nyuuzyou/wikis huggingface dataset:
Fundamental Solution to 2D Laplace's Equation

== Theorem ==
Let $\delta_{\tuple {0, 0}} \in \map {\DD'} {\R^2}$ be the Dirac delta distribution.
Let $\Delta = \dfrac {\partial^2} {\partial x^2} + \dfrac {\partial^2} {\partial y^2}$ be the Laplacian in $2$-dimensional Euclidean space.
Then in the distributional sense:
:$\ds \map \Delta {\frac {\ln r} {2 \pi}} = \delta_{\tuple {0, 0}}$
where $r = \sqrt {x^2 + y^2}$.
== Proof ==
Let $\norm {\,\cdot\,}_2$ be the 2-norm.
=== Local integrability ===
Let $u : \R^2 \to \R$ be a radial real function, say $\map u {\mathbf x} = \map f r$, where $r = \norm {\mathbf x}_2$.
By Laplacian in Polar Coordinates:
:$\ds \Delta f = \map {f} r + \frac {\map {f'} r} r$
Thus:
:$\Delta \map \ln r = 0$
Furthermore, for all $R > 0$ we have:
Hence, $\ln r$ is a locally integrable function:
:$\paren {\mathbf x \mapsto \ln r} \in \map {L^1_{loc}} {\R^2}$
=== Fundamental Solution ===
Let $\phi \in \map \DD {\R^2}$ be a test function with support on the open ball $\map {B_R} {\mathbf 0}$.
Then:
Let $\Omega := \set {\mathbf x \in \R^2 : \epsilon < \norm {\mathbf x}_2 < R}$ with the boundary $\partial \Omega = \map S {\epsilon} \cup \map S {R}$ where
:$\map S \epsilon = \set {\mathbf x : \norm {\mathbf x}_2 = \epsilon}$
:$\map S R = \set {\mathbf x : \norm {\mathbf x}_2 = R}$
By Green's Identities and the definition of test function:
:$\ds \int_{\epsilon \mathop < \norm {\mathbf x}_2 \mathop < R} \ln r \Delta \phi \rd \mathbf x = \int_{\norm {\mathbf x}_2 \mathop = \epsilon} \ln r \dfrac {\partial \phi}{\partial n} \rd s - \int_{\norm {\mathbf x}_2 \mathop = \epsilon} \dfrac {\partial \ln r}{\partial n} \phi \rd s$
where:
:$\dfrac {\partial \phi} {\partial n} = \nabla \phi \cdot \map {\mathbf n} {\mathbf x}$
and $\map n {\mathbf x}$ is the unit normal pointing outward the boundary, while $\rd s$ is a small element of the boundary contour $\partial\Omega$.
For the first integral we have the following estimate:
Furthermore:
 
Thus:
:$\ds \lim_{\epsilon \to 0} \size {\int_{\norm {\mathbf x}_2 \mathop = \epsilon} \ln r \dfrac {\partial \phi}{\partial n} \rd \map s {\mathbf x} } \le 0$
For the second integral we have:
By definition, the test function $\phi$ is continuous.
Let $\eta > 0$.
Then there is $\epsilon_0 > 0$ such that $\epsilon_0 > \epsilon > 0$ and:
:$\norm {\mathbf x}_2 < \epsilon_0 \implies \size {\map \phi {\mathbf x} - \map \phi {\mathbf 0} } < \eta$
Hence:
Since $\eta$ was arbitrary:
:$\ds \forall \eta \in \R_{>0} : \exists \epsilon_0 > \epsilon : \forall \mathbf x \in \R^2 : \norm {\mathbf x}_2 < \epsilon_0 \implies \size {\frac 1 {2 \pi \epsilon} \int_{\norm {\mathbf x}_2 \mathop = \epsilon} \map \phi {\mathbf x} \rd \map s {\mathbf x} - \map \phi {\mathbf 0} } < \eta$
Thus:
So:
:$\map {T_{\Delta \ln r} } \phi = 2 \pi \map {\delta_{\tuple {0, 0} } } \phi$
By linearity of the distribution:
:$T_{\Delta \frac {\ln r} {2 \pi} } = \delta_{\tuple {0, 0} }$
== Sources ==
* : Chapter $\S 6.2$: A glimpse of distribution theory. Derivatives in the distributional sense
Category:Laplace's Equation
Category:Examples of Fundamental Solutions
Fundamental Solution to Nth Derivative

