nyuuzyou/wikis huggingface dataset:
Cauchy's Theorem

* Cauchy's Lemma (Group Theory), also known as Cauchy's Group Theorem, concerning elements of prime order in a finite group
* Cauchy's Lemma (Number Theory) (on representation of integers as sums of $4$ integers or squares)
* Cauchy's (Group) Theorem or Cauchy's Theorem for Groups, concerning existence of a subgroup of prime order in a finite group
* Cauchy's Theorem (Geometry): concerning congruent polytopes
* Cauchy Mean Value Theorem: extension to the Mean Value Theorem
* Cauchy-Goursat Theorem: value of integrals in complex analysis
* Cauchy-Hadamard Theorem: radius of convergence of a power series
* Cauchy-Kovalevsky Theorem: concerning existence and uniqueness of solutions of partial differential equations
* Cauchy-Lipschitz Theorem: concerning existence of solutions to certain differential equations (also known as Picard's Existence Theorem)
* Cauchy-Peano Theorem: concerning existence of solutions to certain initial value problems
Cauchy-Binet Formula

== Theorem ==
Let $\mathbf A$ be an $m \times n$ matrix.
Let $\mathbf B$ be an $n \times m$ matrix.
Let $1 \le j_1, j_2, \ldots, j_m \le n$.
Let $\mathbf A_{j_1 j_2 \ldots j_m}$ denote the $m \times m$ matrix consisting of columns $j_1, j_2, \ldots, j_m$ of $\mathbf A$.
Let $\mathbf B_{j_1 j_2 \ldots j_m}$ denote the $m \times m$ matrix consisting of rows $j_1, j_2, \ldots, j_m$ of $\mathbf B$.
Then:
:$\ds \map \det {\mathbf A \mathbf B} = \sum_{1 \mathop \le j_1 \mathop < j_2 \mathop < \cdots \mathop < j_m \mathop \le n} \map \det {\mathbf A_{j_1 j_2 \ldots j_m} } \map \det {\mathbf B_{j_1 j_2 \ldots j_m} }$
where $\det$ denotes the determinant.
== Proof ==
Let $\tuple {k_1, k_2, \ldots, k_m}$ be an ordered $m$-tuple of integers.
Let $\map \epsilon {k_1, k_2, \ldots, k_m}$ denote the sign of $\tuple {k_1, k_2, \ldots, k_m}$.
Let $\tuple {l_1, l_2, \ldots, l_m}$ be the same as $\tuple {k_1, k_2, \ldots, k_m}$ except for $k_i$ and $k_j$ having been transposed.
Then from Transposition is of Odd Parity:
:$\map \epsilon {l_1, l_2, \ldots, l_m} = -\map \epsilon {k_1, k_2, \ldots, k_m}$
Let $\tuple {j_1, j_2, \ldots, j_m}$ be the same as $\tuple {k_1, k_2, \ldots, k_m}$ by arranged into non-decreasing order.
That is:
:$j_1 \le j_2 \le \cdots \le j_m$
Then it follows that:
:$\map \det {\mathbf B_{k_1 \cdots k_m} } = \map \epsilon {k_1, k_2, \ldots, k_m} \map \det {\mathbf B_{j_1 \cdots j_m} }$
Hence:
If two $j$s are equal:
:$\map \det {\mathbf A_{j_1 \cdots j_m} } = 0$
== Examples ==
== Also known as ==
The Cauchy-Binet Formula is also known, confusingly, as the Binet-Cauchy Identity, which is a direct consequence of this.
== Historical Note ==
== Sources ==
* 
* 
* : $\S 1.2.3$: Sums and Products: Exercise $46$
Category:Matrix Theory
Category:Determinants
Category:Cauchy-Binet Formula
Cauchy-Binet Formula/Example

