nyuuzyou/wikis huggingface dataset:
Sequential Continuity is Equivalent to Continuity in the Reals/Sufficient Condition

== Theorem ==
Let $A \subseteq \R$ be a subset of the real numbers.
Let $c \in A$.
Let $f: A \to \R$ be a real function.
Then if $f$ is continuous at $c$:
:for each sequence $\sequence {x_n}$ in $A$ that converges to $c$, the sequence $\sequence {\map f {x_n} }$ converges to $\map f c$.
== Proof ==
It suffices to show that if $f$ is discontinuous at $c$:
:there exists a real sequence $\sequence {x_n}$ in $A$ such that $\sequence {x_n}$ converges to $c$ but $\sequence {\map f {x_n} }$ does not converge to $\map f c$.
As $f$ is discontinuous, there exists some $\varepsilon > 0$ such that for all $\delta > 0$:
:there exists $x \in A$ with $\size {x - c} < \delta$ such that $\size {\map f x - \map f c} \ge \varepsilon$.
Using this property, we can construct a sequence $\sequence {x_n}$ as follows:
:for each $n \in \N$, pick $x_n \in A$ such that $\size {x_n - c} \le \dfrac 1 n$ and $\size {\map f {x_n} - \map f c} \ge \varepsilon$
Note that since: 
:$\ds \lim_{n \mathop \to \infty} \frac 1 n = 0$
We have by the Squeeze Theorem for Real Sequences: 
:$\ds \lim_{n \mathop \to \infty} \size {x_n - c} = 0$
so $\sequence {x_n}$ converges to $c$.
However $\sequence {\map f {x_n} }$ cannot converge to $\map f c$ since:
:$\size {\map f {x_n} - \map f c} \ge \varepsilon > 0$
for all $n \in \N$.
Therefore, our $\sequence {x_n}$ satisfies our original demand.
Category:Sequential Continuity is Equivalent to Continuity in the Reals
Sequentially Compact Metric Space is Compact/Proof 1

== Theorem ==
== Proof ==
Let $\struct {A, d}$ be a sequentially compact metric space.
From Sequentially Compact Metric Space is Lindelöf, every open cover of $A$ has a countable subcover.
Let $C$ be an open cover $A$.
Extract from it a countable subcover $\set {U_1, U_2, \ldots}$.
 there exists no finite subcover of $C$.
Then, for all $n \in \N_{\ge 1}$, the set $\set {U_1, \ldots, U_n}$ does not cover $A$.
Hence it is possible to choose $x_n \in A$ such that:
:$x_n \notin U_1 \cup \cdots \cup U_n$.
Thus we construct an infinite sequence $\sequence {x_n}_{n \mathop \ge 1}$ of points of $A$.
By assumption $A$ is sequentially compact.
Thus $\sequence {x_n}_{n \mathop \ge 1}$ has a subsequence which converges to some $x \in A$.
But because the $U_i: i \ge 1$ forms a cover for $A$, there exists some $U_m$ such that $x \in U_m$.
Hence by Limit of Sequence is Accumulation Point, there is an infinite number of terms in the sequence $\sequence {x_i}$ which are contained in $U_m$.
But from the method of construction of $\sequence {x_i}$, each $U_n$ can contain only points $x_i$ with $i < n$.
That is, each $U_n$ can only contain a finite number of the terms of $\sequence {x_i}$.
This is a contradiction.
Thus the supposition that there exists no finite subcover of $C$ was false.
Hence the result.
Category:Sequentially Compact Metric Space is Compact
Sequentially Compact Metric Space is Compact/Proof 2

== Theorem ==
== Proof ==
Let $M = \struct {A, d}$ be a sequentially compact metric space.
Let $\UU$ be any open cover of $M$.
By Lebesgue's Number Lemma, there exists a Lebesgue number for $\UU$.
Let $\set {x_1, x_2, \ldots, x_n}$ be a finite $\epsilon$-net for $M$, where $\epsilon$ is this same Lebesgue number.
This exists by Sequentially Compact Metric Space is Totally Bounded.
Let $\map {B_\epsilon} {x_i}$ be the open $\epsilon$-ball of $x_i$.
By definition of Lebesgue number, $\map {B_\epsilon} {x_i}$ is contained in some $U_i \in \UU$.
Since:
:$\ds M \subseteq \bigcup_{i \mathop = 1}^n \map {B_\epsilon} {x_i} \subseteq \bigcup_{i \mathop = 1}^n U_i$
we have a finite subcover $\set {U_1, U_2, \ldots, U_n}$ of $\UU$ for $M$.
Hence the result.
== Sources ==
* : $7.2$: Sequential compactness: Theorem $7.2.13$
Category:Sequentially Compact Metric Space is Compact
Sequentially Compact Metric Space is Complete

