nyuuzyou/wikis huggingface dataset:
Gauss's Lemma on Irreducible Polynomials

== Theorem ==
Let $\Z$ be the ring of integers.
Let $\Z \sqbrk X$ be the ring of polynomials over $\Z$.
Let $h \in \Z \sqbrk X$ be a polynomial.
:$(1): \quad h$ is irreducible in $\Q \sqbrk X$ and primitive
:$(2): \quad h$ is irreducible in $\Z \sqbrk X$.
== Proof ==
=== 1 implies 2 ===
Suppose first that $h$ is not irreducible in $\Z \sqbrk X$.
Let $\ds h = a_0 + a_1 X + \cdots + a_n X^n$.
If $\deg h = 0$, then the content of $h$ is:
:$\cont h = \gcd \set {a_0} = \size {a_0}$
Since $h$ is primitive , we have $h = \pm 1$.
Now by Units of Ring of Polynomial Forms over Field, the units of $\Q \sqbrk X$ are the units of $\Q$.
Thus $h$ is a unit of $\Q \sqbrk X$.
Therefore $h$ is irreducible.
If $\deg h \ge 1$, then by Units of Ring of Polynomial Forms over Integral Domain, the units of $\Z \sqbrk X$ are the units of $\Z$.
Therefore $h$ is not a unit of $\Z \sqbrk X$.
Thus since $h$ is reducible, there is a non-trivial factorization $h = f g$ in $\Z \sqbrk X$, with $f$ and $g$ both not units.
If $\deg f = 0$, that is, $f \in \Z$, then $f$ divides each coefficient of $h$.
Since $h$ is primitive, this means that $f$ divides $\cont h = 1$.
But the divisors of $1$ are $\pm 1$, so $f = \pm 1$.
But then $f$ is a unit in $\Z \sqbrk X$, a contradiction.
Therefore $\deg f \ge 1$, so $f$ is a non-unit in $\Q \sqbrk X$.
Similarly, $g$ is a non-unit in $\Q \sqbrk X$.
Therefore $h = fg$ is a non-trivial factorization in $\Q \sqbrk X$.
=== 2 implies 1 ===
Suppose now that $h$ is not irreducible in $\Q \sqbrk X$.
That is, $h$ has a non-trivial factorization in $\Q \sqbrk X$.
Since the units of $\Q \sqbrk X$ are the units of $\Q$, this means that $h = f g$, with $f$ and $g$ both of positive degree.
Let $c_f$ and $c_g$ be the contents of $f$ and $g$ respectively.
Define $\tilde f = c_f^{-1} f$ and $\tilde g = c_g^{-1} g$. 
By Content of Scalar Multiple, it follows that $\cont {\tilde f} = \cont {\tilde g} = 1$.
Moreover by Polynomial has Integer Coefficients iff Content is Integer we have $\tilde f, \tilde g \in \Z \sqbrk X$.
Now we have:
:$\tilde f \tilde g = \dfrac {f g} {c_f c_g} = \dfrac h {c_f c_g}$
Taking the content, and using Content of Scalar Multiple we have:
:$\cont {\tilde f \tilde g} = \dfrac 1 {c_f c_g} \cont h$
By Gauss's Lemma on Primitive Rational Polynomials we know that $\cont {\tilde f \tilde g} = 1$.
Moreover, by Irreducible Integer Polynomial is Primitive, $\cont h = 1$.
Therefore we must have $c_f c_g = 1$.
Thus we have a factorization in $\Z \sqbrk X$:
:$\tilde f \tilde g = h$
This is a non-trivial factorization of $h$, as both $f$ and $g$ have positive degree.
Thus $h$ is not irreducible in $\Z \sqbrk X$.
Category:Gauss's Lemma (Polynomial Theory)
Gauss's Lemma on Primitive Rational Polynomials

