avewright/proofwiki-math huggingface dataset:
Theorem: Sequence of Imaginary Reciprocals/Limit Points

The set $S$ has exactly one limit point, and that is $z = 0$.

Proof:
We have that $S$ is countably infinite and also bounded.
Hence by the Bolzano-Weierstrass Theorem $S$ has at least one limit point.
Let $\epsilon \in \R_{>0}$.
Let $\map {N_\epsilon} z$ be the $\epsilon$-neighborhood of $z$.
Let $n \in \N$ such that $n > \dfrac 1 \epsilon$.
Then $\cmod {\dfrac i n} < \epsilon$ and so:
: $\dfrac i n \in \map {N_\epsilon} z$
where $\map {N_\epsilon} {z_0}$ denotes the $\epsilon$-neighborhood of $z$.
Hence, by definition, $z$ is a limit point of $S$.
Now let $z_1 \in \C$ such that $z_1 \ne 0$.
Let $z \in S$ such that $z \ne z_1$.
Then:
:$\cmod {z_1 - z} > 0$
Let $\epsilon \in \R_{>0}$ such that $\epsilon < \cmod {z_1 - z}$.
Then:
:$z \notin \map {N_\epsilon} {z_1}$
By making $\epsilon$ smaller than the minimum $\cmod {z_1 - z}$ for $z \in S$, it is seen that:
:$\forall z \in S: z \ne z_1 \implies z \notin \map {N_\epsilon} {z_1}$
and so $z_1$ is not a limit point of $S$.
{{qed}}
Theorem: Sequence of Imaginary Reciprocals/Closedness

The set $S$ is not closed.

Proof:
From Sequence of Imaginary Reciprocals: Limit Points, $S$ has one limit point $z = 0$.
But:
:$\nexists n \in \N: \dfrac i n = 0$
so $0 \notin S$.
As $S$ does not contain (all) its limit point(s), it follows by definition that $S$ is not closed.
{{qed}}
Theorem: Valuation Ring of P-adic Norm on Rationals

Let $\norm {\,\cdot\,}_p$ be the $p$-adic norm on the rationals $\Q$ for some prime $p$.
The induced valuation ring on $\struct {\Q,\norm {\,\cdot\,}_p}$ is the set:
:$\OO = \Z_{\paren p} = \set {\dfrac a b \in \Q : p \nmid b}$

Proof:
Let $\nu_p: \Q \to \Z \cup \set {+\infty}$ be the $p$-adic valuation on $\Q$.
Then:
{{begin-eqn}}
{{eqn | l = \OO
      | r = \set {\dfrac a b \in \Q : \norm {\dfrac a b}_p \le 1}
      | c = {{Defof|Valuation Ring Induced by Non-Archimedean Norm}}
}}
{{eqn | o = }}
{{eqn | r = \set{\dfrac a b \in \Q : \map {\nu_p} {\dfrac a b} \ge 0}
      | c = {{Defof|P-adic Norm}}
}}
{{eqn | o = }}
{{eqn | r = \set {\dfrac a b \in \Q : \map {\nu_p} a - \map {\nu_p} b \ge 0}
      | c = {{Defof|P-adic Valuation|$p$-adic Valuation on Rationals}}
}}
{{eqn | o = }}
{{eqn | r = \set {\dfrac a b \in \Q : \map {\nu_p} a \ge \map {\nu_p} b}
}}
{{end-eqn}}
Let $\dfrac a b \in \Q$ be in canonical form. 
Then $a \perp b$
Suppose $p \divides b$. 
Then $p \nmid a$.
Hence:
:$\map {\nu_p} b \gt 0 = \map {\nu_p} a$
Suppose $p \nmid b$. 
Then:
:$\map {\nu_p} a \ge 0 = \map {\nu_p} b$
So: 
:$\map {\nu_p} a \ge \map {\nu_p} b$ {{iff}} $p \nmid b$  
Hence:
:$\OO = \set {\dfrac a b \in \Q : p \nmid b }$
{{qed}}
Theorem: Valuation Ideal of P-adic Norm on Rationals