== Theorem ==
Let $M = \struct {A, d}$ be a metric space which is sequentially compact.
Then $M$ is complete.
== Proof ==
Let $\sequence {x_n}_{n \mathop \in \N}$ be a Cauchy sequence in $A$.
As $M$ is sequentially compact, $\sequence {x_n}$ has a convergent subsequence.
By Convergent Subsequence of Cauchy Sequence in Metric Space, this implies that the entire sequence $\sequence {x_n}$ is convergent.
Hence, $M$ is complete.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $5$: Metric Spaces: Complete Metric Spaces
Category:Sequentially Compact Spaces
Category:Complete Metric Spaces
Sequentially Compact Metric Space is Lindelöf

== Theorem ==
Let $M = \struct {X, d}$ be a sequentially compact metric space.
Then $M$ is also a Lindelöf space.
That is, from every open cover of $M$, it is possible to extract a countable subcover.
== Proof ==
Take any open cover $C$ of $M$.
We need to find a countable subset of $C$ which still covers $X$.
We have that a Sequentially Compact Metric Space is Second-Countable.
Thus, by definition, the topology of $M$ has a countable basis.
Let $\BB$ be a countable basis for the topology induced by $d$ of $M$.
Let $x \in X$.
As $C$ covers $X$:
:$\exists U_x \in C: x \in U_x$
As $\BB$ is a basis:
:$\exists B_x \in \BB: x \in B_x \subseteq U_x$
Thus:
: $(1): \quad \forall x \in X: \exists B_x \in \BB: x \in B_x$
Consider the set $\Sigma := \set {B_x: x \in X}$.
$\Sigma$ is a subset of $\BB$.
Hence $\Sigma$ is countable.
As $\Sigma$ contains every $x \in X$ from $(1)$, $\Sigma$ covers $X$.
By construction of $\Sigma$, every open set in $\Sigma$ is contained in some $U \in C$.
For each open set $B \in \Sigma$, choose one $U_B \in C$ such that $B \subseteq U_B$.
Let $\UU$ be the set defined as:
:$\UU = \set {U_B: B \in \Sigma}$
As $\UU$ does not have more sets than $\Sigma$, $\UU$ is countable.
We have that:
: $\forall B \in \Sigma: B \subseteq U_B$
Thus it follows that $\UU$ covers $X$.
As:
:$\forall U_B \in \UU: U_B \in C$
$U$ is a countable subcover of $C$.
Thus a countable subcover has been obtained from $C$.
As $C$ is arbitrary, it follows that $M$ is a Lindelöf space.
== Also see ==
* Sequentially Compact Metric Space is Compact 
Category:Lindelöf Spaces
Category:Metric Spaces
Category:Sequentially Compact Spaces
Sequentially Compact Metric Space is Totally Bounded

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Let $M$ be sequentially compact.
Then $M$ is totally bounded.
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Sequentially Compact Metric Space is Compact
* Compact Metric Space is Totally Bounded
Category:Metric Spaces
Category:Sequentially Compact Spaces
Category:Totally Bounded Metric Spaces
Category:Sequentially Compact Metric Space is Totally Bounded
Sequentially Compact Metric Space is Totally Bounded/Proof 2