== Examples of Cauchy-Binet Formula ==
=== Cauchy-Binet Formula: $m = n$ ===
=== Cauchy-Binet Formula: $m = 1$ ===
=== Cauchy-Binet Formula: Matrix by Transpose ===
=== Cauchy-Binet Formula: $m > n$ ===
=== Cauchy-Binet Formula: $m = 2$ ===
Category:Cauchy-Binet Formula
Cauchy-Binet Formula/Example/Matrix by Transpose

== Theorem ==
Let $\mathbf A$ be an $m \times n$ matrix.
Let $\mathbf A^\intercal$ be the transpose $\mathbf A$.
Let $1 \le j_1, j_2, \ldots, j_m \le n$.
Let $\mathbf A_{j_1 j_2 \ldots j_m}$ denote the $m \times m$ matrix consisting of columns $j_1, j_2, \ldots, j_m$ of $\mathbf A$.
Let $\mathbf A^\intercal_{j_1 j_2 \ldots j_m}$ denote the $m \times m$ matrix consisting of rows $j_1, j_2, \ldots, j_m$ of $\mathbf A^\intercal$.
Then:
:$\ds \map \det {\mathbf A \mathbf A^\intercal} = \sum_{1 \mathop \le j_1 \mathop < j_2 \mathop < \cdots \mathop < j_m \le n} \paren {\map \det {\mathbf A_{j_1 j_2 \ldots j_m} } }^2$
where $\det$ denotes the determinant.
== Proof ==
The Cauchy-Binet Formula gives:
:$\ds \map \det {\mathbf A \mathbf B} = \sum_{1 \mathop \le j_1 \mathop < j_2 \mathop < \cdots \mathop < j_m \le n} \map \det {\mathbf A_{j_1 j_2 \ldots j_m} } \map \det {\mathbf B_{j_1 j_2 \ldots j_m} }$
where:
:$\mathbf A$ is an $m \times n$ matrix
:$\mathbf B$ is an $n \times m$ matrix.
:For $1 \le j_1, j_2, \ldots, j_m \le n$:
::$\mathbf A_{j_1 j_2 \ldots j_m}$ denotes the $m \times m$ matrix consisting of columns $j_1, j_2, \ldots, j_m$ of $\mathbf A$.
::$\mathbf B_{j_1 j_2 \ldots j_m}$ denotes the $m \times m$ matrix consisting of rows $j_1, j_2, \ldots, j_m$ of $\mathbf B$.
From the definition of transpose $\mathbf A^\intercal$ is an $n \times m$ matrix.
Hence the Cauchy-Binet Formula can be applied directly:
:$\ds \map \det {\mathbf A \mathbf A^\intercal} = \sum_{1 \mathop \le j_1 \mathop < j_2 \mathop < \cdots \mathop < j_m \le n} \map \det {\mathbf A_{j_1 j_2 \ldots j_m} } \map \det {\mathbf A^\intercal_{j_1 j_2 \ldots j_m} }$
Note that by construction:
:$\mathbf A_{j_1 j_2 \ldots j_m}$ is a square matrix
Also, by definition of transpose:
:$\mathbf A^\intercal_{j_1 j_2 \ldots j_m} = \paren {\mathbf A_{j_1 j_2 \ldots j_m} }^\intercal$
The result follows from Determinant of Transpose:
:$\map \det {\mathbf A} = \map \det {\mathbf A^\intercal}$
== Sources ==
* : $\S 1.2.3$: Sums and Products: Exercise $46 \ \text{(iii)}$
Category:Cauchy-Binet Formula
Cauchy-Binet Formula/Example/m equals 1

