nyuuzyou/wikis huggingface dataset:
Steinitz's Theorem

== Theorem ==
Let $E / F$ be a finite field extension.
:$(1): \quad E / F$ is simple: there exists $\alpha \in E$ such that $E = \map F \alpha$
:$(2): \quad$ there are only finitely many intermediate fields between $E$ and $F$. 
== Proof ==
=== $1$ implies $2$ ===
Suppose $E = F \sqbrk \alpha$ for some $\alpha \in F$.
Let $M$ be any intermediate field between $E$ and $F$.
Let $g$ be the minimal polynomial of $\alpha$ over $M$. Write
:$\ds g = x^ d + \sum _{i \mathop = 0}^{d \mathop - 1} a_ i x^ i$
Let $M'$ be the field extension of $F$ generated by all the coefficients of $g$.
:$M' = F \paren {a_0, \dots, a_{d-1}}$
Then $M' \subseteq M$ by definition of the minimal polynomial over $M$.
Since $g \in M' \sqbrk x$,
$g$ is the minimal polynomial of $\alpha$ over $M'$,
then the degree of $\alpha$ over $M'$ is $d$, but $E = M' \sqbrk \alpha$, so
:$\index E {M'} = d$
similarly, we have $\index E M = d$,
we conclude by Tower Law,
:$\index M {M'} = 1$
hence
:$M = M' = F \paren {a_0, \dots, a_{d-1}}$
Thus it suffices to show there are at most finitely many possibilities for the polynomial $g$. 
Let $f$ be the minimal polynomial of $\alpha$ over $F$,
since $f \in M \sqbrk x$, and $\map f \alpha = 0$,
by definition of $g$,
:$g \divides f$
By unique factorization in $E \sqbrk x$,
the polynomial $f$ has only finitely many divisors in $E \sqbrk x$,
so there are only finitely many intermediate fields between $E$ and $F$.
=== $2$ implies $1$ ===
Suppose there are only finitely many intermediate fields between $E$ and $F$.
If $F$ is a finite field (equivalently $E$ is a finite field), this follows from Finite Extension of $\F_p$ is Generated By a Single Element, since the generator of $E / \F_p$ also generates $E / F$.
If $F$ is infinite, then each intermediate field between $E$ and $F$ is a proper $F$-subspace of $E$,
by Vector Space over an Infinite Field is not equal to the Union of Proper Subspaces their union can't be all of $E$.
Thus any element outside this union must generate $E$, since it is not contained in any proper subfield of $E$.
== Also see ==
* Primitive Element Theorem
== Sources ==
* 
* : Chapter $1 Galois Theory: $\S 5$ The Fundamental Theorem of Galois Theory: Proposition $5.6$
* : Chapter $\text V}: $\S 4$ Separable Extensions: Theorem $4.6$
Category:Field Extensions
Stem-and-Leaf Display/Examples/Arbitrary Example 1

== Example of Stem-and-Leaf Display ==
Consider the set of data:
:$\set {10, 27, 19, 11, 14, 41, 38, 59, 7, 21}$
We consider:
:the tens digits as the stems
:the units digits as the leaves.
Then we arrange the data into a stem-and-leaf display thus:
:$\begin {array} {l|l}
\text {Stem} & \text {Leaf} \\
\hline
0 & 7 \\
1 & 0 \ 1 \ 4 \ 9 \\
2 & 1 \ 7 \\
3 & 8 \\
4 & 1 \\
5 & 9
\end {array}$
Hence the data have been completely ordered and arranged into what is effectively a histogram on its side.
Writing a five-number summary is subsequently easy to do.
== Sources ==
* 
* 
Category:Examples of Stem-and-Leaf Displays
Step Function satisfies Dirichlet Conditions

