nyuuzyou/wikis huggingface dataset:
Densely Ordered/Examples/Arbitrary Densely Ordered

== Example of Densely Ordered Set ==
Let $S$ be the subset of the rational numbers $\Q$ defined as:
:$S = \Q \cap \paren {\openint 0 1 \cup \hointr 2 3}$
Then $\struct {S, \le}$ is a densely ordered set.
Hence $\struct {S, \le}$ is isomorphic to $\struct {\Q, \le}$.
== Proof ==
Let $a, b \in S$ such that $0 < a < b < 1$.
Then from Rational Numbers are Densely Ordered:
:$\exists c \in \Q: a < c < b$
and as $0 < c < 1$ it follows that $c \in S$.
Let $a, b \in S$ such that $2 \le a < b < 3$.
Then from Rational Numbers are Densely Ordered:
:$\exists c \in \Q: a < c < b$
and as $2 < c < 3$ it follows that $c \in S$.
Now let $a, b \in S$ such that $0 < a < 1$ and $2 \le b < 3$.
Then from Rational Numbers are Densely Ordered:
:$\exists c \in \Q: a < c < 1$
and as $0 < c < 1$ it follows that $c \in S$.
So it follows that $\struct {S, \le}$ is a densely ordered set.
Next consider the mapping $\phi: \Q \to S$ defined as:
:$\forall x \in \Q: \map \phi x = \begin {cases} \dfrac {-1} {x - 1} : x < 0 \\ \dfrac {3 x + 2} {x + 1} : x \ge 0 \end {cases}$
It can be seen that:
:$\Img \phi = \Q \cap \paren {\openint 0 1 \cup \hointr 2 3}$
and:
:$\forall x, y \in \Q: x \le y \iff \map \phi x \le \map \phi y$
Hence the result.
== Sources ==
* : Part $1$: Not Entirely Naive Set Theory: Chapter $2$: Partial Order Relations: Exercise $23$
Category:Densely Ordered
Densely Ordered/Examples/Arbitrary Non-Densely Ordered

== Example of Ordered Set which is not Densely Ordered ==
Let $S$ be the subset of the rational numbers $\Q$ defined as:
:$S = \Q \cap \paren {\hointl 0 1 \cup \hointr 2 3}$
Then $\struct {S, \le}$ is not a densely ordered set.
Thus $\struct {S, \le}$ is not isomorphic to $\struct {\Q, \le}$.
== Proof ==
It will be noted that $1 \in S$ and $2 \in S$ but there exists no $c \in S$ such that $1 < c < 2$.
Hence the result from Densely Ordered is Order Property.
== Sources ==
* : Part $1$: Not Entirely Naive Set Theory: Chapter $2$: Partial Order Relations
Category:Densely Ordered
Densely Ordered is Order Property

== Theorem ==
Let $\struct {S_1, \preccurlyeq_1}$ and $\struct {S_2, \preccurlyeq_2}$ be isomorphic ordered sets.
Let $\struct {S_1, \preccurlyeq_1}$ be densely ordered.
Then $\struct {S_2, \preccurlyeq_2}$ is also densely ordered.
That is, the property of being densely ordered is an order property.
== Proof ==
Let $\struct {S_1, \preccurlyeq_1}$ and $\struct {S_2, \preccurlyeq_2}$ be isomorphic ordered sets.
Let $\struct {S_1, \preccurlyeq_1}$ be densely ordered.
Let $a, b \in S_2$ such that $a \prec b$.
Then as $\phi$ is an order isomorphism:
:$\exists p, q \in S_1: a = \map \phi p, b = \map \phi q$
Then by definition of densely ordered:
:$\exists r \in S_1: p \prec r \prec q$
By definition of order isomorphism:
:$\map \phi p \prec \map \phi r \prec \map \phi q$
That is:
:$\exists c \in S_2: a \prec c \prec b$
where $c = \map \phi r$.
The result follows by definition of densely ordered.
== Sources ==
* : Part $1$: Not Entirely Naive Set Theory: Chapter $2$: Partial Order Relations
Category:Examples of Order Properties
Category:Densely Ordered
Denseness Preserved in Coarser Topology

