nyuuzyou/wikis huggingface dataset:
Sign of Permutation on n Letters is Well-Defined

== Theorem ==
Let $n \in \N$ be a natural number.
Let $S_n$ denote the symmetric group on $n$ letters.
Let $\rho \in S_n$ be a permutation in $S_n$.
Let $\map \sgn \rho$ denote the sign of $\rho$.
Then $\map \sgn \rho$ is well-defined, in that it is either $1$ or $-1$. 
== Proof ==
We need to prove that for any permutation $\rho \in S_n$, $\rho$ cannot be expressed as the composite of both an even number and an odd number of transpositions.
Consider the permutation formed by composing $\rho$ with an arbitrary transposition $\tau$.
Let $\rho$ be expressed as the composite of disjoint cycles whose lengths are all greater than $1$.
By Disjoint Permutations Commute, the order in which the various cycles of $\rho$ are composed does not matter.
Let $\tau = \begin {bmatrix} a & b \end {bmatrix}$ for some $a, b \in \set {1, 2, \ldots, n}$ where $a \ne b$.
There are three cases:
$(1): \quad$ Neither $a$ nor $b$ appear in the expression for $\rho$.
That is, $\tau$ and $\rho$ are disjoint.
Then $\rho \circ \tau$ can be expressed as the same set of disjoint cycles as $\rho$, but with an extra cycle $\begin {bmatrix} a & b \end {bmatrix}$ appended.
$(2): \quad$ Just one of $a$ and $b$ occurs in the expression for $\rho$.
, let $a$ appear in the expression for $\rho$.
Let $a$ appear in the cycle $\rho_0$.
Then:
Thus composing $\rho$ with $\tau$ results in adding an extra element to one cycle and leaving the others as they are.
$(3): \quad$ Both $a$ and $b$ occur in the expression for $\rho$.
If $a$ and $b$ both occur in the same cycle $\rho_0$, the operation of composition goes like this:
If $a$ and $b$ appear in different cycles $\rho_1$ and $\rho_2$, we have:
Thus in case $(3)$, composition with $\tau$ results in the number of cycles either increasing or decreasing by $1$, while the total number of elements in those cycles stays the same.
For all $\rho \in S_n$, Let $\rho$ be expressed in cycle notation as a composite of $n$ cycles containing $m_1, m_2, \ldots, m_n$ elements respectively, where each $m_i \ge 2$.
Let the mapping $P: S_n \to \set {1, -1}$ be defined as follows:
:$\forall \rho \in S_n: \map P \rho = \paren {-1}^{m_1 - 1} \paren {-1}^{m_1 - 1} \cdots \paren {-1}^{m_n - 1}$
where $\map P {I_{S_n} } = 1$.
From the above, it can be seen that $\map P {\rho \circ \tau} = -\map P \rho$.
Let $\rho$ be expressible as the composite of $r$ transpositions.
By an inductive proof it can be shown that:
:$\map P \rho = \paren {-1}^r$.
But $\map P \rho$ is independent of the actual transpositions that are used to build $\rho$.
Thus $\map P \rho = 1$ for one such expression $\map P \rho = 1$ for all such expressions.
That is, $\rho$ cannot have an expression in cycle notation as the composite of an even number of transpositions and at the same time have an expression in cycle notation as the composite of an odd number of transpositions.
Hence the result.
== Sources ==
* : Appendix: Elementary set and number theory
Category:Sign of Permutation
Sign of Quadratic Function Between Roots

