast \ast \ast}$ we aim to find $\phi \in X^\ast$ such that: 
:$\map \Phi F = \map {\phi^\wedge} F = \map {\map {J_{X^\ast} } f} F$ for each $F \in X^{\ast \ast}$
where $J_{X^\ast}$ is the evaluation linear transformation $X^\ast \to X^{\ast \ast \ast}$.
Since $X$ is reflexive, for each $F \in X^{\ast \ast}$ there exists $x \in X$ such that:
:$F = x^\wedge = \map {J_X} x$
where $J_X$ is the evaluation linear transformation $X \to X^{\ast \ast}$.
So, it suffices to find $\phi$ such that for each $x \in X$ we have: 
for each $x \in X$. 
We show that this actually defines an element $\phi \in X^\ast$.
From Evaluation Linear Transformation on Normed Vector Space is Linear Transformation from Space to Second Normed Dual, we have that: 
:$x \mapsto x^\wedge$ is a linear transformation from $X$ to $X^{\ast \ast}$.
Further, from Evaluation Linear Transformation on Normed Vector Space is Linear Isometry:
:$x \mapsto x^\wedge$ is a bounded linear transformation from $X$ to $X^{\ast \ast}$.
Since we also have that: 
:$\Phi$ is a bounded linear functional $X^{\ast \ast} \to \Bbb F$
we have: 
:$\phi$ is a bounded linear functional $X \to \Bbb F$
from Composition of Bounded Linear Transformations is Bounded Linear Transformation.
That is: 
:$\phi \in X^\ast$
as required.
=== Sufficient Condition ===
Suppose that $X^\ast$ is reflexive.
 that $X$ is not reflexive.
From Normed Vector Space is Reflexive iff Surjective Evaluation Linear Transformation, this means that: 
:for all $\Phi \in X^{\ast \ast \ast}$ there exists $\phi \in X^\ast$ such that $\Phi = \phi^\wedge$
but:
:for some $F \in X^{\ast \ast}$ there does not exist $x \in X$ such that $F = x^\wedge$.
From Image of Evaluation Linear Transformation on Banach Space is Closed Linear Subspace of Second Dual, we have that: 
:$\map J X = \set {x^\wedge \in X^{\ast \ast} : x \in X}$ is a closed linear subspace of $X^{\ast \ast}$.
From Existence of Distance Functional, there exists $\Phi \in X^{\ast \ast \ast}$ such that: 
:$\norm {\Phi}_{X^{\ast \ast \ast} } = 1$ 
and: 
:$\map \Phi F = 0$ for all $F \in \map J X$.
That is: 
:$\map \Phi {x^\wedge} = 0$ for all $x \in X$.
Since $X^\ast$ is reflexive, there exists $\phi \in X^\ast$ such that: 
:$\Phi = \phi^\wedge$
Then for each $x \in X$ we have: 
Then we have: 
:$\Phi \equiv 0$
so:
:$\norm \Phi_{X^{\ast \ast \ast} } = 0 \ne 1$
a contradiction.
So we have that $X$ is reflexive.
Category:Reflexive Spaces
Category:Normed Dual Spaces
Band Matrix/Examples/Arbitrary Example 1

== Example of Band Matrix ==
This is an example of a band matrix:
:$\begin{pmatrix} 1 & 2 & 3 & 0 & 0 \\ 1 & 1 & 2 & 1 & 0 \\ 0 & 2 & 1 & 4 & 1 \\ 0 & 0 & 2 & 1 & 1 \\ 0 & 0 & 0 & 1 & 1 \end{pmatrix}$
== Sources ==
* 
* 
Category:Band Matrices
Barber Paradox

== Paradox ==
There exists a community, one of whose members is a .
This barber operated under an unusual rule: his task was to shave every man in the community who did not shave himself, and only those men.
Who shaves the barber?
If he does not shave himself, then he must shave himself.
But if he shaves himself, he must not shave himself.
== Analysis 1 ==
== Analysis 2 ==
== Resolution ==
This paradox has been demonstrated to be an antinomy, and can be resolved by several means, for example:
=== Resolution 1 ===
=== Resolution 2 ===
=== Resolution 3 ===
== Historical Note ==
== Sources ==
* : $7.$ The Theory of Types and Symbolism: Classes
* : $\text{I}$: 'NOT' and 'IF': $(2)$
* : Chapter $1$: Elementary, my dear Watson: $\S 1.2$: Elements, my dear Watson: Example $1.2.1$
* : Chapter $1$: General Background: $\S 8$ Russell's paradox
Category:Barber Paradox
Category:Russell's Paradox
Category:Antinomies
Category:Logical Paradoxes
Barber Paradox/Analysis 1

