nyuuzyou/wikis huggingface dataset:
Doubling the Cube/Conchoid of Nicomedes/Historical Note

== Historical Note on Doubling the Cube by Conchoid of Nicomedes ==
The technique for doubling the cube by using a Conchoid of Nicomedes was devised by .
== Sources ==
* : $1 \cdotp 25992 \, 10498 \, 94873 \, 16476 \ldots$
* : $1 \cdotp 25992 \, 10498 \, 94873 \, 16476 \ldots$
Category:Historical Notes
Doubling the Cube/Fallacious Proof/Edw. J. Goodwin

== Fallacious Proof of Doubling the Cube ==
:Doubling the dimensions of the cube octuples its contents, and doubling its contents increases its dimensions twenty-five plus one per cent.
:::: -- 
== Resolution ==
== Sources ==
* 
Category:Doubling the Cube
Category:Fallacies
Doubling the Cube/Historical Note

== Historical Note on Doubling the Cube ==
The exercise to construct a cube double the volume of a given cube, using a compass and straightedge construction, was an exercise that the ancient Greeks failed to succeed in.
This was one of three such problems: the other two being Squaring the Circle and Trisecting the Angle.
There are several techniques available that use other tools, but these were considered unacceptably vulgar to the followers of .
According to one version of the legend, the island of was being ravaged by a plague. The people consulted the , who instructed them that a new altar to had to be built which was twice the size of the existing one, which was cubic in shape.
Another version of the legend suggests that it was not a plague, but civil or political unrest causing the trouble.
The word size was misunderstood, and the new altar had each of the linear dimensions doubled, whereas apparently they should have doubled its volume instead. Hence, unfortunately, the gods were not appeased, and the desired abatement of the plague did not happen.
To quote :
::It must be supposed, not that the god specially wished this problem solved, but that he would have the Greeks desist from war and wickedness and cultivate the Muses, so that their passions being assuaged by philosophy and mathematics, they might live in innocent and mutually helpful intercourse with one another.
== Sources ==
* : Preface
* : Chapter $\text{II}$: Modern Minds in Ancient Bodies
* : $1 \cdotp 25992 \, 10498 \, 94873 \, 16476 \ldots$
* : $1 \cdotp 25992 \, 10498 \, 94873 \, 16476 \ldots$
* 
* 
Category:Historical Notes
Doubling the Cube/Historical Note/Mistake

== Source Work ==
:
:The Dictionary
::$1 \cdotp 25992 \, 10498 \, 94873 \, 16476 \ldots$
:
:The Dictionary
::$1 \cdotp 25992 \, 10498 \, 94873 \, 16476 \ldots$
== Mistake ==
:The legend was told that the Athenians send a deputation to the oracle at to inquire how they might save themselves from a plague that was ravaging the city. They were instructed to double the size of the altar of Apollo.
Other sources are consistent in that the plague (or whatever the cause of unrest) was in fact at itself, and the oracle was the famous .
== Sources ==
* : $1 \cdotp 25992 \, 10498 \, 94873 \, 16476 \ldots$
* : $1 \cdotp 25992 \, 10498 \, 94873 \, 16476 \ldots$
Category:Mistakes/Curious and Interesting Numbers
Doubling the Cube/Intersection of Parabola and Hyperbola

== Theorem ==
The problem of Doubling the Cube can be solved by finding the intersection of a parabola and a hyperbola.
== Proof ==
== Historical Note ==
== Sources ==
* : $1 \cdotp 25992 \, 10498 \, 94873 \, 16476 \ldots$
* : $1 \cdotp 25992 \, 10498 \, 94873 \, 16476 \ldots$
Category:Doubling the Cube
Doubling the Cube/Intersection of Parabola and Hyperbola/Historical Note