== Theorem ==
Let $\Q$ be the field of rational numbers.
Let $\Q \sqbrk X$ be the ring of polynomials over $\Q$ in one indeterminate $X$.
Let $\map f X, \map g X \in \Q \sqbrk X$ be primitive polynomials.
Then their product $f g$ is also a primitive polynomial.
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Content of Rational Polynomial is Multiplicative
Category:Polynomial Theory
Category:Gauss's Lemma (Polynomial Theory)
Gauss's Lemma on Primitive Rational Polynomials/Proof 1

== Theorem ==
== Proof ==
Let $f$ and $g$ be as follows:
From the definition of primitive polynomial, the coefficients of $f$ and $g$ are all integers.
From the definition of polynomial product:
:$\ds f g = \sum_{k \mathop \in \Z} c_k \mathbf X^k$ 
where:
:$\ds c_k = \sum_{\substack {p \mathop + q \mathop = k \\ p, q \mathop \in \Z} } a_p b_q$
it is clear that the coefficients of $f g$ are also all integers.
 $f g$ is not primitive.
Then its coefficients must have a GCD greater than $1$.
Therefore there exists some prime $p$ which divides all the coefficients of $fg$.
Now $p$ can not divides all the coefficients of either $f$ or $g$, because they are primitive polynomials.
So:
:Let $i$ be the smallest integer such that $p$ does not divide $a_i$
:Let $j$ be the smallest integer such that $p$ does not divide $b_j$.
Consider the coefficient $c_{i+j}$ of $f g$:
:$\ds c_{i + j} = \sum_{k \mathop = 0}^{i + j} a_k b_{i + j - k} = a_0 b_{i + j} + a_1 b_{i + j - 1} + \cdots + a_i b_j + \cdots + a_{i + j - 1} b_1 + a_{i + j} b_0$
From the assumption, it follows that $p \divides c_{i + j}$, and so:
:$\ds p \divides \sum_{k \mathop = 0}^{i + j} a_k b_{i + j - k}$
where $\divides$ denotes divisibility.
Also, from the choice of $i$ and $j$, we have:
:$p \divides a_m$ whenever $m < i$
:$p \divides b_n$ whenever $n < j$
Now all the terms of $\ds \sum_{k \mathop = 0}^{i + j} a_k b_{i + j - k}$, except for $a_i b_j$, contain a factor from either $\set {a_0, a_1, \ldots, a_{i - 1} }$ or $\set {b_0, b_1, \ldots, b_{j - 1} }$.
It follows that we have:
:$p \divides \paren {\ds \sum_{k \mathop = 0}^{i - 1} a_k b_{i + j - k} + \sum_{k \mathop = i + 1}^{i + j} a_k b_{i + j - k} }$
But then, as also $p \divides c_{i + j}$, it follows that $p \divides a_i b_j$ as well.
From Euclid's Lemma for Prime Divisors, $p$ has to divide one or the other.
This contradicts the definition of $i$ and $j$.
So $i$ and $j$ cannot both exist.
It follows that $p$ divides at least one of $f$ and $g$, one of which is therefore not primitive.
From this contradiction, we conclude that $f g$ must be primitive.
Hence the result.
== Sources ==
* : Chapter $6$: Polynomials and Euclidean Rings: $\S 31$. Polynomials with Integer Coefficients: Theorem $60$
Category:Gauss's Lemma (Polynomial Theory)
Gauss's Lemma on Primitive Rational Polynomials/Proof 2

