nyuuzyou/wikis huggingface dataset:
Euclidean Topology is Product Topology/Proof 2

== Theorem ==
== Proof ==
The proof proceeds by induction.
For all $n \in \Z_{\ge 2}$, let $\map P n$ be the proposition:
:the Euclidean topology on $\R^n$ and the product topology on $\R^n$ are the same.
=== Basis for the Induction ===
$\map P 2$ is the case:
:the Euclidean topology on $\R^n$ and the product topology on $\R^n$ are the same.
This is demonstrated in Euclidean Topology on Cartesian Plane is Product Topology.
Thus $\map P 2$ is seen to hold.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
So this is the induction hypothesis:
:the Euclidean topology on $\R^k$ and the product topology on $\R^k$ are the same.
from which it is to be shown that:
:the Euclidean topology on $\R^{k + 1}$ and the product topology on $\R^{k + 1}$ are the same.
=== Induction Step ===
This is the induction step:
We have that:
:$\R^{k + 1} = \R^k \times \R$
By the induction hypothesis:
:the Euclidean topology on $\R^k$ and the product topology on $\R^k$ are the same.
By the basis for the induction:
:the Euclidean topology on $\R^k \times \R$ and the product topology on $\R^k \times \R$ are the same.
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\forall n \in \Z_{\ge 0}:$
::the Euclidean topology on $\R^n$ and the product topology on $\R^n$ are the same.
== Sources ==
* : $3$: Continuity generalized: topological spaces: $3.5$: Products
Category:Euclidean Topology is Product Topology
Euclidean Valuation of Non-Unit is less than that of Product

== Theorem ==
Let $\struct {D, +, \times}$ be a Euclidean domain whose zero is $0$, and unity is $1$.
Let the valuation function of $D$ be $\nu$.
Let $b, c \in D_{\ne 0}$.
Then:
:If $c$ is not a unit of $D$ then $\map \nu b < \map \nu {b c}$
== Proof ==
By principal ideal domain, $D$ is a principal ideal domain
Consider the principal ideal $U = \ideal b$ of $D$ generated by $b$.
As $\nu$ is a valuation function, we have:
:$\forall x \in D, x \ne 0: \map \nu b \le \map \nu {b \times x}$
As $U$ is a principal ideal:
:$\forall a \in \ideal b: \exists x \in D: a = x \times b$
and so:
:$\forall a \in U: \map \nu b \le \map \nu a$
Let $\map \nu b = \map \nu {b c}$.
Then from above:
:$\forall a \in U: \map \nu {b c} \le \map \nu a$
From the argument in Euclidean Domain is Principal Ideal Domain, we have that $U$ is also the ideal generated by $b c$.
Thus $a b$ is a divisor of $b$ itself, that is:
:$\exists y \in D: b = b c y$
Thus:
:$c y = 1$
and so $c$ is a unit of $D$.
So if $c$ is not a unit, then it must be the case that $\map \nu b < \map \nu {b c}$.
== Sources ==
* : Chapter $6$: Polynomials and Euclidean Rings: $\S 30$. Unique Factorization: Theorem $58 \ \text{(i)}$
Category:Euclidean Domains
Euler's Conjecture on Orthogonal Latin Squares

