nyuuzyou/wikis huggingface dataset:
Abel's Test for Uniform Convergence

== Theorem ==
Let $\sequence {\map {a_n} z}$ and $\sequence {\map {b_n} z}$ be sequences of complex functions on a compact set $K$.
Let $\sequence {\map {a_n} z}$ be such that:
:$\sequence {\map {a_n} z}$ is bounded in $K$
:$\ds \sum \size {\map {a_n} z - \map {a_{n + 1} } z}$ is convergent with a sum which is bounded in $K$
:$\ds \sum \map {b_n} z$ is uniformly convergent in $K$.
Then $\ds \sum \map {a_n} z \map {b_n} z$ is uniformly convergent on $K$.
== Proof ==
First we modify the statement of Abel's Lemma
=== Lemma ===
Suppose $\sum b_k$ converges.
Let $B_k = b_k + b_{k + 1} + b_{k + 2} + \cdots$
Then:
:$a_n b_n + \cdots + a_{n + k} b_{n + k} = B_n a_n + B_{n + 1} \paren {a_{n + 1} - a_n} + \cdots + B_{n + k} \paren {a_{n + k} - a_{n + k - 1} } - B_{n + k + 1} a_{n + k}$
====Proof of lemma====
=== Proof of theorem ===
We show that $\ds \sum_{j\mathop = n}^{n + k} \map {a_j} z \map {b_j} z$ is uniformly small if $n$ is large enough.
Using the notation of the lemma, since $\map {B_n} z \to 0$ uniformly, let $n$ be so large that $\size {\map {B_N} z} \le \epsilon$ in $K$ for all $N \ge n$.
Since $\sequence {\map {a_n} z}$ is uniformly bounded in $K$, let $\size {\map {a_n} z} \le M$ for all $z \in K$.
Since $\ds \sum \size {\map {a_n} z - \map {a_{n + 1} } z}$ is convergent with a sum which is bounded in $K$, let $n$ be so large that:
:$\forall k \in \N: \size {a_{n + 1} - a_n} + \cdots + \size {a_{n + k} - a_{n + k - 1} } \le \epsilon$ for all $z \in K$
then we have:
:$\forall z \in K: \size {B_{n + 1} \paren {a_{n + 1} - a_n} } + \cdots + \size {B_{n + k} \paren {a_{n + k} - a_{n + k - 1} } } \le M \epsilon$
Therefore:
Therefore, $\ds \sum \map {a_n} z \map {b_n} z$ is uniformly convergent on $K$.
== Sources ==
* 
Category:Abel's Test
Category:Complex Analysis
Abel-Plana Formula

== Theorem ==
Let $\map f z$ be analytic for real part of $\map \Re z \ge 0$.
Suppose that either $\ds \sum_{n \mathop = 0}^\infty \map f n$ converges or $\ds \int_0^\infty \map f x \rd x$ converges.
Assume further that:
:$\ds \lim_{y \mathop \to \infty} \size {\map f {x \pm i y} } e^{-2 \pi y} = 0$ uniformly in $x$ on every finite interval
:$\ds \int_0^{\infty} \size {\map f {x \pm i y} } e^{-2 \pi y} \rd y$ exists for every $x \ge 0$ and tends to $0$ as $x \to \infty$.
Then:
:$\ds \sum_{n \mathop = 0}^\infty \map f n = \int_0^\infty \map f x \rd x + \dfrac 1 2 \map f 0 + i \int_0^\infty \dfrac {\map f {i t} - \map f {-i t} } {e^{2 \pi t} - 1} \rd t$
== Proof ==
== Sources == 
* : $7$: The Process of Analysis
* 
Category:Analytic Number Theory
Abel-Ruffini Theorem

== Theorem ==
There is no general algebraic solution for determining all the roots of a polynomial of degree $5$ or higher.
== Proof ==
The general polynomial:
:$\map f x = x^n - s_1 x^{n - 1} + \cdots + \paren {-1}^n s_n$
has Galois group $S_n$.
From Symmetric Group on Greater than 4 Letters is Not Solvable:
:$S_n$ is not solvable for $n \ge 5$.
By Polynomial is Solvable by Radicals iff Galois Group is Solvable
:the general polynomial is not solvable by radicals.
== Historical Note ==
== Sources ==
* : $4$
* : $5$
* 
* : $4$
* : $5$
Category:Polynomial Theory
Category:Polynomial Equations
Category:5
Abel-Ruffini Theorem/Historical Note