== Historical Note on Doubling the Cube by Intersection of Parabola and Hyperbola ==
The technique for doubling the cube by finding the intersection of a parabola and a hyperbola was discovered by .
== Sources ==
* : $1 \cdotp 25992 \, 10498 \, 94873 \, 16476 \ldots$
* : $1 \cdotp 25992 \, 10498 \, 94873 \, 16476 \ldots$
Category:Historical Notes
Doubling the Cube/Intersection of Parabolas

== Theorem ==
The problem of Doubling the Cube can be solved by finding the intersection of two parabolas.
== Proof ==
== Historical Note ==
== Sources ==
* : $1 \cdotp 25992 \, 10498 \, 94873 \, 16476 \ldots$
* : $1 \cdotp 25992 \, 10498 \, 94873 \, 16476 \ldots$
Category:Doubling the Cube
Doubling the Cube/Intersection of Parabolas/Historical Note

== Historical Note on Doubling the Cube by Intersection of Parabolas ==
The technique for doubling the cube by finding the intersection of two parabolas was discovered by .
== Sources ==
* : $1 \cdotp 25992 \, 10498 \, 94873 \, 16476 \ldots$
* : $1 \cdotp 25992 \, 10498 \, 94873 \, 16476 \ldots$
Category:Historical Notes
Doubling the Cube by Compass and Straightedge Construction is Impossible

== Theorem ==
There is no compass and straightedge construction to allow a cube to be constructed whose volume is double that of a given cube.
== Proof ==
Suppose it is possible.
Then from a cube of edge length $L$ we can construct a new cube with edge length $\sqrt [3] 2 L$.
$\sqrt [3] 2$ is algebraic of degree $3$.
This contradicts Constructible Length with Compass and Straightedge.
== Also known as ==
== Sources ==
* : Chapter $\text{II}$: Modern Minds in Ancient Bodies
* 
* 
* : Chapter $2$: The Logic of Shape: Problems for the Greeks
* 
Category:Doubling the Cube
Category:Compass and Straightedge Constructions
Category:Field Extensions
Doubly Sequenced Generating Function for Binomial Coefficients

== Theorem ==
Let $\sequence {a_{m n} }$ be the doubly subscripted sequence defined as:
:$\forall m, n \in \N_{\ge 0}: a_{m n} = \dbinom n m$
where $\dbinom n m$ denotes a binomial coefficient.
Then the generating function for $\sequence {a_{m n} }$ is given as:
:$\map G {w, z} = \dfrac 1 {1 - z - w z}$
== Proof ==
== Sources ==
* : $\S 1.2.9$: Generating Functions: Exercise $12$
Category:Generating Functions
Category:Binomial Coefficients
Dougall's Hypergeometric Theorem

== Theorem ==
Let $x, y, z, n \in \C$.
Let $n \notin \Z_{\lt 0}$.
Let $\map \Re {x + y + z + n + 1} > 0$.
Then:
:$\ds \map { {}_5 \operatorname F_4} { { {\dfrac n 2 + 1, n, -x, -y, -z} \atop {\dfrac n 2, x + n + 1, y + n + 1, z + n + 1} } \, \middle \vert \, 1} = \dfrac {\map \Gamma {x + n + 1} \map \Gamma {y + n + 1} \map \Gamma {z + n + 1} \map \Gamma {x + y + z + n + 1} } {\map \Gamma {n + 1} \map \Gamma {x + y + n + 1} \map \Gamma {y + z + n + 1} \map \Gamma {x + z + n + 1} }$
where:
:${}_5 \operatorname F_4$ is the generalized hypergeometric function
:$x^{\overline k}$ denotes the $k$th rising factorial power of $x$
:$\map \Gamma {n + 1} = n!$ is the Gamma function.
=== Corollary 1 ===
=== Corollary 2 ===
=== Corollary 3 ===
=== Corollary 4 ===
=== Corollary 5 ===
=== Corollary 6 ===
== Proof ==
By definition of the generalized hypergeometric function
:$\ds \map { {}_5 \operatorname F_4} { { {\dfrac n 2 + 1, n, -x, -y, -z} \atop {\dfrac n 2, x + n + 1, y + n + 1, z + n + 1} } \, \middle \vert \, 1} = \ds \sum_{k \mathop = 0}^\infty \dfrac {\paren {\dfrac n 2 + 1}^{\overline k} n^{\overline k} \paren {-x}^{\overline k} \paren {-y}^{\overline k} \paren {-z}^{\overline k} } {\paren {\dfrac n 2}^{\overline k} \paren {x + n + 1}^{\overline k} \paren {y + n + 1}^{\overline k} \paren {z + n + 1}^{\overline k} } \dfrac {1^k} {k!}$
== Also known as ==
Some sources refer to this theorem as Dougall's Theorem, but that is often used for the Dougall-Ramanujan Identity.
== Examples ==
== Also see ==
* Dixon's Hypergeometric Theorem
* Gauss's Hypergeometric Theorem
* Kummer's Hypergeometric Theorem
* Properties of Generalized Hypergeometric Function
== Sources ==
* : Chapter $\text {10}$. Hypergeometric Series: $\text I$
* 
Category:Dougall's Hypergeometric Theorem
Category:Hypergeometric Functions
Category:Gamma Function
Dougall's Hypergeometric Theorem/Corollary 1

