nyuuzyou/wikis huggingface dataset:
Lp Norm is Norm

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space and let $p \in \hointr 1 \infty$.
Let $\map {L^p} {X, \Sigma, \mu}$ be the $L^p$ vector space on $\struct {X, \Sigma, \mu}$.
Let $\norm \cdot_p$ be the $L^p$ norm.
Then $\norm \cdot_p$ is a norm on $\map {L^p} {X, \Sigma, \mu}$.
== Proof ==
Let $\map {\mathcal L^p} {X, \Sigma, \mu}$ be the Lebesgue $p$-space. 
Let $\sim$ be the $\mu$-almost everywhere equality relation on $\map {\mathcal L^p} {X, \Sigma, \mu}$.
Let $\eqclass f \sim \in \map {L^p} {X, \Sigma, \mu}$. 
Then, we have by the definition of the $L^p$ norm we have: 
:$\norm {\eqclass f \sim}_p = \norm f_p$ 
From P-Seminorm is Seminorm, we have: 
:$\norm f_p \ge 0$
so:
:$\norm {\eqclass f \sim}_p \ge 0$
So $\norm \cdot_p$ is a map from $\map {L^p} {X, \Sigma, \mu}$ to the non-negative real numbers.
It remains to verify the norm axioms.
=== ===
From P-Seminorm of Function Zero iff A.E. Zero, we have: 
:$\norm f_p = 0$ for $f \in \map {\mathcal L^p} {X, \Sigma, \mu}$ $f = 0$ $\mu$-almost everywhere.
That is: 
:$\norm f_p = 0$ $f \sim 0$
That is, from Equivalence Class Equivalent Statements:
:$\norm f_p = 0$ $\eqclass f \sim = \eqclass 0 \sim = 0_{\map {L^p} {X, \Sigma, \mu} }$.
Then, for $\eqclass f \sim \in \map {L^p} {X, \Sigma, \mu}$, we have: 
:$\norm {\eqclass f \sim}_p = 0$ $\norm f_p = 0$ 
from the definition of the $L^p$ norm.
This is in turn equivalent to $\eqclass f \sim = 0_{\map {L^p} {X, \Sigma, \mu} }$
So we have positive definiteness.
=== ===
Let $\eqclass f \sim \in \map {L^p} {X, \Sigma, \mu}$ and $\lambda \in \R$. 
Then: 
So we have positive homogeneity.
=== ===
Let $\eqclass f \sim, \eqclass g \sim \in \map {L^p} {X, \Sigma, \mu}$. 
Then: 
so the triangle inequality holds.
Category:Lp Norms
Category:Normed Vector Spaces
Lp Norm is Well-Defined

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space, and let $p \in \hointr 1 \infty$. 
Let $\map {\LL^p} {X, \Sigma, \mu}$ be the Lebesgue $p$-space of $\struct {X, \Sigma, \mu}$. 
Let $\sim$ be the $\mu$-almost-everywhere equality relation on $\map {\LL^p} {X, \Sigma, \mu}$.
Let $\map {L^p} {X, \Sigma, \mu}$ be the $L^p$ space on $\struct {X, \Sigma, \mu}$.
Let $\eqclass f \sim \in \map {L^p} {X, \Sigma, \mu}$. 
Then the $L^p$ norm: 
:$\ds \norm {\eqclass f \sim}_p = \paren {\norm f_p}^{1/p}$
is well-defined.
== Proof ==
Note that: 
:$\ds \int \size f^p \rd \mu$ 
is well-defined from the definition of the Lebesgue $p$-space. 
We show that for $E \in \map {L^p} {X, \Sigma, \mu}$, $\norm E_p$ is independent of the representative chosen for $E$.
Let:
:$E = \eqclass f \sim = \eqclass g \sim$
for $\eqclass f \sim, \eqclass g \sim \in \map {L^p} {X, \Sigma, \mu}$.
We show that: 
:$\norm f_p = \norm g_p$ 
From Equivalence Class Equivalent Statements, we have: 
:$f \sim g$ 
So, from the definition of the almost-everywhere equality relation, we have: 
:$\map f x = \map g x$ for $\mu$-almost all $x \in X$.
That is, there exists a $\mu$-null set $N$ such that: 
:if $\map f x \ne \map g x$ then $x \in N$.
If $\size {\map f x}^p \ne \size {\map g x}^p$, we have $\map f x \ne \map g x$, so:
:if $\size {\map f x}^p \ne \size {\map g x}^p$ then $x \in N$.
So:
:$\size {\map f x}^p = \size {\map g x}^p$ for $\mu$-almost all $x \in X$. 
So, from A.E. Equal Positive Measurable Functions have Equal Integrals, we have: 
:$\ds \paren {\int \size f^p \rd \mu}^{1/p} = \paren {\int \size g^p \rd \mu}^{1/p}$ 
That is: 
:$\norm f_p = \norm g_p$
as required.
Category:Lp Norms
Lp Space is Subset of Space of Real-Valued Measurable Functions Identified by A.E. Equality