== Theorem ==
Let $\mathbf A = \sqbrk a_{1 n}$ be a row matrix with $n$ columns.
and $\mathbf B = \sqbrk b_{n 1}$ be a column matrix with $n$ rows.
Let $\mathbf A \mathbf B$ be the (conventional) matrix product of $\mathbf A$ and $\mathbf B$.
Then:
:$\ds \map \det {\mathbf A \mathbf B} = \sum_{j \mathop = 1}^n a_j b_j$
where:
:$a_j$ is element $a_{1 j}$ of $\mathbf A$
:$b_j$ is element $b_{j 1}$ of $\mathbf B$.
== Proof ==
The Cauchy-Binet Formula gives:
:$\ds \map \det {\mathbf A \mathbf B} = \sum_{1 \mathop \le j_1 \mathop < j_2 \mathop < \cdots \mathop < j_m \mathop \le n} \map \det {\mathbf A_{j_1 j_2 \ldots j_m} } \map \det {\mathbf B_{j_1 j_2 \ldots j_m} }$
where:
:$\mathbf A$ is an $m \times n$ matrix
:$\mathbf B$ is an $n \times m$ matrix.
:For $1 \le j_1, j_2, \ldots, j_m \le n$:
::$\mathbf A_{j_1 j_2 \ldots j_m}$ denotes the $m \times m$ matrix consisting of columns $j_1, j_2, \ldots, j_m$ of $\mathbf A$.
::$\mathbf B_{j_1 j_2 \ldots j_m}$ denotes the $m \times m$ matrix consisting of rows $j_1, j_2, \ldots, j_m$ of $\mathbf B$.
When $m = 1$, the relation:
:$1 \le j_1 < j_2 < \cdots < j_m \le n$
degenerates to:
:$1 \le j \le n$
By definition of order $1$ determinant:
:$\det \sqbrk {a_j} = a_j$
:$\det \sqbrk {b_j} = b_j$
Hence the result.
== Sources ==
* : $\S 1.2.3$: Sums and Products: Exercise $46 \ \text{(ii)}$
Category:Determinants
Category:Cauchy-Binet Formula
Cauchy-Binet Formula/Example/m greater than n

== Theorem ==
Let $\mathbf A$ be an $m \times n$ matrix.
Let $\mathbf B$ be an $n \times m$ matrix.
Let $m > n$.
Then:
:$\map \det {\mathbf A \mathbf B} = 0$ 
== Proof ==
The Cauchy-Binet Formula gives:
:$(1): \quad \ds \map \det {\mathbf A \mathbf B} = \sum_{1 \mathop \le j_1 \mathop < j_2 \mathop < \cdots \mathop < j_m \le n} \map \det {\mathbf A_{j_1 j_2 \ldots j_m} } \map \det {\mathbf B_{j_1 j_2 \ldots j_m} }$
where:
:$\mathbf A$ is an $m \times n$ matrix
:$\mathbf B$ is an $n \times m$ matrix.
:For $1 \le j_1, j_2, \ldots, j_m \le n$:
::$\mathbf A_{j_1 j_2 \ldots j_m}$ denotes the $m \times m$ matrix consisting of columns $j_1, j_2, \ldots, j_m$ of $\mathbf A$.
::$\mathbf B_{j_1 j_2 \ldots j_m}$ denotes the $m \times m$ matrix consisting of rows $j_1, j_2, \ldots, j_m$ of $\mathbf B$.
But here $m > n$.
Therefore the set $\set {j_1, j_2, \ldots, j_m}$ such that:
:$1 \mathop \le j_1 \mathop < j_2 \mathop < \cdots \mathop < j_m \le n$
is the empty set.
Thus the of $(1)$ is a vacuous summation.
Hence the result.
== Sources ==
* : $\S 1.2.3$: Sums and Products: Exercise $46 \ \text{(iv)}$
Category:Cauchy-Binet Formula
Cauchy-Bunyakovsky-Schwarz Inequality

== Theorem ==
=== Semi-Inner Product Spaces ===
=== Lebesgue $2$-Space ===
=== Complex Numbers ===
=== Definite Integrals ===
== Also known as ==
== Also see ==
The special case of the Cauchy-Bunyakovsky-Schwarz Inequality in a Euclidean space is called Cauchy's Inequality.
It is usually stated as:
* Hölder's Inequality
Category:Cauchy-Bunyakovsky-Schwarz Inequality
Category:Hilbert Spaces
Category:Algebra
Category:Linear Algebra
Category:Analysis
Category:Integral Calculus
Category:Inequalities
Cauchy-Bunyakovsky-Schwarz Inequality/Also known as