== Theorem ==
== Proof ==
Recall Polynomial has Integer Coefficients iff Content is Integer.
By hypothesis $f$ and $g$ have content $1 \in \Z$.
Therefore:
:$f, g \in \Z \sqbrk X$
 that $f g$ is not primitive, say:
:$\cont {f g} = d \ne 1$
By the Fundamental Theorem of Arithmetic we can choose a prime $p$ dividing $d$.
From Quotient Epimorphism from Integers by Principal Ideal, let $\pi : \Z \to \Z / p \Z$ be the quotient epimorphism to the ring of integers modulo $p$.
From Ring of Integers Modulo Prime is Field, $\Z / p \Z$ is a field.
Let $\Pi : \Z \sqbrk X \to \paren {\Z / p \Z} \sqbrk X$ be the induced homomorphism of the polynomial rings.
By construction, $p$ divides each coefficient of $f g$, so:
:$\map \Pi {f g} = \map \Pi f \, \map \Pi g = 0$
From Polynomial Forms over Field form Integral Domain, $\paren {\Z / p \Z} \sqbrk X$ is an integral domain.
Thus:
:$\map \Pi f = 0$ or $\map \Pi g = 0$
After possibly exchanging $f$ and $g$, we may assume that $\map \Pi f = 0$.
Now by Kernel of Induced Homomorphism of Polynomial Forms, if:
:$f = a_0 + a_1 X + \cdots + a_n X^n$
we must have:
:$\map \pi {a_i} = 0$
for $i = 0, \ldots, n$.
That is:
:$a_i \in p \Z$
for $i = 0, \ldots, n$.
But this says precisely that $p$ divides each $a_i$, $i = 0, \ldots, n$.
Therefore $p$ divides the content of $f$, a contradiction.
Hence our assumption that $f g$ is not primitive was invalid.
The result follows by Proof by Contradiction.
Category:Gauss's Lemma (Polynomial Theory)
Gauss's Lemma on Unique Factorization Domains

== Theorem ==
Let $R$ be a unique factorization domain.
Then the ring of polynomials $R \sqbrk X$ is also a unique factorization domain.
== Proof ==
Since a UFD is Noetherian, and a Noetherian Domain is UFD if every irreducible element is prime, it is sufficient to prove that every irreducible element of $R \sqbrk X$ is prime.
etc,
Category:Factorization
Category:Unique Factorization Domains
Gauss-Bonnet Theorem

== Theorem ==
Let $M$ be a compact $2$-dimensional Riemannian manifold with boundary $\partial M$.
Let $\kappa$ be the Gaussian curvature of $M$.
Let $k_g$ be the geodesic curvature of $\partial M$.
Then:
:$\ds \int_M \kappa \rd A + \int_{\partial M} k_g \rd s = 2 \pi \map \chi M$
where:
:$\d A$ is the element of area of the surface
:$\d s$ is the line element along $\partial M$
:$\map \chi M$ is the Euler characteristic of $M$.
== Proof ==
== Historical Note ==
== Also see ==
* Chern-Gauss-Bonnet Theorem
* Riemann-Roch Theorem
* Atiyah-Singer Index Theorem
== Sources ==
* : $\S 1$: What Is Curvature? Surfaces in Space
Category:Differential Geometry
Gauss-Lucas Theorem

== Theorem ==
Let $P$ be a non-constant polynomial in $\C$.
Then all zeroes of its derivative $P'$ belong to the convex hull of the set of zeroes of $P$.
== Proof ==
Over the complex numbers, $P$ is a product of prime factors:
:$\ds \map P z = \alpha \prod_{i \mathop = 1}^n \paren{ z - a_i }$
where:
:$a_1, a_2, \ldots, a_n \in \C$ are the (not necessary distinct) zeroes of $P$
:$\alpha \in \C$ is the leading coefficient of $P$
:$n$ is the degree of $P$.
Let $z$ be any complex number for which $\map P z \ne 0$.
Then we have for the logarithmic derivative:
:$\ds \frac {\map {P'} z} {\map P z} = \sum_{i \mathop = 1}^n \frac 1 {z - a_i}$
In particular, if $z$ is a zero of $P'$ and $\map P z \ne 0$, then:
:$\ds \sum_{i \mathop = 1}^n \frac 1 {z - a_i} = 0$
or:
:$\ds \sum_{i \mathop = 1}^n \frac {\overline z - \overline {a_i} } {\size {z - a_i}^2} = 0$
This may also be written as:
 
