nyuuzyou/wikis huggingface dataset:
Simple Infinite Continued Fraction is Uniquely Determined by Limit/Proof 2

== Theorem ==
== Proof ==
Recall that by Simple Infinite Continued Fraction Converges, they do indeed have a limit.
The result will be achieved by the Second Principle of Mathematical Induction.
Suppose $\sqbrk {a_0, a_1, a_2, \ldots} = \sqbrk {b_0, b_1, b_2, \ldots}$ have the same value.
First we note that if $\sqbrk {a_0, a_1, a_2, \ldots} = \sqbrk {b_0, b_1, b_2, \ldots}$ then:
:$a_0 = b_0$
since both are equal to the integer part of the common value.
This is our basis for the induction.
Now suppose that for some $k \ge 1$, we have:
:$a_0 = b_0, a_1 = b_1, \ldots, a_k = b_k$
Then all need to do is show that $a_{k + 1} = b_{k + 1}$.
Now:
:$\sqbrk {a_0, a_1, a_2, \ldots} = \sqbrk {a_0, a_1, \ldots, a_k, \sqbrk {a_{k + 1}, a_{k + 2}, \ldots} }$
and similarly:
:$\sqbrk {b_0, b_1, b_2, \ldots} = \sqbrk {b_0, b_1, \ldots, b_k, \sqbrk {b_{k + 1}, b_{k + 2}, \ldots} }$
As these have the same value and have the same first $k$ partial denominators, it follows that:
:$\sqbrk {a_{k + 1}, a_{k + 2}, \ldots, } = \sqbrk {b_{k + 1}, b_{k + 2}, \ldots}$
But now $a_{k + 1} = b_{k + 1}$ as each is equal to the integer part of the value of this simple infinite continued fraction.
Hence the result.
Category:Simple Infinite Continued Fraction is Uniquely Determined by Limit
Simple Loop Image Equals Set Homeomorphic to Circle

== Theorem ==
Let $\struct { X, \tau_X }$ be a Hausdorff space.
Let $C \subseteq X$ be a subset of $X$.
Let $\tau_C$ be the subspace topology on $C$ induced by $\tau_X$.
Let $\mathbb S^1$ denote the unit circle whose center is at the origin of the Euclidean space $\R^2$.
Let $\tau_{\mathbb S^1}$ be the subspace topology on $\mathbb S^1$ induced by the Euclidean topology on $\R^2$.
Then $\struct{ C, \tau_C }$ is homeomorphic to $\struct { \mathbb S^1 , \tau_{\mathbb S^1} }$, there exists a simple loop $\gamma : \closedint 0 1 \to X$ with image equal to $C$.
== Proof ==
=== Sufficient condition ===
Let $\sim$ be the equivalence relation on the closed real interval $\closedint 0 1$ defined by:
Let $q : \closedint 0 1 \to \closedint 0 1 / \sim$ be the canonical surjection induced by $\sim$.
Let $\tau_q$ be the quotient topology on the quotient space $\closedint 0 1 / \sim$ induced by $q$.
By definition of quotient topology, $\tau_q$ is the identification topology on $\closedint 0 1 / \sim$ by the identification mapping $q$.
Identification Mapping is Continuous shows that $q$ is continuous.
Parameterization of Unit Circle is Simple Loop shows that there exists a simple loop in $\R^2$ with image equal to $\mathbb S^1$.
Euclidean Space is Complete Metric Space and Metric Space is Hausdorff shows that $\R^2$ is a Hausdorff space.
Simple Loop in Hausdorff Space is Homeomorphic to Quotient Space of Interval shows that there exists a homeomorphism $f: \closedint 0 1 / \sim \to \mathbb S^1$.
By assumption, there exists a homeomorphism $h: \mathbb S^1 \to C$.
Composite of Continuous Mappings is Continuous shows that the mapping:
:$h \circ f \circ q : \closedint 0 1 \to C$
is continuous.
By definition of homeomorphism, $h$ and $f$ are bijections.
Quotient Mapping is Surjection shows that $q$ is a surjective mapping.
Composite of Surjections is Surjection shows that $h \circ f \circ q$ is a surjective mapping, so $\Img {h \circ f \circ q} = C$.
By definition of $\sim$, we see that:
:$\map q {t_1} \ne \map q {t_2}$ for all $t_1 ,t_2 \in \hointr 0 1$ with $t_1 \ne t_2$
:$\map q 0 = \map q 1$
It follows that $h \circ f \circ q$ is a simple loop, as:
:$\map {h \circ f \circ q }{t_1} \ne \map {h \circ f \circ q}{t_2}$ for all $t_1 ,t_2 \in \hointr 0 1$ with $t_1 \ne t_2$
:$\map {h \circ f \circ q} 0 = \map {h \circ f \circ q} 1$
Define $\gamma : \closedint 0 1 \to X$ by $\map \gamma t = \map {h \circ f \circ q} t$.
Continuity of Composite with Inclusion/Inclusion on Mapping shows that $\gamma$ is continuous.
It follows that $\gamma$ is a simple loop.
=== Necessary Condition ===
By assumption, there exists a simple loop $\gamma : \closedint 0 1 \to X$ with $\Img \gamma = C$.
Simple Loop in Hausdorff Space is Homeomorphic to Quotient Space of Interval shows that there exists a homeomorphism $g : \closedint 0 1 / \sim \to C$.
As shown above, there exists a homeomorphism $f: \closedint 0 1 / \sim \to \mathbb S^1$.
Composite of Homeomorphisms is Homeomorphism shows that $f \circ g^{-1} : C \to \mathbb S^1$ is a homeomorphism.
Category:Loops (Topology)
Category:Circles
Category:Examples of Homeomorphisms
Simple Loop in Hausdorff Space is Homeomorphic to Quotient Space of Interval