== Theorem ==
Let $a \in \R_{>0}$ be a (strictly) positive real number.
Let $\alpha$ and $\beta$, where $\alpha < \beta$, be the roots of the quadratic function:
:$\map Q x = a x^2 + b x + c$
whose discriminant $b^2 - 4 a c$ is (strictly) positive.
Then:
:$\begin {cases} \map Q x < 0 & : \text {when $\alpha < x < \beta$} \\ \map Q x > 0 & : \text {when $x < \alpha$ or $x > \beta$} \end {cases}$
== Proof ==
Because $b^2 - 4 a c > 0$, we have from Solution to Quadratic Equation with Real Coefficients that the roots of $\map Q x$ are real and unequal.
This demonstrates the existence of $\alpha$ and $\beta$, where by hypothesis we state that $\alpha < \beta$.
We can express $\map Q x$ as:
:$\map Q x = a \paren {x - \alpha} \paren {x - \beta}$
When $\alpha < x < \beta$ we have that:
:$x - \alpha > 0$
:$x - \beta < 0$
and so:
:$\map Q x = a \paren {x - \alpha} \paren {x - \beta} < 0$
When $x < \alpha$ we have that:
:$x - \alpha < 0$
:$x - \beta < 0$
and so:
:$\map Q x = a \paren {x - \alpha} \paren {x - \beta} > 0$
When $x > \beta$ we have that:
:$x - \alpha > 0$
:$x - \beta > 0$
and so:
:$\map Q x = a \paren {x - \alpha} \paren {x - \beta} > 0$
Hence the result.
== Sources ==
* : $\S 1$: Real Numbers: Exercise $\S 1.12 \ (4)$
Category:Quadratic Functions
Sign of Quotient of Factors of Difference of Squares

== Theorem ==
Let $a, b \in \R$ such that $a \ne b$.
Then
:$\map \sgn {a^2 - b^2} = \map \sgn {\dfrac {a + b} {a - b} } = \map \sgn {\dfrac {a - b} {a + b} }$
where $\sgn$ denotes the signum of a real number.
=== Corollary ===
== Proof ==
Category:Real Analysis
Category:Signum Function
Sign of Quotient of Factors of Difference of Squares/Corollary

== Corollary to Sign of Quotient of Factors of Difference of Squares ==
Let $a, b \in \R$ such that $a \ne b$.
Then
:$-\map \sgn {\dfrac {b - a} {b + a} } = \map \sgn {a^2 - b^2} = -\map \sgn {\dfrac {b + a} {b - a} }$
where $\sgn$ denotes the signum of a real number.
== Proof ==
Category:Real Analysis
Category:Signum Function
Sign of Sine

== Theorem ==
Let $x$ be a real number.
where $\sin$ is the real sine function.
== Proof ==
First the case where $n \ge 0$ is addressed.
The proof proceeds by induction.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$\forall x \in \R:$
::$2 n \pi < x < \paren {2 n + 1} \pi \implies \sin x > 0$
::$\paren {2 n + 1} \pi < x < \paren {2 n + 2} \pi \implies \sin x < 0$
=== Basis for the Induction ===
Let $n = 0$.
From the corollary to Sine and Cosine are Periodic on Reals:
:$\sin x$ is strictly positive on the interval $\openint 0 \pi$
and:
:$\sin x$ is strictly negative on the interval $\openint \pi {2 \pi}$
Thus:
:$2 \cdot 0 \cdot \pi < x < \paren {2 \cdot 0 + 1} \pi \implies \sin x > 0$
:$\paren {2 \cdot 0 \cdot + 1} \pi < x < \paren {2 \cdot 0 + 2} \pi \implies \sin x < 0$
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that if $\map P k$ is true, where $k \ge 0$, then it logically follows that $\map P {k + 1}$ is true.
So this is our induction hypothesis:
:$\forall x \in \R:$
::$2 k \pi < x < \paren {2 k + 1} \pi \implies \sin x > 0$
::$\paren {2 k + 1} \pi < x <\paren {2 k + 2} \pi \implies \sin x < 0$
Then we need to show:
:$\forall x \in \R:$
::$2 \paren {k + 1} \pi < x < \paren {2 \paren {k + 1} + 1} \pi \implies \sin x > 0$
::$\paren {2 \paren {k + 1} + 1} \pi < x < \paren {2 \paren {k + 1} + 2} \pi \implies \sin x < 0$
That is:
:$\forall x \in \R:$
::$\paren {2 k + 2} \pi < x < \paren {2 k + 3} \pi \implies \sin x > 0$
::$\paren {2 k + 3} \pi < x < \paren {2 k + 4} \pi \implies \sin x < 0$
=== Induction Step ===
This is our induction step:
Also:
It follows by induction that:
:$\forall n \in \Z_{\ge 0}: \forall x \in \R:$
::$2 n \pi < x < \paren {2 n + 1} \pi \implies \sin x > 0$
::$\paren {2 n + 1} \pi < x < \paren {2 n + 2} \pi \implies \sin x < 0$
=== Negative $n$ ===
Let $n \in \Z_{\le 0}$ be a negative integer.
Then, by definition, $-\paren {n + 1}$ is a (strictly) positive integer.
So:
Similarly:
== Also see ==
* Sign of Cosine
* Sign of Tangent
* Sign of Cotangent
* Sign of Secant
* Sign of Cosecant
Category:Sine Function
Sign of Tangent