Let $\norm {\,\cdot\,}_p$ be the $p$-adic norm on the rationals $\Q$ for some prime $p$.
The induced valuation ideal on $\struct {\Q,\norm {\,\cdot\,}_p}$ is the set:
:$\PP = p \Z_{\ideal p} = \set {\dfrac a b \in \Q : p \nmid b, p \divides a}$
where $\Z_{\ideal p}$ is the induced valuation ring on $\struct {\Q,\norm {\,\cdot\,}_p}$

Proof:
Let $\nu_p: \Q \to \Z \cup \set {+\infty}$ be the $p$-adic valuation on $\Q$.
Then:
{{begin-eqn}}
{{eqn | l = \PP
      | r = \set {\dfrac a b \in \Q : \norm{\dfrac a b}_p < 1}
      | c = {{Defof|Valuation Ideal Induced by Non-Archimedean Norm}}
}}
{{eqn | o = }}
{{eqn | r = \set {\dfrac a b \in \Q : \map {\nu_p} {\dfrac a b} > 0}
      | c = {{Defof|P-adic Norm|$p$-adic Norm}}
}}
{{eqn | o = }}
{{eqn | r = \set {\dfrac a b \in \Q : \map {\nu_p} a - \map {\nu_p} b > 0}
      | c = {{Defof|P-adic Valuation|$p$-adic Valuation on Rationals}}
}}
{{eqn | o = }}
{{eqn | r = \set {\dfrac a b \in \Q : \map {\nu_p} a > \map {\nu_p} b}
}}
{{end-eqn}}
Let $\dfrac a b \in \Q$ be in canonical form. 
Then $a \perp b$
Suppose $p \divides a$. 
Then $p \nmid b$.
Hence:
:$\map {\nu_p} a > 0 = \map {\nu_p} b$
Suppose $p \nmid a$. 
Then:
:$\map {\nu_p} b \ge 0 = \map {\nu_p} a$
So: 
:$\map {\nu_p} a > \map {\nu_p} b$ {{iff}} $p \nmid b$ and $p \divides a$   
Hence:
:$\PP = \set {\dfrac a b \in \Q : p \nmid b, p \divides a}$
So:
{{begin-eqn}}
{{eqn | l = \dfrac a b \in \PP
      | o = \leadstoandfrom
      | r = p \nmid b, p \divides a
}}
{{eqn | o = \leadstoandfrom
      | r = p \nmid b, \exists a' \in \Z: a = p a' 
}}
{{eqn | o = \leadstoandfrom
      | r = \exists a' \in \Z: a = p a',  \dfrac {a'} b \in \Z_{\ideal p}
      | c = Valuation Ring of P-adic Norm on Rationals
}}
{{eqn | o = \leadstoandfrom
      | r = \dfrac a b \in p \Z_{\ideal p}
}}
{{end-eqn}}
Hence:
:$\PP = p \Z_{\ideal p}$
{{qed}}
Theorem: Residue Field of P-adic Norm on Rationals

Let $\norm {\,\cdot\,}_p$ be the $p$-adic norm on the rationals $\Q$ for some prime $p$.
The induced residue field on $\struct {\Q,\norm {\,\cdot\,}_p}$ is isomorphic to the field $\F_p$ of integers modulo $p$.

Proof:
By Valuation Ring of P-adic Norm on Rationals:
:$\Z_{\ideal p} = \set {\dfrac a b \in \Q : p \nmid b}$
is the induced valuation ring on $\struct {\Q,\norm {\,\cdot\,}_p}$.
By Valuation Ideal of P-adic Norm on Rationals:
:$p \Z_{\ideal p} = \set {\dfrac a b \in \Q : p \nmid b, p \divides a}$
is the induced valuation ideal on $\struct {\Q,\norm {\,\cdot\,}_p}$.
By definition, the induced residue field on $\struct {\Q,\norm {\,\cdot\,}_p}$ is the quotient ring $\Z_{\ideal p} / p \Z_{\ideal p}$.
By Quotient Ring of Integers with Principal Ideal, $\F_p$ is isomorphic to $\Z / p \Z$, where $p \Z$ is the principal ideal of $\Z$ generated by $p$.
To complete the proof it is sufficient to show that $\Z / p \Z$ is isomorphic to $\Z_{\ideal p} / p \Z_{\ideal p}$.
By Integers form Subring of Valuation Ring of P-adic Norm on Rationals then $\Z$ is a subring of $\Z_{\ideal p}$.
Let $\phi : \Z \to \Z_{\ideal p} / p \Z_{\ideal p}$ be the mapping defined by:
:$\forall a \in \Z: \map \phi a = a + p \Z_{\ideal p}$
Theorem: Sequence of Imaginary Reciprocals/Boundary Points

Every point of $S$, along with the point $z = 0$, is a boundary point of $S$.