== Theorem ==
Let $\alpha, \beta \in \R$ be a real numbers such that $\alpha < \beta$.
Let $\map f x$ be a step function defined on the interval $\openint \alpha \beta$.
Then $f$ satisfies the Dirichlet conditions.
== Proof ==
Recall the definition of step function:
:
Recall the Dirichlet conditions:
We inspect the Dirichlet conditions in turn.
=== $(\text D 1)$: Absolute Integrability ===
From Definite Integral of Step Function:
:$\ds \int_\alpha^\beta \map f x \rd x = \sum_{k \mathop = 1}^n \lambda_k \paren {\beta_k - \alpha_k}$
where $\alpha_k, \beta_k$ are the endpoints of $\mathbb I_k$ for $1 \le k \le n$.
When $\lambda_k < 1$, we have that:
:$\size {\lambda_k} = -\lambda_k$
and so:
:$\ds \int_{\alpha_k}^{\beta_k} \lambda_k \rd x = -\lambda_k \paren {\beta_k - \alpha_k}$
Absolute integrability follows.
=== $(\text D 2)$: Finite Number of Local Maxima and Minima ===
By nature of $f$ being a step function, the image set of $f$ is finite:
:$f \openint \alpha \beta = \set {\lambda_1, \lambda_2, \ldots, \lambda n}$
The set of local maxima and the set of local minima must be subsets of $f \openint \alpha \beta$.
Therefore there must be a finite number of each.
=== $(\text D 3)$: Discontinuities are Finite in Number and Nature ===
From Constant Real Function is Continuous, there are no discontinuities except perhaps at the endpoints of $\mathbb I_k$ for $1 \le k \le n$.
Let us inspect those endpoints.
Let $\alpha_k, \beta_k$ be the endpoints of $\mathbb I_k$ for $1 \le k \le n$, where $\alpha_k < \beta_k$.
Let $\mathbb I_1, \mathbb I_2, \ldots, \mathbb I_n$ be ordered such that:
:$\forall k: \alpha_k < \alpha_{k + 1}$
By definition, $\beta_{k - 1} = \alpha_k$.
Thus we have that:
and if $\lambda_k \ne \lambda_{k - 1}$ then this is indeed a finite discontinuity.
By the nature of the step function, there are a finite number of them.
Thus all Dirichlet conditions are satisfied by $f$.
== Sources ==
* : Chapter One: $\S 2$. Fourier Series
Category:Step Functions
Category:Dirichlet Conditions
Stewart's Theorem

== Theorem ==
Let $\triangle ABC$ be a triangle with sides $a, b, c$.
Let $CP$ be a cevian from $C$ to $P$.
:
Then:
:$a^2 \cdot AP + b^2 \cdot PB = c \paren {CP^2 + AP \cdot PB}$
== Proof ==
It is also known as Apollonius's Theorem after .
Category:Triangles
Stieltjes Function of Measure is Stieltjes Function

== Theorem ==
Let $\mu$ be a measure on $\R$ with the Borel $\sigma$-algebra $\map \BB \R$.
Suppose that for every $n \in \N$:
:$\map \mu {\closedint {-n} n} < +\infty$
Then $F_\mu: \R \to \overline \R$, the Stieltjes function of $\mu$, is a Stieltjes function.
== Proof ==
By definition, $F_\mu$ is a Stieltjes function it is increasing and left-continuous.
=== $F_\mu$ is Increasing ===
Let $x, y \in \R$ such that $x \le y$.
It is apparent that for all $z \in \R$:
:$z \le 0 \implies \map {F_\mu} z \le 0$
:$z \ge 0 \implies \map {F_\mu} z \ge 0$
Therefore, only the cases $x \le y \le 0$ and $0 \le x \le y$ remain.
In the first of these cases, note that:
:$\hointr y 0 \subseteq \hointr x 0$
Hence from Measure is Monotone, it follows that:
:$\map {F_\mu} x = - \map \mu {\hointr x 0} \le - \map \mu {\hointr y 0} = \map {F_\mu} y$
The remaining case is analogous, using the observation that:
:$\hointr 0 x \subseteq \hointr 0 y$
Hence $F_\mu$ is increasing.
=== $F_\mu$ is Left-Continuous ===
Suppose that $x > 0$.
Suppose further that $x_k \uparrow x$ is an increasing sequence with limit $x$, and that $0 < x_1 < x$.
Then as an increasing sequence of sets, we have:
:$\hointr 0 {x_k} \uparrow \hointr 0 x$
Now we compute:
Next, suppose $x = 0$.
Suppose further that $x_k \uparrow 0$ is an increasing sequence with limit $0$.
Now, we have the decreasing sequence of sets $\hointr {x_k} 0 \downarrow \O$.
Reasoning as above, but applying Characterization of Measures: $(3)$ instead yields the result in this case.
Finally, suppose $x < 0$ and $x_k \uparrow x$ is an increasing sequence with limit $x$.
Observe the decreasing sequence of sets $\hointr {x_k} 0 \downarrow \hointr x 0$.
This time, applying Characterization of Measures: $(3')$ yields the result.
Note the additional assumption on $\mu$ was in fact used, in combination with Measure is Monotone.
This is necessary to establish the conditions for the last two applications of Characterization of Measures.
From the Trichotomy Law for Real Numbers and Proof by Cases, the result now follows.
== Sources ==
* : $\S 7$: Problem $9 \ \text{(i)}$
Category:Stieltjes Functions
Category:Measure Theory
Stieltjes Function of Measure of Finite Stieltjes Function

