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22573
|
\section{Tangent of Half Angle for Spherical Triangles}
Tags: Spherical Trigonometry, Half Angle Formulas for Spherical Triangles
\begin{theorem}
Let $\triangle ABC$ be a spherical triangle on the surface of a sphere whose center is $O$.
Let the sides $a, b, c$ of $\triangle ABC$ be measured by the angles subtended at $O$, where $a, b, c$ are opposite $A, B, C$ respectively.
Then:
:$\tan \dfrac A 2 = \sqrt {\dfrac {\map \sin {s - b} \, \map \sin {s - c} } {\sin s \, \map \sin {s - a} } }$
where $s = \dfrac {a + b + c} 2$.
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | l = \tan \dfrac A 2
| r = \dfrac {\sqrt {\dfrac {\sin \paren {s - b} \sin \paren {s - c} } {\sin b \sin c} } } {\sqrt {\dfrac {\sin s \, \map \sin {s - a} } {\sin b \sin c} } }
| c = Sine of Half Angle for Spherical Triangles, Cosine of Half Angle for Spherical Triangles
}}
{{eqn | r = \dfrac {\sqrt {\sin \paren {s - b} \sin \paren {s - c} } } {\sqrt {\sin s \, \map \sin {s - a} } }
| c = simplification
}}
{{end-eqn}}
The result follows.
{{qed}}
\end{proof}
|
22574
|
\section{Tangent of Half Angle plus Quarter Pi}
Tags: Trigonometric Identities, Tangent Function, Secant Function
\begin{theorem}
:$\map \tan {\dfrac x 2 + \dfrac \pi 4} = \tan x + \sec x$
\end{theorem}
\begin{proof}
Firstly, we have:
{{begin-eqn}}
{{eqn | n = 1
| l = \tan x
| r = \frac {2 \tan \frac x 2} {1 - \tan ^2 \frac x 2}
| c = Double Angle Formula for Tangent
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = \map \tan {\frac x 2 + \frac \pi 4}
| r = \frac {\tan \frac x 2 + \tan \frac \pi 4} {1 - \tan \frac x 2 \tan \frac \pi 4}
| c = Tangent of Sum
}}
{{eqn | r = \frac {\tan \frac x 2 + 1} {1 - \tan \frac x 2}
| c = Tangent of $45 \degrees$
}}
{{eqn | r = \frac {\paren {\tan \frac x 2 + 1} \paren {\tan \frac x 2 + 1} } {\paren {1 - \tan \frac x 2} \paren {\tan \frac x 2 + 1} }
}}
{{eqn | r = \frac {\tan^2 \frac x 2 + 2 \tan \frac x 2 + 1} {1 - \tan^2 \frac x 2}
| c = Difference of Two Squares, Square of Sum
}}
{{eqn | r = \frac {2 \tan \frac x 2} {1 - \tan^2 \frac x 2} + \frac {\tan^2 \frac x 2 + 1} {1 - \tan^2 \frac x 2}
}}
{{eqn | r = \tan x + \frac {\tan^2 \frac x 2 + 1} {1 - \tan^2 \frac x 2}
| c = Double Angle Formula for Tangent: see $(1)$ above
}}
{{eqn | r = \tan x + \frac {\sin^2 \frac x 2 + \cos^2 \frac x 2} {\cos^2 \frac x 2 - \sin^2 \frac x 2}
| c = multiplying Denominator and Numerator by $\cos^2 \frac x 2$
}}
{{eqn | r = \tan x + \frac {\sin^2 \frac x 2 + \cos^2 \frac x 2} {\cos 2 \frac x 2}
| c = Double Angle Formula for Cosine
}}
{{eqn | r = \tan x + \frac 1 {\cos x}
| c = Sum of Squares of Sine and Cosine
}}
{{eqn | r = \tan x + \sec x
| c = Secant is Reciprocal of Cosine
}}
{{end-eqn}}
{{qed}}
Category:Trigonometric Identities
Category:Tangent Function
Category:Secant Function
\end{proof}
|
22575
|
\section{Tangent of Half Side for Spherical Triangles}
Tags: Half Side Formulas for Spherical Triangles
\begin{theorem}
Let $\triangle ABC$ be a spherical triangle on the surface of a sphere whose center is $O$.
Let the sides $a, b, c$ of $\triangle ABC$ be measured by the angles subtended at $O$, where $a, b, c$ are opposite $A, B, C$ respectively.
Then:
:$\tan \dfrac a 2 = \sqrt {\dfrac {-\cos S \, \map \cos {S - A} } {\map \cos {S - B} \, \map \cos {S - C} } }$
where $S = \dfrac {a + b + c} 2$.
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | l = \tan \dfrac a 2
| r = \dfrac {\sqrt {\dfrac {-\cos S \, \map \cos {S - A} } {\sin B \sin C} } } {\sqrt {\dfrac {\map \cos {S - B} \, \map \cos {S - C} } {\sin B \sin C} } }
| c = Sine of Half Angle for Spherical Triangles, Cosine of Half Angle for Spherical Triangles
}}
{{eqn | r = \sqrt {\dfrac {-\cos S \, \map \cos {S - A} } {\map \cos {S - B} \, \map \cos {S - C} } }
| c = simplification
}}
{{end-eqn}}
Hence the result.
{{qed}}
\end{proof}
|
22576
|
\section{Tangent of Right Angle}
Tags: Tangent Function
\begin{theorem}
:$\tan 90^\circ = \tan \dfrac \pi 2$ is undefined
where $\tan$ denotes tangent.
\end{theorem}
\begin{proof}
From Tangent is Sine divided by Cosine, $\tan \theta = \dfrac {\sin \theta} {\cos \theta}$.
When $\cos \theta = 0$, $\dfrac {\sin \theta} {\cos \theta}$ can be defined only if $\sin \theta = 0$.
But there are no such $\theta$ such that both $\cos \theta = 0$ and $\sin \theta = 0$.
When $\theta = \dfrac \pi 2$, $\cos \theta = 0$.
Thus $\tan \theta$ is undefined at this value.
{{qed}}
\end{proof}
|
22577
|
\section{Tangent of Three Right Angles}
Tags: Tangent Function
\begin{theorem}
:$\tan 270^\circ = \tan \dfrac {3 \pi} 2$ is undefined
where $\tan$ denotes tangent.
\end{theorem}
\begin{proof}
We have:
{{begin-eqn}}
{{eqn | l = \tan 270^\circ
| r = \tan \left({360^\circ - 90^\circ}\right)
| c =
}}
{{eqn | r = -\tan 90^\circ
| c = Tangent of Conjugate Angle
}}
{{end-eqn}}
But from Tangent of Right Angle, $\tan 90^\circ$ is undefined.
Hence so is $\tan 270^\circ$.
{{qed}}
\end{proof}
|
22578
|
\section{Tangent over Secant Plus One}
Tags: Trigonometric Identities
\begin{theorem}
:$\dfrac {\tan x} {\sec x + 1} = \dfrac {\sec x - 1} {\tan x}$
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | l = \frac {\tan x} {\sec x + 1}
| r = \frac {\sin x} {\cos x \paren {\sec x + 1} }
| c = Tangent is Sine divided by Cosine
}}
{{eqn | r = \frac {\sin x} {\cos x \paren {\frac 1 {\cos x} + 1} }
| c = Secant is Reciprocal of Cosine
}}
{{eqn | r = \frac {\sin x} {1 + \cos x}
| c = simplifying
}}
{{eqn | r = \frac {\sin^2 x} {\sin x \paren {1 + \cos x} }
| c = Multiply by $1 = \dfrac {\sin x} {\sin x}$
}}
{{eqn | r = \frac {1 - \cos^2 x} {\sin x \paren {1 + \cos x} }
| c = Sum of Squares of Sine and Cosine
}}
{{eqn | r = \frac {\paren {1 + \cos x} \paren {1 - \cos x} } {\sin x \paren {1 + \cos x} }
| c = Difference of Two Squares
}}
{{eqn | r = \frac {1 - \cos x} {\sin x}
}}
{{eqn | r = \frac {\cos x \paren {\frac 1 {\cos x} - 1} } {\sin x}
}}
{{eqn | r = \frac {\cos x \paren {\sec x - 1} } {\sin x}
| c = Secant is Reciprocal of Cosine
}}
{{eqn | r = \frac {\sec x - 1} {\tan x}
| c = Tangent is Sine divided by Cosine
}}
{{end-eqn}}
{{qed}}
Category:Trigonometric Identities
\end{proof}
|
22579
|
\section{Tangent times Tangent plus Cotangent}
Tags: Trigonometric Identities, Tangent times Tangent plus Cotangent
\begin{theorem}
:$\tan x \paren {\tan x + \cot x} = \sec^2 x$
where $\tan$, $\cot$ and $\sec$ denote tangent, cotangent and secant respectively.
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | l=\tan x \left({\tan x + \cot x}\right)
| r=\tan x \sec x \csc x
| c=Sum of Tangent and Cotangent
}}
{{eqn | r=\frac {\sin x} {\cos^2 x \sin x}
| c=by definition of tangent, secant and cosecant
}}
{{eqn | r=\frac 1 {\cos^2x}
| c=
}}
{{eqn | r=\sec^2x
| c=by definition of secant
}}
{{end-eqn}}
{{qed}}
Or directly:
{{begin-eqn}}
{{eqn | l=\tan x \left({\tan x + \cot x}\right)
| r=\frac {\sin x} {\cos x} \left({\frac {\sin x} {\cos x} + \frac {\cos x} {\sin x} }\right)
| c=by definition of tangent and cotangent
}}
{{eqn | r=\frac {\sin x} {\cos x} \left({\frac {\sin^2 x + \cos^2 x} {\cos x \sin x} }\right)
| c=
}}
{{eqn | r=\frac {\sin x} {\cos x} \left({\frac 1 {\cos x \sin x} }\right)
| c=Sum of Squares of Sine and Cosine
}}
{{eqn | r=\frac 1 {\cos^2x}
| c=
}}
{{eqn | r=\sec^2x
| c=by definition of secant
}}
{{end-eqn}}
{{qed}}
Category:Trigonometric Identities
69372
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2011-11-09T19:38:42Z
Lord Farin
560
69372
wikitext
text/x-wiki
\end{proof}
|
22580
|
\section{Tangent to Astroid between Coordinate Axes has Constant Length}
Tags: Hypocycloids
\begin{theorem}
Let $C_1$ be a circle of radius $b$ roll without slipping around the inside of a circle $C_2$ of radius $a = 4 b$.
Let $C_2$ be embedded in a cartesian plane with its center $O$ located at the origin.
Let $P$ be a point on the circumference of $C_1$.
Let $C_1$ be initially positioned so that $P$ is its point of tangency to $C_2$, located at point $A = \tuple {a, 0}$ on the $x$-axis.
Let $H$ be the astroid formed by the locus of $P$.
The segment of the tangent to $H$ between the $x$-axis and the $y$-axis is constant, immaterial of the point of tangency.
\end{theorem}
\begin{proof}
:400px
From Equation of Astroid, $H$ can be expressed as:
:$\begin{cases}
x & = a \cos^3 \theta \\
y & = a \sin^3 \theta
\end{cases}$
Thus the slope of the tangent to $H$ at $\tuple {x, y}$ is:
{{begin-eqn}}
{{eqn | l = \frac {\d y} {\d x}
| r = \frac {3 a \sin^2 \theta \cos \theta \rd \theta} {-3 a \cos^2 \theta \sin \theta \rd \theta}
| c =
}}
{{eqn | r = -\tan \theta
| c =
}}
{{end-eqn}}
Thus the equation of the tangent to $H$ is given by:
:$y - a \sin^3 \theta = -\tan \theta \paren {x - a \cos^3 \theta}$
{{explain|Find, or post up, the equation of a line of given tangent passing through point $\tuple {x, y}$ as this is what is needed here}}
The $x$-intercept is found by setting $y = 0$ and solving for $x$:
{{begin-eqn}}
{{eqn | l = x
| r = a \cos^3 \theta + a \sin^2 \theta \cos \theta
| c =
}}
{{eqn | r = a \cos \theta \paren {\cos^2 \theta + \sin^2 \theta}
| c =
}}
{{eqn | r = a \cos \theta
| c = Sum of Squares of Sine and Cosine
}}
{{end-eqn}}
Similarly, the $y$-intercept is found by setting $x = 0$ and solving for $y$, which gives:
:$y = a \sin \theta$
The length of the part of the tangent to $H$ between the $x$-axis and the $y$-axis is given by:
{{begin-eqn}}
{{eqn | l = \sqrt {a^2 \cos^2 \theta + a^2 \sin^2 \theta}
| r = a \sqrt {\cos^2 \theta + \sin^2 \theta}
| c =
}}
{{eqn | r = a
| c = Sum of Squares of Sine and Cosine
}}
{{end-eqn}}
which is constant.
{{qed}}
\end{proof}
|
22581
|
\section{Tangent to Cycloid}
Tags: Cycloids
\begin{theorem}
Let $C$ be a cycloid generated by the equations:
:$x = a \paren {\theta - \sin \theta}$
:$y = a \paren {1 - \cos \theta}$
Then the tangent to $C$ at a point $\tuple {x, y}$ on $C$ is given by the equation:
:$y - a \paren {1 - \cos \theta} = \dfrac {\sin \theta} {1 - \cos \theta} \paren {x - a \theta + a \sin \theta}$
\end{theorem}
\begin{proof}
From Slope of Tangent to Cycloid, the slope of the tangent to $C$ at the point $\tuple {x, y}$ is given by:
:$\dfrac {\d y} {\d x} = \cot \dfrac \theta 2$
This tangent to $C$ also passes through the point $\tuple {a \paren {\theta - \sin \theta}, a \paren {1 - \cos \theta} }$.
{{Finish|Find (or write) the result which gives the equation of a line from its slope and a point through which it passes.}}
Hence the result.
{{qed}}
\end{proof}
|
22582
|
\section{Tangent to Cycloid is Vertical at Cusps}
Tags: Cycloids
\begin{theorem}
The tangent to the cycloid whose locus is given by:
:$x = a \paren {\theta - \sin \theta}$
:$y = a \paren {1 - \cos \theta}$
is vertical at the cusps.
\end{theorem}
\begin{proof}
From Slope of Tangent to Cycloid, the slope of the tangent to $C$ at the point $\tuple {x, y}$ is given by:
:$\dfrac {\d y} {\d x} = \cot \dfrac \theta 2$
At the cusps, $\theta = 2 n \pi$ for $n \in \Z$.
Thus at the cusps, the slope of the tangent to $C$ is $\cot n \pi$.
From Shape of Cotangent Function:
:$\ds \lim_{\theta \mathop \to n \pi} \cot \theta \to \infty$
Hence the result by definition of vertical tangent line.
{{qed}}
\end{proof}
|
22583
|
\section{Tangent to Cycloid passes through Top of Generating Circle}
Tags: Cycloids
\begin{theorem}
Let $C$ be a cycloid generated by the equations:
:$x = a \paren {\theta - \sin \theta}$
:$y = a \paren {1 - \cos \theta}$
Then the tangent to $C$ at a point $P$ on $C$ passes through the top of the generating circle of $C$.
\end{theorem}
\begin{proof}
From Tangent to Cycloid, the equation for the tangent to $C$ at a point $P = \tuple {x, y}$ is given by:
:$(1): \quad y - a \paren {1 - \cos \theta} = \dfrac {\sin \theta} {1 - \cos \theta} \paren {x - a \theta + a \sin \theta}$
From Equation of Cycloid, the point at the top of the generating circle of $C$ has coordinates $\tuple {2 a, a \theta}$.
Substituting $x = 2 a$ in $(1)$:
{{begin-eqn}}
{{eqn | l = y - a \paren {1 - \cos \theta}
| r = \dfrac {\sin \theta} {1 - \cos \theta} \paren {a \theta - a \theta + a \sin \theta}
| c =
}}
{{eqn | ll= \leadsto
| l = y
| r = a \paren {1 - \cos \theta} + \dfrac {\sin \theta} {1 - \cos \theta} a \sin \theta
| c =
}}
{{eqn | r = \frac {a \paren {1 - \cos \theta}^2 + a \sin^2 \theta} {1 - \cos \theta}
| c =
}}
{{eqn | r = \frac {a \paren {1 - 2 \cos \theta + \cos^2 \theta} + a \sin^2 \theta} {1 - \cos \theta}
| c =
}}
{{eqn | r = \frac {a \paren {1 - 2 \cos \theta + 1} } {1 - \cos \theta}
| c =
}}
{{eqn | r = \frac {a \paren {2 - 2 \cos \theta} } {1 - \cos \theta}
| c =
}}
{{eqn | r = 2 a
| c =
}}
{{end-eqn}}
That is, the tangent to $C$ passes through $\tuple {a \theta, 2 a}$ as was required.
{{qed}}
\end{proof}
|
22584
|
\section{Tarski's Geometry is Complete/Corollary}
Tags: Tarski's Geometry
\begin{theorem}
Tarski's geometry does not contain minimal arithmetic.
{{explain|contain}}
\end{theorem}
\begin{proof}
Immediate from Tarski's Geometry is Complete and the corollary to Gödel's First Incompleteness Theorem.
{{qed}}
Category:Tarski's Geometry
\end{proof}
|
22585
|
\section{Tarski's Undefinability Theorem}
Tags: Mathematical Logic
\begin{theorem}
Let $\ZZ$ be the standard structure $\struct {\Z, +, \cdot, s, <, 0}$ for the language of arithmetic.
Let $\operatorname {Th}_\ZZ$ be the sentences which are true in $\ZZ$.
Let $\Theta$ be the set of Gödel numbers of those sentences in $\operatorname {Th}_\ZZ$.
$\Theta$ is not definable in $\operatorname {Th}_\ZZ$.
\end{theorem}
\begin{proof}
$\operatorname {Th}_\ZZ$ is easily seen to be a consistent extension of minimal arithmetic. (In fact, the axioms in minimal arithmetic were selected based on the behavior of standard arithmetic.)
Thus, the theorem is a special case of Set of Gödel Numbers of Arithmetic Theorems Not Definable in Arithmetic (and can be seen to follow immediately).
{{qed}}
{{Namedfor|Alfred Tarski|cat = Tarski}}
\end{proof}
|
22586
|
\section{Tartaglia's Formula}
Tags: Tetrahedra, Geometry, Solid Geometry
\begin{theorem}
Let $T$ be a tetrahedron with vertices $\mathbf d_1, \mathbf d_2, \mathbf d_3$ and $\mathbf d_4$.
For all $i$ and $j$, let the distance between $\mathbf d_i$ and $\mathbf d_j$ be denoted $d_{ij}$.
Then the volume $V_T$ of $T$ satisfies:
:$V_T^2 = \dfrac {1} {288} \det \ \begin{vmatrix}
0 & 1 & 1 & 1 & 1\\
1 & 0 & d_{12}^2 & d_{13}^2 & d_{14}^2 \\
1 & d_{12}^2 & 0 & d_{23}^2 & d_{24}^2 \\
1 & d_{13}^2 & d_{23}^2 & 0 & d_{34}^2 \\
1 & d_{14}^2 & d_{24}^2 & d_{34}^2 & 0
\end{vmatrix}$
\end{theorem}
\begin{proof}
A proof of Tartaglia's Formula will be found in a proof of the Value of Cayley-Menger Determinant as a tetrahedron is a $3$-simplex.
{{proof wanted}}
{{Namedfor|Niccolò Fontana Tartaglia|cat = Tartaglia}}
\end{proof}
|
22587
|
\section{Taxicab Metric is Metric}
Tags: Taxicab Metric is Metric, Metric Spaces, Taxicab Metric, Examples of Metrics
\begin{theorem}
The taxicab metric is a metric.
\end{theorem}
\begin{proof}
From the definition, the taxicab metric is as follows:
Let $M_{1'} = \left({A_{1'}, d_{1'}}\right), M_{2'} = \left({A_{2'}, d_{2'}}\right), \ldots, M_{n'} = \left({A_{n'}, d_{n'}}\right)$ be a finite number of metric spaces.
Let $\mathcal A$ be the Cartesian product $\displaystyle \prod_{i \mathop = 1}^n A_{i'}$.
