nyuuzyou/wikis huggingface dataset:
Ampère's Law with Maxwell's Addition/Also presented as

== Ampère's Law with Maxwell's Addition: Also presented as == 
Ampère's Law with Maxwell's Addition can also be seen presented as:
:$\nabla \times \mathbf H = \mathbf J + \dfrac {\partial \mathbf D} {\partial t}$
where:
:$\mathbf D = \varepsilon_0 \mathbf E$ denotes the electric displacement field
:$\mathbf H = \dfrac 1 {\mu_0} \mathbf B$ denotes the magnetic field strength
== Sources ==
* 
* 
Category:Ampère's Law with Maxwell's Addition
Ampère-Maxwell Law

== Theorem ==
Let $\mathbf B$ be a magnetic field due to a steady current $I$ flowing through a wire $s$.
Then:
:$\ds \oint \mathbf B \cdot \rd \mathbf l = \mu_0 I$
where:
:the line integral is taken around a closed path
:$\d \mathbf l$ is an infinitesimal vector associated with $s$
:$\mu_0$ denotes the vacuum permeability.
That is, the line integral of $\mathbf B$ through the area enclosed by the closed path is equal to $\mu_0 I$.
== Proof ==
== Also known as ==
== Also see ==
* Ampère's Law with Maxwell's Addition
* Ampère's Force Law
Category:Ampère-Maxwell Law
Category:Electromagnetism
Analog between Logic and Set Theory

== Theorem ==
The concepts of set theory have directly corresponding concepts in logic:
:
Set Theory 	Logic
Set: $S, T$ 	Statement: $p, q$
Union: $S \cup T$ 	Disjunction: $p \lor q$
Intersection: $S \cap T$ 	Conjunction: $p \land q$
Subset: $S \subseteq T$ 	Conditional: $p \implies q$
Symmetric Difference: $S \symdif T$	Exclusive Or: $p \oplus q$
Complement: $\relcomp {} S$ 	Logical Not: $\lnot p$
Set Equality: $S = T$ 	Biconditional: $p \iff q$
Venn Diagram 	Truth Table
== Proof ==
Let $P$ and $Q$ be propositional functions.
Let $S$ and $T$ be subsets of a universe $\Bbb U$ such that:
:$S = \set {x \in \Bbb U: \map P x}$
:$T = \set {x \in \Bbb U: \map Q x}$
By the following definitions:
== Sources ==
* : Chapter $1$: Elementary, my dear Watson: $\S 1.2$: Elements, my dear Watson
Category:Set Theory
Category:Logic
Analogue Formula for Spherical Law of Cosines

== Theorem ==
Let $\triangle ABC$ be a spherical triangle on the surface of a sphere whose center is $O$.
Let the sides $a, b, c$ of $\triangle ABC$ be measured by the angles subtended at $O$, where $a, b, c$ are opposite $A, B, C$ respectively.
Then:
=== Corollary ===
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Sources ==
* : Chapter $\text I$. Spherical Trigonometry: $7$. The analogue formula.
Category:Spherical Trigonometry
Category:Spherical Law of Cosines
Category:Analogue Formula for Spherical Law of Cosines
Analogue Formula for Spherical Law of Cosines/Corollary

== Theorem ==
Let $\triangle ABC$ be a spherical triangle on the surface of a sphere whose center is $O$.
Let the sides $a, b, c$ of $\triangle ABC$ be measured by the angles subtended at $O$, where $a, b, c$ are opposite $A, B, C$ respectively.
Then:
== Proof ==
Let $\triangle A'B'C'$ be the polar triangle of $\triangle ABC$.
Let the sides $a', b', c'$ of $\triangle A'B'C'$ be opposite $A', B', C'$ respectively.
From Spherical Triangle is Polar Triangle of its Polar Triangle we have that:
:not only is $\triangle A'B'C'$ be the polar triangle of $\triangle ABC$
:but also $\triangle ABC$ is the polar triangle of $\triangle A'B'C'$.
We have:
and:
== Sources ==
* : Chapter $\text I$. Spherical Trigonometry: $11$. Polar formulae.
Category:Analogue Formula for Spherical Law of Cosines
Analogue Formula for Spherical Law of Cosines/Proof 2