Proof:
Consider the point $z = \dfrac i n \in S$.
Let $\delta \in \R_{>0}$.
Let $\map {N_\delta} z$ be the $\delta$-neighborhood of $z$.
Then $\map {N_\delta} z$ contains at least one point of $S$ ($i / n$ itself) as well as points which are not in $S$.
Hence, by definition, $z$ is a boundary point of $S$.
Let $z = 0$.
Similarly, let $\map {N_\delta} z$, 
Let $\delta \in \R_{>0}$.
Let $\map {N_\delta} z$ be the $\delta$-neighborhood of $z$.
Let $n \in \N$ such that $n > \dfrac 1 \delta$.
Then $\cmod {\dfrac i n} < \delta$ and so:
: $\dfrac i n \in \map {N_\delta} z$
Thus every $\delta$-neighborhood of $z = 0$ also contains points of $S$ and points not in $S$.
Hence, by definition, $z = 0$ is also a boundary point of $S$.
{{qed}}
Theorem: Sequence of Imaginary Reciprocals/Interior

No point of $S$ is an interior point.

Proof:
From Sequence of Imaginary Reciprocals: Boundary Points, every $z \in S$ is a boundary point of $S$.
Thus no $z \in S$ is an interior point.
{{qed}}
Theorem: Sequence of Imaginary Reciprocals/Openness

$S$ is not an open set.

Proof:
From Sequence of Imaginary Reciprocals: Interior, no $z \in S$ is an interior point.
Hence $S$ cannot be open.
{{qed}}
Theorem: Sequence of Imaginary Reciprocals/Connectedness

$S$ is not connected.

Proof:
Let $z_1 \in S$ and $z_2 \in S$ be joined by a polygonal path $P$.
Then there are points of $P$ which are not in $S$.
Hence, by definition, $S$ is not connected.
{{qed}}
Theorem: Sequence of Imaginary Reciprocals/Not an Open Region

$S$ is not an open region.

Proof:
From Sequence of Imaginary Reciprocals: Openness, $S$ is not an open set.
From Sequence of Imaginary Reciprocals: Connectedness, $S$ is not connected.
Hence, by definition, $S$ is not an open region of $\C$.
{{qed}}
Theorem: Sequence of Imaginary Reciprocals/Closure

The closure of $S$ is the set:
:$\set {0, i, \dfrac i 2, \dfrac i 3, \dfrac i 3, \ldots}$

Proof:
By definition, the closure of $S$ is the set $S$ along with all its limit points.
Sequence of Imaginary Reciprocals: Limit Points, $S$ has one limit point $z = 0$.
Hence the result.
{{qed}}
Theorem: Sequence of Imaginary Reciprocals/Countability

The set $S$ is countably infinite.

Proof:
Let $\phi: \N \to S$ be the mapping defined as:
:$\forall n \in \N: \map \phi n = \dfrac i n$
$\phi$ is a bijection.
Hence the result by definition of countably infinite.
{{qed}}
Theorem: Sequence of Imaginary Reciprocals/Not Compact

The set $S$ is not compact.

Proof:
From Sequence of Imaginary Reciprocals: Boundedness, $S$ is bounded in $\C$.
But from Sequence of Imaginary Reciprocals: Closedness, $S$ is not closed.
Hence the result by definition of compact.
{{qed}}
Theorem: Sequence of Imaginary Reciprocals/Closure is Compact

The closure $S^-$ of the set $S$ is compact.

