nyuuzyou/wikis huggingface dataset:
Orthogonal Latin Squares of Order 5

== Problem ==
How can two orthogonal Latin squares of order $5$ be constructed?
== Solution ==
The elements of the first Latin square are shown in normal type, while those of the second Latin square are shown in italic type:
$\begin{array} {|c|c|c|c|c|}
\hline 1 \mathit 1 & 2 \mathit 3 & 3 \mathit 5 & 4 \mathit 2 & 5 \mathit 4 \\
\hline 3 \mathit 2 & 4 \mathit 4 & 5 \mathit 1 & 1 \mathit 3 & 2 \mathit 5 \\
\hline 5 \mathit 3 & 1 \mathit 5 & 2 \mathit 2 & 3 \mathit 4 & 4 \mathit 1 \\
\hline 2 \mathit 4 & 3 \mathit 1 & 4 \mathit 3 & 5 \mathit 5 & 1 \mathit 2 \\
\hline 4 \mathit 5 & 5 \mathit 2 & 1 \mathit 4 & 2 \mathit 1 & 3 \mathit 3 \\
\hline \end{array}$
== Sources ==
* : The Thirty-six Officers Problem: $135$
Category:Latin Squares
Orthogonal Latin Squares of Order 5/Mistake

== Source Work ==
: 
:The Puzzles:
::The Thirty-six Officers Problem
:::$135$
== Mistake ==
:How can five each of As, Bs, Cs, Ds and Es be placed in these cells so that no letter is repeated in any row or column?
$\begin{array} {|c|c|c|c|c|}
\hline \ \ & \ \ & \ \ & \ \ & \ \ \\
\hline & & & & \\
\hline & & & & \\
\hline & & & & \\
\hline & & & & \\
\hline \end{array}$
== Correction ==
There is more to it than just arranging the letters so that no letter is repeated in any row or column.
We need to place $2$ letters in each space so that no letter of either pair is repeated in any row or column.
Otherwise we just get something like this boring arrangement:
$\begin{array} {|c|c|c|c|c|}
\hline A & B & C & D & E \\
\hline B & C & D & E & A \\
\hline C & D & E & A & B \\
\hline D & E & A & B & C \\
\hline E & A & B & C & D \\
\hline \end{array}$
which is not what is being asked.
== Sources ==
* : The Thirty-six Officers Problem: $135$
Category:Mistakes/Curious and Interesting Puzzles
Orthogonal Latin Squares of Order 6 do not Exist/Historical Note

== Historical Note on Orthogonal Latin Squares of Order 6 do not Exist ==
This problem was posed by , who couched it as follows:
:Place $36$ officers,
:comprising a colonel, lieutenant-colonel, major, captain, lieutentant and sub-lieutenant
:from each of $6$ regiments,
:in a square array
:so that no rank or regiment will be repeated in any row or column.
This turns out to be impossible to do.
This was not proved until achieved it in $1901$.
== Sources ==
* : The Thirty-six Officers Problem
Category:Latin Squares
Orthogonal Matrix/Examples/2x2 Reflection Matrix

== Example of Orthogonal Matrix ==
The reflection matrix for a plane reflection in the line $y = x \tan \dfrac \theta 2$:
:$\mathbf Q = \begin {pmatrix} \cos \theta & \sin \theta \\ \sin \theta & -\cos \theta \end {pmatrix}$
is an orthogonal matrix.
== Proof ==
From Matrix Equation of Plane Reflection we have that $\mathbf Q$ is indeed the reflection matrix for a plane reflection in the line $y = x \tan \dfrac \theta 2$.
Then we see that:
== Sources ==
* 
* 
Category:Examples of Orthogonal Matrices
Category:Reflection Matrices
Orthogonal Matrix/Examples/2x2 Rotation Matrix

== Example of Orthogonal Matrix ==
The rotation matrix for a plane rotation about the origin through an angle $\theta$:
:$\mathbf P = \begin {pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end {pmatrix}$
is an orthogonal matrix.
== Proof ==
From Matrix Equation of Plane Rotation we have that $\mathbf P$ is indeed the rotation matrix for a plane rotation about the origin through an angle $\theta$.
Then we see that:
== Sources ==
* 
* 
Category:Examples of Orthogonal Matrices
Category:Rotation Matrices
Orthogonal Projection is Bounded