== Theorem ==
== Proof ==
We have that $M$ is a sequentially compact space.
So every infinite sequence in $M$ has a subsequence which converges to a point in $A$.
Let $\epsilon \in \R_{>0}$ be a strictly positive real number.
By definition, $M$ is totally bounded there exists a finite $\epsilon$-net for $M$.
 there exists no finite $\epsilon$-net for $M$.
The aim is to construct an infinite sequence $\sequence {x_n} _{n \mathop \in \N}$ in $A$ that has no convergent subsequence.
Suppose that $x_0, x_1, \ldots, x_r \in A$ have been defined such that:
:$\forall m, n \in \set {0, 1, \ldots, r} : m \ne n \implies \map d {x_m, x_n} \ge \epsilon$
That is, any two distinct elements of $\set {x_0, x_1, \ldots, x_r}$ are at least $\epsilon$ apart.
We have that there exists no finite $\epsilon$-net for $M$.
Specifically, $\set {x_0, x_1, \ldots, x_r}$ is therefore not an $\epsilon$-net for $M$.
So, by definition of a $\epsilon$-net:
:$\ds A \nsubseteq \bigcup_{i \mathop = 0}^r \map {B_\epsilon} {x_i}$
where $\map {B_\epsilon} {x_i}$ denotes the open $\epsilon$-ball of $x_i$ in $M$.
Thus there must exist $x_{r + 1} \in A$ such that:
:$\ds x_{r + 1} \notin \bigcup_{i \mathop = 0}^r \map {B_\epsilon} {x_i}$
That is:
:$\exists x_{r + 1} \in A: \forall m, n \in \set {0, 1, \ldots, r, r + 1} : m \ne n \implies \map d {x_m, x_n} \ge \epsilon$
Thus, by induction, the infinite sequence $\sequence {x_n}$ in $A$ has been constructed such that:
:$\forall m, n \in \N: m \ne n \implies \map d {x_m, x_n} \ge \epsilon$
Thus $\sequence {x_n}$ has no Cauchy subsequence.
From Convergent Sequence in Metric Space is Cauchy Sequence, $\sequence {x_n}$ has no convergent subsequence either. 
Thus, by definition, $M$ is not sequentially compact.
But $M$ is sequentially compact.
Hence by Proof by Contradiction there exists a finite $\epsilon$-net for $M$.
Hence $M$ is totally bounded.
== Also see ==
* Sequentially Compact Metric Space is Compact
* Compact Metric Space is Totally Bounded
== Sources ==
* : $7.2$: Sequential compactness: Proposition $7.2.9$
Category:Sequentially Compact Metric Space is Totally Bounded
Category:Proofs by Contraposition
Sequentially Compact Metric Subspace is Sequentially Compact in Itself iff Closed

== Theorem ==
Let $M$ be a metric space.
Let $C \subseteq M$ be a subspace of $M$ which is sequentially compact in $M$.
Then $C$ is sequentially compact in itself $C$ is closed in $M$.
== Proof ==
Follows directly from Closure of Subset of Metric Space by Convergent Sequence.
Category:Metric Subspaces
Category:Sequentially Compact Spaces
Sequentially Compact Space is Countably Compact

== Theorem ==
A sequentially compact topological space is also countably compact.
== Proof ==
Let $T = \struct {S, \tau}$ be a sequentially compact topological space.
By definition of countably compact space, it suffices to show that every infinite sequence in $S$ has an accumulation point in $S$.
By the definition of a sequentially compact space, every infinite sequence in $S$ has a subsequence which converges in $S$.
The result follows from Limit of Sequence is Accumulation Point.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $3$: Compactness: Global Compactness Properties
Category:Sequentially Compact Spaces
Category:Countably Compact Spaces
Serial Relation is not Null

== Theorem ==
Let $S$ be a set such that $S \ne \O$.
Let $\RR$ be a serial relation on $S$.
Then $\RR$ is not a null relation.
== Proof ==
As $S$ is non-empty set:
:$\exists x: x \in S$
As $\RR$ be a serial relation on $S$:
:$\exists y \in S: \tuple {x, y} \in \RR$
That is:
:$\RR \ne \O$
Hence the result by definition of null relation.
== Sources ==
* : Chapter $4$: The Predicate Calculus $2$: $5$ Properties of Relations: Exercise $2 \ \text{(b)}$
Category:Serial Relations
Category:Null Relation
Series Expansion for Pi Cosecant of Pi Lambda