Proof:
From Topological Closure is Closed, $S^-$ is closed.
From Sequence of Imaginary Reciprocals: Boundedness, $S$ is bounded in $\C$.
It follows trivially that $S^-$ is also bounded in $\C$.
Hence the result by definition of compact.
{{qed}}
Theorem: Product of Complex Conjugates/Examples/3 Arguments

Let $z_1, z_2, z_3 \in \C$ be complex numbers.
Let $\overline z$ denote the complex conjugate of the complex number $z$.
Then:
:$\overline {z_1 z_2 z_3} = \overline {z_1} \cdot \overline {z_2} \cdot \overline {z_3}$

Proof:
From Product of Complex Conjugates: General Result:
:$\ds \overline {\prod_{j \mathop = 1}^n z_j} = \prod_{j \mathop = 1}^n \overline {z_j}$
The result follows by setting $n = 3$.
Theorem: Valuation Ideal is Maximal Ideal of Induced Valuation Ring/Corollary 1

:$(a): \quad \OO$ is a local ring.
:$(b): \quad \PP$ is the unique maximal left ideal of $\OO$
:$(c): \quad \PP$ is the unique maximal right ideal of $\OO$

Proof:
By Valuation Ideal is Maximal Ideal of Induced Valuation Ring then:
:$\PP$ is a maximal left ideal of $\OO$.
Let $J \subsetneq \OO$ be any maximal left ideal of $\OO$.
Let $x \in \OO \setminus \PP$.
{{AimForCont}} $x \in J$.
By Norm of Inverse then:
:$\norm {x^{-1}} = 1 / \norm x = 1 / 1 = 1$ 
Hence:
:$x^{-1} \in \OO$
Since $J$ is a left ideal then:
:$x^{-1} x = 1_R \in J$
Thus:
:$\forall y \in \OO: y \cdot 1_R = y \in J$
That is: 
:$J = \OO$
This contradicts the assumption that $J \ne \OO$.
So:
:$x \notin J$
Hence: 
:$\paren {\OO \setminus \PP} \cap J = \O$
That is:
:$J \subseteq \PP$
Since $J$ and $\PP$ are both maximal left ideals then:
:$J = \PP$
Hence:
:$\PP$ is the unique maximal left ideal of $\OO$.
By {{Defof|Local Ring}}:
:$\OO$ is a local ring
From Maximal Left and Right Ideal iff Quotient Ring is Division Ring:
:every maximal right ideal of $\OO$ is a maximal left ideal
Hence:
:$\PP$ is the unique maximal right ideal of $\OO$.
The result follows.
{{qed}}
Theorem: Residue Field of P-adic Norm on Rationals/Lemma 1

:$\phi$ is a homomorphism.

Proof:
Since $p \nmid 1$ then for all $a \in \Z$, $a = \dfrac a 1 \in \Z_{\paren p}$.
Hence $\Z \subset \Z_{\paren p}$ is a subring of $\Z_{\paren p}$.
Let $i: \Z \to \Z_{\paren p}$ be the inclusion mapping defined by:
:$\map i a = a$
By Inclusion Mapping is Monomorphism then $i$ is a ring monomorphism.
Let $q: \Z_{\paren p} \to \Z_{\paren p} / p \, \Z_{\paren p}$ be the quotient ring epimorphism from $\Z_{\paren p}$ to $\Z_{\paren p} / p \, \Z_{\paren p}$.
By Quotient Ring Epimorphism is Epimorphism, $q$ is a epimorphism.
By definition, $\phi = q \circ i$ is the composition of $i$ with $q$.
By Composition of Ring Homomorphisms is Ring Homomorphism, $\phi$ is a homomorphism.
{{qed}}
Theorem: Residue Field of P-adic Norm on Rationals/Lemma 2

:$p \Z = \map \ker \phi$

Proof:
Let $\map \ker \phi$ denote the kernel of $\phi$.
Then:
{{begin-eqn}}
{{eqn | l = a
      | o = \in
      | r = \map \ker \phi
      | c =
}}
{{eqn | ll= \leadstoandfrom
      | l = \map \phi a
      | r = p \Z_{\ideal p}
      | c = {{Defof|Kernel of Ring Homomorphism}}
}}
{{eqn | ll= \leadstoandfrom
      | l = a + p \Z_{\ideal p}
      | r = p \Z_{\ideal p}
      | c = Definition of $\phi$
}}
{{eqn | ll= \leadstoandfrom
      | l = a = a - 0 
      | o = \in
      | r = p \Z_{\ideal p}
      | c = Element in Right Coset iff Product with Inverse in Subgroup
}}
{{eqn | ll= \leadstoandfrom
      | q = \exists a' \in \Z
      | l = a = p a'
      | r = p a'
}}
{{eqn | ll= \leadstoandfrom
      | l = a
      | o = \in
      | r = p \Z
}}
{{end-eqn}}
Hence:
:$p \Z = \map \ker \phi$
{{qed}}
Theorem: Residue Field of P-adic Norm on Rationals/Lemma 3