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $T' = \struct {S, \tau'}$ be another topological space on $S$ such that $\tau'$ is coarser than $\tau$.
Let $H \subseteq S$ be everywhere dense in $T$.
Then $H$ is also everywhere dense in $T'$.
== Proof ==
Let $H \subseteq S$ be everywhere dense in $T$. 
As $\tau'$ is coarser than $\tau$, by definition $\tau' \subseteq \tau$.
Let $x \in S$.
As $H \subseteq S$ is everywhere dense in $T$, either $x \in H$ or $x$ is a limit point of $H$ in $T$.
Suppose $x \notin H$.
Then $x$ is a limit point of $H$ in $T$.
Then by definition of limit point of $H$:
:$\forall U \in \tau, x \in U: \exists y \in H: y \in U, y \ne x$ 
As $\tau' \subseteq \tau$ we have that $U \in \tau' \implies U \in \tau$.
So:
:$\forall U \in \tau', x \in U: \exists y \in H: y \in U, y \ne x$ 
That is, $x$ is a limit point of $H$ in $T\ '$.
As $x$ is arbitrary, this follows for all $x \in S, x \notin H$.
So, for all $x \in S$, either $x \in H$ or $x$ is a limit point of $H$ in $T'$.
Thus, by definition, $H \subseteq S$ is everywhere dense in $T'$.
Category:Denseness
Densest Packing of Identical Circles

== Theorem ==
The densest packing of identical circles in the plane obtains a density of $\dfrac \pi {2 \sqrt 3} = \dfrac \pi {\sqrt {12} }$:
:$\dfrac \pi {2 \sqrt 3} = 0 \cdotp 90689 \, 96821 \ldots$
This happens when they are packed together in a hexagonal array, with each circle touching $6$ others.
== Proof ==
Consider the rectangular area $ABCD$ of the densest packing of circles.
:
Let the radius of one circle be $1$.
The length $AB$ is $2$.
The length $AC$ is $2 \sqrt 3$.
Thus, from Area of Rectangle, the area of $\Box ABCD$ is $4 \sqrt 3$.
Within $ABCD$ there is one complete circle and one quarter of each of $4$ other circles.
That makes a total of $2$ circles.
Thus, from Area of Circle, the area of $ABCD$ which is inside a circle is $2 \pi$.
So the density is:
as required.
== Sources ==
* : $0,90689 9682 \ldots$
* : $0 \cdotp 9068 \ldots$
* : $0 \cdotp 9068 \ldots$
Category:Geometry
Density not greater than Weight

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Then
:$\map d T \le \map w T$
where
:$\map d T$ denotes the density of $T$,
:$\map w T$ denotes the weight of $T$.
== Proof ==
By definition of weight there exists a basis $\BB$ of $T$:
:$\map w T = \card \BB$
where $\card \BB$ denotes the cardinality of $\BB$.
By Axiom of Choice define a mapping $f: \paren {\BB \setminus \O} \to S$:
:$\forall U \in \BB: U \ne \O \implies f \sqbrk U \in U$
We will prove that
:$\forall U \in \tau: U \ne \O \implies U \cap \Img f \ne \O$
where $\Img f$ denotes the image of $f$.
Let $U \in \tau$ such that
:$U \ne \O$
By definition of empty set:
:$\exists x: x \in U$
Then by definition of basis:
:$\exists V \in \BB: x \in V \subseteq U$
By definition of image:
:$f \sqbrk V \in \Img f$
By definition of $f$:
:$f \sqbrk V \in V$
By definition of subset:
:$f \sqbrk V \in U$
Then by definition of intersection:
:$f \sqbrk V \in U \cap \Img f$
Thus by definition of empty set:
:$U \cap \Img f \ne \O$ 
Then:
:$\forall x \in S: \forall U \in \tau: x \in U \implies U \cap \Img f \ne \O$
Hence by Condition for Point being in Closure:
:$\forall x \in S: x \in \paren {\Img f}^-$
where $^-$ denotes the topological closure.
Then by definition of subset:
:$S \subseteq \paren {\Img f}^- \subseteq S$
Thus by definition of set equality:
:$S = \paren {\Img f}^-$
Thus by definition:
:$\Img f$ is dense
By definition of density:
:$\map d T \le \card {\Img f}$
$f$ as $\paren {\BB \setminus \O} \to \Img f$ by definition is surjection.
Therefore by Surjection iff Cardinal Inequality:
:$\card {\Img f} \le \card {\BB \setminus \O}$
From Set Difference is Subset:
:$\BB \setminus \O \subseteq \BB$
Then by Subset implies Cardinal Inequality:
:$\card {\BB \setminus \O} \le \card \BB$
Thus:
:$\map d T \le \map w T$
== Sources ==
* 
* 
Category:Denseness
Denumerable Class is Set

