avewright/proofwiki-math huggingface dataset:
Theorem: Approximate Motion of Simple Pendulum

Consider a simple pendulum consisting of a bob whose mass is $m$, at the end of a rod of negligible mass of length $a$.
Let the bob be pulled to one side so that the rod is at a small angle $\alpha$ (less than about $10 \degrees$ or $15 \degrees$) from the vertical and then released.
Let $T$ be the period of the pendulum, that is, the time through which the bob takes to travel from one end of its path to the other, and back again.
Then:
:$T \approx 2 \pi \sqrt {\dfrac a g}$
where $g$ is the Acceleration Due to Gravity.

Proof:
At a time $t$, let:
:the rod be at an angle $\theta$ to the the vertical
:the bob be travelling at a speed $v$
:the displacement of the bob from where it is when the rod is vertical, along its line of travel, be $s$.
:350px
From Motion of Simple Pendulum, the equation of motion of the bob is given by:
:$\dfrac {a^2} 2 \paren {\dfrac {\d \theta} {\d t} }^2 = g a \paren {a \cos \theta - a \cos \alpha}$
Differentiating {{WRT|Differentiation}} $t$ and simplifying yields:
:$a \dfrac {\d^2 \theta} {\d t^2} = -g \sin \theta$
For small angles, $\sin \theta \approx \theta$ and so:
:$\dfrac {\d^2 \theta} {\d t^2} + \dfrac g a \theta = 0$
This is an instance of a $2$nd order linear ODE.
From Linear Second Order ODE: $y' ' + k^2 y = 0$, this has the solution:
:$\theta = c_1 \sin \sqrt {\dfrac g a} t + c_1 \cos \sqrt {\dfrac g a} t$
We have at $t = 0$ that $\theta = \alpha$ and $\dfrac {\d \theta} {\d t} = 0$.
This yields:
:$\theta = \alpha \cos \sqrt {\dfrac g a} t$
and so the period of this simple pendulum is:
:$T = 2 \pi \sqrt {\dfrac a g}$
{{qed}}
Theorem: Space is Separable iff Density not greater than Aleph Zero

Let $T$ be a topological space.
Then:
:$T$ is separable {{iff}} $\map d T \le \aleph_0$
where
:$\map d T$ denotes the density of $T$,
:$\aleph$ denotes the aleph mapping.

Proof:
:$T$ is separable
{{iff}}:
:there exists a countable subset of $T$ which is dense by definition of separable space
{{iff}}:
:there exists a subset $A$ of $T$ such that $A$ is dense and exists an injection $A \to \N$ by definition of countable set
{{iff}}:
:there exists a subset $A$ of $T$ such that $A$ is dense and $\card A \le \card \N$ by Injection iff Cardinal Inequality
{{iff}}:
:there exists a subset $A$ of $T$ such that $A$ is dense and $\card A \le \aleph_0$ by Aleph Zero equals Cardinality of Naturals
{{iff}}:
:$\map d T \le \aleph_0$ by definition of density
where $\card A$ denotes the cardinality of $A$.
{{qed}}
Theorem: Boundary of Union of Separated Sets equals Union of Boundaries

Let $T$ be a topological space.
Let $A, B$ be subsets of $T$.
Let $A$ and $B$ are separated.
Then:
:$\map \partial {A \cup B} = \partial A \cup \partial B$
where:
:$\partial A$ denotes the boundary of $A$
:$A \cup B$ denotes the union of $A$ and $B$.