:$\phi : \Z \to \Z_{\paren p} / p \Z_{\paren p}$ is a surjection.

Proof:
Let $a / b \in \Z_{\paren p}$, where $a / b$ are in canonical form.
Then $p \nmid b$
Let $\F_p$ be the field of integers modulo $p$.
By the definition of a field:
:$\exists b' \in \Z: b b' \equiv 1 \pmod p$
By the definition of congruence modulo $p$:
:$p \divides b b' - 1$
By Divisor Divides Multiple:
:$\forall a \in \Z: p \divides a b b' - a$
By Valuation Ideal of P-adic Norm on Rationals then:
:$a b' - \dfrac a b = \dfrac {a b b' - a} b \in p \Z_{\paren p}$
By Element in Left Coset iff Product with Inverse in Subgroup:
:$\map \phi {a b'} = a b' + p \Z_{\paren p} = a / b + p \Z_{\paren p}$
It follows that:
:$\phi : \Z \to \Z_{\paren p} / p \Z_{\paren p}$ is a surjection.
{{qed}}
Theorem: Equation for Line through Two Points in Complex Plane/Parametric Form 2

$L$ can be expressed by the equations:
{{begin-eqn}}
{{eqn | l = x - x_1
      | r = t \paren {x_2 - x_1}
}}
{{eqn | l = y - y_1
      | r = t \paren {y_2 - y_1}
}}
{{end-eqn}}
These are the '''parametric equations of $L$''', where $t$ is the parameter.

Proof:
From Equation for Line through Two Points in Complex Plane: Parametric Form 1:
:$z = z_1 + t \paren {z_2 - z_1}$
Letting:
{{begin-eqn}}
{{eqn | l = z
      | r = x + i y
}}
{{eqn | l = z_1
      | r = x_1 + i y_1
}}
{{eqn | l = z_2
      | r = x_2 + i y_2
}}
{{end-eqn}}
the parametric equations follow by equating real parts and imaginary parts.
{{qed}}
Theorem: Equation for Perpendicular Bisector of Two Points in Complex Plane/Parametric Form 1

$L$ can be expressed by the equation:
:$z - \dfrac {z_1 + z_2} 2 = i t\paren {z_2 - z_1}$
or:
:$z = \dfrac {z_1 + z_2} 2 + i t\paren {z_2 - z_1}$
This form of $L$ is known as the '''parametric form''', where $t$ is the '''parameter'''.

Proof:
Let $z_1$ and $z_2$ be represented by the points $A = \tuple {x_1, y_1}$ and $B = \tuple {x_2, y_2}$ respectively in the complex plane.
Let $z$ be an arbitrary point on $L$ represented by the point $P$.
:400px
We have that $L$ passes through the point:
:$\dfrac {z_1 + z_2} 2$
and is perpendicular to the straight line:
:$z = z_1 + t \paren {z_2 - z_1}$
By Multiplication by Imaginary Unit is Equivalent to Rotation through Right Angle:
:the vector $i \paren {z_2 - z_1}$ is perpendicular to the vector $\paren {z_2 - z_1}$
Hence the perpendicular bisector can be written as:
:$z = \dfrac {z_1 + z_2} 2 + t i \paren {z_2 - z_1}$
{{qed}}
Category:Equation for Perpendicular Bisector of Two Points in Complex Plane
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Theorem: Equation of Hyperbola in Complex Plane

Let $\C$ be the complex plane.
Let $H$ be a hyperbola in $\C$ whose major axis is $d \in \R_{>0}$ and whose foci are at $\alpha, \beta \in \C$.
Then $C$ may be written as:
:$\cmod {z - \alpha} - \cmod {z - \beta} = d$
where $\cmod {\, \cdot \,}$ denotes complex modulus.