== Theorem ==
Let $A$ be a class.
Let $\N$ be the natural numbers.
Suppose that $F: \N \to A$ is a bijection.
Then $A$ is a set.
== Proof ==
By the Axiom of Infinity, $\N$ is a set.
Thus by the Axiom of Replacement, $A$ is also a set.
Category:Set Theory
Denying the Antecedent

== Fallacy ==
Let $p \implies q$ be a conditional statement.
Let its antecedent $p$ be false.
Then it is a fallacy to assert that the consequent $q$ is also necessarily false.
That is:
This fallacy is called denying the antecedent.
=== Corollary ===
== Proof ==
We apply the Method of Truth Tables to the proposition.
$\begin{array}{|ccc|cc||cc|} \hline
p & \implies & q & \neg & p & \neg & q\\
\hline
\F & \T & \F & \T & \F & \T & \F \\
\F & \T & \T & \T & \F & \F & \T \\
\T & \F & \F & \F & \T & \T & \F \\
\T & \T & \T & \F & \T & \F & \T \\
\hline
\end{array}$
As can be seen, when $p \implies q$ is true, and so is $\neg p$, then it is not always the case that $\neg q$ is also true.
== Also see ==
* Affirming the Consequent
* Conditional and Inverse are not Equivalent
== Sources ==
* : Chapter $1$: The Propositional Calculus $1$: $2$ Conditionals and Negation
* : $2$ Arguments Containing Compound Statements: $2.3$: Argument Forms and Truth Tables
* 
* 
Category:Denying the Antecedent
Category:Formal Fallacies
Category:Fallacies
Category:Conditional
Denying the Antecedent/Corollary

== Corollary to Denying the Antecedent ==
Let $p \implies q$ be a conditional statement.
Let its antecedent $p$ be false.
Then nothing can be inferred about the truth value of $q$.
== Proof ==
By applying the Method of Truth Tables to the proposition, it is seen that $q$ can be either true or false, and both are consistent with both $\neg p$ and $p \implies q$.
$\begin{array}{|ccc|cc||cc|} \hline
p & \implies & q & \neg & p & \neg & q\\
\hline
\F & \T & \F & \T & \F & \T & \F \\
\F & \T & \T & \T & \F & \F & \T \\
\T & \F & \F & \F & \T & \T & \F \\
\T & \T & \T & \F & \T & \F & \T \\
\hline
\end{array}$
== Sources ==
* : Chapter $1$: Sets, Functions, and Relations: $\S 2$: Some Remarks on the Use of the Connectives and, or, implies
Category:Denying the Antecedent
Dependent Choice (Fixed First Element)

== Theorem ==
Let $\RR$ be a binary relation on a non-empty set $S$.
Suppose:
:$\forall a \in S: \exists b \in S: a \mathrel \RR b$
that is, that $\RR$ is a left-total relation (that is, a serial relation).
Let $s \in S$.
Then there exists a sequence $\sequence {x_n}_{n \mathop \in \N}$ in $S$ such that:
:$x_0 = s$
:$\forall n \in \N: x_n \mathrel \RR x_{n + 1}$
== Proof ==
Let $S' = \set {y \in S: s \mathrel {\RR^+} y}$, where $\RR^+$ is the transitive closure of $\RR$.
Let $\RR'$ be the restriction of $\RR$ to $S'$. 
For each $x \in S'$, there is a $y \in S$ such that $x \mathrel \RR y$. But then $s \mathrel {\RR^+} y$, so $y \in S'$, so $x \mathrel {\RR'} y$.
Thus $\RR'$ is a left-total relation on $S'$.
$S'$ is non-empty: since $\RR$ is left-total, there is a $t \in S$ such that $s \mathrel \RR t$, so $s \mathrel {\RR^+} t$, so $t \in S'$.
By Axiom of Dependent Choice, there is a infinite sequence $\sequence {y_n}_{n \mathop \in \N}$ in $S'$ such that for each $n \in \N$, $y_n \mathrel {\RR'} y_{n + 1}$.
Then by definition of restriction, $y_n \mathrel \RR y_{n + 1}$ for each $n \in \N$.
By definition of $S'$, $s \mathrel {\RR^+} y_0$.
By definition of transitive closure, there are elements $x_0, \dots, x_m$ such that $s = x_0 \mathrel \RR x_1 \mathrel \RR \cdots \mathrel \RR x_m \mathrel \RR x_m = y_0$.
Then for $n > m$, define $x_n$ as $y_{n - m}$.
This sequence then meets the requirements.
Category:Set Theory
Dependent Choice for Finite Sets