== Conjecture ==
Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Two orthogonal Latin squares do not exist for order $4 n + 2$.
== Refutation ==
While the conjecture is true for order $6$, it fails for order $10, 14, \ldots$.
The following is an example of two orthogonal Latin squares presented in the same array.
The elements of the first Latin square are shown in normal type, while those of the second Latin square are shown in italic type:
$\begin{array} {|c|c|c|c|c|c|c|c|c|c|}
\hline 4 \mathit 6 & 5 \mathit 7 & 6 \mathit 8 & 7 \mathit 0 & 8 \mathit 1 & 0 \mathit 2 & 1 \mathit 3 & 2 \mathit 4 & 3 \mathit 5 & 9 \mathit 9 \\
\hline 7 \mathit 1 & 9 \mathit 4 & 3 \mathit 7 & 6 \mathit 5 & 1 \mathit 2 & 4 \mathit 0 & 2 \mathit 9 & 0 \mathit 6 & 8 \mathit 8 & 5 \mathit 3 \\
\hline 9 \mathit 3 & 2 \mathit 6 & 5 \mathit 4 & 0 \mathit 1 & 3 \mathit 8 & 1 \mathit 9 & 8 \mathit 5 & 7 \mathit 7 & 6 \mathit 0 & 4 \mathit 2 \\
\hline 1 \mathit 5 & 4 \mathit 3 & 8 \mathit 0 & 2 \mathit 7 & 0 \mathit 9 & 7 \mathit 4 & 6 \mathit 6 & 5 \mathit 8 & 9 \mathit 2 & 3 \mathit 1 \\
\hline 3 \mathit 2 & 7 \mathit 8 & 1 \mathit 6 & 8 \mathit 9 & 6 \mathit 3 & 5 \mathit 5 & 4 \mathit 7 & 9 \mathit 1 & 0 \mathit 4 & 2 \mathit 0 \\
\hline 6 \mathit 7 & 0 \mathit 5 & 7 \mathit 9 & 5 \mathit 2 & 4 \mathit 4 & 3 \mathit 6 & 9 \mathit 0 & 8 \mathit 3 & 2 \mathit 1 & 1 \mathit 8 \\
\hline 8 \mathit 4 & 6 \mathit 9 & 4 \mathit 1 & 3 \mathit 3 & 2 \mathit 5 & 9 \mathit 8 & 7 \mathit 2 & 1 \mathit 0 & 5 \mathit 6 & 0 \mathit 7 \\
\hline 5 \mathit 9 & 3 \mathit 0 & 2 \mathit 2 & 1 \mathit 4 & 9 \mathit 7 & 6 \mathit 1 & 0 \mathit 8 & 4 \mathit 5 & 7 \mathit 3 & 8 \mathit 6 \\
\hline 2 \mathit 8 & 1 \mathit 1 & 0 \mathit 3 & 9 \mathit 6 & 5 \mathit 0 & 8 \mathit 7 & 3 \mathit 4 & 6 \mathit 2 & 4 \mathit 9 & 7 \mathit 5 \\
\hline 0 \mathit 0 & 8 \mathit 2 & 9 \mathit 5 & 4 \mathit 8 & 7 \mathit 6 & 2 \mathit 3 & 5 \mathit 1 & 3 \mathit 9 & 1 \mathit 7 & 6 \mathit 4 \\
\hline \end{array}$
Note that all integers from $0 \mathit 0$ to $9 \mathit 9$ appear exactly once each.
== Historical Note ==
== Sources ==
* 
* : $10$
* : $10$
Category:Famous False Conjectures
Category:Latin Squares
Euler's Conjecture on Orthogonal Latin Squares/Historical Note

== Historical Note on Euler's Conjecture on Orthogonal Latin Squares ==
 made this conjecture in $1782$.
It was refuted $1959$ by , and .
This discovery was significant enough to be reported in the mainstream news of the time.
As a result, , and were given the nickname Euler's spoilers.
== Sources ==
* 
* : $10$
* : $10$
Category:Historical Notes
Euler's Continued Fraction Formula

