hcju/proofwikips huggingface dataset:
Alexandroff Extension of Rational Number Space is Biconnected
Let $\struct {\Q, \tau_d}$ be the rational number space under the Euclidean topology $\tau_d$. Let $p$ be a new element not in $\Q$. Let $\Q^* := \Q \cup \set p$. Let $T^* = \struct {\Q^*, \tau^*}$ be the Alexandroff extension on $\struct {\Q, \tau_d}$. Then $T^*$ is a biconnected space.
Hilbert Sequence Space is Complete Metric Space
Let $A$ be the set of all real sequences $\left\langle{x_i}\right\rangle$ such that the series $\displaystyle \sum_{i \mathop \ge 0} x_i^2$ is convergent. Let $\ell^2 = \struct {A, d_2}$ be the Hilbert sequence space on $\R$. Then $\ell^2$ is a complete metric space.
Hilbert Sequence Space is Separable
Let $A$ be the set of all real sequences $\sequence {x_i}$ such that the series $\displaystyle \sum_{i \mathop \ge 0} x_i^2$ is convergent. Let $\ell^2 = \struct {A, d_2}$ be the Hilbert sequence space on $\R$. Then $\ell^2$ is a separable space.
Hilbert Sequence Space is Second-Countable
Let $\ell^2$ be the Hilbert sequence space on $\R$. Then $\ell^2$ is a second-countable space.
Hilbert Sequence Space is Lindelöf
Let $\ell^2$ be the Hilbert sequence space on $\R$. Then $\ell^2$ is a Lindelöf space.
Compact Subset of Hilbert Sequence Space is Closed
Let $A$ be the set of all real sequences $\sequence {x_i}$ such that the series $\displaystyle \sum_{i \mathop \ge 0} x_i^2$ is convergent. Let $\ell^2 = \struct {A, d_2}$ be the Hilbert sequence space on $\R$. Let $H$ be a compact subset of $\ell^2$ Then $H$ is closed in $\ell^2$.
Subset of Hilbert Sequence Space with Non-Empty Interior is not Compact
Let $A$ be the set of all real sequences $\left\langle{x_i}\right\rangle$ such that the series $\displaystyle \sum_{i \mathop \ge 0} x_i^2$ is convergent. Let $\ell^2 = \struct {A, d_2}$ be the Hilbert sequence space on $\R$. Let $H$ be a subset of $\ell^2$ whose interior  is non-empty. Then $H$ is not compact in $\ell^2$.
Point in Hilbert Sequence Space has no Compact Neighborhood
Let $A$ be the set of all real sequences $\left\langle{x_i}\right\rangle$ such that the series $\displaystyle \sum_{i \mathop \ge 0} x_i^2$ is convergent. Let $\ell^2 = \struct {A, d_2}$ be the Hilbert sequence space on $\R$. Then no point of $\ell^2$ has a compact neighborhood.
Compact Subset of Hilbert Sequence Space is Nowhere Dense
Let $A$ be the set of all real sequences $\sequence {x_i}$ such that the series $\displaystyle \sum_{i \mathop \ge 0} x_i^2$ is convergent. Let $\ell^2 = \struct {A, d_2}$ be the Hilbert sequence space on $\R$. Let $H$ be a compact subset of $\ell^2$. Then $H$ is nowhere dense in $\ell^2$.
Hilbert Sequence Space is Arc-Connected
Let $A$ be the set of all real sequences $\sequence {x_i}$ such that the series $\displaystyle \sum_{i \mathop \ge 0} x_i^2$ is convergent. Let $\ell^2 = \struct {A, d_2}$ be the Hilbert sequence space on $\R$. Then $\ell^2$ is arc-connected.
Hilbert Sequence Space is Homeomorphic to Countable Infinite Product of Real Number Spaces
Let $A$ be the set of all real sequences $\sequence {x_i}$ such that the series $\displaystyle \sum_{i \mathop \ge 0} x_i^2$ is convergent. Let $\ell^2 = \struct {A, d_2}$ be the Hilbert sequence space on $\R$. Let $\struct {\R, \tau_d}$ denote the real number line under the Euclidean topology. Let $\R^\omega = \displaystyle \prod_{i \mathop \in \N} \struct {\R, \tau_d}$ denote the countable-dimensional real Cartesian space under the Tychonoff topology. Then $\ell^2$ is homeomorphic to $\R^\omega$.