== Theorem ==
Let $f: \R \to \R$ be a finite Stieltjes function.
Let $\mu_f$ be the measure of $f$.
Let $f_{\mu_f}$ be the Stieltjes function of $\mu_f$.
Then $f_{\mu_f} = f$.
== Proof ==
Category:Measure Theory
Stirling's Formula

== Theorem ==
The factorial function can be approximated by the formula:
:$n! \sim \sqrt {2 \pi n} \paren {\dfrac n e}^n$
where $\sim$ denotes asymptotically equal.
=== Refinement ===
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Also defined as ==
== Also known as ==
== Examples ==
== Also see ==
* Limit of Error in Stirling's Formula
* Stirling's Formula for Gamma Function
== Sources ==
* : Chapter $1$: The Laplace Transform: Some Special Functions: $\text {I}$. The Gamma function: $4$
* : $\S 16$: Asymptotic Expansions for the Gamma Function: $16.16$
* : $1.1$ What is asymptotics? $(1.1.1)$
* : $2 \cdotp 506 \, 628 \ldots$
* : $24$
* : $\S 1.2.5$: Permutations and Factorials: $(7)$
* : $2 \cdotp 50662 \, 8 \ldots$
* : $24$
* 
* 
* : $\S 25$: The Gamma Function: Asymptotic Expansions for the Gamma Function: $25.16.$
Category:Stirling's Formula
Category:Factorials
Category:Asymptotics
Category:Special Functions
Stirling's Formula/Also defined as

== Stirling's Formula: Also defined as == 
Stirling's formula can also be seen reported variously as:
Other variants are sometimes encountered, for example:
:$\ds \lim_{n \mathop \to \infty} \dfrac {n!} {e^{-n} n^n \sqrt {2 \pi n} } = 1$
== Sources ==
* : Chapter $1$: The Laplace Transform: Some Special Functions: $\text {I}$. The Gamma function: $4$
* : $\S 16$: Asymptotic Expansions for the Gamma Function: $16.16$
* : $1.1$ What is asymptotics? $(1.1.1)$
* : $\S 17.2$
* : $24$
* : $24$
* : $\S 25$: The Gamma Function: Asymptotic Expansions for the Gamma Function: $25.16.$
Category:Stirling's Formula
Stirling's Formula/Also known as

== Stirling's Formula: Also known as == 
Stirling's formula is otherwise known as Stirling's approximation.
Some refer to it as Stirling's asymptotic formula.
== Sources ==
* : $2 \cdotp 506 \, 628 \ldots$
* : $\S 1.2.5$: Permutations and Factorials: $(7)$
* : $2 \cdotp 50662 \, 8 \ldots$
Category:Stirling's Formula
Stirling's Formula/Examples/8

== Example of Use of Stirling's Formula ==
The factorial of $8$ is given by Stirling's Formula as:
:$8! \approx 39 \ 902$
which shows an error of about $1 \%$.
== Proof ==
We have that by the usual calculation:
:$8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 40 \ 320$
Hence:
== Also see == 
* Stirling's Formula/Refinement/Examples/8
* Stirling's Formula for Gamma Function/Examples/First 10
== Sources ==
* : $\S 1.2.5$: Permutations and Factorials
Category:Stirling's Formula
Stirling's Formula/Proof 1