== Theorem ==
Let $\lambda \in \R \setminus \Z$ be a real number which is not an integer.
Then:
:$\ds \pi \csc \pi \lambda = \sum_{n \mathop = 1}^\infty \paren {-1}^n \paren {\frac 1 {n + \lambda} + \frac 1 {n - 1 - \lambda} }$
== Proof ==
Let $\map f x$ be the real function defined on $\openint 0 \pi$ as:
:$\map f x = \cos \lambda x$
From Half-Range Fourier Cosine Series: $\cos \lambda x$ over $\openint 0 \pi$ its Fourier series can be expressed as:
:$\ds \cos \lambda x \sim \frac {2 \lambda \sin \lambda \pi} \pi \paren {\frac 1 {2 \lambda^2} + \sum_{n \mathop = 1}^\infty \paren {-1}^n \frac {\cos n x} {\lambda^2 - n^2} }$
Because of the nature of this expansion, we have that:
:$\map f {0^+} = \map f {0^-}$
and so the expansion holds for $x = 0$.
So, setting $x = 0$:
This expression is fine as is, but to obtain the form we set out to prove, we observe:
and thus:
The result follows.
== Sources ==
* : Exercises on Chapter $\text I$: $5. \, \text{(ii)}$
Category:Cosecant Function
Series Expansion for Pi Cotangent of Pi Lambda

== Theorem ==
Let $\lambda \in \R \setminus \Z$ be a real number which is not an integer.
Then:
:$\ds \pi \cot \pi \lambda = \frac 1 \lambda + \sum_{n \mathop = 1}^\infty \frac {2 \lambda} {\lambda^2 - n^2}$
== Proof ==
Let $\map f x$ be the real function defined on $\openint 0 \pi$ as:
:$\map f x = \cos \lambda x$
From Half-Range Fourier Cosine Series: $\cos \lambda x$ over $\openint 0 \pi$ its Fourier series can be expressed as:
:$\ds \cos \lambda x \sim \frac {2 \lambda \sin \lambda \pi} \pi \paren {\frac 1 {2 \lambda^2} + \sum_{n \mathop = 1}^\infty \paren {-1}^n \frac {\cos n x} {\lambda^2 - n^2} }$
Because of the nature of this expansion, we have that:
:$\map f \pi = \map f {-\pi}$
and so the expansion holds for $x = \pi$.
Also note that because $\lambda$ is not an integer, $\sin \lambda \pi \ne 0$ and so $\cot \pi \lambda$ is defined.
So, setting $x = \pi$:
== Sources ==
* : Exercises on Chapter $\text I$: $5. \, \text{(i)}$
Category:Cotangent Function
Series Expansion for Pi over 8 Root 2

== Theorem ==
:$\ds \frac \pi {8 \sqrt 2} = \sum_{n \mathop = 1}^\infty \paren {-1}^{n - 1} \frac {2 n - 1} {\paren {4 n - 1} \paren {4 n - 3} }$
== Proof ==
Let $\map f x$ be the real function defined on $\openint {-\pi} \pi$ as:
:$\map f x = \begin {cases}
\cos x & : -\pi < x < 0 \\
-\cos x & : 0 < x < \pi \end {cases}$
From Fourier Series: $\cos x$ over $\openint {-\pi} 0$, $-\cos x$ over $\openint 0 \pi$, we have:
:$\ds \map f x \sim -\frac 8 \pi \sum_{r \mathop = 1}^\infty \frac {r \sin 2 r x} {4 r^2 - 1}$
Setting $x = \dfrac \pi 4$, we have:
When $r$ is even, $\dfrac {r \pi} 2$ is an integer multiple of $\pi$.
Hence, in this case, from Sine of Multiple of Pi:
:$\sin \dfrac {r \pi} 2 = 0$
When $r$ is odd it can be expressed as $r = 2 n - 1$ for $n \ge 1$.
Hence we have:
== Sources ==
* : Exercises on Chapter $\text I$: $3$.
Category:Formulas for Pi
Series Expansion for Pi over Root 2

== Theorem ==
:$\ds \frac \pi {\sqrt 2} = \sum_{r \mathop = 1}^\infty \paren {-1}^{r - 1} \frac {r - \frac 1 2} {r^2 - r + \frac 3 {16} }$
== Proof 1 ==
== Proof 2 ==
Category:Formulas for Pi
Category:Series Expansion for Pi over Root 2
Series Expansion for Pi over Root 2/Mistake