Fréchet Space (Functional Analysis) is Metric Space
Let $\struct {\R^\omega, d}$ be the '''Fréchet space on $\R^\omega$'''. Then $\struct {\R^\omega, d}$ is a metric space.
Countable Infinite Product of Real Number Spaces is Homeomorphic to Fréchet Metric Space
Let $\struct {\R, \tau_d}$ denote the real number line under the Euclidean topology. Let $T = \struct {\R^\omega, \tau} = \displaystyle \prod_{i \mathop \in \N} \struct {\R, \tau_d}$ denote the countable-dimensional real Cartesian space under the product topology $\tau$. Let $\struct {\R^\omega, d}$ be the '''Fréchet space on $\R^\omega$''', where: :$\map d {x, y} = \displaystyle \sum_{i \mathop \in \N} \dfrac {2^{-i} \size {x_i - y_i} } {1 + \size {x_i - y_i} }$ Then the topology induced by $d$ is exactly the Tychonoff topology $\tau$.
Separable Metric Space is Homeomorphic to Subspace of Fréchet Metric Space
Let $M = \struct {A, d}$ be a metric space whose induced topology is separable. Then $M$ is homeomorphic to a subspace of the Fréchet space $\struct {\R^\omega, d}$ on the countable-dimensional real Cartesian space $\R^\omega$.
Hilbert Cube is Metric Space
Let $M = \struct {I^\omega, d_2}$ be the Hilbert cube. Then $M$ is a metric space.
Hilbert Cube is Homeomorphic to Countable Infinite Product of Real Number Unit Intervals
Let $M_1 = \struct {I^\omega, d_2}$ be the Hilbert cube: :$M_1 = \displaystyle \prod_{k \mathop \in \N} \closedint 0 {\dfrac 1 k}$ under the same metric as that of the Hilbert sequence space: :$\displaystyle \forall x = \sequence {x_i}, y = \sequence {y_i} \in I^\omega: \map {d_2} {x, y} := \paren {\sum_{k \mathop \ge 0} \paren {x_k - y_k}^2}^{\frac 1 2}$ Let $M_2$ be the metric space defined as: :$M_2 = \displaystyle \prod_{k \mathop \in \N} \closedint 0 1$ under the Tychonoff topology. Then $M_1$ and $M_2$ are homeomorphic.
Hilbert Cube is Completely Normal
Let $M = \struct {I^\omega, d_2}$ be the Hilbert cube. Then $M$ is a completely normal space.
Hilbert Cube is Separable
Let $M = \struct {I^\omega, d_2}$ be the Hilbert cube. Then $M$ is a separable space.
Hilbert Cube is Second-Countable
Let $M = \struct {I^\omega, d_2}$ be the Hilbert cube. Then $M$ is a second-countable space.
Hilbert Cube is Arc-Connected
Let $M = \struct {I^\omega, d_2}$ be the Hilbert cube. Then $M$ is an arc-connected space.
Interval of Totally Ordered Set is Convex
Let $\struct {S, \preccurlyeq}$ be a totally ordered set. Let $I \subseteq S$ be an interval in $S$. Then $I$ is convex.
Convex Set of Ordered Set is not necessarily Interval
Let $\struct {S, \preccurlyeq}$ be an ordered set. Let $C$ be a convex set of $S$. Then it is not necessarily the case that $C$ is an interval of $S$.
Union of Non-Disjoint Convex Sets is Convex Set
Let $\struct {S, \preccurlyeq}$ be an ordered set. Let $\CC$ be a set of convex sets of $S$ such that their intersection is non-empty: :$\displaystyle \bigcap \CC \ne \O$ Then the union $\displaystyle \bigcup \CC$ is also convex.
Subset of Convex Set can be Uniquely Expressed as Partition of Maximal Convex Sets
Let $\struct {S, \preccurlyeq}$ be an ordered set. Then $S$ can be uniquely expressed as the partition whose components are maximally convex sets.
Separated Subsets of Linearly Ordered Space under Order Topology
Let $T = \struct {S, \preceq, \tau}$ be a linearly ordered space. Let $A$ and $B$ be separated sets of $T$. Let $A^*$ and $B^*$ be defined as: :$A^* := \ds \bigcup \set {\closedint a b: a, b \in A, \closedint a b \cap B^- = \O}$ :$B^* := \ds \bigcup \set {\closedint a b: a, b \in B, \closedint a b \cap A^- = \O}$ where $A^-$ and $B^-$ denote the closure of $A$ and $B$ in $T$. Then $A^*$ and $B^*$ are themselves separated sets of $T$.