The taxicab metric on $\displaystyle \mathcal A$ is:
: $\displaystyle d_1 \left({x, y}\right) = \sum_{i \mathop = 1}^n d_{i'} \left({x_{i'}, y_{i'}}\right)$
for $x = \left({x_1, x_2, \ldots, x_n}\right), y = \left({y_1, y_2, \ldots, y_n}\right) \in \mathcal A$.
\end{proof}
|
22588
|
\section{Taxicab Metric is Topologically Equivalent to Chebyshev Distance on Real Vector Space}
Tags: Maximum Metric, Taxicab Metric, Chebyshev Distance
\begin{theorem}
For $n \in \N$, let $\R^n$ be a real vector space.
Let $d_1$ be the taxicab metric on $\R^n$.
Let $d_\infty$ be the Chebyshev distance on $\R^n$.
Then
:$\forall x, y \in \R^n: \map {d_\infty} {x, y} \le \map {d_1} {x, y} \le n \cdot \map {d_\infty} {x, y}$
It follows that $d_1$ and $d_\infty$ are Lipschitz equivalent.
\end{theorem}
\begin{proof}
By definition of the Chebyshev distance on $\R^n$, we have:
:$\ds \map {d_\infty} {x, y} = \max_{i \mathop = 1}^n {\size {x_i - y_i} }$
where $x = \tuple {x_1, x_2, \ldots, x_n}$ and $y = \tuple {y_1, y_2, \ldots, y_n}$.
Let $j$ be chosen so that:
:$\ds \size {x_j - y_j} = \max_{i \mathop = 1}^n {\size {x_i - y_i} }$
Then:
{{begin-eqn}}
{{eqn | l = \map {d_\infty} {x, y}
| r = \size {x_j - y_j}
| c = {{Defof|Chebyshev Distance on Real Vector Space}}
}}
{{eqn | o = \le
| r = \sum_{i \mathop = 1}^n \size {x_i - y_i}
}}
{{eqn | r = \map {d_1} {x, y}
}}
{{eqn | o = \le
| r = n \size {x_j - y_j}
}}
{{eqn | r = n \cdot \map {d_\infty} {x, y}
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
22589
|
\section{Taxicab Metric on Metric Space Product is Continuous}
Tags: Continuous Mappings on Metric Spaces, Taxicab Metric
\begin{theorem}
Let $M = \struct {A, d}$ be a metric space.
Let $\AA$ be the Cartesian product $A \times A$.
Let $d_1$ be the taxicab metric on $\AA$:
:$\ds \map {d_1} {x, y} = \sum_{i \mathop = 1}^n \map {d_{i'} } {x_{i'}, y_{i'} }$
for $x = \tuple {x_1, x_2, \ldots, x_n}, y = \tuple {y_1, y_2, \ldots, y_n} \in \AA$.
Then $d_1: \AA \to \AA$ is a continuous function.
\end{theorem}
\begin{proof}
Recall the definition of continuous mapping in this context.
Given metric spaces $M_X = \struct {X, d_X}$ and $M_Y = \struct {Y, d_Y}$, and a mapping $f : X \to Y$, we say that $f$ is $\struct {X, d_X} \to \struct {Y, d_Y}$-continuous {{iff}}:
:$\forall x_0 \in X: \forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: \forall x \in X: \map {d_X} {x, x_0} < \delta \implies \map d {\map f x, \map f {x_0} } < \epsilon$
Hence it is necessary to prove that $d: \struct {A \times A, d_1} \to \struct {\R, \size {\,\cdot \,} }$ is continuous.
That is, given $\tuple {x_0, y_0} \in A \times A$ and $\epsilon>0$, it is necessary to find $\delta \in \R_{>0}$ such that:
:$\forall x, y \in A: \map {d_1} {\tuple {x, y}, \tuple {x_0, y_0} } < \delta \implies \size {\map d {x, y} - \map d {x_0, y_0} } < \epsilon$
Hence, let $\tuple {x_0, y_0} \in A \times A$.
If $\map {d_1} {\tuple {x, y}, \tuple {x_0, y_0} } = \map d {x, x_0} + \map d {y, y_0} < \epsilon$, then:
:$\size {\map d {x, y} - \map d {x_0, y_0} } \le \size {\map d {x, y} - \map d {x, y_0} } + \size {\map d {x, y_0} - \map d {x_0, y_0} } \le 2 \epsilon$
using {{Metric-space-axiom|2}}.
The result follows.
{{qed}}
\end{proof}
|
22590
|
\section{Taxicab Metric on Real Number Plane is Translation Invariant}
Tags: Translation Mappings, Taxicab Metric
\begin{theorem}
Let $\tau_{\mathbf t}: \R^2 \to \R^2$ denote the translation of the Euclidean plane by the vector $\mathbf t = \begin {pmatrix} a \\ b \end {pmatrix}$.
Let $d_1$ denote the taxicab metric on $\R^2$.
Then $d_1$ is unchanged by application of $\tau$:
:$\forall x, y \in \R^2: \map {d_1} {\map \tau x, \map \tau y} = \map {d_1} {x, y}$
\end{theorem}
\begin{proof}
Let $x = \tuple {x_1, x_2}$ and $y = \tuple {y_1, y_2}$ be arbitrary points in $\R^2$.
Then:
{{begin-eqn}}
{{eqn | l = \map {d_1} {\map \tau x, \map \tau y}
| r = \map {d_1} {x - \mathbf t, y - \mathbf t}
| c = {{Defof|Translation in Euclidean Space}}
}}
{{eqn | r = \size {\paren {x_1 - a} - \paren {y_1 - a} } + \size {\paren {x_2 - b} - \paren {y_2 - b} }
| c = Definition of $\mathbf t$, {{Defof|Taxicab Metric on Real Number Plane}}
}}
{{eqn | r = \size {x_1 - y_1} + \size {x_2 - y_2}
| c = simplification
}}
{{eqn | r = \map {d_1} {x, y}
| c = {{Defof|Taxicab Metric on Real Number Plane}}
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
22591
|
\section{Taxicab Metric on Real Number Plane is not Rotation Invariant}
Tags: Geometric Rotations, Taxicab Metric
\begin{theorem}
Let $r_\alpha: \R^2 \to \R^2$ denote the rotation of the Euclidean plane about the origin through an angle of $\alpha$.
Let $d_1$ denote the taxicab metric on $\R^2$.
Then it is not necessarily the case that:
:$\forall x, y \in \R^2: \map {d_1} {\map {r_\alpha} x, \map {r_\alpha} y} = \map {d_1} {x, y}$
\end{theorem}
\begin{proof}
Proof by Counterexample:
Let $x = \tuple {0, 0}$ and $y = \tuple {0, 1}$ be arbitrary points in $\R^2$.
Then:
{{begin-eqn}}
{{eqn | l = \map {d_1} {x, y}
| r = \map {d_1} {\tuple {0, 0}, \tuple {0, 1} }
| c = Definition of $x$ and $y$
}}
{{eqn | r = \size {0 - 0} + \size {0 - 1}
| c = {{Defof|Taxicab Metric on Real Number Plane}}
}}
{{eqn | r = 1
| c =
}}
{{end-eqn}}
Now let $\alpha = \dfrac \pi 4 = 45 \degrees$.
{{begin-eqn}}
{{eqn | l = \map {d_1} {\map {r_\alpha} x, \map {r_\alpha} y}
| r = \map {d_1} {\tuple {0, 0}, \tuple {\dfrac {\sqrt 2} 2, \dfrac {\sqrt 2} 2} }
| c = {{Defof|Plane Rotation}}
}}
{{eqn | r = \size {0 - \dfrac {\sqrt 2} 2} + \size {0 - \dfrac {\sqrt 2} 2}
| c = {{Defof|Taxicab Metric on Real Number Plane}}
}}
{{eqn | r = \sqrt 2
| c = simplification
}}
{{eqn | o = \ne
| r = 1
| c =
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
22592
|
\section{Taxicab Metric on Real Vector Space is Metric}
Tags: Taxicab Metric on Real Vector Space is Metric, Taxicab Metric
\begin{theorem}
The taxicab metric on the real vector space $\R^n$ is a metric.
\end{theorem}
\begin{proof}
This is an instance of the taxicab metric on the general cartesian product of $A_{1'}, A_{2'}, \ldots, A_{n'}$.
This is proved in Taxicab Metric is Metric.
{{qed}}
\end{proof}
|
22593
|
\section{Taxicab Norm is Norm}
Tags: Taxicab Norm, Norm Theory, Examples of Norms
\begin{theorem}
The taxicab norm is a norm on the real and complex numbers.
\end{theorem}
\begin{proof}
By P-Norm is Norm, $\norm {\, \cdot \,}_p$ is a norm.
By definition, the taxicab norm is $\norm {\, \cdot \,}_1$.
Therefore, the taxicab norm is a norm.
{{qed}}
\end{proof}
|
22594
|
\section{Taylor's Theorem/One Variable with Two Functions}
Tags: Taylor's Theorem
\begin{theorem}
Let $f$ and $g$ be real functions satisfying following conditions:
:$(1): \quad f$ is $n + 1$ times differentiable on the open interval $\openint a x$
:$(2): \quad f$ is of differentiability class $C^n$ on the closed interval $\closedint a x$
:$(3): \quad g$ is $k + 1$ times differentiable on the open interval $\openint a x$
:$(4): \quad g$ is of differentiability class $C^k$ on the closed interval $\closedint a x$
:$(5): \quad \map {g^{\paren {k + 1}}} t \ne 0$ for any $t \in \openint a x$
Then the following equation holds for some real number $\xi \in \openint a x$:
:$\dfrac {\map {f^{\paren {n + 1} } } \xi /n!} {\map {g^{\paren {k + 1} } } \xi /k!} \paren {x - \xi}^{n - k} = \dfrac {\map f x - \map f a - \map {f'} a \paren {x - a} - \dfrac {\map {f''} a} {2!} \paren {x - a}^2 - \dotsb - \dfrac {\map {f^{\paren n} } a} {n!} \paren {x - a}^n} {\map g x - \map g a - \map {g'} a \paren {x - a} - \dfrac {\map {g''} a} {2!} \paren {x - a}^2 - \dotsb - \dfrac {\map {g^{\paren k} } a} {k!} \paren {x - a}^k}$
or equivalently:
{{begin-eqn}}
{{eqn | l = \map f x
| r = \map f a + \map {f'} a \paren {x - a} + \dfrac {\map {f''} a} {2!} \paren {x - a}^2 + \dotsb + \dfrac {\map {f^{\paren n} } a} {n!} \paren {x - a}^n + R_n
}}
{{eqn | l = R_n
| r = \dfrac {\map {f^{\paren {n + 1} } } \xi / n!} {\map {g^{\paren {k + 1} } } \xi / k!} \paren {x - \xi}^{n - k} \paren {\map g x - \map g a - \map {g'} a \paren {x - a} - \dfrac {\map {g''} a} {2!} \paren {x - a}^2 - \dotsb - \dfrac {\map {g^{\paren k} } a} {k!} \paren {x - a}^k}
}}
{{end-eqn}}
\end{theorem}
\begin{proof}
We define $F$ and $G$ as follows:
{{begin-eqn}}
{{eqn | l = \map F t
| r = \map f t + \map {f'} t \paren {x - t} + \dfrac {\map {f''} t} {2!} \paren {x - t}^2 + \dotsb + \dfrac {\map {f^{\paren n} } t} {n!} \paren {x - t}^n
}}
{{eqn | l = \map G t
| r = \map g t + \map {g'} t \paren {x - t} + \dfrac {\map {g''} t} {2!} \paren {x - t}^2 + \dotsb + \dfrac {\map {g^{\paren k} } t} {k!} \paren {x - t}^k
}}
{{end-eqn}}
Then $F$ and $G$ are continuous on $\closedint a x$ and differentiable on $\openint a x$.
Differentiating {{WRT|Differentiation}} $t$:
{{begin-eqn}}
{{eqn | l = \map {F'} t
| r = \map {f'} t + \paren {\map {f''} t \paren {x - t} - \map {f'} t} + \paren {\frac {\map {f^{\paren 3} } t} {2!} \paren {x - t}^2 - \frac {\map {f^{\paren 2} } t} {1!} \paren {x - t} } + \dotsb + \paren {\frac {\map {f^{\paren {n + 1} } } t} {n!} \paren {x - t}^n - \frac {\map {f^{\paren n} } t} {\paren {n - 1}!} \paren {x - t}^{n - 1} }
}}
{{eqn | r = \frac {\map {f^{\paren {n + 1} } } t} {n!} \paren {x - t}^n
}}
{{eqn | l = \map {G'} t
| r = \frac {\map {g^{\paren {k + 1} } } t} {k!} \paren {x - t}^k
}}
{{end-eqn}}
By condition $(5)$ of the statement of the theorem, it follows that $G$ does not vanish on $\openint a x$.
By Cauchy Mean Value Theorem, there exists a real number $\xi \in \openint a x$ such that:
:$\dfrac {\map {F'} \xi} {\map {G'} \xi} = \dfrac {\map F x - \map F a} {\map G x - \map G a}$
That is:
:$\dfrac {\map {f^{\paren {n + 1} } } \xi / n!} {\map {g^{\paren {k + 1} } } \xi /k!} \paren {x - \xi}^{n - k} = \dfrac {\map f x - \map f a - \map {f'} a \paren {x - a} - \dfrac {\map {f''} a} {2!} \paren {x - a}^2 - \dotsb - \dfrac {\map {f^{\paren n} } a} {n!} \paren {x - a}^n} {\map g x - \map g a - \map {g'} a \paren {x - a} - \dfrac {\map {g''} a} {2!} \paren {x - a}^2 - \dotsb - \dfrac {\map {g^{\paren k}} a} {k!} \paren {x - a}^k}$
{{qed}}
\end{proof}
|
22595
|
\section{Taylor Series of Analytic Function has infinite Radius of Convergence}
Tags: Taylor Series, Real Analysis
\begin{theorem}
Let $F$ be a complex function.
Let $F$ be analytic everywhere.
Let the restriction of $F$ to $\R \to \C$ be a real function $f$.
This means:
:$\forall x \in \R: \map f x = \map \Re {\map F {x, 0} }, 0 = \map \Im {\map F {x, 0} }$
where $\tuple {x, 0}$ denotes the complex number with real part $x$ and imaginary part $0$.
Let $x_0$ be a point in $\R$.
Then:
:the Taylor series of $f$ about $x_0$ converges to $f$ at every point in $\R$.
\end{theorem}
\begin{proof}
The result follows by Convergence of Taylor Series of Function Analytic on Disk for the case $R = \infty$.
{{qed}}
\end{proof}
|
22596
|
\section{Taylor Series of Holomorphic Function}
Tags: Complex Analysis
\begin{theorem}
Let $a \in \C$ be a complex number.
Let $r > 0$ be a real number.
Let $f$ be a function holomorphic on some open ball, $D = B \paren {a, r}$.
Then:
:$\ds \map f z = \sum_{n \mathop = 0}^\infty \frac {\map {f^n} a} {n!} \paren {z - a}^n$
for all $z \in D$.
\end{theorem}
\begin{proof}
In Holomorphic Function is Analytic, it is shown that:
:$\ds \map f z = \sum_{n \mathop = 0}^\infty \paren {\frac 1 {2 \pi i} \int_{\partial D} \frac {\map f t} {\paren {t - a}^{n + 1} } \rd t} \paren {z - a}^n$
for all $z \in D$.
From Cauchy's Integral Formula: General Result, we have:
:$\ds \frac 1 {2 \pi i} \int_{\partial D} \frac {\map f t} {\paren {t - a}^{n + 1} } \rd t = \frac {\map {f^n} a} {n!}$
Hence the result.
{{qed}}
Category:Complex Analysis
\end{proof}
|
22597
|
\section{Taylor Series of Logarithm of Gamma Function}
Tags: Gamma Function, Natural Logarithm, Natural Logarithms
\begin{theorem}
Let $\gamma$ denote the Euler-Mascheroni constant.
Let $\map \zeta s$ denote the Riemann zeta function.
Let $\map \Gamma z$ denote the gamma function.
Let $\Log$ denote the natural logarithm.
Then $\map \Log {\map \Gamma z}$ has the power series expansion:
{{begin-eqn}}
{{eqn | l = \map \Log {\map \Gamma z}
| r = -\map \gamma {z - 1} + \sum_{k \mathop = 2}^\infty \frac {\paren {-1}^k \map \zeta k} k \paren {z - 1}^k
}}
{{end-eqn}}
which is valid for all $z \in \C$ such that $\cmod {z - 1} < 1$.