Proof:
By definition of complex modulus:
:$\cmod {z - \alpha}$ is the distance from $z$ to $\alpha$
:$\cmod {z - \beta}$ is the distance from $z$ to $\beta$.
Thus $\cmod {z - \alpha} - \cmod {z - \beta}$ is the difference of the distance from $z$ to $\alpha$ and from $z$ to $\beta$.
This is precisely the equidistance property of the hyperbola.
From Equidistance of Hyperbola equals Transverse Axis, the constant distance $d$ is equal to the transverse axis of $H$.
{{qed}}
Theorem: Geodesic Equation/2d Surface Embedded in 3d Euclidean Space

Let $\sigma: U \subset \R^2 \to V \subset \R^3$ be a smooth surface specified by a vector-valued function:
:$\mathbf r = \map {\mathbf r} {u, v}$
Then a geodesic of $\sigma$ satisfies the following system of differential equations:
:$\dfrac {E_u u'^2 + 2 F_u u' v' + G_u v'^2} {\sqrt{E u'^2 + 2 F u' v' + G v'^2} } - \dfrac \d {\d t} \dfrac {2 \paren {E u' + F v'} } {\sqrt{E u'^2 + 2 F u' v' + G v'^2} } = 0$
:$\dfrac {E_v u'^2 + 2 F_v u' v' + G_v v'^2} {\sqrt{E u'^2 + 2 F u' v' + G v'^2} } - \dfrac \d {\d t} \dfrac {2 \paren {F u' + G v'} } {\sqrt{E u'^2 + 2 F u' v' + G v'^2} } = 0$
where $E, F, G$ are the functions of the first fundamental form:
:$E = {\mathbf r}_u \cdot {\mathbf r}_u, F = {\mathbf r}_u \cdot {\mathbf r}_v, G = {\mathbf r}_v \cdot {\mathbf r}_v$

Proof:
A curve on the surface $\mathbf r$ can be specified as $u = \map u t$, $v = \map v t$
The arc length between the points corresponding to $t_0$ and $t_1$ equals:
:$\ds J \sqbrk {u, v} = \int_{t_0}^{t_1} \sqrt {E u'^2 + 2 F u'v' + G v'^2} \rd t$
The following derivatives will appear in Euler's Equations:
{{begin-eqn}}
{{eqn | l = \dfrac {\partial} {\partial u'} \sqrt {E u'^2 + 2 F u'v' + G v'^2}
      | r = \frac 1 {2 \sqrt {E u'^2 + 2 F u'v' + G v'^2} } \map {\dfrac {\partial} {\partial u'} } {E u'^2 + 2 F u'v' + G v'^2}
}}
{{eqn | r = \frac {2 E u' + 2 F_u u'} {2 \sqrt {E u'^2 + 2 F u'v' + G v'^2} }
}}
{{eqn | r = \frac {E u' + F u'} {\sqrt {E u'^2 + 2 F u'v' + G v'^2} }
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \dfrac {\partial} {\partial u} \sqrt {E u'^2 + 2 F u'v' + G v'^2}
      | r = \frac 1 {2 \sqrt {E u'^2 + 2 F u'v' + G v'^2} } \map {\dfrac {\partial} {\partial u} } {E u'^2 + 2 F u'v' + G v'^2}
}}
{{eqn | r = \frac {E_u u'^2 + 2 F_u u' v' + G_u v'^2} {2 \sqrt {E u'^2 + 2 F u'v' + G v'^2} }
}}
{{end-eqn}}
Analogous relations hold for derivatives {{WRT|Differentiation}} $v$ and $v'$.
Euler's Equation together with the results above yield the geodesic equations.
{{qed}}
Theorem: Magnitude and Direction of Equilibrant

Let $\mathbf F_1, \mathbf F_2, \ldots, \mathbf F_n$ be a set of $n$ forces acting on a particle $B$ at a point $P$ in space.
The '''equilibrant''' $\mathbf E$ of $\mathbf F_1, \mathbf F_2, \ldots, \mathbf F_n$ is:
:$\mathbf E = -\ds \sum_{k \mathop = 1}^n \mathbf F_k$
That is, the magnitude and direction of $\mathbf E$ is such as to balance out the effect of $\mathbf F_1, \mathbf F_2, \ldots, \mathbf F_n$.