== Theorem ==
Let $H$ be the Heaviside step function.
Let $n \in \N_{>0}$.
Let $\ds \map {f_n} x = \map H x \frac {x^{n - 1}} {\paren {n - 1}!}$.
Let $T_{f_n}$ be the distribution associated with $f_n$.
Then, in the distributional sense, $T_{f_n}$ is the fundamental solution of
:$\dfrac {\rd^n} {\rd x^n} T_{f_n} = \delta$
== Proof ==
Proof by Principle of Mathematical Induction:
=== Basis for the Induction ===
Let $n = 1$.
By Distributional Derivative of Heaviside Step Function:
:$\dfrac \rd {\rd x} T_{f_1} = \delta$
=== Induction Hypothesis ===
This is our induction hypothesis:
:$\dfrac {\rd^k} {\rd x^k} T_{f_k} = \delta$
Now we need to show true for $n = k + 1$:
:$\dfrac {\rd^{k + 1} } {\rd x^{k + 1} } T_{f_{k + 1} } = \delta$
=== Induction Step ===
This is our induction step:
$\ds x \stackrel f \longrightarrow \map H x \frac {x^k} {k!}$ is a continuously differentiable real function on $\R \setminus \set 0$ and possibly has a discontinuity at $x = 0$.
By Differentiable Function as Distribution we have that $T'_f = T_{f'}$.
Moreover:
:$\ds x < 0 \implies \paren {\map H x \frac {x^k} {k!} }' = 0$.
:$\ds x > 0 \implies \paren {\map H x \frac {x^k} {k!} }' = \frac {x^{k - 1} } {\paren {k - 1}!}$.
Altogether:
:$\ds \forall x \in \R \setminus \set 0 : \paren {\map H x \frac {x^k} {k!} }' = \map H x \frac {x^{k - 1} } {\paren {k - 1}!}$
Furthermore:
:$\ds \map f {0^+} - \map f {0^-} = \map H {0^+} \frac {\paren {0^+}^k} {k!} - \map H {0^-} \frac {\paren {0^-}^k} {k!} = 0$
By the Jump Rule:
:$T'_{\map H x \frac {x^k} {k!} } = T_{\map H x \frac {x^{k - 1} } {\paren {k - 1}!} }$
Suppose:
:$\dfrac {\rd^n} {\rd x^k} T_{f_k} = \delta$
The result follows by induction.
== Sources ==
* : Chapter $\S 6.4$: A glimpse of distribution theory. Multiplication by $C^\infty$ functions
Category:Examples of Fundamental Solutions
Category:Distributional Derivatives
Fundamental Solution to Reduced Linear First Order ODE with Constant Coefficients

== Theorem ==
Let $H$ be the Heaviside step function.
Let $\lambda \in \R$.
Let $\map f x = \map H x \map \exp {\lambda x}$.
Let $T_f$ be the distribution associated with $f$.
Then, in the distributional sense, $T_f$ is the fundamental solution of
:$\paren {\dfrac \d {\d x} - \lambda} T_f = \delta$
== Proof ==
$x \stackrel f {\longrightarrow} \map H x \map \exp {\lambda x}$ is a continuously differentiable real function on $\R \setminus \set 0$ and possibly has a discontinuity at $x = 0$.
By Differentiable Function as Distribution we have that $T'_f = T_{f'}$.
Moreover:
:$x < 0 \implies \paren {\map H x \map \exp {\lambda x}}' = 0$.
:$x > 0 \implies \paren {{\map H x} \map \exp { \lambda x}}' = \lambda \map \exp {\lambda x}$.
Altogether:
:$\forall x \in \R \setminus \set 0 : \paren {{\map H x} \map \sin x}' = \lambda \map H x \map \exp {\lambda x}$
Furthermore:
:$\ds \map f {0^+} - \map f {0^-} = \map H {0^+} \map \exp {0^+} - \map H {0^-} \map \exp {0^-} = 1$
By the Jump Rule:
:$T'_{\map H x \map \exp {\lambda x}} = T_{\lambda \map H x \map \exp {\lambda x}} + \delta$
Rearranging the terms and using the linearity of distribution gives:
:$T'_{\map H x \map \exp {\lambda x}} - \lambda T_{\map H x \map \exp {\lambda x}} = \delta$
== See also ==
* Solution to Linear First Order ODE with Constant Coefficients
== Sources ==
* : Chapter $\S 6.4$: A glimpse of distribution theory. Multiplication by $C^\infty$ functions
Category:Examples of Fundamental Solutions
Category:Distributional Derivatives
Category:First Order ODEs
Fundamental Solution to y'' + w y

