nyuuzyou/wikis huggingface dataset:
Absolutely Convergent Generalized Sum over Union of Disjoint Index Sets

== Theorem ==
Let $V$ be a Banach space.
Let $I$ and $J$ be disjoint indexing sets.
Let $K = I \cup J$.
Let $\family{v_k}_{k \mathop \in K}$ be an indexed family of elements of $V$.
Then:
:the generalized sum $\ds \sum_{k \mathop \in K} v_k$ converges absolutely
:the generalized sums $\ds \paren{\sum_{i \mathop \in I} v_i}$ and $\ds \paren{\sum_{j \mathop \in J} v_j}$ converge absolutely
In which case:
:$\ds \sum_{k \mathop \in K} \norm{v_k} = \paren{\sum_{i \mathop \in I} \norm{v_i}} + \paren{\sum_{j \mathop \in J} \norm{v_j}}$
== Proof ==
=== Necessary Condition ===
Let $\ds \sum_{k \mathop \in K} v_k$ converge absolutely.
By definition of absolute net convergence:
:$\ds \sum_{k \mathop \in K} \norm{v_k}$ converges.
From Generalized Sum over Subset of Absolutely Convergent Generalized Sum is Absolutely Convergent:
:$\ds \paren{\sum_{i \mathop \in I} \norm{v_i}}$ and $\ds \paren{\sum_{j \mathop \in J} \norm{v_j}}$ converge
By definition of absolute net convergence:
:$\ds \paren{\sum_{i \mathop \in I} v_i}$ and $\ds \paren{\sum_{j \mathop \in J} v_j}$ converge absolutely.
From Generalized Sum over Union of Disjoint Index Sets:
:$\ds \sum_{k \mathop \in K} \norm{v_k} = \paren{\sum_{i \mathop \in I} \norm{v_i}} + \paren{\sum_{j \mathop \in J} \norm{v_j}}$
=== Sufficient Condition ===
Let $\ds \paren{\sum_{i \mathop \in I} v_i}$ and $\ds \paren{\sum_{j \mathop \in J} v_j}$ converge absolutely.
By definition of absolute net convergence:
:$\ds \paren{\sum_{i \mathop \in I} \norm{v_i}}$ and $\ds \paren{\sum_{j \mathop \in J} \norm{v_j}}$ converge
From Generalized Sum over Union of Disjoint Index Sets:
:$\ds \sum_{k \mathop \in K} \norm{v_k} = \paren{\sum_{i \mathop \in I} \norm{v_i}} + \paren{\sum_{j \mathop \in J} \norm{v_j}}$
By definition of absolute net convergence:
:$\ds \sum_{k \mathop \in K} v_k$ converges absolutely.
Category:Generalized Sums
Absolutely Convergent Product Does not Diverge to Zero

== Theorem ==
Let $\struct {\mathbb K, \norm {\, \cdot \,} }$ be a valued field.
Let the infinite product $\ds \prod_{n \mathop = 1}^\infty \paren {1 + a_n}$ be absolutely convergent.
Then it is not divergent to $0$.
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Absolutely Convergent Product is Convergent
Category:Infinite Products
Category:Absolutely Convergent Product Does not Diverge to Zero
Absolutely Convergent Product Does not Diverge to Zero/Proof 1

== Theorem ==
== Proof ==
By Factors in Absolutely Convergent Product Converge to One, $\norm {a_n} < 1$ for $n \ge n_0$.
Let $n_1 \ge n_0$.
 the product diverges to $0$.
Then:
:$\ds \prod_{n \mathop = n_1}^\infty \paren {1 + a_n} = 0$
By Norm of Limit:
:$\ds \prod_{n \mathop = n_1}^\infty \norm {1 + a_n} = 0$
By the Triangle Inequality and Squeeze Theorem:
:$\ds \prod_{n \mathop = n_1}^\infty \paren {1 - \norm {a_n} } = 0$
By the Weierstrass Product Inequality, we have for $N \ge n_1$:
:$\ds \prod_{n \mathop = n_1}^N \paren {1 - \norm {a_n} } \ge 1 - \sum_{n \mathop = n_1}^N \norm{a_n}$
Taking limits:
:$0 \ge 1 - \ds \sum_{n \mathop = n_1}^\infty \norm {a_n}$
Because $\ds \prod_{n \mathop = 1}^\infty \paren {1 + a_n}$ is absolutely convergent:
:$\ds \sum_{n \mathop = n_1}^\infty \norm {a_n} < 1$
for $n_1$ sufficiently large.
This is a contradiction.
Category:Absolutely Convergent Product Does not Diverge to Zero
Absolutely Convergent Product Does not Diverge to Zero/Proof 2