== Corollary to Dougall's Hypergeometric Theorem ==
Let $\map \Re {x + y + 1} > 0$.
Then:
:$\ds \map { {}_5 \operatorname F_4} { { {\dfrac n 2 + 1, n, n, -x, -y} \atop {\dfrac n 2, x + n + 1, y + n + 1, 1} } \, \middle \vert \, 1} = \dfrac {\map \Gamma {x + n + 1} \map \Gamma {y + n + 1} \map \Gamma {x + y + 1} } { \map \Gamma {n + 1} \map \Gamma {x + y + n + 1} \map \Gamma {y + 1} \map \Gamma {x + 1} } $
== Proof ==
Set $z = -n$ in Dougall's Hypergeometric Theorem
Before substitution:
:$\ds \map { {}_5 \operatorname F_4} { { {\dfrac n 2 + 1, n, -x, -y, -z} \atop {\dfrac n 2, x + n + 1, y + n + 1, z + n + 1} } \, \middle \vert \, 1} = \dfrac {\map \Gamma {x + n + 1} \map \Gamma {y + n + 1} \map \Gamma {z + n + 1} \map \Gamma {x + y + z + n + 1} } { \map \Gamma {n + 1} \map \Gamma {x + y + n + 1} \map \Gamma {y + z + n + 1} \map \Gamma {x + z + n + 1} } $
After substitution:
:$\ds \map { {}_5 \operatorname F_4} { { {\dfrac n 2 + 1, n, -x, -y, n} \atop {\dfrac n 2, x + n + 1, y + n + 1, -n + n + 1} } \, \middle \vert \, 1} = \dfrac {\map \Gamma {x + n + 1} \map \Gamma {y + n + 1} \map \Gamma {-n + n + 1} \map \Gamma {x + y - n + n + 1} } { \map \Gamma {n + 1} \map \Gamma {x + y + n + 1} \map \Gamma {y - n + n + 1} \map \Gamma {x - n + n + 1} } $
:$\ds \map { {}_5 \operatorname F_4} { { {\dfrac n 2 + 1, n, n, -x, -y} \atop {\dfrac n 2, x + n + 1, y + n + 1, 1} } \, \middle \vert \, 1} = \dfrac {\map \Gamma {x + n + 1} \map \Gamma {y + n + 1} \map \Gamma {x + y + 1} } { \map \Gamma {n + 1} \map \Gamma {x + y + n + 1} \map \Gamma {y + 1} \map \Gamma {x + 1} } $
== Sources ==
* : Chapter $\text {10}$. Hypergeometric Series: $\text I$
Category:Dougall's Hypergeometric Theorem
Category:Hypergeometric Functions
Category:Gamma Function
Dougall's Hypergeometric Theorem/Corollary 2