== Theorem ==
Let $H$ be the Heaviside step function.
Let $\omega \in \R : \omega \ne 0$.
Let $\ds \map f x = \map H x \frac {\map \sin {\omega x}} \omega$.
Let $T_f$ be the distribution associated with $f$.
Then, in the distributional sense, $T_f$ is the fundamental solution of
:$\paren {\dfrac {\d^2}{\d x^2} + \omega^2} T_f = \delta$
== Proof ==
$x \stackrel f {\longrightarrow} \map H x \frac {\map \sin {\omega x}} \omega$ is a continuously differentiable real function on $\R \setminus \set 0$ and possibly has a discontinuity at $x = 0$.
By Differentiable Function as Distribution we have that $T'_f = T_{f'}$.
Moreover:
:$\ds x < 0 \implies \paren {\map H x \frac {\map \sin {\omega x}}{\omega}}' = 0$.
:$\ds x > 0 \implies \paren {\map H x \frac {\map \sin {\omega x}}{\omega}}' = \map \cos {\omega x}$.
Altogether:
:$\forall x \in \R \setminus \set 0 : \paren {{\map H x} \frac {\map \sin {\omega x}} \omega}' = \map H x \map \cos {\omega x}$
Furthermore:
:$\ds \map f {0^+} - \map f {0^-} = \map H {0^+} \frac {\map \sin {\omega \cdot 0^+}} \omega - \map H {0^-} \frac {\map \sin {\omega \cdot 0^-}} \omega = 0$
By the Jump Rule:
:$T'_{\map H x \frac {\map \sin {\omega x}} \omega} = T_{\map H x \map \cos {\omega x}}$
$x \stackrel g {\longrightarrow} \map H x \map \cos {\omega x}$ is a continuously differentiable real function on $\R \setminus \set 0$ and possibly has a discontinuity at $x = 0$.
By Differentiable Function as Distribution we have that $T'_g = T_{g'}$.
Moreover:
:$\ds x < 0 \implies \paren {\map H x \map \cos \omega}' = 0$.
:$\ds x > 0 \implies \paren {\map H x \map \cos \omega}' = - \omega \map \sin {\omega x}$.
Altogether:
:$\forall x \in \R \setminus \set 0 : \paren {\map H x \map \cos \omega}' = - \map H x \omega \map \sin {\omega x}$
Furthermore:
:$\ds \map g {0^+} - \map g {0^-} = \map H {0^+} \map \cos {\omega \cdot 0^+} - \map H {0^-} \map \cos {\omega \cdot 0^-} = 1$
By the Jump Rule:
:$T'_{\map H x \map \cos {\omega x}} = T_{- \omega \map H x \map \sin {\omega x}} + \delta$
Hence:
:$T_{\map H x \frac {\map \sin {\omega x}} \omega} = T_{- \omega^2 \map H x \frac {\map \sin {\omega x}} \omega} + \delta$
Rearranging the terms and using the linearity of distribution gives:
:$T_{\map H x \frac {\map \sin {\omega x} } \omega} + \omega^2 T_{\map H x \frac {\map \sin {\omega x}} \omega} = \delta$
== Sources ==
* : Chapter $\S 6.4$: A glimpse of distribution theory. Multiplication by $C^\infty$ functions
Category:Examples of Fundamental Solutions
Category:Distributional Derivatives
Category:Second Order ODEs
Fundamental Solutions to Distributional Homogeneous ODE with Constant Coefficients differ by Classical Solution