== Theorem ==
== Proof ==
We have that $\ds \prod_{n \mathop = 1}^\infty \paren {1 - \norm {a_n} }$ is absolutely convergent.
By Factors in Absolutely Convergent Product Converge to One, $\norm {a_n} < 1$ for $n \ge n_0$.
Thus $\ds \sum_{n \mathop = n_0}^\infty \map \ln {1 - \norm {a_n} }$ is absolutely convergent.
 the product diverges to $0$.
Then:
:$\ds \prod_{n \mathop = n_0}^\infty \paren {1 + a_n} = 0$
By Norm of Limit:
:$\ds \prod_{n \mathop = n_0}^\infty \norm {1 + a_n} = 0$
By the Triangle Inequality and Squeeze Theorem:
:$\ds \prod_{n \mathop = n_0}^\infty \paren {1 - \norm {a_0} } = 0$
By Logarithm of Infinite Product of Real Numbers:
:$\ds \sum_{n \mathop = n_0}^\infty \map \ln {1 - \norm {a_n} }$
diverges to $-\infty$.
This is a contradiction.
Category:Absolutely Convergent Product Does not Diverge to Zero
Absolutely Convergent Product is Convergent

== Theorem ==
Let $\struct {\mathbb K, \norm {\,\cdot\,} }$ be a valued field.
Let $\mathbb K$ be complete.
Let the infinite product $\ds \prod_{n \mathop = 1}^\infty \paren {1 + a_n}$ be absolutely convergent.
Then it is convergent.
== Proof ==
Let $P_n$ and $Q_n$ denote the $n$th partial products of $\ds \prod_{n \mathop = 1}^\infty \paren {1 + a_n}$ and $\ds \prod_{n \mathop = 1}^\infty \paren {1 + \norm {a_n} }$ respectively.
We show that $\sequence {P_n}$ is Cauchy.
We have, for $m > n$:
Because $\sequence {Q_n}$ converges, $\sequence {Q_n}$ is Cauchy.
By the above inequality, $\sequence {P_n}$ is Cauchy.
Because $\mathbb K$ is complete, $\sequence {P_n}$ converges to some $a \in \mathbb K$.
By Absolutely Convergent Product Does not Diverge to Zero, the product converges.
Category:Infinite Products
Absolutely Convergent Series in Normed Vector Space is Convergent iff Space is Banach

== Theorem ==
Let $\struct {X, \norm {\, \cdot \,} }$ be a normed vector space.
Then $X$ is a Banach space :
:every absolutely convergent series $\ds \sum_{n \mathop = 1}^\infty a_n$ is convergent.
== Proof ==
=== Necessary Condition ===
=== Sufficient Condition ===
== Sources ==
* : $\S 1$: Normed and Banach spaces
Category:Absolutely Convergent Series in Normed Vector Space is Convergent iff Space is Banach
Category:Absolute Convergence
Category:Banach Spaces
Category:Absolutely Convergent Series is Convergent
Category:Convergent Sequences (Normed Vector Spaces)
Absolutely Convergent Series in Normed Vector Space is Convergent iff Space is Banach/Necessary Condition