== Theorem ==
Let $\struct { X, \tau_X }$ be a Hausdorff space.
Let $\gamma : \closedint 0 1 \to X$ be a simple loop.
Let $\sim$ be an equivalence relation on $\closedint 0 1$ defined by:
Let $q : \closedint 0 1 \to \closedint 0 1 / \sim$ be the canonical surjection induced by $\sim$.
Let $\tau_q$ be the quotient topology on the quotient space $\closedint 0 1 / \sim$ induced by $q$.
Let $\tau_\gamma$ be the subspace topology on $\Img \gamma$ induced by $\tau_X$, where $\Img \gamma$ denotes the image of $\gamma$.
Then $\struct {\Img \gamma , \tau_\gamma}$ is homeomorphic to $\struct { \closedint 0 1 / \sim , \tau_q}$.
== Proof ==
Define $\tilde \gamma : \closedint 0 1 \to \Img \gamma$ as the restriction of $\gamma$ to $\closedint 0 1 \times \Img \gamma$.
Restriction of Mapping to Image is Surjection shows that $\tilde \gamma$ is surjective.
Subspace of Hausdorff Space is Hausdorff shows that $\struct {\Img \gamma , \tau_\gamma}$ is a Hausdorff space.
Closed Real Interval is Compact shows that $\closedint 0 1$ is compact.
Continuous Mapping from Compact Space to Hausdorff Space is Closed Mapping shows that $\tilde \gamma$ is a closed mapping.
Continuous Closed Surjective Mapping is Quotient Mapping shows that $\tilde \gamma$ is a quotient mapping.
By definition of simple loop, the equivalence relation $\sim$ is the equivalence induced by $\tilde \gamma$.
Continuous Surjection Induces Continuous Bijection from Quotient Space/Corollary 1 shows that there exists a homeomorphism $f: 
\closedint 0 1 / \sim \to \Img \gamma$.
::$\begin{xy} \xymatrix@L+2mu@+1em{
 \closedint 0 1 \ar[r]^*{q}
 \ar[rd]_*{\tilde \gamma}
&
 \closedint 0 1 / \sim \ar@{-->}[d]^*{f}
\\ &
 \Img \gamma
}\end{xy}$
Category:Quotient Spaces
Category:Loops (Topology)
Simple Order Product/Examples/Unit Square with Open Side

