nyuuzyou/wikis huggingface dataset:
Euler Triangle Formula/Lemma 1

:
== Lemma to Euler Triangle Formula ==
:
Let the incenter of $\triangle ABC$ be $I$.
Let the circumcenter of $\triangle ABC$ be $O$.
Let $OI$ be produced to the circumcircle at $G$ and $J$.
Let $CI$ be produced to the circumcircle at $P$.
Let $F$ be the point where the incircle of $\triangle ABC$ meets $BC$.
We are given that:
:the distance between the incenter and the circumcenter is $d$
:the inradius is $\rho$
:the circumradius is $R$.
Then
:$IP \cdot CI = \paren {R + d} \paren {R - d}$
== Proof ==
:$OI = d$
:$OG = OJ = R$
Therefore:
:$IJ = R + d$
:$GI = R - d$
By the Intersecting Chords Theorem:
:$GI \cdot IJ = IP \cdot CI$
Substituting:
:$IP \cdot CI = \paren {R + d} \paren {R - d}$
Category:Euler Triangle Formula
Euler Triangle Formula/Lemma 2

== Lemma to Euler Triangle Formula ==
:
Let the bisector of angle $C$ of triangle $\triangle ABC$ be produced to the circumcircle at $P$.
Let $I$ be the incenter of $\triangle ABC$.
Then:
:$AP = BP = IP$
== Proof 1 ==
== Proof 2 ==
Category:Euler Triangle Formula
Euler Triangle Formula/Lemma 2/Proof 1

== Lemma to Euler Triangle Formula ==
== Proof ==
, it will be demonstrated that $BP = IP$.
Let $CP$ be the bisector of $\angle ACB$.
:$\angle ACP$ and $\angle ICB$ both subtend $AP$.
Hence indirectly by the Inscribed Angle Theorem:
:$\angle ACP = \angle ICB$
From Angles on Equal Arcs are Equal:
:$\angle ACP = \angle ABP$
and so:
:$\angle ABP = \angle ICB$
By the construction of the incircle, $IB$ is the bisector of $B$.
Then:
:$\angle IBA = \angle IBC$
and so:
By Sum of Angles of Triangle equals Two Right Angles:
:$\angle CIB + \angle IBC + \angle ICB = \angle CIB + \angle PIB$
as $\angle CIB$ and $\angle PIB$ are supplementary.
Thus:
and from Triangle with Two Equal Angles is Isosceles:
:$IP = BP$
The proof that $IP = AP$ follows the same lines.
Category:Euler Triangle Formula
Euler Triangle Formula/Lemma 2/Proof 2

== Lemma to Euler Triangle Formula ==
== Proof ==
We have :
:$AI$, $BI$ and $CI$ bisect their respective angles.
Let the half-angles be:
:$\alpha = \dfrac 1 2 \angle CAB$
:$\beta = \dfrac 1 2 \angle ABC$
:$\gamma = \dfrac 1 2 \angle BCA$
:
By Triangle with Two Equal Angles is Isosceles
:$\triangle AIP$ is isosceles
$\leadsto$:
:$AP = BP = IP$
Category:Euler Triangle Formula
Euler Triangle Formula/Proof 1

== Theorem ==
== Proof ==
=== Lemma 1 ===
=== Lemma 2 ===
By :
:$GI \cdot IJ = IP \cdot CI$
substituting:
:$IP \cdot CI = \paren {R + d} \paren {R - d}$
By :
:$IP = PB$
and so:
:$GI \cdot IJ = PB \cdot CI$
Now using the Extension of Law of Sines in $\triangle CPB$:
:$\dfrac {PB} {\map \sin {\angle PCB} } = 2 R$
and so:
:$GI \cdot IJ = 2 R \map \sin {\angle PCB} \cdot CI$
By the $4$th of Euclid's common notions:
:$\angle PCB = \angle ICF$
and so:
:$(1): \quad GI \cdot IJ = 2 R \map \sin {\angle ICF} \cdot CI$
We have that:
:$IF = \rho$
and by Radius at Right Angle to Tangent:
:$\angle IFC$ is a right angle.
By the definition of sine:
:$\map \sin {\angle ICF} = \dfrac {\rho} {CI}$
and so:
:$\map \sin {\angle ICF} \cdot CI = \rho$
Substituting in $(1)$:
Category:Euler Triangle Formula
Eulerian Circuit is Eulerian Trail