== Theorem ==
Let $H$ be a Hilbert space with inner product $\innerprod \cdot \cdot$ and inner product norm $\norm \cdot$.
Let $K$ be a closed linear subspace of $H$.
Let $P_K$ denote the orthogonal projection on $K$.
Then $P_K$ is bounded. 
That is: 
:$\norm {\map {P_K} h} \le \norm h$
for each $h \in H$. 
== Proof ==
Let $h \in H$. 
Note that we can write: 
:$h = \paren {h - \map {P_K} h} + \map {P_K} h$ 
We have, by the definition of orthogonal projection: 
:$\map {P_K} h \in K$
From Unique Point of Minimal Distance to Closed Convex Subset of Hilbert Space, we have: 
:$h - \map {P_K} h \in K^\bot$ 
so that: 
:$\innerprod {\map {P_K} h} {h - \map {P_K} h} = 0$
By Pythagoras's Theorem (Inner Product Space), we therefore have: 
:${\norm h}^2 = \norm {h - \map {P_K} h}^2 + \norm {\map {P_K} h}^2$ 
We have that: 
:$\norm {h - \map {P_K} h}^2 \ge 0$
so:
:${\norm h}^2 \ge \norm {\map {P_K} h}^2$
giving: 
:$\norm {\map {P_K} h} \le \norm h$ 
== Sources ==
* : $\text{I}$ Hilbert Spaces: $\S 2.$ Orthogonality: Theorem $2.7 \text {(b)}$
Category:Orthogonal Projections
Orthogonal Projection is Linear Transformation

== Theorem ==
Let $\GF \in \set {\R, \C}$.
Let $H$ be a Hilbert space over $\mathbb F$ with inner product $\innerprod \cdot \cdot$.
Let $K$ be a closed linear subspace of $H$.
Let $P_K$ denote the orthogonal projection on $K$.
Then $P_K$ is a linear transformation on $H$.
== Proof ==
Let $x, y \in H$. 
Let $\alpha, \beta \in \GF$. 
Let $k \in \KK$. 
Since the inner product is linear in its first argument, we have: 
:$\innerprod {\paren {\alpha x + \beta y} - \paren {\alpha \map {P_K} x + \beta \map {P_K} y} } k = \alpha \innerprod {x - \map {P_K} x} k + \beta \innerprod {y - \map {P_K} y} k$
From Unique Point of Minimal Distance to Closed Convex Subset of Hilbert Space, we have: 
:$\alpha \innerprod {x - \map {P_K} x} k = 0$
and:
:$\beta \innerprod {y - \map {P_K} y} k = 0$
Since $k$ was arbitrary, we have: 
:$\innerprod {\paren {\alpha x + \beta y} - \paren {\alpha \map {P_K} x + \beta \map {P_K} y} } k = 0$
for all $k \in K$. 
Note that from Unique Point of Minimal Distance to Closed Convex Subset of Hilbert Space: 
:$\map {P_K} {\alpha x + \beta y}$ is the unique point in $H$ such that $\innerprod {\paren {\alpha x + \beta y} - \map {P_K} {\alpha x + \beta y} } k = 0$ for each $k \in K$. 
So we therefore have: 
:$\alpha \map {P_K} x + \beta \map {P_K} y = \map {P_K} {\alpha x + \beta y}$
for each $x, y \in H$ and $\alpha, \beta \in \mathbb F$.
So $P_K$ is a linear transformation, as required.
== Sources ==
* : $\text{I}$ Hilbert Spaces: $\S 2.$ Orthogonality: Theorem $2.7 \text{(a)}$
Category:Orthogonal Projections
Orthogonal Projection is Mapping

== Theorem ==
Let $H$ be a Hilbert space.
Let $K$ be a closed linear subspace of $H$.
Let $P_K: H \to H$ be the orthogonal projection on $K$.
Then $P_K$ is a mapping.
== Proof ==
For $P_K$ to be a mapping we need to show that:
:$\forall h \in H: \map{P_K} h$ exists and is unique
By definition of $\map{P_K} h$, this amounts to:
:There is a unique $k \in K$ such that $\norm{ h - k } = \map d {h, K}$
This is precisely the statement of Unique Point of Minimal Distance to Closed Convex Subset of Hilbert Space.
== Sources ==
* : $\text{I}$ Hilbert Spaces: $\S 2.$ Orthogonality
Category:Orthogonal Projections
Orthogonal Projection is Projection