== Cauchy-Bunyakovsky-Schwarz Inequality: Also known as ==
The Cauchy-Bunyakovsky-Schwarz Inequality in its various form is also known as:
:the Cauchy-Schwarz-Bunyakovsky Inequality
:the Cauchy-Schwarz Inequality
:Schwarz's Inequality or the Schwarz Inequality
:Bunyakovsky's Inequality or Buniakovski's Inequality.
For brevity, it is sometimes referred to by the abbreviations CS inequality or CBS inequality.
== Sources ==
* : $3$: Elementary Analytic Methods: $3.2$ Inequalities: Schwarz's Inequality: $3.2.11$
* : $\S 36$: Inequalities: $36.13$: Cauchy-Schwarz Inequality for Integrals
* 
* 
* 
* 
* 
* 
* 
* 
Category:Cauchy-Bunyakovsky-Schwarz Inequality
Cauchy-Bunyakovsky-Schwarz Inequality/Definite Integrals

== Theorem ==
Let $f$ and $g$ be real functions which are continuous on the closed interval $\closedint a b$.
Then:
:$\ds \paren {\int_a^b \map f t \, \map g t \rd t}^2 \le \int_a^b \paren {\map f t}^2 \rd t \int_a^b \paren {\map g t}^2 \rd t$
== Proof ==
where:
The quadratic equation $A x^2 + 2 B x + C$ is non-negative for all $x$.
It follows (using the same reasoning as in Cauchy's Inequality) that the discriminant $\paren {2 B}^2 - 4 A C$ of this polynomial must be non-positive.
Thus:
:$B^2 \le A C$
and hence the result.
== Also known as ==
== Historical Note ==
== Sources ==
* : $3$: Elementary Analytic Methods: $3.2$ Inequalities: Schwarz's Inequality: $3.2.11$
* : $\S 36$: Inequalities: $36.13$: Cauchy-Schwarz Inequality for Integrals
* : $2$: Continuity generalized: metric spaces: $2.2$: Examples: Example $2.2.11$
* : $\S 13.25$
* : $(1)$
* : $(1)$
* 
* 
Category:Analysis
Category:Definite Integrals
Category:Cauchy-Bunyakovsky-Schwarz Inequality
Cauchy-Bunyakovsky-Schwarz Inequality/Inner Product Spaces

== Theorem ==
Let $\mathbb K$ be a subfield of $\C$.
Let $V$ be a semi-inner product space over $\mathbb K$.
Let $x, y$ be vectors in $V$.
Then:
:$\size {\innerprod x y}^2 \le \innerprod x x \innerprod y y$
== Proof 1 ==
== Proof 2 ==
== Also known as ==
== Sources ==
* 
* : $\text{I}$ Hilbert Spaces: $\S 1.$ Elementary Properties and Examples: $1.4$ The Cauchy-Bunyakovsky-Schwarz Inequality
* 
Category:Semi-Inner Product Spaces
Category:Cauchy-Bunyakovsky-Schwarz Inequality
Category:Inner Product Spaces
Cauchy-Bunyakovsky-Schwarz Inequality/Inner Product Spaces/Proof 1