== Theorem ==
Let $G$ be a graph.
Let $C$ be an Eulerian circuit for $G$.
Then $C$ is also an Eulerian trail for $G$. 
== Proof ==
Recall the definition of Eulerian circuit:
Recall the definition of Eulerian trail:
Recall the definition of circuit:
Hence an Eulerian circuit is an instance of an Eulerian trail.
Category:Eulerian Circuits
Category:Eulerian Trails
Eulerian Graph is Semi-Eulerian

== Theorem ==
Let $G$ be an Eulerian graph.
Then $G$ is also a semi-Eulerian graph.
== Proof ==
Recall the definition of Eulerian graph:
Recall the definition of semi-Eulerian graph:
So, let $G$ be an Eulerian graph.
Let $C$ be an Eulerian circuit for $G$.
From Eulerian Circuit is Eulerian Trail, $C$ is also an Eulerian trail for $G$. 
Thus $G$ has an Eulerian trail.
The result follows by definition of semi-Eulerian graph.
Category:Eulerian Graphs
Category:Semi-Eulerian Graphs
Eulerian Logarithmic Integral is Asymptotic to Prime-Counting Function

== Theorem ==
Let $x \in \R$ be a real number such that $x > 2$.
Let $\map \Li x$ denote the Eulerian logarithmic integral of $x$.
Let $\map \pi x$ denote the prime-counting function of $x$.
Then $\map \Li x$ is asymptotically equal to $\map \pi x$.
== Proof ==
== Examples ==
== Sources ==
* 
Category:Eulerian Logarithmic Integral is Asymptotic to Prime-Counting Function
Category:Eulerian Logarithmic Integral
Category:Prime-Counting Function
Eulerian Logarithmic Integral is Asymptotic to Prime-Counting Function/Examples/10^6

== Examples of Use of Eulerian Logarithmic Integral is Asymptotic to Prime-Counting Function ==
Let $\map \Li x$ denote the Eulerian logarithmic integral of $x$.
Let $\map \pi x$ denote the prime-counting function of $x$.
Then:
== Sources ==
* 
Category:Eulerian Logarithmic Integral is Asymptotic to Prime-Counting Function
Eulerian Logarithmic Integral is less than Prime-Counting Function for Infinite Number of Integers

== Theorem ==
Let $\map \Li x$ denote the Eulerian logarithmic integral for $x \in \Z$.
Let $\map \pi x$ denote the prime-counting function for $x \in \Z$.
Then there exist an infinite set of integers $x$ such that:
:$\map \Li x < \map \pi x$
== Proof ==
== Also see ==
* Sign of Difference between Prime-Counting Function and Eulerian Logarithmic Integral Changes Infinitely Often
== Historical Note ==
== Sources ==
* 
Category:Prime-Counting Function
Category:Eulerian Logarithmic Integral
Evaluation Isomorphism is Isomorphism