== Theorem ==
Let $\HH$ be a Hilbert space.
Let $K$ be a closed linear subspace of $H$.
Let $P_K$ denote the orthogonal projection onto $K$.
Then $P_K$ is a projection.
== Proof ==
Let $h \in H$.
From the definition of the orthogonal projection, we have:
:$\map {P_K} h \in K$
So, from Fixed Points of Orthogonal Projection, we have: 
:$\map {P_K^2} h = \map {P_K} {\map {P_K} h} = \map {P_K} h$
Since $h$ was arbitrary, we have:
:$P_K^2 = P_K$
So $P_K$ is an idempotent.
Further, from Kernel of Orthogonal Projection, we have:
:$\ker {P_K} = K^\bot$
where $K^\bot$ is the orthocomplement of $K$.
From Range of Orthogonal Projection, we have:
:$K = \Img {P_K}$
Hence we conclude:
:$\ker {P_K} = \paren {\Img {P_K} }^\bot$
Hence $P_K$ is a projection.
== Sources ==
* : $\text{I}$ Hilbert Spaces: $\S 2.$ Orthogonality: Theorem $2.7 \text{(c)}$
Category:Orthogonal Projections
Orthogonal Projection onto Closed Linear Span

== Theorem ==
Let $H$ be a Hilbert space with inner product $\innerprod \cdot \cdot$ and inner product norm $\norm \cdot$. 
Let $E = \set {e_1, \ldots, e_n}$ be an orthonormal subset of $H$.
Let $M = \vee E$, where $\vee E$ is the closed linear span of $E$. 
Let $P$ be the orthogonal projection onto $M$.
Then:
:$\forall h \in H: P h = \ds \sum_{k \mathop = 1}^n \innerprod h {e_k} e_k$
== Proof ==
Let $h \in H$. 
Let: 
:$\ds u = \sum_{k \mathop = 1}^n \innerprod h {e_k} e_k$ 
We have that:
:$u \in \map \span E$
and from the definition of closed linear span:
:$M = \paren {\map \span E}^-$
We therefore have, by the definition of closure: 
:$u \in M$ 
Let $v = h - u$ 
We want to show that $v \in M^\bot$. 
From Intersection of Orthocomplements is Orthocomplement of Closed Linear Span, it suffices to show that: 
:$v \in E^\bot$
Note that for each $l$ we have: 
:$\innerprod v {e_l} = \innerprod h {e_l} - \innerprod u {e_l}$
since the inner product is linear in its first argument. 
We have: 
so:
:$\innerprod v {e_l} = 0$ 
That is: 
:$v \in E^\bot$
so, by Intersection of Orthocomplements is Orthocomplement of Closed Linear Span, we have: 
:$v \in M^\bot$
We can therefore decompose each $h \in H$ as: 
:$h = u + v$
with $u \in M$ and $v \in M^\bot$. 
So we have: 
for each $h \in H$. 
== Sources ==
* : $\text{I}$ Hilbert Spaces: $\S 4.$ Orthonormal Sets of Vectors and Bases: Proposition $4.7$
Category:Orthogonal Projections
Orthogonal Projection onto Orthocomplement

== Theorem ==
Let $H$ be a Hilbert space.
Let $A$ be a closed linear subspace of $H$.
Let $P_A: H \to H$ be the orthogonal projection onto $A$.
Let $P_{A^\perp}: H \to H$ be the orthogonal projection onto $A^\perp$, the orthocomplement of $A$.
Let $I: H \to H$ be the identity operator on $H$.
Then:
:$P_{A^\perp} = I - P_A$
== Proof ==
Let $h \in H$.
By definition of orthogonal projection, $P_A h = a \in A$ such that $h - a \in A^\perp$.
Then the claim is that for all $h \in H$:
:$\paren {I - P_A } h = h - P_A h$
is in $A^\perp$, and:
:$h - \paren {h - P_A h} \in \paren {A^\perp}^\perp$
That $h - P_A h \in A^\perp$ is part of the definition of $P_A$.
For the other criterion, note:
:$h - \paren {h - P_A h} = P_A h \in A$
By :
:$\paren{ A^\perp }^\perp = A$
== Sources ==
* : $\text{I}$ Hilbert Spaces: $\S 2.$ Orthogonality: Exercise $2$
Category:Orthogonal Projections
Category:Orthocomplements
Orthogonal Series of Legendre Polynomials