== Historical Note on Abel-Ruffini Theorem ==
The Abel-Ruffini Theorem, on the general insolubility of the quintic by radicals, was stated by , who built an incomplete proof in $1799$.
This was published in his two-volume work .
 provided the first complete proof in $1823$.
He published this in a small pamphlet ($1824$) at his own expense.
He sent a copy of this to , but for some reason put it aside and never opened it.
Thus never had cause to visit and the pair never met.
It later transpired that had independently proved this theorem some years earlier, in a work that was not published in $1846$, some $25$ years after his death.
Moreover, analysis of the problem also gave a complete answer to the question of which equations are solvable in radicals and which are not.
== Sources ==
* : Introduction
* : $5$
* : Chapter $\text {A}.27$: Abel ()
* : $5$
Category:Historical Notes
Abelian Group Factored by Prime

== Theorem ==
Let $G$ be a finite abelian group.
Let $p$ be a prime.
Factor the order of $G$ as:
:$\order G = m p^n$
such that $p$ does not divide $m$.
Then:
:$G = H \times K$
where:
:$H = \set {x \in G : x^{p^n} = e}$
and:
:$K = \set {x \in G : x^m = e}$
Furthermore:
:$\order H = p^n$
=== Corollary ===
== Proof ==
From Subgroup of Elements whose Order Divides Integer, both $H$ and $K$ are subgroups of $G$.
Also, because $G$ is abelian, $H$ and $K$ are normal by Subgroup of Abelian Group is Normal.
In order to prove $G = H \times K$, by the Internal Direct Product Theorem it suffices to show that $G = H K$ and $H \cap K = \set e$.
Since we have $\gcd \set {m, p^n} = 1$, there are integers $s$ and $t$ such that $1 = s m + t p^n$ by Bézout's Identity. 
Let $s$ and $t$ be two such integers.
So:
:$\forall x \in G: x = x^{s m + t p^n} = x^{s m} x^{t p^n}$
From Element to Power of Group Order is Identity:
:$x^{\order G} = e$
Therefore:
:$\paren {x^{s m} }^{p^n} = \paren {x^{p^n m} }^s = e^s = e$
and:
:$\paren {x^{t p^n} }^m = \paren {x^{p^n m} }^t = e^t = e$
By definition, $x^{s m} \in H$ and $x^{t p^n} \in K$.
We conclude that $G = H K$.
Now suppose that some $x \in H \cap K$.
Then $x^{p^n} = e = x^m$, so from Element to Power of Multiple of Order is Identity $\order x$ divides both $p^n$ and $m$.
Since $p$ does not divide $m$, it follows that $\order x = 1$.
Therefore, by Identity is Only Group Element of Order 1:
:$x = e$
Thus:
:$H \cap K = \set e$
It follows that $G = H \times K$.
Suppose that $p \divides \order K$.
From Cauchy's Lemma (Group Theory), this implies that there is some element (call it $k$) of $K$ with order $p$.
But we also have $k^m = e$ from the definition of $K$, so by Element to Power of Multiple of Order is Identity we must have $p \divides m$, a contradiction.
We conclude that:
:$p$ does not divide $\order K$
It follows that:
:$p^n \divides \order H$
 a prime $q$ (with $q \ne p$) divides $\order H$.
Again from Cauchy's Lemma (Group Theory), this implies that there is some element $h$ of $H$ with order $q$.
But since $h^{p^n} = e$ from the definition of $H$, Element to Power of Multiple of Order is Identity gives us $q \divides p^n$, a contradiction.
It follows by Proof by Contradiction that $q$ does not divide $\order H$.
We conclude $\order H = p^n$, as desired.
Category:Abelian Groups
Abelian Group Factored by Prime/Corollary