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space, and let $p \in \closedint 1 \infty$.
Let $\map {\mathcal M} {X, \Sigma, \R}/\sim_\mu$ be the space of real-valued measurable functions identified by $\mu$-A.E. equality.
Let $\map {L^p} {X, \Sigma, \mu}$ be the $L^p$ space of $\struct {X, \Sigma, \mu}$.
Then: 
:$\map {L^p} {X, \Sigma, \mu} \subseteq \map {\mathcal M} {X, \Sigma, \R}/\sim_\mu$
In particular:
:if $f \in \map {\LL^p} {X, \Sigma, \mu}$ and $f \sim_\mu F$ we have $F \in \map {\LL^p} {X, \Sigma, \mu}$
== Proof ==
Let $f \in \map {\LL^p} {X, \Sigma, \mu}$.
Denote by $\eqclass f {\sim_\mu}$ the $\sim_\mu$-equivalence class of $f$ in $\map {\LL^p} {X, \Sigma, \mu}$.
Let $\eqclass f {\sim_\mu}^\ast$ be the $\sim_\mu$-equivalence class of $f$ in $\map {\mathcal M} {X, \Sigma, \R}$.
We aim to show that: 
:$\eqclass f {\sim_\mu} = \eqclass f {\sim_\mu}^\ast$
Since $\map {\LL^p} {X, \Sigma, \mu} \subseteq \map {\mathcal M} {X, \Sigma, \R}$, we have:
:$\eqclass f {\sim_\mu} \subseteq \eqclass f {\sim_\mu}^\ast$
Now let $F \in \eqclass f {\sim_\mu}^\ast$.
Then: 
:$F \sim_\mu f$ 
From Pointwise Exponentiation preserves A.E. Equality, we have: 
:$\size F^p \sim_\mu \size f^p$
Then, from A.E. Equal Positive Measurable Functions have Equal Integrals: Corollary 1, we have: 
:$\ds \int \size F^p \rd \mu = \int \size f^p \rd \mu$
so that: 
:$\ds \paren {\int \size F^p \rd \mu}^{1/p} = \paren {\int \size f^p \rd \mu}^{1/p}$
so:
:$\norm F_p = \norm f_p < \infty$
We therefore finally have: 
:$F \in \map {\LL^p} {X, \Sigma, \mu}$
giving:
:$\eqclass f {\sim_\mu} = \eqclass f {\sim_\mu}^\ast$
Category:Lp Spaces
Category:Space of Real-Valued Measurable Functions Identified by A.E. Equality
Lp Space is Vector Space