== Theorem ==
Let $\map {P_n} x$ denote the Legendre polynomial of order $n$.
Let $f: \Bbb I \to \R$ be a real function where $I$ is the closed real interval $\closedint {-1} 1$.
Then:
where:
:$\ds A_k = \int_{-1}^1 \map f x \map {P_k} x \rd x$
== Proof ==
== Sources ==
* : $\S 25$: Legendre Functions: Orthogonal Series of Legendre Polynomials: $25.27$ and $25.28$
* : $\S 28$: Legendre and Associated Legendre Functions: Orthogonal Series of Legendre Polynomials: $28.27$ and $28.28$
Category:Orthogonality of Legendre Polynomials
Orthogonal Set is Linearly Independent Set

== Theorem ==
Let $\struct {V, \innerprod \cdot \cdot}$ be an inner product space over a real or complex subfield $\mathbb F$.
Let $S \subseteq V$ be an orthogonal set.
Let $\mathbf 0 \notin S$, where $\mathbf 0$ denotes the zero vector of $V$.
Then $S$ is a linearly independent set.
== Proof ==
Let $n \in \N$.
Let $\lambda_1, \ldots, \lambda_n \in \mathbb F$, and let $u_1, \ldots, u_n \in S$ such that:
:$\ds \sum_{i \mathop= 1}^n \lambda_i u_1 = \mathbf 0$
To prove that $S$ is a linearly independent set, we must show that $\lambda_k = 0$ for all $k \in \set { 1, \ldots, n}$.
We calculate:
As $u_k \ne \mathbf 0$ by assumption, it follows by Inner Product Axiom $\text N 3$: Non-Negative Definiteness that $\innerprod {u_k}{u_k} \ne 0$.
From Field has no Proper Zero Divisors, it follows that $0 = \lambda_k$ for all $k \in \set {1, \ldots, n}$.
== Sources ==
* : $\S 4.4$
Category:Orthogonality (Linear Algebra)
Category:Linear Independence
Orthogonal Trajectories/Examples

== Examples of Families of Orthogonal Trajectories ==
=== Concentric Circles ===
=== Circles Tangent to $y$-axis ===
=== Exponential Functions ===
=== Parabolas Tangent to $x$-axis ===
=== Parabolas with Focus at Origin ===
=== Rectangular Hyperbolas ===
=== Cardioids ===
=== Example: $x + C e^{-x}$ ===
Category:Examples of Orthogonal Trajectories
Orthogonal Trajectories/Examples/Cardioids

== Theorem ==
Consider the one-parameter family of curves of cardioids given in polar form as:
:$(1): \quad r = c \paren {1 + \cos \theta}$
Its family of orthogonal trajectories is given by the equation:
:$r = c \paren {1 - \cos \theta}$
:
== Proof ==
We use the technique of formation of ordinary differential equation by elimination.
Differentiating $(1)$ $r$ gives:
Thus from Orthogonal Trajectories of One-Parameter Family of Curves, the family of orthogonal trajectories is given by:
:$r \dfrac {\d \theta} {\d r} = \dfrac {\sin \theta} {1 + \cos \theta}$
So:
Hence the result.
== Sources ==
* : $1$: The Nature of Differential Equations: $\S 3$: Families of Curves. Orthogonal Trajectories: Problem $1 \ \text c$
Category:Examples of Orthogonal Trajectories
Category:Cardioids
Orthogonal Trajectories/Examples/Circles Tangent to Y Axis

== Theorem ==
Consider the one-parameter family of curves:
:$(1): \quad x^2 + y^2 = 2 c x$
which describes the loci of circles tangent to the $y$-axis at the origin.
Its family of orthogonal trajectories is given by the equation:
:$x^2 + y^2 = 2 c y$
which describes the loci of circles tangent to the $x$-axis at the origin.
:
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $1$: The Nature of Differential Equations: $\S 3$: Families of Curves. Orthogonal Trajectories
Category:Examples of Orthogonal Trajectories
Category:Circles
Category:Circles Tangent to Y Axis
Orthogonal Trajectories/Examples/Circles Tangent to Y Axis/Proof 1