== Theorem ==
Let $R$ be a commutative ring with unity.
Let $G$ be a unitary $R$-module whose dimension is finite.
Then the evaluation linear transformation $J: G \to G^{**}$ is an isomorphism.
== Proof ==
Let $\sequence {a_n}$ be an ordered basis of $G$.
Then by definition, every $x \in G$ can be written in the form:
:$\ds \sum_{k \mathop = 1}^n \lambda_k a_k$
where $\lambda_k \in R$.
From Unique Representation by Ordered Basis, this representation is unique.
From Existence of Ordered Dual Basis:
:$\sequence {\map J {a_n} }$ is the ordered basis of $G^{**}$ dual to the ordered basis of $G^*$ which itself is dual to $\sequence {a_n}$.
From Unique Linear Transformation Between Modules, there exists a unique linear transformation $\phi: G \to H$ satisfying:
:$\forall k \in \closedint 1 n: \map \phi {a_k} = b_k$
From the preceding it follows that this unique $\phi$ has to be $J$.
As every element of $G^{**}$ is so representable, it follows that $J$ is surjective.
Now let $x, y \in G$ such that $\map J x = \map J y$.
We have that:
:$x = \ds \sum_{k \mathop = 1}^n \lambda_k a_k$
for some unique $\sequence {\lambda_n}$ in $R$
:$y = \ds \sum_{k \mathop = 1}^n \mu_k a_k$
for some unique $\sequence {\mu_n}$ in $R$
Hence from Unique Linear Transformation Between Modules:
:$\map J x = \ds \sum_{k \mathop = 1}^n \lambda_k b_k$
for some unique $\sequence {\lambda_n}$ in $R$
:$\map J y = \ds \sum_{k \mathop = 1}^n \mu_k b_k$
for some unique $\sequence {\mu_n}$ in $R$
But as $\map J x = \map J y$ it must be the case that:
:$\forall k \in \set {1, 2, \ldots, n}: \lambda_k = \mu_k$
and it follows that $x = y$.
Thus $J$ is injective.
It has been shown that $J$ is both injective and surjective.
Hence, by definition, $J$ is bijective.
By definition, a linear transformation is a homomorphism.
Hence, again by definition, $J$ is an isomorphism.
== Also see ==
* Definition:Evaluation Isomorphism: what an evaluation linear transformation is called when its dimension is finite.
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 28$. Linear Transformations
Category:Linear Transformations
Category:Module Isomorphisms
Category:Evaluation Linear Transformations (Module Theory)
Evaluation Linear Transformation is Bilinear

== Theorem ==
Let $R$ be a commutative ring.
Let $G$ be an $R$-module.
Let $G^*$ be the algebraic dual of $G$.
Let $\innerprod x t$ be the evaluation linear transformation from $G$ to $G^{**}$.
Then the mapping $\phi: G \times G^* \to R$ defined as:
:$\forall \tuple {x, t} \in G \times G^*: \map \phi {x, t} = \innerprod x t$
satisfies the following bilinearity properties:
== Proof ==
and:
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 28$. Linear Transformations
Category:Linear Transformations
Category:Evaluation Linear Transformations (Module Theory)
Evaluation Linear Transformation is Linear Transformation

== Theorem ==
Let $R$ be a commutative ring with unity.
Let $G$ be an $R$-module.
Let $G^*$ be the algebraic dual of $G$.
Let $G^{**}$ be the double dual of $G^*$.
Let the mapping $J: G \to G^{**}$ be the evaluation linear transformation from $G$ into $G^{**}$ defined as:
:$\forall x \in G: \map J x = x^\wedge$
where for each $x \in G$, $x^\wedge: G^* \to R$ is defined as:
:$\forall t \in G^*: \map {x^\wedge} t = \map t x$
 
Then $J$ is a linear transformation.
== Proof ==
From Underlying Mapping of Evaluation Linear Transformation is Element of Double Dual, we have that:
:$x^\wedge \in G^{**}$
Hence $x^\wedge$ a fortiori is a linear transformation.
It remains to be shown that $J: G \to G^{**}$ is a linear transformation.
That is, that the following conditions are satisfied by $J$:
:$(1): \quad \forall x, y \in G: \map J {x + y} = \map J x + \map J y$
:$(2): \quad \forall x \in G: \forall \lambda \in R: \map J {\lambda \times x} = \lambda \times J x$
Hence:
and:
Hence the result.
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 28$. Linear Transformations
Category:Linear Transformations
Category:Evaluation Linear Transformations (Module Theory)
Evaluation Linear Transformation on Normed Vector Space is Linear Isometry