== Theorem ==
== Proof ==
The proof proceeds by induction.
For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:
:$\ds \sum_{k \mathop = 0}^n \prod_{j \mathop = 0}^k a_j = \cfrac {a_0} {1 - \cfrac {a_1} {1 + a_1 - \cfrac {a_2} {1 + a_2 - \cfrac {a_3} {1 + a_3 - \cfrac {\ddots} {\ddots - \cfrac {a_{n - 1} } {1 + a_{n - 1} - \cfrac {a_n} {1 + a_n} } } } } } } $
=== Basis for the Induction ===
$\map P 1$ is the case:
Since $\dfrac {a_1} {1 + a_1} \ne 1$
Thus $\map P 1$ is seen to hold.
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $\map P n$ is true, where $n \ge 1$, then it logically follows that $\map P {n + 1}$ is true.
So this is the induction hypothesis:
:$\ds \sum_{k \mathop = 0}^n \prod_{j \mathop = 0}^k a_j = \cfrac {a_0} {1 - \cfrac {a_1} {1 + a_1 - \cfrac {a_2} {1 + a_2 - \cfrac {a_3} {1 + a_3 - \cfrac {\ddots} {\ddots - \cfrac {a_{n - 1} } {1 + a_{n - 1} - \cfrac {a_n} {1 + a_n} } } } } } } $
from which it is to be shown that:
:$\ds \sum_{k \mathop = 0}^{n + 1} \prod_{j \mathop = 0}^k a_j = \cfrac {a_0} {1 - \cfrac {a_1} {1 + a_1 - \cfrac {a_2} {1 + a_2 - \cfrac {a_3} {1 + a_3 - \cfrac {\ddots} {\ddots - \cfrac {a_n } {1 + a_n - \cfrac {a_{n + 1} } {1 + a_{n + 1} } } } } } } }$
=== Induction Step ===
This is the induction step:
We first note that:
Before continuing, we define x as follows:
:$\ds x := \cfrac {-a_2} {1 + a_2 - \cfrac {a_3} {1 + a_3 - \cfrac {\ddots} {\ddots - \cfrac {a_n } {1 + a_n - \cfrac {a_{n + 1} } {1 + a_{n + 1} } } } } }$
Next, we apply the induction hypothesis to the values $a_1$, $a_2$, $\cdots$ $a_{n + 1}$
So $\map P n \implies \map P {n + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\forall n \in \Z_{>0}:$
:$\ds \sum_{k \mathop = 0}^n \prod_{j \mathop = 0}^k a_j = \cfrac {a_0} {1 - \cfrac {a_1} {1 + a_1 - \cfrac {a_2} {1 + a_2 - \cfrac {a_3} {1 + a_3 - \cfrac {\ddots} {\ddots - \cfrac {a_{n - 1} } {1 + a_{n - 1} - \cfrac {a_n} {1 + a_n} } } } } } } $
== Examples ==
== Sources ==
* : Volume $\text {1}$ Chapter $\text {18}$. De Fractionibus Continuis
Category:Continued Fractions
Euler's Continued Fraction Formula/Examples

== Examples of Euler's Continued Fraction Formula ==
=== Example: $\map {\arcsin} x$ ===
=== Example: $\map {\arctan} x$ ===
=== Example: $e^x$ ===
=== Example: $\map {\log} {1 + x}$ ===
Category:Examples of Euler's Continued Fraction Formula
Category:Continued Fractions
Euler's Cosecant Identity

== Theorem ==
:$\csc z = \dfrac {2 i} {e^{i z} - e^{-i z} }$
where:
:$z \in \C$ is a complex number
:$\csc z$ denotes the cosecant function
:$i$ denotes the imaginary unit: $i^2 = -1$
== Proof ==
== Also see ==
* Euler's Sine Identity
* Euler's Cosine Identity
* Euler's Tangent Identity
* Euler's Cotangent Identity
* Euler's Secant Identity
* Arccosecant Logarithmic Formulation
== Sources ==
* : $\S 7$: Relationship between Exponential and Trigonometric Functions: $7.22$
* : $2$: Functions, Limits and Continuity: The Elementary Functions: $4$
Category:Euler's Identities
Category:Cosecant Function
Euler's Cosine Identity

== Theorem ==
For any complex number $z \in \C$:
:$\cos z = \dfrac {e^{i z} + e^{-i z} } 2$
where:
:$\exp z$ denotes the exponential function
:$\cos z$ denotes the complex cosine function
:$i$ denotes the imaginary unit.
=== Real Domain ===
This result is often presented and proved separately for arguments in the real domain:
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Also presented as ==
== Also see ==
* Euler's Sine Identity
* Euler's Tangent Identity
* Euler's Cotangent Identity
* Euler's Secant Identity
* Euler's Cosecant Identity
* Arccosine Logarithmic Formulation
== Sources ==
* : $2$: Functions, Limits and Continuity: The Elementary Functions: $4$
* 
* 
Category:Euler's Identities
Category:Euler's Cosine Identity
Category:Cosine Function
Euler's Cosine Identity/Real Domain

== Theorem ==
For any real number $x \in \R$:
:$\cos x = \dfrac {e^{i x} + e^{-i x} } 2$
where:
:$e^{i x}$ denotes the exponential function
:$\cos x$ denotes the real cosine function
:$i$ denotes the inaginary unit.
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Proof 4 ==
== Also presented as ==
== Also see ==
* Euler's Sine Identity
* Euler's Tangent Identity
* Euler's Cotangent Identity
* Euler's Secant Identity
* Euler's Cosecant Identity
* Arccosine Logarithmic Formulation
== Sources ==
* : $\S 7$: Relationship between Exponential and Trigonometric Functions: $7.18$
Category:Euler's Cosine Identity
Category:Cosine Function
Euler's Cosine Identity/Real Domain/Proof 4