== Paradox ==
== Analysis ==
This is an application of Russell's Paradox.
Let $\mathbb U$ be the set of all the men of the community.
Thus $\mathbb U$ is considered to be the universe.
Let $S: \mathbb U \to \set {T, F}$ be the propositional function:
:$\forall x \in \mathbb U: \map S x \iff x \text { is shaved by $x$}$
Let $b \in \mathbb U$ be the barber.
Let $B: \mathbb U \to \set {T, F}$ be the propositional function:
:$\forall x \in \mathbb U: \map B x \iff x \text { is shaved by $b$}$
The initial premises can be coded:
:$(1): \quad \forall x \in \mathbb U: \paren {\neg \map S x} \iff \map B x$
:$(2): \quad \map B b \iff \map S b$
Hence:
:$\map S b \iff \map B b \iff \paren {\neg \map S b}$
and so from Biconditional is Transitive:
:$\map S b \iff \paren {\neg \map S b}$
So from either case there derives a contradiction.
Thus the initial premises are contradictory and cannot both hold.
== Sources ==
* 
Category:Barber Paradox
Barber Paradox/Analysis 2

== Paradox ==
== Analysis ==
Let $\map M x$ be defined as:
:$x$ is a man in the community. 
Let $\map S {x, y}$ be defined as:
:$x$ shaves $y$.
Let $b$ be the barber.
Suppose $\map M b$.
Suppose that:
:$\forall x, y: \paren {\map S {x, y} \implies \map M x, \map M y}$
Suppose to the contrary that:
:$\forall x: \paren {\map M x \implies \paren {\map S {b, x} \iff \neg \map S {x, x} } }$
For $x = b$ we obtain the contradiction:
:$\map S {b, b} \iff \neg \map S {b, b}$
Therefore, it must be false that:
:$\forall x: \paren {\map M x \implies \paren {\map S {b, x} \iff \neg \map S {x, x} } }$
Category:Barber Paradox
Barber Paradox/Resolution 1

== Paradox ==
== Analysis ==
== Resolution ==
Let $b$ be defined so that $b \notin \mathbb U$.
That is, suppose $b$ is not one of the men of the community.
This could be the case by, for example:
:$(1): \quad$ The barber is a woman
:$(2): \quad$ The barber is a boy too young to shave.
Then as $b \notin \mathbb U$, it is not necessarily the case that:
:$\paren {\neg \map S b} \implies \map B b$
Thus $b$ is allowed not to be shaved, by himself or anyone else.
Category:Barber Paradox
Barber Paradox/Resolution 2

== Paradox ==
== Analysis ==
== Resolution ==
Let the only condition above be relaxed, and rewrite it as:
:his task was to shave every man in the community who did not shave himself.
The initial premises would be coded:
:$(1): \quad \forall x \in \mathbb U: \paren {\neg \map S x} \implies \map B x$
:$(2): \quad \map B b \iff \map S b$
Thus it is not the case that:
:$\map B x \implies \paren {\neg \map S x}$
and so the barber is allowed to shave at least one man who does shave himself, the barber himself necessarily being one such.
Category:Barber Paradox
Barber Paradox/Resolution 3

== Paradox ==
== Analysis ==
== Resolution ==
Let the every condition above be relaxed, and rewrite it as:
:his task was that he may shave only men in the community who did not shave themselves.
The initial premises can be coded:
:$(1): \quad \forall x \in \mathbb U: \map B x \implies \paren {\neg \map S x}$
:$(2): \quad \map B b \iff \map S b$
Thus it is not the case that:
:$\paren {\neg \map S x} \implies \map B x$
However, from $\map B b \iff \map S b$ this then means that $\neg \map B b$.
So at least one person in the community is not shaven by $b$ at all, the barber himself necessarily being one such.
This of course does not preclude the possibility that some other person, who is not the barber, may also shave people.
These may or may not include the barber, who may retain his beard.
Category:Barber Paradox
Barbier's Theorem

== Theorem ==
Let $K$ be a closed curve of constant diameter.
Let the circumference of $K$ be $c$.
Let the diameter of $K$ be $d$.
Then:
:$\dfrac c d = \pi$
== Proof ==
== Sources ==
* 
* : $3 \cdotp 14159 \, 26535 \, 89793 \, 23846 \, 26433 \, 83279 \, 50288 \, 41971 \ldots$
Category:Geometry
Barlow's Prediction