== Theorem ==
== Proof ==
We use the technique of formation of ordinary differential equation by elimination.
Differentiating $(1)$ $x$ gives:
:$2 x + 2 y \dfrac {\d y} {\d x} = 2 c$
from which:
:$\dfrac {\d y} {\d x} = \dfrac {y^2 - x^2} {2 x y}$
Thus from Orthogonal Trajectories of One-Parameter Family of Curves, the family of orthogonal trajectories is given by:
:$\dfrac {\d y} {\d x} = \dfrac {2 x y} {x^2 - y^2}$
Let:
:$\map M {x, y} = 2 x y$
:$\map N {x, y} = x^2 - y^2$
Put $t x, t y$ for $x, y$:
Thus both $M$ and $N$ are homogeneous functions of degree $2$.
Thus, by definition, $(1)$ is a homogeneous differential equation.
By Solution to Homogeneous Differential Equation, its solution is:
:$\ds \ln x = \int \frac {\d z} {\map f {1, z} - z} + C$
where:
:$\map f {x, y} = \dfrac {2 x y} {x^2 - y^2}$
Thus:
== Sources ==
* : $\S 2.7$: Homogeneous Equations: Problem $2$
Category:Homogeneous Differential Equations
Category:Circles Tangent to Y Axis
Orthogonal Trajectories/Examples/Circles Tangent to Y Axis/Proof 2

== Theorem ==
== Proof ==
We use the technique of formation of ordinary differential equation by elimination.
Expressing $(1)$ in polar coordinates, we have:
:$(2): \quad r = 2 c \cos \theta$
Differentiating $(1)$ $\theta$ gives:
:$(3): \quad \dfrac {\d r} {\d \theta} = -2 c \sin \theta$
Eliminating $c$ from $(2)$ and $(3)$:
:$r \dfrac {\d \theta} {\d r} = -\dfrac {\cos \theta} {\sin \theta}$
Thus from Orthogonal Trajectories of One-Parameter Family of Curves, the family of orthogonal trajectories is given by:
:$r \dfrac {\d \theta} {\d r} = \dfrac {\sin \theta} {\cos \theta}$
Using the technique of Solution to Separable Differential Equation:
:$\ds \int \frac {\d r} r = \int \dfrac {\cos \theta} {\sin \theta} \rd \theta$
which by Primitive of Reciprocal and various others gives:
:$\ln r = \map \ln {\sin \theta} + \ln 2 c$
or:
:$r = 2 c \sin \theta$
This can be expressed in Cartesian coordinates as:
:$x^2 + y^2 = 2 c y$
Hence the result.
== Sources ==
* : $1$: The Nature of Differential Equations: $\S 3$: Families of Curves. Orthogonal Trajectories
Category:Circles Tangent to Y Axis
Orthogonal Trajectories/Examples/Concentric Circles

== Theorem ==
Consider the one-parameter family of curves:
:$(1): \quad x^2 + y^2 = c$
Its family of orthogonal trajectories is given by the equation:
:$y = c x$
:
== Proof ==
We use the technique of formation of ordinary differential equation by elimination.
Differentiating $(1)$ $x$ gives:
:$2 x + 2 y \dfrac {\d y} {\d x} = 0$
from which:
:$\dfrac {\d y} {\d x} = -\dfrac x y$
Thus from Orthogonal Trajectories of One-Parameter Family of Curves, the family of orthogonal trajectories is given by:
:$\dfrac {\d y} {\d x} = \dfrac y x$
By Solution to Separable Differential Equation:
:$\ds \int \frac {\d y} y = \int \frac {\d x} x$
which by Primitive of Reciprocal gives:
:$\ln y = \ln x + \ln c$
or:
:$y = c x$
Hence the result.
== Sources ==
* : $1$: The Nature of Differential Equations: $\S 3$: Families of Curves. Orthogonal Trajectories
Category:Examples of Orthogonal Trajectories
Category:Circles
Orthogonal Trajectories/Examples/Exponential Functions