== Theorem ==
== Proof ==
Let $a_n = \dfrac {n!} {\sqrt {2 n} \paren {\frac n e}^n}$.
=== Part 1 ===
It will be shown that:
:$\ds \lim_{n \mathop \to \infty} a_n = a$
for some constant $a$.
This will imply that:
:$\ds \lim_{n \mathop \to \infty} \frac {n!} {a \sqrt{2 n} \paren {\frac n e}^n} = 1$
By applying Power Series Expansion for $\map \ln {1 + x}$:
Let $b_n = \ln a_n$.
so the sequence $\sequence {b_n}$ is (strictly) decreasing.
From $(1)$:
Hence:
and so: 
and:
:$b_1 - b_n < \dfrac 1 4$
Therefore:
:$b_n > b_1 - \dfrac 1 4 = \dfrac 3 4 - \dfrac 1 2 \map \ln 2$
Thus by definition $\sequence {b_n}$ is bounded below. 
By Monotone Convergence Theorem, it follows that $\sequence {b_n}$ is convergent.
Let $b$ denote its limit. 
Then:
:$\ds \lim_{n \mathop \to \infty} a_n = e^{\lim_{n \mathop \to \infty} b_n} = e^b = a$
as required.
=== Part 2 ===
By Wallis's Product, we have:
In Part 1 it was proved that:
:$n! \sim a \sqrt {2 n} \paren {\dfrac n e}^n$
Substituting for $n!$ in $(2)$ yields:
Hence the result.
Category:Stirling's Formula
Stirling's Formula/Proof 2

== Theorem ==
== Proof ==
Consider the sequence $\sequence {d_n}$ defined as:
:$d_n = \map \ln {n!} - \paren {n + \dfrac 1 2} \ln n + n$
From Lemma 2 it is seen that $\sequence {d_n}$ is a decreasing sequence.
From Lemma 3 it is seen that the sequence:
:$\sequence {d_n - \dfrac 1 {12 n} }$
is increasing.
In particular:
:$\forall n \in \N_{>0}: d_n - \dfrac 1 {12 n} \ge d_1 - \dfrac 1 {12}$
and so $\sequence {d_n}$ is bounded below.
From the Monotone Convergence Theorem (Real Analysis), it follows that $\sequence {d_n}$ is convergent.
Let $d_n \to d$ as $n \to \infty$.
From Exponential Function is Continuous, we have:
:$\exp d_n \to \exp d$ as $n \to \infty$
Let $C = \exp d$.
Then:
:$\dfrac {n!} {n^n \sqrt n e^{-n} } \to C$ as $n \to \infty$
It remains to be shown that $C = \sqrt {2 \pi}$.
Let $I_n$ be defined as:
:$\ds I_n = \int_0^{\frac \pi 2} \sin^n x \rd x$
Then from Lemma 4:
:$\ds \lim_{n \mathop \to \infty} \frac {I_{2 n} } {I_{2 n + 1} } = 1$
So:
Hence the result.
== Sources ==
* : $\S 17.2$
Category:Stirling's Formula
Stirling's Formula/Proof 2/Lemma 1

== Lemma ==
Let $\map f x$ be the real function defined on the open interval $\openint {-1} 1$ as:
:$\map f x := \dfrac 1 {2 x} \map \ln {\dfrac {1 + x} {1 - x} } - 1$
Then:
:$\ds \map f x = \sum_{k \mathop = 1}^\infty \dfrac {x^{2 n} } {2 n + 1}$
== Proof ==
== Sources ==
* : $\S 17.2$
Category:Stirling's Formula
Stirling's Formula/Proof 2/Lemma 2

== Theorem ==
The sequence $\sequence {d_n}$ defined as:
:$d_n = \map \ln {n!} - \paren {n + \dfrac 1 2} \ln n + n$
is decreasing.
== Proof ==
The proof strategy is to demonstrate that the sign of $d_n - d_{n + 1}$ is positive.
Let:
:$\map f x := \dfrac 1 {2 x} \map \ln {\dfrac {1 + x} {1 - x} } - 1$
for $\size x < 1$.
Then from Lemma 1:
:$(2): \quad \ds \map f x = \sum_{k \mathop = 1}^\infty \frac {x^{2 n} } {2 n + 1}$
Thus $\map f x > 0$ for $\size x < 1$.
Putting $x = \dfrac 1 {2 n + 1}$ it can be seen that $(1)$ is $\map f {\dfrac 1 {2 n + 1} }$.
As $-1 < \dfrac 1 {2 n + 1} < 1$ it can be seen that $(2)$ can be applied and so:
:$\forall n \in \N: d_n - d_{n + 1} \ge 0$
Thus $\sequence {d_n}$ is a decreasing sequence.
== Sources ==
* : $\S 17.2$
Category:Stirling's Formula
Stirling's Formula/Proof 2/Lemma 3