== Theorem ==
Let $x$ be a real number.
Then:
where $\tan$ denotes the tangent function.
== Proof ==
From Tangent is Sine divided by Cosine:
:$\tan x = \dfrac {\sin x} {\cos x}$
Since $n$ is an integer, $n$ is either odd or even.
=== Case 1 ===
Let $n$ be odd.
Hence let $m$ be an integer such that $n = 2 m + 1$.
==== Case 1.1 ====
:$n \pi < x < \paren {n + \dfrac 1 2} \pi \implies \paren {2 m + 1} \pi < x < \paren {2 m + \dfrac 3 2} \pi$
From Sign of Sine, $\sin x$ is negative.
From Sign of Cosine, $\cos x$ is negative.
Therefore:
:$\tan x = \dfrac {\sin x} {\cos x} > 0$
==== Case 1.2 ====
:$\paren {n + \dfrac 1 2} \pi < x < \paren {n + 1} \pi \implies \paren {2 m + \dfrac 3 2} \pi < x < \paren {2 m + 2} \pi$
From Sign of Sine, $\sin x$ is negative.
From Sign of Cosine, $\cos x$ is positive.
Therefore:
:$\tan x = \dfrac {\sin x} {\cos x} < 0$
=== Case 2 ===
Let $n$ be even.
Hence let $m$ be an integer such that $n = 2 m$.
==== Case 2.1 ====
:$n \pi < x < \paren {n + \dfrac 1 2} \pi \implies 2 m \pi < x < \paren {2 m + \dfrac 1 2} \pi$
From Sign of Sine, $\sin x$ is positive.
From Sign of Cosine, $\cos x$ is positive.
Therefore:
:$\tan x = \dfrac {\sin x} {\cos x} > 0$
==== Case 2.2 ====
:$\paren {n + \dfrac 1 2} \pi < x < \paren {n + 1} \pi \implies \paren {2 m + \dfrac 1 2} \pi < x < \paren {2 m + 1} \pi$
From Sign of Sine, $\sin x$ is positive.
From Sign of Cosine, $\cos x$ is negative.
Therefore:
:$\tan x = \dfrac {\sin x} {\cos x} < 0$
Therefore:
== Also see ==
* Sign of Sine
* Sign of Cosine
* Sign of Cotangent
* Sign of Secant
* Sign of Cosecant
Category:Tangent Function
Sign of Trigonometric Functions/Mnemonic

== Mnemonic ==
It has been suggested that it is difficult to remember which of the three main trigonometric functions (sine, cosine and tangent) are positive and which are negative in which quadrants of the Cartesian plane.
Hence the invention of mnemonics to aid their correct identification.
:
It is noted that:
:in the first quadrant, all of sine, cosine and tangent) are positive.
:in the second quadrant, of these only the sine function is positive.
:in the third quadrant, of these only the tangent function is positive.
:in the fourth quadrant, of these only the cosine function is positive.
Hence, by placing each of the initial letters of $\text {Cosine}$, $\text {All}$, $\text {Sine}$ and $\text {Tangent}$ in each of the four quadrants, starting in the second quadrant and moving anticlockwise around the origin, the word $\text {CAST}$ is spelled.
== Sources ==
* : $\text V$. Trigonometry: Angles larger than $90 \degrees$
Category:Trigonometric Functions
Signature of Embedded Pseudo-Riemannian Submanifold

== Theorem ==
Let $\struct {M, g}$ is a pseudo-Riemannian manifold of signature $\tuple {r, s}$.
Let $f \in \map {C^\infty} {\tilde M}$ be a smooth mapping.
Let $M = \map {f^{-1}} c$ for some $c \in \R$.
Suppose:
:$\forall p \in M : \map {\tilde g_p} {\grad f, \grad f} > 0$
Then $M$ is an embedded pseudo-Riemannian submanifold of $\tilde M$ of signature $\tuple {r - 1, s}$.
Otherwise suppose: 
:$\forall p \in M : \map {\tilde g_p} {\grad f, \grad f} < 0$ 
Then $M$ is an embedded pseudo-Riemannian submanifold of $\tilde M$ of signature $\tuple {r, s - 1}$. 
Furthermore, in either case, $\grad f$ is everywhere normal to $M$.
== Proof ==
== Sources ==
* : $\S 2$: Riemannian Metrics. Pseudo-Riemannian Metrics
Category:Pseudo-Riemannian Geometry
Signature of Pseudo-Riemannian Submanifold