== Theorem ==
== Proof ==
Let $x, y \in V$.
Let $\lambda \in \mathbb K$.
Then:
where $\lambda^*$ denotes the complex conjugate of $\lambda$.
(If $\mathbb K$ is a subfield of $\R$ , then $\lambda^* = \lambda$.)
First, suppose $\innerprod y y \ne 0$.
Insert $\lambda = \innerprod x y \innerprod y y^{-1}$ in the inequality:
Reorder the inequality to get:
:$\size {\innerprod x y}^2 \le \innerprod x x \innerprod y y$
Next, suppose $\innerprod y y = 0$.
By Number Field has Rational Numbers as Subfield, $\Q \subset \mathbb K$.
Let $n \in \N \subset \Q \subset \mathbb K$ be a natural number.
Insert $\lambda = n \innerprod x y$ in the inequality:
Rearrange the inequality to get:
:$\innerprod x x \ge 2 n \size {\innerprod x y}^2$
If $\size {\innerprod x y}^2 \ne 0$, then:
:$\dfrac {\innerprod x x} {2 \size {\innerprod x y}^2} \ge n$
for every $n \in \N$, contradicting the Axiom of Archimedes.
Therefore, by Proof by Contradiction:
:$\size {\innerprod x y}^2 = 0$.
Then:
:$\size {\innerprod x y}^2 \le 0 = \innerprod x x \innerprod y y$
Category:Cauchy-Bunyakovsky-Schwarz Inequality
Cauchy-Bunyakovsky-Schwarz Inequality/Inner Product Spaces/Proof 2

== Theorem ==
== Proof ==
This proof assumes that $V$ is a semi-inner product space over $\R$.
Then for all $x, y \in V$, we have $\innerprod x y = \innerprod y x$ by property $(1')$ of semi-inner products.
Define $f_{x, y}: \R \to \R_{\ge 0}$ by:
:$\map {f_{x, y} } \lambda = \innerprod {x - \lambda y} {x - \lambda y}$
Then by property $(4)$ of semi-inner product:
:$\forall \lambda \in \R: \map {f_{x, y} } \lambda \ge 0$
For all $\lambda \in \R$, it follows that:
where we have put $a = \innerprod y y$, $b = -2 \innerprod x y$, and $c = \innerprod x x$.
Then $f_{x, y}$ is a quadratic polynomial which satisfies $\map {f_{x, y} } \lambda \ge 0$.
Hence $f_{x, y}$ has at most one distinct real root.
From Solution to Quadratic Equation, it follows that the discriminant $\Delta$ satisfies:
:$\Delta = b^2 - 4 a c \le 0$
Therefore:
:$4 \innerprod x y^2 - 4 \innerprod x x \innerprod y y \le 0$
which we rearrange as:
:$\size {\innerprod x y}^2 = \innerprod x y^2 \le \innerprod x x \innerprod y y$
Category:Cauchy-Bunyakovsky-Schwarz Inequality
Cauchy-Bunyakovsky-Schwarz Inequality/Lebesgue 2-Space

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $f, g: X \to \R$ be $\mu$-square integrable functions, that is $f, g \in \map {\LL^2} \mu$, Lebesgue $2$-space.
Then:
:$\ds \int \size {f g} \rd \mu \le \norm f_2^2 \cdot \norm g_2^2$
where $\norm {\, \cdot \,}_2$ is the $2$-norm.
=== Equality ===
Equality in the above, that is:
:$\ds \int \size {f g} \rd \mu \le \norm f_2^2 \cdot \norm g_2^2$
holds for almost all $x \in X$:
:$\dfrac {\size {\map f x}^2} {\norm f_2^2} = \dfrac {\size {\map g x}^2} {\norm g_2^2}$
== Proof ==
Follows directly from Hölder's Inequality for Integrals with $p = q = 2$.
== Also known as ==
== Sources ==
* : $(1)$
* : $12.3$
* : $(1)$
Category:Cauchy-Bunyakovsky-Schwarz Inequality
Category:Lebesgue Spaces
Cauchy-Bunyakovsky-Schwarz Inequality for Definite Integrals/Historical Note

== Historical Note on Cauchy-Bunyakovsky-Schwarz Inequality for Definite Integrals ==
The Cauchy-Bunyakovsky-Schwarz Inequality for Definite Integrals was first stated in this form by in $1859$, and later rediscovered by in $1888$.
Category:Historical Notes
Cauchy-Goursat Theorem