== Theorem ==
== Proof ==
:
Suppose $c$ is less than $\dfrac \pi 2$.
Let $BA$ be produced to $D$ so that $BD = \dfrac \pi 2$.
Then:
:$AD = \dfrac \pi 2 - c$
and:
:$\angle CAD = pi - A$
Let $C$ and $D$ be joined by an arc of a great circle, denoted $x$.
From the triangle $\sphericalangle DAC$, using the Spherical Law of Cosines:
From the triangle $\sphericalangle DBC$, using the Spherical Law of Cosines:
Hence:
:$\sin a \cos B = \sin c \cos b - \cos c \sin b \cos A$
The case where $c > \dfrac \pi 2$ is worked similarly, but by making $D$ the point between $A$ and $B$ such that $BD$ is $\dfrac \pi 2$.
== Sources ==
* : Chapter $\text I$. Spherical Trigonometry: $7$. The analogue formula.
Category:Analogue Formula for Spherical Law of Cosines
Analogue Formula for Spherical Law of Cosines/Proof 3

== Theorem ==
== Proof ==
:
Let $A$, $B$ and $C$ be the vertices of a spherical triangle on the surface of a sphere $S$.
By definition of a spherical triangle, $AB$, $BC$ and $AC$ are arcs of great circles on $S$.
By definition of a great circle, the center of each of these great circles is $O$.
Let $O$ be joined to each of $A$, $B$ and $C$.
Let $P$ be an arbitrary point on $OC$.
Construct $PQ$ perpendicular to $OA$ meeting $OA$ at $Q$.
Construct $PR$ perpendicular to $OB$ meeting $OB$ at $R$.
In the plane $OAB$:
:construct $QS$ perpendicular to $OA$
:construct $RS$ perpendicular to $OB$
where $S$ is the point where $QS$ and $RS$ intersect.
Let $OS$ and $PS$ be joined.
Let tangents be constructed at $A$ to the arcs of the great circles $AC$ and $AB$.
These tangents contain the spherical angle $A$.
But by construction, $QS$ and $QP$ are parallel to these tangents.
Hence $\angle PQS = \sphericalangle A$.
Similarly, $\angle PRS = \sphericalangle B$.
Also we have:
It is to be proved that $PS$ is perpendicular to the plane $AOB$.
By construction, $OQ$ is perpendicular to both $PQ$ and $QS$.
Thus $OQ$ is perpendicular to the plane $PQS$.
Similarly, $OR$ is perpendicular to the plane $PRS$.
Thus $PS$ is perpendicular to both $OQ$ and $OR$.
Thus $PS$ is perpendicular to every line in the plane of $OQ$ and $OR$.
That is, $PS$ is perpendicular to the plane $OAB$.
In particular, $PS$ is perpendicular to $OS$, $SQ$ and $SR$
It follows that $\triangle PQS$ and $\triangle PRS$ are right triangles.
From the right triangles $\triangle OQP$ and $\triangle ORP$, we have:
Let us denote the angle $\angle SOQ$ by $x$.
Then:
:$\angle ROS = c - x$
From the right triangles $\triangle OSQ$ and $\triangle OSR$, we have:
We have:
and:
and:
from which it follows that:
:$\sin a \cos B = \cos b \sin c - \sin b \cos c \cos A$
The result follows.
== Sources ==
* : Chapter $\text I$. Spherical Trigonometry: $9$. Alternative proofs of the formulae $\mathbf A$, $\mathbf B$ and $\mathbf C$.
Category:Analogue Formula for Spherical Law of Cosines
Analysis of Card Game with Bluffing