:$\ds \paren {\sum_{i \mathop = 1}^n \frac 1 {\size {z - a_i}^2} } \overline z = \paren {\sum_{i \mathop = 1}^n \frac 1 {\size {z - a_i}^2} \overline {a_i} }$
Taking their conjugates, we see that $z$ is a weighted sum with positive coefficients that sum to one, or the barycenter on affine coordinates, of the complex numbers $a_i$ (with different mass assigned on each root whose weights collectively sum to $1$).
On the other hand, if $\map P z = \map {P'} z = 0$, then:
:$z = 1 \cdot z + 0 \cdot a_i$
and is still a convex combination of the zeroes of $P$.
Category:Complex Analysis
Gauss-Ostrogradsky Theorem

== Theorem ==
Let $U$ be a subset of $\R^3$ which is compact and has a piecewise smooth boundary $\partial U$.
Let $\mathbf V: \R^3 \to \R^3$ be a smooth vector field defined on a neighborhood of $U$.
Then:
:$\ds \iiint \limits_U \paren {\nabla \cdot \mathbf V} \rd v = \iint \limits_{\partial U} \mathbf V \cdot \mathbf n \rd S$
where $\mathbf n$ is the unit normal to $\partial U$, directed outwards.
== Informal Proof ==
== Formal Proof ==
== Intuitive Illustration ==
== Also known as ==
== Also see ==
* Green's Theorem
== Historical Note ==
== Sources ==
* : $\S 22$: Integrals involving Vectors: $22.59$
* : Appendix $\text B$: Fields and differential operators: $\text B.1$ The Operators Div, Grad and Curl
* 
* 
* 
* 
* 
Category:Vector Calculus
Category:Gauss-Ostrogradsky Theorem
Gauss-Ostrogradsky Theorem/Also known as

== Gauss-Ostrogradsky Theorem Also known as ==
The Gauss-Ostrogradsky Theorem is also known as:
:the Divergence Theorem
:Gauss's Theorem
:Gauss's Divergence Theorem or Gauss's Theorem of Divergence
:Ostrogradsky's Theorem
:the Ostrogradsky-Gauss Theorem.
== Sources ==
* 
* 
* 
* 
* 
* 
Category:Gauss-Ostrogradsky Theorem
Gauss-Ostrogradsky Theorem/Formal Proof

== Theorem ==
== Proof ==
It suffices to prove the theorem for rectangular prisms.
The Riemann-sum nature of the triple integral then guarantees the theorem for arbitrary regions.
Let:
:$R = \set {\tuple {x, y, z}: a_1 \le x \le a_2, b_1 \le y \le b_2, c_1 \le z \le c_2}$ 
and let $S = \partial R$, oriented outward. 
Then:
:$S = A_1 \cup A_2 \cup A_3 \cup A_4 \cup A_5 \cup A_6$
where:
:$A_1, A_2$ are those sides perpendicular to the $x$-axis
:$A_3, A_4$ are those sides perpendicular to the $y$ axis
and
:$A_5, A_6$ are those sides perpendicular to the $z$-axis
and in all cases the lower subscript indicates a side closer to the origin.
Let:
:$\mathbf V = M \mathbf i + N \mathbf j + P \mathbf k$
where $M, N, P: \R^3 \to \R$.
Then:
We turn now to examine $\mathbf n$: 
Hence:
We also have:
where $\d S$ is the area element.
This is true because each side is perfectly flat, and constant with respect to one coordinate.
Hence:
and in general:
Hence the result:
:$\ds \iiint_R \nabla \cdot \mathbf V \rd v = \iint_{\partial R} \mathbf V \cdot \mathbf n \rd S$
== Also known as ==
== Also see ==
* Green's Theorem
Category:Gauss-Ostrogradsky Theorem
Gauss-Ostrogradsky Theorem/Historical Note