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | l = \map \Gamma {x + 1}
| r = x \map \Gamma x
| c = Gamma Difference Equation
}}
{{eqn | r = \paren {x } \paren {x - 1 } \map \Gamma {x - 1 }
| c =
}}
{{eqn | r = \paren {x } \paren {x - 1 } \paren {x - 2 } \map \Gamma {x - 2 }
| c =
}}
{{eqn | r = \paren {x } \paren {x - 1 } \paren {x - 2 } \cdots \paren {x - \floor x } \map \Gamma {x -\floor x }
| c = For $\floor x$, see the {{Defof|Floor Function}}
}}
{{end-eqn}}
Hence:
{{begin-eqn}}
{{eqn | l = \map \Log {\map \Gamma {x + 1} }
| r = \map \Log { \paren {x } \paren {x - 1 } \paren {x - 2 } \cdots \paren {x - \floor x } \map \Gamma {x - \floor x } }
| c =
}}
{{eqn | r = \map \Log x + \map \Log {x - 1 } + \map \Log {x - 2 } + \cdots + \map \Log {x - \floor x } + \map \Log {\map \Gamma {x - \floor x } }
| c = Sum of Logarithms
}}
{{end-eqn}}
Therefore:
{{begin-eqn}}
{{eqn | l = \frac \d {\d x} \map \Log {\map \Gamma {x + 1} }
| r = \frac \d {\d x} \paren {\map \Log x + \map \Log {x - 1} + \map \Log {x - 2 } + \cdots + \map \Log {x - \floor x } + \map \Log {\map \Gamma {x - \floor x} } }
| c =
}}
{{eqn | r = \frac \d {\d x} \map \Log x + \frac \d {\d x} \map \Log {x - 1} + \frac \d {\d x} \map \Log {x - 2} + \cdots + \frac \d {\d x} \map \Log {x - \floor x} + \frac \d {\d x} \map \Log {\map \Gamma {x - \floor x} }
| c = Sum Rule for Derivatives
}}
{{eqn | r = \frac 1 x + \frac 1 {x - 1} + \frac 1 {x - 2} + \cdots + \frac 1 {x - \floor x} + \frac \d {\d x} \map \Log {\map \Gamma {x - \floor x} }
| c = Derivative of Natural Logarithm Function
}}
{{eqn | n = 1
| r = \sum_{k \mathop = 0}^{\floor x} \frac 1 {x - k} + \frac \d {\d x} \map \Log {\map \Gamma {x - \floor x} }
| c =
}}
{{end-eqn}}
Setting:
:$z = \paren {x - \floor x} \leadsto \d z = \d x$
:$M = \floor x + 1$:
we have:
:$z + M = x + 1$
Hence:
{{begin-eqn}}
{{eqn | l = \frac \d {\d x} \map \Log {\map \Gamma {x - \floor x} }
| r = \frac \d {\d x} \map \Log {\map \Gamma {x + 1} } - \sum_{k \mathop = 0}^{\floor x} \frac 1 {x - k}
| c = rearranging $(1)$
}}
{{eqn | l = \frac \d {\d z} \map \Log {\map \Gamma z}
| r = \frac \d {\d z} \map \Log {\map \Gamma {z + M} } - \sum_{k \mathop = 0}^{M - 1 } \frac 1 {\paren {z + M - 1} - k}
| c = substituting with the values above
}}
{{eqn | n = 2
| l = \frac \d {\d z} \map \Log {\map \Gamma z}
| r = \frac \d {\d z} \map \Log {\map \Gamma {z + M} } - \sum_{k \mathop = 0}^{M - 1 } \frac 1 {z + k}
| c =
}}
{{eqn | ll= \leadsto
| l = \frac {\d^n} {\d z^n} \map \Log {\map \Gamma z}
| r = \frac {\d^n} {\d z^n} \map \Log {\map \Gamma {z + M} } + \sum_{k \mathop = 0}^{M - 1} \frac {\paren {-1}^n \paren {n - 1}!} {\paren {z + k}^n}
| c = Sum Rule for Derivatives and Nth Derivative of Reciprocal of Mth Power
}}
{{end-eqn}}
Recall Stirling's Formula for Gamma Function:
{{begin-eqn}}
{{eqn | l = \map \Gamma {z + 1}
| r = \sqrt {2 \pi z} \, z^z e^{-z} \paren {1 + \dfrac 1 {12 z} + \dfrac 1 {288 z^2} - \dfrac {139} {51 \, 480 z^3} + \cdots}
| c =
}}
{{eqn | l = \map \Log {\map \Gamma {z + 1} }
| r = \frac 1 2 \map \Log {2 \pi } + \frac 1 2 \map \Log z + z \map \Log z - z + \map \log {1 + \dfrac 1 {12 z} + \dfrac 1 {288 z^2} - \dfrac {139} {51 \, 480 z^3} + \cdots}
| c = Sum of Logarithms and Logarithm of Power/Natural Logarithm
}}
{{eqn | l = \frac \d {\d z} \map \Log {\map \Gamma {z + 1} }
| r = 0 + \dfrac 1 {2 z} + \paren {\map \Log z + \dfrac z z } - 1 + \map \OO {\frac 1 z}
| c = Product Rule for Derivatives and Derivative of Natural Logarithm Function
}}
{{eqn | r = \map \Log z + \map \OO {\frac 1 z}
| c = For $\map \OO {\frac 1 z}$, see {{Defof|Big-O Notation}}
}}
{{eqn | ll = \leadsto
| l = \frac {\d^n} {\d z^n} \map \Log {\map \Gamma {z + 1} }
| r = \map \OO {\frac 1 z}
| c = For $n > 1$
}}
{{end-eqn}}
We now have:
:$\ds \frac \d {\d z} \map \Log {\map \Gamma {z + 1} } = \map \Log z + \map \OO {\frac 1 z}$
Therefore:
{{begin-eqn}}
{{eqn | l = \frac \d {\d z} \map \Log {\map \Gamma {z + M } }
| r = \map \Log {z + M - 1} + \map \OO {\frac 1 {z + M - 1} }
| c =
}}
{{eqn | l = \frac \d {\d z} \map \Log {\map \Gamma {z + M} } - \sum_{k \mathop = 0}^{M - 1} \frac 1 {z + k}
| r = \map \Log {z + M - 1} - \sum_{k \mathop = 0}^{M - 1} \frac 1 {z + k} + \map \OO {\frac 1 {z + M - 1} }
| c = subtracting $\ds \sum_{k \mathop = 0}^{M - 1} \frac 1 {z + k}$ from both sides of the equation
}}
{{eqn | l = \lim_{M \mathop \to \infty} \paren {\frac \d {\d z} \map \Log {\map \Gamma {z + M} } - \sum_{k \mathop = 0}^{M - 1} \frac 1 {z + k} }
| r = \lim_{M \mathop \to \infty} \paren {\map \Log {z + M} - \sum_{k \mathop = 0}^{M - 1} \frac 1 {z + k} } + \lim_{M \mathop \to \infty} \map \OO {\frac 1 {z + M} }
| c = {{Defof|Limit of Real Function}}, taking the limit as M goes to $\infty$ on both sides of the equation
}}
{{eqn | r = \lim_{M \mathop \to \infty} \paren {\map \Log {z + M} - \sum_{k \mathop = 0}^{M - 1} \frac 1 {z + k} }
| c = $\ds \lim_{M \mathop \to \infty} \map \OO {\frac 1 {z + M} } = 0$
}}
{{end-eqn}}
We now have:
:$\ds \lim_{M \mathop \to \infty} \frac \d {\d z} \map \Log {\map \Gamma {z + M} } = \lim_{M \mathop \to \infty} \map \Log {z + M} $
Setting $z = 1$ into $(2)$:
{{begin-eqn}}
{{eqn | l = \frac \d {\d z} \map \Log {\map \Gamma 1}
| r = \lim_{M \mathop \to \infty} \frac \d {\d z} \map \Log {\map \Gamma {1 + M} } - \sum_{k \mathop = 0}^{M - 1} \frac 1 {1 + k}
| c =
}}
{{eqn | r = \lim_{M \mathop \to \infty} \map \Log {1 + M} - \sum_{k \mathop = 0}^{M - 1} \frac 1 {1 + k}
| c =
}}
{{eqn | r = \lim_{M \mathop \to \infty} \map \Log {1 + M} - \sum_{j \mathop = 1}^{M} \frac 1 j
| c = setting $j = 1 + k$
}}
{{eqn | r = -\gamma
| c = {{Defof|Euler-Mascheroni Constant}}
}}
{{end-eqn}}
Also:
:$\ds \frac {\d^{1 + k} } {\d z^{1 + k}} \map \Log {\map \Gamma {z + 1} } = \map \OO {\frac 1 z}$
shows that:
:$\ds \lim_{M \mathop \to \infty} \frac {\d^{1 + k} } {\d z^{1 + k} } \map \Log {\map \Gamma {M + 1} } = 0$
thus for $n > 1$:
{{begin-eqn}}
{{eqn | l = \frac {\d^n} {\d z^n} \map \Log {\map \Gamma 1}
| r = \frac {\d^n} {\d z^n} \map \Log {\map \Gamma {1 + M} } + \sum_{k \mathop = 1}^M \frac {\paren {-1}^n \paren {n - 1}!} {k^n}
| c =
}}
{{eqn | l = \frac {\d^n} {\d z^n} \map \Log {\map \Gamma 1}
| r = \lim_{M \mathop \to \infty} \paren {\frac {\d^n} {\d z^n} \map \Log {\map \Gamma {1 + M} } + \sum_{k \mathop = 1}^M \frac {\paren {-1}^n \paren {n - 1}!} {k^n} }
}}
{{eqn | l = \frac {\d^n} {\d z^n} \map \Log {\map \Gamma 1}
| r = \paren {-1}^n \paren {n - 1}! \sum_{k \mathop = 1}^{\infty} \frac 1 {k^n}
}}
{{eqn | r = \paren {-1}^n \paren {n - 1}! \map \zeta n
| c = {{Defof|Riemann Zeta Function}}
}}
{{end-eqn}}
Thus by definition of Taylor series:
{{begin-eqn}}
{{eqn | l = \map f z
| r = \sum_{n \mathop = 0}^\infty \map {f^{\paren n} } a \frac {\paren {z - a}^n} {n!}
}}
{{eqn | r = \map f a + \map {f^{\paren 1} } a \frac {\paren {z - a}^1} {1!} + \map {f^{\paren 2} } a \frac {\paren {z - a}^2} {2!} + \cdots
}}
{{eqn | l = \map \Log {\map \Gamma z}
| r = \map \Log {\map \Gamma 1} - \gamma \paren {z - 1} + \sum_{k \mathop = 2}^\infty \frac {\paren {-1}^k \map \zeta k \paren {k - 1}!} {k!} \paren {z - 1}^k
}}
{{eqn | r = -\gamma \paren {z - 1} + \sum_{k \mathop = 2}^\infty \frac{\paren {-1}^k \map \zeta k} k \paren {z - 1}^k
| c = Gamma of One is One and Natural Logarithm of 1 is 0
}}
{{end-eqn}}
From Zeroes of Gamma Function, we see that $\map \Gamma z$ is non-zero everywhere.
Thus $\map \Log {\map \Gamma z}$ has poles only where $\Gamma$ does, that is, the negative integers.
Since the radius of convergence of a power series is equal to the distance of its center to the closest point where the function is not analytic:
The radius of convergence of $\map \Log {\map \Gamma z}$ is $\cmod {1 - 0} = 1$.
{{qed}}
Category:Gamma Function
Category:Natural Logarithms
\end{proof}
|
22598
|
\section{Taylor Series reaches closest Singularity}
Tags: Taylor Series, Real Analysis
\begin{theorem}
Let $F$ be a complex function.
Let $F$ be analytic everywhere except at a finite number of singularities.
Let a singularity of $F$ be one of the following:
:a pole
:an essential singularity
:a branch point
In the latter case $F$ is a restriction of a multifunction to one of its branches.
Let $x_0$ be a real number.
Let $F$ be analytic at the complex number $\tuple {x_0, 0}$.
Let $R \in \R_{>0}$ be the distance from the complex number $\tuple {x_0, 0}$ to the closest singularity of $F$.
Let the restriction of $F$ to $\R \to \C$ be a real function $f$.
This means:
:$\forall x \in \R: \map f x = \map \Re {\map F {x, 0} }, 0 = \map \Im {\map F {x, 0} }$
where $\tuple {x, 0}$ denotes the complex number with real part $x$ and imaginary part $0$.
Then:
:the Taylor series of $f$ about $x_0$ converges to $f$ at every point $x \in \R$ satisfying $\size {x - x_0} < R$
\end{theorem}
\begin{proof}
We have that $F$ is analytic everywhere except at its singularities.
Also, the distance from the complex number $\tuple {x_0, 0}$ to the closest singularity of $F$ is $R$.
Therefore:
:$F$ is analytic at every point $z \in \C$ satisfying $\size {z - \tuple {x_0, 0} } < R$
where $\tuple {x_0 , 0}$ denotes the complex number with real part $x_0$ and imaginary part $0$.
The result follows by Convergence of Taylor Series of Function Analytic on Disk.
{{qed}}
\end{proof}
|
22599
|
\section{Telescoping Series/Example 1}
Tags: Telescoping Series, Series
\begin{theorem}
Let $\sequence {b_n}$ be a sequence in $\R$.
Let $\sequence {a_n}$ be a sequence whose terms are defined as:
:$a_k = b_k - b_{k + 1}$
Then:
:$\ds \sum_{k \mathop = 1}^n a_k = b_1 - b_{n + 1}$
If $\sequence {b_n}$ converges to zero, then:
:$\ds \sum_{k \mathop = 1}^\infty a_k = b_1$
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | l = \ds \sum_{k \mathop = 1}^n a_k
| r = \sum_{k \mathop = 1}^n \paren {b_k - b_{k + 1} }
| c =
}}
{{eqn | r = \sum_{k \mathop = 1}^n b_k - \sum_{k \mathop = 1}^n b_{k + 1}
| c =
}}
{{eqn | r = \sum_{k \mathop = 1}^n b_k - \sum_{k \mathop = 2}^{n + 1} b_k
| c = Translation of Index Variable of Summation
}}
{{eqn | r = b_1 + \sum_{k \mathop = 2}^n b_k - \sum_{k \mathop = 2}^n b_k - b_{n + 1}
| c =
}}
{{eqn | r = b_1 - b_{n + 1}
| c =
}}
{{end-eqn}}
If $\sequence {b_k}$ converges to zero, then $b_{n + 1} \to 0$ as $n \to \infty$.
Thus:
:$\ds \lim_{n \mathop \to \infty} s_n = b_1 - 0 = b_1$
So:
:$\ds \sum_{k \mathop = 1}^\infty a_k = b_1$
{{Qed}}
\end{proof}
|
22600
|
\section{Temperature of Body under Newton's Law of Cooling}
Tags: Thermodynamics, Heat
\begin{theorem}
Let $B$ be a body in an environment whose ambient temperature is $H_a$.
Let $H$ be the temperature of $B$ at time $t$.
Let $H_0$ be the temperature of $B$ at time $t = 0$.
Then:
:$H = H_a - \paren {H_0 - H_a} e^{-k t}$
where $k$ is some positive constant.
\end{theorem}
\begin{proof}
By Newton's Law of Cooling:
:The rate at which a hot body loses heat is proportional to the difference in temperature between it and its surroundings.
We have the differential equation:
:$\dfrac {\d H} {\d t} \propto - \paren {H - H_a}$
That is:
:$\dfrac {\d H} {\d t} = - k \paren {H - H_a}$
where $k$ is some positive constant.
This is an instance of the Decay Equation, and so has a solution:
:$H = H_a + \paren {H_0 - H_a} e^{-k t}$
{{qed}}
{{Namedfor|Isaac Newton|cat = Newton}}
\end{proof}
|
22601
|
\section{Tempered Distribution Space is Proper Subset of Distribution Space}
Tags: Distributions
\begin{theorem}
Let $\map {\DD'} \R$ be the distribution space.
Let $\map {\SS'} \R$ be the tempered distribution space.
Then $\map {\SS'} \R$ is a proper subset of $\map {\DD'} \R$:
:$\map {\SS'} \R \subsetneqq \map {\DD'} \R$
\end{theorem}
\begin{proof}
By Convergence of Sequence of Test Functions in Test Function Space implies Convergence in Schwartz Space we have that $\map {\SS'} \R \subseteq \map {\DD'} \R$.
{{Research|how?}}
Consider the real function $\map f x = e^{x^2}$.
We have that:
:Real Power Function for Positive Integer Power is Continuous
:Exponential Function is Continuous/Real Numbers
:Composite of Continuous Mappings is Continuous
Thus, $f$ is a continuous real function.
Also:
:$\forall x \in \R : e^{x^2} < \infty$
Hence, $f$ is locally integrable.
By Locally Integrable Function defines Distribution, $T_f \in \map {\DD'} \R$.
{{AimForCont}} $T_f$ is a tempered distribution.
We have that $e^{-x^2}$ is a Schwartz test function.
Then:
{{begin-eqn}}
{{eqn | l = \map {T_f} {e^{-x^2} }
| r = \int_{- \infty}^\infty e^{x^2} e^{-x^2} \rd x
}}
{{eqn | r = \int_{- \infty}^\infty 1 \rd x
}}
{{eqn | r = \infty
}}
{{end-eqn}}
Hence, $\map {T_f} {e^{-x^2} } \notin \R$.
This is a contradiction.
Therefore, $T_f \notin \map {\SS'} \R$ while at the same time $T_f \in \map {\DD'} \R$.
{{qed}}
\end{proof}
|
22602
|
\section{Tensor Product of Free Modules is Free}
Tags: Commutative Algebra, Homological Algebra, Free Modules, Module Theory
\begin{theorem}
Let $A$ be a commutative ring with unity.
Let $F$ and $F'$ be free $A$-modules.
Then the tensor product $F \otimes_A F'$ is a free $A$-module.
\end{theorem}
\begin{proof}
By Free Module is Isomorphic to Free Module on Set there are sets $I$ and $I'$ and isomorphisms $\Psi : A^{\paren I} \to F$ and $\Psi' : A^{\paren {I'} } \to F'$.
By Tensor Product Distributes over Direct Sum, there is an isomorphism:
:$\ds A^{\paren I} \otimes_A A^{\paren {I'} } \cong \bigoplus_{i \mathop \in I} \bigoplus_{i' \mathop \in I'} A$
By Direct Sum of Direct Sums is Direct Sum, there is an isomorphism:
:$\ds \bigoplus_{i \mathop \in I} \bigoplus_{i' \mathop \in I'} A \cong \bigoplus_{\tuple {i, i'} \mathop \in I \times I'} A$
Hence $A^{\paren I} \otimes_A A^{\paren {I'} } \cong A^{\paren {I \times I'} }$ by definition of $A^{\paren {I \times I'} }$.
By Free Module on Set is Free $A^{\paren {I \times I'} }$ is free.
By Module Isomorphic to Free Module is Free $A^{\paren I} \otimes_A A^{\paren {I'} }$ is free.
{{qed}}
Category:Commutative Algebra
Category:Free Modules
Category:Homological Algebra
Category:Module Theory
\end{proof}
|
22603
|
\section{Tensor Product of Projective Modules is Projective}
Tags: Commutative Algebra
\begin{theorem}
Let $A$ be a commutative ring with unity.
Let $P$ and $Q$ be projective $A$-modules.
Then the tensor product $P \otimes_A Q$ is a projective $A$-module.
\end{theorem}
\begin{proof}
By Projective iff Direct Summand of Free Module, there exist $A$-modules $P'$ and $Q'$, such that $P \oplus P'$ and $Q \oplus Q'$ are free.
By Tensor Product Distributes over Direct Sum, there is an isomorphism
: $\paren {P \oplus P'} \otimes_A \paren {Q \oplus Q'} \cong \paren {P \otimes_A Q} \oplus \paren {P' \otimes_A Q} \oplus \paren {P \otimes_A Q'} \oplus \paren {P' \otimes_A Q'}$
By Tensor Product of Free Modules is Free the {{LHS}} is free.
Hence $P \otimes_A Q$ is a direct summand of a free module.
By Projective iff Direct Summand of Free Module $P \otimes_A Q$ is projective.
{{qed}}
Category:Commutative Algebra
\end{proof}
|
22604
|
\section{Tensor with Zero Element is Zero in Tensor}
Tags: Tensor Algebra, Homological Algebra
\begin{theorem}
Let $R$ be a ring.
Let $M$ be a right $R$-module.
Let $N$ be a left $R$-module.
Let $M \otimes_R N$ denote their tensor product.
Then:
:$0\otimes_R n = m \otimes_R 0 = 0 \otimes_R 0$
is the zero in $M \otimes_R N$.
\end{theorem}
\begin{proof}
Let $m \in M$ and $n \in N$
Then
{{begin-eqn}}
{{eqn | l = m \otimes_R n
| r = \paren {m + 0} \otimes_R n
| c = {{GroupAxiom|2}}
}}
{{eqn | r = m \otimes_R n + 0 \otimes_R n
| c = {{Defof|Tensor Equality}}
}}
{{eqn | r = m \otimes_R \paren {n + 0}
| c = {{GroupAxiom|2}}
}}
{{eqn | r = m \otimes_R n + m\otimes_R 0
| c = {{Defof|Tensor Equality}}
}}
{{eqn | r = m \otimes_R n + 0 \otimes_R \paren {n + 0}
| c = {{GroupAxiom|2}}
}}
{{eqn | r = m \otimes_R n + m \otimes_R 0 + 0 \otimes_R n + 0 \otimes_R 0
| c = {{Defof|Tensor Equality}}
}}
{{eqn | r = m \otimes_R n + 0 \otimes_R 0
| c = {{Defof|Tensor Equality}}
}}
{{end-eqn}}
Hence $0 \otimes_R n$, $m \otimes_R 0$ and $0 \otimes_R 0$ must all be identity elements for $M \otimes_R N$ as a left module.
{{qed}}
Category:Tensor Algebra
Category:Homological Algebra
\end{proof}
|
22605
|
\section{Termial on Real Numbers is Extension of Integers}
Tags: Number Theory, Termial Function
\begin{theorem}
The termial function as defined on the real numbers is an extension of its definition on the integers $\Z$.
\end{theorem}
\begin{proof}
From the definition of the termial function on the integers:
:$\ds n? = \sum_{k \mathop = 1}^n k = 1 + 2 + \cdots + n$
From Closed Form for Triangular Numbers, we have that:
:$\ds \forall n \in \Z_{> 0}: \sum_{k \mathop = 1}^n k = \dfrac {n \paren {n + 1} } 2$
This agrees with the definition of the termial function on the real numbers.
Hence the result, by definition of extension.
{{qed}}
\end{proof}
|
22606
|
\section{Terminal Object is Unique}
Tags: Category Theory
\begin{theorem}
Let $\mathbf C$ be a metacategory.
Let $1$ and $1'$ be two terminal objects of $\mathbf C$.
Then there is a unique isomorphism $u: 1 \to 1'$.
Hence, terminal objects are unique up to unique isomorphism.
\end{theorem}
\begin{proof}
Consider the following commutative diagram:
::$\begin{xy}
<-4em,0em>*+{1} = "M",
<0em,0em> *+{1'}= "N",
<0em,-4em>*+{1} = "M2",
<4em,-4em>*+{1'}= "N2",
"M";"N" **@{-} ?>*@{>} ?*!/_.6em/{u},
"M";"M2" **@{-} ?>*@{>} ?*!/^.6em/{\operatorname{id}_1},
"N";"M2" **@{-} ?>*@{>} ?*!/_.6em/{v},
"N";"N2" **@{-} ?>*@{>} ?*!/_1em/{\operatorname{id}_{1'}},
"M2";"N2"**@{-} ?>*@{>} ?*!/^.6em/{u},
\end{xy}$
It commutes as each of the morphisms in it points to a terminal object, and hence is unique.