== Theorem ==
== Proof ==
Consider the differential equation:
:$(1): \quad D^2_x \map f x = -\map f x$
subject to the initial conditions:
:$(2): \quad \map f 0 = 1$
:$(3): \quad D_x \map f 0 = 0$
=== Step 1 ===
We will prove that $y = \cos x$ is a particular solution of $(1)$.
Thus $y = \cos x$ fulfils $(1)$.
Then from Cosine of Zero is One:
:$\cos 0 = 1$
Thus $y = \cos x$ fulfils $(2)$.
Then:
Thus $y = \cos x$ fulfils $(3)$.
So $y = \cos x$ is a particular solution of $(1)$.
=== Step 2 ===
We will prove that $z = \dfrac {e^{i x} + e^{-i x} } 2$ is a particular solution of $(1)$.
Thus $z = \dfrac {e^{i x} + e^{-i x} } 2$ fulfils $(1)$.
Then:
Thus $z = \dfrac {e^{i x} + e^{-i x} } 2$ fulfils $(2)$.
Then:
Thus $z = \dfrac {e^{i x} + e^{-i x} } 2$ fulfils $(3)$.
So $z = \dfrac {e^{i x} + e^{-i x} } 2$ is a particular solution of $(1)$.
We have shown that $y$ and $z$ are both particular solutions of $(1)$.
But a particular solution to a differential equation is unique.
Therefore $y = z$.
That is:
:$\cos x = \dfrac {e^{i x} + e^{-i x} } 2$
Category:Euler's Cosine Identity
Euler's Cotangent Identity

== Theorem ==
:$\cot z = i \dfrac {e^{i z} + e^{-i z} } {e^{i z} - e^{-i z} }$
where:
:$z \in \C$ is a complex number such that $\forall k \in \Z: z \ne k \pi$
:$\cot z$ denotes the cotangent function
:$i$ denotes the imaginary unit: $i^2 = -1$
== Proof 1 ==
== Proof 2 ==
== Also presented as ==
== Also see ==
* Euler's Sine Identity
* Euler's Cosine Identity
* Euler's Tangent Identity
* Euler's Secant Identity
* Euler's Cosecant Identity
* Arccotangent Logarithmic Formulation
== Sources ==
* : $\S 7$: Relationship between Exponential and Trigonometric Functions: $7.20$
* : $2$: Functions, Limits and Continuity: The Elementary Functions: $4$
Category:Euler's Cotangent Identity
Category:Euler's Identities
Category:Cotangent Function
Euler's Cotangent Identity/Also presented as

== Euler's Cotangent Identity: Also presented as == 
Euler's Cotangent Identity can also be presented as:
:$\cot z = \dfrac {i \paren {e^{i z} + e^{-i z} } } {e^{i z} - e^{-i z} }$
== Sources ==
* : $2$: Functions, Limits and Continuity: The Elementary Functions: $4$
Category:Euler's Cotangent Identity
Euler's Cotangent Identity/Proof 1

== Theorem ==
== Proof ==
We have, , that $z$ is a complex number such that:
:$\forall k \in \Z: z \ne k \pi$
Therefore:
:$\sin z \ne 0$
It follows from the definition of the complex cotangent function that:
:$\cot z$
is well-defined.
Hence:
Category:Euler's Cotangent Identity
Euler's Cotangent Identity/Proof 2

== Theorem ==
== Proof ==
We have, , that $z$ is a complex number such that:
:$\forall k \in \Z: z \ne k \pi$
Therefore:
:$\sin z \ne 0$
It follows from the definition of the complex cotangent function that:
:$\cot z$
is well-defined.
Hence:
Category:Euler's Cotangent Identity
Euler's Criterion

== Theorem ==
Let $a$ be a residue order $n$ of $m$, where $a$ and $m$ are coprime.
Then:
:$a^{\map \phi m / d} \equiv 1 \pmod m$
where:
:$\map \phi m$ denotes the Euler $\phi$ function of $m$
:$d$ denotes the gretest common divisor of $\map \phi m$ and $n$
:$\equiv$ denotes modulo congruence.
=== Euler's Criterion for Quadratic Residue ===
== Proof ==
== Sources ==
* 
* 
* 
* 
Category:Euler's Criterion
Category:Residues (Number Theory)
Category:Number Theory
Euler's Criterion/Quadratic Residue/Proof 1