Proof:
By definition of separated sets:
:$(1): \quad A^- \cap B = A \cap B^- = \O$
By Separated Sets are Disjoint:
:$A \cap B = \O$
{{begin-eqn}}
{{eqn | l = \partial A \cup \partial B
      | r = \map \partial {A \cup B} \cup \map \partial {A \cap B} \cup \paren {\partial A \cap \partial B}
      | c = Union of Boundaries
}}
{{eqn | r = \map \partial {A \cup B}\cup \O \cup \paren {\partial A \cap \partial B}
      | c = Boundary of Empty Set is Empty
}}
{{eqn | r = \map \partial {A \cup B} \cup \paren {\partial A \cap \partial B}
      | c = Union with Empty Set
}}
{{end-eqn}}
We will prove that:
:$\partial A \cap \partial B \subseteq \map \partial {A \cup B}$
Let $x \in \partial A \cap \partial B$.
Then by definition of intersection:
:$x \in \partial A \land x \in \partial B$
Hence by Boundary is Intersection of Closure with Closure of Complement:
:$x \in A^- \cap \paren {\relcomp T A}^- \land x \in B^- \cap \paren {\relcomp T B}^-$
where:
:$A^-$ denotes the closure of $A$
:$\relcomp T A = T \setminus A$ denotes the relative complement of $A$ in $T$.
Then by definition of intersection:
:$x \in A^- \land x \in B^-$
Therefore by definition of empty set, intersection, and $(1)$:
:$x \notin B \land x \notin A$
Then by definition of union:
:$x \notin A \cup B$
By definition of relative complement:
:$x \in \relcomp T {A \cup B}$
By definition of closure:
:$\relcomp T {A \cup B} \subseteq \paren {\relcomp T {A \cup B} }^-$
Then by definition of subset:
:$x \in \paren {\relcomp T {A \cup B} }^-$
By Closure of Finite Union equals Union of Closures:
:$A^- \cup B^- = \paren {A \cup B}^-$
By definition of union:
:$x \in \paren {A \cup B}^-$
Then by definition of intersection:
:$x \in \paren {A \cup B}^- \cap \paren {\relcomp T {A \cup B} }^-$
Thus  by Boundary is Intersection of Closure with Closure of Complement:
:$x \in \map \partial {A \cup B}$
This ends the proof of inclusion.
Thus by Union with Superset is Superset the result:
:$\partial A \cup \partial B = \map \partial {A \cup B}$
{{qed}}
Theorem: Boundary of Empty Set is Empty

Let $T$ be a topological space.
Then: 
:$\partial_T \O = \O$
where $\partial_T \O$ denotes the boundary in topology $T$ of $\O$.

Proof:
By Boundary is Intersection of Closure with Closure of Complement:
:$\partial_T \O = \O^- \cap \relcomp T \O^-$
where $\O^-$ denotes the closure of $\O$.
By Closure of Empty Set is Empty Set:
:$\O^- = \O$
Thus the result follows by Intersection with Empty Set.
{{qed}}
Theorem: Boundary of Empty Set is Empty

Let $T$ be a topological space.
Then: 
:$\partial_T \O = \O$
where $\partial_T \O$ denotes the boundary in topology $T$ of $\O$.

Proof:
From Open and Closed Sets in Topological Space, $\O$ is clopen in $T$.
The result follows from Set is Clopen iff Boundary is Empty.
{{qed}}
Theorem: Terminal Speed of Body under Fall Retarded Proportional to Square of Speed

Let $B$ be a body falling in a gravitational field.
Let $B$ be falling through a medium which exerts a resisting force of magnitude $k v^2$ upon $B$ which is proportional to the square of the speed of $B$ relative to the medium.
Then the terminal speed of $B$ is given by:
:$v =  \sqrt {\dfrac {g m} k}$