Thus, $v$ is an inverse to $u$, and so $u$ is an isomorphism.
{{qed}}
\end{proof}
|
22607
|
\section{Terminal Velocity of Body under Fall Retarded Proportional to Square of Velocity}
Tags: Mechanics
\begin{theorem}
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 velocity of $B$ relative to the medium.
Then the terminal velocity of $B$ is given by:
:$v = \sqrt {\dfrac {g m} k}$
\end{theorem}
\begin{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}}
\end{proof}
|
22608
|
\section{Terminal Velocity of Body under Fall Retarded Proportional to Velocity}
Tags: Mechanics
\begin{theorem}
Let $B$ be a body falling in a gravitational field.
Let $B$ be falling through a medium which exerts a resisting force $k \mathbf v$ upon $B$ which is proportional to the velocity of $B$ relative to the medium.
Then the terminal velocity of $B$ is given by:
:$v = \dfrac {g m} k$
\end{theorem}
\begin{proof}
Let $B$ start from rest.
From Motion of Body Falling through Air, the differential equation governing the motion of $B$ is given by:
:$m \dfrac {\d^2 \mathbf s} {\d t^2} = m \mathbf g - k \dfrac {\d \mathbf s} {\d t}$
Dividing through by $m$ and setting $c = \dfrac k m$ gives:
:$\dfrac {\d^2 \mathbf s} {\d t^2} = \mathbf g - c \dfrac {\d \mathbf s} {\d t}$
By definition of velocity:
:$\dfrac {\d \mathbf v} {\d t} = \mathbf g - c \mathbf v$
and so:
{{begin-eqn}}
{{eqn | l = \int \dfrac {\d \mathbf v} {\mathbf g - c \mathbf v}
| r = \int \rd t
| c =
}}
{{eqn | ll= \leadsto
| l = -\dfrac 1 c \map \ln {\mathbf g - c \mathbf v}
| r = t + c_1
| c =
}}
{{eqn | ll= \leadsto
| l = \mathbf g - c \mathbf v
| r = \mathbf c_2 e^{-c t}
| c =
}}
{{end-eqn}}
When $t = 0$ we have that $\mathbf v = 0$ and so:
:$\mathbf c_2 = \mathbf g$
Hence by taking magnitudes:
:$v = \dfrac g c \paren {1 - e^{-c t} }$
Since $c > 0$ it follows that $v \to \dfrac g c$ as $t \to \infty$.
Thus in the limit:
:$v = \dfrac g c = \dfrac {g m} k$
{{qed}}
\end{proof}
|
22609
|
\section{Terms in Convergent Series Converge to Zero}
Tags: Convergent Series, Series
\begin{theorem}
Let $\sequence {a_n}$ be a sequence in any of the standard number fields $\Q$, $\R$, or $\C$.
Suppose that the series $\ds \sum_{n \mathop = 1}^\infty a_n$ converges in any of the standard number fields $\Q$, $\R$, or $\C$.
Then:
:$\ds \lim_{n \mathop \to \infty} a_n = 0$
{{expand|Expand (on a different page) to Banach spaces}}
\end{theorem}
\begin{proof}
Let $\ds s = \sum_{n \mathop = 1}^\infty a_n$.
Then $\ds s_N = \sum_{n \mathop = 1}^N a_n \to s$ as $N \to \infty$.
Also, $s_{N - 1} \to s$ as $N \to \infty$.
Thus:
{{begin-eqn}}
{{eqn | l = a_N
| r = \paren {a_1 + a_2 + \cdots + a_{N - 1} + a_N} - \paren {a_1 + a_2 + \cdots + a_{N - 1} }
| c =
}}
{{eqn | r = s_N - s_{N - 1}
| c =
}}
{{eqn | o = \to
| r = s - s = 0 \text{ as } N \to \infty
| c =
}}
{{end-eqn}}
Hence the result.
{{qed}}
\end{proof}
|
22610
|
\section{Test Function/Examples/Exponential of One over x Squared minus One}
Tags: Examples of Test Functions
\begin{theorem}
Let $\phi : \R \to \R$ be a real function with support on $x \in \closedint {-1} 1$ such that:
:$\map \phi x = \begin{cases}
\ds \map \exp {\frac 1 {x^2 - 1}} & : \size x < 1 \\
0 & : \size x \ge 1
\end{cases}$
Then $\phi$ is a test function.
\end{theorem}
\begin{proof}
Consider a real function $f : \R \to \R$ such that:
:$\map f x = \begin {cases}
\map \exp {-\dfrac 1 x} & : x > 0 \\
0 & : x \le 0
\end {cases}$
From Nth Derivative of Exponential of Minus One over x:
:$\dfrac {\d^n} {\d x^n} \map \exp {-\dfrac 1 x} = \dfrac {\map {P_n} x} {x^{2 n} } \map \exp {-\dfrac 1 x}$
where $\map {P_n} x$ is a real polynomial of degree $n$.
Then the {{RHS}} can be rewritten in terms of at most $n + 1$ terms of the form $\dfrac 1 {x^m} \map \exp {-\dfrac 1 x}$ where $m \in \N$.
Let us take the limit $x \to 0$ from the right:
{{begin-eqn}}
{{eqn | l = \lim_{x \mathop \to 0^+} \frac 1 {x^m} \map \exp {-\frac 1 x}
| r = \lim_{z \mathop \to \infty} \frac {z^m} {\map \exp z}
| c = Substitution $\ds z = \frac 1 x$
}}
{{eqn | r = 0
| c = Limit at Infinity of Polynomial over Complex Exponential
}}
{{end-eqn}}
Therefore:
:$\ds \lim_{x \mathop \to 0^+} \frac {\map {P_n} x} {x^{2 n} } \map \exp {-\frac 1 x} = 0$
By construction:
:$\ds \map \phi x = \map f {1 - x^2} = \begin {cases}
\map \exp {-\dfrac 1 {1 - x^2} } & : 1 - x^2 > 0 \\
0 & : 1 - x^2 \le 0
\end {cases}$
where:
{{begin-eqn}}
{{eqn | l = 1 - x^2
| o = >
| r = 0
}}
{{eqn | ll= \leadsto
| l = 1
| o = >
| r = x^2
}}
{{eqn | ll= \leadsto
| l = \size x
| o = <
| r = 1
}}
{{end-eqn}}
Furthermore:
{{begin-eqn}}
{{eqn | l = \dfrac {\d \map \phi x} {\d x}
| r = \dfrac {\d \map f y} {\d y} \paren {-2 x}
}}
{{eqn | r = \frac {\map {P_1} y} {y^2} \map \exp {-\frac 1 y} \paren {-2 x}
}}
{{eqn | r = \frac {\map {M_3} x} {y^2} \map \exp {-\frac 1 y}
}}
{{end-eqn}}
where $y = 1 - x^2$ and $\map {M_3} x$ is a real polynomial of degree $3$.
Similarly, any higher derivative of $\map \phi x$ will have $\map {M_k} x$ instead of $\map {P_n} x$ with $k \ge n$.
Thus:
:$\dfrac {\d^n} {\d x^n} \map \phi x = \dfrac {\map {M_k} x} {y^{2 n} } \map \exp {-\dfrac 1 y}$
Consequently:
{{begin-eqn}}
{{eqn | l = \lim_{x \mathop \to -1^+} \frac {\map {M_k} x} {y^{2 n} } \map \exp {-\frac 1 y}
| r = \lim_{x \mathop \to -1^+} \map {M_k} x \lim_{x \mathop \to -1^+} \frac {\map \exp {- \frac 1 y} } {y^{2n} }
| c = Product Rule for Limits of Real Functions
}}
{{eqn | r = C \lim_{y \mathop \to 0^+} \frac {\map \exp {- \frac 1 y} } {y^{2n} }
| c = $C \in \R$
}}
{{eqn | r = 0
}}
{{end-eqn}}
Analogously:
:$\ds \lim_{x \mathop \to 1^-} \frac {\map {M_k} x} {y^{2 n} } \map \exp {-\frac 1 y} = 0$
Since outside of the support we have that $\map \phi x = 0$, the limit coming from outside is also $0$.
Therefore, $\map \phi x$ is smooth at the boundaries of its support.
Also:
:$\forall x \in \openint {-1} 1 : \map \phi x \in \CC^\infty$
By definition, $\phi$ is a test function.
{{qed}}
\end{proof}
|
22611
|
\section{Test Function Space with Pointwise Addition and Pointwise Scalar Multiplication forms Vector Space}
Tags: Functional Analysis, Examples of Vector Spaces
\begin{theorem}
Let $\map \DD {\R^d}$ be the test function space.
Let $\struct {\C, +_\C, \times_\C}$ be the field of complex numbers.
Let $\paren +$ be the pointwise addition of test functions.
Let $\paren {\, \cdot \,}$ be the pointwise scalar multiplication of test functions over $\C$.
Then $\struct {\map \DD {\R^d}, +, \, \cdot \,}_\C$ is a vector space.
\end{theorem}
\begin{proof}
Let $f, g, h \in \map \DD {\R^d}$ be test functions with the compact support $K$.
Let $\lambda, \mu \in \C$.
Let $\map 0 x$ be a real-valued function such that:
:$\map 0 x : \R^d \to 0$.
Let us use real number addition and multiplication.
$\forall x \in \R^d$ define pointwise addition as:
:$\map {\paren {f + g}} x := \map f x +_\C \map g x$.
Define pointwise scalar multiplication as:
:$\map {\paren {\lambda \cdot f}} x := \lambda \times_\C \map f x$
Let $\map {\paren {-f} } x := -\map f x$.
\end{proof}
|
22612
|
\section{Test Function with Vanishing Partial Derivative}
Tags: Partial Differentiation, Examples of Test Functions
\begin{theorem}
Let $\phi \in \map \DD {\R^2}$ be a test function such that:
:$\tuple {x, y} \stackrel \phi {\longrightarrow} \map \phi {x, y}$
Suppose $\phi$ is a solution to the following partial differential equation:
:$\ds \dfrac {\partial \phi}{\partial x} = 0$
Then $\phi$ is identically $0$.
\end{theorem}
\begin{proof}
$\ds \dfrac {\partial \phi}{\partial x} = 0$ implies that:
:$\forall x \in \R : \map \phi {x, y} = \map C y$
where $C : \R \to \C$ is a complex-valued function.
By definition, $\phi$ is a test function.
Hence, $\phi$ must have a compact support $\Omega \subset \R^2$.
Let $\map {B^-_\epsilon} 0 \subset \R^2$ be a closed ball in Euclidean space such that:
:$\Omega \subseteq \map {B^-_\epsilon} 0$
By Closed Ball in Euclidean Space is Compact, $\map {B^-_\epsilon} 0$ is a compact.
Then $\map {B^-_\epsilon} 0$ also qualifies as a compact support of $\phi$.
By definition of a test function:
:$\forall y \in \R : \size y > \epsilon \implies \map \phi {x, y} = 0$
But for each $y \in \R$ we have that $\map \phi {x, y} = \map C y$ is a constant.
By smoothness of test functions, this constant has to be the same for all $y \in \R$.
Hence, $\map \phi {x, y} = 0$.
{{qed}}
\end{proof}
|
22613
|
\section{Test for Submonoid}
Tags: Abstract Algebra, Monoids
\begin{theorem}
To show that $\struct {T, \circ}$ is a submonoid of a monoid $\struct {S, \circ}$, we need to show that:
:$(1): \quad T \subseteq S$
:$(2): \quad \struct {T, \circ}$ is a magma (that is, that it is closed)
:$(3): \quad \struct {T, \circ}$ has an identity.
\end{theorem}
\begin{proof}
From Subsemigroup Closure Test, $(1)$ and $(2)$ are sufficient to show that $\struct {T, \circ}$ is a subsemigroup of $\struct {S, \circ}$.
Demonstrating the presence of an identity is then sufficient to show that it is a monoid.
{{qed}}
Category:Monoids
\end{proof}
|
22614
|
\section{Tetrahedral Number as Sum of Squares}
Tags: Tetrahedral Numbers, Tetrahedral Number as Sum of Squares, Square Numbers
\begin{theorem}
:$H_n = \ds \sum_{k \mathop = 0}^{n / 2} \paren {n - 2 k}^2$
where $H_n$ denotes the $n$th tetrahedral number.
\end{theorem}
\begin{proof}
Let $n$ be even such that $n = 2 m$.
We have:
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 0}^{n / 2} \paren {n - 2 k}^2
| r = \sum_{k \mathop = 0}^m \paren {2 m - 2 k}^2
| c =
}}
{{eqn | r = \sum_{k \mathop = 0}^m \paren {2 k}^2
| c = Permutation of Indices of Summation
}}
{{eqn | r = \frac {2 m \paren {m + 1} \paren {2 m + 1} } 3
| c = Sum of Sequence of Even Squares
}}
{{eqn | r = \frac {2 m \paren {2 m + 1} \paren {2 m + 2} } 6
| c =
}}
{{eqn | r = H_{2 m}
| c = Closed Form for Tetrahedral Numbers
}}
{{end-eqn}}
Let $n$ be odd such that $n = 2 m + 1$.
We have:
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 0}^{n / 2} \paren {n - 2 k}^2
| r = \sum_{k \mathop = 0}^m \paren {2 m + 1 - 2 k}^2
| c =
}}
{{eqn | r = \sum_{k \mathop = 0}^m \paren {2 k + 1}^2
| c = Permutation of Indices of Summation
}}
{{eqn | r = \frac {\paren {m + 1} \paren {2 m + 1} \paren {2 m + 3} } 3
| c = Sum of Sequence of Odd Squares: Formulation 1
}}
{{eqn | r = \frac {\paren {2 m + 1} \paren {2 m + 2} \paren {2 m + 3} } 6
| c =
}}
{{eqn | r = H_{2 m + 1}
| c = Closed Form for Tetrahedral Numbers
}}
{{end-eqn}}
{{qed}}
Category:Tetrahedral Numbers
Category:Square Numbers
Category:Tetrahedral Number as Sum of Squares
\end{proof}
|
22615
|
\section{Thabit's Rule}
Tags: Amicable Pairs, Amicable Numbers
\begin{theorem}
Let $n$ be a positive integer such that:
{{begin-eqn}}
{{eqn | l = a
| r = 3 \times 2^n - 1
| c =
}}
{{eqn | l = b
| r = 3 \times 2^{n - 1} - 1
| c =
}}
{{eqn | l = c
| r = 9 \times 2^{2 n - 1} - 1
| c =
}}
{{end-eqn}}
are all prime.
Then:
:$\tuple {2^n a b, 2^n c}$
forms an amicable pair.
\end{theorem}
\begin{proof}
Let $r = 2^n a b, s = 2^n c$.
Let $\map {\sigma_1} k$ denote the divisor sum of an integer $k$.
From Divisor Sum of Power of 2:
:$\map {\sigma_1} {2^n} = 2^{n + 1} - 1$
From Divisor Sum of Prime Number:
{{begin-eqn}}
{{eqn | l = \map {\sigma_1} a
| r = 3 \times 2^n
| c =
}}
{{eqn | l = \map {\sigma_1} b
| r = 3 \times 2^{n - 1}
| c =
}}
{{eqn | l = \map {\sigma_1} c
| r = 9 \times 2^{2 n - 1}
| c =
}}
{{end-eqn}}
From Divisor Sum Function is Multiplicative:
{{begin-eqn}}
{{eqn | l = \map {\sigma_1} r
| r = \map {\sigma_1} {2^n} \map {\sigma_1} a \map {\sigma_1} b
| c =
}}
{{eqn | r = \paren {2^{n + 1} - 1} \paren {3 \times 2^n} \paren {3 \times 2^{n - 1} }
| c =
}}
{{eqn | r = \paren {2^{n + 1} - 1} \paren {9 \times 2^{2 n - 1} }
| c =
}}
{{end-eqn}}
and:
{{begin-eqn}}
{{eqn | l = \map {\sigma_1} s
| r = \map {\sigma_1} {2^n} \map {\sigma_1} c
| c =
}}
{{eqn | r = \paren {2^{n + 1} - 1} \paren {9 \times 2^{2 n - 1} }
| c =
}}
{{eqn | r = \map {\sigma_1} r
| c =
}}
{{end-eqn}}
Thus it is seen that:
:$\map {\sigma_1} r = \map {\sigma_1} s$
Now we have:
{{begin-eqn}}
{{eqn | l = r + s
| r = 2^n a b + 2^n c
| c =
}}
{{eqn | r = 2^n \paren {\paren {3 \times 2^n - 1} \paren {3 \times 2^{n - 1} - 1} + 9 \times 2^{2 n - 1} - 1 }
| c =
}}
{{eqn | r = 2^n \paren {9 \times 2^{2 n - 1} - 3 \times 2^n - 3 \times 2^{n - 1} + 1 + 9 \times 2^{2 n - 1} - 1}
| c =
}}
{{eqn | r = 2^n \paren {2 \times 9 \times 2^{2 n - 1} - 3 \times 2 \times 2^{n - 1} - 3 \times 2^{n - 1} }
| c =
}}
{{eqn | r = 2^{n + 1} \paren {9 \times 2^{2 n - 1} - 9 \times 2^{n - 2} }
| c =
}}
{{eqn | r = 2^{n + 1} \paren {9 \times 2^{2 n - 1} } - 2^{n + 1} \paren {9 \times 2^{n - 2} }
| c =
}}
{{eqn | r = 2^{n + 1} \paren {9 \times 2^{2 n - 1} } - \paren {9 \times 2^{2 n - 1} }
| c =
}}
{{eqn | r = \paren {2^{n + 1} - 1} \paren {9 \times 2^{2 n - 1} }
| c =
}}
{{end-eqn}}
and so it is seen that:
:$r + s = \map {\sigma_1} r = \map {\sigma_1} s$
Hence the result, by definition of amicable pair.
{{qed}}
\end{proof}
|
22616
|
\section{Thales' Theorem}
Tags: Circles, Euclidean Geometry, Thales' Theorem
\begin{theorem}
Let $A$ and $B$ be two points on opposite ends of the diameter of a circle.
Let $C$ be another point on the circle such that $C \ne A, B$.
Then the lines $AC$ and $BC$ are perpendicular to each other.
:400px
\end{theorem}
\begin{proof}
:400px
Let $O$ be the center of the circle, and define the vectors $\mathbf u = \overrightarrow{OC}$, $\mathbf v = \overrightarrow{OB}$ and $\mathbf w = \overrightarrow{OA}$.
If $AC$ and $BC$ are perpendicular, then $\left({ \mathbf u - \mathbf w}\right) \cdot \left({\mathbf u - \mathbf v}\right) = 0$ (where $\cdot$ is the dot product).
Notice that since $A$ is directly opposite $B$ in the circle, $\mathbf w = - \mathbf v$.
Our expression then becomes
:$\left({\mathbf u + \mathbf v}\right) \cdot \left({\mathbf u - \mathbf v}\right)$
From the distributive property of the dot product,
:$\left({ \mathbf u + \mathbf v}\right) \cdot \left({\mathbf u - \mathbf v}\right) = \mathbf u \cdot \mathbf u - \mathbf u \cdot \mathbf v + \mathbf v \cdot \mathbf u - \mathbf v \cdot \mathbf v$
From the commutativity of the dot product and Dot Product of a Vector with Itself, we get
:$\mathbf u \cdot \mathbf u - \mathbf u \cdot \mathbf v + \mathbf v \cdot \mathbf u - \mathbf v \cdot \mathbf v = \left|{\mathbf u}\right|^2 - \mathbf u \cdot \mathbf v + \mathbf u \cdot \mathbf v - \left|{\mathbf v}\right|^2 = \left|{\mathbf u}\right|^2 - \left|{\mathbf v}\right|^2$
Since the vectors $\mathbf u$ and $\mathbf v$ have the same length (both go from the centre of the circle to the circumference), we have that $|\mathbf u| = |\mathbf v|$, so our expression simplifies to
:$\left|{\mathbf u}\right|^2 - \left|{\mathbf v}\right|^2 = \left|{\mathbf u}\right|^2 - \left|{\mathbf u}\right|^2 = 0$
The result follows.