== Theorem ==
== Proof ==
Trivially, any $a \not \equiv 0 \pmod p$ is either a quadratic residue or a quadratic non-residue, modulo $p$.
Therefore, it suffices to check the sufficient condition for both of the equations (i.e., the if parts from the iffs).
So let $a$ be a quadratic non-residue of $p$.
Also, let $b \in \set {1, 2, \ldots, p - 1}$.
The congruence $b x \equiv a \pmod p$ has (modulo $p$) a unique solution $b'$ by Solution of Linear Congruence.
Note that $b' \not\equiv b$, because otherwise we would have $b^2 \equiv a \pmod p$ and $a$ would be a quadratic residue of $p$.
It follows that the residue classes $\set {1, 2, \ldots, p - 1}$ modulo $p$ fall into $\dfrac {p - 1} 2$ pairs $b, b'$ such that $b b' \equiv a \pmod p$.
Therefore, we have:
:$\paren {p - 1}! = 1 \times 2 \times \cdots \times \paren {p - 1} \equiv a \times a \times \cdots \times a \equiv a^{\frac {p - 1} 2} \pmod p$
From Wilson's Theorem, we also have:
:$\paren {p - 1}! \equiv -1 \pmod p$
And so, for any quadratic non-residue of $p$:
:$a^{\frac {p - 1} 2} \equiv -1 \pmod p$
Subsequently, let $a$ be a quadratic residue of $p$.
By definition of a quadratic residue, the congruence $x^2 \equiv a \pmod p$ has a solution $x$.
Suppose also $y$ is a solution. Then we have:
So either $x + y \equiv 0 \pmod p$ or $x - y \equiv 0 \pmod p$ from Product is Zero Divisor means Zero Divisor.
It follows that when $c \pmod p$ is one solution, $-c \pmod p$ is, too.
Also, these solutions are distinct as $p$ is odd.
Furthermore, we conclude that these are the only two solutions.
Now, remove $c$ and $p - c$ from $\set {1, 2, \ldots, p - 1}$.
The remaining integers fall, modulo $p$, into $\dfrac {p - 3} 2$ pairs $b, b'$ such that $b b' \equiv a \pmod p$ by Solution of Linear Congruence.
Therefore, we can compute the following:
Again applying Wilson's Theorem, we conclude:
:$-a^{\frac {p-1} 2} \equiv - 1 \pmod p$
The assertion for quadratic residues $a$ of $p$ follows.
Category:Euler's Criterion
Euler's Criterion/Quadratic Residue/Proof 2

== Theorem ==
== Proof ==
First note that the square roots of $1$ are $1, -1 \pmod p$.
Also, we have that $a^{p - 1} \equiv 1 \pmod p$ by Fermat's Little Theorem.
Combining these two observations, we find:
:$a^{\frac {p - 1} 2} \equiv 1 \text{ or } -1 \pmod p$
The theorem is therefore equivalent to stating that $a$ is a quadratic residue modulo $p$ $a^{\frac{p - 1} 2} \equiv 1 \pmod p$.
Namely, considering the above, we see this also implies that all quadratic non-residues will be congruent to $-1 \pmod p$.
We prove each direction of the equivalent statement separately:
=== Sufficient Condition ===
Assume $a$ is a quadratic residue modulo $p$.
We pick $k$ such that $k^2 \equiv a \pmod p$.
Then by Congruence of Powers and Fermat's Little Theorem:
:$a^{\frac{p-1} 2} \equiv k^{p-1} \equiv 1 \pmod p$
=== Necessary Condition ===
Now assume $a^{\frac{p - 1} 2} \equiv 1 \pmod p$.
Then let $y$ be a primitive root modulo $p$, so that $a$ can be written as $y^j$.
In particular:
:$y^{j \frac {p - 1} 2} \equiv 1 \pmod p$
From the definition of $y$, it has order $p-1$.
It follows that $p - 1 \divides j \dfrac {p - 1} 2$ from Element to Power of Multiple of Order is Identity.
We conclude that $j$ is necessarily an even integer, and denote $j' = \dfrac j 2$.
Let $k$ be such that $k \equiv y^{j'} \pmod p$.
By construction, we have:
:$k^2 \equiv y^{2 j'} \equiv y^j \equiv a \pmod p$
Hence $a$ is a quadratic residue modulo $p$.
Category:Euler's Criterion
Euler's Equation/Independent of x