== Source Work ==
* : Exercises on Chapter $\text I$: $2$.
== Mistake ==
:Deduce that
::$\ds \sum_{n \mathop = 1}^\infty \paren {-1}^{r - 1} \frac {r - \frac 1 2} {r^2 - r + \frac 3 {16} } = \frac \pi {\sqrt 2}$
== Correction==
That lower index should of course be $r$:
::$\ds \sum_{r \mathop = 1}^\infty \paren {-1}^{r - 1} \frac {r - \frac 1 2} {r^2 - r + \frac 3 {16} } = \frac \pi {\sqrt 2}$
otherwise it makes no sense.
== Sources ==
* : Exercises on Chapter $\text I$: $2$.
Category:Mistakes/Fourier Series/Sneddon
Series Expansion for Pi over Root 2/Proof 1

== Theorem ==
== Proof ==
Let $\map f x$ be the real function defined on $\openint 0 {2 \pi}$ as:
:$\map f x = \begin{cases} \sin \dfrac x 2 & : 0 \le x < \pi \\ -\sin \dfrac x 2 & : \pi < x \le 2 \pi \end{cases}$
From Half-Range Fourier Sine Series: $\sin \dfrac x 2$ over $\closedint 0 \pi$, $-\sin \dfrac x 2$ over $\closedint 0 {2 \pi}$, we have:
:$\map f x \sim \ds \frac 8 \pi \sum_{n \mathop = 1}^\infty \paren {-1}^{n - 1} \frac {n \sin n x} {4 n^2 - 1}$
Setting $x = \dfrac {\pi} 2$, we have:
When $n$ is even, $\dfrac {n \pi} 2$ is an integer multiple of $\pi$.
Hence, in this case, from Sine of Multiple of Pi:
:$\sin \dfrac {n \pi} 2 = 0$
When $n$ is odd it can be expressed as $n = 2 r - 1$ for $r \ge 1$.
Hence we have:
== Sources ==
* : Exercises on Chapter $\text I$: $2$.
Category:Series Expansion for Pi over Root 2
Series Expansion of Bessel Function of the First Kind

== Theorem ==
Let $n \in \Z_{\ge 0}$ be a (strictly) non-negative integer.
Let $\map {J_n} x$ denote the Bessel function of the first kind of order $n$.
Then:
=== Negative Index ===
== Proof ==
We employ Frobenius's method to find the solutions to the Bessel's Equation:
:$x^2 \dfrac {\d^2 y} {\d x^2} + x \dfrac {\d y} {\d x} + \paren {x^2 - n^2} y = 0$
for $n \ge 0$, in the form:
:$\ds \map y x = \sum_{k \mathop = 0}^\infty A_k x^{k + r}$
defined on $x > 0$, for some constants $r, A_i$, with $A_0 \ne 0$, which are to be determined.
Differentiating the expression $x$:
Substituting $y, y', y' '$ into Bessel's Equation:
Comparing the constant term on both sides:
Take $r = n$.
Comparing the rest of the coefficients:
From the recurrence relation above, we see that $A_k = 0$ for odd $k$, and:
Substituting this result to our original equation: 
== Also see ==
* Bessel Function of the First Kind of Negative Integer Order for when $n \in \set {-1, -2, -3, \ldots}$
== Sources ==
* : Chapter $1$: The Laplace Transform: Some Special Functions: $\text {II}$. Bessel functions
* : $\S 24$: Bessel Functions: Bessel Function of the First Kind of Order $n$: $24.2$
* 
* 
* : $\S 27$: Bessel Functions: Bessel Function of the First Kind of Order $n$: $27.2.$
* 
Category:Bessel Functions
Series Expansion of Bessel Function of the First Kind/Negative Index

== Theorem ==
Let $n \in \Z_{\ge 0}$ be a (strictly) non-negative integer.
Let $\map {J_n} x$ denote the Bessel function of the first kind of order $n$.
Then:
== Proof ==
From Series Expansion of Bessel Function of the First Kind:
The result follows by substituting $-n$ for $n$ in $1$ and simplifying.
== Also see ==
* Bessel Function of the First Kind of Negative Integer Order for when $n \in \set {-1, -2, -3, \ldots}$
== Sources ==
* : $\S 24$: Bessel Functions: Bessel Function of the First Kind of Order $n$: $24.3$
* : $\S 27$: Bessel Functions: Bessel Function of the First Kind of Order $n$: $27.3.$
Category:Bessel Functions
Series Expansion of Function over Complete Orthonormal Set