== Corollary to Dougall's Hypergeometric Theorem ==
Let $\map \Re {x + y + n} > 0$.
Then:
:$\ds \map { {}_4 \operatorname F_3} { { {\dfrac n 2 + 1, -x, -y, 1} \atop {\dfrac n 2, x + n + 1, y + n + 1} } \, \middle \vert \, 1} = \dfrac {\paren {x + n} \paren {y + n} } {n \paren {x + y + n} } $
== Proof ==
Set $z = -1$ in Dougall's Hypergeometric Theorem
Before substitution:
After substitution:
== Sources ==
* : Chapter $\text {10}$. Hypergeometric Series: $\text I$
Category:Dougall's Hypergeometric Theorem
Category:Hypergeometric Functions
Category:Gamma Function
Dougall's Hypergeometric Theorem/Corollary 3

== Corollary to Dougall's Hypergeometric Theorem ==
Let $\map \Re {2x + 2y + n + 2} > 0$.
Let $n \notin \Z_{\lt 0}$
Then:
:$\ds \map { {}_4 \operatorname F_3} { { {\dfrac n 2 + 1, n, -x, -y} \atop {\dfrac n 2, x + n + 1, y + n + 1} } \, \middle \vert \, -1} = \dfrac {\map \Gamma {x + n + 1} \map \Gamma {y + n + 1} } {\map \Gamma {n + 1} \map \Gamma {x + y + n + 1} } $
== Proof ==
Two lemmata:
=== Kummer's Hypergeometric Theorem: Lemma 1 ===
=== Lemma ===
Let $z \to \infty$ in Dougall's Hypergeometric Theorem
We have:
== Sources ==
* : Chapter $\text {10}$. Hypergeometric Series: $\text I$
Category:Dougall's Hypergeometric Theorem
Category:Hypergeometric Functions
Category:Gamma Function
Dougall's Hypergeometric Theorem/Corollary 3/Lemma

== Lemma for Dougall's Hypergeometric Theorem/Corollary 3 ==
:$\ds \lim_{z \mathop \to \infty} \dfrac {\paren {x + z + n + 1}^{\overline y} } {\paren {z+ n + 1}^{\overline y} } = 1$
where $x^{\overline y}$ denotes the $y$th rising factorial of $x$.
== Proof ==
From Properties of Limit at Infinity of Real Function: Product Rule, we have:
From L'Hôpital's Rule:Corollary 2, we have:
:$\ds \lim_{z \mathop \to a^+} \frac {\map f z} {\map g z} = \lim_{z \mathop \to a^+} \frac {\map {f'} z} {\map {g'} z}$
In the present example, for the kth limit, we have $\ds \map {f_k} z = \paren {x + z + n + y}$ and $\ds \map {g_k} z = \paren {z + n + y}$
Therefore taking the derivative of the numerator $\map {f_k} z$ and denominator $\map {g_k} z$ $z$, we obtain:
:$\ds \lim_{z \mathop \to \infty}\paren {\dfrac {\paren {x + z + n + y} } {\paren {z + n + y} } } = \lim_{z \mathop \to \infty} \dfrac 1 1 = 1$
Therefore:
Category:Dougall's Hypergeometric Theorem
Category:Rising Factorials
Dougall's Hypergeometric Theorem/Corollary 4

== Corollary to Dougall's Hypergeometric Theorem ==
Let $\map \Re {x - n + 1} > 0$.
Then:
:$\ds \map { {}_5 \operatorname F_4} { { {\dfrac n 2 + 1, n, n, n, -x} \atop {\dfrac n 2, x + n + 1, 1, 1} } \, \middle \vert \, 1} = \dfrac {\map \sin {\pi n} \map \Gamma {x + n + 1} \map \Gamma {x - n + 1} } {\pi n \paren {\map \Gamma {x + 1} }^2 } $
== Proof ==
Set $y = z = -n$ in Dougall's Hypergeometric Theorem
Before substitution:
After substitution:
== Sources ==
* : Chapter $\text {10}$. Hypergeometric Series: $\text I$
Category:Dougall's Hypergeometric Theorem
Category:Hypergeometric Functions
Category:Gamma Function
Dougall's Hypergeometric Theorem/Corollary 5