== Theorem ==
Let $\struct {X, \norm {\, \cdot \,} }$ be a normed vector space.
Let $\ds \sum_{n \mathop = 1}^\infty a_n$ be an absolutely convergent series in $X$.
Suppose $X$ is a Banach space.
Then $\ds \sum_{n \mathop = 1}^\infty a_n$ is convergent.
== Proof ==
That $\ds \sum_{n \mathop = 1}^\infty a_n$ is absolutely convergent means that $\ds \sum_{n \mathop = 1}^\infty \norm {a_n}$ converges in $\R$.
Hence by Convergent Sequence in Normed Vector Space is Cauchy Sequence:
:the sequence of partial sums is a Cauchy sequence.
Now let $\epsilon > 0$.
Let $N \in \N$ such that for all $m, n \in \N$, $m \ge n \ge N$ implies that:
:$\ds \sum_{k \mathop = n + 1}^m \norm {a_k} = \size {\sum_{k \mathop = 1}^m \norm {a_k} - \sum_{k \mathop = 1}^n \norm {a_k} } < \epsilon$
This $N$ exists because the sequence is Cauchy.
Now observe that, for $m \ge n \ge N$, one also has:
It follows that the sequence of partial sums of $\ds \sum_{n \mathop = 1}^\infty a_n$ is Cauchy.
As $X$ is a Banach space, this implies that $\ds \sum_{n \mathop = 1}^\infty a_n$ converges.
== Also see ==
* Absolutely Convergent Generalized Sum Converges
== Sources ==
* : $\S 1.4$: Normed and Banach spaces. Sequences in a normed space; Banach spaces
Category:Absolutely Convergent Series in Normed Vector Space is Convergent iff Space is Banach
Absolutely Convergent Series in Normed Vector Space is Convergent iff Space is Banach/Sufficient Condition

== Theorem ==
Let $\struct {X, \norm {\, \cdot \,}}$ be a normed vector space.
Let $\ds \sum_{n \mathop = 1}^\infty a_n$ be an absolutely convergent series in $X$.
Suppose $\ds \sum_{n \mathop = 1}^\infty a_n$ is convergent.
Then $X$ is a Banach space.
== Proof ==
Let $\sequence {x_n}_{n \mathop \in \N}$ be a Cauchy sequence in $X$.
We have that:
:$\forall \epsilon \in \R_{>0}: \exists N \in \N: \forall m, n \in \N: m, n \ge N: \norm {x_n - x_m} < \epsilon$
We will prove the existence of a subsequence $\sequence {x_{n_k} }_{k \mathop \in \N}$ such that:
:$n > n_k \implies \norm {x_n - x_{n_k} } < \dfrac 1 {2^k}$
=== Basis for the Induction ===
Let $\epsilon = \dfrac 1 2$.
Choose $n_1$ such that:
:$n, m \ge n_1 \implies \norm {x_n - x_m} < \dfrac 1 2$
In particular, when $m = n_1$:
:$n \ge n_1 \implies \norm {x_n - x_{n_1} } < \dfrac 1 2$
=== Induction Step ===
Suppose $x_{n_1}, \dots x_{n_k}$ have been constructed.
Let $\epsilon = \dfrac 1 {2^{k + 1} }$.
Choose $n_{k + 1}$ such that $n_{k + 1} > n_k$ and:
:$n, m \ge n_{k + 1} \implies \norm {x_n - x_m} < \dfrac 1 {2^{k + 1} }$
In particular, when $m = n_{k + 1}$:
:$n \ge n_{k + 1} \implies \norm {x_n - x_{n_{k + 1} } } < \dfrac 1 {2^{k + 1}}$
Define:
:$u_1 := x_{n_1}$
:$u_{k + 1} := x_{n_{k + 1} } - x_{n_k}$
Now we have a sequence $\sequence {u_k}_{k \mathop \in \N}$.
Consider the series $\ds \sum_{k \mathop = 1}^\infty \norm {u_k}$:
Thus, $\ds \sum_{k \mathop = 1}^\infty u_k$ is absolutely convergent.
By assumption in the theorem, $\ds \sum_{k \mathop = 1}^\infty u_k$ is convergent.
In other words:
:$\ds \lim_{k \mathop \to \infty} \sum_{j \mathop = 1}^k u_j = u$.
On the other hand:
Therefore:
:$\ds \lim_{k \mathop \to \infty} x_{n_k} = u =: x$
So $\sequence {x_{n_k} }_{k \mathop \in \N}$ converges in $X$.
We have that $\sequence {x_{n_k} }_{k \mathop \in \N}$ is a convergent subsequence of a Cauchy sequence$\sequence {x_n}_{n \mathop \in \N}$.
By Convergent Subsequence of Cauchy Sequence in Normed Vector Space, $\sequence {x_n}_{n \mathop \in \N}$ is convergent with the same limit $x$.
By definition, the underlying space is Banach.
== Sources ==
* : $\S 1.4$: Normed and Banach spaces. Sequences in a normed space; Banach spaces
Category:Absolutely Convergent Series in Normed Vector Space is Convergent iff Space is Banach
Absolutely Convergent Series is Convergent