Generators of Special Linear Group of Order 2 over Integers
Let: :$ S = \begin{pmatrix} 0 & - 1 \\ 1 & 0 \end{pmatrix}$ and: :$T = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$ Then $S$ and $T$ are generators for the special linear group of order $2$ over $\Z$.
Partition of Linearly Ordered Space by Convex Components is Linearly Ordered Set
Let $T = \struct {S, \preceq, \tau}$ be a linearly ordered space. Let $A$ and $B$ be separated sets of $T$. Let $A^*$ and $B^*$ be defined as: :$A^* := \displaystyle \bigcup \set {\closedint a b: a, b \in A, \closedint a b \cap B^- = \O}$ :$B^* := \displaystyle \bigcup \set {\closedint a b: a, b \in B, \closedint a b \cap A^- = \O}$ where $A^-$ and $B^-$ denote the closure of $A$ and $B$ in $T$. Let $A^*$, $B^*$ and $\relcomp S {A^* \cup B^*}$ be expressed as the union of convex components of $S$: :$\displaystyle A^* = \bigcup A_\alpha, \quad B^* = \bigcup B_\beta, \quad \relcomp S {A^* \cup B^*} = \bigcup C_\gamma$ where $\relcomp S X$ denotes the complement of $X$ with respect to $S$. Then the set $M = \set {A_\alpha, B_\beta, C_\gamma}$ inherits a linear ordering from $S$, and so is a linearly ordered set.
Successor Sets of Linearly Ordered Set Induced by Convex Component Partition
Let $T = \struct {S, \preceq, \tau}$ be a linearly ordered space. Let $A$ and $B$ be separated sets of $T$. Let $A^*$ and $B^*$ be defined as: :$A^* := \displaystyle \bigcup \set {\closedint a b: a, b \in A, \closedint a b \cap B^- = \O}$ :$B^* := \displaystyle \bigcup \set {\closedint a b: a, b \in B, \closedint a b \cap A^- = \O}$ where $A^-$ and $B^-$ denote the closure of $A$ and $B$ in $T$. Let $A^*$, $B^*$ and $\relcomp S {A^* \cup B^*}$ be expressed as the union of convex components of $S$: :$\displaystyle A^* = \bigcup A_\alpha, \quad B^* = \bigcup B_\beta, \quad \relcomp S {A^* \cup B^*} = \bigcup C_\gamma$ where $\relcomp S X$ denotes the complement of $X$ with respect to $S$. Let $M$ be the linearly ordered set: :$M = \set {A_\alpha, B_\beta, C_\gamma}$ as defined in Partition of Linearly Ordered Space by Convex Components is Linearly Ordered Set. Then each of the sets $A_\alpha \in M$ has an immediate successor in $M$ if $A_\alpha$ intersects the closure of $S_\alpha$, the set of strict upper bounds for $A_\alpha$. Similarly for $B_\beta$. That immediate successor ${C_\alpha}^+$ to $A_\alpha$ is an element in $\set {C_\gamma}$.
Linearly Ordered Space is T5
Let $T = \struct {S, \preceq, \tau}$ be a linearly ordered space. Then $T$ is a $T_5$ space.
Linearly Ordered Space is T1
Let $T = \struct {S, \preceq, \tau}$ be a linearly ordered space. Then $T$ is a $T_1$ (Fréchet) space.
Linearly Ordered Space is Completely Normal
Let $T = \struct {S, \preceq, \tau}$ be a linearly ordered space. Then $T$ is a completely normal space.
Linearly Ordered Space is Compact iff Complete
Let $T = \struct {S, \preceq, \tau}$ be a linearly ordered space. Then $T$ is a compact space  it is complete.
Basis for Open Ordinal Topology
Let $\Gamma$ be a limit ordinal. Let $\hointr 0 \Gamma$ denote the open ordinal space on $\Gamma$. Consider the set $\BB$ of subsets of $\hointr 0 \Gamma$ of the form: :$\openint \alpha {\beta + 1} = \hointl \alpha \beta = \set {x \in \hointr 0 \Gamma: \alpha < x < \beta + 1}$ for $\alpha, \beta \in \hointr 0 \Gamma$. {{explain|It's unclear from the source work used whether this is the basis for the Open Ordinal Topology or Closed Ordinal Topology.}} Then $\BB$ forms a basis for $\hointr 0 \Gamma$.