== Theorem ==
Let $\RR$ be a binary relation on a non-empty set $S$.
For each $a \in S$, let $C_a = \set {b \in S: a \mathrel \RR b }$
Suppose that:
:For all $a \in S$, $C_a$ is a non-empty finite set.
Let $s \in S$.
Then there exists a sequence $\sequence {x_n}_{n \mathop \in \N}$ in $S$ such that:
:$x_0 = s$
:$\forall n \in \N: x_n \mathrel \RR x_{n+1}$
== Remark ==
This theorem follows trivially from Dependent Choice (Fixed First Element), a form of the Axiom of Dependent Choice.
The proof below shows that it follows from the weaker Axiom of Countable Choice for Finite Sets.
== Proof ==
Define $\sequence {D_n}$ recursively:
Let $D_0 = \set s$.
For each $n \in \N$ let $D_{n + 1} = \map \RR {D_n}$.
Now, for each $n \in \N$ let $E_n$ be the set of all enumerations of $D_n$.
Then $E_n$ is non-empty and finite for each $n$.
By the Axiom of Countable Choice for Finite Sets, there is a sequence $\sequence {e_n}$ such that for each $n$, $e_n \in E_n$.
Now recursively define $\sequence {x_n}$:
:Define $x_0$ as $s$.
:Define $x_{n + 1}$ as the first element in the enumeration $e_n$ which is in $\map \RR {x_n}$.
Then $\sequence {x_n}$ is the required sequence.
== Also see ==
* Axiom:Axiom of Dependent Choice which differs from this in that there is no constraint that each element is related to only a finite number of other elements. 
Category:Set Theory
Dependent Subset Contains a Circuit

== Theorem ==
Let $M = \struct {S, \mathscr I}$ be a matroid.
Let $\mathscr C$ denote the set of all circuits of $M$.
Let $A$ be a dependent subset.
Then:
:$\exists C \in \mathscr C : C \subseteq A$
== Proof ==
Consider the ordered set $\struct {\powerset S \setminus \mathscr I, \subseteq}$.
From Element of Finite Ordered Set is Between Maximal and Minimal Elements:
:$\exists C \in \mathscr I : C \subseteq A$ and $A$ is minimal in $\struct {\powerset S \setminus \mathscr I, \subseteq}$.
By definition of a circuit:
:$C \in \mathscr C$
Category:Matroid Dependent Subsets
Category:Matroid Circuits
Dependent Subset of Independent Set Union Singleton Contains Singleton

== Theorem ==
Let $M = \struct {S, \mathscr I}$ be a matroid.
Let $X$ be an independent subset of $M$.
Let $x \in S$.
Let $C$ be a dependent subset of $M$ such that:
:$C \subseteq X \cup \set x$.
Then:
:$x \in C$
== Proof ==
From the contrapositive statement of Superset of Dependent Set is Dependent:
:$C \nsubseteq X$
From the contrapositive statement of Set Difference with Superset is Empty Set:
:$C \setminus X \ne \O$
From Set Difference over Subset:
:$C \setminus X \subseteq \paren {X \cup \set x} \setminus X = \set x$
From Power Set of Singleton:
:$C \setminus X = \set x$
From Set Difference is Subset:
:$\set x \subseteq C$
From Singleton of Element is Subset:
:$x \in C$
Category:Matroid Dependent Subsets
Category:Matroid Independent Subsets
Derivation of Auxiliary Equation to Constant Coefficient LSOODE