== Corollary to Abelian Group Factored by Prime ==
A finite abelian group $G$ can be factored as follows:
Let $\order G = \ds \prod_{i \mathop = 1}^k p_i^{n_i}$ be the prime factorization of the order of $G$.
Then we have:
:$G = \ds \prod_{i \mathop = 1}^k H_i$, where $H_i = \set {x \in G : x^{p_i^{n_i} } = e}$
This factorization is unique up to ordering of the factors.
== Proof ==
Let $\ds \prod_{i \mathop = 1}^k p_i^{n_i}$ be the prime factorization of $\order G$.
We proceed by induction on $k$.
=== Basis for the induction ===
For $n = 1$, the statement is trivial.
=== Induction Hypothesis ===
Now we assume the theorem is true for abelian groups whose order has $k - 1$ distinct prime factors.
=== Induction Step ===
Apply Abelian Group Factored by Prime to $G$ and $p_1$. 
By definition, $H = H_1$.
Also, the resulting $K$ has $\order K = \ds \prod_{i \mathop = 2}^k p_i^{n_i}$.
Therefore, it satisfies the induction hypothesis.
It follows that $G = H_1 \times \ds \prod_{i \mathop = 2}^k H_i$.
From:
:Subgroup of Abelian Group is Normal
:the definition of Sylow $p$-subgroup
all the $H_i$ are normal Sylow $p$-subgroups.
From Sylow $p$-Subgroup is Unique iff Normal, the factorization is unique up to ordering of the factors.
Category:Abelian Groups
Abelian Group Induces Commutative B-Algebra

== Theorem ==
Let $\struct {G, \circ}$ be an abelian group whose identity element is $e$.
Let $*$ be the product inverse operation on $G$:
:$\forall a, b \in G: a * b = a \circ b^{-1}$
where $b^{-1}$ is the inverse element of $b$ under the operation $\circ$.
Then the algebraic structure $\struct {G, *}$ is a commutative $B$-algebra.
That is:
:$\forall a, b \in G: a * \paren {0 * b} = b * \paren {0 * a}$
== Proof ==
From Group Induces $B$-Algebra, $\struct {G, *}$ is a $B$-Algebra.
As in the proof Group Induces $B$-Algebra, we let:
:$0 := e$
Now we demonstrate $0$-commutativity.
Let $x, y \in G$:
Hence the result.
Category:B-Algebras
Category:Abelian Groups
Abelian Group Induces Entropic Structure

== Theorem ==
Let $\struct {G, \circ}$ be an abelian group.
Let the operation $*$ be defined on $G$ such that:
:$\forall x, y \in G: x * y = x \circ y^{-1}$
Then $\struct {G, *}$ is an entropic structure.
== Proof ==
We need to prove that $\forall a, b, c, d \in G: \paren {a * b} * \paren {c * d} = \paren {a * c} * \paren {b * d}$.
So:
== Also presented as ==
This is usually presented in the form:
Let the operation $-$ be defined on $\struct {G, +}$ such that:
:$\forall x, y \in G: x - y = x + \paren {-y}$
Then $\struct {G, -}$ is an entropic structure.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 13$: Compositions Induced on Cartesian Products and Function Spaces: Exercise $13.12 \ \text{(b)}$
Category:Abelian Groups
Category:Entropic Structures
Abelian Group is Simple iff Prime

== Theorem ==
Let $G$ be a non-trivial abelian group.
Then $G$ is simple $G$ is a prime group.
== Proof ==
First we note that the trivial group is (trivially) simple, but not prime because $1$ is not prime.
Hence the specification of $G$ as being non-trivial.
=== Necessary Condition ===
Let $G$ be a simple non-trivial abelian group.
 $G$ is an infinite group.
Then by Infinite Group has Infinite Number of Subgroups $G$ has an infinite number of distinct subgroups.
But then by Subgroup of Abelian Group is Normal, $G$ cannot be simple.
So by contradiction $G$ must be finite.
Let $n$ be the order of $G$.
By definition of simple, $G$ has no non-trivial proper normal subgroups.
From Subgroup of Abelian Group is Normal, it follows that $G$ can have no non-trivial proper subgroups at all.
From Cauchy's Group Theorem, if $p$ is a prime number which is a divisor of $n$, then $G$ has a subgroup of order $p$.
It follows that if $G$ is simple, there can be no prime number less than $n$ which is a divisor of $n$.
It follows that $n$ is prime.
=== Sufficient Condition ===
Suppose $G$ is a prime group.
By Prime Group is Simple, it follows that $G$ is simple.
Because $G$ is prime, its order is greater than $1$.
Thus $G$ is non-trivial.
== Sources ==
* : $1$: Introduction to Finite Group Theory: $1.11$
* 
Category:Simple Groups
Category:Abelian Groups
Abelian Group of Order Twice Odd has Exactly One Order 2 Element