== Historical Note on Gauss-Ostrogradsky Theorem ==
The Gauss-Ostrogradsky Theorem was first discovered by in $1762$.
It was the later independently rediscovered by in $1813$ during the course of his investigations into electromagnetism.
, who also gave the first proof of the general theorem, rediscovered it in $1826$.
It was also rediscovered by in $1828$, in $1824$ and in $1828$.
== Sources ==
* : Chapter $\text {A}.25$: Gauss ()
Category:Historical Notes
Gauss-Ostrogradsky Theorem/Informal Proof

== Theorem ==
== Proof ==
Let $S$ be the surface of $U$.
By definition, the surface integral of $\mathbf V$ over $S$ is therefore defined as:
:$\ds \iint_S \mathbf V \cdot \mathbf n \rd S = \iint_S \mathbf V \cdot \rd \mathbf S$
where $\rd \mathbf S$ is the differential of vector area.
:
Let $\d v$ be an element of volume within $U$.
In the above diagram, a small cuboid has been used for convenience.
The total flux emerging from $\d v$ is:
:$\operatorname {div} \mathbf V \rd v$
where in this context $\mathbf V$ is the vector quantity at the center of $\d v$.
For the face $abcd$, the positive direction of the $x$ component of $\mathbf V$ and the outward normal are in the same sense, and so the flux is positive.
Consider another small cuboid contiguous to $\d v$, shown dotted in the above diagram.
This has the same component of $\mathbf V$, but this is acting inward, and so is negative.
Hence the flux through the face of one cuboid cancels out the flux through the adjacent face of the next cuboid.
This argument can be applied by imagining a sequence of such cuboids until you reach the surface of $U$.
This is the only part of the construction we have just built which contributes to the flux through $S$.
By applying the same treatment to the surfaces of all such small elements of volume, we arrive at a total flux:
:$\ds \iint_S \mathbf V \cdot \rd \mathbf S$
But at the same time we have integrated $\operatorname {div} \mathbf V \rd v$ throughout the whole of the volume of $U$.
This also measures the total flux through $S$.
Hence:
== Also known as ==
== Also see ==
* Green's Theorem
== Sources ==
* : Chapter $\text {VI}$: The Theorems of Gauss and Stokes: $1$. The Divergence Theorem of Gauss: $(6.1)$
Category:Gauss-Ostrogradsky Theorem
Gauss-Ostrogradsky Theorem/Intuitive Illustration

== Theorem ==
== Proof ==
Consider a closed surface $S$ within a body of fluid which at an arbitrary point $P$ is in motion with velocity $\mathbf V$.
The total rate of flow of fluid passing through $S$ can be found in $2$ different ways:
:$(1): \quad$ By calculating $\mathbf V \cdot \rd \mathbf S$, which is the product of a small element of $S$ and the component of velocity which is perpendicular to it, and then adding all these contributions
:$(2): \quad$ By investigating the divergence of an element of volume, which is the excess of the sources of fluid over its sinks per unit volume and integrating $\operatorname {div} \mathbf V \rd v$ throughout the volume enclosed by $S$.
These $2$ results are physically equivalent, because the excess fluid leaving $S$ over that entering must be because of the fluid injected into $B$ by the aggregate of sources and sinks.
== Also known as ==
== Also see ==
* Green's Theorem
== Sources ==
* : Chapter $\text {VI}$: The Theorems of Gauss and Stokes: $1$. The Divergence Theorem of Gauss
Category:Gauss-Ostrogradsky Theorem
Gauss Lemma for Riemannian Manifolds