== Analysis of Card Game with Bluffing ==
The card game with bluffing is analysed as follows.
First we recall the game mechanics:
=== Strategies of $A$ ===
If $A$ receives the high card, the bid is compulsory.
If $A$ receives the low card, the bid is optional.
Let $A_1$ be the strategy that $A$ pays $1$ credit.
Let $A_2$ be the strategy that $A$ bids $2$ credits.
=== Strategies of $B$ ===
If $A$ pays, $B$ has no option but to receive $1$ credit.
If $A$ bids, then $B$ has $2$ strategies.
Let $B_1$ be the strategy that $B$ challenges $A$'s bid.
Let $B_2$ be the strategy that $B$ pays $1$ credit.
=== The outcome ===
There are $4$ possible combinations of strategies:
:$\left({A_1, B_1}\right)$:
If $A$ had the low card, $A$ pays $1$ credit to $B$.
If $A$ had the high card, $B$ challenges $A$'s bid and so pays $2$ credits to $A$.
There is a $50 \%$ chance of either one happening.
The mean payoff to $A$ is therefore $\dfrac {2 - 1} 2 = \dfrac 1 2$ credit.
:$\left({A_1, B_2}\right)$:
If $A$ had the low card, $A$ pays $1$ credit to $B$.
If $A$ had the high card, $B$ pays $1$ credit to $A$.
There is a $50 \%$ chance of either one happening.
The mean payoff to $A$ is therefore $\dfrac {1 - 1} 2 = 0$ credits.
:$\left({A_2, B_1}\right)$:
If $A$ had the low card, $A$ bids $2$ credits, $B$ challenges $A$'s bid and wins, and so $A$ pays $2$ credits to $B$.
If $A$ had the high card, $A$ bids $2$ credits, $B$ challenges $A$'s bid and loses, and so $B$ pays $2$ credits to $A$.
There is a $50 \%$ chance of either one happening.
The mean payoff to $A$ is therefore $\dfrac {2 - 2} 2 = 0$ credits.
:$\left({A_2, B_2}\right)$:
If $A$ had the low card, $A$ bids $2$ credits, $B$ accepts $A$'s bid and $B$ pays $1$ credit to $A$.
If $A$ had the high card, $A$ bids $2$ credits, $B$ accepts $A$'s bid and $B$ pays $1$ credit to $A$.
In both cases the payoff to $A$ is $1$ credit.
The mean payoff to $A$ is therefore $1$ credit.
== Sources ==
* : Chapter $\text{I}$: An Outline of the Theory of Games: $2$
Category:Examples of Games
Analysis of Prisoner's Dilemma

== Analysis of the Prisoner's Dilemma ==
== Analysis ==
Prisoner $A$ reasons as follows:
:$B$ will either confess or not.
:If he confesses, then if I also confess, I get $3$ years rather than $4$.
:On the other hand, if $B$ keeps silent, then if I confess I go free, rather than serve $1$ year.
:Therefore, whatever $B$ does, whether that be to confess or to keep silent, it is in my best interests to confess.
By using the same line of reasoning, prisoner $B$ also reaches the conclusion that it is also in his best interests to confess.
It follows that each prisoner will serve $3$ years, rather than the $1$ they will serve by both keeping silent.
Thus it is seen that by using careful and logical reasoning, both prisoners settle on a strategy which does not yield the best outcome for either of them.
== Also see ==
== Historical Note ==
== Sources ==
* 
Category:Prisoner's Dilemma
Analytic Continuation of Dirichlet L-Function