Proof:
Let $B$ start from rest.
The differential equation governing the motion of $B$ is given by:
:$m \dfrac {\d^2 \mathbf s} {\d t^2} = m \mathbf g - k \paren {\dfrac {\d \mathbf s} {\d t} }^2$
Dividing through by $m$ and setting $c = \dfrac k m$ gives:
:$\dfrac {\d^2 \mathbf s} {\d t^2} = \mathbf g - c \paren {\dfrac {\d \mathbf s} {\d t} }^2$
By definition of velocity:
:$\dfrac {\d \mathbf v} {\d t} = \mathbf g - c \mathbf v^2$
for some constant $c$.
and so taking magnitudes of the vector quantities:
{{begin-eqn}}
{{eqn | l = \int \dfrac {\d v} {g - c v^2}
      | r = \int \rd t
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \dfrac 1 {2 c} \sqrt {\frac c g} \ln \paren {\frac {\sqrt {\frac c g} + v} {\sqrt {\frac c g} - v} }
      | r = t + c_1
      | c = Primitive of $\dfrac 1 {a^2 - x^2}$: Logarithm Form
}}
{{end-eqn}}
We have $v = 0$ when $t = 0$ which leads to $\ln 1 = 0 + c_1$ and thus $c_1 = 0$.
Hence:
{{begin-eqn}}
{{eqn | l = \ln \paren {\frac {\sqrt {\frac c g} + v} {\sqrt {\frac c g} - v} }
      | r = 2 t \sqrt {c g}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\sqrt {\frac c g} + v} {\sqrt {\frac c g} - v}
      | r = e^{2 t \sqrt {c g} }
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \sqrt {\frac c g} + v
      | r = \paren {\sqrt {\frac c g} - v} e^{2 t \sqrt {c g} }
      | c = 
}}
{{eqn | ll= \leadsto
      | l = v \paren {e^{2 t \sqrt {c g} } + 1}
      | r = v \paren {e^{2 t \sqrt {c g} } - 1} \sqrt {\frac g c}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = v
      | r = \sqrt {\frac g c} \paren {\frac {e^{2 t \sqrt {c g} } - 1} {e^{2 t \sqrt {c g} } + 1} }
      | c = 
}}
{{eqn | r = \sqrt {\frac g c} \paren {\frac {1 - e^{-2 t \sqrt {c g} } } {1 + e^{-2 t \sqrt {c g} } } }
      | c = 
}}
{{end-eqn}}
Since $c > 0$ it follows that $v \to \sqrt {\dfrac g c}$ as $t \to \infty$.
Thus in the limit:
:$v = \sqrt {\dfrac g c} =  \sqrt {\dfrac {g m} k}$
{{qed}}
Theorem: Escape Speed of Projectile Fired Upwards

Let $P$ be a planet.
Let $P$ have an acceleration due to gravity at its surface of $g$.
Let $P$ have a radius of $R$.
Then the escape speed of $P$ is given by:
:$V = \sqrt {2 g R}$

Proof:
Let a projectile $B$ of mass $m$ be fired vertically upwards from the surface of $P$ at such a speed that it escapes the gravitational field of $P$ completely.
$F$ be the force exerted on $B$ by the gravitational field of $P$.
Let $x$ be the distance of $B$ from the surface of $P$ at time $t$.
We have:
:$F = -\dfrac k {x^2}$
where $k$ can be calculated at the surface of $P$ as:
{{begin-eqn}}
{{eqn | l = F
      | r = m g
      | c = 
}}
{{eqn | r = \dfrac k {R^2}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = k
      | r = R^2 m g
      | c = 
}}
{{end-eqn}}
Thus:
:$F = -\dfrac {m g R^2} {x^2}$
Let $B$ be travelling away from $P$ with speed $v$ at time $t$.
Thus:
{{begin-eqn}}
{{eqn | l = F
      | r = -m \frac {\d v} {\d t}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = -\frac {m g R^2} {x^2}
      | r = m v \frac {\d v} {\d x}
      | c = 
}}
{{end-eqn}}
So:
:$\ds -\int \frac {R^2 g} {x^2} \rd x = \int v \rd v$
This gives:
:$\dfrac {R^2 g} x = \dfrac {v^2} 2 + C$
Now when $x = R$, $v = v_0$, the launch speed, which leads to:
:$C = R g - \dfrac {v_0^2} 2$
and so:
:$\dfrac {v^2} 2 = \dfrac {R^2 g} x - R g + \dfrac {v_0^2} 2$
For $\dfrac {v^2} 2 > 0$ we need:
:$\dfrac {v_0^2} 2 > R g \paren {1 - \dfrac R x}$
whatever $x$ may be.
So when $x \to \infty$ we have:
:$\dfrac {v_0^2} 2 > R g$
Hence:
:$v_0 > \sqrt {2 R g}$
{{qed}}
Theorem: Length of Arch of Sine Function

The length of one arch of the sine function:
:$y = \sin x$
is given by:
:$L = 2 \sqrt 2 \map E {\dfrac {\sqrt 2} 2}$
where $E$ denotes the incomplete elliptic integral of the second kind.