Omega is Closed in Uncountable Closed Ordinal Space but not G-Delta Set
Let $\Omega$ denote the first uncountable ordinal. Let $\closedint 0 \Omega$ denote the closed ordinal space on $\Omega$. Then $\set \Omega$ is a closed set of $\closedint 0 \Omega$ but not a $G_\delta$ set.
Uncountable Closed Ordinal Space is not First-Countable
Let $\Omega$ denote the first uncountable ordinal. Let $\closedint 0 \Omega$ denote the closed ordinal space on $\Omega$. Then $\closedint 0 \Omega$ is not a first-countable space.
Omega as Limit Point of Intervals of Uncountable Closed Ordinal Space
Let $\Omega$ denote the first uncountable ordinal. Let $\closedint 0 \Omega$ denote the closed ordinal space on $\Omega$. Then $\Omega$ is a limit point of the set $\openint a \Omega$, but not the limit point of any sequence of points in $\openint a \Omega$.
Uncountable Open Ordinal Space is not Separable
Let $\Omega$ denote the first uncountable ordinal. Let $\hointr 0 \Omega$ denote the open ordinal space on $\Omega$. Then $\hointr 0 \Omega$ is not a separable space.
Uncountable Closed Ordinal Space is not Separable
Let $\Omega$ denote the first uncountable ordinal. Let $\closedint 0 \Omega$ denote the closed ordinal space on $\Omega$. Then $\closedint 0 \Omega$ is not a separable space.
Uncountable Open Ordinal Space is First-Countable
Let $\Omega$ denote the first uncountable ordinal. Let $\hointr 0 \Omega$ denote the open ordinal space on $\Omega$. Then $\hointr 0 \Omega$ is a first-countable space.
Ordinal Space is Completely Normal
Let $\Gamma$ denote a limit ordinal. Let $\hointr 0 \Gamma$ denote the open ordinal space on $\Gamma$. Let $\closedint 0 \Gamma$ denote the closed ordinal space on $\Gamma$. Then $\hointr 0 \Gamma$ and $\closedint 0 \Gamma$ are both completely normal.
Uncountable Closed Ordinal Space is not Perfectly Normal
Let $\Omega$ denote the first uncountable ordinal. Let $\closedint 0 \Omega$ denote the closed ordinal space on $\Omega$. Then $\closedint 0 \Omega$ is not a perfectly normal space.
Uncountable Closed Ordinal Space is not Second-Countable
Let $\Omega$ denote the first uncountable ordinal. Let $\closedint 0 \Omega$ denote the closed ordinal space on $\Omega$. Then $\closedint 0 \Omega$ is not a second-countable space.
Uncountable Open Ordinal Space is not Second-Countable
Let $\Omega$ denote the first uncountable ordinal. Let $\hointr 0 \Omega$ denote the open ordinal space on $\Omega$. Then $\hointr 0 \Omega$ is not a second-countable space.
Ring Element is Unit iff Unit in Integral Extension
let $A$ be a commutative ring with unity. Let $a \in A$. Let $B$ be an integral ring extension of $A$.  :$(1): \quad a$ is a unit of $A$ :$(2): \quad a$ is a unit of $B$
Separable Elements Form Field
Let $E/F$ be an algebraic field extension. Then the subset of separable elements of $E$ form a intermediate field, called the '''relative separable closure'''.
Decomposition of Field Extension as Separable Extension followed by Purely Inseparable
Let $E/F$ be an algebraic field extension. Then the relative separable closure $K=F^{sep}$ in $E$ is the unique intermediate field with the following properties: * $K/F$ is separable. * $E/K$ is purely inseparable.
Subextensions of Separable Field Extension are Separable
Let $E/K/F$ be a tower of fields. Let $E/F$ be separable. Then $E/K$ and $K/F$ are separable.
Finite Orbit under Group of Automorphisms of Field implies Separable over Fixed Field
Let $E$ be a field. Let $G \le \Aut E$ be a subgroup of its automorphism group. Let $F = \map {\operatorname {Fix}_E} G$ be its fixed field. Let $\alpha \in E$ have a finite orbit under $G$. Then $\alpha$ is separable over $F$.