== Theorem ==
Let $\chi : G := \paren {\Z / q \Z}^\times \to \C^\times$ be a Dirichlet character modulo $q$.
Let $\map L {s, \chi}$ be the Dirichlet $L$-function for $\chi$.
Let $\chi$ be the trivial character.
Then $\map L {s, \chi}$ has an analytic continuation to $\C$ except for a simple pole at $s = 1$.
Let $\chi$ be non-trivial.
Then $\map L {s, \chi}$ is analytic on $\map \Re s > 0$.
== Proof ==
Let $\chi$ be the trivial character.
Then by Dirichlet L-Function from Trivial Character:
:$\ds \map L {s, \chi} = \map \zeta s \cdot \prod_{p \mathop \divides q} \paren {1 - p^{-s} }$
where $\divides$ denotes divisibility.
Also, by Poles of Riemann Zeta Function, $\zeta$ is analytic on $\C$ except for a simple pole at $s = 1$.
Since $\map L {s, \chi}$ is just $\zeta$ times some constant, the same holds for this function.
If $\chi$ is non-trivial, then by the Orthogonality Relations for Characters:
:$\ds \sum_{x \mathop \in G} \map \chi x = 0$
By definition, $\chi$ is $q$-periodic, and zero on integers not coprime to $q$.
So for any $M \in \N$:
:$\ds \sum_{n \mathop = M + 1}^{M + Q} \map \chi n = 0$
Let $M, N \in \N$ be arbitrary.
Let $d$ be such that $M + q d \le N \le M + q \paren {d + 1}$.
Then:
So the coefficients $\map \chi n$ of $\map L {s, \chi}$ have bounded partial sums.
Therefore, by Convergence of Dirichlet Series with Bounded Partial Sums, $\map L {s, \chi}$ converges locally uniformly to an analytic function on $\map \Re s > 0$.
Category:Analytic Number Theory
Analytic Continuation of Generating Function of Dirichlet Series

== Theorem ==
Let $\ds \map \lambda s = \sum_{n \mathop = 1}^\infty \frac {a_n} {n^s}$ be a Dirichlet series 
Let $c \in \R$ be greater than the abscissa of absolute convergence of $\lambda$ and greater than $0$.
Let $\ds \map g z = \sum_{k \mathop = 1}^\infty \map \lambda {c k} z^k $ be the generating function of $\map \lambda {c k}$
Then the analytic continuation of $g$ to $\C$ is equal to:
:$\ds \sum_{n \mathop = 1}^\infty a_n \frac z {n^c - z}$
== Proof ==
We will first show that the series is meromorphic functions on $\C$ with simple poles at $n^c$.
For $\cmod z < R$, pick $M$ such that:
:$M^c > 2 R$
This is always possible, as:
:$c > 0$
Then for $n > M$: 
Therefore, for $\cmod z \le R$, pick $M$ as described above.
For all $K > M$:
Because $c$ is chosen to be greater than the abscissa of absolute convergence of $\lambda$, the above series converges.
Therefore, for all $\epsilon > 0$ we can pick $N$ such that for $K > N$:
Because the summands are meromorphic functions, the series is an absolutely uniformly convergent series of analytic functions on the region $\cmod z < R, z \ne k^c$ for all $k \in \N$.
By Uniform Limit of Analytic Functions is Analytic, the series converges to an analytic function on the domain $z \in \C$:
:$z \ne k^c$ 
for all $k \in \N$. 
Also, each $k^c$ is a pole at exactly one summand, where it is a simple pole.
Therefore, the series has simple poles at each $k^c$, meaning that the function is meromorphic on $\C$.
We now show that the generating function is equal to the series for $\cmod z < 1$:
By Fubini-Tonelli Theorem, if:
:$\ds \sum_{n \mathop = 1}^\infty \sum_{k \mathop = 1}^\infty \cmod {a_n \paren {\frac z {n^c} }^k}$
converges, then:
:$\ds \sum_{k \mathop = 1}^\infty \sum_{n \mathop = 1}^\infty a_n \paren {\frac z {n^c} }^k = \sum_{n \mathop = 1}^\infty \sum_{k \mathop = 1}^\infty a_n \paren {\frac z {n^c} }^k$
We see that for $\cmod z < 1$: 
The summands of $(2)$ are equal to the summands of:
:$\ds \sum_{n \mathop = 1}^\infty \cmod {a_n} \frac {\cmod z} {\cmod {n^c - z} }$
for all but finitely n.
By $(1)$, the above series converges, and thus $(2)$ as well.
Thus we may write: 
Thus the generating function is equal to the series on the unit circle.
Thus by Definition:Analytic Continuation, the series is the unique analytic continuation of the generating function. 
Category:Dirichlet Series
Analytic Continuation of Riemann Zeta Function