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space, and let $p \in \closedint 1 \infty$.
Let $\map \MM {X, \Sigma, \R}/\sim_\mu$ be the space of real-valued measurable functions identified by $\mu$-A.E. equality.
Let $\map {L^p} {X, \Sigma, \mu}$ be the $L^p$ space of $\struct {X, \Sigma, \mu}$.
Let $+$ denote pointwise addition on $\map \MM {X, \Sigma, \R}/\sim_\mu$.
Let $\cdot$ be pointwise scalar multiplication on $\map \MM {X, \Sigma, \R}/\sim_\mu$.
Then: 
:$\struct {\map {L^p} {X, \Sigma, \mu}, +, \cdot}_\R$
forms a vector space. 
== Proof ==
It is sufficient to show that $\struct {\map {L^p} {X, \Sigma, \mu}, +, \cdot}_\R$ is a vector subspace of $\struct {\map \MM {X, \Sigma, \R}/\sim_\mu, +, \cdot}_\R$.
From $L^p$ Space is Subset of Space of Measurable Functions Identified by A.E. Equality:
:$\map {L^p} {X, \Sigma, \mu} \subseteq \map \MM {X, \Sigma, \R}/\sim_\mu$
Let:
:$\eqclass f {\sim_\mu}, \eqclass g {\sim_\mu} \in \map {L^p} {X, \Sigma, \mu}$
and:
:$\lambda \in \R$
Then:
:$f, g \in \map {\LL^p} {X, \Sigma, \mu}$
where $\map {\LL^p} {X, \Sigma, \mu}$ is the Lebesgue $p$-space $\struct {X, \Sigma, \mu}$. 
From the definition of pointwise addition on $\map \MM {X, \Sigma, \R}/\sim_\mu$: 
:$\eqclass f {\sim_\mu} + \lambda \eqclass g {\sim_\mu} = \eqclass {f + \lambda g} {\sim_\mu}$
From Lebesgue Space is Vector Space:
:$f + \lambda g \in \map {\LL^p} {X, \Sigma, \mu}$
So, from $L^p$ Space is Subset of Space of Measurable Functions Identified by A.E. Equality:
:$\eqclass {f + \lambda g} {\sim_\mu} \in \map {L^p} {X, \Sigma, \mu}$.
So, from the One-Step Vector Subspace Test:
:$\struct {\map {L^p} {X, \Sigma, \mu}, +, \cdot}_\R$ is a vector subspace of $\struct {\map \MM {X, \Sigma, \R}/\sim_\mu, +, \cdot}_\R$.
So $\struct {\map {L^p} {X, \Sigma, \mu}, +, \cdot}_\R$ forms a vector space as required.
Category:Lp Spaces
Category:Vector Spaces
Lucas' Theorem

== Theorem ==
Let $p$ be a prime number.
Let $n, k \in \Z_{\ge 0}$.
Then:
:$\dbinom n k \equiv \dbinom {\floor {n / p} } {\floor {k / p} } \dbinom {n \bmod p} {k \bmod p} \pmod p$
where:
:$\dbinom n k$ denotes a binomial coefficient
:$n \bmod p$ denotes the modulo operation
:$\floor \cdot$ denotes the floor function.
=== Corollary ===
== Proof ==
First we show that:
:$\dbinom n k \equiv \dbinom {\floor {n / p} } {\floor {k / p} } \dbinom {n \bmod p} {k \bmod p} \pmod p$
Consider $\dbinom n k$ as the fraction:
:$\dfrac {n \paren {n - 1} \paren {n - 2} \cdots \paren {n - k + 1} } {k \paren {k - 1} \paren {k - 2} \cdots 1}$
This can be expressed as:
:$(1): \quad \dbinom n k = \paren {\dfrac n k} \paren {\dfrac {n - 1} {k - 1} } \paren {\dfrac {n - 2} {k - 2} } \cdots \paren {\dfrac {n - k + 1} 1}$
Let $k = s p + t$ from the Division Theorem.
Thus:
: $t = k \bmod p$
The denominators of the first $t$ factors in $(1)$ do not have $p$ as a divisor.
Now let $n = u p + v$, again from the Division Theorem.
Thus:
:$v = n \bmod p$
Now, when dealing with non-multiples of $p$, we can work modulo $p$ in both the numerator and denominator, from Common Factor Cancelling in Congruence.
So we consider the first $t$ factors of $(1)$ modulo $p$.
These are:
:$\paren {\dfrac {u p + v} {s p + t} } \paren {\dfrac {u p + v - 1} {s p + t - 1} } \cdots \paren {\dfrac {u p + v - t + 1} {s p + 1} } \equiv \paren {\dfrac v t} \paren {\dfrac {v - 1} {t - 1} } \cdots \paren {\dfrac {v - t + 1} 1} \pmod p$
So, these first $t$ terms of $(1)$ taken together are congruent modulo $p$ to the corresponding terms of:
:$\dbinom {n \bmod p} {k \bmod p}$
These differ by multiples of $p$.
So we are left with $k - k \bmod p$ factors.
These fall into $\floor {k / p}$ groups, each of which has $p$ consecutive values.
Each of these groups contains exactly one multiple of $p$.
The other $p - 1$ factors in a given group are congruent (modulo $p$) to $\paren {p - 1}!$ so they cancel out in numerator and denominator.
We now need to investigate the $\left \lfloor {k / p} \right \rfloor$ multiples of $p$ in the numerator and denominator.
We divide each of them by $p$ and we are left with the binomial coefficient:
:$\dbinom {\floor {\paren {n - k \bmod p} / p} } {\floor {k / p} }$
Now, if $k \bmod p \le n \bmod p$, this equals:
:$\dbinom {\floor {n / p} } {\floor {k / p} }$
Otherwise, if $k \bmod p > n \bmod p$, the other factor:
:$\dbinom {n \bmod p} {k \bmod p}$
is zero.
So the formula holds in general.
== Sources ==
* 
* : $\S 1.2.6$: Binomial Coefficients: Exercise $10 \ \text{(e)}$
Category:Binomial Coefficients
Category:Number Theory
Category:Prime Numbers
Lucas' Theorem/Corollary