== Theorem ==
Let $E_*, E, T \in \map {\DD'} \R$ be distributions.
Let $D$ be an ordinary differential operator with constant coefficients.
Let $f$ be a function differentiable by $D$.
Let $T_f \in \map {\DD'} \R$ be a distribution associated with $f$.
Let $\delta$ be the Dirac delta distribution.
Let $E_*$ be the fundamental solution to $D E_* = \delta$
Then $E$ is a fundamental solution to $DE = \delta$ $E = E_* + T_F$ where:
:$DE_* = \delta$
:$DF = 0$
== Proof ==
=== Necessary Condition ===
Suppose both $E_*$ and $E$ are fundamental solutions:
:$DE = \delta$
:$DE_* = \delta$
Taking the difference yields:
:$D \paren {E - E_*} = \mathbf 0$
where $\mathbf 0 \in \map {\DD'} \R$ is the zero distribution.
By Solution to Distributional Ordinary Differential Equation with Constant Coefficients:
:$E - E_* = T_F$
where $DF = 0$.
Hence:
:$E = E_* + T_F$
=== Sufficient condition ===
Suppose $F$ is a classical solution to $DF = 0$.
That is, suppose:
:$DT_F = \mathbf 0$
Suppose $E_* \in \map {\DD'} \R$ is a fundamental solution to $DE_* = \delta$.
Let $E$ be a distribution such that $E:= E_* + T_F$.
Then:
== Sources ==
* : Chapter $\S 6.4$: A glimpse of distribution theory. Multiplication by $C^\infty$ functions
Category:Examples of Distributional Solutions
Category:Distributional Derivatives
Category:Fundamental Solutions
Fundamental Theorem of Algebra

== Theorem ==
Every non-constant polynomial with coefficients in $\C$ has a root in $\C$.
That is, the field of complex numbers is algebraically closed.
== Proof using Algebraic Topology ==
== Proof using Liouville's Theorem ==
== Proof using Cauchy-Goursat Theorem ==
== Proof using Symmetric Polynomials ==
== Elementary Proof ==
== Proof using Cauchy's Integral Formula ==
== Also known as ==
== Historical Note ==
== Sources ==
* : Introduction
* : $1$: Complex Numbers: Polynomial Equations
* 
* 
* 
* 
* 
* 
Category:Fundamental Theorem of Algebra
Category:Polynomial Theory
Category:Algebra
Category:Analysis
Category:Fundamental Theorems
Fundamental Theorem of Algebra/Also known as

== Fundamental Theorem of Algebra: Also known as ==
The Fundamental Theorem of Algebra is known in France as d'Alembert's Theorem, for .
It is also known as the d'Alembert-Gauss Theorem, for and .
== Sources ==
* : Introduction
* 
* 
* 
* 
Category:Fundamental Theorem of Algebra
Fundamental Theorem of Algebra/Historical Note

== Historical Note on Fundamental Theorem of Algebra ==
The Fundamental Theorem of Algebra was first conjectured by , who published about it in $1629$, recognising the existence of imaginary roots.
A proof, geometric in nature, of the Fundamental Theorem of Algebra was published in $1746$ by . It was for some time called D'Alembert's Theorem.
However, it was later discovered that 's proof was incorrect.
The first correct proof was published by in his doctoral dissertation in :
:Demonstratio nova theorematis omnem functionem algebraicam rationalem integram unius variabilis in factores reales primi vel secundi gradus resolvi posse (A new proof of the theorem that every integral rational algebraic function of one variable can be resolved into real factors of the first or second degree)
in which he gave the Fundamental Theorem of Algebra its name.
During the course of his career, he gave a total of four proofs of this theorem.
The first full and rigorous proof in the field of complex numbers was published in $1814$ by .
== Sources ==
* : Introduction
* : Chapter $\text {A}.25$: Gauss ()
* 
* 
Category:Historical Notes
Fundamental Theorem of Algebra/Proof 1