Proof:
Let $L$ be the length of one arch of $y = \sin x$.
Then:
{{begin-eqn}}
{{eqn | l = L
      | r = 2 \int_0^{\pi/2} \sqrt {1 + \paren {\map {\frac \d {\d x} } {\sin x} }^2} \rd x
      | c = {{Defof|Length of Curve}}
}}
{{eqn | r = 2 \int_0^{\pi/2} \sqrt {1 + \cos^2 x} \rd x
      | c = Derivative of Sine Function
}}
{{eqn | r = 2 \int_0^{\pi/2} \sqrt {2 - \sin^2 x} \rd x
      | c = Sum of Squares of Sine and Cosine
}}
{{eqn | r = 2 \sqrt 2 \int_0^{\pi/2} \sqrt {1 - \paren {\frac {\sqrt 2} 2} \sin^2 x} \rd x
      | c = extracting $\sqrt 2$
}}
{{eqn | r = 2 \sqrt 2 \map E {\dfrac {\sqrt 2} 2}
      | c = {{Defof|Complete Elliptic Integral of the Second Kind}}
}}
{{end-eqn}}
{{qed}}
Theorem: Length of Lemniscate of Bernoulli

The total length of the lemniscate of Bernoulli given in polar coordinates as:
:$r^2 = a^2 \cos 2 \theta$
is given by:
{{begin-eqn}}
{{eqn | l = L
      | r = 4 a \map F {\sqrt 2, \dfrac \pi 4}
      | c = 
}}
{{eqn | r = \dfrac 1 {\sqrt {2 \pi} } \paren {\map \Gamma {\dfrac 1 4} }^2
      | c = 
}}
{{end-eqn}}
where $F$ denotes the incomplete elliptic integral of the first kind.

Proof:
The arc length of a small length increment $\d s$ is given in polar co-ordinates by:
:$\paren {\d s}^2 = \paren {r \rd \theta}^2 + \paren {\d r}^2$
from which:
:$\dfrac {\d s} {\d \theta} = \sqrt {r^2 + \paren {\dfrac {\d r} {\d \theta} }^2}$
Half of one lobe of the lemniscate is achieved when $\theta$ goes from $0$ to $\pi / 4$.
Therefore the total length of the lemniscate of Bernoulli is given by:
:$\ds L = 4 \int_0^{\pi/4} \sqrt {r^2 + \paren {\dfrac {\d r} {\d \theta} }^2} \rd \theta$
First we show:
{{begin-eqn}}
{{eqn | l = r^2
      | r = a^2 \cos 2 \theta
      | c = 
}}
{{eqn | ll= \leadsto
      | l = 2 r \frac {\d r} {\d \theta}
      | r = -2 a^2 \sin 2 \theta
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\d r} {\d \theta}
      | r = -\frac {a^2 \sin 2 \theta} {a \sqrt {\cos 2 \theta} }
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \paren {\dfrac {\d r} {\d \theta} }^2
      | r = \frac {a^2 \sin^2 2 \theta} {\cos 2 \theta}
      | c = 
}}
{{end-eqn}}
So:
{{begin-eqn}}
{{eqn | l = r^2 + \paren {\dfrac {\d r} {\d \theta} }^2
      | r = a^2 \cos 2 \theta + \frac {a^2 \sin^2 2 \theta} {\cos 2 \theta}
      | c = 
}}
{{eqn | r = a^2 \frac {\cos^2 2 \theta + \sin^2 2 \theta} {\cos 2 \theta}
      | c = 
}}
{{eqn | r = \frac {a^2} {\cos 2 \theta}
      | c = Sum of Squares of Sine and Cosine
}}
{{end-eqn}}
Thus:
{{begin-eqn}}
{{eqn | l = L
      | r = 4 a \int_0^{\pi / 4} \frac {\d \theta} {\sqrt {\cos 2 \theta} }
      | c = 
}}
{{eqn | r = 4 a \int_0^{\pi / 4} \frac {\d \theta} {\sqrt {1 - 2 \sin^2 \theta} }
      | c = 
}}
{{eqn | r = 4 a \map F {\sqrt 2, \dfrac \pi 4}
      | c = {{Defof|Incomplete Elliptic Integral of the First Kind}}
}}
{{end-eqn}}
{{qed}}
Theorem: Boundary of Union is Subset of Union of Boundaries

Let $T = \struct {S, \tau}$ be a topological space.
Let $A, B$ be subsets of $S$.
Then:
:$\map \partial {A \cup B} \subseteq \partial A \cup \partial B$
where $\partial A$ denotes the boundary of $A$.