== Theorem ==
Let $V$ be a Banach space with norm $\norm {\, \cdot \,}$.
Let $\ds \sum_{n \mathop = 1}^\infty a_n$ be an absolutely convergent series in $V$.
Then $\ds \sum_{n \mathop = 1}^\infty a_n$ is convergent.
== Proof ==
That $\ds \sum_{n \mathop = 1}^\infty a_n$ is absolutely convergent means that $\ds \sum_{n \mathop = 1}^\infty \norm {a_n}$ converges in $\R$.
Hence the sequence of partial sums is a Cauchy sequence by Convergent Sequence is Cauchy Sequence.
Now let $\epsilon > 0$.
Let $N \in \N$ such that for all $m, n \in \N$, $m \ge n \ge N$ implies that:
:$\ds \sum_{k \mathop = n + 1}^m \norm {a_k} = \size {\sum_{k \mathop = 1}^m \norm {a_k} - \sum_{k \mathop = 1}^n \norm {a_k} } < \epsilon$
This $N$ exists because the sequence is Cauchy.
Now observe that, for $m \ge n \ge N$, one also has:
It follows that the sequence of partial sums of $\ds \sum_{n \mathop = 1}^\infty a_n$ is Cauchy.
As $V$ is a Banach space, this implies that $\ds \sum_{n \mathop = 1}^\infty a_n$ converges.
== Proof for Real Numbers ==
== Proof for Complex Numbers ==
== Also see ==
* Absolutely Convergent Generalized Sum Converges
== Sources ==
* : Chapter $1$: Normed and Banach spaces
Category:Absolute Convergence
Category:Banach Spaces
Category:Absolutely Convergent Series is Convergent
Absolutely Convergent Series is Convergent/Complex Numbers

== Theorem ==
Let $\ds \sum_{n \mathop = 1}^\infty z_n$ be an absolutely convergent series in $\C$.
Then $\ds \sum_{n \mathop = 1}^\infty z_n$ is convergent.
== Proof ==
Let $z_n = u_n + i v_n$.
We have that:
and similarly:
:$\cmod {z_n} > \size {v_n}$
From the Comparison Test, the series $\ds \sum_{n \mathop = 1}^\infty u_n$ and $\ds \sum_{n \mathop = 1}^\infty v_n$ are absolutely convergent.
From Absolutely Convergent Real Series is Convergent, $\ds \sum_{n \mathop = 1}^\infty u_n$ and $\ds \sum_{n \mathop = 1}^\infty v_n$ are convergent.
By Convergence of Series of Complex Numbers by Real and Imaginary Part, it follows that $\ds \sum_{n \mathop = 1}^\infty z_n$ is convergent.
== Sources ==
* : $\S 4.3$. Series: Theorem
Category:Absolutely Convergent Series is Convergent
Absolutely Convergent Series is Convergent/Real Numbers