== Theorem ==
Let $\struct {\tilde M, \tilde g}$ be a pseudo-Riemannian manifold of signature $\tuple {r, s}$.
Let $M$ be a smooth hypersurface in $\tilde M$.
Let $N_p M$ be the normal space at $p \in M$.
Suppose:
:$\forall p \in M : \forall v \in N_p M : \map {\tilde g} {v, v} > 0$
Then $M$ is a pseudo-Riemannian submanifold of signature $\tuple {r - 1; s}$.
Otherwise, suppose:
:$\forall p \in M : \forall v \in N_p M : \map {\tilde g} {v, v} < 0$
Then $M$ is a pseudo-Riemannian submanifold of signature $\tuple {r; s - 1}$.
== Proof ==
== Sources ==
* : $\S 2$: Riemannian Metrics. Pseudo-Riemannian Metrics
Category:Definitions/Pseudo-Riemannian Geometry
Signed Measure Finite iff Finite Total Variation

== Theorem ==
Let $\struct {X, \Sigma}$ be a measurable space.
Let $\mu$ be a signed measure on $\struct {X, \Sigma}$. 
Let $\size \mu$ be the variation of $\mu$. 
Then $\mu$ is finite :
:$\map {\size \mu} X < \infty$
== Proof ==
=== Sufficient Condition ===
Suppose that: 
:$\map {\size \mu} X < \infty$
Then, from Absolute Value of Signed Measure Bounded Above by Variation, we have: 
:$\size {\map \mu X} \le \map {\size \mu} X$
so:
:$\size {\map \mu X} < \infty$
So $\mu$ is finite.
=== Necessary Condition ===
Suppose that $\mu$ is finite.
Let $\tuple {\mu^+, \mu^-}$ be the Jordan decomposition of $\mu$.
From Jordan Decomposition of Finite Signed Measure, $\mu^+$ and $\mu^-$ are finite measures.
So:
:$\map {\mu^+} X < \infty$ and $\map {\mu^-} X < \infty$.
Then, we have: 
Category:Signed Measures
Category:Total Variation of Signed Measure
Signed Measure may not be Monotone

== Theorem ==
Let $\struct {X, \Sigma}$ be measurable space. 
Let $\mu$ be a signed measure on $\struct {X, \Sigma}$. 
Then $\mu$ may not be monotone.
== Proof ==
Let:
:$\struct {X, \Sigma} = \struct {\R, \map \BB \R}$
where $\map \BB \R$ is the Borel $\sigma$-algebra on $\R$.
Define: 
:$\mu = \delta_1 - 2 \delta_2$
where $\delta_1$ and $\delta_2$ are the Dirac measures at $1$ and $2$ respectively.
Since $\delta_1$ and $\delta_2$ are both finite measures, we have: 
:$\mu$ is a signed measure
from Linear Combination of Signed Measures is Signed Measure.
Then, we have: 
:$\closedint 0 1 \subseteq \closedint 0 2$
with: 
and:
So:
:$\closedint 0 1 \subseteq \closedint 0 2$ and $\map \mu {\closedint 0 2} \le \map \mu {\closedint 0 1}$
So:
:$\mu$ is not monotone.
Category:Signed Measures
Signed Measure of Limit of Decreasing Sequence of Measurable Sets