== Theorem ==
Let $\struct {X, \norm \cdot_X}$ be a normed vector space.
Let $\struct {X^{\ast \ast}, \norm \cdot_{X^{\ast \ast} } }$ be the second normed dual of $\struct {X, \norm \cdot_X}$.
Let $J : X \to X^{\ast \ast}$ be the evaluation linear transformation for $X$. 
Then: 
:$J$ is a linear isometry.
== Proof ==
From Evaluation Linear Transformation on Normed Vector Space is Linear Transformation from Space to Second Normed Dual, we have that: 
:$J$ is a linear transformation.
It remains to show that:
:$\norm {\map J x}_{X^{\ast \ast} } = \norm x_X$
for each $x \in X$.
For each $x \in X$, we denote: 
:$\map J x = x^\wedge$
We then aim to show: 
:$\norm {x^\wedge}_{X^{\ast \ast} } = \norm x_X$
for each $x \in X$. 
As shown in Evaluation Linear Transformation on Normed Vector Space is Linear Transformation from Space to Second Normed Dual, we have:
:$\cmod {\map {x^\wedge} f} \le \norm f_{X^\ast} \norm x_X$
for each $f \in X^\ast$.
So, we have: 
:$\norm x_X \in \set {c > 0 : \cmod {\map {x^\wedge} f} \le c \norm f_{X^\ast} \text { for all } f \in X^\ast}$
So, from the definition of the norm on bounded functionals and the definition of infimum, we have: 
:$\norm {x^\wedge}_{X^{\ast \ast} } \le \norm x_X$
We now aim to show that: 
:$\norm x_X \le \norm {x^\wedge}_{X^{\ast \ast} }$
From Existence of Support Functional, there exists $f \in X^\ast$ such that: 
:$\map f x = \norm x_X$
and:
:$\norm f_{X^\ast} = 1$ 
Then we have: 
So we have: 
:$\norm x_X \in \set {\cmod {\map {x^\wedge} f} : \norm f_{X^\ast} = 1}$
So from the definition of the norm on bounded functionals and the definition of supremum, we have: 
:$\norm {x^\wedge}_{X^{\ast \ast} } \ge \norm x_X$
So we have: 
:$\norm {\map {x^\wedge} f} = \norm x_X$
That is: 
:$\norm {\map J x}_{X^{\ast \ast} } = \norm x_X$
Category:Linear Isometries
Category:Evaluation Linear Transformations (Normed Vector Spaces)
Evaluation Linear Transformation on Normed Vector Space is Linear Transformation from Space to Second Normed Dual

== Theorem ==
Let $\Bbb F \in \set {\R, \C}$. 
Let $\struct {X, \norm \cdot_X}$ be a normed vector space over $\Bbb F$.
Let $\struct {X^\ast, \norm \cdot_{X^\ast} }$ be the normed dual of $\struct {X, \norm \cdot_X}$.
Let $\map L {X^\ast, \Bbb F}$ be the set of linear functionals on $X^\ast$. 
For each $x \in X$, define $x^\wedge : X^\ast \to \Bbb F$ by: 
:$\map {x^\wedge} f = \map f x$
Define $J : X \to \map L {X^\ast, \Bbb F}$ by: 
:$\map J x = x^\wedge$
for each $x \in X$.
Then: 
:$J$ is a linear transformation $X \to X^{\ast \ast}$. 
where $X^{\ast \ast}$ denotes the second normed dual.
== Proof ==
We first show that $J$ is a linear transformation.
Let $x, y \in X$ and $\alpha, \beta \in \Bbb F$.
Then, we have, for each $f \in X^\ast$:
So:
:$\map J {\alpha x + \beta y} = \alpha \map J x + \beta \map J y$
So $J$ is linear.
We now show that, for each $x \in X$, we have: 
:$\map J x = x^\wedge \in X^{\ast \ast}$
where $X^{\ast \ast}$ is the second normed dual.
We first show that $x^\wedge$ is a linear functional for each $x \in X$.
Let $x \in X$, $f, g \in X^\ast$ and $\alpha, \beta \in \Bbb F$.
Then: 
so $x^\wedge$ is a linear functional for each $x \in X$.
We now show that $x^\wedge$ is bounded for each $x \in X$.
Let $x \in X$.
Then, for each $f \in X^\ast$, we have: 
So: 
:$x^\wedge$ is a bounded linear functional.
That is:
:$\map J x \in X^{\ast \ast}$
== Sources ==
* $26.1$: The Second Dual
Category:Evaluation Linear Transformations (Normed Vector Spaces)
Evaluation Linear Transformation on Normed Vector Space is Weak to Weak-* Continuous Embedding into Second Normed Dual