{{Qed}}
\end{proof}
|
22617
|
\section{Theorem of Even Perfect Numbers/Necessary Condition}
Tags: Perfect Numbers, Number Theory, Mersenne Numbers, Theorem of Even Perfect Numbers
\begin{theorem}
Let $a \in \N$ be an even perfect number.
Then $a$ is in the form:
:$2^{n - 1} \paren {2^n - 1}$
where $2^n - 1$ is prime.
\end{theorem}
\begin{proof}
Let $a \in \N$ be an even perfect number.
We can extract the highest power of $2$ out of $a$ that we can, and write $a$ in the form:
:$a = m 2^{n - 1}$
where $n \ge 2$ and $m$ is odd.
Since $a$ is perfect and therefore $\map {\sigma_1} a = 2 a$:
{{begin-eqn}}
{{eqn| l = m 2^n
| r = 2 a
| c =
}}
{{eqn| r = \map {\sigma_1} a
| c =
}}
{{eqn| r = \map {\sigma_1} {m 2^{n - 1} }
| c =
}}
{{eqn| r = \map {\sigma_1} m \map {\sigma_1} {2^{n - 1} }
| c = Divisor Sum Function is Multiplicative
}}
{{eqn| r = \map {\sigma_1} m {2^n - 1}
| c = Divisor Sum of Power of Prime
}}
{{end-eqn}}
So:
:$\map {\sigma_1} m = \dfrac {m 2^n} {2^n - 1}$
But $\map {\sigma_1} m$ is an integer and so $2^n - 1$ divides $m 2^n$.
From Consecutive Integers are Coprime, $2^n$ and $2^n - 1$ are coprime.
So from Euclid's Lemma $2^n - 1$ divides $m$.
Thus $\dfrac m {2^n - 1}$ divides $m$.
Since $2^n - 1 \ge 3$ it follows that:
:$\dfrac m {2^n - 1} < m$
Now we can express $\map {\sigma_1} m$ as:
:$\map {\sigma_1} m = \dfrac {m 2^n} {2^n - 1} = m + \dfrac m {2^n - 1}$
This means that the sum of all the divisors of $m$ is equal to $m$ itself plus one other divisor of $m$.
Hence $m$ must have exactly two divisors, so it must be prime by definition.
This means that the other divisor of $m$, apart from $m$ itself, must be $1$.
That is:
:$\dfrac m {2^n - 1} = 1$
Hence the result.
{{qed}}
\end{proof}
|
22618
|
\section{Theorem of Even Perfect Numbers/Sufficient Condition}
Tags: Perfect Numbers, Number Theory, Mersenne Numbers, Theorem of Even Perfect Numbers
\begin{theorem}
Let $n \in \N$ be such that $2^n - 1$ is prime.
Then $2^{n - 1} \paren {2^n - 1}$ is perfect.
{{:Euclid:Proposition/IX/36}}
\end{theorem}
\begin{proof}
Suppose $2^n - 1$ is prime.
Let $a = 2^{n - 1} \paren {2^n - 1}$.
Then $n \ge 2$ which means $2^{n - 1}$ is even and hence so is $a = 2^{n - 1} \paren {2^n - 1}$.
Note that $2^n - 1$ is odd.
Since all divisors (except $1$) of $2^{n - 1}$ are even it follows that $2^{n - 1}$ and $2^n - 1$ are coprime.
Let $\map {\sigma_1} n$ be the divisor sum of $n$, that is, the sum of all divisors of $n$ (including $n$).
From Divisor Sum Function is Multiplicative, it follows that $\map {\sigma_1} a = \map {\sigma_1} {2^{n - 1} } \map {\sigma_1} {2^n - 1}$.
But as $2^n - 1$ is prime, $\map {\sigma_1} {2^n - 1} = 2^n$ from Divisor Sum of Prime Number.
Then we have that $\map {\sigma_1} {2^{n - 1} } = 2^n - 1$ from Divisor Sum of Power of Prime.
Hence it follows that $\map {\sigma_1} a = \paren {2^n - 1} 2^n = 2 a$.
Hence from the definition of perfect number it follows that $2^{n - 1} \paren {2^n - 1}$ is perfect.
{{qed}}
{{Euclid Note|36|IX}}
\end{proof}
|
22619
|
\section{Theoretical Justification for Cycle Notation}
Tags: Permutation Theory
\begin{theorem}
Let $\N_k$ be used to denote the initial segment of natural numbers:
:$\N_k = \closedint 1 k = \set {1, 2, 3, \ldots, k}$
Let $\rho: \N_n \to \N_n$ be a permutation of $n$ letters.
Let $i \in \N_n$.
Let $k$ be the smallest (strictly) positive integer for which $\map {\rho^k} i$ is in the set:
:$\set {i, \map \rho i, \map {\rho^2} i, \ldots, \map {\rho^{k - 1} } i}$
Then:
:$\map {\rho^k} i = i$
\end{theorem}
\begin{proof}
{{AimForCont}} $\map {\rho^k} i = \map {\rho^r} i$ for some $r > 0$.
As $\rho$ has an inverse in $S_n$:
:$\map {\rho^{k - r} } i = i$
This contradicts the definition of $k$, because $k - r < k$
Thus:
:$r = 0$
The result follows.
{{qed}}
\end{proof}
|
22620
|
\section{Theories with Infinite Models have Models with Order Indiscernibles}
Tags: Model Theory
\begin{theorem}
Let $T$ be an $\LL$-theory with infinite models.
Let $\struct {I, <$ be an infinite strict linearly ordered set.
There is a model $\MM \models T$ containing an order indiscernible set $\set {x_i : i \in I}$.
\end{theorem}
\begin{proof}
We will construct the claimed model using the Compactness Theorem.
Let $LL^*$ be the language obtained by adding constant symbols $c_i$ to $\LL$ for each $i \in I$.
Let $T^*$ be the $\LL^*$-theory obtained by adding to $T$ the $\LL^*$-sentences:
:$c_i \ne c_j$ for each $i \ne j$ in $I$
and:
:$\map \phi {c_{i_1}, \dotsc, c_{i_n} } \leftrightarrow \map \phi {c_{j_1}, \dotsc, c_{j_n} }$ for each $n \in \N$, each $\LL$-formula $\phi$ with $n$ free variables, and each pair of chains $i_1 < \cdots < i_n$ and $j_1 < \cdots < j_n$ in $I$.
For future reference, we will refer to these last sentences using as those which '''assert indiscernibility with respect to $\phi$'''.
If we can find a model of $T^*$, then its interpretations of the constants $c_i$ for $i\in I$ will be order indiscernibles.
Suppose $\Delta$ is a finite subset of $T^*$.
{{refactor|Extract this section as a lemma}}
:We will show that there is a model of $\Delta$ using the Infinite Ramsey's Theorem.
:Let $I_\Delta$ be the finite subset of $I$ containing those $i$ for which $c_i$ occurs in some sentence in $\Delta$.
:Let $\psi_1, \dots, \psi_k$ be the finitely many $\LL$-formulas in $\Delta$ which assert indiscernibility with respect to $\phi_1, \dots, \phi_k$.
:Let $m$ be the maximum number of free variables in the formulas $\phi_1, \dots, \phi_k$.
:Let $\MM$ be an infinite model of $T$ (which exists by assumption).
:Let $X$ be any infinite subset of the universe of $\MM$ such that $X$ is linearly ordered by some relation $\prec$.
:Define a partition $P$ of $X^{\paren m} = \set {X' \subseteq X: \card {X'} = m}$ into $2^k$ components $S_A$ for each $A \subseteq \set {1, 2, \dotsc, k}$ by:
::$\set {x_1, \dots, x_m} \in S_A \iff x_1 \prec \cdots \prec x_m \text { and } A = \set {i: \MM \models \map {\phi_i} {x_1, \dotsc, x_m} }$
:By Infinite Ramsey's Theorem, there is an infinite subset $Y\subseteq X$ such that each element of $Y^{\paren m} = \set {Y' \subseteq Y: \card {Y'} = m}$ is in the same component $S_A$ of $P$.
:Note that $Y$ is indexed by some infinite subset $J_Y$ of $J$ and hence still linearly ordered by $\prec$.
:We now show that the constants $c_i$ for $i \in I_\Delta$ can be interpreted as elements of $Y$ in $\MM$, and that the sentences $\psi_1, \dots, \psi_k$ will be satisfied using this interpretation.
:Since $I_\Delta$ is finite and linearly ordered, and $J_Y$ is infinite and linearly ordered, there is clearly an increasing function $f:I_\Delta \to J_Y$.
:For each $h = 1, \dotsc, k$ and any pair of chains $i_1 < \cdots < i_n$ and $j_1 < \cdots < j_n$ in $I_\Delta$, we thus have:
::$\MM \models \map {\phi_h} {y_{\map f {i_1} }, \dotsc, y_{\map f {i_n} } }$ {{iff}} $h \in A$ {{iff}} $\MM \models \map {\phi_h} {y_{\map f {j_1} }, \dotsc, y_{\map f {j_n} } }$
:and hence:
::$\MM \models \map {\phi_h} {y_{\map f {i_1} }, \dotsc, y_{\map f {i_n} } } \iff \map {\phi_h} {y_{\map f {j_1} }, \dotsc, y_{\map f {j_n} } }$
:So, if we interpret each $c_i$ for $i \in I_\Delta$ as $y_{\map f i}$, we will have $\MM \models \psi_h$ for each $h = 1, \dotsc, k$.
:Thus, we have shown that $\MM$ can be extended to an $\LL^*$-structure which models $\Delta$.
Hence all finite subsets $\Delta$ of $T^*$ are satisfiable.
By the Compactness Theorem, we have that $T^*$ is satisfiable.
Let $\NN$ be any model of $T^*$.
Let $n_i$ be the interpretation of $c_i$ in $\NN$ for each $i \in I$.
Then $\set {n_i: i \in I}$ is easily seen to be an order indiscernible set, since $T^*$ was defined to include sentences guaranteeing such.
{{qed}}
Category:Model Theory
\end{proof}
|
22621
|
\section{Theory of Algebraically Closed Fields of Characteristic p is Complete}
Tags: Mathematical Logic, Model Theory
\begin{theorem}
Let $p$ be either $0$ or a prime number.
Let $ACF_p$ be the theory of algebraically closed fields of characteristic $p$ in the language $\LL_r = \set {0, 1, +, -, \cdot}$ for rings, where:
:$0, 1$ are constants
and:
:$+, -, \cdot$ are binary functions.
Then $ACF_p$ is complete.
\end{theorem}
\begin{proof}
By the Łoś-Vaught Test, it suffices to show that $ACF_p$ is satisfiable, has no finite models, and is $\kappa$-categorical for some uncountable $\kappa$.
\end{proof}
|
22622
|
\section{Theory of Set of Formulas is Theory}
Tags: Formal Semantics
\begin{theorem}
Let $\LL$ be a logical language.
Let $\mathscr M$ be a formal semantics for $\LL$.
Let $\FF$ be a set of $\LL$-formulas.
Let $\map T \FF$ be the $\LL$-theory of $\FF$.
Then $\map T \FF$ is a theory.
\end{theorem}
\begin{proof}
Let $\phi$ be a $\LL$-formula.
Suppose that $\map T \FF \models_{\mathscr M} \phi$.
By definition of $\map T \FF$:
:$\FF \models_{\mathscr M} \psi$
for every $\psi \in \map T \FF$.
Hence by definition of semantic consequence:
:$\FF \models_{\mathscr M} \map T \FF$
By Semantic Consequence is Transitive, it follows that:
:$\FF \models_{\mathscr M} \phi$
Finally, by definition of $\map T \FF$:
:$\phi \in \map T \FF$.
Since $\phi$ was arbitrary, the result follows.
{{qed}}
\end{proof}
|
22623
|
\section{There Exists No Universal Set}
Tags: Naive Set Theory, There Exists No Universal Set
\begin{theorem}
There exists no set which is an absolutely universal set.
That is:
:$\map \neg {\exists \, \UU: \forall T: T \in \UU}$
where $T$ is any arbitrary object at all.
That is, a set that contains ''everything'' cannot exist.
\end{theorem}
\begin{proof}
{{AimForCont}} such a $\UU$ exists.
Using the Axiom of Specification, we can create the set:
:$R = \set {x \in \UU: x \notin x}$
But from Russell's Paradox, this set cannot exist.
Thus:
:$R \notin \UU$
and so $\UU$ cannot contain everything.
{{qed}}
\end{proof}
|
22624
|
\section{There are no Odd Unitary Perfect Numbers}
Tags: Unitary Perfect Numbers
\begin{theorem}
No unitary perfect numbers exist which are odd.
\end{theorem}
\begin{proof}
Let $n$ be an odd number with prime decomposition $n = p_1^{a_1} \cdots p_k^{a_k}$.
By Sum of Unitary Divisors of Integer, the sum of its unitary divisors is $\ds \prod_{1 \mathop \le i \mathop \le k} \paren {1 + p_i^{a_i} }$.
To be a unitary perfect number, this must be equal to $2 n$.
Since each $p$ is odd, each $1 + p_i^{a_i}$ is even.
Hence $\ds \prod_{1 \mathop \le i \mathop \le k} \paren {1 + p_i^{a_i} }$ is divisible by $2^k$.
Since $n$ is odd, $2 n$ is divisible by $2$ but not $4$.
Thus $k = 1$.
So $n$ is a prime power.
By Sum of Unitary Divisors of Power of Prime, the sum of its unitary divisors is $1 + n$.
This cannot be equal to $2 n$, since $n > 1$.
Therefore there are no unitary perfect numbers which are odd.
{{qed}}
\end{proof}
|
22625
|
\section{Third Hyperoperation is Integer Power Operation}
Tags: Hyperoperation
\begin{theorem}
The '''$3$rd hyperoperation''' is the integer power operation restricted to the positive integers:
:$\forall x, y \in \Z_{\ge 0}: H_3 \left({x, y}\right) = x^y$
\end{theorem}
\begin{proof}
By definition of the hyperoperation sequence:
:$\forall n, x, y \in \Z_{\ge 0}: H_n \left({x, y}\right) = \begin{cases}
y + 1 & : n = 0 \\
x & : n = 1, y = 0 \\
0 & : n = 2, y = 0 \\
1 & : n > 2, y = 0 \\
H_{n - 1} \left({x, H_n \left({x, y - 1}\right)}\right) & : n > 0, y > 0 \end{cases}$
Thus the $3$rd hyperoperation is defined as:
:$\forall x, y \in \Z_{\ge 0}: H_3 \left({x, y}\right) = \begin{cases}
1 & : y = 0 \\
H_2 \left({x, H_3 \left({x, y - 1}\right)}\right) & : y > 0 \end{cases}$
From Second Hyperoperation is Multiplication Operation:
:$(1): \quad \forall x, y \in \Z_{\ge 0}: H_3 \left({x, y}\right) = \begin{cases}
1 & : y = 0 \\
x \times H_3 \left({x, y - 1}\right) & : y > 0 \end{cases}$
The proof proceeds by induction.
For all $y \in \Z_{\ge 0}$, let $P \left({y}\right)$ be the proposition:
:$\forall x \in \Z_{\ge 0}: H_3 \left({x, y}\right) = x^y$
\end{proof}
|
22626
|
\section{Third Isomorphism Theorem/Groups}
Tags: Isomorphism Theorems, Quotient Groups, Third Isomorphism Theorem, Normal Subgroups, Isomorphisms, Group Isomorphisms, Group Homomorphisms
\begin{theorem}
Let $G$ be a group, and let:
:$H, N$ be normal subgroups of $G$
:$N$ be a subset of $H$.
Then:
:$(1): \quad H / N$ is a normal subgroup of $G / N$
:::where $H / N$ denotes the quotient group of $H$ by $N$
:$(2): \quad \dfrac {G / N} {H / N} \cong \dfrac G H$
:::where $\cong$ denotes group isomorphism.
\end{theorem}
\begin{proof}
We define a mapping:
:$\phi: G / N \to G / H$ by $\map \phi {g N} = g H$
Since $\phi$ is defined on cosets, we need to check that $\phi$ is well-defined.
Suppose $x N = y N$.
Then:
:$y^{-1} x \in N$
Then:
:$N \le H \implies y^{-1} x \in H$
and so:
:$x H = y H$
So:
:$\map \phi {x N} = \map \phi {y N}$
and $\phi$ is indeed well-defined.
Now $\phi$ is a homomorphism, from:
{{begin-eqn}}
{{eqn | l = \map \phi {x N} \map \phi {y N}
| r = \paren {x H} \paren {y H}
| c =
}}
{{eqn | r = x y H
| c =
}}
{{eqn | r = \map \phi {x y N}
| c =
}}
{{eqn | r = \map \phi {x N y N}
| c =
}}
{{end-eqn}}
Also, since $N \subseteq H$, it follows that:
:$\order N \le \order H$
So:
:$\order {G / N} \ge \order {G / H}$, indicating $\phi$ is surjective.
So:
{{begin-eqn}}
{{eqn | l = \map \ker \phi
| r = \set {g N \in G / N: \map \phi {g N} = e_{G / H} }
| c =
}}
{{eqn | r = \set {g N \in G / N: g H = H}
| c =
}}
{{eqn | r = \set {g N \in G / N: g \in H}
| c =
}}
{{eqn | r = H / N
| c =
}}
{{end-eqn}}
The result follows from the First Isomorphism Theorem.
{{qed}}
\end{proof}
|
22627
|
\section{Third Isomorphism Theorem/Groups/Corollary}
Tags: Isomorphism Theorems, Normal Subgroups, Quotient Groups, Group Isomorphisms
\begin{theorem}
Let $G$ be a group.
Let $N$ be a normal subgroup of $G$.
Let $q: G \to \dfrac G N$ be the quotient epimorphism from $G$ to the quotient group $\dfrac G N$.
Let $K$ be the kernel of $q$.
Then:
:$\dfrac G N \cong \dfrac {G / K} {N / K}$
\end{theorem}
\begin{proof}
From Kernel is Normal Subgroup of Domain we have that $K$ is a normal subgroup of $G$.
Thus the Third Isomorphism Theorem for Groups can be applied directly.
{{qed}}
\end{proof}
|
22628
|
\section{Third Isomorphism Theorem/Rings}
Tags: Isomorphism Theorems, Isomorphisms, Ring Isomorphisms, Ideal Theory, Ring Homomorphisms
\begin{theorem}
Let $R$ be a ring, and let:
:$J, K$ be ideals of $R$
:$J$ be a subset of $K$.
Then:
:$(1): \quad K / J$ is an ideal of $R / J$
:where $K / J$ denotes the quotient ring of $K$ by $J$
:$(2): \quad \dfrac {R / J} {K / J} \cong \dfrac R K$
:where $\cong$ denotes ring isomorphism.
This result is also referred to by some sources as the '''first isomorphism theorem''', and by others as the '''second isomorphism theorem'''.
\end{theorem}
\begin{proof}
In Ring Homomorphism whose Kernel contains Ideal, take $\phi: R \to R / K$ to be the quotient epimorphism.
Then (from the same source) its kernel is $K$.
Thus we have that:
:$\phi = \psi \circ \nu$
where $\psi : R / J \to R / K$ is a homomorphism.
This can be illustrated by means of the following commutative diagram:
::$\begin{xy}\xymatrix@L+2mu@+1em{
R \ar[dr]_*{\phi}
\ar[r]^*{\nu}
& R / J \ar@{-->}[d]^*{\psi}
\\
&
R / K
}\end{xy}$
As $\phi$ is an epimorphism then from Surjection if Composite is Surjection we have that $\psi$ is a surjection.
So $\Img \psi = \Img \phi = R / K$ and the First Isomorphism Theorem applies.