== Theorem ==
Consider the one-parameter family of curves of graphs of the exponential function:
:$(1): \quad y = c e^x$
Its family of orthogonal trajectories is given by the equation:
:$y^2 = -2 x + c$
:
== Proof ==
We use the technique of formation of ordinary differential equation by elimination.
Differentiating $(1)$ $x$ gives:
:$\dfrac {\d y} {\d x} = c e^x$
Thus from Orthogonal Trajectories of One-Parameter Family of Curves, the family of orthogonal trajectories is given by:
:$\dfrac {\d y} {\d x} = -\dfrac 1 y$
So:
Hence the result.
== Sources ==
* : $1$: The Nature of Differential Equations: $\S 3$: Families of Curves. Orthogonal Trajectories: Problem $1 \ \text d$
Category:Examples of Orthogonal Trajectories
Orthogonal Trajectories/Examples/Parabolas Tangent to X Axis

== Theorem ==
Consider the one-parameter family of curves of parabolas which are tangent to the $x$-axis at the origin:
:$(1): \quad y = c x^2$
Its family of orthogonal trajectories is given by the equation:
:$x^2 + 2 y^2 = c$
:
== Proof ==
We use the technique of formation of ordinary differential equation by elimination.
Differentiating $(1)$ $x$ gives:
:$x \dfrac {\d y} {\d x} + y = 0$
Thus from Orthogonal Trajectories of One-Parameter Family of Curves, the family of orthogonal trajectories is given by:
:$\dfrac {\d y} {\d x} = -\dfrac x {2 y}$
So:
Hence the result.
== Sources ==
* : $1$: The Nature of Differential Equations: $\S 3$: Families of Curves. Orthogonal Trajectories: Problem $1 \ \text b$
Category:Examples of Orthogonal Trajectories
Category:Parabolas
Orthogonal Trajectories/Examples/Parabolas with Focus at Origin

== Theorem ==
Consider the one-parameter family of curves of parabolas whose focus is at the origin and whose axis is the $x$-axis:
:$(1): \quad y^2 = 4 c \paren {x + c}$
Its family of orthogonal trajectories is given by the equation:
:$y^2 = 4 c \paren {x + c}$
:
== Proof ==
We use the technique of formation of ordinary differential equation by elimination.
Differentiating $(1)$ $x$ gives:
Thus from Orthogonal Trajectories of One-Parameter Family of Curves, the family of orthogonal trajectories is given by:
== Sources ==
* : $1$: The Nature of Differential Equations: $\S 3$: Families of Curves. Orthogonal Trajectories: Problem $2$
Category:Examples of Orthogonal Trajectories
Orthogonal Trajectories/Examples/Rectangular Hyperbolas

== Theorem ==
Consider the one-parameter family of curves of rectangular hyperbolas:
:$(1): \quad x y = c$
Its family of orthogonal trajectories is given by the equation:
:$x^2 - y^2 = c$
:
== Proof ==
We use the technique of formation of ordinary differential equation by elimination.
Differentiating $(1)$ $x$ gives:
:$x \dfrac {\d y} {\d x} + y = 0$
Thus from Orthogonal Trajectories of One-Parameter Family of Curves, the family of orthogonal trajectories is given by:
:$\dfrac {\d y} {\d x} = \dfrac x y$
So:
Hence the result.
== Sources ==
* : $1$: The Nature of Differential Equations: $\S 3$: Families of Curves. Orthogonal Trajectories: Problem $1 \ \text a$
Category:Examples of Orthogonal Trajectories
Category:Hyperbolas
Orthogonal Trajectories/Examples/x + C exp -x

== Theorem ==
Consider the one-parameter family of curves:
:$(1): \quad y = x + C e^{-x}$
Its family of orthogonal trajectories is given by the equation:
:$x = y - 2 + C e^{-y}$
:
== Proof ==
We use the technique of formation of ordinary differential equation by elimination.
Differentiating $(1)$ $x$ gives:
:$\dfrac {\d y} {\d x} = 1 - C e^{-x}$
Eliminating $C$:
Thus from Orthogonal Trajectories of One-Parameter Family of Curves, the family of orthogonal trajectories is given by:
:$-\dfrac {\d y} {\d x} = - x = 1 - y$
The integrating factor is $e^y$, giving:
:$\ds e^y x = \int y e^y - e^y \rd y$
Using Primitive of $x e^{a x}$:
:$\ds \int y e^y \rd y = y e^y - e^y$
Thus we get:
:$e^y x = y e^y - e^y - e^y + C$
which gives us:
:$x = y - 2 + C e^{-y}$
== Sources ==
* : $\S 2.10$: Problem $5$
Category:Examples of Orthogonal Trajectories
Category:Exponential Function
Orthogonal Trajectories of One-Parameter Family of Curves