== Corollary to Lucas' Theorem ==
Let $p$ be a prime number.
Let $n, k \in \Z$.
Let the representations of $n$ and $k$ to the base $p$ be given by:
:$n = a_r p^r + \cdots + a_1 p + a_0$
:$k = b_r p^r + \cdots + b_1 p + b_0$
Then:
:$\ds \dbinom n k \equiv \prod_{j \mathop = 0}^r \dbinom {a_j} {b_j} \pmod p$
where $\dbinom n k$ denotes a binomial coefficient.
== Proof ==
Consider the representations of $n$ and $k$ to the base $p$:
:$n = a_r p^r + \cdots + a_1 p + a_0$
:$k = b_r p^r + \cdots + b_1 p + b_0$
Let:
:$n_1 = \floor {n / p}$
:$k_1 = \floor {k / p}$
We have that:
:$n \bmod p = a_0$
:$k \bmod p = b_0$
:$n_1 = a_r p^{r - 1} + a_{r - 1} p^{r - 2} \cdots + a_1$
:$k_1 = b_r p^{r - 1} + b_{r - 1} p^{r - 2} \cdots + b_1$
It follows from Lucas' Theorem that:
:$\dbinom n k \equiv \dbinom {n_1} {k_1} \dbinom {a_0} {b_0} \pmod p$
Now we do the same again to the representation to the base $p$ of $n_1$ and $n_2$.
Thus:
:$\dbinom n k \equiv \dbinom {\floor {n_1 / p}} {\floor {k_1 / p} } \dbinom {a_1} {b_1} \dbinom {a_0} {b_0} \pmod p$
and so on until:
:$\floor {n_r / p}$
and:
:$\floor {k_r / p}$
Hence the result.
== Sources ==
* 
* : $\S 1.2.6$: Binomial Coefficients: Exercise $10 \ \text{(f)}$
Category:Binomial Coefficients
Category:Number Theory
Category:Prime Numbers
Lucas-Carmichael Number/Examples/2015

== Example of Lucas-Carmichael Number ==
$2015$ is a Lucas-Carmichael number:
:$p \divides 2015 \implies \paren {p + 1} \divides 2016$
== Proof ==
We have that:
:$2015 = 5 \times 13 \times 31$
Then:
Category:Lucas-Carmichael Numbers
Category:2015
Lucas-Carmichael Number/Examples/2915

== Example of Lucas-Carmichael Number ==
$2915$ is a Lucas-Carmichael number:
:$p \divides 2915 \implies \paren {p + 1} \divides 2916$
== Proof ==
We have that:
:$2915 = 5 \times 11 \times 53$
Then:
Category:Lucas-Carmichael Numbers
Category:2915
Lucas-Carmichael Number/Examples/399

== Example of Lucas-Carmichael Number ==
$399$ is a Lucas-Carmichael number:
:$p \divides 399 \implies \paren {p + 1} \divides 400$
== Proof ==
We have that:
:$399 = 3 \times 7 \times 19$
Then:
Category:Lucas-Carmichael Numbers
Category:399
Lucas-Carmichael Number/Examples/935

== Example of Lucas-Carmichael Number ==
$935$ is a Lucas-Carmichael number:
:$p \divides 935 \implies \paren {p + 1} \divides 936$
== Proof ==
We have that:
:$935 = 5 \times 11 \times 17$
Then:
Category:Lucas-Carmichael Numbers
Category:935
Lucas Number 2n in terms of Square of Lucas Number n