Proof:
By Intersection is Subset:
:$\relcomp S A \cap \relcomp S B \subseteq \relcomp S A \land \relcomp S A \cap \relcomp S B \subseteq \relcomp S B$
Then by Topological Closure of Subset is Subset of Topological Closure:
:$\paren {\relcomp S A \cap \relcomp S B}^- \subseteq \paren {\relcomp S A}^- \land \paren {\relcomp S A \cap \relcomp S B}^- \subseteq \paren {\relcomp S B}^-$
Hence by Boundary is Intersection of Closure with Closure of Complement:
:$\paren {\relcomp S A \cap \relcomp S B}^- \cap A^- \subseteq \partial A \land \paren {\relcomp S A \cap \relcomp S B}^- \cap B^- \subseteq \partial B$
Thus
{{begin-eqn}}
{{eqn | l = \map \partial {A \cup B}
      | r = \paren {\relcomp S {A \cup B} }^- \cap \paren {A \cup B}^-
      | c = Boundary is Intersection of Closure with Closure of Complement
}}
{{eqn | r = \paren {\relcomp S A \cap \relcomp S B}^- \cap \paren {A \cup B}^-
      | c = Complement of Union
}}
{{eqn | r = \paren {\relcomp S A \cap \relcomp S B}^- \cap \paren {A^- \cup B^-}
      | c = Closure of Finite Union equals Union of Closures
}}
{{eqn | r = \paren {\relcomp S A \cap \relcomp S B}^- \cap A^- \cup \paren {\relcomp S A \cap \relcomp S B}^- \cap B^-
      | c = Intersection Distributes over Union
}}
{{eqn | o = \subseteq
      | r = \partial A \cup \partial B
      | c = Set Union Preserves Subsets
}}
{{end-eqn}}
{{qed}}
Theorem: Boundary of Intersection is Subset of Union of Boundaries

Let $T = \struct {S, \tau}$ be a topological space.
Let $A, B$ be subsets of $S$.
Then:
:$\map \partial {A \cap B} \subseteq \partial A \cup \partial B$
where $\partial A$ denotes the boundary of $A$.

Proof:
By Intersection is Subset:
:$A \cap B \subseteq A \land A \cap B \subseteq B$
Then by Topological Closure of Subset is Subset of Topological Closure:
:$\paren {A \cap B}^- \subseteq A^- \land \paren {A \cap B}^- \subseteq B^-$
Hence by Boundary is Intersection of Closure with Closure of Complement:
:$\paren {A \cap B}^- \cap \paren {\relcomp S A}^- \subseteq \partial A \land \paren {A \cap B}^- \cap \paren {\relcomp S B}^- \subseteq \partial B$
Thus
{{begin-eqn}}
{{eqn | l = \map \partial {A \cap B}
      | r = \paren {A \cap B}^- \cap \paren {\relcomp S {A \cap B} }^-
      | c = Boundary is Intersection of Closure with Closure of Complement
}}
{{eqn | r = \paren {A \cap B}^- \cap \paren {\relcomp S A \cup \relcomp S B}^-
      | c = Complement of Intersection
}}
{{eqn | r = \paren {A \cap B}^- \cap \paren {\paren {\relcomp S A}^- \cup \paren {\relcomp S B}^-}
      | c = Closure of Finite Union equals Union of Closures
}}
{{eqn | r = \paren {A \cap B}^- \cap \paren {\relcomp S A}^- \cup \paren {A \cap B}^- \cap \paren {\relcomp S B}^-
      | c = Intersection Distributes over Union
}}
{{eqn | o = \subseteq
      | r = \partial A \cup \partial B
      | c = Set Union Preserves Subsets
}}
{{end-eqn}}
{{qed}}
Theorem: Complete Elliptic Integral of the First Kind as Power Series