== Theorem ==
Every one-parameter family of curves has a unique family of orthogonal trajectories.
== Proof ==
Let $\map f {x, y, z}$ define a one-parameter family of curves $\FF$.
From One-Parameter Family of Curves for First Order ODE‎, there is a corresponding first order ODE:
:$\map F {x, y, \dfrac {\d y} {\d x} }$
whose solution is $\FF$.
From Slope of Orthogonal Curves, the slope of one curve is the negative reciprocal of any curve orthogonal to it.
So take the equation:
:$\map F {x, y, \dfrac {\d y} {\d x} }$
and from it create the equation:
:$\map F {x, y, -\dfrac {\d x} {\d y} }$
that is, replace $\dfrac {\d y} {\d x}$ with $-\dfrac {\d x} {\d y}$.
This is also a first order ODE, which corresponds with a one-parameter family of curves $\GG$ defined by the implicit function $\map f {x, y, z}$.
There is clearly one way of doing the above.
Hence the result.
== Sources ==
* : $1$: The Nature of Differential Equations: $\S 3$: Families of Curves. Orthogonal Trajectories
Category:Orthogonal Trajectories
Category:One-Parameter Families
Orthogonality Relations for Characters

== Theorem ==
Let $G$ be a finite abelian group with identity $e$.
Let $G^*$ be the dual group of characters $\chi : G \to \C_{\ne 0}$.
Let $\chi_0$ be the trivial character on $G$.
Let $\psi: G \to \C_{\ne 0}$ be any character.
Let $y \in G$ be arbitrary.
Then:
:$\ds \sum_{x \mathop \in G} \map \psi x = \begin {cases}
\order G & : \psi = \chi_0 \\
0 & : \psi \ne \chi_0 \end {cases}$
and:
:$\ds \sum_{\chi \mathop \in G^*} \map \chi y = \begin {cases}
\order {G^*} & : y = e \\
0 & : y \ne e \end {cases}$
== Proof ==
If $\psi = \chi_0$, then it is straightforward that:
If $\psi \ne \chi_0$, then $\exists y \in G$ such that $\map \psi y \ne 1$.
As $x$ runs through $G$ in the summation, $y x$ also runs through $G$.
So:
Since by assumption $\map \psi y \ne 1$, it must be true that:
:$\ds \sum_{x \mathop \in G} \map \psi x = 0$
Category:Analytic Number Theory
Orthogonality of Eigenfunctions of Compact Riemannian Manifold without Boundary

== Theorem ==
Let $\struct {M, g}$ be a compact Riemannian manifold without boundary.
Let $u, v \in \map {C^\infty} M : M \to \R$ be smooth mappings on $M$.
Suppose $u$ and $v$ are eigenfunctions of $M$ with distinct eigenvalues.
Then:
:$\ds \int_M u v \rd V_g = 0$
where $\rd V_g$ is the Riemannian volume form.
== Proof ==
== Sources ==
* : $\S 2$: Riemannian Metrics. Problems
Category:Riemannian Geometry
Orthogonality of Legendre Polynomials

== Theorem ==
Let $\map {P_n} x$ denote the Legendre polynomial of order $n$.
=== Inequality ===
=== Equality ===
== Proof ==
== Sources ==
* : $\S 25$: Legendre Functions: Orthogonality of Legendre Polynomials
* : $\S 28$: Legendre and Associated Legendre Functions: Orthogonality of Legendre Polynomials
Category:Orthogonality of Legendre Polynomials
Category:Legendre Polynomials
Category:Orthogonal
Orthogonality of Legendre Polynomials/Equality

== Theorem ==
Let $\map {P_n} x$ denote the Legendre polynomial of order $n$.
:$\ds \int_{-1}^1 \paren {\map {P_n} x}^2 \rd x = \dfrac 2 {2 n + 1}$
== Proof ==
== Sources ==
* : $\S 25$: Legendre Functions: Orthogonality of Legendre Polynomials: $25.26$
* : $\S 28$: Legendre and Associated Legendre Functions: Orthogonality of Legendre Polynomials: $28.26.$
Category:Orthogonality of Legendre Polynomials