== Example of Simple Order Product ==
Consider the simple order product of the real intervals $\hointr 0 1$ and $\closedint 0 1$ under the usual ordering:
:$\struct {T, \preccurlyeq_s} := \struct {\hointr 0 1, \le} \otimes^s \struct {\closedint 0 1, \le}$
$\struct {T, \preccurlyeq_s}$ has one minimal element:
:$\tuple {0, 0}$
which is also the smallest element: of $\struct {T, \preccurlyeq_s}$.
$\struct {T, \preccurlyeq_s}$ has no greatest element and no maximal elements.
== Proof ==
Recall the definition of the ordering in the simple order product:
::$\forall \tuple {a, b}, \tuple {c, d} \in T: \tuple {a, b} \preccurlyeq_s \tuple {c, d} \iff a \le c \text { and } b \le d$
Let $\tuple {a, b} \in T$ be a minimal element of $T$.
:$\forall \tuple {x, y} \in T: \tuple {x, y} \preccurlyeq_s \tuple {a, b} \implies \tuple {x, y} = \tuple {a, b}$
Suppose $a \ne 0$.
Then:
:$\tuple {0, b} \preccurlyeq_s \tuple {a, b}$ but $\tuple {0, b} \ne \tuple {a, b}$
Similarly, suppose $b \ne 0$.
Then:
:$\tuple {a, 0} \preccurlyeq_s \tuple {a, b}$ but $\tuple {a, 0} \ne \tuple {a, b}$
Hence if $\tuple {a, b}$ is a minimal element of $T$ it follows that:
:$\tuple {a, b} = \tuple {0, 0}$
Next we note that:
:$\forall \tuple {c, d} \in T: 0 \le c, 0 \le d$
and so:
:$\forall \tuple {c, d} \in T: \tuple {0, 0} \preccurlyeq_s \tuple {c, d}$
Hence by definition $\tuple {0, 0}$ is a smallest element of $T$.
By Smallest Element is Unique, $\tuple {0, 0}$ is then the smallest element of $T$.
 $T$ has a maximal element $\tuple {u, v}$.
Then we have:
:$\tuple {u, v} \preccurlyeq_s \tuple {x, y} \implies \tuple {u, v} = \tuple {x, y}$
In particular, this means:
:$u \le x \implies u = x$
But we have that:
:$u \in \hointr 0 1$
and so:
:$u < 1$
So take:
:$x = \dfrac u 1 + \dfrac 1 1 = \dfrac {u + 1} 2$
Hence from Mediant is Between:
:$u < x$
and so it is not the case that $u \le x \implies u = x$.
Hence it is not the case that:
:$\tuple {u, v} \preccurlyeq_s \tuple {x, y} \implies \tuple {u, v} = \tuple {x, y}$
and so $\tuple {u, v}$ is not a maximal element of $T$.
So by Proof by Contradiction it follows that there is no maximal element of $T$.
 $\tuple {u, v}$ is a greatest element of $T$.
From Greatest Element is Maximal, $\tuple {u, v}$ would have to be a maximal element of $T$.
But from above $T$ has no maximal element.
Hence by Proof by Contradiction it follows that there is no greatest element of $T$.
Hence the result.
== Sources ==
* : Part $1$: Not Entirely Naive Set Theory: Chapter $2$: Partial Order Relations: Exercise $31$
Category:Examples of Order Products
Simple Order Product of Pair of Ordered Semigroups is Ordered Semigroup