== Corollary to Dougall's Hypergeometric Theorem ==
Let $\map \Re {n} < \dfrac 1 2$.
Then:
:$\ds \map { {}_5 \operatorname F_4} { { {\dfrac n 2 + 1, n, n, n, n} \atop {\dfrac n 2, 1, 1, 1} } \, \middle \vert \, 1} = \dfrac {\paren {\map \Gamma n}^2 \map \sin {\pi n} \map \tan {\pi n} } {\pi^2 \map \Gamma {2n + 1} } $
== Proof ==
Set $x = y = z = -n$ in Dougall's Hypergeometric Theorem
Before substitution:
After substitution:
== Sources ==
* : Chapter $\text {10}$. Hypergeometric Series: $\text I$
Category:Dougall's Hypergeometric Theorem
Category:Hypergeometric Functions
Category:Gamma Function
Dougall's Hypergeometric Theorem/Corollary 6

== Corollary to Dougall's Hypergeometric Theorem ==
Let $\map \Re {n} < \dfrac 2 3$.
Then:
:$\ds \map { {}_4 \operatorname F_3} { { {\dfrac n 2 + 1, n, n, n} \atop {\dfrac n 2, 1, 1} } \, \middle \vert \, -1} = \dfrac {\map \sin {\pi n} } {\pi n } $
== Proof ==
Let $x = y = -n$ in Dougall's Hypergeometric Theorem: Corollary 3
Before substitution:
After substitution:
== Sources ==
* : Chapter $\text {10}$. Hypergeometric Series: $\text I$
Category:Dougall's Hypergeometric Theorem
Category:Hypergeometric Functions
Category:Gamma Function
Dougall's Hypergeometric Theorem/Examples

== Examples of Use of Dougall's Hypergeometric Theorem==
=== Example: $\map { {}_5 \operatorname F_4} {\dfrac 9 8, \dfrac 1 4, \dfrac 1 4, \dfrac 1 4, \dfrac 1 4; \dfrac 1 8, 1, 1, 1; 1}$ ===
=== Example: $\map { {}_4 \operatorname F_3} {\dfrac 5 4, \dfrac 1 2, \dfrac 1 2, \dfrac 1 2; \dfrac 1 4, 1, 1; -1}$ ===
Category:Examples of Use of Dougall's Hypergeometric Theorem
Dougall's Hypergeometric Theorem/Examples/4F3(1.25,0.5,0.5,0.5;0.25,1,1;-1)