== Prediction ==
: ascertained that $2^{31} - 1 = 2147483647$ is a prime number; and this is the greatest at present known to be such, and, consequently, the last of the above perfect numbers [that is, $2^{30}\left({2^{31} - 1}\right)$], which depends upon this, is the greatest perfect number known at present, and probably the greatest that ever will be discovered; for, as they are merely curious without being useful, it is not likely that any person will attempt to find one beyond it.
This statement was made by , in his $1811$ work .
He repeated this statement word for word in his $1814$ work .
See the definition of Mersenne prime to follow up on exactly how inaccurate that prediction was.
== Also see ==
* Theorem of Even Perfect Numbers
== Sources ==
* 
* : $28$
* : $28$
Category:Mistakes
Barycenter Exists and is Well Defined

== Theorem ==
Let $\EE$ be an affine space over a field $k$.
Let $p_1, \ldots, p_n \in \EE$ be points.
Let $\lambda_1, \ldots, \lambda_n \in k$ such that $\ds \sum_{i \mathop = 1}^n \lambda_i = 1$.
Then the barycentre of $p_1, \ldots, p_n$ with weights $\lambda_1, \ldots, \lambda_n$ exists and is unique.
== Proof ==
Let $r$ be any point in $\EE$.
Set:
:$\ds q = r + \sum_{i \mathop = 1}^n \lambda_i \vec{r p_i}$
We are required to prove that for any other point $m \in \EE$:
:$\ds q = m + \sum_{i \mathop = 1}^n \lambda_i \vec{m p_i}$
So:
Hence the result.
Category:Affine Geometry
Basel Problem

== Theorem ==
:$\ds \map \zeta 2 = \sum_{n \mathop = 1}^\infty {\frac 1 {n^2} } = \frac {\pi^2} 6$
where $\zeta$ denotes the Riemann zeta function.
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Proof 4 ==
== Proof 5 ==
== Proof 6 ==
== Proof 7 ==
== Proof 8 ==
== Proof 9 ==
== Proof 10 ==
== Historical Note ==
== Sources ==
* : $\S 19$: Series involving Reciprocals of Powers of Positive Integers: $19.19$
* : $3 \cdotp 14159 \, 26535 \, 89793 \, 23846 \, 26433 \, 83279 \, 50288 \, 41972 \ldots$
* 
* 
* : Preface to first edition: Prerequisites: $\text {(iv)}$
* : $3 \cdotp 14159 \, 26535 \, 89793 \, 23846 \, 26433 \, 83279 \, 50288 \, 41971 \ldots$
* 
* 
* : $\S 21$: Series of Constants: Series Involving Reciprocals of Powers of Positive Integers: $21.19.$
* 
Category:Basel Problem
Category:Riemann Zeta Function at Even Integers
Category:Examples of Riemann Zeta Function
Category:Formulas for Pi
Category:Named Theorems
Basel Problem/Historical Note

== Historical Note on Basel Problem ==
The Basel Problem was first posed by in $1644$.
Its solution is generally attributed to , who solved it in $1734$ and delivered a proof in $1735$.
However, it has also been suggested that it was in fact first solved by .
 had earlier established that the series was convergent, but had failed to work out what to.
The problem is named after Basel, the home town of as well as of the .
:If only my brother were alive now.
:::: -- 
== Sources ==
* : Chapter $\text {A}.20$: The Bernoulli Brothers
* : Chapter $\text {A}.21$: Euler ()
Category:Historical Notes
Basel Problem/Proof 1