The '''complete elliptic integral of the first kind''':
:$\ds \map K k = \int_0^{\pi / 2} \frac {\rd \phi} {\sqrt {1 - k^2 \sin^2 \phi} } = \int_0^1 \frac {\rd v} {\sqrt {\paren {1 - v^2} \paren {1 - k^2 v^2} } }$
can be expressed as the power series:
{{begin-eqn}}
{{eqn | l = \map K k
      | r = \frac \pi 2 \sum_{i \mathop \ge 0} \paren {\prod_{j \mathop = 1}^i \frac {2 j - 1} {2 j} }^2 k^{2 i}
      | c = 
}}
{{eqn | r = \frac \pi 2 \paren {1 + \paren {\frac 1 2}^2 k^2 + \paren {\frac {1 \cdot 3} {2 \cdot 4} }^2 k^4 + \paren {\frac {1 \cdot 3 \cdot 5} {2 \cdot 4 \cdot 6} }^2 k^6 + \cdots}
      | c = 
}}
{{end-eqn}}

Proof:
From Reduction Formula for Integral of Power of Sine, $\forall i \in \N$:
{{begin-eqn}}
{{eqn | l = \int_0^{\pi / 2} \sin^{2 i} \phi \rd \phi
      | r = \frac {2 i - 1} {2 i} \int_0^{\pi / 2} \sin^{2 i - 2} \phi \rd \phi - \intlimits {\frac {\sin^{2 i - 1} x \cos x} i} {x = 0} {x = \frac \pi 2}
}}
{{eqn | r = \frac {2 i - 1} {2 i} \int_0^{\pi / 2} \sin^{2 i - 2} \phi \rd \phi
      | c = $\sin 0 = \cos \dfrac \pi 2 = 0$
}}
{{eqn | r = \cdots
}}
{{eqn | r = \prod_{j \mathop = 1}^i \frac {2 j - 1} {2 j} \int_0^{\pi / 2} 1 \rd \phi
}}
{{eqn | r = \frac \pi 2 \prod_{j \mathop = 1}^i \frac {2 j - 1} {2 j}
}}
{{end-eqn}}
Hence:
{{begin-eqn}}
{{eqn | l = \map K k
      | r = \int_0^{\pi / 2} \frac {\rd \phi} {\sqrt{1 - k^2 \sin^2 \phi} }
}}
{{eqn | r = \int_0^{\pi / 2} \sum_{i \mathop \ge 0} \binom {- \frac 1 2} i \paren {- k^2 \sin^2 \phi}^i \rd \phi
      | c = General Binomial Theorem
}}
{{eqn | r = \sum_{i \mathop \ge 0} k^{2 i} \paren {\prod_{j \mathop = 1}^i \frac {\frac 1 2 - j} j} \paren {-1}^i \int_0^{\pi / 2} \sin^{2 i} \phi \rd \phi
      | c = {{Defof|Binomial Coefficient|subdef = Real Numbers}}
}}
{{eqn | r = \sum_{i \mathop \ge 0} k^{2 i} \paren {\prod_{j \mathop = 1}^i \frac {1 - 2 j} {2 j} } \paren {\prod_{j \mathop = 1}^i \paren {-1} }\paren {\frac \pi 2 \prod_{j \mathop = 1}^i \frac {2 j - 1} {2 j} }
      | c = from above
}}
{{eqn | r = \frac \pi 2 \sum_{i \mathop \ge 0} \paren {\prod_{j \mathop = 1}^i \frac {2 j - 1} {2 j} }^2 k^{2 i}
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Complete Elliptic Integral of the Second Kind as Power Series