== Theorem ==
Let $\struct {X, \Sigma}$ be a measurable space. 
Let $E \in \Sigma$. 
Let $\sequence {E_n}_{n \mathop \in \N}$ be an decreasing sequence of $\Sigma$-measurable sets such that: 
:$E_n \downarrow E$ 
where $E_n \downarrow E$ denotes limit of decreasing sequence of sets, and:
:there exists $m \in \N$ such that $\map \mu {E_m}$ is finite.
Then: 
:$\ds \map \mu E = \lim_{n \mathop \to \infty} \map \mu {E_n}$
== Proof ==
Let $\tuple {\mu^+, \mu^-}$ be the Jordan decomposition of $\mu$.
Then $\mu^+$ and $\mu^-$ are measures with: 
:$\mu = \mu^+ - \mu^-$
where at least one of $\mu^+$ and $\mu^-$ is finite.
Then we have: 
:$\map \mu E = \map {\mu^+} E - \map {\mu^-} E$
and: 
:$\map \mu {E_m} = \map {\mu^+} {E_m} - \map {\mu^-} {E_m}$
Since $\map \mu {E_m}$ is finite, we have that: 
:$\map {\mu^+} {E_m}$ and $\map {\mu^-} {E_m}$ are finite.
From , we have: 
:$\ds \map {\mu^+} E = \lim_{n \mathop \to \infty} \map {\mu^+} {E_n}$
and:
:$\ds \map {\mu^-} E = \lim_{n \mathop \to \infty} \map {\mu^-} {E_n}$
with at most one of these limits infinite.
So:
== Also see ==
* Measure of Limit of Decreasing Sequence of Measurable Sets
Category:Signed Measures
Signed Measure of Limit of Increasing Sequence of Measurable Sets

== Theorem ==
Let $\struct {X, \Sigma}$ be a measurable space.
Let $\mu$ be a signed measure on $\struct {X, \Sigma}$. 
Let $E \in \Sigma$. 
Let $\sequence {E_n}_{n \mathop \in \N}$ be an increasing sequence of $\Sigma$-measurable sets such that: 
:$E_n \uparrow E$ 
where $E_n \uparrow E$ denotes the limit of an increasing sequence of sets.
Then: 
:$\ds \map \mu E = \lim_{n \mathop \to \infty} \map \mu {E_n}$
== Proof ==
Let $\tuple {\mu^+, \mu^-}$ be the Jordan decomposition of $\mu$.
Then $\mu^+$ and $\mu^-$ are measures with: 
:$\mu = \mu^+ - \mu^-$
where at least one of $\mu^+$ and $\mu^-$ is finite.
Then we have: 
:$\map \mu E = \map {\mu^+} E - \map {\mu^-} E$
From Measure of Limit of Increasing Sequence of Measurable Sets, we have: 
:$\ds \map {\mu^+} E = \lim_{n \mathop \to \infty} \map {\mu^+} {E_n}$
and:
:$\ds \map {\mu^-} E = \lim_{n \mathop \to \infty} \map {\mu^-} {E_n}$
with at most one of these limits infinite.
So:
Category:Signed Measures
Category:Increasing Sequences of Sets
Signed Stirling Number of the First Kind of 0

== Theorem ==
:$\map s {0, n} = \delta_{0 n}$
where:
:$\map s {0, n}$ denotes a signed Stirling number of the first kind
:$\delta_{0 n}$ denotes the Kronecker delta.
== Proof ==
By definition of signed Stirling number of the first kind:
$\ds x^{\underline 0} = \sum_k \map s {0, k} x^k$
Thus we have:
Thus, in the expression:
:$\ds x^{\underline 0} = \sum_k \map s {0, k} x^k$
we have:
:$\map s {0, 0} = 1$
and for all $k \in \Z_{>0}$:
:$\map s {0, k} = 0$
That is:
:$\map s {0, k} = \delta_{0 k}$
== Also see ==
* Unsigned Stirling Number of the First Kind of 0
* Stirling Number of the Second Kind of 0
* Particular Values of Signed Stirling Numbers of the First Kind
Category:Examples of Stirling Numbers of the First Kind
Signed Stirling Number of the First Kind of 1

== Theorem ==
:$\map s {1, n} = \delta_{1 n}$
where:
:$\map s {1, n}$ denotes a signed Stirling number of the first kind
:$\delta_{1 n}$ is the Kronecker delta.
== Proof ==
== Also see ==
* Unsigned Stirling Number of the First Kind of 1
* Stirling Number of the Second Kind of 1
* Particular Values of Signed Stirling Numbers of the First Kind
Category:Examples of Stirling Numbers of the First Kind
Signed Stirling Number of the First Kind of Number with Greater