== Theorem ==
Let $G$ be an abelian group whose identity element is $e$.
Let the order of $G$ be $2 n$ such that $n$ is odd.
Then there exists exactly one $g \in G$ with $g \ne e$ such that $g = g^{-1}$.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $\S 1.9$: Exercise $5.7$
Category:Abelian Groups
Category:Abelian Group of Order Twice Odd has Exactly One Order 2 Element
Abelian Group of Order Twice Odd has Exactly One Order 2 Element/Proof 1

== Theorem ==
== Proof ==
By Abelian Group Factored by Prime, the subgroup $H_2$ defined as:
:$H_2 := \set {g \in G: g^2 = e}$
has precisely two elements.
One of them has to be $e$, since $e^2 = e$.
The result follows.
Category:Abelian Group of Order Twice Odd has Exactly One Order 2 Element
Abelian Group of Order Twice Odd has Exactly One Order 2 Element/Proof 2

== Theorem ==
== Proof ==
By Even Order Group has Order 2 Element, $G$ has an element $x$ of order $2$.
 $y$ is another element of order $2$.
Then $x y = y x$ is another element of order $2$.
The subset $H = \set {g \in G: g^2 = e} = \set {e, x, y, x y}$ of $G$ forms a subgroup of $G$.
Thus $\order H = 4$.
But as $n$ is odd, it follows that $\order H$ is not a divisor of $2 n$.
This contradicts Lagrange's Theorem (Group Theory).
The result follows.
== Sources ==
Category:Abelian Group of Order Twice Odd has Exactly One Order 2 Element
Abelian Group of Prime-power Order is Product of Cyclic Groups/Corollary

== Theorem ==
Let $G$ be an abelian group of prime-power order.
Then $G$ can be written as an internal direct product of cyclic groups of prime-power order.
== Proof ==
From Lagrange's Theorem, we see that in the proof of Abelian Group of Prime-power Order is Product of Cyclic Groups, also $K$ satisfies the induction hypothesis.
The result follows as $\left\langle{a}\right\rangle$ is cyclic and of prime-power order.
Category:Abelian Groups
Abelian Group of Semiprime Order is Cyclic

== Theorem ==
Let $p$ and $q$ be distinct prime numbers.
Let $G$ be an abelian group such that:
:$\order G = p q$
where $\order G$ denotes the order of $G$.
Then $G$ is cyclic.
== Proof ==
By Order of Element Divides Order of Finite Group, the order of elements of $G$ are all in $\set {1, p, q, p q}$.
We have that Identity is Only Group Element of Order 1.
Suppose $G$ were to contain more than $p - 1$ elements of order $p$.
Then by Order of Finite Abelian Group with $p+$ Order $p$ Elements is Divisible by $p^2$:
:$p^2 \divides \order G$
where $\divides$ denotes divisibility.
As $p^2 \nmid p q$ it follows that $G$ contains no more than $p - 1$ elements of order $p$.
Similarly it follows that $G$ contains no more than $q - 1$ elements of order $q$.
These, along with $e$, account for $\paren {p - 1} + \paren {q - 1} + 1 = p + q - 1$ elements.
Thus the number of elements of order $p q$ is:
:$p q - \paren {p + q - 1} = \paren {p - 1} \paren {q - 1}$
which is strictly positive.
The result follows from Group whose Order equals Order of Element is Cyclic.
== Sources ==
* : Chapter $7$: Cosets and Lagrange's Theorem: Exercise $20$
Category:Abelian Groups
Category:Finite Groups
Category:Cyclic Groups
Abelianization of Free Group is Free Abelian Group