== Theorem ==
Let $\ds \sum_{n \mathop = 1}^\infty a_n$ be an absolutely convergent series in $\R$.
Then $\ds \sum_{n \mathop = 1}^\infty a_n$ is convergent.
== Proof ==
By the definition of absolute convergence, we have that $\ds \sum_{n \mathop = 1}^\infty \size {a_n}$ is convergent.
From the Comparison Test we have that $\ds \sum_{n \mathop = 1}^\infty a_n$ converges :
:$\forall n \in \N_{>0}: \size {a_n} \le b_n$
where $\ds \sum_{n \mathop = 1}^\infty b_n$ is convergent.
So substituting $\size {a_n}$ for $b_n$ in the above, the result follows.
== Sources ==
* : $\S 6.21$
Category:Absolutely Convergent Series is Convergent
Absolutely Symmetric Function/Examples/Arbitrary Example 1

== Examples of Absolutely Symmetric Functions ==
Let $f: \R^3 \to \R$ be the real-valued function defined as:
:$\forall \tuple {x, y, z} \in \R^3: \map f {x, y, z} = x^2 + y^2 + 2 x y z$
Then $f$ is an absolutely symmetric function.
== Sources ==
* 
* 
Category:Examples of Absolutely Symmetric Functions
Absolutely Symmetric Function/Examples/Arbitrary Example 2

== Examples of Absolutely Symmetric Functions ==
Let $f: \R^2 \to \R$ be the real-valued function defined as:
:$\forall \tuple {x, y} \in \R^2: \map f {x, y} = x^2 + 2 x y + y^2$
Then $f$ is an absolutely symmetric function.
== Sources ==
* 
* 
Category:Examples of Absolutely Symmetric Functions
Absorbing Set in Vector Space contains Zero Vector

== Theorem ==
Let $\GF \in \set {\R, \C}$.
Let $X$ be a vector space over $\GF$. 
Let $A \subseteq X$ be an absorbing subset of $X$. 
Then:
:$\mathbf 0_X \in A$
where $\mathbf 0_X$ denotes the zero vector in $X$.
== Proof ==
From the definition of an absorbing subset, there exists $t \in \R_{>0}$ such that ${\mathbf 0}_X \in t A$.
So $\mathbf 0_X \in A$.
Category:Absorbing Sets
Absorption Laws (Boolean Algebras)

== Theorem ==
Let $\struct {S, \vee, \wedge}$ be a Boolean algebra, defined as in Definition 1.
Then:
That is, $\vee$ absorbs $\wedge$, and $\wedge$ absorbs $\vee$.
== Proof ==
Let $a, b \in S$.
Then:
as desired.
The result:
:$a = a \wedge \paren {a \vee b}$
follows from the Duality Principle.
== Sources ==
* : $\S 1.5$: Lemma $1.1$
* 
* 
Category:Boolean Algebras
Absorption Laws (Logic)/Conjunction Absorbs Disjunction

== Theorem ==
:$p \land \paren {p \lor q} \dashv \vdash p$
This can be expressed as two separate theorems:
=== Forward Implication ===
=== Reverse Implication ===
== Proof by Truth Table ==
== Proof 2 ==
== Sources ==
* : Chapter $1$: The Propositional Calculus $1$: $5$ Further Proofs: Résumé of Rules: Theorem $31$
* : $\S 2.10$: Exercise $2.5$
Category:Absorption Laws (Logic)
Absorption Laws (Logic)/Conjunction Absorbs Disjunction/Forward Implication

== Theorem ==
:$p \land \paren {p \lor q} \vdash p$
== Proof ==
== Sources ==
* : Chapter $1$: The Propositional Calculus $1$: $5$ Further Proofs: Résumé of Rules: Theorems $31 \ \text{(a)}$
Category:Absorption Laws (Logic)
Absorption Laws (Logic)/Conjunction Absorbs Disjunction/Proof by Truth Table