== Theorem ==
Let $L_n$ denote the $n$th Lucas number.
Then:
:$L_{2 n} = {L_n}^2 + 2 \left({-1}\right)^n$
== Proof ==
Let:
:$\phi = \dfrac {1 + \sqrt 5} 2$
:$\hat \phi = \dfrac {1 - \sqrt 5} 2$
Note that we have:
Then:
Hence the result.
== Sources ==
* : $11$
* : $11$
Category:Lucas Numbers
Lucas Number as Element of Recursive Sequence

== Theorem ==
Let $L_k$ be the $k$th Lucas number, defined as the sum of two Fibonacci numbers:
:$L_n = F_{n - 1} + F_{n + 1}$
Then $L_n$ can be defined as the $n$th element of the recursive sequence:
:$L_n = \begin{cases}
2 & : n = 0 \\
1 & : n = 1 \\
L_{n - 1} + L_{n - 2} & : \text{otherwise} \end{cases}$
== Proof ==
Proof by induction:
Let $L_n$ be the Lucas number defined as the sum of two Fibonacci numbers:
:$L_n = F_{n - 1} + F_{n + 1}$
For all $n \in \N$, let $\map P n$ be the proposition:
:$L_n = \begin{cases}
2 & : n = 0 \\
1 & : n = 1 \\
L_{n - 1} + L_{n - 2} & : \text{otherwise} \end{cases}$
=== Basis for the Induction ===
We have that:
:$L_0 = F_{-1} + F_1 = 1 + 1 = 2$
:$L_1 = F_0 + F_2 = 0 + 1 = 1$
:$L_2 = F_1 + F_3 = 1 + 2 = 3$
Thus $\map P 0$, $\map P 1$ and $\map P 2$ hold.
$\map P 3$ is the case:
: $L_3 = F_2 + F_4 = 1 + 3 = 4$
So $\map P 3$, as $L_3 = L_1 + L_2$.
This is our basis for the induction.
=== Induction Hypothesis ===
Let us make the supposition that, for some $k \in \N: k \ge 1$, the proposition $\map P j$ holds for all $j \in \N: 1 \le j \le k$.
We shall show that it logically follows that $\map P {k + 1}$ is true.
So this is our induction hypothesis:
:$\forall 1 \le j \le k: L_j = L_{j - 1} + L_{j - 2}$
Then we need to show:
:$L_{k + 1} = L_k + L_{k - 1}$
=== Induction Step ===
This is our induction step:
Hence $L_n = L_{n - 2} + L_{n - 1}$ follows by the Second Principle of Mathematical Induction.
That is: $\sequence {L_n}$ is the sequence defined as:
:$L_n = \begin{cases}
2 & : n = 0 \\
1 & : n = 1 \\
L_{n - 1} + L_{n - 2} & : \text{otherwise} \end{cases}$
== Sources ==
* : $\text {1-1}$ Principle of Mathematical Induction: Exercise $13$
Category:Lucas Numbers
Lucas Number as Sum of Fibonacci Numbers

== Theorem ==
Let $L_k$ be the $k$th Lucas number, defined as:
:$L_n = \begin{cases}
2 & : n = 0 \\
1 & : n = 1 \\
L_{n - 1} + L_{n - 2} & : \text{otherwise} \end{cases}$
Then:
:$L_n = F_{n - 1} + F_{n + 1}$
where $F_k$ is the $k$th Fibonacci number.
== Proof ==
Proof by induction:
For all $n \in \N_{>0}$, let $\map P n$ be the proposition:
:$L_n = F_{n - 1} + F_{n + 1}$
=== Basis for the Induction ===
$\map P 1$ is true, as this just says:
:$L_1 = 1 = F_0 + F_2$
which holds by definition of the Fibonacci numbers.
This is our basis for the induction.
=== Induction Hypothesis ===
Let us make the supposition that, for some $k \in \N: k \ge 1$, the proposition $\map P j$ holds for all $j \in \N: 1 \le j \le k$.
We shall show that it logically follows that $\map P {k + 1}$ is true.
So this is our induction hypothesis:
:$L_j = F_{j - 1} + F_{j + 1}$
Then we need to show:
:$L_{k + 1} = F_k + F_{k + 2}$
=== Induction Step ===
This is our induction step:
So $\map P {k + 1}$ and the result follows by the Second Principle of Mathematical Induction.
Therefore:
: $\forall n \in \N, n \ge 1: L_n = F_{n - 1} + F_{n + 1}$
Category:Lucas Numbers
Category:Fibonacci Numbers
Lune of Hippocrates