== Theorem ==
Consider the linear Second Order ODE with Constant Coefficients:
:$(1): \quad \dfrac {\d^2 y} {\d x^2} + p \dfrac {\d y} {\d x} + q y = \map R x$
and its auxiliary equation:
:$(2): \quad m^2 + p m + q = 0$
The fact that the solutions of $(2)$ dictate the general solution of $(1)$ can be derived.
== Proof ==
Let the reduced equation of $(1)$ be expressed in the form:
:$(3): \quad D^2 y + p D y + q y = 0$
where $D$ denotes the derivative operator $x$:
:$D := \dfrac \d {\d x}$
Thus:
:$D^2 := \dfrac {\d^2} {\d x^2}$
We can express $(3)$ in the form:
:$(4): \quad \paren {D^2 + p y + q} y = 0$
Consider the expression:
:$(5): \quad \paren {D - k_1} \paren {D - k_2} y$
for constants $k_1$ and $k_2$ (not necessarily real).
We have:
Thus $(3)$ can be written:
:$(6): \quad \paren {D - k_1} \paren {D - k_2} y = 0$
From Sum of Roots of Quadratic Equation and Product of Roots of Quadratic Equation, we recognise that $k_1$ and $k_2$ are the solutions of $(2)$.
Let $z := \paren {D - k_2} y$.
Then from $(6)$ we have:
:$\paren {D - k_1} z = 0$
That is:
:$(7): \quad \dfrac {\d z} {\d x} - k_1 z = 0$
From Solution to Linear First Order ODE with Constant Coefficients, $(7)$ has the general solution:
:$z = C e^{k_1 x}$
Thus we have:
:$\dfrac {\d y} {\d x} - k_2 y = z = C_1 e^{k_1 x}$
From Solution to Linear First Order ODE with Constant Coefficients:
:$(8): \quad y e^{-k_1 x} = C_2 \ds \int e^{\paren {k_1 - k_2} x} \rd x + C_2$
Suppose $k_1 \ne k_2$.
Then the of $(8)$ evaluates to:
:$\dfrac C {k_1 - k_2} e^{\paren {k_1 - k_2} x}$
If $k_1 = k_2$ then it is merely:
:$\ds \int C e^{0 \cdot x} \rd x = \int C \rd x = C x$
We can of course replace $\dfrac C {k_1 - k_2}$ with another constant.
It follows that the general solution of $(3)$ can be expressed as a linear combination of :
:$e^{k_1 x}$ and $e^{k_2 x}$
if $k_1 \ne k_2$, and:
:$e^{k_1 x}$ and $x e^{k_1 x}$
if $k_1 = k_2$.
If $k_1$ and $k_2$ are complex conjugates, we have that:
:$k_1 = \alpha + i \omega$
:$k_2 = \alpha - i \omega$
for some real $\alpha$ and $\omega$.
This leads to the corresponding solutions:
:$e^{\paren {\alpha \pm i \omega} x} = e^{\alpha x} \paren {\cos \omega x \pm i \sin \omega x}$
Hence any linear combination of $e^{\alpha x} \paren {\cos \omega x \pm i \sin \omega x}$ can be expressed as:
:$e^{\alpha x} \paren {A \cos \omega x + B \sin \omega x}$
and the task is complete.
== Sources ==
* : Chapter $1$: Linear Differential Equations with Constant Coefficients: $\S 2$. The second order equation: $\S 2.1$ The reduced equation
Category:Constant Coefficient LSOODEs
Derivation of Fourier Series over General Range