== Theorem ==
Let $\Bbb F \in \set {\R, \C}$.
Let $X$ be a normed vector space over $\Bbb F$. 
Let $X^\ast$ be the normed dual of $X$. 
Let $X^{\ast \ast}$ be the second normed dual of $X$. 
Let $w$ be the weak topology on $X$.
Let $w^\ast$ be the weak-$\ast$ topology on $X^{\ast \ast}$.
Let $\iota : X \to X^{\ast \ast}$ be the evaluation linear transformation.
Then $\iota$ is $\struct {w, w^\ast}$-continuous.
== Proof ==
From the definition of the initial topology, $w^\ast$ is generated by the mappings $f^\wedge : X^{\ast \ast} \to \Bbb F$ defined by: 
:$\map {f^{\wedge} } \Phi = \map \Phi f$
for each $f \in X^\ast$ and $\Phi \in X^{\ast \ast}$.
From Continuity in Initial Topology, it therefore suffices to verify that $f^\wedge \circ \iota : \struct {X, w} \to \Bbb F$ is continuous for each $f \in X^\ast$.
From Characterization of Continuity of Linear Functional in Weak Topology, we can equivalently show that $f^\wedge \circ \iota \in X^\ast$ for each $f \in X^\ast$.
For each $x \in X$ and $f \in X^\ast$, we have: 
Hence we have: 
:$f^\wedge \circ \iota = f \in X^\ast$
and we are done. 
Category:Weak Topologies on Topological Vector Spaces
Category:Weak-* Topologies
Category:Evaluation Linear Transformations (Normed Vector Spaces)
Evaluation Linear Transformation on Normed Vector Space is Weak to Weak-* Homeomorphism onto Image

== Theorem ==
Let $\Bbb F \in \set {\R, \C}$.
Let $X$ be a normed vector space over $\Bbb F$. 
Let $X^\ast$ be the normed dual of $X$. 
Let $X^{\ast \ast}$ be the second normed dual of $X$. 
Let $w$ be the weak topology on $X$.
Let $w^\ast$ be the weak-$\ast$ topology on $X^{\ast \ast}$.
Let $\iota : X \to X^{\ast \ast}$ be the evaluation linear transformation.
Then $\iota : \struct {X, w} \to \struct {\iota X, w^\ast}$ is a homeomorphism.
== Proof ==
From Evaluation Linear Transformation on Normed Vector Space is Weak to Weak-* Continuous Embedding into Second Normed Dual, $\iota$ is continuous.
From Evaluation Linear Transformation on Normed Vector Space is Linear Isometry, $\iota$ is bijective.
It remains to show that $\iota^{-1} : \struct {\iota X, w^\ast} \to \struct {X, w}$ is continuous.
From Continuity in Initial Topology, it is enough to show that: 
:$f \circ \iota^{-1} : \struct {\iota X, w^\ast} \to \Bbb F$ is continuous.
From Characterization of Continuity of Linear Functional in Weak-* Topology‎ and Subspace Topology on Initial Topology is Initial Topology on Restrictions, there exists $g \in X^\ast$ such that: 
:$f \circ \iota^{-1} = g^\wedge \restriction_{\iota X}$
We have, for $x^\wedge \in \iota X$:
:$\map {\paren {f \circ \iota^{-1} } } {x^\wedge} = \map f x$
Now note that: 
:$\map f x = \map {x^\wedge} f = \map {f^\wedge} {x^\wedge}$ 
from the definition of the evaluation linear transformation.
So we have: 
:$f^\wedge \restriction_{\iota X} = f \circ \iota^{-1}$ 
So $f \circ \iota^{-1} : \struct {\iota X, w^\ast} \to \Bbb F$ is continuous.
So by Continuity in Initial Topology $\iota^{-1} : \struct {\iota X, w^\ast} \to \struct {X, w}$ is continuous.
So $\iota : \struct {X, w} \to \struct {\iota X, w^\ast}$ is a homeomorphism.
Category:Evaluation Linear Transformations
Category:Weak Topologies on Topological Vector Spaces
Category:Weak-* Topologies
Evaluation Mapping is Injective iff Mappings Separate Points