== Example of Use of Dougall's Hypergeometric Theorem ==
:$1 - 5 \paren {\dfrac 1 2}^3 + 9 \paren {\dfrac {1 \times 3} {2 \times 4} }^3 - 13 \paren {\dfrac {1 \times 3 \times 5} {2 \times 4 \times 6} }^3 + \cdots = \dfrac 2 \pi$
== Proof ==
From Dougall's Hypergeometric Theorem: Corollary 6:
:$\ds \map { {}_4 \operatorname F_3} { { {\dfrac n 2 + 1, n, n, n} \atop {\dfrac n 2, 1, 1} } \, \middle \vert \, -1} = \dfrac {\map \sin {\pi n} } {\pi n } $
where:
:$\ds \map { {}_4 \operatorname F_3} { { {\dfrac n 2 + 1, n, n, n} \atop {\dfrac n 2, 1, 1} } \, \middle \vert \, -1}$ is the generalized hypergeometric function of $1$: $\ds \sum_{k \mathop = 0}^\infty \dfrac { \paren {\dfrac n 2 + 1}^{\overline k} \paren {n^{\overline k} }^3 } { \paren {\dfrac n 2}^{\overline k} \paren {1^{\overline k} }^2 } \dfrac {\paren {-1}^k} {k!}$
:$x^{\overline k}$ denotes the $k$th rising factorial power of $x$
:$\map \Gamma {n + 1} = n!$ is the Gamma function.
On the we have:
On the we have:
:$\ds \dfrac {\map \sin {\dfrac \pi 2} } {\dfrac \pi 2 } = \dfrac 2 \pi$
Therefore:
:$1 - 5 \paren {\dfrac 1 2}^3 + 9 \paren {\dfrac {1 \times 3} {2 \times 4} }^3 - 13 \paren {\dfrac {1 \times 3 \times 5} {2 \times 4 \times 6} }^3 + \cdots = \dfrac 2 \pi$
== Sources ==
* : Chapter $\text {10}$. Hypergeometric Series: $\text I$
Category:Examples of Use of Dougall's Hypergeometric Theorem‎
Dougall's Hypergeometric Theorem/Examples/5F4(1.125,0.25,0.25,0.25,0.25;0.125,1,1,1;1)

== Example of Use of Dougall's Hypergeometric Theorem ==
:$1 + 9 \paren {\dfrac 1 4}^4 + 17 \paren {\dfrac {1 \times 5} {4 \times 8} }^4 + 25 \paren {\dfrac {1 \times 5 \times 9} {4 \times 8 \times 12} }^4 + \cdots = \dfrac {2 \sqrt 2 } { \sqrt \pi \paren {\map \Gamma {\dfrac 3 4} }^2 }$
== Proof ==
From Dougall's Hypergeometric Theorem:
:$\ds \map { {}_5 \operatorname F_4} { { {\dfrac n 2 + 1, n, -x, -y, -z} \atop {\dfrac n 2, x + n + 1, y + n + 1, z + n + 1} } \, \middle \vert \, 1} = \dfrac {\map \Gamma {x + n + 1} \map \Gamma {y + n + 1} \map \Gamma {z + n + 1} \map \Gamma {x + y + z + n + 1} } {\map \Gamma {n + 1} \map \Gamma {x + y + n + 1} \map \Gamma {y + z + n + 1} \map \Gamma {x + z + n + 1} }$
where:
:$\ds \map { {}_5 \operatorname F_4} { { {\dfrac n 2 + 1, n, -x, -y, -z} \atop {\dfrac n 2, x + n + 1, y + n + 1, z + n + 1} } \, \middle \vert \, 1}$ is the generalized hypergeometric function of $1$: $\ds \sum_{k \mathop = 0}^\infty \dfrac { \paren {\dfrac n 2 + 1}^{\overline k} n^{\overline k} \paren {-x}^{\overline k} \paren {-y}^{\overline k} \paren {-z}^{\overline k} } {\paren {\dfrac n 2}^{\overline k} \paren {x + n + 1}^{\overline k} \paren {y + n + 1}^{\overline k} \paren {z + n + 1}^{\overline k} } \dfrac {1^k} {k!}$
:$x^{\overline k}$ denotes the $k$th rising factorial power of $x$
:$\map \Gamma {n + 1} = n!$ is the Gamma function.
We have:
and:
Recall the Euler's Reflection Formula:
:$\map \Gamma z \map \Gamma {1 - z} = \dfrac \pi {\map \sin {\pi z} }$
Therefore:
Substituting these results back into our equation above:
Therefore:
:$1 + 9 \paren {\dfrac 1 4}^4 + 17 \paren {\dfrac {1 \times 5} {4 \times 8} }^4 + 25 \paren {\dfrac {1 \times 5 \times 9} {4 \times 8 \times 12} }^4 + \cdots = \dfrac {2 \sqrt 2} {\sqrt \pi \paren {\map \Gamma {\dfrac 3 4} }^2}$
== Sources ==
* : Chapter $\text {10}$. Hypergeometric Series: $\text I$
Category:Examples of Use of Dougall's Hypergeometric Theorem‎
Dougall-Ramanujan Identity