== Theorem ==
== Proof ==
We apply the Method of Truth Tables.
As can be seen by inspection, the appropriate truth values match for all boolean interpretations.
$\begin{array}{|ccccc||c|} \hline
p & \land & (p & \lor & q) & p \\
\hline
\F & \F & \F & \F & \F & \F \\
\F & \F & \F & \T & \T & \F \\
\T & \T & \T & \T & \F & \T \\
\T & \T & \T & \T & \T & \T \\
\hline
\end{array}$
Category:Truth Table Proofs
Category:Absorption Laws (Logic)
Absorption Laws (Logic)/Conjunction Absorbs Disjunction/Reverse Implication

== Theorem ==
:$p \vdash p \land \paren {p \lor q}$
== Proof ==
== Sources ==
* : Chapter $1$: The Propositional Calculus $1$: $5$ Further Proofs: Résumé of Rules: Theorems $31 \ \text{(b)}$
Category:Absorption Laws (Logic)
Absorption Laws (Logic)/Disjunction Absorbs Conjunction

== Theorem ==
:$p \lor \paren {p \land q} \dashv \vdash p$
This can be expressed as two separate theorems:
=== Forward Implication ===
=== Reverse Implication ===
== Proof by Truth Table ==
== Proof 2 ==
== Also see ==
* Absorption Laws (Set Theory)
* Absorption Laws (Lattice Theory)
== Sources ==
* : Chapter $1$: The Propositional Calculus $1$: $5$ Further Proofs: Résumé of Rules: Theorem $32$
* : $\S 2.10$: Exercise $2.5$
Category:Absorption Laws (Logic)
Category:Disjunction
Category:Conjunction
Absorption Laws (Logic)/Disjunction Absorbs Conjunction/Forward Implication

== Theorem ==
:$p \lor \paren {p \land q} \vdash p$
== Proof ==
== Sources ==
* : Chapter $1$: The Propositional Calculus $1$: $5$ Further Proofs: Résumé of Rules: Theorem $32 \ \text{(a)}$
* : $\S 1.2.1$: Rules for natural deduction: Exercises $1.4: \ 2 \ \text{(n)}$
Category:Absorption Laws (Logic)
Absorption Laws (Logic)/Disjunction Absorbs Conjunction/Proof by Truth Table

== Theorem ==
== Proof ==
We apply the Method of Truth Tables.
As can be seen by inspection, the appropriate truth values match for all boolean interpretations.
$\begin{array}{|ccccc||c|} \hline
p & \lor & (p & \land & q) & p \\
\hline
\F & \F & \F & \F & \F & \F \\
\F & \F & \F & \F & \T & \F \\
\T & \T & \T & \F & \F & \T \\
\T & \T & \T & \T & \T & \T \\
\hline
\end{array}$
Category:Truth Table Proofs
Category:Absorption Laws (Logic)
Absorption Laws (Logic)/Disjunction Absorbs Conjunction/Reverse Implication

== Theorem ==
:$p \vdash p \lor \paren {p \land q}$
== Proof ==
== Sources ==
* : Chapter $1$: The Propositional Calculus $1$: $5$ Further Proofs: Résumé of Rules: Theorem $32 \ \text{(b)}$
Category:Absorption Laws (Logic)
Absorption Laws (Set Theory)

== Theorem ==
These two results together are known as the absorption laws, corresponding to the equivalent results in logic.
=== Union Absorbs Intersection ===
=== Intersection Absorbs Union ===
=== Corollary ===
== Sources ==
* : $\S 1$. Sets; inclusion; intersection; union; complementation; number systems: Exercise $3$
* 
* 
* 
* 
Category:Absorption Laws
Category:Set Intersection
Category:Set Union
Absorption Laws (Set Theory)/Intersection Absorbs Union

== Theorem ==
:$S \cap \paren {S \cup T} = S$
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Union Absorbs Intersection, where it is proved that $S \cup \paren {S \cap T} = S$
These two results together are known as the Absorption Laws, corresponding to the equivalent results in logic.
== Sources ==
* : Chapter $1$. Sets: Exercise $3$
* : $\S 1$. Sets; inclusion; intersection; union; complementation; number systems: Exercise $3$
* 
* 
* 
* 
Category:Absorption Laws
Absorption Laws (Set Theory)/Union Absorbs Intersection