== Theorem ==
Let $n, k \in \Z_{\ge 0}$
Let $k > n$.
Let $\map s {n, k}$ denote a signed Stirling number of the first kind.
Then:
:$\map s {n, k} = 0$
== Proof ==
By definition, the signed Stirling numbers of the first kind are defined as the polynomial coefficients $\map s {n, k}$ which satisfy the equation:
:$\ds x^{\underline n} = \sum_k \map s {n, k} x^k$
where $x^{\underline n}$ denotes the $n$th falling factorial of $x$.
Both of the expressions on the and are polynomials in $x$ of degree $n$.
Hence the coefficient $\map s {n, k}$ of $x^k$ where $k > n$ is $0$.
== Also see ==
* Unsigned Stirling Number of the First Kind of Number with Greater
* Stirling Number of the Second Kind of Number with Greater
Category:Stirling Numbers
Signed Stirling Number of the First Kind of Number with Self

== Theorem ==
:$\map s {n, n} = 1$
where $\map s {n, n}$ denotes a signed Stirling number of the first kind.
== Proof ==
From Relation between Signed and Unsigned Stirling Numbers of the First Kind:
:$\ds {n \brack n} = \paren {-1}^{n + n} \map s {n, n}$
We have that:
:$\paren {-1}^{n + n} = \paren {-1}^{2 n} = 1$
and so:
:$\ds {n \brack n} = \map s {n, n}$
The result follows from Unsigned Stirling Number of the First Kind of Number with Self.
== Also see ==
* Unsigned Stirling Number of the First Kind of Number with Self
* Stirling Number of the Second Kind of Number with Self
* Particular Values of Signed Stirling Numbers of the First Kind
Category:Examples of Stirling Numbers of the First Kind
Signed Stirling Number of the First Kind of n+1 with 0

== Theorem ==
Let $n \in \Z_{\ge 0}$.
Then:
:$\map s {n + 1, 0} = 0$
where $\map s {n + 1, 0}$ denotes a signed Stirling number of the first kind.
== Proof ==
By definition of signed Stirling number of the first kind:
:$\map s {n, k} = \delta_{n k}$
where $\delta_{n k}$ is the Kronecker delta.
Thus
Hence the result.
== Also see ==
* Unsigned Stirling Number of the First Kind of n+1 with 0
* Stirling Number of the Second Kind of n+1 with 0
* Particular Values of Signed Stirling Numbers of the First Kind
Category:Examples of Stirling Numbers of the First Kind
Signed Stirling Number of the First Kind of n+1 with 1

== Theorem ==
Let $n \in \Z_{\ge 0}$.
Then:
:$\map s {n + 1, 1} = \paren {-1}^n n!$
where $\map s {n + 1, 1}$ denotes a signed Stirling number of the first kind.
== Proof ==
By Relation between Signed and Unsigned Stirling Numbers of the First Kind:
:$\ds {n + 1 \brack 1} = \paren {-1}^{n + 1 + 1} \map s {n + 1, 1}$
where $\ds {n + 1 \brack 1}$ denotes an unsigned Stirling number of the first kind.
We have that:
:$\paren {-1}^{n + 1 + 1} = \paren {-1}^n$
and so:
:$\ds {n + 1 \brack 1} = \paren {-1}^n \map s {n + 1, 1}$
The result follows from Unsigned Stirling Number of the First Kind of Number with Self.
== Also see ==
* Unsigned Stirling Number of the First Kind of n+1 with 1
* Stirling Number of the Second Kind of n+1 with 1
* Particular Values of Signed Stirling Numbers of the First Kind
Category:Examples of Stirling Numbers of the First Kind
Signed Stirling Number of the First Kind of n with n-1

== Theorem ==
Let $n \in \Z_{> 0}$ be an integer greater than $0$.
Then:
:$\map s {n, n - 1} = -\dbinom n 2$
where:
:$\map s {n, k}$ denotes a signed Stirling number of the first kind
:$\dbinom n 2$ denotes a binomial coefficient.
== Proof ==
From Relation between Signed and Unsigned Stirling Numbers of the First Kind:
:$\ds {n \brack n - 1} = \paren {-1}^{n + n - 1} \map s {n, n - 1}$
where $\ds {n \brack n - 1}$ denotes an unsigned Stirling number of the first kind.
We have that:
:$\paren {-1}^{n + n - 1} = \paren {-1}^{2 n - 1} = -1$
and so:
:$\ds {n \brack n} = -\map s {n, n}$
The result follows from Unsigned Stirling Number of the First Kind of Number with Self.
== Also see ==
* Unsigned Stirling Number of the First Kind of n with n-1
* Stirling Number of the Second Kind of n with n-1
* Particular Values of Signed Stirling Numbers of the First Kind
Category:Examples of Stirling Numbers of the First Kind
Significant Figures/Ambiguous Presentation/Examples/186,000,000