== Theorem ==
Let $\alpha \in \R$ be a real number.
Let $\lambda \in \R_{>0}$ be a strictly positive real number.
Let $f: \R \to \R$ be a function such that $\ds \int_{\mathop \to \alpha}^{\mathop \to \alpha + 2 \lambda} \map f x \rd x$ converges absolutely.
Let:
:$\ds f \sim \frac {a_0} 2 + \sum_{n \mathop = 1}^\infty \paren {a_n \cos \frac {n \pi x} \lambda + b_n \sin \frac {n \pi x} \lambda}$
The Fourier coefficients for $f$ are calculated by:
== Proof ==
By definition of Fourier series over the range of integration $\openint \alpha {\alpha + 2 \pi}$:
:$(1): \quad \dfrac {a_0} 2 + \ds \sum_{n \mathop = 1}^\infty \paren {a_n \cos n x + b_n \sin n x}$
where:
Let $\xi = \dfrac {\pi x} \lambda$.
Then let:
:$\map \phi \xi \equiv \map f x$
We have that $\ds \int_{\mathop \to \alpha}^{\mathop \to \alpha + 2 \lambda} \map f x \rd x$ converges absolutely.
Thus we are ensured of the existence of:
:the limit from the right of $f$ at $\alpha$
:the limit from the left of $f$ at $\alpha + 2 \lambda$
Thus $f$ is defined and bounded on $\openint \alpha {\alpha + 2 \lambda}$.
Then:
Setting $\beta = \dfrac {\pi \alpha} \lambda$, this allows us:
Thus we have that $\phi$ is defined and bounded on $\openint \beta {\beta + 2 \pi}$.
Then:
:$\map \phi \xi \sim \ds \frac {a_0} 2 + \sum_{n \mathop = 1}^\infty \paren {a_n \cos \xi + b_n \sin \xi}$
where:
We have that:
:$\dfrac {\d \xi} {\d x} = \dfrac \pi \lambda$
and so:
and so:
:$\ds \map f x \sim \frac {a_0} 2 + \sum_{n \mathop = 1}^\infty \paren {a_n \cos \frac {n \pi x} \lambda + b_n \sin \frac {n \pi x} \lambda}$
== Sources ==
* : Chapter One: $\S 3$. Other Types of Whole-Range Series: $(1)$
Category:Fourier Series
Derivative Form of Associated Legendre Function of the First Kind

== Theorem ==
Let $\map { {P_n}^m} x$ be an associated Legendre function of the first kind.
Then:
:$\map { {P_n}^m} x = \dfrac {\paren {1 - x^2}^{m / 2} } {2^n n!} \dfrac {\d^{m + n} } {\d x^{m + n} } \paren {x^2 - 1}^n$
== Proof ==
== Sources ==
* : $\S 26$: Associated Legendre Functions: Associated Legendre Functions of the First Kind: $26.2$
* : $\S 28$: Legendre and Associated Legendre Functions: Associated Legendre Functions of the First Kind: $28.50.$
Category:Associated Legendre Functions
Derivative Function is not Invertible

== Theorem ==
Let $\Bbb I = \closedint a b$ be a closed interval on the set of real numbers $\R$ such that $a < b$.
Let $A$ denote the set of all continuous real functions $f: \Bbb I \to \R$.
Let $B \subseteq A$ denote the set of all functions differentiable on $\Bbb I$ whose derivative is continuous on $\Bbb I$.
Let $d: B \to A$ denote the mapping defined as:
:$\forall \map f x \in B: \map d f = \map {D_x} f$
where $D_x$ denotes the derivative of $f$ $x$.
Then $d$ is not an invertible mapping.
== Proof ==
By definition, $d$ is invertible $d$ is a bijection.
It is sufficient to demonstrate that $d$ is not an injection.
Hence a fortiori $d$ is shown to not be a bijection.
Consider a differentiable function $f \in B$.
Then consider the function $g \in B$ defined as:
:$\forall x \in \Bbb I: \map g x = \map f x + C$
where $C \in \R$ is a constant.
Then we have that:
demonstrating that $d$ is not an injection.
The result follows.
== Sources ==
* : Chapter $1$: Theory of Sets: $\S 9$: Inverse Functions, Extensions, and Restrictions: Exercise $1$ 
Category:Differential Calculus
Derivative Function on Set of Functions induces Equivalence Relation

== Theorem ==
Let $X$ be the set of real functions $f: \R \to \R$ which possess continuous derivatives.
Let $\RR \subseteq X \times X$ be the relation on $X$ defined as:
:$\RR = \set {\tuple {f, g} \in X \times X: D f = D g}$
where $D f$ denotes the first derivative of $f$.
Then $\RR$ is an equivalence relation.
== Proof ==
== Sources ==
* : Chapter $1$: Theory of Sets: $\S 7$: Relations: Exercise $4$
Category:Differential Calculus
Category:Equivalence Relations
Derivative Operator is Linear Mapping