== Theorem ==
Let $f: D \to \C$ be a holomorphic function, where $D \subseteq \C$ is a simply connected domain.
Let $C$ be a closed contour in $D$.
Then:
:$\ds \oint_C \map f z \rd z = 0$
=== Corollary ===
== Proof 1 ==
== Proof 2 ==
== Examples ==
== Also known as ==
== Also see ==
* Definition:Contour Integral
== Historical Note ==
== Sources ==
* 
* 
* : $\S 3.1$
* 
* 
* 
Category:Cauchy-Goursat Theorem
Category:Complex Contour Integrals
Cauchy-Goursat Theorem/Also known as

== Cauchy-Goursat Theorem: Also known as ==
The Cauchy-Goursat Theorem is also known as Cauchy's Integral Theorem or the Cauchy Integral Theorem.
Some sources refer to it just as Cauchy's Theorem, but there are a number of theorems named such.
== Sources ==
* 
* 
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Category:Cauchy-Goursat Theorem
Cauchy-Goursat Theorem/Corollary 1

== Corollary ==
Let $f: U \to \C$ be a holomorphic function, where $U \subseteq \C$ is an open set.
Let $C$ be a simple closed contour in $U$.
Let $\Int C \subseteq U$, where $\Int C$ denotes the interior of $C$.
Then:
:$\ds \oint_C \map f z \rd z = 0$
== Proof ==
By Interior of Simply Closed Contour Extends to Simply Connected Domain, there exists a simply connected domain $V$ such that $\Int C \subseteq V \subseteq U$, and $C$ is a contour in $V$.
The result now follows from the main Cauchy-Goursat Theorem.
== Sources ==
* : $\S 3.1$
Category:Cauchy-Goursat Theorem
Cauchy-Goursat Theorem/Examples/z squared over e to the it

== Example of Use of Cauchy-Goursat Theorem ==
Let $\map \gamma t = e^{i t}$.
Let the domain of $\gamma$ be $\hointr 0 {2 \pi}$.
Let $\map f z = z^2$.
Then:
:$\ds \oint_\gamma \map f z \rd z = 0$
== Proof ==
Category:Cauchy-Goursat Theorem
Cauchy-Goursat Theorem/Historical Note

== Historical Note on Cauchy-Goursat Theorem ==
The Cauchy-Goursat Theorem was actually first investigated and proved by , but it was just one of the things that he failed to get round to publishing.
He stated it in a letter of $1811$ to his friend , where he says:
:This is a very beautiful theorem whose fairly simple proof I will give on a suitable occasion. It is connected with other beautiful truths which are concerned with series expansions.
 was, however, the first to publish a proof ca. $1825$, under the assumption that the derivative $f'$ is continuous.
 independently discovered this theorem during his exercise to rebuild the theory of complex analysis from first principles.
 published a general proof in $1884$.
== Sources ==
* : Chapter $\text {A}.25$: Gauss ()
* : Chapter $\text {A}.26$: Cauchy ()
* : Chapter $\text {A}.33$: Weierstrass ()
Category:Historical Notes
Cauchy-Goursat Theorem/Proof 1