The '''complete elliptic integral of the second kind''':
:$\ds \map E k = \int_0^{\pi / 2} \sqrt {1 - k^2 \sin^2 \phi} \, \rd \phi = \int_0^1 \dfrac {\sqrt {1 - k^2 v^2} } {\sqrt {1 - v^2}} \, \rd v$
can be expressed as the power series:
{{begin-eqn}}
{{eqn | l = \map E k
      | r = \frac \pi 2 \sum_{i \mathop \ge 0} \paren {\prod_{j \mathop = 1}^i \frac {2 j - 1} { 2 j} }^2 \frac {k^{2 i} } {1 - 2 i}
      | c = 
}}
{{eqn | r = \frac \pi 2 \paren {1 - \paren {\frac 1 2}^2 k^2 - \paren {\frac {1 \cdot 3} {2 \cdot 4} }^2 \frac {k^4} 3 - \paren {\frac {1 \cdot 3 \cdot 5} {2 \cdot 4 \cdot 6} }^2 \frac {k^6} 5 - \cdots}
      | c = 
}}
{{end-eqn}}

Proof:
From Reduction Formula for Integral of Power of Sine, $\forall i \in \N$:
{{begin-eqn}}
{{eqn | l = \int_0^{\pi / 2} \sin^{2 i} \phi \rd \phi
      | r = \frac {2 i - 1} {2 i} \int_0^{\pi / 2} \sin^{2 i - 2} \phi \rd \phi - \intlimits {\frac {\sin^{2 i - 1} x \cos x} i} {x = 0} {x = \frac \pi 2}
}}
{{eqn | r = \frac {2 i - 1} {2 i} \int_0^{\pi / 2} \sin^{2 i - 2} \phi \rd \phi
      | c = $\sin 0 = \cos \dfrac \pi 2 = 0$
}}
{{eqn | r = \cdots
}}
{{eqn | r = \prod_{j \mathop = 1}^i \frac {2 j - 1} {2 j} \int_0^{\pi / 2} 1 \rd \phi
}}
{{eqn | r = \frac \pi 2 \prod_{j \mathop = 1}^i \frac {2 j - 1} {2 j}
}}
{{end-eqn}}
Hence:
{{begin-eqn}}
{{eqn | l = \map E k
      | r = \int_0^{\pi / 2} \sqrt {1 - k^2 \sin^2 \phi} \, \rd \phi
}}
{{eqn | r = \int_0^{\pi / 2} \sum_{i \mathop \ge 0} \binom {\frac 1 2} i \paren {- k^2 \sin^2 \phi}^i \rd \phi
      | c = General Binomial Theorem
}}
{{eqn | r = \sum_{i \mathop \ge 0} k^{2 i} \paren {\prod_{j \mathop = 1}^i \frac {\frac 3 2 - j} j} \paren {-1}^i \int_0^{\pi / 2} \sin^{2 i} \phi \rd \phi
      | c = {{Defof|Binomial Coefficient|subdef = Real Numbers}}
}}
{{eqn | r = \sum_{i \mathop \ge 0} k^{2 i} \paren {\prod_{j \mathop = 1}^i \frac {3 - 2 j} {2 j} } \paren {\prod_{j \mathop = 1}^i \paren {-1} }\paren {\frac \pi 2 \prod_{j \mathop = 1}^i \frac {2 j - 1} {2 j} }
      | c = From above
}}
{{eqn | r = \frac \pi 2 \sum_{i \mathop \ge 0} \paren {\prod_{j \mathop = 1}^i \frac {2 j - 3} {2 j} } \paren {\prod_{j \mathop = 1}^i \frac {2 j - 1} { 2 j} } k^{2 i}
      | c = 
}}
{{eqn | r = \frac \pi 2 + \frac \pi 2 \sum_{i \mathop \ge 1} \paren {\prod_{j \mathop = 1}^i \frac {2 j - 3} {2 j} } \paren {\prod_{j \mathop = 1}^i \frac {2 j - 1} { 2 j} } k^{2 i}
      | c = 
}}
{{eqn | r = \frac \pi 2 + \frac \pi 2 \sum_{i \mathop \ge 1} \paren {\frac {2 - 3} {2 i - 1} } \paren {\prod_{j \mathop = 1}^i \frac {2 j - 1} { 2 j} }^2 k^{2 i}
      | c = 
}}
{{eqn | r = \frac \pi 2 + \frac \pi 2 \sum_{i \mathop \ge 1} \paren {\prod_{j \mathop = 1}^i \frac {2 j - 1} { 2 j} }^2 \frac {k^{2 i} } {1 - 2 i}
      | c = 
}}
{{eqn | r = \frac \pi 2 \sum_{i \mathop \ge 0} \paren {\prod_{j \mathop = 1}^i \frac {2 j - 1} { 2 j} }^2 \frac {k^{2 i} } {1 - 2 i}
      | c = Product equals $1$ when $i = 0$
}}
{{end-eqn}}
{{qed}}
Theorem: Discrete Space is Separable iff Countable