{{qed}}
\end{proof}
|
22629
|
\section{Third Partial Derivatives of x^y}
Tags: Examples of Partial Derivatives
\begin{theorem}
Let:
:$u = x^y$
Then:
:$\dfrac {\partial^3 u} {\partial x^2 \partial y} = \dfrac {\partial^3 u} {\partial x \partial y \partial x}$
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | l = \dfrac {\partial u} {\partial y}
| r = x^y \ln x
| c = Derivative of General Logarithm Function keeping $x$ constant
}}
{{eqn | ll= \leadsto
| l = \dfrac {\partial^2 u} {\partial x \partial y}
| r = \map {\dfrac \partial {\partial x} } {\dfrac {\partial u} {\partial y} }
| c = {{Defof|Second Partial Derivative}}
}}
{{eqn | r = y x^{y - 1} \ln x + \dfrac 1 x x^y
| c = Power Rule for Derivatives, Derivative of Natural Logarithm keeping $y$ constant
}}
{{eqn | r = x^{y - 1} \paren {y \ln x + 1}
| c = rearranging
}}
{{eqn | ll= \leadsto
| l = \dfrac {\partial^3 u} {\partial x^2 \partial y}
| r = \map {\dfrac \partial {\partial x} } {\dfrac {\partial^2 u} {\partial x \partial y} }
| c = {{Defof|Third Partial Derivative}}
}}
{{eqn | r = x^{y - 1} \map {\dfrac \partial {\partial x} } {y \ln x + 1} + \paren {y \ln x + 1} \map {\dfrac \partial {\partial x} } {x^{y - 1} }
| c = Product Rule for Derivatives
}}
{{eqn | r = x^{y - 1} \paren {\dfrac y x} + \paren {y \ln x + 1} \paren {y - 1} x^{y - 2}
| c = Derivative of Natural Logarithm, Power Rule for Derivatives keeping $y$ constant
}}
{{eqn | r = x^{y - 2} \paren {y + \paren {y \ln x + 1} \paren {y - 1} }
| c = simplifying
}}
{{eqn | r = x^{y - 2} \paren {y + y^2 \ln x + y -y \ln x - 1}
| c = multiplying out
}}
{{eqn | r = x^{y - 2} \paren {2 y + y \paren {y - 1} \ln x - 1}
| c = simplifying
}}
{{end-eqn}}
{{qed|lemma}}
{{begin-eqn}}
{{eqn | l = \dfrac {\partial u} {\partial x}
| r = y x^{y - 1}
| c = Power Rule for Derivatives keeping $y$ constant
}}
{{eqn | ll= \leadsto
| l = \dfrac {\partial^2 u} {\partial y \partial x}
| r = \map {\dfrac \partial {\partial y} } {\dfrac {\partial u} {\partial x} }
| c = {{Defof|Second Partial Derivative}}
}}
{{eqn | r = x^{y - 1} \map {\dfrac \partial {\partial y} } y + y \map {\dfrac \partial {\partial y} } {x^{y - 1} }
| c = Product Rule for Derivatives
}}
{{eqn | r = x^{y - 1} + y x^{y - 1} \ln x
| c = Derivative of Identity Function, Derivative of General Logarithm Function keeping $x$ constant
}}
{{eqn | ll= \leadsto
| l = \dfrac {\partial^3 u} {\partial x \partial y \partial x}
| r = \map {\dfrac \partial {\partial x} } {\dfrac {\partial^2 u} {\partial y \partial x} }
| c = {{Defof|Third Partial Derivative}}
}}
{{eqn | r = \map {\dfrac \partial {\partial x} } {x^{y - 1} + y x^{y - 1} \ln x}
| c =
}}
{{eqn | r = \paren {y - 1} x^{y - 2} + y \paren {x^{y - 1} \map {\dfrac \partial {\partial x} } {\ln x} + \ln x \map {\dfrac \partial {\partial x} } {x^{y - 1} } }
| c = Power Rule for Derivatives, Product Rule for Derivatives
}}
{{eqn | r = \paren {y - 1} x^{y - 2} + y x^{y - 1} \paren {\dfrac 1 x} + y \ln x \paren {y - 1} x^{y - 2}
| c = Derivative of Natural Logarithm, Power Rule for Derivatives
}}
{{eqn | r = x^{y - 2} \paren {\paren {y - 1} + y + y \ln x \paren {y - 1} }
| c = factoring
}}
{{eqn | r = x^{y - 2} \paren {2 y + y \paren {y - 1} \ln x - 1}
| c = simplifying
}}
{{end-eqn}}
{{qed|lemma}}
The two expressions are equal.
Hence the result.
{{qed}}
\end{proof}
|
22630
|
\section{Third Principle of Mathematical Induction}
Tags: Number Theory, Proofs by Induction, Mathematical Induction, Named Theorems, Principle of Mathematical Induction, Proof Techniques
\begin{theorem}
Let $\map P n$ be a propositional function depending on $n \in \N$.
If:
:$(1): \quad \map P n$ is true for all $n \le d$ for some $d \in \N$
:$(2): \quad \forall m \in \N: \paren {\forall k \in \N, m \le k < m + d: \map P k} \implies \map P {m + d}$
then $\map P n$ is true for all $n \in \N$.
\end{theorem}
\begin{proof}
Let $A = \set {n \in \N: \map P n}$.
We show that $A$ is an inductive set.
By $(1)$:
:$\forall 1 \le i \le d: i \in A$
Let:
:$\forall x \ge d: \set {1, 2, \dotsc, x} \subset A$
Then by definition of $A$:
:$\forall k \in \N: x - \paren {d - 1} \le k < x + 1: \map P k$
Thus $\map P {x + 1} \implies x + 1 \in A$
Thus $A$ is an inductive set.
Thus by the fifth axiom of Peano:
:$\forall n \in \N: A = \N \implies \map P n$
{{qed}}
{{Proofread}}
\end{proof}
|
22631
|
\section{Third Sylow Theorem}
Tags: P-Groups, Sylow Theorems, Subgroups, Third Sylow Theorem, Group Theory, Named Theorems
\begin{theorem}
All the Sylow $p$-subgroups of a finite group are conjugate.
\end{theorem}
\begin{proof}
Suppose $P$ and $Q$ are Sylow $p$-subgroups of $G$.
By the Second Sylow Theorem, $Q$ is a subset of a conjugate of $P$.
But since $\order P = \order Q$, it follows that $Q$ must equal a conjugate of $P$.
{{qed}}
\end{proof}
|
22632
|
\section{Thirteen Catalan Polyhedra}
Tags: Catalan Polyhedra
\begin{theorem}
There exist exactly $13$ distinct Catalan polyhedra:
:Triakis tetrahedron
:Triakis octahedron
:Disdyakis dodecahedron
:Tetrakis hexahedron
:Triakis icosahedron
:Disdyakis triacontahedron
:Pentakis dodecahedron
:Rhombic dodecahedron
:Rhombic triacontahedron
:Deltoidal icositetrahedron
:Deltoidal hexecontahedron
:Pentagonal icositetrahedron
:Pentagonal hexecontahedron
\end{theorem}
\begin{proof}
By definition, the Catalan polyhedra are the dual polyhedra of the Archimedean polyhedra.
There are $13$ Archimedean polyhedra, and so there are $13$ Catalan polyhedra.
{{ProofWanted|Include links to specific results linking the Archimedean polyhedra with their duals.}}
\end{proof}
|
22633
|
\section{Three-Way Exclusive Or and Equivalence}
Tags: Propositional Logic, Biconditional, Exclusive Or, Logic
\begin{theorem}
Let $p \iff q$ be the biconditional operator, and $p \oplus q$ be the exclusive or operator.
Then:
: $p \iff q \iff r \dashv \vdash p \oplus q \oplus r$
\end{theorem}
\begin{proof}
We apply the Method of Truth Tables to the proposition.
As can be seen by inspection, in each case, the truth values under the main connectives match for all boolean interpretations.
$\begin{array}{|ccccc||ccccc|} \hline
(p & \iff & q) & \iff & r & (p & \oplus & q) & \oplus & r \\
\hline
F & T & F & F & F & F & F & F & F & F \\
F & T & F & T & T & F & F & F & T & T \\
F & F & T & T & F & F & T & T & T & F \\
F & F & T & F & T & F & T & T & F & T \\
T & F & F & T & F & T & T & F & T & F \\
T & F & F & F & T & T & T & F & F & T \\
T & T & T & F & F & T & F & T & F & F \\
T & T & T & T & T & T & F & T & T & T \\
\hline
\end{array}$
{{qed}}
From Biconditional is Associative and Exclusive Or is Associative, we have that both $\iff$ and $\oplus$ are associative, which justifies the rendition of this result without parentheses.
\end{proof}
|
22634
|
\section{Three Eighths as Pandigital Fraction}
Tags: Pandigital Fractions
\begin{theorem}
$\dfrac 3 8$ cannot be expressed as a pandigital fraction.
\end{theorem}
\begin{proof}
Can be verified by brute force.
Category:Pandigital Fractions
\end{proof}
|
22635
|
\section{Three Fifths as Pandigital Fraction}
Tags: Pandigital Fractions
\begin{theorem}
$\dfrac 3 5$ cannot be expressed as a pandigital fraction.
\end{theorem}
\begin{proof}
Can be verified by brute force.
Category:Pandigital Fractions
\end{proof}
|
22636
|
\section{Three Non-Collinear Planes have One Point in Common}
Tags: Projective Geometry
\begin{theorem}
Three planes which are not collinear have exactly one point in all three planes.
\end{theorem}
\begin{proof}
Let $A$, $B$ and $C$ be the three planes in question.
From Two Planes have Line in Common, $A$ and $B$ share a line, $p$ say.
From Propositions of Incidence: Plane and Line, $p$ meets $C$ in one point.
{{qed}}
\end{proof}
|
22637
|
\section{Three Points Describe a Circle}
Tags: Circles
\begin{theorem}
Let $A$, $B$ and $C$ be points which are not collinear.
Then there exists exactly one circle whose circumference passes through all $3$ points $A$, $B$ and $C$.
\end{theorem}
\begin{proof}
As $A$, $B$ and $C$ are not collinear, the triangle $ABC$ can be constructed by forming the lines $AB$, $BC$ and $CA$.
The result follows from Circumscribing Circle about Triangle.
{{qed}}
\end{proof}
|
22638
|
\section{Three Points in Ultrametric Space have Two Equal Distances}
Tags: Metric Spaces
\begin{theorem}
Let $\struct {X, d}$ be an ultrametric space.
Let $x, y, z \in X$ with $\map d {x, z} \ne \map d {y, z}$.
Then:
:$\map d {x, y} = \max \set {\map d {x, z}, \map d {y, z} }$
\end{theorem}
\begin{proof}
{{WLOG}}, let $\map d {x, z} > \map d {y, z}$.
Then:
{{begin-eqn}}
{{eqn | l = \map d {x, y}
| o = \le
| r = \max \set {\map d {x, z}, \map d {y, z} }
| c = {{Defof|Non-Archimedean Metric}}
}}
{{eqn | r = \map d {x, z}
| c = since $\map d {x, z} > \map d {y, z}$
}}
{{end-eqn}}
On the other hand:
{{begin-eqn}}
{{eqn | l = \map d {x, z}
| o = \le
| r = \max \set {\map d {x, y}, \map d {y, z} }
| c = {{Defof|Non-Archimedean Metric}}
}}
{{eqn | r = \map d {x, y}
| c = since $\map d {x, z} > \map d {y, z}$
}}
{{end-eqn}}
Putting the two inequalities together then:
:$\map d {x, y} = \map d {x, z} = \max \set {\map d {x, z}, \map d {y, z} }$
{{qed}}
\end{proof}
|
22639
|
\section{Three Points in Ultrametric Space have Two Equal Distances/Corollary}
Tags: Metric Spaces
\begin{theorem}
Let $\struct {X, d}$ be an ultrametric space.
Let $x, y, z \in X$.
Then:
:at least two of the distances $\map d {x, y}$, $\map d {x, z}$ and $\map d {y, z}$ are equal.
\end{theorem}
\begin{proof}
Either:
:$\map d {x, z} = \map d {y, z}$
or:
:$\map d {x, z} \ne \map d {y, z}$
By Three Points in Ultrametric Space have Two Equal Distances:
:$\map d {x, z} = \map d {y, z}$ or $\map d {x, y} = \max \set {\map d {x, z}, \map d {y, z} }$
In either case two of the distances are equal.
{{Qed}}
\end{proof}
|
22640
|
\section{Three Points in Ultrametric Space have Two Equal Distances/Corollary 2}
Tags: Normed Division Rings, Definitions: Normed Division Rings
\begin{theorem}
Let $\struct {R, \norm {\,\cdot\,} }$ be a normed division ring with non-Archimedean norm $\norm{\,\cdot\,}$,
Let $x, y \in R$ and $\norm x \ne \norm y$.
Then:
:$\norm {x + y} = \norm {x - y} = \norm {y - x} = \max \set {\norm x, \norm y}$
\end{theorem}
\begin{proof}
Let $d$ be the metric induced by the norm $\norm {\,\cdot\,}$.
By Non-Archimedean Norm iff Non-Archimedean Metric then $d$ is a non-Archimedean metric and $\struct {R, d}$ is an ultrametric space.
Let $x, y \in R$ and $\norm x \ne \norm y$.
By the definition of the non-Archimedean metric $d$ then:
:$\norm x = \norm {x - 0} = \map d {x, 0}$
and similarly:
:$\norm y = \map d {y, 0}$
By assumption then:
:$\map d {x, 0} \ne \map d {y, 0}$
By Three Points in Ultrametric Space have Two Equal Distances then:
:$\norm {x - y} = \map d {x, y} = \max \set {\map d {x, 0}, \map d {y, 0} } = \max \set {\norm x, \norm y}$
By Norm of Negative then:
:$\norm {y - x} = \norm {x - y} = \max \set {\norm x, \norm y}$
By the definition of the non-Archimedean metric $d$ then:
:$\norm y = \norm {0 - \paren {-y} } = \map d {0, -y} = \map d {-y, 0}$
By assumption then:
:$\map d {x, 0} \ne \map d {-y, 0}$
By Three Points in Ultrametric Space have Two Equal Distances then:
:$\map d {x, -y} = \max \set {\map d {x, 0}, \map d {-y, 0} } = \max \set {\norm x, \norm y}$
By the definition of the non-Archimedean metric $d$ then:
:$\map d {x, -y} = \norm {x - \paren {-y} } = \norm {x + y}$
So $\norm {x + y} = \max \set {\norm x, \norm y}$.
The result follows.
{{qed}}
\end{proof}
|
22641
|
\section{Three Points in Ultrametric Space have Two Equal Distances/Corollary 3}
Tags: Normed Division Rings
\begin{theorem}
Let $\struct {R, \norm {\,\cdot\,} }$ be a normed division ring with non-Archimedean norm $\norm{\,\cdot\,}$,
Let $x, y \in R$ and $\norm x \lt \norm y$.
Then:
:$\norm {x + y} = \norm {x - y} = \norm {y - x} = \norm y$
\end{theorem}
\begin{proof}
By Corollary 2 then:
:$\norm {x + y} = \norm {x - y} = \norm {y - x} = \max \set {\norm x, \norm y} = \norm y$
{{qed}}
\end{proof}
|
22642
|
\section{Three Points in Ultrametric Space have Two Equal Distances/Corollary 4}
Tags: Normed Division Rings
\begin{theorem}
Let $\struct {R, \norm {\,\cdot\,} }$ be a normed division ring with non-Archimedean norm $\norm{\,\cdot\,}$,
Let $x, y \in R$.
Then:
:* $\norm {x + y} \lt \norm y \implies \norm x = \norm y$
:* $\norm {x - y} \lt \norm y \implies \norm x = \norm y$
:* $\norm {y - x} \lt \norm y \implies \norm x = \norm y$
\end{theorem}
\begin{proof}
The contrapositive statements are proved.
Let $\norm x \ne \norm y$
By Corollary 2 then:
:$\norm {x + y} = \norm {x - y} = \norm {y - x} = \max \set {\norm x, \norm y} \ge \norm y$
The result follows.
{{qed}}
\end{proof}
|
22643
|
\section{Three Points in Ultrametric Space have Two Equal Distances/Corollary 5}
Tags: Normed Division Rings, P-adic Number Theory
\begin{theorem}
Let $\norm {\, \cdot \,}$ be a non-trivial non-Archimedean norm on the rational numbers $\Q$.
Let $a, b \in \Z_{\ne 0}$ be coprime, $a \perp b$
Then:
:$\norm a = 1$ or $\norm b = 1$
\end{theorem}
\begin{proof}
By Bézout's Identity then:
:$\exists n, m \in \Z : m a + n b = 1$
By Norm of Unity then:
:$\norm {m a + n b} = 1$
By Corollary 5 of Characterisation of Non-Archimedean Division Ring Norms then:
:$\norm a, \norm b, \norm n, \norm m \le 1$
Let $\norm a \lt 1$.
By Norm axiom $(\text N 2)$: Multiplicativity:
:$\norm {m a} = \norm m \norm a \lt 1$
Hence:
:$\norm {m a} < \norm {m a + n b}$
By Corollary 4:
:$\norm {n b} = \norm {m a + n b} = 1$
By Norm axiom $(\text N 2)$: Multiplicativity:
:$\norm n \norm b = 1$.
Hence $\norm b = 1$.
The result follows.
{{qed}}
\end{proof}
|
22644
|
\section{Three Points on Sphere in Same Hemisphere}
Tags: Spherical Geometry
\begin{theorem}
Let $S$ be a sphere.
Let $A$, $B$ and $C$ be points on $S$ which do not all lie on the same great circle.
Then it is possible to divide $S$ into two hemispheres such that $A$, $B$ and $C$ all lie on the same hemisphere.
\end{theorem}
\begin{proof}
Because $A$, $B$ and $C$ do not lie on the same great circle, no two of these points are the endpoints of the same diameter of $S$.
Otherwise it would be possible to construct a great circle passing through all $3$ points $A$, $B$ and $C$.
Let a great circle $E$ be constructed through $A$ and $B$.
Then as $C$ is not on $E$, it is on either one side or the other.
Hence there is a finite spherical angle $\phi$ between $E$ and $C$.
Let a diameter $PQ$ of $S$ be constructed whose endpoints are on $E$ such that:
:neither $P$ nor $Q$ coincides with $A$ or $B$
:both $A$ and $B$ are on the same semicircle of $E$ into which $PQ$ divides $E$.
Let two great circles $F$ and $G$ be constructed through $PQ$ which are at a spherical angle $\dfrac \phi 2$ from $E$, one in one direction and one in the other.
Then $F$ and $G$ both divide $S$ into two hemispheres:
:one such division has a hemisphere which contains $A$ and $B$ but does not contain $E$
:the other such division has a hemisphere which contains $A$, $B$ and $E$.
Hence the result.
{{qed}}
\end{proof}
|
22645
|
\section{Three Quarters as Pandigital Fraction}
Tags: Pandigital Fractions
\begin{theorem}
$\dfrac 3 4$ cannot be expressed as a pandigital fraction.
\end{theorem}
\begin{proof}
Can be verified by brute force.
Category:Pandigital Fractions
\end{proof}
|
22646
|
\section{Three Regular Tessellations}
Tags: Tessellations
\begin{theorem}
There exist exactly $3$ regular tessellations of the plane.
\end{theorem}
\begin{proof}
Let $m$ be the number of sides of each of the regular polygons that form the regular tessellation.
Let $n$ be the number of those regular polygons which meet at each vertex.
From Internal Angles of Regular Polygon, the internal angles of each polygon measure $\dfrac {\paren {m - 2} 180^\circ} m$.
The sum of the internal angles at a point is equal to $360^\circ$ by Sum of Angles between Straight Lines at Point form Four Right Angles
So:
{{begin-eqn}}
{{eqn | l = n \paren {\dfrac {\paren {m - 2} 180^\circ} m}
| r = 360^\circ
| c =
}}
{{eqn | ll= \leadsto
| l = \dfrac {m - 2} m
| r = \dfrac 2 n
| c =
}}
{{eqn | ll= \leadsto
| l = 1 - \dfrac 2 m
| r = \dfrac 2 n
| c =
}}
{{eqn | ll= \leadsto
| l = \dfrac 1 m + \dfrac 1 n
| r = \dfrac 1 2
| c =
}}
{{end-eqn}}
But $m$ and $n$ are both greater than $2$.
So:
:if $m = 3$, $n = 6$.