== Theorem ==
== Proof ==
Let $\map p z$ be a polynomial in $\C$:
:$\map p z = z^m + a_1 z^{m-1} + \cdots + a_m$
where not all of $a_1, \ldots, a_m$ are zero.
Define a homotopy:
:$\map {p_t} z = t \map p z + \left({1-t}\right) z^m$
Then:
:$\dfrac {\map {p_t} z} {z^m} = 1 + t \paren {a_1 \dfrac 1 z + \cdots + a_m \dfrac 1 {z^m} }$
The terms in the parenthesis go to $0$ as $z \to \infty$.
Therefore, there is an $r \in \R_{>0}$ such that:
:$\forall z \in \C: \size z = r: \forall t \in \closedint 0 1: \map {p_t} z \ne 0$
Hence the homotopy:
:$\dfrac {p_t} {\size {p_t} }: S \to \Bbb S^1$
is well-defined for all $t$.
This shows that for any complex polynomial $\map p z$ of order $m$, there is a circle $S$ of sufficiently large radius in $\C$ such that $\dfrac {\map p z} {\size {\map p z}}$ and $\dfrac {z^m} {\size {z^m} }$ are freely homotopic maps $S \to \Bbb S^1$. 
Hence $\dfrac {\map p z} {\size {\map p z} }$ must have the same degree of $\paren {z / r}^m$, which is $m$. 
When $m > 0$, that is $p$ is non-constant, this result and the Extendability Theorem for Intersection Numbers imply $\dfrac {\map p z} {\size {\map p z} }$ does not extend to the disk $\Int S$, implying $\map p z = 0$ for some $z \in \Int S$.
Category:Fundamental Theorem of Algebra
Fundamental Theorem of Algebra/Proof 2

== Theorem ==
== Proof ==
Let $\map P z = a_n z^n + \dots + a_1 z + a_0, \ a_n \ne 0$.
 that $\map P z$ is not zero for any $z \in \C$.
It follows that $1 / \map P z$ must be entire; and is also bounded in the complex plane.
In order to see that it is indeed bounded, we recall that $\exists R \in \R_{>0}$ such that:
   $\cmod {\dfrac 1 {\map P z} } < \dfrac 2 {\cmod {a_n} R^n}, \text{whenever} \ \cmod z > R$.
Hence, $1 / \map P z$ is bounded in the region outside the disk $\cmod z \le R$.
However, $1 / \map P z$ is continuous on that closed disk, and thus it is bounded there as well. 
Furthermore, we observe that $1 / \map P x$ must be bounded in the whole plane.
Through Liouville's Theorem, $1 / \map P x$, and thus $\map P x$, is constant. 
This is a contradiction.
Category:Fundamental Theorem of Algebra
Fundamental Theorem of Algebra/Proof 3

== Theorem ==
== Proof ==
Let $p: \C \to \C$ be a complex, non-constant polynomial.
 that $\map p z \ne 0$ for all $z \in \C$.
Now consider the closed contour integral:
:$\ds \oint \limits_{\gamma_R} \frac 1 {z \cdot \map p z} \rd z$
where $\gamma_R$ is a circle with radius $R$ around the origin.
By Derivative of Complex Polynomial, the polynomial $z \cdot \map p z$ is holomorphic.
Since $\map p z$ is assumed to have no zeros, the only zero of $z \cdot \map p z$ is $0 \in \C$.
Therefore by Reciprocal of Holomorphic Function $\dfrac 1 {z \cdot \map p z}$ is holomorphic in $\C \setminus \set 0$.
Hence the Cauchy-Goursat Theorem implies that the value of this integral is independent of $R > 0$.
On the one hand, one can calculate the value of this integral in the limit $R \to 0$ (or use Cauchy's Residue Theorem), using the parameterization $z = R e^{i \phi}$ of $\gamma_R$:
which is non-zero.
On the other hand, we have the following upper bound for the absolute value of the integral:
But this goes to zero for $R \to \infty$.
We have arrived at a contradiction.
Hence by Proof by Contradiction the assumption that $\map p z \ne 0$ for all $z \in \C$ must be wrong.
Category:Fundamental Theorem of Algebra
Fundamental Theorem of Algebra/Proof 5