Countable Closed Ordinal Space is Second-Countable
Let $\Omega$ denote the first uncountable ordinal. Let $\Gamma$ be a limit ordinal which strictly precedes $\Omega$. Let $\closedint 0 \Gamma$ denote the closed ordinal space on $\Gamma$. Then $\closedint 0 \Gamma$ is a second-countable space.
Countable Open Ordinal Space is Second-Countable
Let $\Omega$ denote the first uncountable ordinal. Let $\Gamma$ be a limit ordinal which strictly precedes $\Omega$. Let $\hointr 0 \Gamma$ denote the open ordinal space on $\Gamma$. Then $\hointr 0 \Gamma$ is a second-countable space.
Minimal Polynomial of Element with Finite Orbit under Group of Automorphisms over Fixed Field in terms of Orbit
Let $E$ be a field. Let $G \le \Aut E$ be a subgroup of its automorphism group. Let $F = \map {\operatorname {Fix_E} } G$ be its fixed field. Let $\alpha \in E$ have a finite orbit under $G$. Then $\alpha$ is algebraic over $F$ and the product of polynomials :$\map p x = \displaystyle \prod_{\beta \mathop \in \Lambda} \paren {x - \beta}$ is the minimal polynomial of $\alpha$ over $F$.
Countable Closed Ordinal Space is Metrizable
Let $\Omega$ denote the first uncountable ordinal. Let $\Gamma$ be a limit ordinal which strictly precedes $\Omega$. Let $\closedint 0 \Gamma$ denote the closed ordinal space on $\Gamma$. Then $\closedint 0 \Gamma$ is a metrizable space.
Closed Ordinal Space is Complete Order Space
Let $\Gamma$ be a limit ordinal. Let $\closedint 0 \Gamma$ denote the closed ordinal space on $\Gamma$. Then $\closedint 0 \Gamma$ is a complete order space.
Closed Ordinal Space is Compact
Let $\Gamma$ be a limit ordinal. Let $\closedint 0 \Gamma$ denote the closed ordinal space on $\Gamma$. Then $\closedint 0 \Gamma$ is a compact space.
Ordinal Space is Strongly Locally Compact
Let $T$ denote an ordinal space on a limit ordinal $\Gamma$. Then $T$ is a strongly locally compact space.
Open Ordinal Space is not Compact in Closed Ordinal Space
Let $\Gamma$ be a limit ordinal. Let $\hointr 0 \Gamma$ denote the open ordinal space on $\Gamma$. Consider the compact subspace $\hointr 0 \Gamma$. Then $\hointr 0 \Gamma$ is not compact in $\closedint 0 \Gamma$.
Integral Transform is Mapping
Let $\map F p$ be an integral transform: :$\map F p = \displaystyle \int_a^b \map f x \map K {p, x} \rd x$ Let $T$ be the integral operator associated with $\map F p$. Then $T$ is a mapping from the domain of $T$ to its image. That is, for every $\map f x$ there exists a unique $\map F p$.
Integral Operator is Linear
Let $T$ be an integral operator. Let $f$ and $g$ be integrable real functions on a domain appropriate to $T$. Then $T$ is a linear operator: :$\forall \alpha, \beta \in \R: \map T {\alpha f + \beta g} = \alpha \map T f + \beta \map T g$
Inverse Integral Operator is Linear if Unique
Let $T$ be an integral operator. Let $f$ be an integrable real function on a domain appropriate to $T$. Let $F = \map T f$ and $G = \map T g$. Let $T$ have a unique inverse $T^{-1}$. Then $T^{-1}$ is a linear operator: :$\forall p, q \in \R: \map {T^{-1} } {p F + q G} = p \map {T^{-1} } F + q \map {T^{-1} } G$
Trigonometric Series is Convergent if Sum of Absolute Values of Coefficients is Convergent
Let $\map S x$ be a trigonometric series: :$\map S x = \dfrac {a_0} 2 + \displaystyle \sum_{n \mathop = 1}^\infty \paren {a_n \cos n x + b_n \sin n x}$ Let the series: :$\displaystyle \sum_{n \mathop = 1}^\infty \paren {\size {a_n} + \size {b_n} }$ be convergent. Then $S$ is a convergent series.
