Symmetry Group of Equilateral Triangle/Cayley Table 0 38649 387966 387965 2019-01-16T07:15:29Z Prime.mover 59 wikitext text/x-wiki == Cayley Table of Symmetry Group of Equilateral Triangle == The Cayley table of the symmetry group of the equilateral triangle can be written: <onlyinclude> :$\begin{array}{c|ccc|ccc} \circ & e & p & q & r & s & t \\ \hline e & e & p & q & r & s & t \\ p & p & q & e & s & t & r \\ q & q & e & p & t & r & s \\ \hline r & r & t & s & e & q & p \\ s & s & r & t & p & e & q \\ t & t & s & r & q & p & e \\ \end{array}$ </onlyinclude> where: {{begin-eqn}} {{eqn | l = e       | o = :       | r = (A) (B) (C)       | c = Identity mapping }} {{eqn | l = p       | o = :       | r = (ABC)       | c = Rotation of $120 \degrees$ anticlockwise about center }} {{eqn | l = q       | o = :       | r = (ACB)       | c = Rotation of $120 \degrees$ clockwise about center }} {{eqn | l = r       | o = :       | r = (BC)       | c = Reflection in line $r$ }} {{eqn | l = s       | o = :       | r = (AC)       | c = Reflection in line $s$ }} {{eqn | l = t       | o = :       | r = (AB)       | c = Reflection in line $t$ }} {{end-eqn}} :400px == Sources == * {{BookReference|Modern Algebra|1965|Seth Warner|prev = Symmetry Group of Square/Cayley Table|next = Symmetry Group of Rectangle/Cayley Table}}: Exercise $2.11$ * {{BookReference|Elements of Abstract Algebra|1966|Richard A. Dean|prev = Cayley Table for Commutative Operation is Symmetrical about Main Diagonal|next = Symmetry Group of Square/Cayley Table}}: $\S 1.4$: Exercise $2.1$ * {{BookReference|An Introduction to Abstract Algebra|1978|Thomas A. Whitelaw|prev = Definition:Symmetry Group of Equilateral Triangle|next = Definition:Finite Group}}: $\S 34$. Examples of groups: $(5)$ * {{BookReference|A Course in Group Theory|1996|John F. Humphreys|prev = Definition:Symmetry Group of Equilateral Triangle|next = Symmetry Group of Equilateral Triangle is Group}}: Chapter $1$: Definitions and Examples: Example $1.9$ Category:Examples of Cayley Tables Category:Symmetry Group of Equilateral Triangle f61utmd07tf309erhwam2gxgb1ihniy


Symmetry Group of Rectangle/Cayley Table 0 38688 388187 372564 2019-01-17T16:32:55Z Ascii 3554 wikitext text/x-wiki == Cayley Table of Symmetry Group of Rectangle == === Definition === {{:Definition:Symmetry Group of Rectangle}} === Cayley Table === <onlyinclude> The Cayley table of the symmetry group of the (non-square) rectangle can be written: :$\begin{array}{c|cccc}   & e & r & h & v \\ \hline e & e & r & h & v \\ r & r & e & v & h \\ h & h & v & e & r \\ v & v & h & r & e \\ \end{array}$ </onlyinclude> == Sources == * {{BookReference|Modern Algebra|1965|Seth Warner|prev = Symmetry Group of Equilateral Triangle/Cayley Table|next = Max is Associative}}: Exercise $2.11$ Category:Examples of Cayley Tables rr9cqxbt1gykhasan8wkbrzy27b2c0s


Parity Ring/Cayley Tables

Parity Ring/Cayley Tables 0 38730 371035 371033 2018-10-14T07:02:20Z Prime.mover 59 wikitext text/x-wiki == Cayley Tables for Parity Ring == The parity ring can be described completely by showing its Cayley tables: <onlyinclude> :$\begin{array}{r|rr} + & \text{even} &  \text{odd} \\ \hline \text{even} & \text{even} & \text{odd} \\ \text{odd} & \text{odd} & \text{even} \\ \end{array} \qquad \begin{array}{r|rr} \times & \text{even} &  \text{odd} \\ \hline \text{even} & \text{even} & \text{even} \\ \text{odd} & \text{even} & \text{odd} \\ \end{array}$ </onlyinclude> == Sources == * {{BookReference|Modern Algebra|1965|Seth Warner|prev = Definition:Parity Ring|next = Parity Addition is Associative}}: $\S 2$: Example $2.2$ Category:Examples of Cayley Tables Category:Parity Ring 5sx34n05ujc4mptjwuhrcpy1r9p0e4e