== Theorem ==
Take the circle whose center is $A$ and whose radius is $AB = AC = AD = AE$.
Let $C$ be the center of a circle whose radius is $CD = CF = CE$.
:
Consider the lune $DFEB$.
Its area is equal to that of the square $AEGC$.
== Proof ==
:
The chords $DB$ and $EB$ are tangent to the arc $DFE$. They divide the lune into three regions: yellow, green and blue.
From Pythagoras's Theorem, $CD = \sqrt 2 AD$.
The green and blue areas are of equal area as each subtend a right angle.
The orange area also subtends a right angle.
So the area of the orange area is $\paren {\sqrt 2}^2$ the area of either the green or blue areas.
That is, the orange area equals the sum of the green and blue areas together.
Thus the area of the lune $DFEB$ is equal to the area of $\triangle DEB$.
It is a simple matter then to show that the area of $\triangle DEB$ is the same as the area of the square $AEGC$.
== Historical Note ==
== Sources ==
* : Chapter $\text {B}.4$: The Lunes of Hippocrates
Category:Lunes (Plane Geometry)
Category:Plane Geometry
Lévy's Continuity Theorem

== Theorem ==
Let $\sequence {X_n}_{n \mathop \ge 1}$ be a sequence of discrete random variables with characteristic functions $\map {\phi_n} t := E \sqbrk {e^{i t X_n} }$.
Let the sequence $\sequence {\phi_n}$ converge to some real function $\phi$:
:$\forall t \in \R: \map {\phi_n} t \to \map \phi t$.
Then the following statements are equivalent:
:$(1): \quad$ The $\sequence {X_n}$ converges in distribution to some random variable $X$:
::$X_n \stackrel {dist} {\to} X$ with characteristic function $\map {\phi_X} t := \map \phi t$
:$(2): \quad$ The sequence $\sequence {X_n}$ is tight, that is:
::$\ds \lim_{M \mathop \to \infty} \sup_{n \mathop \ge 1} P \sqbrk {\size {X_n} \ge M} = 0$
== Proof ==
== Sources ==
* 
Category:Probability Theory
Lévy's Inversion Formula

== Theorem ==
Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.
Let $X$ be a real-valued random variable on $\struct {\Omega, \Sigma, \Pr}$.
Let $ F_X$ be the distribution function of $X$.
Let $\phi : \R \to \C$ be the characteristic function of $X$.
Then for all $a < b$ such that:
:$\map \Pr {X \in \set {a,b} } = 0$
we have:
=== Integrable Characteristic Function ===
== Proof ==
In fact, the first equality holds for all $a < b$, as:
In the following, we shall show the second equality.
Let $\mu$ be the probability distribution of $X$.
Let $m$ be the Lebesgue measure on $\R$.
For $T > 0$ let $m_T$ be the restriction of $m$ to $\closedint {-T} T$, i.e.:
:$\forall A \in \map \BB \R : \map {m_T} A := \map m { A \cap \closedint {-T} T}$
Let $\mu \times m_T$ be the product measure.
Then:
where:
:$\ds \map f {x,t} := \dfrac {e^{it \paren {x - a} } - e^{it \paren {x - b} } }{it}$
The $f$ is essentially bounded with respect to $\mu \times m_T$ , since:
:$\forall \struct {x, t} \in \R \times \R_{\ne 0} : \cmod {\dfrac {e^{it \paren {x - a} } - e^{it \paren {x - b} } }{it} } \le \dfrac 3 2$
by Bounds for Complex Exponential.
In particular, $f$ is $\mu \times m_T$-integrable, as $\mu \times m_T$ is finite.
Thus by Fubini's Theorem:
where:
Observe:
where:
:$\sgn : \R \to \set {-1, 0, 1}$ is the signum function
:$\Si : \R \to \R$ is the sine integral function
Similarly:
:$\ds \int _{-T}^T \dfrac {e^{it \paren {x - b} } }{it} \rd t = 2 \; \map \sgn {x - b} \; \map \Si {T \size {x - b} }$
Thus we have:
:$\ds \map {F_T} x = 2 \; \map \sgn {x - a} \; \map \Si {T \size {x - a} } - 2 \; \map \sgn {x - b} \; \map \Si {T \size {x - b} }$
By Limit at Infinity of Sine Integral Function, for all $x \in \R \setminus \set {a,b}$:
As $\map \mu {\set {a, b} } = 0$ , this means in particular:
:$\ds \lim _{T \mathop \to +\infty} F_T = \chi _{\hointl a b}\quad$ $\mu$-almost surely
On the other hand, Sine Integral Function is Bounded:
:$\ds \sup_{T > 0, \; x \in \R} \size {\map {F_T} x} \le 4 \norm \Si_\infty < +\infty$
Note that the constant function $4 \norm \Si_\infty$ is $\mu$-integrable.
Therefore:
== Sources ==
* : $26$: Characteristic Functions
Category:Lévy's Inversion Formula
Category:Probability Theory
Lévy's Inversion Formula/Integrable Characteristic Function