== Theorem ==
Let $X$ be a set.
Let $\struct {F_X, \iota}$ be a free group on $X$.
Let $F_X^{\mathrm {ab} }$ be its abelianization.
Let $\pi : F_X \to F_X^{\mathrm {ab} }$ be the quotient group epimorphism.
Then $\struct {F_X^{\mathrm {ab} }, \pi \circ \iota}$ is a free abelian group on $X$.
== Proof ==
:$\xymatrix {
X \ar[d]_\iota \ar[rd]^{\forall f} & \\
F_X \ar[d]_\pi \ar[r]^{\exists ! g} & H \\
F_X^{\mathrm {ab} } \ar[ru]_{\exists ! h} }$
Lef $H$ be any abelian group and $f:X \to H$ be any mapping.
By Universal Property of Free Group on Set, there exists a unique group homomorphism $g:F_X \to H$ such that $g \circ \iota = f$.
By Universal Property of Abelianization of Group, there exists a unique group homomorphism $h:F_X^{\mathrm {ab} } \to H$ such that $h \circ \pi = g$.
Therefore, $h \circ (\pi \circ \iota) = f$.
Hence, $F_X^{\mathrm {ab} }$ satisfies the Universal Property of Free Abelian Group on Set.
Hence $F_X^{\mathrm {ab} }$ is a Free Abelian Group on $X$.
Category:Group Theory
Abelianization of Group is Abelian

== Theorem ==
Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $G^{\mathrm {ab} }$ denote the abelianization of $G$.
Then $G^{\mathrm {ab} }$ is an abelian group.
== Proof ==
From Derived Subgroup is Normal, $\sqbrk {G, G}$ is a normal subgroup of $G$.
Hence the above definition is valid.
By definition of derived subgroup:
:$\forall x, y \in G: \sqbrk {x, y} \in \sqbrk {G, G}$
where $\sqbrk {x, y}$ denotes the commutator of $x$ and $y$.
Hence from Quotient Group is Abelian iff All Commutators in Divisor it follows that $G^{\mathrm {ab} }$ is an abelian group.
== Sources ==
* 
Category:Abelianizations of Groups
Category:Abelian Groups
Abi-Khuzam Inequality

== Theorem ==
Let $\triangle ABC$ be a triangle.
Then:
:$\sin A \cdot \sin B \cdot \sin C \le k A \cdot B \cdot C$
where:
:$A, B, C$ are measured in radians
:$k = \paren {\dfrac {3 \sqrt 3} {2 \pi} }^3 \approx 0 \cdotp 56559 \, 56245 \ldots$
=== Condition for Equality ===
:$A=B=C$
== Proof ==
For $0 < \alpha_i < \pi, \alpha_1 + \alpha_2 + \alpha_3 = \pi$ we consider the function:
:$\map F {\alpha_1, \alpha_2, \alpha_3} = \dfrac {\sin \alpha_1 \sin \alpha_2 \sin \alpha_3} {\alpha_1 \alpha_2 \alpha_3}$
$F$ is defined on the region $G$ in the $\tuple {\alpha_1, \alpha_2, \alpha_3}$-space consisting of the points inside the triangle with the vertices:
We define $F$ on the boundary of the triangle by:
:$\map F {0, \alpha_2, \alpha_3} = \dfrac {\sin \alpha_2 \sin \alpha_3} {\alpha_2 \alpha_3}, \alpha_2 \neq 0, \alpha_3 \neq 0, \alpha_2 + \alpha_3 = \pi$
and analogously:
:$\map F {\alpha_1, 0, \alpha_3}$ and $\map F{\alpha_1, \alpha_2, 0}$
furthermore:
:$\map F {\pi, 0, 0} = \map F {0, \pi, 0} = \map F {0, 0, \pi} = 0$
$F$ is now defined on a closed region $\bar G$; it is continuous and differentiable on $\bar G$.
Moreover, as $0 \le F < 1$ there is (at least) one point in $\bar G$ where $F$ has its maximum value.
By the usual procedure, in view of $\alpha_1 + \alpha_2 + \alpha_3 = \pi$, a maximum satisfies:
In $G$ we have:
:$\ds \frac {\partial F} {\partial \alpha_1} = \frac {\sin \alpha_2 \sin \alpha_3} {\alpha_2 \alpha_3} \cdot \frac {\alpha_1 \cos \alpha_1 - \sin \alpha_1} {\alpha_1^2} = \map F {\cot \alpha_1 - \alpha_1^{-1} }$
and, as $F \ne 0$, $(6)$ implies:
For $f = \cot \alpha - \alpha^{-1}$ we obtain:
:$f' = -\sin^{-2} \alpha + \alpha^{-2} < 0$
$f$ is therefore a decreasing function of $\alpha$ (we have $0 > f > -\infty$)
hence $(7)$ implies
:$\ds \alpha_1 = \alpha_2 = \alpha_3 (= \pi / 3)$
in this point we have $F = k$.
We must verify whether larger values appear on the boundary of $\bar G$.
Between $P_2$ and $P_3$ yields
:$\ds F = \frac {\sin \alpha_2 \sin \alpha_3} {\alpha_2 \alpha_3}, \alpha_2 + \alpha_3 = \pi$
and by an argument analogous to the former, but now with two factors instead of three, it follows that for the maximum on $P_2 P_3$ we have $\alpha_2 = \alpha_3 = \pi / 2$ and $F = 4 / \pi^2$, that is less than $k$.
Hence $F \le k$ on $\bar G$, which concludes the proof.
== Sources ==
* 
* 
* : $0,56559 56245 \ldots$
* 
* 
Category:Triangles
Abnormal Subgroup is Self-Normalizing Subgroup