== Lemma ==
Let $\sequence {d_n}$ be the sequence defined as:
:$d_n = \map \ln {n!} - \paren {n + \dfrac 1 2} \ln n + n$
Then the sequence:
:$\sequence {d_n - \dfrac 1 {12 n} }$
is increasing.
== Proof ==
We have:
Let:
:$\map f x := \dfrac 1 {2 x} \map \ln {\dfrac {1 + x} {1 - x} } - 1$
for $\size x < 1$.
Then:
As $-1 < \dfrac 1 {2 n + 1} < 1$ it can be substituted for $x$ in $(2)$:
Thus the sequence:
:$\sequence {d_n - \dfrac 1 {12 n} }$
is increasing.
== Sources ==
* : $\S 17.2$
* 
Category:Stirling's Formula
Stirling's Formula/Proof 2/Lemma 4

== Theorem ==
Let $I_n$ be defined as:
:$\ds I_n = \int_0^{\frac \pi 2} \sin^n x \rd x$
Then:
:$\ds \lim_{n \mathop \to \infty} \frac {I_{2 n} } {I_{2 n + 1} } = 1$
== Proof ==
== Sources ==
* : $\S 17.4 \ (1)$
Category:Sine Function
Category:Stirling's Formula
Stirling's Formula/Proof 2/Lemma 5

== Theorem ==
:$\dfrac {n!} {n^n \sqrt n e^{-n} } \to \sqrt {2 \pi}$ as $n \to \infty$
== Proof ==
By previous work done in Stirling's Formula: Proof 2 it is noted that $\sequence {\dfrac {n!} {n^n \sqrt n e^{-n} } }_{n \mathop \in \N}$ is a convergent sequence.
Let $\dfrac {n!} {n^n \sqrt n e^{-n} } \to C$ as $n \to \infty$.
Let $\ds I_n = \int_0^{\frac \pi 2} \sin^n x \rd x$.
Then:
From Lemma 4:
:$\ds \lim_{n \mathop \to \infty} \frac {I_{2 n} } {I_{2 n + 1} } = 1$
from which:
Hence the result.
== Sources ==
* : $\S 17.4 \ (2)$
Category:Stirling's Formula
Stirling's Formula/Proof 3

== Theorem ==
== Proof ==
From Poisson Distribution Approximated by Normal Distribution, we have:
Let $X$ be a discrete random variable which has the Poisson distribution $\Poisson \lambda$.
Then for large $\lambda$:
:$\Poisson \lambda \approx \Gaussian \lambda \lambda$
where $\Gaussian \lambda \lambda$ denotes the normal distribution.
Since the Poisson Distribution with parameter $\lambda$ converges to the Normal Distribution with mean $\lambda$ and variance $\lambda$, their density functions will be asymptotically equal.
:$\dfrac 1 {k!} \lambda^k e^{-\lambda} \approx \dfrac 1 {\sqrt {2 \pi \lambda} } \map \exp {-\dfrac {\paren {k - \lambda}^2} {2 \lambda} }$
Evaluating this expression at the mean simplifies to the expression:
:$\dfrac 1 {\lambda!} \lambda^\lambda e^{-\lambda} \approx \dfrac 1 {\sqrt {2 \pi \lambda} }$
Rearranging, we obtain:
:$\lambda! \approx \sqrt {2 \pi \lambda} \lambda^\lambda e^{-\lambda}$
Category:Stirling's Formula
Stirling's Formula/Refinement