:if $m = 4$, $n = 4$.
:if $m = 5$, $n = \dfrac {10} 3$, which is not an integer.
:if $m = 6$, $n = 3$.
Now suppose $m > 6$.
We have:
{{begin-eqn}}
{{eqn | l = \dfrac 1 m + \dfrac 1 n
| r = \dfrac 1 2
| c =
}}
{{eqn | ll= \leadsto
| l = \dfrac 1 6 + \dfrac 1 n
| o = >
| r = \dfrac 1 2
| c =
}}
{{eqn | ll= \leadsto
| l = \dfrac 1 n
| o = >
| r = \dfrac 1 3
| c =
}}
{{eqn | ll= \leadsto
| l = 3
| o = >
| r = n
| c =
}}
{{end-eqn}}
But there are no integers between $2$ and $3$, so $m \not > 6$.
There are $3$ possibilities in all.
Therefore all regular tessellations have been accounted for.
{{qed}}
\end{proof}
|
22647
|
\section{Three Regular Tessellations/Squares}
Tags: Tessellations, Squares
\begin{theorem}
Squares form a regular tessellation:
:400px
\end{theorem}
\begin{proof}
{{proof wanted}}
Category:Tessellations
Category:Squares
\end{proof}
|
22648
|
\section{Three Regular Tessellations/Triangles}
Tags: Equilateral Triangles, Tessellations, Triangles
\begin{theorem}
Equilateral triangles form a regular tessellation:
:400px
\end{theorem}
\begin{proof}
{{proof wanted}}
Category:Tessellations
Category:Equilateral Triangles
\end{proof}
|
22649
|
\section{Three Sevenths as Pandigital Fraction}
Tags: Pandigital Fractions
\begin{theorem}
$\dfrac 3 7$ cannot be expressed as a pandigital fraction.
\end{theorem}
\begin{proof}
Can be verified by brute force.
Category:Pandigital Fractions
\end{proof}
|
22650
|
\section{Three Times Sum of Cubes of Three Indeterminates Plus 6 Times their Product}
Tags: Cube Roots of Unity, Algebra
\begin{theorem}
:$3 \paren {a^3 + b^3 + c^3 + 6 a b c} = \paren {a + b + c}^3 + \paren {a + b \omega + c \omega^2}^3 + \paren {a + b \omega^2 + c \omega}^3$
where:
: $\omega = -\dfrac 1 2 + \dfrac {\sqrt 3} 2$
\end{theorem}
\begin{proof}
Multiplying out:
{{begin-eqn}}
{{eqn | l = \paren {a + b + c}^3
| r = \paren {a + b + c} \paren {a^2 + b^2 + c^2 + 2 a b + 2 a c + 2 b c}
| c =
}}
{{eqn | n = 1
| r = a^3 + b^3 + c^3 + 3 a^2 b + 3 a^2 c + 3 a b^2 + 3 b^2 c + 3 a c^2 + 3 b c^2 + 6 a b c
| c =
}}
{{end-eqn}}
Replacing $b$ with $b \omega$ and $c$ with $c \omega^2$ in $(1)$:
{{begin-eqn}}
{{eqn | l = \paren {a + b \omega + c \omega^2}^3
| r = a^3 + b^3 \omega^3 + c^3 \omega^6 + 3 a^2 b \omega + 3 a^2 c \omega^2 + 3 a b^2 \omega^2 + 3 b^2 c \omega^4 + 3 a c^2 \omega^4 + 3 b c^2 \omega^5 + 6 a b c \omega^3
| c =
}}
{{eqn | n = 2
| r = a^3 + b^3 + c^3 + 3 a^2 b \omega + 3 a^2 c \omega^2 + 3 a b^2 \omega^2 + 3 b^2 c \omega + 3 a c^2 \omega + 3 b c^2 \omega^2 + 6 a b c
| c = as $\omega^3 = 1$
}}
{{end-eqn}}
Replacing $b$ with $b \omega^2$ and $c$ with $c \omega$ in $(1)$:
{{begin-eqn}}
{{eqn | l = \paren {a + b \omega^2 + c \omega}^3
| r = a^3 + b^3 \omega^6 + c^3 \omega^3 + 3 a^2 b \omega^2 + 3 a^2 c \omega + 3 a b^2 \omega^4 + 3 b^2 c \omega^5 + 3 a c^2 \omega^2 + 3 b c^2 \omega^4 + 6 a b c \omega^3
| c =
}}
{{eqn | n = 3
| r = a^3 + b^3 + c^3 + 3 a^2 b \omega^2 + 3 a^2 c \omega + 3 a b^2 \omega + 3 b^2 c \omega^2 + 3 a c^2 \omega^2 + 3 b c^2 \omega + 6 a b c
| c = as $\omega^3 = 1$
}}
{{end-eqn}}
Adding together $(1)$, $(2)$ and $(3)$:
{{begin-eqn}}
{{eqn | r = \paren {a + b + c}^3 + \paren {a + b \omega + c \omega^2}^3 + \paren {a + b \omega^2 + c \omega}^3
| o =
| c =
}}
{{eqn | r = a^3 + b^3 + c^3 + 3 a^2 b + 3 a^2 c + 3 a b^2 + 3 b^2 c + 3 a c^2 + 3 b c^2 + 6 a b c
| c =
}}
{{eqn | o =
| ro= +
| r = a^3 + b^3 + c^3 + 3 a^2 b \omega + 3 a^2 c \omega^2 + 3 a b^2 \omega^2 + 3 b^2 c \omega + 3 a c^2 \omega + 3 b c^2 \omega^2 + 6 a b c
| c =
}}
{{eqn | o =
| ro= +
| r = a^3 + b^3 + c^3 + 3 a^2 b \omega^2 + 3 a^2 c \omega + 3 a b^2 \omega + 3 b^2 c \omega^2 + 3 a c^2 \omega^2 + 3 b c^2 \omega + 6 a b c
| c =
}}
{{eqn | r = 3 \paren {a^3 + b^3 + c^3 + 6 a b c}
| c =
}}
{{eqn | o =
| ro= +
| r = \paren {3 a^2 b + 3 a^2 c + 3 a b^2 + 3 b^2 c + 3 a c^2 + 3 b c^2} \paren {1 + \omega + \omega^2}
| c =
}}
{{eqn | r = 3 \paren {a^3 + b^3 + c^3 + 6 a b c}
| c = Sum of Cube Roots of Unity: $1 + \omega + \omega^2 = 0$
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
22651
|
\section{Three Tri-Automorphic Numbers for each Number of Digits}
Tags: Tri-Automorphic Numbers
\begin{theorem}
Let $d \in \Z_{>0}$ be a (strictly) positive integer.
Then there exist exactly $3$ tri-automorphic numbers with exactly $d$ digits.
These tri-automorphic numbers all end in $2$, $5$ or $7$.
\end{theorem}
\begin{proof}
Let $n$ be a tri-automorphic number with $d$ digits.
Let $n = 10 a + b$.
Then:
:$3 n^2 = 300a^2 + 60 a b + 3 b^2$
As $n$ is tri-automorphic, we have:
:$(1): \quad 300 a^2 + 60 a b + 3 b^2 = 1000 z + 100 y + 10 a + b$
and:
:$(2): \quad 3 b^2 - b = 10 x$
where $x$ is an integer.
This condition is only satisfied by $b = 2$, $b = 5$, or $b = 7$
{{ProofWanted|Guess: Try proving for $n {{=}} 10 a + b$ and then by induction.}}
Substituting $b = 2$ in equation $(1)$:
:$a = 9$
Substituting $b = 5$ in equation $(1)$:
:$a = 7$
Substituting $b = 7$ in equation $(1)$:
:$a = 6$
{{qed}}
<!--
Corollary:
For di-automorphic numbers, equation (5) becomes
2b^2-b=10x and b=8
For tetra-automorphic numbers, equation (5) becomes
4b^2-b=10x and b=4
For penta-automorphic numbers, equation (5) becomes
5b^2-b=10x and b=5
-->
\end{proof}
|
22652
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\section{Three times Number whose Divisor Sum is Square}
Tags: Numbers whose Sigma is Square, Three times Number whose Sigma is Square, Sigma Function, Three times Number whose Divisor Sum is Square, Numbers whose Divisor Sum is Square
\begin{theorem}
Let $n \in \Z_{>0}$ be a positive integer.
Let the divisor sum $\map {\sigma_1} n$ of $n$ be square.
Let $3$ not be a divisor of $n$.
Then the divisor sum of $3 n$ is square.
\end{theorem}
\begin{proof}
Let $\sigma \left({n}\right) = k^2$.
We have from Numbers whose $\sigma$ is Square:
:{{:Numbers whose Sigma is Square/Examples/3}}
As $3$ is not a divisor of $n$, it follows that $3$ and $n$ are coprime.
Thus:
{{begin-eqn}}
{{eqn | l = \sigma \left({3 n}\right)
| r = \sigma \left({3 n}\right) \sigma \left({3 n}\right)
| c = Sigma Function is Multiplicative
}}
{{eqn | r = 2^2 k^2
| c = from above
}}
{{eqn | r = \left({2 k}\right)^2
| c = from above
}}
{{end-eqn}}
{{qed}}
Category:Sigma Function
Category:Numbers whose Sigma is Square
286280
286276
2017-02-25T08:51:28Z
Prime.mover
59
286280
wikitext
text/x-wiki
\end{proof}
|
22653
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\section{Time Taken for Body to Fall at Earth's Surface}
Tags: Gravity, Mechanics
\begin{theorem}
Let an object $m$ be released above ground from a point near the Earth's surface and allowed to fall freely.
Let $m$ fall a distance $s$ in time $t$.
Then:
:$s = \dfrac 1 2 g t^2$
or:
:$t = \sqrt {\dfrac {2 s} g}$
where $g$ is the Acceleration Due to Gravity at the height through which $m$ falls.
It is supposed that the distance $s$ is small enough that $g$ can be considered constant throughout.
\end{theorem}
\begin{proof}
From Body under Constant Acceleration: Distance after Time:
:$\mathbf s = \mathbf u t + \dfrac {\mathbf a t^2} 2$
Here the body falls from rest, so:
:$\mathbf u = \mathbf 0$
Thus:
:$\mathbf s = \dfrac {\mathbf g t^2} 2$
and so taking magnitudes:
:$s = \dfrac {g t^2} 2$
It follows by multiplying by $\dfrac 2 g$ that:
:$t^2 = \dfrac {2 s} g$
whence:
:$t = \sqrt {\dfrac {2 s} g}$
{{qed}}
\end{proof}
|
22654
|
\section{Time of Travel down Brachistochrone}
Tags: Cycloids
\begin{theorem}
Let a wire $AB$ be curved into the shape of a brachistochrone.
Let $AB$ be embedded in a constant and uniform gravitational field where Acceleration Due to Gravity is $g$.
Let a bead $P$ be released at $A$ to slide down without friction to $B$.
Then the time taken for $P$ to slide from $A$ to $B$ is:
:$T = \pi \sqrt {\dfrac a g}$
where $a$ is the radius of the generating circle of the cycloid which forms $AB$.
\end{theorem}
\begin{proof}
That the curve $AB$ is indeed a cycloid is demonstrated in Brachistochrone is Cycloid.
Let $A$ be located at the origin of a cartesian plane.
We have the equations of the cycloid:
{{begin-eqn}}
{{eqn | l = x
| r = a \paren {\theta - \sin \theta}
}}
{{eqn | l = y
| r = a \paren {1 - \cos \theta}
}}
{{end-eqn}}
Differentiating with respect to theta:
{{begin-eqn}}
{{eqn | l = \frac {\d x} {\d \theta}
| r = \paren {1 - \cos \theta}
| c =
}}
{{eqn | l = \frac {\d y} {\d \theta}
| r = a \sin \theta
}}
{{end-eqn}}
Let $s$ be the distance along $AB$ from the origin.
Then:
{{begin-eqn}}
{{eqn | l = \paren {\frac {\d s} {\d \theta} }^2
| r = \paren {\frac {\d x} {\d \theta} }^2 + \paren {\frac {\d y} {\d \theta} }^2
| c =
}}
{{eqn | r = a^2 \paren {1 - \cos \theta}^2 + a^2 \sin^2 \theta
| c =
}}
{{eqn | r = a^2 \paren {1 - 2 \cos \theta + \cos^2 \theta + \sin^2 \theta}
| c =
}}
{{eqn | r = 2 a^2 \paren {1 - \cos \theta}
| c =
}}
{{eqn | r = 4 a^2 \map {\sin^2} {\frac \theta 2}
| c =
}}
{{end-eqn}}
Thus:
:$\dfrac {\d s} {\d \theta} = 2 a \map \sin {\dfrac \theta 2}$
Now:
{{begin-eqn}}
{{eqn | l = v
| r = \dfrac {\d s} {\d t}
| c =
}}
{{eqn | ll= \leadsto
| l = \int \rd t
| r = t
| c =
}}
{{eqn | r = \int \frac {\d s} v
| c =
}}
{{end-eqn}}
We also have, from the Principle of Conservation of Energy, that:
{{begin-eqn}}
{{eqn | l = m g y
| r = \frac {m v^2} 2
| c =
}}
{{eqn | ll= \leadsto
| l = v
| r = \dfrac {\d s} {\d t}
| c =
}}
{{eqn | r = \sqrt {2 g y}
| c =
}}
{{eqn | r = \sqrt {2 g a \paren {1 - \cos \theta} }
| c =
}}
{{end-eqn}}
Thus:
{{begin-eqn}}
{{eqn | l = \frac {\d t} {\d \theta}
| r = \frac {\d t} {\d s} \theta
| c =
}}
{{eqn | r = \frac {a \sqrt {2 \paren {1 - \cos \theta} } } {\sqrt {2 g a \paren {1 - \cos \theta} } }
| c =
}}
{{eqn | r = \sqrt {\frac a g}
| c =
}}
{{end-eqn}}
Let $T$ be the time taken to slide down the brachistochrone.
At the top:
:$\theta = 0$
and at the bottom:
:$\theta = \pi$
Then:
{{begin-eqn}}
{{eqn | l = T
| r = \int \rd t
| c =
}}
{{eqn | r = \sqrt {\frac a g} \int_0^\pi \rd \theta
| c =
}}
{{eqn | r = \pi \sqrt {\frac a g}
| c =
}}
{{end-eqn}}
So the time to slide down this brachistochrone from the top to the bottom is:
:$T = \pi \sqrt {\dfrac a g}$
{{qed}}
\end{proof}
|
22655
|
\section{Time of Travel down Brachistochrone/Corollary}
Tags: Cycloids
\begin{theorem}
Let a wire $AB$ be curved into the shape of a brachistochrone.
Let $AB$ be embedded in a constant and uniform gravitational field where Acceleration Due to Gravity is $g$.
Let a bead $P$ be released from anywhere on the wire between $A$ and $B$ to slide down without friction to $B$.
Then the time taken for $P$ to slide to $B$ is:
:$T = \pi \sqrt{\dfrac a g}$
where $a$ is the radius of the generating circle of the cycloid which forms $AB$.
\end{theorem}
\begin{proof}
That the curve $AB$ is indeed a cycloid is demonstrated in Brachistochrone is Cycloid.
Let $A$ be located at the origin of a cartesian plane.
Let the bead slide from an intermediate point $\theta_0$.
We have:
:$v = \dfrac {\d s} {\d t} = \sqrt {2 g \paren {y - y_0} }$
which leads us, via the same route as for Time of Travel down Brachistochrone, to:
{{begin-eqn}}
{{eqn | l = T
| r = \int_{\theta_0}^\pi \sqrt {\frac {2 a^2 \paren {1 - \cos \theta} } {2 g a \paren {\cos \theta_0 - \cos \theta} } } \rd \theta
| c =
}}
{{eqn | r = \sqrt {\frac a g} \int_{\theta_0}^\pi \sqrt {\frac {1 - \cos \theta} {\cos \theta_0 - \cos \theta} } \rd \theta
| c =
}}
{{end-eqn}}
Using the Half Angle Formula for Cosine and Half Angle Formula for Sine, this gives:
:$\ds T = \sqrt {\frac a g} \int_{\theta_0}^\pi \frac {\map \sin {\theta / 2} } {\sqrt {\map \cos {\theta_0 / 2} - \map \cos {\theta / 2} } } \rd \theta$
Now we make the substitution:
{{begin-eqn}}
{{eqn | l = u
| r = \frac {\map \cos {\theta / 2} } {\map \cos {\theta_0 / 2} }
| c =
}}
{{eqn | ll= \leadsto
| l = \frac {\d u} {\d \theta}
| r = -\frac {\map \sin {\theta / 2} } {2 \map \cos {\theta_0 / 2} }
| c =
}}
{{end-eqn}}
Recalculating the limits:
:when $\theta = \theta_0$ we have $u = 1$
:when $\theta = \pi$ we have $u = 0$.
So:
{{begin-eqn}}
{{eqn | l = T
| r = -2 \sqrt {\frac a g} \int_1^0 \frac {\d u} {\sqrt {1 - u^2} }
| c =
}}
{{eqn | r = \intlimits {2 \sqrt {\frac a g} \sin^{-1} u} 0 1
| c =
}}
{{eqn | r = \pi \sqrt {\frac a g}
| c =
}}
{{end-eqn}}
Thus the time to slide down a brachistochrone from any arbitrary point $\theta_0$ is:
:$T = \pi \sqrt {\dfrac a g}$
{{qed}}
\end{proof}
|
22656
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\section{Time when Hour Hand and Minute Hand at Right Angle}
Tags: Clocks
\begin{theorem}
Let the time of day be such that the hour hand and minute hand are at a right angle to each other.
Then the time happens $22$ times in every $12$ hour period:
:when the minute hand is $15$ minutes ahead of the hour hand
:when the minute hand is $15$ minutes behind of the hour hand.
In the first case, this happens at $09:00$ and every $1$ hour, $5$ minutes and $27 . \dot 2 \dot 7$ seconds after
In the second case, this happens at $03:00$ and every $1$ hour, $5$ minutes and $27 . \dot 2 \dot 7$ seconds after.
Thus the times are, to the nearest second:
$\begin {array}
09:00:00 & 03:00:00 \\
10:05:27 & 04:05:27 \\
11:10:55 & 05:10:55 \\
12:16:22 & 06:16:22 \\
13:21:49 & 07:21:49 \\
14:27:16 & 08:27:16 \\
15:32:44 & 09:32:44 \\
16:38:11 & 10:38:11 \\
17:43:38 & 11:43:38 \\
18:49:05 & 12:49:05 \\
19:54:33 & 13:54:33 \\
\end{array}$
and times $12$ hours different.
\end{theorem}
\begin{proof}
Obviously the hands are at right angles at $3$ and $9$ o'clock.
Thus we only need to show that the angle between the hands will be the same after every $1$ hour, $5$ minutes and $27 . \dot 2 \dot 7$ seconds.
Note that:
{{begin-eqn}}
{{eqn | l = 1 h \ 5 m \ 27. \dot 2 \dot 7 s
| r = 65 m \ 27 \tfrac {27}{99} s
}}
{{eqn | r = 65 m \ 27 \tfrac 3 {11} s
}}
{{eqn | r = 65 m \ \frac {300} {11} s
}}
{{eqn | r = 65 m + \frac 5 {11} m
}}
{{eqn | r = \frac {720} {11} m
}}
{{eqn | r = \frac {12} {11} h
}}
{{end-eqn}}
In $\dfrac {12} {11}$ hours:
:The minute hand has rotated $\dfrac {12} {11} \times 360^\circ$
:The hour hand has rotated $\dfrac {12} {11} \times 30^\circ$
Thus the angle between the hands has changed by:
{{begin-eqn}}
{{eqn | l = \frac {12} {11} \times 360^\circ - \frac {12} {11} \times 30^\circ
| r = \frac {12} {11} \times 330^\circ
}}
{{eqn | r = 360^\circ
}}
{{end-eqn}}
which is a full rotation.
Hence after $\dfrac {12} {11}$ hours the angle between the hands would remain unchanged.