== Theorem ==
Let $I := \closedint a b$ be a closed real interval.
Let $C \closedint a b$ be the space of real-valued functions continuous on $I$.
Let $C^1 \closedint a b$ be the space of real-valued functions continuously differentiable on $I$.
Let $D$ be the derivative operator such that:
:$D : \map {C^1} I \to \map C I$
and $Dx := x'$.
Then $D$ is a linear mapping.
== Proof ==
=== Distributivity ===
=== Positive homogenity ===
By definition, $D$ is a linear mapping
== Sources ==
* : Chapter $\S 2.1$: Continuous and linear maps. Linear transformations
Category:Linear Transformations
Category:Differentiability Classes
Derivative Operator on Continuously Differentiable Function Space with C^1 Norm is Continuous

== Theorem ==
Let $I = \closedint 0 1$ be a closed real interval.
Let $\map C I$ be the real-valued, continuous on $I$ function space.
Let $\map {C^1} I$ be the continuously differentiable function space.
Let $x \in \map {C^1} I$ be a continuoulsly differentiable real-valued function.
Let $D : \map {C^1} I \to \map \CC I$ be the derivative operator such that:
:$\forall t \in \closedint 0 1 : \map {D x} t := \map {x'} t$
Suppose $\map C I$ and $\map {C^1} I$ are equipped with $C^0$ and $C^1$ norms respectively
Then $D$ is continuous.
== Proof ==
We have that the Derivative Operator is Linear Mapping.
By definition and Continuity of Linear Transformation between Normed Vector Spaces, $D$ is continuous.
== Sources ==
* : Chapter $\S 2.3$: The normed space $\map {CL} {X,Y}$
Category:Operator Theory
Category:Continuous Mappings
Category:Differentiability Classes
Derivative Operator on Continuously Differentiable Function Space with Supremum Norm is not Continuous

== Theorem ==
Let $I = \closedint 0 1$ be a closed real interval.
Let $\map C I$ be the real-valued, continuous on $I$ function space.
Let $\map {C^1} I$ be the continuously differentiable function space.
Let $x \in \map {C^1} I$ be a continuously differentiable real-valued function.
Let $D : \map {C^1} I \to \map C I$ be the derivative operator such that:
:$\forall t \in \closedint 0 1 : \map {Dx} t := \map {x'} t$
Suppose $\map C I$ and $\map {C^1} I$ are equipped with the supremum norm.
Then $D$ is not continuous.
== Proof ==
 $D$ is continuous.
We have that the Derivative Operator is Linear Mapping.
By Continuity of Linear Transformation between Normed Vector Spaces:
:$\exists M \in \R_{> 0} : \forall x \in \map {C^1} I : \norm {\map D x}_\infty \le M \norm x_\infty$
Suppose $x = t^n$ with $n \in \N$.
Then:
:$\norm x_\infty = \norm {t^n}_\infty = 1$
:$\norm {x'}_\infty = \norm {n t^{n-1}}_\infty = n$
Hence:
In other words:
:$\forall n \in \N : n \le M$
But $M$ is finite.
This is a contradiction.
Hence, $D$ is not continuous.
== Sources ==
* : Chapter $\S 2.3$: The normed space $\map {CL} {X,Y}$
Category:Operator Theory
Category:Continuous Mappings
Category:Differentiability Classes
Derivative at Maximum or Minimum

== Theorem ==
Let $f$ be a real function which is differentiable on the open interval $\openint a b$.
Let $f$ have a local minimum or local maximum at $\xi \in \openint a b$.
Then:
:$\map {f'} \xi = 0$
== Proof 1 ==
== Proof 2 ==
== Historical Note ==
== Sources ==
* : $\text {II}$. Calculus: Differentiation: Maximum, Minimum and Point of Inflection
* : $3$: Elementary Analytic Methods: $3.4$ Limits, Maxima and Minima: $3.4.2 \ (1)$
* : Chapter $\text {A}.13$: Fermat ()
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Category:Derivative at Maximum or Minimum
Category:Local Maxima
Category:Local Minima
Category:Differential Calculus
Derivative at Maximum or Minimum/Historical Note

== Historical Note on Derivative at Maximum or Minimum ==
The Derivative at Maximum or Minimum was first arrived at by , anticipating the more formal techniques used in differential calculus.
This result was given by in his $1684$ article Nova Methodus pro Maximis et Minimis, published in .
== Sources ==
* : Chapter $\text{IV}$: The Prince of Amateurs
* : Chapter $\text {A}.13$: Fermat ()
* : Chapter $\text {A}.19$: Leibniz ()
Category:Historical Notes