Proof:
From Newton's First Law of Motion, the total force on $B$ must equal zero in order for $B$ to remain stationary.
That is, $\mathbf E$ must be such that:
:$\mathbf E + \ds \sum_{k \mathop = 1}^n \mathbf F_k = \bszero$
That is:
:$\mathbf E = -\ds \sum_{k \mathop = 1}^n \mathbf F_k$
{{qed}}
Category:Force
Category:Equilibrants
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Theorem: Modulus of Exponential of i z where z is on Circle

Let $C$ be the circle embedded in the complex plane given by the equation:
:$z = R e^{i \theta}$
Then:
:$\cmod {e^{i z} } = e^{-R \sin \theta}$

Proof:
{{begin-eqn}}
{{eqn | l = \cmod {e^{i z} }
      | r = \cmod {\map \exp {i R \, \map \exp {i \theta} } }
      | c = 
}}
{{eqn | r = \cmod {\map \exp {i R \paren {\cos \theta + i \sin \theta} } }
      | c = 
}}
{{eqn | r = \cmod {\map \exp {R \paren {-\sin \theta + i \cos \theta} } }
      | c = 
}}
{{eqn | r = \cmod {\map \exp {- R \sin \theta} \, \map \exp {i \cos \theta} }
      | c = 
}}
{{eqn | r = \map \exp {- R \sin \theta}
      | c = Modulus and Argument of Complex Exponential
}}
{{end-eqn}}
{{qed}}
Theorem: Bias of Sample Variance

Let $X_1, X_2, \ldots, X_n$ form a random sample from a population  with mean $\mu$ and variance $\sigma^2$.
Let: 
:$\ds \bar X = \frac 1 n \sum_{i \mathop = 1}^n X_i$
Then:
:$\ds {S_n}^2 = \frac 1 n \sum_{i \mathop = 1}^n \paren {X_i - \bar X}^2$
is a biased estimator of $\sigma^2$, with: 
:$\map {\operatorname {bias} } { {S_n}^2} = -\dfrac {\sigma^2} n$