== Theorem ==
Let $\struct {M, g}$ be a Riemannian manifold.
Let $U = \map {\exp_p} {\map {B_\epsilon} 0 }$ be a geodesic ball centered at $p \in M$.
Let $\partial_r$ be the radial vector field on $U \setminus \set p$, where $\setminus$ denotes the set difference.
Then $\partial_r$ is a unit vector field orthogonal to the geodesic spheres in $U \setminus \set p$.
== Proof ==
== Sources ==
* : $\S 6$: Geodesics and Distance. Geodesics Are Locally Minimizing
Category:Riemannian Manifolds
Gauss Multiplication Formula

== Theorem == 
Let $\Gamma$ denote the Gamma Function.
Let $n \in \N_{>0}$ where $\N_{>0}$ denotes the non-zero natural numbers.
Then:
:$\ds \forall z \notin \set {-\frac m n: m \in \N}: \prod_{k \mathop = 0}^{n - 1} \map \Gamma {z + \frac k n} = \paren {2 \pi}^{\paren {n - 1} / 2} n^{1/2 - n z} \map \Gamma {n z}$
== Proof ==
Taking the product for $k = 0$ to $n - 1$, we have:
== Also see ==
* Digamma Additive Formula
* Product Formula for Sine
== Sources ==
* : $\S 16$: Relationships Among Gamma Functions: $16.10$
* : $\S 25$: The Gamma Function: Relationships Among Gamma Functions: $25.10.$
* 
Category:Gamma Function
Gaussian Binomial Coefficient of 1

== Theorem ==
:$\dbinom 1 m_q = \delta_{0 m} + \delta_{1 m}$
That is:
:$\dbinom 1 m_q = \begin{cases} 1 & : m = 0 \text { or } m = 1 \\0 & : \text{otherwise} \end{cases}$
where $\dbinom 1 m_q$ denotes a Gaussian binomial coefficient.
== Proof ==
By definition of Gaussian binomial coefficient:
:$\dbinom 1 m_q = \ds \prod_{k \mathop = 0}^{m - 1} \dfrac {1 - q^{1 - k} } {1 - q^{k + 1} }$
When $m = 0$ the continued product on the is vacuous, and so:
:$\dbinom 1 0_q = 1$
Let $m > 1$.
Then:
When $m > 1$ there exists a factor $1 - q^0 = 0$ in the numerator of the .
Hence when $m > 1$ we have that $\dbinom 1 m_q = 0$.
We are left with:
Category:Gaussian Binomial Coefficients
Gaussian Binomial Theorem

== Theorem ==
Let $n \in \Z_{>0}$ be a (strictly) positive integer.
where $\dbinom n k_q$ denotes a Gaussian binomial coefficient.
=== Real Numbers ===
=== Negation of Upper Index ===
== Proof 1 ==
== Proof 2 ==
== Also known as ==
== Sources ==
* 
* : $\S 1.2.6$: Binomial Coefficients: Exercise $58$
Category:Gaussian Binomial Coefficients
Category:Binomial Theorem
Category:Gaussian Binomial Theorem
Gaussian Binomial Theorem/Also known as

== Gaussian Binomial Theorem: Also known as == 
The Gaussian Binomial Theorem is also known as:
:the $q$-nomial theorem
:the $q$-binomial theorem.
== Sources ==
* 
* : $\S 1.2.6$: Binomial Coefficients: Exercise $58$
Category:Gaussian Binomial Theorem
Gaussian Binomial Theorem/Negation of Upper Index

== Theorem ==
Let $r \in \R$ be a real number.
where:
:$\dbinom r k_q$ denotes a Gaussian binomial coefficient.
:$x \in \R: \left\lvert{x}\right\rvert < 1$
:$q \in \R: \left\lvert{q}\right\rvert < 1$.
== Proof ==
== Sources ==
* 
* 
* : $\S 1.2.6$: Binomial Coefficients: Exercise $58$
Category:Gaussian Binomial Theorem
Gaussian Binomial Theorem/Proof 1