== Theorem ==
== Proof ==
Let $\map p z$ be an arbitrary non-constant polynomial with coefficients in $\C$.
This means that for some $n \in \Z_{ge 1}$ and $a_n \ne 0$:
:$\map p z = a_n z^n + a_{n - 1}z^{n - 1} + \dots + a_0$
We have:
:$\map p z = a_n z^n \paren {1 + \dfrac {a_{n - 1} } {a_n}z^{-1} + \dots + \dfrac 1 {a_n}z^{-n} } = a_n z^n \paren {1 + \map \OO {z^{-1} } }$
as $\size z \to \infty$.
Hence:
:$\ds \lim_{\size z \mathop \to \infty} \size {\map p z} = \infty$
Thus we can pick $R > 0$ large enough so that:
:$\ds \inf_{\size z \mathop \ge R} \size {\map p z} > \size {\map p 0}$
It follows that:
:$\ds \inf_{z \mathop \in \C} \size {\map p z} = \inf_{\size z \mathop \le R} \size {\map p z}$
Since $D_R = \set {z \in \C : \size z \le R}$ is compact and $p$ is continuous, there exists $z_0 \in D_R$ such that:
:$\ds \inf_{\size z \mathop \le R} \size {\map p z} = \size {\map p {z_0} }$
Hence:
:$\ds \size {\map p {z_0} } = \inf_{z \mathop \in \C} \size {\map p z}$
It remains to show that $\map p {z_0} = 0$.
:
:$\map p {z_0} = a \ne 0$
Write:
:$\map {z_0 + \zeta} = a + \map q \zeta$
Note that $\map q \zeta$ is a nonconstant polynomial in $\zeta$ satisfying $\map q 0 = 0$.
Hence for some $k \ge 1$ and $b \ne 0$, we have:
:$\map q \zeta = b \zeta^k + b_{k + 1} \zeta^{k + 1} + \dots + b_n \zeta^n$
That is:
:$\map q \zeta = b \zeta^k + \map \OO {\zeta^{k + 1} }$
as $\zeta \to 0$.
Thus, uniformly in $\omega \in S^1 = \set {\omega \in \C : \size {\omega} = 1}$:
:$\map p {z_0 + \epsilon \omega} = a + b \omega^k \epsilon^k + \map \OO {\epsilon^{k + 1} }$
as $\epsilon \searrow 0$.
Since $\size {-\dfrac {a / \size a} {b / \size b} } = 1$, by Euler's theorem, we can pick $\omega \in S^1$ such that:
:$\omega^k = -\dfrac {a / \size a} {b / \size b}$
Then:
:$\ds \map p {z_0 + \epsilon \omega} = a - a \dfrac {\size b} {\size a} \epsilon^k + \map \OO {\epsilon^{k + 1} } = a \paren {1 - \dfrac {\size b} {\size a} \epsilon^k} + \map \OO {\epsilon^{k + 1} }$
Thus:
:$\size {\map p {z_0 + \epsilon\omega} } = \size a \size {1 - \dfrac {\size b} {\size a} \epsilon^k} + \map \OO {\epsilon^{k + 1} }$
Thus for $\epsilon > 0$ small enough:
But this contradicts that:
:$\size a = \inf_{z \mathop \in \C} \size {\map p z}$
Thus:
:$\map p {z_0} = a = 0$
Category:Fundamental Theorem of Algebra
Fundamental Theorem of Algebra/Proof 6

== Theorem ==
== Proof ==
Let $p: \C \to \C$ be a complex, non-constant polynomial.
 that $\map p z \ne 0$ for all $z \in \C$.
Therefore by Reciprocal of Holomorphic Function $\dfrac 1 {\map p z}$ is entire.
This is a contradiction.
Category:Fundamental Theorem of Algebra
Fundamental Theorem of Arithmetic