== Theorem ==
Let $\struct {S_1, \circ_1, \preccurlyeq_1}$ and $\struct {S_2, \circ_2, \preccurlyeq_2}$ be ordered semigroups.
Let $\struct {S_1 \times S_2, \odot} := \struct {S_1, \circ_1} \times \struct {S_2, \circ_2}$ denote the external direct product of $\struct {S_1, \circ_1}$ and $\struct {S_2, \circ_2}$.
Let $\struct {S_1 \times S_2, \preccurlyeq_s} := \struct {S_1, \preccurlyeq_1} \otimes^s \struct {S_2, \preccurlyeq_2}$ denote the simple (order) product of $\struct {S_1, \preccurlyeq_1}$ and $\struct {S_2, \preccurlyeq_2}$.
Then $\struct {S_1 \times S_2, \odot, \preccurlyeq_s}$ is also an ordered semigroup.
== Proof ==
From Simple Order Product of Pair of Ordered Sets is Ordered Set, $\struct {S_1 \times S_2, \otimes^s}$ is an ordered set.
From External Direct Product of Semigroups, $\struct {S_1 \times S_2, \odot}$ is a semigroup.
It remains to be shown that $\preccurlyeq_s$ is compatible with $\odot$.
Let $\tuple {x_1, x_2}, \tuple {y_1, y_2} \in S_1 \times S_2$ be arbitrary such that $\tuple {x_1, x_2} \preccurlyeq_s \tuple {y_1, y_2}$.
and the proof is complete.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 15$: Ordered Semigroups: Exercise $15.5$
Category:Simple Order Product
Category:Ordered Semigroups
Simple Order Product of Pair of Ordered Sets is Ordered Set

== Theorem ==
Let $\struct {S_1, \preccurlyeq_1}$ and $\struct {S_2, \preccurlyeq_2}$ be ordered sets.
Let $\struct {S_1, \preccurlyeq_1} \otimes^s \struct {S_2, \preccurlyeq_2}$ denote the simple (order) product of $\struct {S_1, \preccurlyeq_1}$ and $\struct {S_2, \preccurlyeq_2}$.
Then $\struct {S_1, \preccurlyeq_1} \otimes^s \struct {S_2, \preccurlyeq_2}$ is also an ordered set.
== Proof ==
Let $\struct {T, \preccurlyeq_s} := \struct {S_1, \preccurlyeq_1} \otimes^s \struct {S_2, \preccurlyeq_2}$.
By definition of simple product:
:$T := S_1 \times S_2$ where $\times$ denotes Cartesian product
:$\forall \tuple {a, b}, \tuple {c, d} \in T: \tuple {a, b} \preccurlyeq_s \tuple {c, d} \iff a \preccurlyeq_1 c \text { and } b \preccurlyeq_2 d$
We check in turn each of the criteria for an ordering:
=== Reflexivity ===
We have by definition of Cartesian product:
:$\forall \tuple {a, b} \in S_1 \times S_2: a \in S_1 \land b \in S_2$
As $\preccurlyeq_1$ and $\preccurlyeq_2$ are both orderings, they are themselves both reflexive.
Hence both $a \preccurlyeq_1 a$ and $b \preccurlyeq_2 b$.
Hence by definition of simple product:
:$\forall \tuple {a, b} \in S: \tuple {a, b} \preccurlyeq_s \tuple {a, b}$
So $\preccurlyeq_s$ is reflexive.
=== Transitivity ===
Suppose $\tuple {a, b} \preccurlyeq_s \tuple {c, d}$ and $\tuple {c, d} \preccurlyeq_s \tuple {e, f}$.
Then by definition of simple product:
:$a \preccurlyeq_1 c$ and $b \preccurlyeq_2 d$
:$c \preccurlyeq_1 e$ and $d \preccurlyeq_2 f$
As $\preccurlyeq_1$ and $\preccurlyeq_2$ are both orderings, they are themselves both transitive.
Hence:
:$a \preccurlyeq_1 e$ and $b \preccurlyeq_2 f$
By definition of simple product:
:$\tuple {a, b} \preccurlyeq_s \tuple {e, f}$
So $\preccurlyeq_s$ is transitive.
=== Antisymmetry ===
Suppose $\tuple {a, b} \preccurlyeq_s \tuple {c, d}$ and $\tuple {c, d} \preccurlyeq_s \tuple {a, b}$.
Then by definition of simple product:
:$a \preccurlyeq_1 c$ and $b \preccurlyeq_2 d$
:$c \preccurlyeq_1 a$ and $d \preccurlyeq_2 b$
As $\preccurlyeq_1$ and $\preccurlyeq_2$ are both orderings, they are themselves both antisymmetric.
Hence:
:$a = c$ and $b = d$
So by Equality of Ordered Pairs:
:$\tuple {a, c} = \tuple {b, d}$
So $\preccurlyeq_s$ is antisymmetric.
So we have shown that $\preccurlyeq_s$ is reflexive, transitive and antisymmetric.
Thus by definition, $\preccurlyeq_s$ is an ordering and so $\struct {S_1, \preccurlyeq_1} \otimes^s \struct {S_2, \preccurlyeq_2}$ is an ordered set.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Exercise $14.18 \ \text {(a)}$
* 
* : Part $1$: Not Entirely Naive Set Theory: Chapter $2$: Partial Order Relations: Exercise $29 \ \text {(a)}$
* 
* 
* 
Category:Simple Order Product
Simple Order Product of Totally Ordered Sets may not be Totally Ordered