== Theorem ==
== Proof ==
By Riemann Zeta Function as a Multiple Integral, 
:$\ds \map \zeta 2 = \int_0^1 \int_0^1 \frac 1 {1 - x y} \rd x\rd y$ 
Let $\tuple {u, v} = \tuple {\dfrac {x + y} 2, \dfrac{y - x} 2}$ so that:
:$\tuple {x, y} = \tuple {u - v, u + v}$
The Jacobian is:
:$\size J = \size {\dfrac {\partial \tuple {x, y} } {\partial \tuple {u, v} } } = 2$
Then, by Change of Variables Theorem (Multivariable Calculus):
:$\map \zeta 2 = \ds 2 \iint \limits_S \frac {\d u \rd v} {1 - u^2 + v^2}$
where $S$ is the square defined by the coordinates:
:$\tuple {0, 0}, \ \tuple {\dfrac 1 2, -\dfrac 1 2}, \ \tuple {1, 0}, \ \tuple {\dfrac 1 2, \dfrac 1 2}$
Exploiting the symmetry of the square and the function over the $u$-axis, we have:
:$\map \zeta 2 = \ds 4 \paren {\int_0^{\frac 1 2} \! \int_0^u \frac {\d v \rd u} {1 - u^2 + v^2} + \int_{\frac 1 2}^1 \! \int_0^{1 - u} \frac {\d v \rd u} {1 - u^2 + v^2} }$
Factoring $1 - u^2$ gives us:
:$\map \zeta 2 = \ds 4 \paren {\int_0^{\frac 1 2} \! \int_0^u \frac 1 {1 - u^2} \frac {\d v \rd u} {\frac {v^2} {1 - u^2} + 1} + \int_{\frac 1 2}^1 \! \int_0^{1 - u} \frac 1 {1 - u^2} \frac {\d v \rd u} {\frac {v^2} {1 - u^2} + 1} }$
and letting:
:$s = \dfrac v {\sqrt {1 - u^2} }, \rd s = \dfrac 1 {\sqrt {1 - u^2} }$
allows us to make a substitution into each integral, giving:
:$\map \zeta 2 = \ds 4 \paren {\int_0^{\frac 1 2} \frac 1 {\sqrt {1 - u^2} } \map \arctan {\frac u {\sqrt {1 - u^2} } } \rd u + \int_{\frac 1 2}^1 \frac 1 {\sqrt {1 - u^2} } \map \arctan {\frac {1 - u} {\sqrt {1 - u^2} } } \rd u}$
Consider the right triangle with sides $1$, $x$ and $\sqrt {1 - x^2}$.
Applying Pythagoras's Theorem:
:$\arcsin x = \arctan \dfrac x {\sqrt {1 - x^2} }$
Let:
This allows us to convert the arctangents from the integrals into arcsines:
:$\map \zeta 2 = \ds 4 \paren {\int_0^{\frac 1 2} {\frac {\arcsin u} {\sqrt {1 - u^2} } \rd u} + \int_{\frac 1 2}^1 {\frac 1 {\sqrt {1 - u^2} } \paren {\frac \pi 4 - \frac {\arcsin u} 2} \rd u} }$
Substituting:
:$s = \arcsin u$, $\rd s = \dfrac 1 {\sqrt{1 - u^2} }$
into the arcsines, and splitting the second integral:
:$\map \zeta 2 = 4 \paren {\dfrac {\pi^2} {72} + \dfrac {\pi^2} {36} } = \dfrac {\pi^2} 6$
== Historical Note ==
== Sources ==
* : Chapter $\text {B}.24$: Euler's Formula $\ds \sum \limits_1^\infty \frac 1 {n^2} = \frac {\pi^2} 6$ by Double Integration
Category:Basel Problem
Basel Problem/Proof 10

== Theorem ==
== Proof ==
For $z \ne 0$, from Mittag-Leffler Expansion for Hyperbolic Cotangent Function, we have: 
:$\ds \frac 1 {2 z} \paren {\pi \map \coth {\pi z} - \frac 1 z} = \sum_{n \mathop = 1}^\infty \frac 1 {z^2 + n^2}$
Consider the set:
:$A = \set {z \in \C : \size z \le \dfrac 1 2}$. 
Then for each $n \in \N$ and $z \in A$, we have: 
For all $n \in \N$, we have: 
:$n^2 - \dfrac 1 4 \ge \dfrac 1 2 n^2$ 
So that: 
:$\dfrac 1 {n^2 - \dfrac 1 4} \le \dfrac 2 {n^2}$ 
We have: 
So, from the Cauchy Integral Test, we have: 
:$\ds \sum_{n \mathop = 1}^\infty \frac 2 {n^2}$ converges.
So, from the Comparison Test: Corollary, we have: 
:$\ds \sum_{n \mathop = 1}^\infty \frac 1 {n^2 - \frac 1 4}$ converges.
So by the Weierstrass M-Test, the series:
:$\ds \sum_{n \mathop = 1}^\infty \frac 1 {z^2 + n^2}$
converges uniformly on $A$. 
Then from Uniformly Convergent Series of Continuous Functions is Continuous, the function $f : A \to \C$ defined by: 
:$\ds \map f z = \sum_{n \mathop = 1}^\infty \frac 1 {z^2 + n^2}$
is continuous.
So:
:$\ds \lim_{z \mathop \to 0} \sum_{n \mathop = 1}^\infty \frac 1 {z^2 + n^2} = \sum_{n \mathop = 1}^\infty \frac 1 {n^2}$
We can write: 
so that: 
:$\ds \sum_{n \mathop = 1}^\infty \frac 1 {n^2} = \lim_{z \mathop \to 0} \paren {\frac {\pi z \paren {e^{2 \pi z} + 1} - e^{2 \pi z} + 1} {2 z^2 \paren {e^{2 \pi z} - 1} } }$
We have at $z = 0$: 
and: 
:$2 z^2 \paren {e^{2 \pi z} + 1} = 0$
So by L'Hopital's Rule:
giving: 
:$\ds \sum_{n \mathop = 1}^\infty \frac 1 {n^2} = \frac {\pi^2} 6$
Category:Basel Problem
Basel Problem/Proof 2