== Theorem ==
Let $\map f x$ be a real function defined over the interval $\openint a b$.
Let $\map f x$ be able to be expressed in terms of a complete orthonormal set of real functions $S := \family {\map {\phi_i} x}_{i \mathop \in I}$ for some indexing set $I$:
:$\map f x = \ds \sum_{i \mathop \in I} a_i \map {\phi_i} x$
Then the coefficients $\family {a_i}_{i \mathop \in I}$ can be determined as:
:$\forall i \in I: a_i = \ds \int_a^b \map f x \map {\phi_i} x \rd x$
== Proof ==
== Sources ==
* : Chapter One: $\S 8$. Orthonormal Sets of Functions: $(3)$
Category:Orthonormal Sets
Series Expansion of Matrix Exponential

== Theorem ==
Let $\mathbf A$ be a square matrix.
Let $t \in \R$ be a real number.
Let $e^{\mathbf A t}$ denote the matrix exponential of $\mathbf A$.
Then:
:$\ds e^{\mathbf A t} = \sum_{n \mathop = 0}^\infty \frac {t^n} {n!} \mathbf A^n$
== Proof ==
Category:Matrix Exponential
Series Law for Extremal Length

== Theorem ==
Let $X$ be a Riemann surface.
Let $\Gamma_1$, $\Gamma_2$ and $\Gamma$ be families of rectifiable curves (or, more generally, families of unions of rectifiable curves) on $X$.
Let every $\gamma \in \Gamma$ contain a $\gamma_1 \in \Gamma_1$ and a $\gamma_2 \in \Gamma_2$ such that $\gamma_1 \cap \gamma_2 = \O$.
Then the extremal lengths of $\Gamma_1$, $\Gamma_2$ and $\Gamma$ satisfy:
:$\map \lambda \Gamma \ge \map \lambda {\Gamma_1} + \map \lambda {\Gamma_2}$
== Proof ==
Let $\rho_1 = \map {\rho_1} z \size {\d z}$ and $\rho_2 = \map {\rho_2} z \size {\d z}$ be conformal metrics as in the definition of extremal length.
It can be assumed that these are normalized:
:$\map A {\rho_j} = \map L {\Gamma_j, \rho_j}$ for $j \in \set {1, 2}$. 
We define another metric $\rho = \map \rho z \size {\d z}$ by:
:$\map \rho z := \map \max {\map {\rho_1} z, \map {\rho_2} z}$
Note that this is a well-defined metric.
By definition, the area form $\map {\rho^2} z \size {\d z}$ satisfies:
:$\map {\rho^2} z \size {\d z}^2 = \map \max {\map {\rho_1} z^2, \map {\rho_2} z^2} \size {\d z}^2 \le \paren {\map {\rho_1} z^2 + \map {\rho_2} z^2} \size {\d z}^2$
Hence:
:$\map A \rho \le \map A {\rho_1} + \map A {\rho_2} = \map L {\Gamma_1, \rho_1} + \map L {\Gamma_2, \rho_2}$
On the other hand, let $\gamma \in \Gamma$.
Let $\gamma_1$, $\gamma_2$ be as in the assumption.
Then:
Thus 
:$\map L {\Gamma, \rho} \ge \map L {\Gamma_1, \rho_1} + \map L {\Gamma_2, \rho_2}$
Combining this with the inequality for the area:
:$\dfrac {\map L {\Gamma, \rho}^2} {\map A \rho} \ge \dfrac {\paren {\map L {\Gamma_1, \rho_1} + \map L {\Gamma_2, \rho_2} }^2} {\map L {\Gamma_1, \rho_1} + \map L {\Gamma_2, \rho_2} } = \map L {\Gamma_1, \rho_1} + \map L {\Gamma_2, \rho_2}$
Taking the supremum over all metrics $\rho_1$ and $\rho_2$ as above:
:$\map L \Gamma \ge \map L {\Gamma_1} + \map L {\Gamma_2}$
as claimed.
== Also known as ==
The series law and the parallel law are also referred to collectively as the composition laws of extremal length.
== Sources ==
* 
Category:Geometric Function Theory
Series Law for Extremal Length/Rho is Well Defined