Proof:
By definition, ${S_n}^2$ is a biased estimator of $\sigma^2$ {{iff}}: 
:$\expect { {S_n}^2} \ne \sigma^2$
We have: 
{{begin-eqn}}
{{eqn | l = \expect { {S_n}^2}
      | r = \expect {\frac 1 n \sum_{i \mathop = 1}^n \paren {X_i - \bar X}^2}
}}
{{eqn | r = \expect {\frac 1 n \sum_{i \mathop = 1}^n \paren {\paren {X_i - \mu} - \paren {\bar X - \mu} }^2}
      | c = writing $X_i - \bar X = X_i - \bar X - \mu + \mu$
}}
{{eqn | r = \expect {\frac 1 n \sum_{i \mathop = 1}^n \paren {\paren {X_i - \mu}^2 - 2 \paren {\bar X - \mu} \paren {X_i -\mu} + \paren {\bar X - \mu}^2} }
      | c = Square of Difference
}}
{{eqn | r = \expect {\frac 1 n \sum_{i \mathop = 1}^n \paren {X_i - \mu}^2 - \frac 2 n \paren {\bar X - \mu} \sum_{i \mathop = 1}^n \paren {X_i -\mu} + \frac 1 n \paren {\bar X - \mu}^2 \sum_{i \mathop = 1}^n 1}
      | c = Summation is Linear
}}
{{end-eqn}}
We have that: 
{{begin-eqn}}
{{eqn | l = \frac 1 n \sum_{i \mathop = 1}^n \paren {X_i - \mu}
      | r = \frac 1 n \sum_{i \mathop = 1}^n X_i - \frac n n \mu
      | c = from $\ds \sum_{i \mathop = 1}^n 1 = n$, noting that $\mu$ is independent of $i$. 
}}
{{eqn | n = 1
      | r = \bar X - \mu
      | c = {{Defof|Arithmetic Mean}}
}}
{{end-eqn}}
So:
{{begin-eqn}}
{{eqn | o = 
      | r = \expect {\frac 1 n \sum_{i \mathop = 1}^n \paren {X_i - \mu}^2 - \frac 2 n \paren {\bar X - \mu} \sum_{i \mathop = 1}^n \paren {X_i -\mu} + \frac 1 n \paren {\bar X - \mu}^2 \sum_{i \mathop = 1}^n 1}
}}
{{eqn | r = \expect {\frac 1 n \sum_{i \mathop = 1}^n \paren {X_i - \mu}^2 - 2 \paren {\bar X - \mu}^2 + \paren {\bar X - \mu}^2}
      | c = from $(1)$
}}
{{eqn | r = \frac 1 n \expect {\sum_{i \mathop = 1}^n \paren {X_i - \mu}^2} - \expect {\paren {\bar X - \mu}^2}
      | c = Expectation is Linear
}}
{{eqn | r = \frac 1 n \sum_{i \mathop = 1}^n \expect {\paren {X_i - \mu}^2} - \var {\bar X}
      | c = {{Defof|Variance}}, Expectation is Linear
}}
{{eqn | r = \frac 1 n \sum_{i \mathop = 1}^n \var {X_i} - \frac {\sigma^2} n
      | c = {{Defof|Variance}}, Variance of Sample Mean
}}
{{eqn | r = \frac n n \sigma^2 - \frac {\sigma^2} n
      | c = $\var {X_i} = \sigma^2$, $\ds \sum_{i \mathop = 1}^n 1 = n$
}}
{{eqn | r = \sigma^2 - \frac {\sigma^2} n
}}
{{eqn | o = \ne
      | r = \sigma^2
}}
{{end-eqn}}
So ${S_n}^2$ is a biased estimator of $\sigma^2$. 
Further, we have: 
:$\map {\operatorname {bias} } { {S_n}^2} = \sigma^2 - \dfrac {\sigma^2} n - \sigma^2 = -\dfrac {\sigma^2} n$
{{qed}}
Category:Inductive Statistics
tocivfjfefo4empu8f42azmzjfjbopv
Theorem: Bessel's Correction

Let $X_1, X_2, \ldots, X_n$ form a random sample from a population with mean $\mu$ and variance $\sigma^2$.
Let: 
:$\ds \bar X = \frac 1 n \sum_{i \mathop = 1}^n X_i$
Then:
:$\ds \hat {\sigma^2} = \frac 1 {n - 1} \sum_{i \mathop = 1}^n \paren {X_i - \bar X}^2$
is an unbiased estimator of $\sigma^2$.

Proof:
We know that $\hat {\sigma^2}$ is an unbiased estimator of $\sigma^2$ {{iff}}: 
:$\expect {\hat {\sigma^2} } = \sigma^2$
In Bias of Sample Variance, it is shown that: 
:$\ds \expect {\frac 1 n \sum_{i \mathop = 1}^n \paren {X_i - \bar X}^2} = \paren {1 - \frac 1 n} \sigma^2$
By Expectation is Linear: 
{{begin-eqn}}
{{eqn | l = n \expect {\frac 1 n \sum_{i \mathop = 1}^n \paren {X_i - \bar X}^2}
      | r = \expect {\sum_{i \mathop = 1}^n \paren {X_i - \bar X}^2}
}}
{{eqn | r = \paren {n - 1} \sigma^2
}}
{{end-eqn}}
So: 
{{begin-eqn}}
{{eqn | l = \sigma^2
      | r = \frac 1 {n - 1} \expect {\sum_{i \mathop = 1}^n \paren {X_i - \bar X}^2}
}}
{{eqn | r = \expect {\frac 1 {n - 1} \sum_{i \mathop = 1}^n \paren {X_i - \bar X}^2}
      | c = Expectation is Linear
}}
{{eqn | r = \expect {\hat {\sigma^2} }
}}
{{end-eqn}}
So $\hat {\sigma^2}$ is an unbiased estimator of $\sigma^2$. 
{{qed}}
{{Namedfor|Friedrich Wilhelm Bessel|cat = Bessel}}
Category:Inductive Statistics
ihvdoo80jfpq1z1jbb2kl8q8pfnzp1r