Convergent Trigonometric Series is Periodic
Let $\map S x$ be a trigonometric series: :$\map S x = \dfrac {a_0} 2 + \displaystyle \sum_{n \mathop = 1}^\infty \paren {a_n \cos n x + b_n \sin n x}$ Let $S$ be convergent. Then $S$ is periodic: :$\forall r \in \Z: \map S {x + 2 r \pi} = \map S x$
Integral over 2 pi of Sine of m x by Sine of n x
Let $m, n \in \Z$ be integers. Let $\alpha \in \R$ be a real number. Then: :$\displaystyle \int_\alpha^{\alpha + 2 \pi} \sin m x \sin n x \rd x = \begin{cases} 0 & : m \ne n \\ \pi & : m = n \end{cases}$ That is: :$\displaystyle \int_\alpha^{\alpha + 2 \pi} \sin m x \sin n x \rd x = \pi \delta_{m n}$ where $\delta_{m n}$ is the Kronecker delta.
Integral over 2 pi of Cosine of m x by Cosine of n x
Let $m, n \in \Z$ be integers. Let $\alpha \in \R$ be a real number. Then: :$\ds \int_\alpha^{\alpha + 2 \pi} \cos m x \cos n x \rd x = \begin {cases} 0 & : m \ne n \\ \pi & : m = n \end {cases}$ That is: :$\ds \int_\alpha^{\alpha + 2 \pi} \cos m x \cos n x \rd x = \pi \delta_{m n}$ where $\delta_{m n}$ is the Kronecker delta.
Integral over 2 pi of Sine of m x by Cosine of n x
Let $m, n \in \Z$ be integers. Let $\alpha \in \R$ be a real number. Then: :$\displaystyle \int_\alpha^{\alpha + 2 \pi} \sin m x \cos n x \rd x = 0$
Integral over 2 pi of Sine of n x
Let $m \in \Z$ be an integer. Then: :$\displaystyle \int_\alpha^{\alpha + 2 \pi} \sin m x \rd x = 0$
Integral over 2 pi of Cosine of n x
Let $m \in \Z$ be an integer. Then: :$\displaystyle \int_\alpha^{\alpha + 2 \pi} \cos m x \rd x = \begin {cases} 0 & : m \ne 0 \\ 2 \pi & : m = 0 \end {cases}$
Coefficients of Sine Terms in Convergent Trigonometric Series
Let $\map S x$ be a trigonometric series which converges to $\map f x$ on the interval $\openint \alpha {\alpha + 2 \pi}$: :$\map f x = \dfrac {a_0} 2 + \displaystyle \sum_{m \mathop = 1}^\infty \left({a_m \cos m x + b_m \sin m x}\right)$ Then: :$\forall n \in \Z_{\ge 0}: b_n = \dfrac 1 \pi \displaystyle \int_\alpha^{\alpha + 2 \pi} \map f x \sin n x \rd x$
Sine of Angle plus Full Angle/Corollary
Let $n \in \Z$ be an integer. Then: :$\map \sin {x + 2 n \pi} = \sin x$
Cosine of Angle plus Full Angle/Corollary
Let $n \in \Z$ be an integer. Then: : $\cos \left({x + 2 n \pi}\right) = \cos x$
Fourier's Theorem
Let $\alpha \in \R$ be a real number. Let $\map f x$ be a real function which is defined and bounded on the interval $\openint \alpha {\alpha + 2 \pi}$. Let $f$ satisfy the Dirichlet conditions on $\openint \alpha {\alpha + 2 \pi}$: {{:Definition:Dirichlet Conditions}} Outside the interval $\openint \alpha {\alpha + 2 \pi}$, let $f$ be periodic and defined such that: :$\map f x = \map f {x + 2 \pi}$ Let $f$ be defined by the Fourier series: :$(1): \quad \displaystyle \frac {a_0} 2 + \sum_{n \mathop = 1}^\infty \paren {a_n \cos n x + b_n \sin n x}$ such that: :$\displaystyle a_n = \dfrac 1 \pi \int_\alpha^{\alpha + 2 \pi} \map f x \cos n x \rd x$ :$\displaystyle b_n = \dfrac 1 \pi \int_\alpha^{\alpha + 2 \pi} \map f x \sin n x \rd x$ Then for all $a \in \R$, $(1)$ converges to the sum: :$\displaystyle \frac 1 2 \paren {\lim_{x \mathop \to a^+} \map f x + \lim_{x \mathop \to a^-} \map f x}$ where the $\lim$ symbols denote the limit from the right and the limit from the left.