== Theorem ==
Let $x, y, z, u, n \in \C$.
Let at least one of $x, y, z, u, -x - y - z - u - 2n - 1 \in \Z_{>0}$.
Then:
where:
:${}_7 \operatorname F_6$ is the generalized hypergeometric function 
:$x^{\overline k}$ denotes the $k$th rising factorial power of $x$
:$\map \Gamma {n + 1} = n!$ is the Gamma function.
== Proof ==
By definition of generalized hypergeometric function of $1$:
:$\ds \map { {}_7 \operatorname F_6} { { {n, 1 + \dfrac n 2, -x, -y, -z, -u, x + y + z + u + 2n + 1} \atop {\dfrac n 2, x + n + 1, y + n + 1, z + n + 1, u + n + 1, -x - y - z - u - n} } \, \middle \vert \, 1} = \sum_{k \mathop = 0}^\infty \dfrac { n^{\overline k} \paren {1 + \dfrac n 2}^{\overline k} \paren {-x}^{\overline k} \paren {-y}^{\overline k} \paren {-z}^{\overline k} \paren {-u}^{\overline k} \paren {x + y + z + u + 2n + 1}^{\overline k} } { \paren {\dfrac n 2}^{\overline k} \paren {x + n + 1}^{\overline k} \paren {y + n + 1}^{\overline k} \paren {z + n + 1}^{\overline k} \paren {u + n + 1}^{\overline k} \paren {-x - y - z - u - n}^{\overline k} } \dfrac {1^k} {k!}$
== Also known as ==
Some sources refer to this theorem as Dougall's Theorem.
== Also see ==
* Dixon's Hypergeometric Theorem
* Dougall's Hypergeometric Theorem
* Gauss's Hypergeometric Theorem
* Kummer's Hypergeometric Theorem
* Properties of Generalized Hypergeometric Function
== Sources ==
* : Chapter $\text {5}$. Methods of Obtaining Transformations of Hypergeometric Series; (2) By Dougall's method and Carlson's theorem
* : Chapter $\text {10}$. Hypergeometric Series: $\text I$
* 
Category:Hypergeometric Functions
Category:Gamma Function
Down Mapping is Generated by Approximating Relation

== Theorem ==
Let $L = \struct {S, \wedge, \preceq}$ be a bounded below meet-continuous meet semilattice.
Let $\map {\operatorname{Ids} } L$ be the set of all ideals in $L$.
Let $I$ be an ideal in $L$.
Let $f: S \to \map {\operatorname{Ids} } L$ be a mapping such that:
:$\forall x \in S: x \preceq \sup I \implies \map f x = \set {x \wedge i: i \in I}$
and
:$\forall x \in S: x \npreceq \sup I \implies \map f x = x^\preceq$
where $x^\preceq$ denotes the lower closure of $x$.
Then
:there exists an auxiliary approximating relation $\RR$ on $S$ such that
::$\forall s \in S: \map f s = s^\RR$
where $s^\RR$ denotes the $\RR$-segment of $s$.
== Proof ==
Define relation $\RR$ on $S$:
:$\forall x, y \in S: \tuple {x, y} \in \RR \iff x \in \map f y$
By definition of $\RR$-segment:
:$\forall x \in S: \map f x = x^\RR$
We will prove that:
:$\RR$ is an approximating relation.
Let $x \in S$.
Suppose the case holds that:
:$x \preceq \sup I$
Thus:
Suppose the case holds that
:$x \npreceq \sup I$
Thus
== Sources ==
* 
* 
Category:Meet-Continuous Lattices
Category:Approximating Relations
Draft:Union of Set of Sets is Greatest Element under Subset Relation