== Theorem ==
== Proof ==
Let:
:$\ds P_k = x \prod_{n \mathop = 1}^k \paren {1 - \frac {x^2} {n^2 \pi^2} }$
We note that:
By Telescoping Series we find that the coefficient of $x^3$ in $P_k$ is given by:
:$(1): \quad \ds -\frac 1 {\pi^2} \sum_{n \mathop = 1}^k \frac 1 {n^2}$
We also have the following two representations of the Sine of x:
So by taking the limit as $k \to \infty$ in $(1)$ and equating with the coefficient of $x^3$ in the Power Series Expansion for Sine Function, we can deduce:
:$\ds -\frac 1 {\pi^2} \sum_{n \mathop = 1}^\infty \frac 1 {n^2} = -\frac 1 {3!}$
hence:
:$\ds \sum_{n \mathop = 1}^{\infty} \frac 1 {n^2} = \frac {\pi^2} 6$
== Historical Note ==
== Sources ==
* : Chapter $\text {B}.14$: Euler's Discovery of the Formula $\ds \sum_1^\infty \frac 1 {n^2} = \frac {\pi^2} 6$
* : Chapter $9$: The Extraordinary Sums of Leonhard Euler (1734): Great Theorem
Category:Basel Problem
Basel Problem/Proof 3

== Theorem ==
== Proof ==
Let $x \in \openint 0 {\dfrac \pi 2}$ and let $n$ be a non-negative integer.
Let $x_k = \dfrac {k \pi} {2 n + 1}$ for $k = 1, 2, \ldots, n$. 
Then:
:$\sin \paren {2 n + 1} x_k = 0$
So we have:
:$\ds \sum_{r \mathop = 0}^n \binom {2 n + 1} {2 r + 1} \paren {-1}^r \cot^{2 \paren {n - r} } x_k = 0$
for $k = 1, 2, \ldots, n$.
The numbers $x_k$ are all distinct and in the interval $\openint 0 {\dfrac \pi 2}$. 
By Shape of Cotangent Function, $\cot x$ is positive and injective in the interval $\openint 0 {\dfrac \pi 2}$.
Therefore $\cot^2 x$ is also injective in this interval. 
Hence the numbers $c_k = \cot^2 x_k$ are distinct for $k = 1, 2, \ldots, n$. 
These numbers are the $n$ distinct roots of the $n$th degree polynomial:
:$\ds \map f c := \sum_{r \mathop = 0}^n \binom {2 n + 1} {2 r + 1} (-1)^r c^{n - r}$
By Viète's Formulas, we can calculate the sum of the roots:
:$\ds \sum_{k \mathop = 1}^n \cot^2 x_k = \frac {\binom {2 n + 1} 3} {\binom {2 n + 1} 1} = \frac {2 n \paren {2 n - 1} } 6$
From the Difference of Squares of Cosecant and Cotangent $\cot^2 x = \csc^2 x - 1$ we can similarly deduce:
:$\ds \sum_{k \mathop = 1}^n \csc^2 x_k = \frac {2 n \paren {2 n - 1} } 6 + n = \frac {2 n \paren {2 n + 2} } 6$
By Shape of Sine Function, $\sin x$ is positive in the interval $\openint 0 {\dfrac \pi 2}$. 
So from the Tangent Inequality $\sin x < x < \tan x$ for $x \in \openint 0 {\dfrac \pi 2}$, we can deduce that $\cot^2 x < 1 / x^2 < \csc^2 x$ in the same interval.
Summing this inequality from $x_1$ to $x_n$ gives:
or equivalently:
By Combination Theorem for Limits of Real Functions, the and both tend to $\dfrac {\pi^2} 6$ as $n$ tends to infinity. 
Therefore by the Squeeze Theorem:
:$\map \zeta 2 = \ds \sum_{k \mathop = 1}^\infty \frac 1 {k^2} = \frac {\pi^2} 6$
== Historical Note ==
Category:Basel Problem
Basel Problem/Proof 5