== Theorem ==
Let $G$ be a group.
Let $H$ be an abnormal subgroup of $G$.
Then $H$ is a self-normalizing subgroup of $G$.
== Proof ==
We have:
* Abnormal Subgroup is Weakly Abnormal Subgroup
* Weakly Abnormal Subgroup is Self-Normalizing Subgroup
Hence the result.
Category:Abnormal Subgroups
Category:Self-Normalizing Subgroups
Abridged Multiplication/Examples/Arbitrary Example

== Example of Abridged Multiplication ==
Let:
Let it be required that $z = x y$ is to be rounded to $2$ decimal places.
Then, when we calculate:
:$z = 5.6982 \times 23$
we can ignore the $4$th decimal place in the partial products of $x y$.
== Demonstration ==
Performing the full multiplication:
 5.6982
 x 23
----------
 17.0946
 113.9640
----------
 131.0586
----------
which rounds to $131.06$.
Ignoring the $4$th decimal place:
 5.6982
 x 23
----------
 17.0946
 113.9640
----------
 17.094 -- dropping the 4th decimal place
 113.964
----------
 131.058
---------
which again rounds to $131.06$.
== Sources ==
* 
* 
Category:Examples of Abridged Multiplication
Absolute Continuity of Measures is Transitive Relation

== Theorem ==
Let $\struct {X, \Sigma}$ be a measurable space.
Let $\mu$, $\nu$ and $\lambda$ be measures on $\struct {X, \Sigma}$ such that: 
:$\mu$ is absolutely continuous with respect to $\nu$
and:
:$\nu$ is absolutely continuous with respect to $\lambda$.
That is: 
:$\mu \ll \nu$
and:
:$\nu \ll \lambda$
Then:
:$\mu$ is absolutely continuous with respect to $\lambda$.
That is: 
:$\mu \ll \lambda$
In other words:
:$\ll$ is a transitive relation.
== Proof ==
Let $A \in \Sigma$ be such that $\map \lambda A = 0$. 
Then, since $\nu \ll \lambda$ we have $\map \nu A = 0$ from the definition of absolute continuity.
Since $\mu \ll \nu$ we similarly have $\map \mu A = 0$ again applying the definition of absolute continuity.
So, whenever $A \in \Sigma$ is such that $\map \lambda A = 0$ we have $\map \mu A = 0$.
So $\mu$ is absolutely continuous with respect to $\lambda$.
Category:Absolutely Continuous Measures
Category:Transitive Relations
Absolute Continuity of Signed Measure in terms of Jordan Decomposition