== Theorem ==
:$S \cup \paren {S \cap T} = S$
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Intersection Absorbs Union, where it is proved that $S \cap \paren {S \cup T} = S$
These two results together are known as the Absorption Laws, corresponding to the equivalent results in logic.
== Sources ==
* : Chapter $1$. Sets: Exercise $3$
* : $\S 1$. Sets; inclusion; intersection; union; complementation; number systems: Exercise $3$
* 
* 
* 
* 
Category:Absorption Laws
Abstract Model of Algebraic Extensions

== Theorem ==
Let $K$ be a field and $\alpha \in \overline{K}$ be an element of the algebraic closure of $K$ which is algebraic over $K$.
Let $m_\alpha$ be the minimal polynomial of $\alpha$ over $K$.
Then:
:$K[\alpha] \cong K[x]/ \langle m_\alpha\rangle$
== Proof ==
Let $\phi: K[x] \to K [\alpha]$ be the homomorphism which fixes $K$ and maps $x$ to $\alpha$.
The kernel of $\phi$ consists of all polynomials which vanish at $\alpha$, which is precisely the ideal $\langle m_\alpha\rangle$.
The result follows by the First Fundamental Theorem on Ring Homomorphisms.
Category:Field Extensions
Abundancy Index/Sequence

== Sequence ==
The sequence of abundancy indices for the first few positive integers is as follows.
The rational non-integer numbers have been rounded to $3$ decimal places.
:$\begin{array} {|r|r|l|}
\hline
n & \map {\sigma_1} n & \dfrac {\map {\sigma_1} n} n \\
\hline
1 & 1 & 1 \\
2 & 3 & 1 \cdotp 5 \\
3 & 4 & 1 \cdotp 333 \\
4 & 7 & 1 \cdotp 75 \\
5 & 6 & 1 \cdotp 2 \\
6 & 12 & 2 \\
7 & 8 & 1 \cdotp 143 \\
8 & 15 & 1 \cdotp 875 \\
9 & 13 & 1 \cdotp 444 \\
10 & 18 & 1 \cdotp 8 \\
11 & 12 & 1 \cdotp 091 \\
12 & 28 & 2 \cdotp 333 \\
13 & 14 & 1 \cdotp 077 \\
14 & 24 & 1 \cdotp 714 \\
15 & 24 & 1 \cdotp 6\\
16 & 31 & 1 \cdotp 938 \\
17 & 18 & 1 \cdotp 059 \\
18 & 39 & 2 \cdotp 167 \\
19 & 20 & 1 \cdotp 053 \\
20 & 42 & 2 \cdotp 1 \\
\hline
\end{array} \qquad
\begin{array} {|r|r|l|}
\hline
n & \map {\sigma_1} n & \dfrac {\map {\sigma_1} n} n \\
\hline
21 & 32 & 1 \cdotp 524 \\
22 & 36 & 1 \cdotp 636 \\
23 & 24 & 1 \cdotp 043 \\
24 & 60 & 2 \cdotp 5 \\
25 & 31 & 1 \cdotp 24 \\
26 & 42 & 1 \cdotp 615 \\
27 & 40 & 1 \cdotp 481 \\
28 & 56 & 2 \\
29 & 30 & 1 \cdotp 034 \\
30 & 72 & 2 \cdotp 4 \\
31 & 32 & 1 \cdotp 032 \\
32 & 63 & 1 \cdotp 969 \\
33 & 48 & 1 \cdotp 455 \\
34 & 54 & 1 \cdotp 588 \\
35 & 48 & 1 \cdotp 371 \\
36 & 91 & 2 \cdotp 528 \\
37 & 38 & 1 \cdotp 027 \\
38 & 60 & 1 \cdotp 679 \\
39 & 56 & 1 \cdotp 436 \\
40 & 90 & 2 \cdotp 25 \\
\hline
\end{array}$
Category:Abundancy
Abundancy Index of Product is greater than Abundancy Index of Proper Factors