== Theorem ==
Let $X$ be a real-valued random variable.
Let $P_X$ be the probability distribution of $X$.
Let $\phi : \R \to \C$ be the characteristic function of $X$.
Suppose that $\phi$ is Lebesgue integrable, i.e.:
:$\ds \int _\R \cmod {\map \phi t} \rd t < + \infty$
Then $P_X$ is absolutely continuous the Lebesgue measure.
Moreover, the Radon-Nikodym derivative is given by:
:$\ds \map g x := \dfrac 1 {2 \pi} \int_\R \map \phi t e^{- i t x} \rd t$
== Proof ==
Let $a < b$ be such that:
:$\map {P_X} {\set {a,b} } = 0$
Then:
Now:
:$\ds \forall t \in \R : \quad \cmod {\map \phi t \frac {e^{- i t a} - e^{- i t b} }{i t} } \le \cmod {\map \phi t} \paren {b - a}$
Thus:
Category:Lévy's Inversion Formula
Lüroth's Theorem

== Theorem ==
Let $K$ be a field.
Let $\map K X$ be the rational function field, for some indeterminate $X$.
Let $M$ be an intermediate field between $K$ and $\map K X$.
Then there exists a rational function $\map f X \in \map K X$ such that:
:$M = \map K {\map f X}$
In other words, every intermediate extension between $K$ and $\map K X$ is a simple extension.
== Proof ==
By Gauss's Lemma (Polynomial Theory),
== Source ==
* : Ch. $\text {VIII}.1$: Transcendence bases
Category:Definitions/Field Extensions
Machin's Formula for Pi

== Theorem ==
:$\dfrac \pi 4 = 4 \arctan \dfrac 1 5 - \arctan \dfrac 1 {239} \approx 0 \cdotp 78539 \, 81633 \, 9744 \ldots$
The calculation of $\pi$ (pi) can then proceed using the Gregory Series:
:$\arctan \dfrac 1 x = \dfrac 1 x - \dfrac 1 {3 x^3} + \dfrac 1 {5 x^5} - \dfrac 1 {7 x^7} + \dfrac 1 {9 x^9} - \cdots$
which is valid for $x \ge 1$.
== Proof 1 ==
== Proof 2 ==
== Also known as ==
Some sources give this as Machin's Identity.
== Historical Note ==
== Sources ==
* : $3 \cdotp 14159 \, 26535 \, 89793 \, 23846 \, 26433 \, 83279 \, 50288 \, 41972 \ldots$
* : $3 \cdotp 14159 \, 26535 \, 89793 \, 23846 \, 26433 \, 83279 \, 50288 \, 41971 \ldots$
* 
Category:Formulas for Pi
Category:Machin's Formula for Pi
Machin's Formula for Pi/Historical Note

== Historical Note on Machin's Formula for Pi ==
 devised his formula for $\pi$ in $1706$.
It allowed him to calculate $\pi$ to over $100$ decimal places.
This greatly surpassed the work of .
== Sources ==
* : $3 \cdotp 14159 \, 26535 \, 89793 \, 23846 \, 26433 \, 83279 \, 50288 \, 41972 \ldots$
* : $3 \cdotp 14159 \, 26535 \, 89793 \, 23846 \, 26433 \, 83279 \, 50288 \, 41971 \ldots$
Category:Historical Notes