== Theorem ==
A refinement of Stirling's Formula is:
:$n! = \sqrt {2 \pi n} \paren {\dfrac n e}^n \paren {1 + \dfrac 1 {12 n} + \map \OO {\dfrac 1 {n^2} } }$
where $\map \OO \cdot$ is the big-$\OO$ notation.
== Proof ==
Let $\ds c_n = \ln \map \Gamma {z + n}$
We begin by observing:
Then from $\paren 1$ above, we have:
:$\ds c_n = \ln \map \Gamma {z + 1} + \sum_{k \mathop = 1}^{n - 1} \map \ln {z + k}$
Therefore:
By the analogy between the derivative and the finite difference, we consider $c_n$ to be approximately $\ds \int \map \ln {z + n} \rd z$.
With this in mind, based upon Primitive of Logarithm of x, we set:
$\ds c_n = \paren {z + n} \map \ln {z + n} - \paren {z + n} + d_n$
Quick recap:
Therefore:
Let $\ds d_n = e_n - \dfrac 1 2 \map \ln {z + n}$
Then:
We now observe that
Let $\ds \map {K_1} z = \lim_{n \mathop \to \infty} \paren {e_n - e_0} $
Then:
Therefore, we have:
Exponentiating both sides, we obtain:
We claim $\map C z$ is independent of $z$
Therefore, $\map C z$ is a constant and
:$\map \Gamma z \sim C z^{z - \frac 1 2} e^{-z}$ as $\map \Re z \to \infty$
Therefore:
To determine C, we use Wallis's Product:
Substituting for $n!$ in $(8)$ yields:
Putting all of the pieces together, we have:
== Examples ==
== Also see == 
* Stirling's Formula for Gamma Function
== Sources ==
* : $\S 1.2.5$: Permutations and Factorials: Exercise $5$
* : Chapter $1$. The Gamma and Beta Functions
Category:Stirling's Formula
Stirling's Formula/Refinement/Examples/8

== Example of Use of the Refinement to Stirling's Formula ==
The factorial of $8$ is given by the refinement to Stirling's formula as:
:$8! \approx 40 \, 318$
which shows an error of about $0.005 \%$.
== Proof ==
We have that by the usual calculation:
:$8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 40 \ 320$
Hence:
== Also see == 
* Stirling's Formula/Examples/8
* Stirling's Formula for Gamma Function/Examples/First 10
== Sources ==
* : $\S 1.2.5$: Permutations and Factorials: Exercise $5$
Category:Stirling's Formula
Stirling's Formula for Gamma Function

== Theorem ==
Let $\Gamma$ denote the Gamma function.
Let $z \in \C$ with a strictly positive real part and $\size {\arg z} < \dfrac \pi 2$.
Then:
:$\map \Gamma {z + 1} = \sqrt {2 \pi z} \, z^z e^{-z} \paren {1 + \dfrac 1 {12 z} + \dfrac 1 {288 z^2} - \dfrac {139} {51 \, 840 z^3} - \dfrac {571} {2\, 488 \, 320 z^4} + \dfrac {163 \, 879} {209\, 018 \, 880 z^5} + \dfrac {5\, 246\, 819} {75\, 246\, 796\, 800 z^6} + \cdots}$
== Proof ==
From Logarithmic Approximation of Error Term of Stirling's Formula for Gamma Function, we have:
:$\ds \map \Ln {\map \Gamma z} = \paren {z - \dfrac 1 2} \map \Ln z - z + \dfrac {\ln 2 \pi} 2 + \sum_{n \mathop = 1}^d \frac {B_{2 n} } {2 n \paren {2 n - 1} z^{2 n - 1} } + \OO \paren {\dfrac 1 {z^{2 d + 1} } }$
Taking the first $3$ terms in the sum, we get:
Exponentiating both sides, we obtain:
== Also known as ==
== Examples ==
== Also see == 
* Logarithmic Approximation of Error Term of Stirling's Formula for Gamma Function
* Limit of Error in Stirling's Formula
* Stirling's Formula
== Sources ==
* : $12$: The Transcendental Functions
* : $\S 16$: Asymptotic Expansions for the Gamma Function: $16.15$
* : $\S 25$: The Gamma Function: Asymptotic Expansions for the Gamma Function: $25.15.$
Category:Stirling's Formula
Category:Asymptotic Expansions
Category:Gamma Function
Stirling's Formula for Gamma Function/Also known as

== Stirling's Formula for Gamma Function: Also known as == 
Stirling's Formula for Gamma Function is also known as Stirling's asymptotic series.
== Sources ==
* : $\S 16$: Asymptotic Expansions for the Gamma Function: $16.15$
* : $\S 25$: The Gamma Function: Asymptotic Expansions for the Gamma Function: $25.15.$
Category:Stirling's Formula
Stirling's Formula for Gamma Function/Examples/First 10