== Theorem ==
Let $X$ be a topological space.
Let $\family {Y_i}_{i \mathop \in I}$ be an indexed family of topological spaces for some indexing set $I$.
Let $\family {f_i : X \to Y_i}_{i \mathop \in I}$ be an indexed family of continuous mappings.
Let $\ds Y = \prod_{i \mathop \in I} Y_i$ be the product space of $\family {Y_i}_{i \mathop \in I}$. 
Let $f : X \to Y$ be the evaluation mapping induced by $\family{f_i}_{i \mathop \in I}$.
Then:
:$f$ is an injection $\family {f_i : X \to Y_i}_{i \mathop \in I}$ separates points
== Proof ==
We have:
:$f$ is an injection
Category:Evaluation Mappings (Topological Spaces)
Evaluation Mapping on T1 Space is Embedding if Mappings Separate Points from Closed Sets

== Theorem ==
Let $X$ be a $T_1$ topological space.
Let $\family {Y_i}_{i \mathop \in I}$ be an indexed family of topological spaces for some indexing set $I$.
Let $\family {f_i : X \to Y_i}_{i \mathop \in I}$ be an indexed family of continuous mappings.
Let $\family {f_i}_{i \mathop \in I}$ separate points from closed sets.
Let $\ds Y = \prod_{i \mathop \in I} Y_i$ be the product space of $\family {Y_i}_{i \mathop \in I}$. 
Let $f : X \to Y$ be the evaluation mapping induced by $\family{f_i}_{i \mathop \in I}$.
Then:
:$f$ is an embedding
== Proof ==
Let $\BB = \set{f_i^{-1} \sqbrk V : i \in I, V \text{ is open in } Y_i}$.
From Preimage of Open Sets forms Basis if Continuous Mappings Separate Points from Closed Sets:
:$\BB$ is a basis for $X$
From Analytic Basis is Analytic Sub-Basis:
:$\BB$ is a sub-basis for $X$
By definition of a $T_1$ space:
:all points of $X$ are closed
Since $\family {f_i}_{i \mathop \in I}$ separate points from closed sets then:
:$\family {f_i}_{i \mathop \in I}$ separates points
From Characterization for Topological Evaluation Mapping to be Embedding:
:the evaluation mapping $f$ is an embedding
== Sources ==
* : Chapter $4$: Embedding and Metrization, $\S$Embedding in Cubes, $5$ Embedding Lemma
* : Chapter $3$: New Space from Old: $\S8$: Product Spaces, Weak Topologies: Theorem $8.16$
Category:Continuous Mappings
Category:Embeddings (Topology)
Evaluation of Integral using Laplace Transform

== Theorem ==
Let $\laptrans {\map f t} = \map F s$ denote the Laplace transform of the real function $f$.
Then:
:$\ds \int_0^{\to \infty} \map f t \rd t = \map F 0$
assuming the integral is convergent.
== Proof ==
By definition of Laplace transform:
:$\ds \int_0^{\to \infty} e^{-s t} \map f t \rd t = \map F s$
The result follows by taking the limit as $s \to 0$.
== Examples ==
== Sources ==
* : Chapter $1$: The Laplace Transform: Evaluation of Integrals
Category:Laplace Transforms
Category:Integral Calculus
Evaluation of Integral using Laplace Transform/Examples/Example 1

== Examples of Use of Evaluation of Integral using Laplace Transform ==
:$\ds \int_0^\infty \dfrac {e^{-t} - e^{-3 t} } t \rd t = \ln 3$
== Proof ==
Let $\map f t = e^{-t} - e^{-3 t}$.
Then from Laplace Transform of Exponential:
:$\laptrans {\map f t} = \dfrac 1 {s + 1} - \dfrac 1 {s + 3}$
Hence:
== Sources ==
* : Chapter $1$: The Laplace Transform: Solved Problems: Evaluation of Integrals: $45 \ \text{(b)}$
Category:Examples of Laplace Transforms
Even Convergent of Simple Continued Fraction is Strictly Smaller than Odd Convergent