{{qed}}
Category:Clocks
\end{proof}
|
22657
|
\section{Titanic Prime consisting of 111 Blocks of each Digit plus Zeroes}
Tags: Titanic Primes
\begin{theorem}
The integer defined as:
:$\paren {123456789}_{111} \paren 0_{2284} 1$
where $\paren a_b$ means $b$ instances of $a$ in a string, is a titanic prime.
\end{theorem}
\begin{proof}
It is noted that it has $9 \times 111 + 2284 + 1 = 3284$ digits, making it titanic.
It can be expressed arithmetically as:
:$123456789 \times \dfrac {10^{999} - 1} {10^9 - 1} \times 10^{2285} + 1$
{{Alpertron-factorizer|date = $6$th March $2022$|time = $50.2$ seconds}}
\end{proof}
|
22658
|
\section{Titanic Prime whose Digits are all 0 or 1}
Tags: Titanic Primes
\begin{theorem}
The integer defined as:
:$10^{641} \times \dfrac {10^{640} - 1} 9 + 1$
is a titanic prime all of whose digits are either $0$ or $1$.
That is:
:$\paren 1_{640} \paren 0_{640} 1$
where $\paren a_b$ means $b$ instances of $a$ in a string.
\end{theorem}
\begin{proof}
It is clear that the digits are instances of $0$ and $1$.
It is also noted that it has $640 \times 2 + 1 = 1281$ digits, making it titanic.
{{Alpertron-factorizer|date = $6$th March $2022$|time = $2.5$ seconds}}
\end{proof}
|
22659
|
\section{Titanic Prime whose Digits are all 9 except for one 1}
Tags: Titanic Primes
\begin{theorem}
The integer defined as:
:$2 \times 10^{3020} - 1$
is a titanic prime all of whose digits are $9$ except one, which is $1$.
That is:
:$1 \left({9}\right)_{3020}$
where $\left({a}\right)_b$ means $b$ instances of $a$ in a string.
\end{theorem}
\begin{proof}
It is clear that the digits are instances of $9$ except for the first $1$.
It is also noted that it has $3020 + 1 = 3021$ digits, making it titanic.
{{Alpertron-factorizer|date = $6$th March $2022$|time = $25.8$ seconds}}
\end{proof}
|
22660
|
\section{Titanic Prime whose Digits are all Odd}
Tags: Titanic Primes
\begin{theorem}
The integer defined as:
:$1358 \times 10^{3821} - 1$
is a titanic prime all of whose digits are odd.
That is:
:$1357 \paren 9_{3821}$
where $\paren a_b$ means $b$ instances of $a$ in a string.
\end{theorem}
\begin{proof}
It is clear that the digits are all instances of $9$ except for the initial $1357$, all of which are odd.
It is also noted that it has $4 + 3821 = 3825$ digits, making it titanic.
{{Alpertron-factorizer|date = $6$th March $2022$|time = $34.4$ seconds}}
\end{proof}
|
22661
|
\section{Titanic Prime whose Digits are all Prime}
Tags: Prime Numbers, Titanic Primes
\begin{theorem}
The integer defined as:
:$7352 \times \dfrac {10^{1104} - 1} {10^4 - 1} + 1$
is a titanic prime all of whose digits are themselves prime.
That is:
:$\underbrace{7352}_{275} 7353$
\end{theorem}
\begin{proof}
It is clear that the digits are instances of $7$, $3$, $5$ and $2$, and so are all prime.
It is also noted that it has $275 \times 4 + 4 = 1104$ digits, making it titanic.
{{Alpertron-factorizer|date = $6$th March $2022$|time = $2.1$ seconds}}
\end{proof}
|
22662
|
\section{Titanic Sophie Germain Prime}
Tags: Titanic Primes, Sophie Germain Primes
\begin{theorem}
The integer defined as:
:$39 \, 051 \times 2^{6001} - 1$
is a titanic prime which is also a Sophie Germain prime:
{{begin-eqn}}
{{eqn | o =
| r = 11820 \, 50794 \, 19125 \, 52383 \, 74423 \, 53078 \, 56017 \, 05024 \, 84819 \, 01689
}}
{{eqn | o =
| r = 74975 \, 95139 \, 68621 \, 89553 \, 48654 \, 81137 \, 72841 \, 27658 \, 52217 \, 40999
}}
{{eqn | o =
| r = 04778 \, 71896 \, 78015 \, 63535 \, 94741 \, 82340 \, 68638 \, 88011 \, 18130 \, 14219
}}
{{eqn | o =
| r = 81435 \, 50235 \, 73607 \, 51980 \, 74200 \, 04306 \, 58030 \, 53360 \, 79821 \, 16678
}}
{{eqn | o =
| r = 32541 \, 21729 \, 72493 \, 53731 \, 27605 \, 59447 \, 95967 \, 46064 \, 11137 \, 07858
}}
{{eqn | o =
| r = 37078 \, 27755 \, 33462 \, 32179 \, 66482 \, 80947 \, 33386 \, 65681 \, 87582 \, 11189
}}
{{eqn | o =
| r = 25630 \, 83169 \, 50526 \, 70023 \, 66301 \, 83449 \, 99960 \, 25913 \, 90035 \, 61496
}}
{{eqn | o =
| r = 03726 \, 62661 \, 50693 \, 56343 \, 90085 \, 30468 \, 46645 \, 69888 \, 03202 \, 50070
}}
{{eqn | o =
| r = 38139 \, 19172 \, 69637 \, 71838 \, 13812 \, 48256 \, 38384 \, 37787 \, 83423 \, 06357
}}
{{eqn | o =
| r = 09062 \, 96393 \, 13908 \, 65400 \, 30048 \, 07291 \, 64958 \, 29772 \, 97828 \, 35273
}}
{{eqn | o =
| r = 02603 \, 73947 \, 05739 \, 46904 \, 93564 \, 50661 \, 00172 \, 36892 \, 20285 \, 60354
}}
{{eqn | o =
| r = 58830 \, 25332 \, 20848 \, 80128 \, 32451 \, 94645 \, 21648 \, 78503 \, 66425 \, 73281
}}
{{eqn | o =
| r = 55405 \, 94426 \, 29476 \, 00573 \, 05011 \, 86259 \, 25148 \, 08537 \, 31389 \, 24832
}}
{{eqn | o =
| r = 90593 \, 45279 \, 70389 \, 89332 \, 87614 \, 90279 \, 77417 \, 70009 \, 37843 \, 56718
}}
{{eqn | o =
| r = 78965 \, 55090 \, 40413 \, 05491 \, 45610 \, 39734 \, 55313 \, 36378 \, 82326 \, 51747
}}
{{eqn | o =
| r = 26323 \, 96872 \, 58800 \, 36097 \, 85595 \, 50576 \, 58179 \, 78961 \, 56439 \, 38001
}}
{{eqn | o =
| r = 61356 \, 42993 \, 82918 \, 89157 \, 64818 \, 24068 \, 61810 \, 98754 \, 13407 \, 25598
}}
{{eqn | o =
| r = 81076 \, 88939 \, 65566 \, 79970 \, 94454 \, 12508 \, 20606 \, 03037 \, 82723 \, 11003
}}
{{eqn | o =
| r = 86445 \, 85147 \, 95431 \, 68421 \, 48123 \, 63910 \, 96321 \, 63833 \, 76594 \, 77873
}}
{{eqn | o =
| r = 36044 \, 25100 \, 46756 \, 76942 \, 21197 \, 98655 \, 69863 \, 08993 \, 13991 \, 54810
}}
{{eqn | o =
| r = 29955 \, 71299 \, 30916 \, 19908 \, 66968 \, 53268 \, 78801 \, 17165 \, 95377 \, 09390
}}
{{eqn | o =
| r = 12417 \, 99779 \, 38952 \, 06419 \, 62790 \, 94932 \, 21996 \, 15477 \, 09894 \, 18755
}}
{{eqn | o =
| r = 79741 \, 05192 \, 62661 \, 21081 \, 92384 \, 45257 \, 78675 \, 87928 \, 74768 \, 12218
}}
{{eqn | o =
| r = 63148 \, 68786 \, 76854 \, 53862 \, 69957 \, 63612 \, 71978 \, 31119 \, 74476 \, 86496
}}
{{eqn | o =
| r = 45065 \, 87748 \, 91053 \, 15072 \, 63384 \, 65410 \, 90174 \, 27502 \, 19115 \, 20006
}}
{{eqn | o =
| r = 99485 \, 86281 \, 23536 \, 18641 \, 48374 \, 90557 \, 49920 \, 15285 \, 92211 \, 19416
}}
{{eqn | o =
| r = 75209 \, 57766 \, 75409 \, 22211 \, 29543 \, 79999 \, 81129 \, 89523 \, 59262 \, 62800
}}
{{eqn | o =
| r = 46942 \, 15484 \, 08243 \, 63610 \, 64351 \, 53563 \, 01617 \, 42451 \, 12051 \, 59183
}}
{{eqn | o =
| r = 34354 \, 13049 \, 42449 \, 46301 \, 59875 \, 51181 \, 09280 \, 53716 \, 57952 \, 29658
}}
{{eqn | o =
| r = 01206 \, 92006 \, 20396 \, 63689 \, 45859 \, 75910 \, 58626 \, 38955 \, 88424 \, 79023
}}
{{eqn | o =
| r = 70325 \, 29477 \, 90965 \, 29020 \, 39505 \, 24422 \, 75678 \, 32327 \, 27410 \, 18290
}}
{{eqn | o =
| r = 15226 \, 89958 \, 01677 \, 48481 \, 42430 \, 49977 \, 81717 \, 47239 \, 67104 \, 08734
}}
{{eqn | o =
| r = 21063 \, 13953 \, 69197 \, 18416 \, 66197 \, 78782 \, 49199 \, 73757 \, 81152 \, 15777
}}
{{eqn | o =
| r = 88246 \, 98396 \, 88365 \, 29090 \, 59197 \, 96301 \, 79613 \, 87838 \, 71578 \, 75079
}}
{{eqn | o =
| r = 17192 \, 38121 \, 06694 \, 45136 \, 51899 \, 17332 \, 26537 \, 65466 \, 92624 \, 57805
}}
{{eqn | o =
| r = 18650 \, 91862 \, 60159 \, 38818 \, 25424 \, 40894 \, 26520 \, 87364 \, 29048 \, 52293
}}
{{eqn | o =
| r = 88924 \, 40043 \, 51
}}
{{end-eqn}}
\end{theorem}
\begin{proof}
At $1812$ digits, it is clear by definition that this prime is titanic.
{{Alpertron-factorizer|date = $6$th March $2022$|time = $4.7$ seconds}}
To show that it is in fact a Sophie Germain prime, we also need to check that:
:$2 \paren {39 \, 051 \times 2^{6001} - 1} + 1 = 39 \, 051 \times 2^{6002} - 1$
is also prime.
{{Alpertron-factorizer|date = $6$th March $2022$|time = $4.5$ seconds}}
\end{proof}
|
22663
|
\section{Tonelli's Theorem}
Tags: Measure Theory, Tonelli's Theorem
\begin{theorem}
Let $\struct {X, \Sigma_X, \mu}$ and $\struct {Y, \Sigma_Y, \nu}$ be $\sigma$-finite measure spaces.
Let $\struct {X \times Y, \Sigma_X \otimes \Sigma_Y, \mu \times \nu}$ be the product measure space of $\struct {X, \Sigma_X, \mu}$ and $\struct {Y, \Sigma_Y, \nu}$.
Let $f: X \times Y \to \overline \R_{\ge 0}$ be a positive $\Sigma_X \otimes \Sigma_Y$-measurable function.
Then:
:$\ds \int_{X \times Y} f \map \rd {\mu \times \nu} = \int_Y \int_X \map f {x, y} \map {\d \mu} x \map {\d \nu} y = \int_X \int_Y \map f {x, y} \map {\d \nu} y \map {\d \mu} x$
\end{theorem}
\begin{proof}
We rewrite the demand as:
:$\ds \int_{X \times Y} f \map \rd {\mu \times \nu} = \int_Y \paren {\int_X f^y \rd \mu} \rd \nu = \int_X \paren {\int_Y f_x \rd \nu} \rd \mu$
where $f_x$ is the $x$-vertical section of $f$, and $f^y$ is the $y$-horizontal section of $f$.
From Horizontal Section of Measurable Function is Measurable, we have:
:$f^y$ is $\Sigma_X$-measurable.
From Vertical Section of Measurable Function is Measurable, we have:
:$f_x$ is $\Sigma_Y$-measurable.
From Integral of Horizontal Section of Measurable Function gives Measurable Function, we then have:
:$\ds y \mapsto \int_X f^y \rd \mu$ is $\Sigma_Y$-measurable
and from Integral of Vertical Section of Measurable Function gives Measurable Function, we have:
:$\ds x \mapsto \int_Y f_x \rd \nu$ is $\Sigma_X$-measurable.
So, both:
:$\ds \int_Y \paren {\int_X f^y \rd \mu} \rd \nu$
and:
:$\ds \int_X \paren {\int_Y f_x \rd \nu} \rd \mu$
are well-defined.
We first prove the case of:
:$f = \chi_E$
where $E$ is a $\Sigma_X \otimes \Sigma_Y$-measurable set.
\end{proof}
|
22664
|
\section{Top in Filter}
Tags: Order Theory
\begin{theorem}
Let $\left({S, \preceq}\right)$ be a bounded above ordered set.
Let $F$ be a filter on $S$.
Then $\top \in F$
where $\top$ denotes the greatest element of $S$.
\end{theorem}
\begin{proof}
By definition of filter in ordered set:
:$F$ is non-empty and upper.
By definition of non-empty set:
:$\exists x: x \in F$
By definition of greatest element:
:$x \preceq \top$
Thus by definition of upper set:
:$\top \in F$
{{qed}}
\end{proof}
|
22665
|
\section{Top in Ordered Set of Topology}
Tags: Topology, Order Theory
\begin{theorem}
Let $T = \left({S, \tau}\right)$ be a topological space.
Let $P = \left({\tau, \subseteq}\right)$ be an inclusion ordered set of $\tau$.
Then $P$ is bounded above and $\top_P = S$
where $\top_P$ denotes the greatest element in $P$.
\end{theorem}
\begin{proof}
By definition of topological space:
:$S \in \tau$
By definition of subset:
:$\forall A \in \tau: A \subseteq S$
Hence $P$ is bounded above.
Thus by definition of the greatest element:
:$\top_P = S$
{{qed}}
\end{proof}
|
22666
|
\section{Top is Complete}
Tags: Category Theory, Topological Spaces
\begin{theorem}
The category of topological spaces is complete.
\end{theorem}
\begin{proof}
Let $\II$ be a small category.
Let $D : \II \to \mathbf {Top}$ be a diagram in the category of topological spaces $\mathbf {Top}$.
Let $\family {\lim D, \family {\pi_i}_{i \mathop \in \II}}$ be the limit of topological spaces of $D$.
By Limit of Topological Spaces is Limit $\family {\lim D, \family {\pi_i}_{i \mathop \in \II}}$ is a categorical limit of $D$.
{{qed}}
Category:Topological Spaces
Category:Category Theory
\end{proof}
|
22667
|
\section{Top is Meet Irreducible}
Tags: Order Theory, Meet Irreducible
\begin{theorem}
Let $\left({S, \wedge, \preceq}\right)$ be a bounded above meet semilattice.
Then $\top$ is meet irreducible
where $\top$ denotes the greatest element in $S$.
\end{theorem}
\begin{proof}
Let $x, y \in S$ such that
:$\top = x \wedge y$
By Meet Precedes Operands
:$\top \preceq x$ and $\top \preceq y$
By definition of greatest element:
:$x \preceq \top$
Thus by definition of antisymmetry:
:$\top = x$
{{qed}}
\end{proof}
|
22668
|
\section{Top is Prime Element}
Tags: Prime Elements
\begin{theorem}
Let $L = \struct {S, \wedge, \preceq}$ be a bounded above meet semilattice.
Then $\top$ is a prime element
where $\top$ denotes the greatest element of $L$.
\end{theorem}
\begin{proof}
Let $x, y \in S$ such that
:$x \wedge y \preceq \top$
Thus by definition of greatest element:
:$x \preceq \top$ or $y \preceq \top$
{{qed}}
\end{proof}
|
22669
|
\section{Top is Unique}
Tags: Lattice Theory
\begin{theorem}
Let $\struct {S, \vee, \wedge, \preceq}$ be a lattice.
Then $S$ has at most one top.
\end{theorem}
\begin{proof}
By definition, a top for $S$ is a greatest element.
The result follows from Greatest Element is Unique.
{{qed}}
Category:Lattice Theory
\end{proof}
|
22670
|
\section{Topological Closure is Closed}
Tags: Topology, Set Closures, Closed Sets
\begin{theorem}
Let $T$ be a topological space.
Let $H \subseteq T$.
Then the closure $\map \cl H$ of $H$ is closed in $T$.
\end{theorem}
\begin{proof}
From Closure of Topological Closure equals Closure:
:$\map \cl {\map \cl H} = \map \cl H$
From Set is Closed iff Equals Topological Closure, it follows that $\map \cl H$ is closed.
{{qed}}
\end{proof}
|
22671
|
\section{Topological Closure of Singleton is Irreducible}
Tags: Irreducible Spaces, Topology, Topological Closure of Singleton is Irreducible
\begin{theorem}
Let $T = \struct {S, \tau}$ be a topological space.
Let $x$ be a point of $T$.
Then:
:$\set x^-$ is irreducible
where $\set x^-$ denotes the topological closure of $\set x$.
\end{theorem}
\begin{proof}
{{AimForCont}} that
:$\left\{ {x}\right\}^-$ is not irreducible.
By Set is Subset of its Topological Closure:
:$\left\{ {x}\right\} \subseteq \left\{ {x}\right\}^-$
By definitions of singleton and Subset:
:$x \in \left\{ {x}\right\}^-$
By definition of irreducible:
:$\exists X_1, X_2 \subseteq S: X_1, X_2$ are closed $\land \left\{ {x}\right\}^- = X_1 \cup X_2 \land \left\{ {x}\right\}^- \ne X_1 \land \left\{ {x}\right\}^- \ne X_2$
By definition of union:
:$x \in X_1$ or $x \in X_2$
By definitions of singleton and subset:
:$\left\{ {x}\right\} \subseteq X_1$ or $\left\{ {x}\right\} \subseteq X_2$
By definition of Definition:Closure (Topology)/Definition 3:
:$\left\{ {x}\right\}^- \subseteq X_1$ or $\left\{ {x}\right\}^- \subseteq X_2$
By Set is Subset of Union:
:$X_1 \subseteq \left\{ {x}\right\}$ abd $X_2 \subseteq \left\{ {x}\right\}$
Thus by definition of set equality:
:this contradicts $\left\{ {x}\right\}^- \ne X_1 \land \left\{ {x}\right\}^- \ne X_2$
{{qed}}
\end{proof}
|
22672
|
\section{Topological Closure of Subset is Subset of Topological Closure}
Tags: Set Closures, Topological Closure of Subset is Subset of Topological Closure, Subsets, Topology, Subset
\begin{theorem}
Let $T = \struct {S, \tau}$ be a topological space.
Let $H \subseteq K$ and $K \subseteq S$.
Then:
:$\map \cl H \subseteq \map \cl K$
where $\map \cl H$ denotes the closure of $H$.
\end{theorem}
\begin{proof}
From the definition of closure:
: $\operatorname{cl}\left({H}\right)$ is the union of $H$ and its limit points.
Let $x \in \operatorname{cl}\left({H}\right)$.
If $x \in H$ then $x \in K \implies x \in \operatorname{cl}\left({K}\right)$.
Otherwise $x$ is a limit point of $H$.
That is, every open set $U$ of $T$ such that $x \in U$ contains $y \in H$ such that $y \ne x$.
But as $y \in H$ it follows that $y \in K$.
So every open set $U$ of $T$ such that $x \in U$ contains $y \in K$ such that $y \ne x$.
This is the definition for a limit point of $K$.
Thus $x \in \operatorname{cl}\left({K}\right)$.
{{WIP|Extract the above paragraph into something like "Limit Point of Subset is Limit Point" (which proves that $H \subseteq K \implies H' \subseteq K'$). Then use Set Union Preserves Subsets to prove this result.}}
{{qed}}
\end{proof}
|
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