== Theorem ==
== Proof ==
Let the function $f_n$ be defined as:
:$(1): \quad \map {f_n} x := \dfrac 1 2 + \cos x + \cos 2 x + \cdots + \cos n x$
By Lagrange's Cosine Identity:
:$(2): \quad \map {f_n} x = \dfrac {\map \sin {\paren {2 n + 1} x / 2} } {2 \map \sin {x / 2} }$
Let $E_n$ be defined as:
The terms for even $k$ on the { are zero, so:
:$\ds (4): \quad \dfrac 1 2 E_{2 n - 1} = \dfrac {\pi^2} 8 - \sum_{k \mathop = 1}^n \dfrac 1 {\paren {2 k - 1}^2}$
It remains to be shown that:
:$\ds \lim_{n \mathop \to \infty} E_{2 n - 1} = 0$
which will establish:
:$\ds (5): \quad \dfrac {\pi^2} 8 = \sum_{k \mathop = 1}^n \dfrac 1 {\paren {2 k - 1}^2}$
Using $(2)$, let $\map g x$ be the function defined as:
:$\map g x := \map {\dfrac \d {\d x} } {\dfrac {x / 2} {\map \sin {x /2} } }$
Using Integration by Parts, we obtain:
:$\ds (6): \quad E_{2 n - 1} = \dfrac 1 {4 n - 1} \paren {2 + 2 \int_0^\pi \map g x \cos \dfrac {\paren {4 n - 1} x} 2 \rd x}$
during which we use Limit of $\dfrac {\sin x} x$: Corollary:
:$\ds \lim_{x \mathop \to 0} \dfrac {x / 2} {\map \sin {x / 2} } = 1$
We have that $\map g x$ is increasing on the interval of integration.
Therefore $\map g x$ is bounded on the interval $\closedint 0 \pi$ by $\map g \pi = \dfrac \pi 2$.
Hence $(5)$ has been established as being true.
Now we divide the (strictly) positive integers into even and odd, and use $(5)$ to obtain:
== Historical Note ==
== Sources ==
* 
* : Chapter $\text {B}.15$: A Rigorous Proof of Euler's Formula $\ds \sum_1^\infty \frac 1 {n^2} = \frac {\pi^2} 6$
Category:Basel Problem
Basel Problem/Proof 7

== Theorem == 
== Proof ==
By Fourier Series of $x^2$, for $x \in \openint {-\pi} \pi$:
:$\ds x^2 = \frac {\pi^2} 3 + \sum_{n \mathop = 1}^\infty \paren {\paren {-1}^n \frac 4 {n^2} \cos n x}$
Letting $x \to \pi$ from the left:
Category:Basel Problem
Basel Problem/Proof 8

== Theorem == 
== Proof == 
By Fourier Series of Identity Function over $-\pi$ to $\pi$, for $x \in \openint {-\pi} \pi$:
:$\ds x \sim 2 \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^{n + 1} } n \map \sin {n x}$
Hence:
 
Category:Basel Problem
Basel Problem/Proof 9

== Theorem == 
== Proof == 
Let $\map f x$ be the real function defined on $\openint 0 {2 \pi}$ as:
:$\map f x = \begin {cases} \paren {x - \pi}^2 & : 0 < x \le \pi \\ \pi^2 & : \pi < x < 2 \pi \end {cases}$
From Fourier Series: Square of x minus pi, Square of pi, its Fourier series can be expressed as:
:$\map f x \sim \ds \dfrac {2 \pi^2} 3 + \sum_{n \mathop = 1}^\infty \paren {\frac {2 \cos n x} {n^2} + \paren {\frac {\paren {-1}^n \pi} n + \frac {2 \paren {\paren {-1}^n - 1} } {\pi n^3} } \sin n x}$
Setting $x = 0$:
== Sources ==
* : Chapter One: $\S 2$. Fourier Series: Example $1$
Category:Basel Problem