== Example of Use of Stirling's Formula for Gamma Function ==
:
From Stirling's Formula/Examples/8, we have
From Stirling's Formula/Refinement/Examples/8, we have
These values can be found in the tables above in the row $z = 8$
== Also see == 
* Stirling's Formula/Examples/8
* Stirling's Formula/Refinement/Examples/8
Category:Stirling's Formula
Category:Factorials
Category:Gamma Function
Stirling Number of n with n-m is Polynomial in n of Degree 2m/Second Kind

== Theorem ==
Let $m \in \Z_{\ge 0}$.
The Stirling number of the second kind $\ds {n \brace n - m}$ is a polynomial in $n$ of degree $2 m$.
== Proof ==
== Sources ==
* : $\S 1.2.6$: Binomial Coefficients
Category:Stirling Numbers
Stirling Number of n with n-m is Polynomial in n of Degree 2m/Unsigned First Kind

== Theorem ==
Let $m \in \Z_{\ge 0}$.
The unsigned Stirling number of the first kind $\ds {n \brack n - m}$ is a polynomial in $n$ of degree $2 m$.
== Proof ==
The proof proceeds by induction over $m$.
For all $m \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$\ds {n \brack n - m}$ is a polynomial in $n$ of degree $2 m$.
=== Basis for the Induction ===
$\map P 0$ is the case:
:$\ds {n \brack n} = 1$
which is a polynomial in $n$ of degree $0$
Thus $\map P 0$ is seen to hold.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown 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 the induction hypothesis:
:$\ds {n \brack n - k}$ is a polynomial in $n$ of degree $2 k$.
from which it is to be shown that:
:$\ds {n \brack n - \paren {k + 1} }$ is a polynomial in $n$ of degree $2 \paren {k + 1}$.
=== Induction Step ===
This is the induction step:
Let $\map f {n, d}$ denote an arbitrary polynomial in $n$ of degree $d$.
Then:
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\ds {n \brack n - m}$ is a polynomial in $n$ of degree $2 m$.
== Sources ==
* : $\S 1.2.6$: Binomial Coefficients
Category:Stirling Numbers
Stirling Number of the Second Kind of 0

== Theorem ==
:$\ds {0 \brace n} = \delta_{0 n}$
where:
:$\ds {0 \brace n}$ denotes a Stirling number of the second kind
:$\delta_{0 n}$ denotes the Kronecker delta.
== Proof ==
By definition of Stirling numbers of the second kind:
$\ds x^{\underline 0} = \sum_k {0 \brace k} x^k$
Thus we have:
Thus, in the expression:
:$\ds x^0 = \sum_k {0 \brace k} x^{\underline k}$
we have:
:$\ds {0 \brace 0} = 1$
and for all $k \in \Z_{>0}$:
:$\ds {0 \brace k} = 0$
That is:
:$\ds {0 \brace k} = \delta_{0 k}$
== Also see ==
* Unsigned Stirling Number of the First Kind of 0
* Signed Stirling Number of the First Kind of 0
* Particular Values of Stirling Numbers of the Second Kind
== Sources ==
* : $\S 1.2.6$: Binomial Coefficients: $(48)$
Category:Examples of Stirling Numbers of the Second Kind
Stirling Number of the Second Kind of 1

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

== Theorem ==
Let $n, k \in \Z_{\ge 0}$.
Let $k > n$.
Let $\ds {n \brace k}$ denote a Stirling number of the second kind.
Then:
:$\ds {n \brace k} = 0$
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Unsigned Stirling Number of the First Kind of Number with Greater
* Signed Stirling Number of the First Kind of Number with Greater
Category:Stirling Numbers
Category:Stirling Number of the Second Kind of Number with Greater
Stirling Number of the Second Kind of Number with Greater/Proof 1

== Theorem ==
Let $n, k \in \Z_{\ge 0}$ such that $k > n$.
== Proof ==
By definition, the Stirling numbers of the second kind are defined as the coefficients $\ds {n \brace k}$ which satisfy the equation:
:$\ds x^n = \sum_k {n \brace k} x^{\underline k}$
where $x^{\underline k}$ denotes the $k$th falling factorial of $x$.
Both of the expressions on the and are polynomials in $x$ of degree $n$.
Hence the coefficient $\ds {n \brace k}$ of $x^{\underline k}$ where $k > n$ is $0$.
Category:Stirling Number of the Second Kind of Number with Greater