== Theorem ==
The Riemann zeta function is meromorphic on $\C$.
== Proof ==
The (yet to be confirmed) meromorphic continuation of the Riemann zeta function to the half-plane $\set {s: \map \Re s > 0}$ is given by Integral Representation of Riemann Zeta Function in terms of Fractional Part:
:$(1): \quad \ds \map \zeta s = \frac s {s - 1} - s \int_1^\infty \fractpart x x^{-s - 1} \rd x$
where $\fractpart x$ is the fractional part of $x$.
Let $\map \Re s \le 0$.
Then the value of $\map \zeta s$ can be computed from the relation:
:$\map \Gamma {\dfrac s 2} \pi^{-s/2} \map \zeta s = \map \Gamma {\dfrac {s - 1} 2} \pi^{\frac {1 - s} 2} \map \zeta {1 - s}$
That is:
:$\map \xi s = \map \xi {1 - s}$
where $\xi$ is the completed zeta function.
First we show that $(1)$ is analytic for $\map \Re s > 0$.
For $n \ge 1$, let:
Here $\ll$ is the order notation.
By the Mean Value Theorem, for some $n \le \theta \le n + 1$:
:$\paren {n + 1}^s - n^s = s \theta^{s - 1} \le s \paren {n + 1}^{s-1}$
Thus if $s = \sigma + i t$:
:$\size {a_n} \le \size {\dfrac s {n^{s + 1} } } = \dfrac {\sigma^2 + t^2} {n^{\sigma + 1} }$
Since:
:$\ds \map \zeta s = \frac s {s - 1} - \sum_{n \mathop \ge 1} a_n$
it follows that this representation converges absolutely uniformly on $\map \Re s > 0$.
Thus by Uniform Limit of Analytic Functions is Analytic $\map \zeta s$ is analytic for $\map \Re s > 0$ and $s \ne 1$.
For all $s$ with $\map \Re s < \dfrac 1 2$, $\map \zeta s$ is simply the reflection of $\zeta$ in the upper half-plane.
Therefore, $\zeta$ is also analytic for all $s$ with $\map \Re s < 0$.
Category:Riemann Zeta Function
Analytic Continuation of Riemann Zeta Function using Dirichlet Eta Function

== Theorem ==
Let $\zeta$ be the Riemann zeta function.
Let $\eta$ be the Dirichlet Eta Function.
Then:
:$\dfrac 1 {1 - 2^{1 - s} } \map \eta s$
defines an analytic continuation of $\zeta$ to the half-plane $\map \Re s > 0$ minus $s = 1$.
== Proof ==
By Riemann Zeta Function in terms of Dirichlet Eta Function, it coincides with $\zeta$ for $\map \Re s > 1$.
By Dirichlet Eta Function is Analytic, it is analytic for $\map \Re s > 0$, except at $s = 1$.
== Also see ==
* Analytic Continuations of Riemann Zeta Function to Right Half-Plane
Category:Riemann Zeta Function
Category:Dirichlet Eta Function
Analytic Continuation of Riemann Zeta Function using Jacobi Theta Function

== Theorem ==
Let $\zeta$ be the Riemann zeta function.
Then
:$\ds \frac {\pi^{s / 2} } {\map \Gamma {\frac s 2}} \cdot \paren {-\frac 1 {s \paren {1 - s} } + \int_1^\infty \paren {x^{s / 2 - 1} + x^{-\paren {s + 1} / 2} } \map \omega x \rd x}$
defines an analytic continuation of $\zeta$ to the half-plane $\map \Re s > 0$ minus $s = 1$.
== Proof ==
By Integral Representation of Riemann Zeta Function in terms of Jacobi Theta Function, it coincides with $\map \zeta s$ for $\map \Re s > 1$.
Interchanging integral and derivative, one shows that the integral is analytic for $\map \Re s > 0$.
== Also see ==
* Analytic Continuations of Riemann Zeta Function to Right Half-Plane
Category:Riemann Zeta Function
Analytic Continuation of Riemann Zeta Function using Mellin Transform of Fractional Part