== Theorem ==
== Proof ==
Suppose that $C$ is a simple closed staircase contour.
Then $C$ is a concatenation of $n$ directed smooth curves that can be parameterized as line segments, where $n \in \N_{ \ge 4}$.
The image of $C$ is equal to the boundary of a polygon embedded in the complex plane.
Denote this polygon as $P_n$, where $n$ will be equal to the number of sides of $P_n$, and denote the boundary of $P_n$ as $\partial P_n$.
From Boundary of Polygon is Topological Boundary, it follows that $\partial P_n$ is the boundary of $\Int C$, where $\Int C$ denotes the interior of $C$.
From Complex Plane is Homeomorphic to Real Plane, it follows that we can identify the complex plane $\C$ with the real plane $\R^2$ by the homeomorphism $\map \phi {x, y} = x + i y$.
By Interior of Simple Closed Contour is Well-Defined, there exists a Jordan curve $f: \closedint 0 1 \to \R^2$ with $\Img f = \phi^{-1} \sqbrk { \Img C }$.
From Simple Connectedness is Preserved under Homeomorphism, it follows that $\phi^{-1} \sqbrk D$ is simply connected.
From Jordan Curve Characterization of Simply Connected Set, it follows that $\Int f \subseteq \phi^{-1} \sqbrk D$.
It follows that $\Int C = \phi \sqbrk { \Int f } \subseteq D$.
Two Ears Theorem shows that $P_n$ has an ear, a triangle $\triangle_n$ with two sides in common with $P_n$.
The two sides will be represented by the directed smooth curves $C_k , C_{k+1}$ for some $k \in \set { 1, \ldots, n }$, or possibly $C_n , C_1$.
The third side is a chord of $P_n$ that dissects $P_n$ into $\triangle_n$ and another polygon $P_{n-1}$.
The polygon $P_{n-1}$ has $n-1$ sides.
Let $\tilde C$ be the directed smooth curve that can be parameterized as the line segment equal to the chord, with start point equal to the start point of $C_k$, and end point equal to the end point of $C_{k+1}$.
We define two contours with images equal to the boundaries of $\triangle_n$, respectively $P_{n-1}$, as follows:
As $\Int { \triangle_n } \subseteq \Int C \subseteq D$, we can use Goursat's Integral Lemma to get:
We continue by finding an ear $\triangle_{n-1}$ of $P_{n-1}$, which results in a dissection of $P_{n-1}$ into $\triangle_{n-1}$ and $P_{n-2}$, which is a polygon with $n-2$ sides.
From Zero Simple Staircase Integral Condition for Primitive, it follows that $f$ has a primitive defined in $D$.
From Primitive of Function on Connected Domain, it follows that in the general case where $C$ is a closed contour, we have:
:$\ds \oint_C \map f z \rd z = 0$
== Sources ==
* : $\S 3.1$
Category:Cauchy-Goursat Theorem
Cauchy-Hadamard Theorem/Complex Case

== Theorem ==
Let $\xi \in \C$ be a complex number.
Let $\ds \map S z = \sum_{n \mathop = 0}^\infty a_n \paren {z - \xi}^n$ be a (complex) power series about $\xi$.
Then the radius of convergence $R$ of $\map S z$ is given by:
:$\ds \dfrac 1 R = \limsup_{n \mathop \to \infty} \cmod {a_n}^{1/n}$
If:
:$\ds \limsup_{n \mathop \to \infty} \cmod {a_n}^{1/n} = 0$
then the radius of convergence is infinite, and $\map S z$ is absolutely convergent for all $z \in \C$.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* 
* 
Category:Cauchy-Hadamard Theorem
Category:Complex Power Series
Cauchy-Hadamard Theorem/Complex Case/Proof 1

== Theorem ==
== Proof ==
Let $\epsilon \in \R_{>0}$, and let $z \in \C$.
Suppose that $\cmod {z - \xi} = R - \epsilon$.
By definition of radius of convergence, it follows that $S \paren z$ is absolutely convergent.
From the $n$th Root Test:
:$\ds \limsup_{n \mathop \to \infty} \cmod {a_n \paren {z - \xi}^n}^{1/n} \le 1$
By Multiple Rule for Complex Sequences, this inequality can be rearranged to obtain:
:$\ds \limsup_{n \mathop \to \infty} \cmod {a_n}^{1/n} \le \dfrac 1 {\cmod {z - \xi} } = \dfrac 1 {R - \epsilon}$
As $\epsilon > 0$ was arbitrary, it follows that:
:$\ds \limsup_{n \mathop \to \infty} \cmod {a_n}^{1/n} \le \dfrac 1 R$
Now, suppose that $\cmod {z - \xi} = R + \epsilon$.
Then $S \paren z$ is divergent, so the $n$th Root Test shows that:
:$\ds \limsup_{n \mathop \to \infty} \cmod {a_n \paren {z - \xi}^n}^{1/n} \ge 1$
which we can rearrange to obtain:
:$\ds \limsup_{n \mathop \to \infty} \cmod {a_n}^{1/n} \ge \dfrac 1 {\cmod {z - \xi} } = \dfrac 1 {R - \epsilon}$
As $\epsilon > 0$ was arbitrary, it follows that:
:$\ds \limsup_{n \mathop \to \infty} \cmod {a_n}^{1/n} \ge \dfrac 1 R$
Thus:
:$\ds \limsup_{n \mathop \to \infty} \cmod {a_n}^{1/n} = \dfrac 1 R$
Hence the result.
Category:Cauchy-Hadamard Theorem
Cauchy-Hadamard Theorem/Complex Case/Proof 2