== Theorem ==
Let $y$ be a mapping.
Let $J$ be a functional such that:
:$\ds J \sqbrk y = \int_a^b \map F {y, y'} \rd x$
Then the corresponding Euler's Equation can be reduced to:
:$F - y' F_{y'} = C$ 
where $C$ is an arbitrary constant.
== Proof ==
Assume that:
:$\ds J \sqbrk y = \int_a^b \map F {y, y'} \rd x$
Then:
Multiply this differential equation by $y'$.
This gives:
Integration yields the desired result.
== Sources ==
* : $\S 1.4$: The Simplest Variational Problem. Euler's Equation
Category:Calculus of Variations
Euler's Equation/Independent of y

== Theorem ==
Let $y$ be a mapping
Let $J$ be a functional such that
:$\ds J \sqbrk y = \int_a^b \map F {x,y'} \rd x$
Then the corresponding Euler's equation can be reduced to:
:$F_{y'} = C$ 
where $C$ is an arbitrary constant.
== Proof ==
Assume that:
:$\ds J \sqbrk y = \int_a^b \map F {x,y'} \rd x$
Euler's equation for $J$ is:
:$\dfrac \d {\d x} F_{y'} = 0$
Integration yields:
:$F_{y'} = C$ 
== Sources ==
* : $\S 1.4$: The Simplest Variational Problem. Euler's Equation
Category:Calculus of Variations
Euler's Equation/Independent of y'

== Theorem ==
Let $y$ be a mapping.
Let $J$ a functional be such that
:$\ds J \sqbrk y = \int_a^b \map F {x,y} \rd x$
Then the corresponding Euler's Equation can be reduced to:
:$F_y = 0$ 
Furthermore, this is an algebraic equation.
== Proof ==
Assume that:
:$\ds J \sqbrk y = \int_a^b \map F {x,y} \rd x$
Then Euler's Equation for $J$ is:
:$F_y = 0$
Since $F$ is independent of $y'$, the equation is algebraic.
== Sources ==
* : $\S 1.4$: The Simplest Variational Problem. Euler's Equation
Category:Calculus of Variations
Euler's Equation/Integrated wrt Length Element

== Theorem ==
Let $y$ be a real mapping belonging to $C^2$ differentiability class.
Assume that:
:$\ds J \sqbrk y = \int_a^b \map f {x, y, y'} \rd s$
where
:$\rd s = \sqrt {1 + y'^2} \rd x$
Then Euler's Equation can be reduced to:
:$f_y - f_x y' - f_{y'} y' y - f \dfrac {y} {\paren {1 + y'^2}^{\frac 3 2} } = 0$
== Proof ==
Substitution of $\rd s$ into $J$ results in the following functional:
:$\ds J \sqbrk y = \int_a^b \map f {x, y, y'} \sqrt {1 + y'^2} \rd x$
We can consider this as a functional with the following effective $F$:
:$F = \map f {x, y, y'} \sqrt {1 + y'^2}$
Find Euler's Equation:
Due to assumptions on $y$, the prefactor does not vanish.
By Euler's Equation, the last expression vanishes.
== Sources ==
* : $\S 1.4$: The Simplest Variational Problem. Euler's Equation
Category:Calculus of Variations
Euler's Equation for Vanishing Variation in Canonical Variables

== Theorem ==
Consider the following system of differential equations:
:$(1): \quad \begin {cases} F_{y_i} - \dfrac \d {\d x} F_{y_i'} = 0 \\
\dfrac {\d {y_i} } {\d x} = y_i'\end{cases}$
where $i \in \set {1, \ldots, n}$.
Let the coordinates $\tuple {x, \family {y_i}_{1 \mathop \le i \mathop \le n}, \family {y_i'}_{1 \mathop \le i \mathop \le n}, F}$ be transformed to canonical variables:
:$\tuple {x, \family {y_i}_{1 \mathop \le i \mathop \le n}, \family {p_i}_{1 \mathop \le i \mathop \le n}, H}$
Then the system $(1)$ is transformed into:
:$\begin {cases}
\dfrac {\d y_i} {\d x} = \dfrac {\partial H} {\partial p_i} \\
\dfrac {\d p_i} {\d x} = -\dfrac {\partial H} {\partial y_i}
\end {cases}$
== Proof ==
Find the full differential of Hamiltonian:
By equating coefficients of differentials in last two equations we find that:
:$\dfrac {\partial H} {\partial x} = -\dfrac {\partial F} {\partial x}$
:$\dfrac {\partial H} {\partial y_i} = -\dfrac {\partial F} {\partial y_i}$
:$\dfrac {\partial H} {\partial p_i} = y_i'$
From the third identity it follows that:
:$\paren {\dfrac {\d y_i} {\d x} = y_i} \implies \paren {\dfrac {\d y_i} {\d x} = \dfrac {\partial H} {\partial p_i} }$
while the second identity together with the definition of $p_i$ assures that:
:$\paren {\dfrac {\partial F} {\partial y_i} - \dfrac \d {\d x} \dfrac {\partial F} {\partial y_i} = 0} \implies \paren {\dfrac {\d p_i} {\d x} = -\dfrac {\partial H} {\partial y_i} }$
== Sources ==
* : $\S 4.16$: The Canonical Form of the Euler's Equations
Category:Calculus of Variations
Euler's Equation for Vanishing Variation is Invariant under Coordinate Transformations