== Theorem ==
Let $M$ be a set.
Let $\bigcup M \in M$.
Let $(M, \subseteq)$ be the ordered set formed on $M$ by the subset relation (see Subset Relation is Ordering).
Then $\bigcup M$ is the greatest set by set inclusion ($M$ corresponds to $\TT$) of $(M, \subseteq)$.
== Proof ==
By Set is Subset of Union:
:$\forall \paren {N \in M}: N \subseteq \bigcup M$. 
Therefore, $\bigcup M$ is the greatest element under the subset relation.
Category:Set Union
Drop in EMF caused by Capacitance is proportional to Accumulated Charge

== Physical Law ==
The drop in EMF $E_C$ caused by a capacitor is proportional to the electric charge $Q$ that has accumulated on the capacitor:
:$E_C = \dfrac 1 C Q$
The reciprocal of the constant of proportion $\dfrac 1 C$ is known as the capacitance of the capacitor.
== Sources ==
* : $\S 2.13$: Simple Electric Circuits
Category:Capacitance
Drop in EMF caused by Inductance is proportional to Rate of Change of Current

== Physical Law ==
The drop in EMF $E_L$ caused by an inductor is proportional to the rate of change of current $I$ through the inductor:
:$E_L = L \dfrac {\d I} {\d t}$
== Also see ==
The constant of proportion $L$ is known as the inductance of the inductor.
== Sources ==
* : $\S 2.13$: Simple Electric Circuits
Category:Inductance
Du Bois-Reymond Constants/Example/First

== Example of Du Bois-Reymond Constant ==
The first du Bois-Reymond constant $C_1$ does not exist.
This is because:
:$\ds \int_0^\infty \size {\map {\dfrac \d {\d t} } {\dfrac {\sin t} t}^n} \rd t - 1$
does not converge for $n = 1$.
== Proof ==
Category:Examples of Du Bois-Reymond Constants
Dual Category is Category

== Theorem ==
Let $\mathbf C$ be a metacategory.
Let $\mathbf C^{\text{op} }$ be its dual category.
Then $\mathbf C^{\text{op} }$ is also a metacategory.
== Proof ==
Let us verify the axioms $(C1)$ up to $(C3)$ for a metacategory.
Let $f^{\text{op} }: C^{\text{op} } \to D^{\text{op} }$ and $g^{\text{op} }: D^{\text{op} } \to E^{\text{op} }$ be morphisms in $\mathbf C^{\text{op} }$.
Then $f: D \to C$ and $g: E \to D$ are morphisms in $\mathbf C$, and so is $f \circ g: E \to C$.
Therefore, also $g^{\text{op} } \circ f^{\text{op} }: C^{\text{op} } \to E^{\text{op} }$ is a morphism in $\mathbf C^{\text{op} }$, and $(C1)$ is shown to hold.
For $(C2)$, observe that for $f^{\text{op} }: C^{\text{op} } \to D^{\text{op} }$, we have:
Similarly:
Hence $(C2)$ is shown to hold.
To show $(C3)$, reason as follows:
Hence $\mathbf C^{\text{op} }$ is a metacategory.
Category:Dual Categories
Dual Distributive Lattice is Distributive

== Theorem ==
Let $L = \struct {S, \vee, \wedge, \preceq}$ be a lattice.
Then
:$L$ is a distributive lattice
:$L^{-1}$ is a distributive lattice
where $L^{-1} = \struct {S, \succeq}$ denotes the dual of $L$.
== Proof ==
=== Sufficient Condition ===
Let $L$ be a distributive lattice.
By Dual of Lattice Ordering is Lattice Ordering:
:$L^{-1}$ is lattice.
Let $x, y, z \in S$.
$\vee'$ and $\wedge'$ denotes join and meet in $L^{-1}$.
Thus
=== Necessary Condition ===
This follows by .
== Sources ==
* 
Category:Distributive Lattices
Category:Dual Orderings