== Theorem ==
Let $\struct {X, \Sigma}$ be a measurable space.
Let $\mu$ be a measure on $\struct {X, \Sigma}$.
Let $\nu$ be a signed measure on $\struct {X, \Sigma}$.
Let $\tuple {\nu^+, \nu^-}$ be the Jordan decomposition of $\nu$.
Then $\nu$ is absolutely continuous with respect to $\mu$ :
:$\nu^+$ and $\nu^-$ are absolutely continuous with respect to $\mu$.
== Proof ==
We have that $\nu$ is absolutely continuous with respect to $\mu$ :
:$\size \nu$ is absolutely continuous with respect to $\mu$
where $\size \nu$ is the variation of $\nu$.
From the definition of variation, we have: 
:$\size \nu = \nu^+ + \nu^-$ 
Suppose that $\nu$ is absolutely continuous with respect to $\mu$. 
Then: 
:$\nu^+ + \nu^-$ absolutely continuous with respect to $\mu$.
That is:
:for all $A \in \Sigma$ with $\map \mu A = 0$ we have $\map {\nu^+} A + \map {\nu^-} A = 0$.
Since $\nu^+ \ge 0$ and $\nu^-$, we have that: 
:$\map {\nu^+} A + \map {\nu^-} A = 0$ implies that $\map {\nu^+} A = 0$ and $\map {\nu^-} A = 0$.
So:
:for all $A \in \Sigma$ with $\map \mu A = 0$ we have $\map {\nu^+} A = 0$
so:
:$\nu^+$ is absolutely continuous with respect to $\mu$.
We also have: 
:for all $A \in \Sigma$ with $\map \mu A = 0$ we have $\map {\nu^-} A = 0$
so:
:$\nu^-$ is absolutely continuous with respect to $\mu$.
So:
:if $\nu$ is absolutely continuous with respect to $\mu$ then $\nu^+$ and $\nu^-$ are absolutely continuous with respect to $\mu$.
Suppose that: 
:$\nu^+$ and $\nu^-$ are absolutely continuous with respect to $\mu$.
Then: 
:for all $A \in \Sigma$ with $\map \mu A = 0$ we have $\map {\nu^+} A = 0$ and $\map {\nu^-} A = 0$.
So:
:for all $A \in \Sigma$ with $\map \mu A = 0$ we have $\map {\nu^+} A + \map {\nu^-} A = 0$
giving: 
:$\nu^+ + \nu^-$ is absolutely continuous with respect to $\mu$.
That is: 
:$\size \nu$ is absolutely continuous with respect to $\mu$.
So:
:if $\nu^+$ and $\nu^-$ are absolutely continuous with respect to $\mu$ then $\nu$ is absolutely continuous with respect to $\mu$.
== Sources ==
* : $4.2$: Absolute Continuity
Category:Signed Measures
Absolute Cumulative Frequency/Examples/Arbitrary Example 1

== Example of Absolute Cumulative Frequency ==
Consider the sample:
:$2, 5, 3, 3, 3, 5, 3, 6, 2, 3, 9, 5$
The absolute cumulative frequency of the observation $5$ is $10$.
== Sources ==
* 
* 
Category:Examples of Absolute Cumulative Frequency
Absolute Difference Function is Primitive Recursive

== Theorem ==
The absolute difference function $\operatorname {adf}: \N^2 \to \N$, defined as:
:$\map {\operatorname {adf} } {n, m} = \size {n - m}$
where $\size a$ is defined as the absolute value of $a$, is primitive recursive‎.
== Proof ==
We note that:
:$\size {n - m} = \paren {n \mathop {\dot -} m} + \paren {m \mathop {\dot -} n} = \map {\operatorname {add} } {\paren {n \mathop {\dot -} m}, \paren {m \mathop {\dot -} n} }$
Next we note that:
:$m \mathop {\dot -} n = \map {\pr^2_2} {n, m} \mathop {\dot -} \map {\pr^2_1} {n, m}$
where $\pr^2_k$ is the projection function.
Then:
Hence we see that $\operatorname {adf}$ is obtained by substitution from:
:the primitive recursive function $n \mathop {\dot -} m$
:the primitive recursive function $\map {\operatorname {add} } {n, m}$
:the projection function.
Hence the result.
Category:Primitive Recursive Functions