== Theorem ==
Let $\struct {S_1, \preccurlyeq_1}$ and $\struct {S_2, \preccurlyeq_2}$ be totally ordered sets.
Let $\struct {S_1, \preccurlyeq_1} \otimes^s \struct {S_2, \preccurlyeq_2}$ denote the simple (order) product of $\struct {S_1, \preccurlyeq_1}$ and $\struct {S_2, \preccurlyeq_2}$.
Then it is not necessarily the case that $\struct {S_1, \preccurlyeq_1} \otimes^s \struct {S_2, \preccurlyeq_2}$ is also a totally ordered set.
== Proof ==
From Simple Order Product of Pair of Ordered Sets is Ordered Set, we do have that $\struct {S_1, \preccurlyeq_1} \otimes^s \struct {S_2, \preccurlyeq_2}$ is an ordered set.
Let us take the simple product of the ordered set that is the natural numbers under the usual ordering with itself:
:$\struct {\N, \le} \otimes^s \struct {\N, \le}$
We have from the Well-Ordering Principle that $\struct {\N, \le}$ is a well-ordered set, and so a totally ordered set.
Consider the ordered pairs of natural numbers:
:$s = \tuple {1, 4}$
:$t = \tuple {2, 3}$
Both $s$ and $t$ are elements of $\N \times \N$.
We have that:
:$1 \le 2$
but:
:$4 \not \le 3$
So it is not the case either that:
:$1 \le 2$ and $4 \le 3$
or:
:$2 \le 1$ and $3 \le 4$
Hence $s$ and $t$ are non-comparable elements.
The result follows by definition of totally ordered set.
== Sources ==
* : Part $1$: Not Entirely Naive Set Theory: Chapter $2$: Partial Order Relations: Exercise $29 \ \text {(b)}$
Category:Simple Order Product
Category:Total Orderings
Simple Variable End Point Problem