== Theorem == 
== Proof ==
The proof proceeds by induction.
For all $n \in \Z_{\ge 1}$, let $\map P n$ be the proposition:
:$\ds \prod_{k \mathop = 1}^n \paren {1 + q^{k - 1} x} = \sum_{k \mathop = 0}^n \dbinom n k_q q^{k \paren {k - 1} / 2} x^k$
=== Basis for the Induction ===
$\map P 1$ is the case:
Thus $\map P 1$ is seen to hold.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that, if $\map P r$ is true, where $r \ge 1$, then it logically follows that $\map P {r + 1}$ is true.
So this is the induction hypothesis:
:$\ds \prod_{k \mathop = 1}^r \paren {1 + q^{k - 1} x} = \sum_{k \mathop = 0}^r \dbinom r k_q q^{k \paren {k - 1} / 2} x^k$
from which it is to be shown that:
:$\ds \prod_{k \mathop = 1}^{r + 1} \paren {1 + q^{k - 1} x} = \sum_{k \mathop = 0}^{r + 1} \dbinom {r + 1} k_q q^{k \paren {k - 1} / 2} x^k$
=== Induction Step ===
This is the induction step:
So $\map P r \implies \map P {r + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\forall n \in \Z_{\ge 1}: \ds \prod_{k \mathop = 1}^n \paren {1 + q^{k - 1} x} = \sum_{k \mathop = 0}^n \dbinom n k_q q^{k \paren {k - 1} / 2} x^k$
Category:Gaussian Binomial Theorem
Gaussian Binomial Theorem/Proof 2

== Theorem == 
== Proof ==
Let:
Then:
:$\map f {q x} = \paren {1 + q x} \paren {1 + q^2 x} \cdots \paren {1 + q^n x}$
Therefore:
:$\paren {1 + x} \map f {q x} = \map f x \paren {1 + q^n x}$
If we write:
:$\ds \map f x = \sum_{k \mathop = 0}^n Q_k x^k$
Then:
:$\ds \paren {1 + x} \sum_{k \mathop = 0}^n Q_k \paren {q x}^k = \paren {1 + q^n x}\sum_{k \mathop = 0}^n Q_k x^k$
Equating coefficients gives us:
Therefore:
Therefore:
:$\forall n \in \Z_{\ge 1}: \ds \prod_{k \mathop = 1}^n \paren {1 + q^{k - 1} x} = \sum_{k \mathop = 0}^n \dbinom n k_q q^{k \paren {k - 1} / 2} x^k$
== Sources ==
* 
Category:Gaussian Binomial Theorem
Gaussian Binomial Theorem/Real Numbers

== Theorem ==
Let $r \in \R$ be a real number.
:$\ds \sum_{k \mathop \in \Z} \dbinom r k_q q^{k \paren {k - 1} / 2} x^k = \prod_{k \mathop \ge 0} \dfrac {1 + q^k x} {1 + q^{r + k} x}$
where:
:$\dbinom r k_q$ denotes a Gaussian binomial coefficient
:$x \in \R: \size x < 1$
:$q \in \R: \size q < 1$.
== Proof ==
== Sources ==
* : $\S 1.2.6$: Binomial Coefficients: Exercise $58$
Category:Gaussian Binomial Theorem
Gaussian Delta Sequence

== Theorem ==
Let $\sequence {\map {\delta_n} x}$ be a sequence such that:
:$\ds \map {\delta_n} x := \frac n {\sqrt \pi} e^{- n^2 x^2}$
Then $\sequence {\map {\delta_n} x}_{n \mathop \in {\N_{>0} } }$ is a delta sequence.
That is, in the distributional sense it holds that:
:$\ds \lim_{n \mathop \to \infty} \map {\delta_n} x = \map \delta x$
or
:$\ds \lim_{n \mathop \to \infty} \int_{-\infty}^\infty \map {\delta_n} x \map \phi x \rd x = \map \delta \phi$
where $\phi \in \map \DD \R$ is a test function, $\delta$ is the Dirac delta distribution, and $\map \delta x$ is the abuse of notation, usually interpreted as an infinitely thin and tall spike with its area equal to $1$.
== Proof 1 ==
== Proof 2 ==
Category:Gaussian Delta Sequence
Category:Examples of Delta Sequences
Category:Dirac Delta Distribution
Gaussian Delta Sequence/Proof 1