== Theorem ==
Let $n \in \Z_{>0}$ be a composite number such that $n = r s$, where $r, s \in \Z_{>1}$.
Then:
:$\dfrac {\map {\sigma_1} n} n > \dfrac {\map {\sigma_1} r} r$
and consequently also:
:$\dfrac {\map {\sigma_1} n} n > \dfrac {\map {\sigma_1} s} s$
where $\sigma_1$ denotes the divisor sum function.
That is, the abundancy index of a composite number is strictly greater than the abundancy index of its proper divisors.
== Proof ==
Consider the divisors of $r$.
Let $d \divides r$, where $\divides$ indicates divisibility.
We have that:
:$d \divides n$
and also that:
:$d s \divides n$
Thus:
Similarly for $s$.
Category:Abundancy
Abundancy of Integers in form 945 + 630n

== Theorem ==
A large number of integers of the form $945 + 630 n$, for $n \in \Z_{\ge 0}$, are abundant.
The first counterexample is for $n = 52$.
== Proof ==
== Sources ==
* 
* : $33,705$
Category:Abundancy
Category:Abundant Numbers
Abundant Number/Examples/12

== Example of Abundant Number ==
The aliquot parts of $12$ are:
:$1, 2, 3, 4, 6$
We have that:
:$1 + 2 + 3 + 4 + 6 = 16 > 12$
demonstrating that $12$ is an abundant number.
== Proof ==
* 
* 
Category:Examples of Abundant Numbers
Abundant Number/Sequence

== Sequence of Abundant Numbers ==
The sequence of abundant numbers begins:
:$12, 18, 20, 24, 30, 36, 40, 42, 48, 54, 56, 60, 66, 70, 72, 78, 80, 84, 88, 90, 96, 100, 102, 104, 108, 112, 114, 120, \ldots$
== Sources ==
* : $12$
* : $12$
Category:Abundant Numbers
Acceleration Due to Gravity

== Physical Law ==
A body $B$ in a uniform gravitational field experiences a force which gives rise to a constant acceleration independent of the mass of the body.
If the force due to the gravitational field is the only force on $B$, it is said to be in free fall.
== Derivation ==
This law can be derived from Newton's Law of Universal Gravitation.
Let the mass of $B$ be $m$.
Let the mass of the body $P$ which gives rise to the gravitational field be $M$.
Then the force on $B$ is given by:
:$F = G \dfrac {M m} {r^2}$
where:
:$G$ is the universal gravitational constant
:$r$ is the distance between the centers of gravity of $B$ and $P$.
The assumption is that $M$ is orders of magnitude greater than $m$, and $r$ is also several orders of magnitude greater than the displacements observed on $B$ in the local frame.
Then:
:$F = m \dfrac {G M} {r^2}$
But from Newton's Second Law of Motion:
:$F = m a$
where $a$ is the magnitude of the acceleration which would be imparted to $B$ if no other force were acting on it.
Hence:
:$a = \dfrac {G M} {r^2}$
which is, as we said, independent of the mass of $B$.
== Gravity on Earth ==
== Historical Note ==
== Sources ==
* : $\text {I}$: Motion, Sound and Heat: Chapter $2$: Falling Bodies: Free Fall
Category:Mechanics
Category:Gravitation
Acceleration Due to Gravity/Earth's Surface

== Physical Law ==
Let a body $B$ be situated in a uniform gravitational field $M$ given rise to by a body $P$.
When the body $P$ of mass $M$ is the Earth, and the body $B$ of mass $m$ is located at or near its surface, it is usual to use $g$ for the local gravitational constant $\dfrac {G M} {r^2}$.
Therefore the force on $B$ can be expressed as:
:$F = m g$
=== Standard Gravity ===
== Historical Note ==
== Sources ==
* : $\text {I}$: Motion, Sound and Heat: Chapter $2$: Falling Bodies: Free Fall
* : $1$: The Nature of Differential Equations: $\S 1$: Introduction: $(2)$
Category:Gravity