== Theorem ==
Let $y$ and $F$ be mappings.
Suppose the endpoints of $y$ lie on two given vertical lines $x = a$ and $x = b$.
Suppose $J$ is a functional of the form
:$(1): \quad J \sqbrk y = \ds \int_a^b \map F {x, y, y'} \rd x$
and has an extremum for a certain function $\hat y$.
Then $y$ satisfies the system of equations
:$\begin {cases}
F_y - \dfrac \d {\d x} F_{y'} = 0 \\
\bigvalueat {F_{y'} } {x \mathop = a} = 0 \\
\bigvalueat {F_{y'} } {x \mathop = b} = 0
\end {cases}$
== Proof ==
From Condition for Differentiable Functional to have Extremum we have
:$\bigvalueat {\delta J \sqbrk {y; h} } {y \mathop = \hat y} = 0$
The variation exists if $J$ is a differentiable functional.
We will start from the increment of a functional:
Using multivariate Taylor's theorem, one can expand $\map F {x, y + h, y' + h'}$ with respect to $h$ and $h'$:
:$\map F {x, y + h, y' + h'} = \bigvalueat {\map F {x, y + h, y' + h'} } {h \mathop = 0, \, h' \mathop = 0} + \valueat {\dfrac {\partial \map F {x, y + h, y' + h'} } {\partial y} } {h \mathop = 0, \, h' \mathop = 0} h + \valueat {\dfrac {\partial {\map F {x, y + h, y' + h'} } } {\partial y'} } {h \mathop = 0, \, h' \mathop = 0} h' + \map \OO {h^2, h h', h'^2}$
Substitute this back into the integral. 
Note that the first term in the expansion and the negative one in the integral will cancel out:
:$\ds \Delta J \paren {y; h} = \int_a^b \paren {\map F {x, y, y'}_y h + \map F {x, y, y'}_{y'} h' + \map \OO {h^2, h h', h'^2} } \rd x$
Terms in $\map \OO {h^2, h'^2}$ represent terms of order higher than 1 with respect to $h$ and $h'$.
Now we expand:
:$\ds \int_a^b\map \OO {h^2, h h', h'^2} \rd x$
Every term in this expansion will be of the form:
:$\ds \int_a^b \map A {m, n} \frac {\partial^{m + n} \map F {x, y, y'} } {\partial y^m \partial {y'}^n} h^m h'^n \rd x$
where $m, n \in \N$ and $m + n \ge 2$
By definition, the integral not counting in $\map \OO {h^2, h h', h'^2}$ is a variation of functional.
:$\ds \delta J \sqbrk {y; h} = \int_a^b \paren {F_y h + F_{y'} h'} \rd x$
Now, integrate by parts and note that $\map h x$ does not necessarily vanish at the endpoints:
Then, for arbitrary $\map h x$, $J$ has an extremum if:
:$ \begin {cases}
F_y - \dfrac \d {\d x} F_{y'} = 0 \\
\bigvalueat {F_{y'} } {x \mathop = a} = 0\\
\bigvalueat {F_{y'} } {x \mathop = b} = 0
\end {cases}$
== Sources ==
* : $\S 1.6$: A Simple Variable End Point Problem
Category:Calculus of Variations
Simple Variable End Point Problem/Endpoints on Curves