== Theorem ==
Euler's Equation for Vanishing Variation is invariant under coordinate transformations.
== Proof ==
Let $J \sqbrk y$ be an integral functional:
:$\ds J \sqbrk y = \int_a^b \map F {x, y, y'} \rd x$
Suppose, we introduce new curvilinear coordinates $u,v$ such that:
:$x = \map x {u, v}$
:$y = \map y {u, v}$
such that:
:$\begin{vmatrix}
\map {\dfrac {\partial x} {\partial u} } {u, v} & \map {\dfrac {\partial x} {\partial v} } {u, v} \\
\map {\dfrac {\partial y} {\partial u} } {u, v} & \map {\dfrac {\partial y} {\partial v} } {u, v}
\end{vmatrix} \ne 0$
After this transformation, a curve $y = \map y x$ in coordinates $\tuple {x, y}$ will be described by an equation $v = \map v u$ in coordinates $\tuple {u, v}$.
Let $\mathscr J \sqbrk v$ be a functional defined according to the following chain of operations:
Now, consider the method of finite differences like it is used here.
For functional $J \sqbrk y$ we have the area $\Delta \sigma$ bounded by functions $\map y x$ and $\map y x + \map h x$.
Likewise, for functional $\mathscr J \sqbrk v$ we have area $\Delta \Sigma$ bounded by functions $\map v u$ and $\map v u + \map \eta u$.
Note that as it was able to construct $\map v u$ from $\map y x$, same can be done for $\map \eta u$ starting with $\map h x$.
As both areas approach zero, they become differential surface elements, which are related by corresponding Jacobian determinant of coordinate transformations:
$\d \sigma = \begin{vmatrix}
x_u & x_v \\
y_u & y_v
\end{vmatrix} \rd \Sigma$
The last thing to be shown is that variational derivative vanishes in new coordinates.
By definition, it has to satisfy
$\ds \lim_{\Delta \sigma \mathop \to 0} \frac {\Delta J \sqbrk {y; h} } {\Delta \sigma} = 0$
Since $J\sqbrk y = \mathscr J \sqbrk v$ with respect to a specific set of coordinates, we have
We see that, if Jacobian determinant is nonvanishing, the condition of vanishing variational derivative is the same.
The equivalence of Euler's equation in both sets of coordinates is the result of same method of finite differences being used in each case.
== Sources ==
* : $\S 1.8$: Invariance of Euler's Equation
Category:Calculus of Variations
Euler's Equations of Motion for Rotation of Rigid Body

== Theorem ==
Let a rigid body $B$ rotate about an axis $\AA$ which is fixed in relation to $B$ and parallel to the principal axis of inertia of $B$.
Then the rotation of $B$ about $\AA$ is described by:
:$\mathbf I \cdot \dot {\boldsymbol \omega} + \boldsymbol \omega \times \paren {\mathbf I \cdot \boldsymbol\omega} = \mathbf M$
where:
:$\mathbf M$ is the torque applied to $B$ about $\AA$
:$\mathbf I$ is the moment of inertia of $B$ with respect to $\AA$
:$\boldsymbol \omega$ is the angular velocity about $\AA$.
== Proof ==
== Also presented as ==
== Sources ==
* : $\S 3$: Appendix $\text A$: Euler
* : Chapter $\text {A}.21$: Euler ()
* 
* 
Category:Euler's Equations of Motion for Rotation of Rigid Body
Category:Classical Mechanics