== Theorem ==
Let $\zeta$ denote the Riemann zeta function.
The analytic continuation of $\zeta$ to the half-plane $\map \Re s > 0$ is given by:
:$\ds \frac s {s - 1} - s \int_1^\infty \fractpart x x^{-s - 1} \rd x$
where:
:$x^{-s - 1}$ takes the principal value $e^{-\map \ln x \paren {s + 1} }$
:$\fractpart x$ is the fractional part of $x$.
== Proof ==
Let $s = \sigma + i t$.
By Integral Representation of Riemann Zeta Function in terms of Fractional Part, the above integral coincides with $\map \zeta s$ for $\sigma > 1$.
We show that it is analytic for $0 < \sigma < 1$.
For $n \ge 1$, let:
By the Mean Value Theorem, for some $n \le \theta \le n + 1$:
:$\paren {n + 1}^\sigma - n^\sigma = \sigma \theta^{\sigma - 1}$
and for $\sigma \le 1$ we have:
:$\sigma \theta^{\sigma - 1} \le \sigma n^{\sigma - 1}$
Thus we have:
Thus on any compact subset of the half-plane $\sigma > 0$ which is bounded away from $0$, we have that the series:
:$\ds \frac s {s - 1} - \sum_{n \mathop = 1}^\infty a_n = \frac s {s - 1} - s \int_1^\infty \fractpart x x^{-s - 1} \rd x$
converges uniformly as:
Since it is a compact region, $\cmod s$ is bounded by some constant $C$.
So:
:$\ds \cmod {\sum_{n \mathop = M}^\infty a_n} \le C \sum_{n \mathop = M}^\infty \frac 1 {n^{\sigma + 1} }$
As Dirichlet series are uniformly convergent on compact regions bounded away from their abscissa of convergence, we may find an $M$ large enough so that the above is smaller than $\epsilon$ for all $\sigma$.
Therefore the series is uniformly convergent.
Thus by Uniform Limit of Analytic Functions is Analytic the series defines an analytic function on the strip $0 < \sigma < 1$.
This is also equal to $\zeta$ on the half-plane $\sigma>1$.
Thus the series defines the unique analytic continuation of $\zeta$ onto the half-plane $\sigma > 0$.
== Also see ==
* Analytic Continuations of Riemann Zeta Function to Right Half-Plane
Category:Riemann Zeta Function
Analytic Continuations of Riemann Zeta Function to Complex Plane

== Theorem ==
The Riemann zeta function $\zeta$ has a unique analytic continuation to $\C \setminus \set 1$.
== Proof ==
Note that by Riemann Zeta Function is Analytic, $\map \zeta s$ is indeed analytic for $\map \Re s > 1$.
By Complex Plane minus Point is Connected, $\C \setminus \set 1$ is connected.
By Uniqueness of Analytic Continuation, there is at most one analytic continuation of $\zeta$ to $\C \setminus \set 1$.
* Functional Equation for Riemann Zeta Function, analyticity is shown at Analytic Continuation of Riemann Zeta Function
* The stepwise method as in Riemann Zeta Function in terms of Dirichlet Eta Function
== Also see ==
* Analytic Continuations of Riemann Zeta Function
== Sources ==
* 
* 
Category:Riemann Zeta Function
Analytic Continuations of Riemann Zeta Function to Right Half-Plane

== Theorem ==
The Riemann zeta function has a unique analytic continuation to $\set {s \in \C : \map \Re s > 0} \setminus \set 1$, the half-plane $\map \Re s > 0$ minus the point $s = 1$.
== Proof ==
Note that by Riemann Zeta Function is Analytic, $\map \zeta s$ is indeed analytic for $\map \Re s > 1$.
By Complex Half-Plane minus Point is Connected, $\set {\sigma > 0} \setminus \set 1$ is connected.
By Uniqueness of Analytic Continuation, there is at most one analytic continuation of $\zeta$ to $\set {\sigma > 0} \setminus \set 1$.
By either:
* Analytic Continuation of Riemann Zeta Function using Dirichlet Eta Function
* Analytic Continuation of Riemann Zeta Function using Mellin Transform of Fractional Part
* Analytic Continuation of Riemann Zeta Function using Jacobi Theta Function
there exists one.
== Also see ==
* Analytic Continuations of Riemann Zeta Function
Category:Riemann Zeta Function
Analytic Function on Banach Space is Continuous