== Theorem ==
Let $y$, $F$, $\phi$ and $\psi$ be smooth real functions.
Let $J = J \sqbrk y$ be a functional of the form:
:$\ds J \sqbrk y = \int_{x_0}^{x_1} \map F {x, y, y'} \rd x$
Let $P_0$, $P_1$ be the endpoints of the curve $y$.
Suppose $P_0$, $P_1$ lie on curves $y = \map {\phi} x$, $y = \map {\psi} x$.
Then the extremum of $J \sqbrk y$ is a curve which satisfies the following system of Euler and transversality equations:
== Proof ==
By general variation of integral functional with $n = 1$:
:$\ds \delta J \sqbrk{y; h} = \int_{x_0}^{x_1} \intlimits {\paren {F_y - \dfrac \d {\d x} F_{y'} } \map h x + F_{y'} \delta y} {x \mathop = x_0} {x \mathop = x_1} + \bigintlimits {\paren {F - y'F_{y'} } \delta x} {x \mathop = x_0} {x \mathop = x_1}$
Since the curve $y = \map y x$ is an extremum of $\map J y$, the integral term vanishes:
:$\ds \delta J \sqbrk {y; h} = \bigintlimits {F_{y'} \delta y} {x \mathop = x_0} {x \mathop = x_1} + \bigintlimits {\paren {F - y'F_{y'} } \delta x} {x \mathop = x_0} {x \mathop = x_1}$
According to Taylor's Theorem:
:$\delta y_0 = \paren {\map {\phi'} {x_0} + \epsilon_0} \delta x_0$
:$\delta y_1 = \paren {\map {\phi'} {x_1} + \epsilon_1} \delta x_1$
where $\epsilon_0 \to 0$ as $\delta x_0 \to 0$ and $\epsilon_1 \to 0$ as $\delta x_1 \to 0$.
Substitution of $\delta y_0$ and $\delta y_1$ into $\delta J$ leads to:
:$\delta J \sqbrk {y; h} = \bigvalueat {\paren {F_{y'} \psi' + F - y' F_{y'} } } {x \mathop = x_1} \delta x_1 - \bigvalueat {\paren {F_{y'} \psi' + F - y' F_{y'} } } {x \mathop = x_0} \delta x_0$
$y = \map y x$ is an extremal of $J \sqbrk y$, thus $\delta J = 0$.
$\delta x_0$ and $\delta x_1$ are independent increments. 
Hence both remaining terms in $\delta J$ vanish independently.
== Sources ==
* : $\S 3.14$: End Points Lying on Two Given Lines or Surfaces
Category:Calculus of Variations
Simplest Form of Non-Negative Number in Golden Mean Number System is Unique

== Theorem ==
Let $x \in \R_{\ge 0}$ be represented in the golden mean number system.
Let $S$ be the representation for $x$ in its simplest form.
Then $S$ is unique in the sense that there exists no other representation of $x$ in simplest form.
== Proof ==
== Sources ==
* : $\S 1.2.8$: Fibonacci Numbers: Exercise $35$
Category:Golden Mean Number System
Simplest Variational Problem

== Problem ==
Let $\map F {x, y, z}$ be a real-valued function of a differentiability class $C^2$ all its arguments. 
Let $y: \R \to \R$ be a continuously differentiable real function for $x \in \sqbrk {a, b}$ such that
:$\map y a = A$
:$\map y b = B$
Then, among all real functions $y$, find the one for which the functional:
:$\ds J \sqbrk y = \int_a^b \map F {x, y, y'} \rd x$
has a weak extremum.
== Sources ==
* : $\S 1.4$: The Simplest Variational Problem. Euler's Equation
Category:Calculus of Variations
Simpson's Dissection

== Theorem ==
Let $\omega = e^{2 i \pi / q}$ be a primitive $q$th root of unity.
Let $p \not \equiv 0 \pmod q$.
Let:
:$\ds \map f x = \sum_{n \mathop = 0}^\infty a_n x^n$
Then:
:$\ds \sum_{n \mathop = 0}^\infty a_{n q + p} x^{n q + p} = \dfrac 1 q \sum_{j \mathop = 0}^{q - 1} \omega^{- j p} \map f {\omega^j x}$
== Proof ==
Expanding the sum on the , we obtain:
Summing down the $q$ values of the term with exponent n in the array shown above, we obtain:
:$\ds \dfrac 1 q a_n x^n \times \paren {1 + \omega^{1 \paren {n - p} } + \omega^{2 \paren {n - p} } + \omega^{3 \paren {n - p} } + \cdots + \omega^{\paren {q - 1} \paren {n - p} } }$
From Sum of Powers of Primitive Complex Roots of Unity, we have:
When $n \equiv p \pmod q$:
:$\ds \dfrac 1 q a_n x^n \times \paren {1 + \omega^{1 \paren {n - p} } + \omega^{2 \paren {n - p} } + \omega^{3 \paren {n - p} } + \cdots + \omega^{\paren {q - 1} \paren {n - p} } } = a_n x^n$
When $n \not \equiv p \pmod q$:
:$\ds \dfrac 1 q a_n x^n \times \paren {1 + \omega^{1 \paren {n - p} } + \omega^{2 \paren {n - p} } + \omega^{3 \paren {n - p} } + \cdots + \omega^{\paren {q - 1} \paren {n - p} } } = 0$
Hence:
== Sources ==
* 
* : Chapter $\text {1}$. The Gamma and Beta Functions
* 
Category:Simpson's Dissection
Category:Series
Simpson's Rule