== Theorem ==
Let $n \in \N \cup \set \infty$ be an extended natural number.
Let $\sqbrk {a_0, a_1, \ldots}$ be a simple continued fraction in $\R$ of length $n$.
Let $p_0, p_1, p_2, \ldots$ and $q_0, q_1, q_2, \ldots$ be its numerators and denominators.
Let $\sequence {C_0, C_1, \ldots}$ be its sequence of convergents.
Every even convergent is strictly smaller than every odd convergent.
== Proof ==
Let $k \ge 1$.
From Denominators of Simple Continued Fraction are Strictly Positive, $q_k q_{k - 1}>0$.
From Difference between Adjacent Convergents of Simple Continued Fraction:
:$C_k - C_{k - 1} = \dfrac {\paren {-1}^{k + 1} } {q_k q_{k - 1} }$
Let $k$ be even.
Then:
:$\exists s \in \N: k = 2 s$
and:
:$C_{2 s} < C_{2 s - 1}$
Let $k$ be odd.
Then:
:$\exists t \in \N: k = 2 t + 1$
and:
:$C_{2 t + 1 } > C_{2 t}$
Now, consider any even convergent $C_{2 s}$ and any odd convergent $C_{2 t + 1}$, with $s, t \ge 0$.
If $2 s < 2 t + 1$, then $2 s \le 2 t$, so:
If $2 s > 2 t + 1$, then $2 s - 1 \ge 2 t + 1$, so:
In either case:
:$C_{2 s} < C_{2 t + 1}$
Hence the result.
== Also see ==
* Properties of Convergents of Continued Fractions
Category:Simple Continued Fractions
Even Convergents of Simple Continued Fraction are Strictly Increasing

== Theorem ==
Let $n \in \N \cup \set \infty$ be an extended natural number.
Let $\sqbrk {a_0, a_1, \ldots}$ be a simple continued fraction in $\R$ of length $n$.
Let $p_0, p_1, p_2, \ldots$ and $q_0, q_1, q_2, \ldots$ be its numerators and denominators.
Let $\sequence {C_0, C_1, \ldots}$ be its sequence of convergents.
The even convergents satisfy $C_0 < C_2 < C_4 \cdots$.
== Proof ==
Let $k \ge 2$ be an even integer.
From Difference between Adjacent Convergents But One of Simple Continued Fraction:
:$C_k - C_{k - 2} = \dfrac {\paren {-1}^k a_k} {q_k q_{k - 2} } = \dfrac {a_k} {q_k q_{k - 2} }$
By definition of simple continued fraction, $a_k > 0$.
By Convergents of Simple Continued Fraction are Rationals in Canonical Form, $q_k > 0$ and $q_{k - 2} > 0$.
Thus $C_k > C_{k - 2}$.
== Also see ==
* Odd Convergents of Simple Continued Fraction are Strictly Decreasing
* Properties of Convergents of Continued Fractions
Category:Simple Continued Fractions
Even Derivatives of Cotangent of Pi Z at One Fourth

== Theorem ==
:$ \ds \valueat {\dfrac {\d^{2 n} } {\d z^{2 n} } \cot \pi z} {z \mathop = \frac 1 4} = \paren {-1}^n \paren {2 \pi}^{2 n} E_{2 n}$
where:
:$E_n$ denotes the $n$th Euler number
:$n$ is a non-negative integer.
== Proof ==
=== Lemma ===
== Sources ==
* 
Category:Euler Numbers
Even Derivatives of Cotangent of Pi Z at One Fourth/Lemma

== Even Derivatives of Cotangent of Pi Z at One Fourth: Lemma ==
Let $z \ne \paren {4 n + 1} \dfrac 1 4$
Then: 
:$\ds \map \tan {\pi z + \dfrac \pi 4} = \map \sec {2 \pi z} + \map \tan {2 \pi z}$
where:
:$\sec$ and $\tan$ are secant and tangent respectively.
== Proof ==
Category:Even Derivatives of Cotangent of Pi Z at One Fourth
Even Function Times Even Function is Even

== Theorem ==
Let $X \subset \R$ be a symmetric set of real numbers:
:$\forall x \in X: -x \in X$
Let $f, g: X \to \R$ be two even functions.
Let $f \cdot g$ denote the pointwise product of $f$ and $g$.
Then $\paren {f \cdot g}: X \to \R$ is also an even function.
== Proof ==
Thus, by definition, $\paren {f \cdot g}$ is an even function.
== Also see ==
* Odd Function Times Even Function is Odd
* Odd Function Times Odd Function is Even
Category:Even Functions