== Theorem ==
Let $U$ be an open subset of $\C$.
Let $\struct {X, \norm {\, \cdot \,} }$ be a Banach space over $\C$.
Let $f : U \to X$ be an analytic function.
Then $f$ is continuous.
== Proof ==
Let $x \in U$. 
Since $f : U \to X$ is analytic function, the limit:
:$\ds \lim_{y \mathop \to x} \frac {\map f y - \map f x} {y - x} = \map {f'} x$ exists.
We have, from :
:$\ds \norm {\map f y - \map f x} = \cmod {y - x} \norm {\frac {\map f y - \map f x} {y - x} }$
From Modulus of Limit: Normed Vector Space, we have:
:$\ds \lim_{y \mathop \to x} \norm {\frac {\map f y - \map f x} {y - x} } = \norm {\map {f'} x}$
We also have:
:$\ds \lim_{y \mathop \to x} \cmod {y - x} = 0$
So from Product Rule for Complex Sequences, we have:
:$\ds \lim_{y \mathop \to x} \norm {\map f y - \map f x} = 0$
So, we have that $f$ is continuous at $x$.
Since $x \in U$ was arbitrary, we have that $f$ is continuous.
Category:Analytic Functions
Category:Banach Spaces
Angle Between Non-Zero Vectors Always Defined

== Theorem ==
Let $\mathbf v$ and $\mathbf w$ be non-zero vectors in real Euclidean space $\R^n$.
Then the angle between $\mathbf v$ and $\mathbf w$ is always defined.
== Proof ==
=== Case 1 ===
Suppose that $\mathbf v$ and $\mathbf w$ are not scalar multiples of each other.
:
From Construction of Triangle from Given Lengths, it is sufficient to show that sum of the lengths of any two sides is greater than the third side.
Consider the side with length $\norm {\mathbf v}$.
From the triangle inequality for vectors:
Note that the equality is a strict inequality because the vectors are not scalar multiples of each other.
Consider the side with length $\norm {\mathbf w}$.
Lastly, Consider the side with length $\norm {\mathbf v - \mathbf w}$.
=== Case 2 ===
Suppose that $\mathbf v$ and $\mathbf w$ are scalar multiples of each other.
Then the existence of $\theta$ follows directly from the definition of the angle between vectors that are scalar multiples of each other.
== Sources ==
* 
Category:Vector Algebra
Category:Analytic Geometry
Angle Between Two Straight Lines described by Homogeneous Quadratic Equation

== Theorem ==
Let $\LL_1$ and $\LL_2$ represent $2$ straight lines in the Cartesian plane which are represented by a homogeneous quadratic equation $E$ in two variables:
:$E: \quad a x^2 + 2 h x y + b y^2$
Then the angle $\psi$ between $\LL_1$ and $\LL_2$ is given by:
:$\tan \psi = \dfrac {2 \sqrt {h^2 - a b} } {a + b}$
== Proof ==
Let us rewrite $E$ as follows:
:$b y^2 + 2 h x y + a x^2 = b \paren {y - \mu_1 x} \paren {y - \mu_2 x}$
Thus from Homogeneous Quadratic Equation represents Two Straight Lines through Origin:
:$\LL_1$ and $\LL_2$ are represented by the equations $y = \mu_1 x$ and $y = \mu_2 x$ respectively.
From Sum of Roots of Quadratic Equation:
:$\mu_1 + \mu_2 = -\dfrac {2 h} b$
From Product of Roots of Quadratic Equation:
:$\mu_1 \mu_2 = \dfrac a b$
From Angle between Straight Lines in Plane:
:$\tan \psi = \dfrac {\mu_1 - \mu_2} {1 + \mu_1 \mu_2}$
We have that:
== Sources ==
* : Chapter $\text {II}$. The Straight Line: $16$. Angle between the two straight lines represented by the homogeneous equation
Category:Straight Lines