== Example of Ambiguous Presentation of Significant Figures ==
Consider the number $n$ presented as:
:$n = 186 \, 000 \, 000$
It is impossible to tell exactly how many significant figures $n$ is presented to.
This could be any number from $3$ to $9$.
Suppose $n$ is presented to $5$ significant figures.
Then $n$ can be written in scientific notation as:
:$n = 1 \cdotp 8600 \times 10^9$
and the matter is immediately clear.
If desired, $n$ can also be presented as:
:$n = 168 \cdotp 00$ million
== Sources ==
* : Chapter $1$: Significant Figures
Category:Examples of Significant Figures
Significant Figures/Examples

== Examples of Significant Figures ==
=== Significant Figures of $64 \cdotp 4$ ===
=== Significant Figures of $4 \cdotp 5300$ ===
=== Significant Figures of $0 \cdotp 0018$ ===
=== Significant Figures of $0 \cdotp 001800$ ===
=== Significant Figures of $149 \cdotp 8 \ \mathrm {mm}$ ===
=== Significant Figures of $149 \cdotp 80 \ \mathrm {mm}$ ===
=== Significant Figures of $0 \cdotp 0028 \ \mathrm m$ ===
=== Significant Figures of $0 \cdotp 002 \, 80 \ \mathrm m$ ===
=== Significant Figures of $1 \cdotp 002 \, 80 \ \mathrm m$ ===
=== Significant Figures of $9 \ \mathrm g$ ===
=== Significant Figures of $9$ houses ===
=== Significant Figures of $4 \cdotp 0 \times 10^3 \ \mathrm g$ ===
=== Significant Figures of $7 \cdotp 584 \, 00 \times 10^{-5} \ \mathrm N$ ===
=== Significant Figures of $12048$ ===
=== $12345$ to $3$ Significant Figures ===
=== Populations of Cities ===
Category:Examples of Significant Figures
Significant Figures/Examples/0.001800

== Example of Significant Figures ==
:$0 \cdotp 001800$ has $4$ significant figures.
Beware: It may appear that there are only $2$ significant figures.
However, although $2$ of the digits given are non-zero, the final $2$ zero digits are significant, because they are specifically stated.
== Sources ==
* : Chapter $1$: Significant Figures: Example 4.
Category:Examples of Significant Figures
Significant Figures/Examples/0.00280 m

== Example of Significant Figures ==
:$0 \cdotp 002 \, 80 \ \mathrm m$ has $3$ significant figures.
== Sources ==
* : Chapter $1$: Variables and Graphs: Solved Problems: Scientific Notation and Significant Figures: $1.6 \ \text {(d)}$
Category:Examples of Significant Figures
Significant Figures/Examples/0.0028 m

== Example of Significant Figures ==
:$0 \cdotp 0028 \ \mathrm m$ has $2$ significant figures.
== Sources ==
* : Chapter $1$: Variables and Graphs: Solved Problems: Scientific Notation and Significant Figures: $1.6 \ \text {(c)}$
Category:Examples of Significant Figures
Significant Figures/Examples/1.00280 m

== Example of Significant Figures ==
:$1 \cdotp 002 \, 80 \ \mathrm m$ has $6$ significant figures.
== Sources ==
* : Chapter $1$: Variables and Graphs: Solved Problems: Scientific Notation and Significant Figures: $1.6 \ \text {(e)}$
Category:Examples of Significant Figures
Significant Figures/Examples/12048

== Example of Significant Figures ==
The numbers:
:$1 \cdotp 2048$
:$1 \cdotp 2040$
:$0 \cdotp 0120 \, 48$
:$0 \cdotp 001 \, 2040$
:$1204 \cdotp 0$
are all reported to $5$ significant figures.
== Sources ==
* 
Category:Examples of Significant Figures
Significant Figures/Examples/12345

== Example of Significant Figures ==
The numbers:
:$1234 \cdotp 5$
:$0 \cdotp 012 \, 345$
reported to $3$ significant figures become:
:$1230$
:$0 \cdotp 0123$
respectively.
== Sources ==
* 
* 
Category:Examples of Significant Figures