Definite Integral of Step Function
Let $\alpha, \beta \in \R$ be a real numbers such that $\alpha < \beta$. Let $f \left({x}\right)$ be a step function defined on the interval $\left[{\alpha \,.\,.\, \beta}\right]$: :$f \left({x}\right) = \lambda_1 \chi_{\mathbb I_1} + \lambda_2 \chi_{\mathbb I_2} + \cdots + \lambda_n \chi_{\mathbb I_n}$ where: :$\lambda_1, \lambda_2, \ldots, \lambda_n$ are real constants :$\mathbb I_1, \mathbb I_2, \ldots, \mathbb I_n$ are intervals, where these intervals partition $\left[{\alpha \,.\,.\, \beta}\right]$ :$\chi_{\mathbb I_1}, \chi_{\mathbb I_2}, \ldots, \chi_{\mathbb I_n}$ are characteristic functions of $\mathbb I_1, \mathbb I_2, \ldots, \mathbb I_n$. Then the definite integral of $f$ {{{WRT|Integration}} $x$ over $\left[{\alpha \,.\,.\, \beta}\right]$ is given by: :$\displaystyle \int_\alpha^\beta f \left({x}\right) \, \mathrm d x = \sum_{k \mathop = 1}^n \lambda_k \left({\beta_k - \alpha_k}\right)$ where $\alpha_k, \beta_k$ are the endpoints of $\mathbb I_k$ for $1 \le k \le n$.
Step Function satisfies Dirichlet Conditions
Let $\alpha, \beta \in \R$ be a real numbers such that $\alpha < \beta$. Let $\map f x$ be a step function defined on the interval $\openint \alpha \beta$. Then $f$ satisfies the Dirichlet conditions.
Parseval's Theorem/Formulation 1
Let $f$ be a real function which is square-integrable over the interval $\openint {-\pi} \pi$. Let $f$ be expressed by the Fourier series:  :$\map f x \sim \dfrac {a_0} 2 + \displaystyle \sum_{n \mathop = 1}^\infty \paren {a_n \cos n x + b_n \sin n x}$ Then: :$\displaystyle \frac 1 \pi \int_{-\pi}^\pi \size {\map {f^2} x} \rd x = \frac { {a_0}^2} 2 + \sum_{n \mathop = 1}^\infty \paren { {a_n}^2 + {b_n}^2}$
Fourier Series/Square of x minus pi, Square of pi
:400pxrightthumb$\map f x$ and $5$th order expansion Let $\map f x$ be the real function defined on $\openint 0 {2 \pi}$ as: :$\map f x = \begin{cases} \paren {x - \pi}^2 & : 0 < x \le \pi \\ \pi^2 & : \pi < x < 2 \pi \end{cases}$ Then its Fourier series can be expressed as: {{begin-eqn}} {{eqn | l = \map f x       | o = \sim       | r = \frac {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 {1 - \paren {-1}^n} } {\pi n^3} } \sin n x}       | c =  }} {{eqn | r = \frac {2 \pi^2} 3 + 2 \paren {\cos x + \frac {\cos 2 x} {2^2} + \frac {\cos 3 x} {3^2} + \cdots}       | c =  }} {{eqn | o =        | ro= -       | r = \pi \sin x + \frac \pi 2 \sin x - \frac \pi 3 \sin x + \frac \pi 4 \sin x - \cdots       | c =  }} {{eqn | o =        | ro= -       | r = \dfrac 4 \pi \sin x - \frac 4 {\pi 3^3} \sin 3 x - \frac 4 {\pi 5^3} \sin 5 x \cdots       | c =  }} {{end-eqn}}
Sum of Reciprocals of Squares Alternating in Sign
{{begin-eqn}} {{eqn | l = \dfrac {\pi^2} {12}       | r = \sum_{n \mathop = 1}^\infty \dfrac {\paren {-1}^{n + 1} } {n^2}       | c =  }} {{eqn | r = \frac 1 {1^2} - \frac 1 {2^2} + \frac 1 {3^2} - \frac 1 {4^2} + \cdots       | c =  }} {{end-eqn}}