== Theorem ==
== Proof ==
Furthermore:
Let $a,b \in \R$.
Then:
Suppose $0 < a < b$.
Then:
Analogously, suppose $a < b < 0$.
Then:
Let $\epsilon \in \R_{> 0}$.
Then:
$\epsilon$ is an arbitrary positive real number.
Hence, for every $\epsilon \in \R_{> 0}$ contributions from expressions with $\map \phi {\xi_+}$ and $\map \phi {\xi_-}$ vanish.
Suppose $\xi_\epsilon \ne 0$.
By Real Numbers are Densely Ordered:
:$\forall \epsilon \in \R_{> 0} : \exists \epsilon' \in \R_{> 0} : 0 < \epsilon' < \epsilon$
Then $\epsilon'$ we have that $\xi_\epsilon = \xi_{+'}$ or $\xi_\epsilon = \xi_{-'}$, where $\xi_{+'} \in \hointr {\epsilon'} \infty$ and $\xi_{-'} \in \hointl {-\infty} {-\epsilon'}$.
But from the result above, for every $\epsilon' \in \R_{> 0}$ contributions from expressions with $\map \phi {\xi_{+'}}$ and $\map \phi {\xi_{-'}}$ vanish.
Therefore, the only nonvanishing contribution can come from $\xi_\epsilon = 0$.
== Sources ==
: Chapter $1$ Mathematical Preliminaries $1.11$ Dirac Delta Function
Category:Gaussian Delta Sequence
Gaussian Delta Sequence/Proof 2

== Theorem ==
== Proof ==
Let $\map g x = \map \phi x - \map \phi 0$.
Then:
:$\ds \int_{- \infty}^\infty \map \phi x \map {\delta_n} x \rd x = \map \phi 0 + \int_{- \infty}^\infty \map g x \map {\delta_n} x \rd x$
Let $A \in \R_{> 0}$.
Then:
By definition, $\map \phi x$ is bounded.
Then:
It follows that:
:$\exists M \in \R_{> 0} : \forall x \in \R : \size {\map g x} < M$ 
Then:
Analogously:
We have that $\map g 0 = 0$.
By definition, $\phi$ is a smooth real function on $\R$.
By Differentiable Function is Continuous, $\map g x$ is continuous at $x = 0$.
Then:
:$\ds \forall \epsilon \in \R_{> 0} : \exists \delta \in \R_{> 0} : 0 < \size x < \delta \implies \size {\map g x} < \epsilon$
Let $A$ be such that $A < \delta$.
Then:
Then:
To sum up:
:$\ds \forall \epsilon \in \R_{>0} : \exists N \in \R_{>0} : \forall n \in \N_{>0} : \forall n > N \implies \size {\int_{-\infty}^\infty \map g x \map {\delta_n} x \rd x} < \epsilon$
By definition of the limit of a real sequence:
:$\ds \lim_{n \mathop \to \infty} \int_{-\infty}^\infty \map g x \map {\delta_n} x \rd x = 0$
However:
== Sources ==
* : Chapter $1$ Mathematical Preliminaries $1.11$ Dirac Delta Function
Category:Gaussian Delta Sequence
Gaussian Elimination/Examples

== Examples of Use of Gaussian Elimination ==
=== Arbitrary Matrix $1$ ===
=== Arbitrary Matrix $2$ ===
=== Arbitrary Matrix $3$ ===
=== Arbitrary Matrix $4$ ===
=== Arbitrary Matrix $5$ ===
=== Arbitrary Matrix $6$ ===
=== Arbitrary Matrix $7$ ===
Category:Examples of Gaussian Elimination