== Theorem ==
For every integer $n$ such that $n > 1$, $n$ can be expressed as the product of one or more primes, uniquely up to the order in which they appear.
== Proof ==
In Integer is Expressible as Product of Primes it is proved that every integer $n$ such that $n > 1$, $n$ can be expressed as the product of one or more primes.
In Prime Decomposition of Integer is Unique, it is proved that this prime decomposition is unique up to the order of the factors.
== Examples ==
== Also known as ==
== Historical Note ==
== Sources ==
* : Chapter $\text{IV}$: The Prince of Amateurs
* : Introduction $\S 6$: The division process in $I$
* : Appendix: Elementary set and number theory
* : Chapter $3$: The Integers: $\S 13$. Unique Factorisation: Theorem $22$
* : $\text {2-4}$ The Fundamental Theorem of Arithmetic: Theorem $\text {2-5}$
* : Chapter $1$: Properties of the Natural Numbers: $\S 24$
* : $\S 13$: The fundamental theorem of arithmetic
* : $\text I$: The Series of Primes: $1.3$ Statement of the fundamental theorem of arithmetic: Theorem $2$
* : Chapter $2$: Integers and natural numbers: $\S 2.2$: Divisibility and factorization in $\mathbf Z$: Theorem $4$
* : Appendix $1$: Some Results from the Elementary Theory of Numbers: Theorem $10$
* : $1$
* : Chapter $\text {B}.2$: More about Numbers: Irrationals, Perfect Numbers and Mersenne Primes
* : Chapter $\text {B}.16$: The Sequence of Primes: Theorem $1$
* : Part $1$: Not Entirely Naive Set Theory: Chapter $2$: Partial Order Relations
* : $\S 1.2.4$: Integer Functions and Elementary Number Theory: Exercise $21$
* : $1$
* 
* 
* 
* 
* 
* 
Category:Fundamental Theorem of Arithmetic
Category:Prime Numbers
Category:Fundamental Theorems
Category:Factorization
Category:Prime Decompositions
Fundamental Theorem of Arithmetic/Also known as

== Fundamental Theorem of Arithmetic: Also known as ==
The fundamental theorem of arithmetic is also known as the unique factorization theorem.
However, this is also sometimes used for the name of the equivalent result for the general Euclidean domain and it can be argued that the names are best kept separate.
== Sources ==
* : Appendix $1$: Some Results from the Elementary Theory of Numbers: Theorem $10$
* : Chapter $\text {B}.2$: More about Numbers: Irrationals, Perfect Numbers and Mersenne Primes
* : Chapter $\text {B}.16$: The Sequence of Primes: Theorem $1$
* 
* 
* 
Category:Fundamental Theorem of Arithmetic
Fundamental Theorem of Calculus

== Theorem ==
=== First Part ===
=== Second Part ===
== Motivation ==
== Historical Note ==
== Sources ==
* : Chapter $\text{VI}$: On the Seashore
* : Chapter $1$: Theory of Sets: $\S 9$: Inverse Functions, Extensions, and Restrictions: Exercise $1$ 
* : $1$: The Nature of Differential Equations: $\S 2$: General Remarks on Solutions
* 
* 
* 
* 
* 
Category:Calculus
Category:Fundamental Theorems
Category:Fundamental Theorem of Calculus
Fundamental Theorem of Calculus/First Part

== Theorem ==
Let $f$ be a real function which is continuous on the closed interval $\closedint a b$.
Let $F$ be a real function which is defined on $\closedint a b$ by:
:$\ds \map F x = \int_a^x \map f t \rd t$
Then $F$ is a primitive of $f$ on $\closedint a b$.
=== Corollary ===
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
By Topological Manifold/Examples/Real Cartesian Space, the closed real interval is a manifold. 
We have that $\F$ is a smooth 0-form with compact support on a smooth $1$-dimensional oriented manifold $\left[{a \,.\,.\, b}\right]$.
We have that the boundary of $\left[{a \,.\,.\, b}\right]$ is $\partial \left[{a \,.\,.\, b}\right]$.
We denote $\d \F$ to be the exterior derivative of $\F$.
Hence, by the General Stokes' Theorem:
:$\ds \int_{\left[{a \,.\,.\, b}\right]} \rd F = \int_{\partial {\left[{a \,.\,.\, b}\right]}} = \F(b) - \F(a)$
Category:Fundamental Theorem of Calculus
Fundamental Theorem of Calculus/First Part/Corollary

== Corollary to Fundamental Theorem of Calculus (First Part) ==
Let $f$ be a real function which is continuous on the closed interval $\closedint a b$.
Let $F$ be a real function which is defined on $\closedint a b$ by:
:$\ds \map F x = \int_a^x \map f t \rd t$
Then:
:$\ds \frac \d {\d x} \int_a^x \map f t \rd t = \map f x$
== Proof ==
Follows from the Fundamental Theorem of Calculus (First Part) and the definition of primitive.
== Sources ==
* : $\S 4.4$
Category:Fundamental Theorem of Calculus