== Lemma for Series Law for Extremal Length ==
== Proof ==
To see that $\rho$ is a well-defined metric, we need to check that it transforms correctly when changing local coordinates.
Let $z = \map z t$ and $w = \map w t$ be charts on the Riemann surface $X$.
Let $\map {\rho_1^z} t$ and $\map {\rho_1^w} t$ be the coefficient functions when $\rho_1$ is expressed in the local coordinates $z$ and $w$, respectively.
We use the analogous notation for $\rho_2$ and $\rho$.
Since $\rho_j$ is a metric for $j \in \set {1, 2}$, we have:
:$\map {\rho_j^w} t = \map {\rho_j^z} t \cdot \size {\dfrac {\d z} {\d w} }$
where $\dfrac {\d z} {\d w}$ denotes, the derivative of the coordinate change $z \circ w^{-1}$.
Thus we have:
This means that $\rho$ transforms correctly and is a metric, as desired.
Series of Measures is Measure

== Theorem ==
Let $\struct {X, \Sigma}$ be a measurable space.
Let $\sequence {\mu_n}_{n \mathop \in \N}$ be a sequence of measures on $\left({X, \Sigma}\right)$.
Let $\sequence {a_n}_{n \mathop \in \N} \subseteq \R_{\ge 0}$ be a sequence of positive real numbers.
Then the series of measures $\mu: \Sigma \to \overline \R$, defined by:
:$\ds \map \mu E := \sum_{n \mathop \in \N} a_n \map {\mu_n} E$
is also a measure on $\struct {X, \Sigma}$.
== Proof ==
Let us verify the conditions for a measure in turn.
Let $E \in \Sigma$.
Then:
:$\forall n \in \N: \map {\mu_n} E \ge 0$
and so:
:$a_n \map {\mu_n} E \ge 0$
Therefore, by Series of Positive Real Numbers has Positive Limit:
:$\ds \map \mu E = \sum_{n \mathop \in \N} a_n \map {\mu_n} E \ge 0$
For every $n \in \N$, also $\map {\mu_n} \O = 0$.
Therefore, it immediately follows that:
:$\ds \map \mu \O = \sum_{n \mathop \in \N} a_n \map {\mu_n} \O = \sum_{n \mathop \in \N} 0 = 0$
Finally, let $\sequence {E_n}_{n \mathop \in \N}$ be a sequence of pairwise disjoint sets in $\Sigma$.
Then:
Therefore, having verified all three axioms, $\mu$ is a measure.
== Sources ==
* : $\S 4$: Problem $6 \ \text{(ii)}$
Category:Measures
Category:Series of Measures
Series of Positive Measurable Functions is Positive Measurable Function

== Theorem ==
Let $\struct {X, \Sigma}$ be a measurable space.
Let $\sequence {f_n}_{n \mathop \in \N} \in \MM_{\overline \R}^+$, $f_n: X \to \overline \R$ be a sequence of positive measurable functions.
Let $\ds \sum_{n \mathop \in \N} f_n: X \to \overline \R$ be the pointwise series of the $f_n$.
Then $\ds \sum_{n \mathop \in \N} f_n$ is also a positive measurable function.
== Proof ==
== Sources ==
* : $9.9$
Category:Measure Theory
Series of Power over Factorial Converges

== Theorem ==
The series $\ds \sum_{n \mathop = 0}^\infty \frac {x^n} {n!}$ converges for all real values of $x$.
== Proof ==
If $x = 0$ the result is trivially true as:
:$\forall n \ge 1: \dfrac {0^n} {n!} = 0$
If $x \ne 0$ we have:
:$\size {\dfrac {\paren {\dfrac {x^{n + 1} } {\paren {n + 1}!} } } {\paren {\dfrac {x^n} {n!} } } } = \dfrac {\size x} {n + 1} \to 0$
as $n \to \infty$.
This follows from the results:
:Sequence of Powers of Reciprocals is Null Sequence, where $\dfrac 1 n \to 0$ as $n \to \infty$
:The Squeeze Theorem for Real Sequences, as $\dfrac 1 {n + 1} < \dfrac 1 n$
:The Multiple Rule for Real Sequences, putting $\lambda = \size x$.
Hence by the Ratio Test: $\ds \sum_{n \mathop = 0}^\infty \frac {x^n} {n!}$ converges.
Alternatively, the Comparison Test could be used but this is more cumbersome in this instance.
Another alternative is to view this as an example of Radius of Convergence of Power Series over Factorial setting $\xi = 0$.
== Also see ==
* Equivalence of Definitions of Exponential Function, where it is shown that this series converges to the exponential function.
== Sources ==
* : $\S 6.19$
Category:Series