Let $T = \struct {S, \tau}$ be a discrete topological space.
Then:
:$T$ is separable {{iff}} $S$ is countable.

Proof:
=== Sufficient Condition ===
Immediate from Separable Discrete Space is Countable.
{{qed|lemma}}
Theorem: Set is Countable iff Cardinality not greater Aleph Zero

Let $X$ be a set.
Then:
:$X$ is countable {{iff}} $\size X \le \aleph_0$
where
:$\size X$ denotes the cardinality of $X$,
:$\aleph$ denotes the aleph mapping.

Proof:
:$X$ is countable
{{iff}}
:there exists an injection $X \to \N$ by definition of countable set
{{iff}}
:$\size X \le \size \N$ by Injection iff Cardinal Inequality
{{iff}}
:$\size X \le \aleph_0$ by Aleph Zero equals Cardinality of Naturals.
{{qed}}
Category:Countable Sets
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Theorem: Bernoulli's Theorem

Let the probability of the occurrence of an event be $p$.
Let $n$ independent trials be made, with $k$ successes.
Then:
:$\ds \lim_{n \mathop \to \infty} \frac k n = p$

Proof:
Let the random variable $k$ have the binomial distribution with parameters $n$ and $p$, that is:
:$k \sim \Binomial n p$
where $k$ denotes the number of successes of the $n$ independent trials of the event with probability $p$.
From Expectation of Binomial Distribution:
:$\expect k = n p \leadsto \dfrac 1 n \expect k = p$
Expectation is Linear gives:
:$ \expect {\dfrac k n} = p =: \mu$
Similarly, from Variance of Binomial Distribution:
:$\var k = n p \paren {1 - p} \leadsto \dfrac 1 {n^2} \var k = \dfrac {p \paren {1 - p} } n$
From Variance of Linear Combination of Random Variables:
:$\var {\dfrac k n} = \dfrac {p \paren {1 - p} } n =: \sigma^2$
By applying the Bienaymé-Chebyshev Inequality to $\dfrac k n$, we have for any $l > 0$:
:$\map \Pr {\size {\dfrac k m - \mu} \ge l \sigma} \le \dfrac 1 {l^2}$
Now, let $\epsilon > 0$ and choose $l = \dfrac \epsilon \sigma$, to get:
:$\map \Pr {\size {\dfrac k m - \mu} \ge \dfrac \epsilon \sigma \cdot \sigma} \le \dfrac {\sigma^2} {\epsilon^2}$
Simplifying and plugging in the values of $\mu$ and $\sigma^2$ defined above yields:
:$\map \Pr {\size {\dfrac k n - p} \ge \epsilon} \le \dfrac {p \paren {1 - p} } {n \epsilon^2}$
Scaling both sides by $-1$ and adding $1$ to both sides yields:
:$1 - \map \Pr {\size {\dfrac k n - p} \ge \epsilon} \ge 1 - \dfrac {p \paren {1 - p} } {n \epsilon^2}$
Applying Union of Event with Complement is Certainty to the {{LHS}}:
:$\map \Pr {\size {\dfrac k n - p} \le \epsilon} \ge 1 - \dfrac {p \paren {1 - p} } {n\epsilon^2}$
Taking the limit as $n$ approaches infinity on both sides, we have:
:$\ds \lim_{n \mathop \to \infty} \map \Pr {\size {\frac k n - p} < \epsilon} = 1$
{{qed}}