== Theorem ==
Let $f$ be a real function which is integrable on the closed interval $\closedint a b$.
:$\ds \int_a^b \map f x \rd x \approx \dfrac {b - a} 6 \paren {\map f a + 4 \map f {\dfrac {a + b} 2} + \map f b}$
Hence the area under the curve is approximated by the area under the quadratic polynomial passing through $\tuple {x, \map f x}$ for the $3$ values $x = a$, $x = \dfrac {a + b} 2$ and $x = b$.
=== Repeated Simpson's Rule ===
== Proof ==
== Also known as ==
== Sources ==
* 
* 
Category:Simpson's Rule
Category:Definite Integrals
Category:Numerical Integration
Simpson's Rule/Repeated

== Theorem ==
Let $f$ be a real function which is integrable on the closed interval $\closedint a b$.
Let $P = \set {x_0, x_1, x_2, \ldots, x_{n - 1}, x_n}$ form a normal subdivision of $\closedint a b$:
:$\forall r \in \set {1, 2, \ldots, n}: x_r - x_{r - 1} = \dfrac {b - a} n$
where $n$ is even.
Then the definite integral of $f$ $x$ from $a$ to $b$ can be approximated as:
where:
:$\forall i \in \set {0, 1, 2, \ldots, n}: y_i = \map f {x_i}$
:$h = \dfrac {b - a} n$
This is known as the repeated Simpson's rule.
== Proof ==
== Also known as ==
== Sources ==
* : $\S 15$: Approximate Formulas for Definite Integrals: $15.17$
* 
* 
* : $\S 18$: Definite Integrals: Approximate Formulas for Definite Integrals: $18.17$
Category:Simpson's Rule
Simson Line Theorem

== Theorem ==
Let $\triangle ABC$ be a triangle.
Let $P$ be a point on the circumcircle of $\triangle ABC$.
Then the feet of the perpendiculars drawn from $P$ to each of the sides of $\triangle ABC$ are collinear.
:
This line is called the Wallace-Simson line.
== Proof ==
In the figure above, construct the lines $BP$ and $CP$.
:
By the converse of Opposite Angles of Cyclic Quadrilateral sum to Two Right Angles, $EPDB$ is cyclic.
By the converse of Angles in Same Segment of Circle are Equal, $EPCF$ is cyclic.
Therefore:
This gives:
:$\angle DEP + \angle PEF = 180 \degrees$
Hence $DEF$ is a straight line.
== Sources ==
* 
Category:Wallace-Simson Line
Category:Triangles
Simultaneous Equation With Two Unknowns

== Theorem ==
A pair of simultaneous linear equations of the form:
where $a e \ne b d$, has as its only solution:
== Proof 1 ==
The solution for $x$ can be found similarly.
When $a e = b d$ we have that $a e - b d = 0$ and hence no solution exists.
== Proof 2 ==
This is an example of Solution to Simultaneous Linear Equations and can be solved using the technique of matrices.
Category:Linear Algebra
Simultaneous Equations/Examples/Arbitrary Example 1

== Example of Simultaneous Equations ==
The simultaneous equations:
has the solution:
This can be interpreted as that the point $\tuple {-2, 8}$ on the Cartesian plane is where the two straight lines $x + y = 6$ and $2 x + y = 4$ intersect.
== Proof ==
:
== Sources ==
* 
* 
Category:Examples of Simultaneous Equations