== Theorem ==
== Proof ==
Let $L = \limsup \cmod {a_n}^{1/n}$.
We will consider only the case $0 < L < \infty$, as the cases $L = 0$ and $L = \infty$ follow quite simply from this one.
We have that:
:$\forall r \in \closedint 0 {\dfrac 1 L}: L < \dfrac 1 r$
Thus there exists $\epsilon \in \R_{> 0}$ such that:
:$L + \epsilon < \dfrac 1 R$
and so:
:$r < \dfrac 1 {L + \epsilon}$
Let $\tilde r = r \paren {L + \epsilon}$.
Then:
:$0 \le \tilde r < 1$
This means that the geometric series $\ds \sum_{n \mathop = 0}^\infty$ is convergent.
From the properties of the limit superior, we know that:
:$\forall n \gg 1: \cmod {a_n}^{1/n} < L + \epsilon$
Thus:
:$\forall n \gg 1: \cmod {a_n r^n} < \paren {L + \epsilon}^n r^n = \tilde r^n$
This means that the series $\ds \sum_{n \mathop = 0}^\infty \cmod {a_n r^n}$ is also convergent.
Thus $\ds \sum_{n \mathop = 0}^\infty a_n r^n$ is absolutely convergent and therefore convergent.
Thus, by the definition of radius of convergence, we have:
:$r \le R$
which holds for all $r \in \closedint 0 {\dfrac 1 L}$.
Hence:
:$\dfrac 1 L \le R$
Let now $\epsilon \in \openint 0 L$ be fixed.
Because of the properties of the limit superior, we can extract from the sequence of the formula a subsequence such that:
:$\cmod {a_{n_k} }^{1 / n_k} > L - \epsilon$
for all $k \in \N$.
Thus:
:$\cmod {a_{n_k} \cdot \dfrac 1 {\paren {L + \epsilon}^{n_k} } } > 1$
for all $k \in \N$.
So the series:
:$\ds \sum_{n \mathop = 0}^\infty a_n \paren {\frac 1 {L + \epsilon} }^n$
is not convergent.
Thus:
:$R \le \dfrac 1 {L + \epsilon}$
This holds for all $\epsilon \in \R_{>0}$.
So:
:$R \le \dfrac 1 L$
Hence:
:$\dfrac 1 L \le R \le \dfrac 1 L$
which means:
:$R = \dfrac 1 L$
as required.
Category:Cauchy-Hadamard Theorem
Cauchy-Hadamard Theorem/Real Case

== Theorem ==
Let $\xi \in \R$ be a real number.
Let $\ds \map S x = \sum_{n \mathop = 0}^\infty a_n \paren {x - \xi}^n$ be a power series about $\xi$.
Then the radius of convergence $R$ of $S \paren x$ is given by:
:$\ds \frac 1 R = \limsup_{n \mathop \to \infty} \size {a_n}^{1/n}$
If:
:$\ds \frac 1 R = \limsup_{n \mathop \to \infty} \size {a_n}^{1/n} = 0$
then the radius of convergence is infinite and therefore the interval of convergence is $\R$.
== Proof ==
From the $n$th root test, $S \paren x$ is convergent if $\ds \limsup_{n \mathop \to \infty} \size {a_n \paren {x - \xi}^n}^{1/n} < 1$.
Thus:
The result follows from the definition of radius of convergence.
== Sources ==
* : $\S 15.2$
Category:Cauchy-Hadamard Theorem
Category:Real Power Series