Ring of Integers Modulo 2/Cayley Tables

Ring of Integers Modulo 2/Cayley Tables 0 38733 371029 371027 2018-10-14T06:56:57Z Prime.mover 59 wikitext text/x-wiki {{refactor|Transclude the separate tables for Addition and Multiplication}} == Cayley Tables for Ring of Integers Modulo $2$ == The Ring of Integers Modulo $2$: :$\struct {\Z_2, +_2, \times_2}$ can be described completely by showing its Cayley tables: <onlyinclude> :$\begin{array} {r|rr} \struct {\Z_2, +_2} & \eqclass 0 2 & \eqclass 1 2 \\ \hline \eqclass 0 2 & \eqclass 0 2 & \eqclass 1 2 \\ \eqclass 1 2 & \eqclass 1 2 & \eqclass 0 2 \\ \end{array} \qquad \begin{array}{r|rr} \struct {\Z_2, \times_2} & \eqclass 0 2 & \eqclass 1 2 \\ \hline \eqclass 0 2 & \eqclass 0 2 & \eqclass 0 2 \\ \eqclass 1 2 & \eqclass 0 2 & \eqclass 1 2 \\ \end{array}$ They can be presented more simply as: :$\begin{array}{r|rr} + & 0 & 1 \\ \hline 0 & 0 & 1 \\ 1 & 1 & 0 \\ \end{array} \qquad \begin{array}{r|rr} \times & 0 & 1 \\ \hline 0 & 0 & 0 \\ 1 & 0 & 1 \\ \end{array}$ </onlyinclude> == Sources == * {{BookReference|Modern Algebra|1965|Seth Warner|prev = Isomorphism between Ring of Integers Modulo 2 and Parity Ring|next = Isomorphism between Roots of Unity under Multiplication and Integers under Modulo Addition}}: $\S 6$: Example $6.1$ Category:Examples of Cayley Tables b2p640i0ujnz408d3pofryxivukop17


Group of Gaussian Integer Units/Cayley Table 0 38755 379140 370979 2018-11-30T08:00:20Z Prime.mover 59 wikitext text/x-wiki == Cayley Table for Group of Gaussian Integer Units == The group of Gaussian integer units: :$\struct {U_\C, \times}$ can be described completely by showing its Cayley table: <onlyinclude> :$\begin{array}{r|rrrr} \times & 1 & i & -1 & -i \\ \hline 1 & 1 & i & -1 & -i \\ i & i & -1 & -i & 1 \\ -1 & -1 & -i & 1 & i \\ -i & -i & 1 & i & -1 \\ \end{array}$ </onlyinclude> == Sources == * {{BookReference|Modern Algebra|1965|Seth Warner|prev = Modulo Addition/Cayley Table/Modulo 4|next = Set of Integers under Addition is Isomorphic to Set of Even Integers under Addition}}: $\S 6$: Example $6.2$ * {{BookReference|An Introduction to Abstract Algebra|1978|Thomas A. Whitelaw|prev = Units of Gaussian Integers form Group|next = Quaternion Group/Complex Matrices}}: $\S 34$. Examples of groups: $(6) \ \text{(i)}$ Category:Examples of Cayley Tables qkddxegk81xc7tizvsj0uc76fk5m7su


Modulo Addition/Cayley Table/Modulo 4

Modulo Addition/Cayley Table/Modulo 4 0 38770 399538 385902 2019-04-09T06:08:50Z Prime.mover 59 wikitext text/x-wiki == Cayley Table for Addition Modulo $4$ == The additive group of integers modulo $4$ can be described by showing its Cayley table: <onlyinclude> :$\begin{array}{r|rrrr} \struct {\Z_4, +_4} & \eqclass 0 4 & \eqclass 1 4 & \eqclass 2 4 & \eqclass 3 4 \\ \hline \eqclass 0 4 & \eqclass 0 4 & \eqclass 1 4 & \eqclass 2 4 & \eqclass 3 4 \\ \eqclass 1 4 & \eqclass 1 4 & \eqclass 2 4 & \eqclass 3 4 & \eqclass 0 4  \\ \eqclass 2 4 & \eqclass 2 4 & \eqclass 3 4 & \eqclass 0 4 & \eqclass 1 4 \\ \eqclass 3 4 & \eqclass 3 4 & \eqclass 0 4 & \eqclass 1 4 & \eqclass 2 4 \\ \end{array}$ </onlyinclude> It can also be presented: :$\begin{array}{r|rrrr} +_4 & 0 & 1 & 2 & 3 \\ \hline 0 & 0 & 1 & 2 & 3  \\ 1 & 1 & 2 & 3 & 0 \\ 2 & 2 & 3 & 0 & 1 \\ 3 & 3 & 0 & 1 & 2 \\ \end{array}$ == Sources == * {{BookReference|Modern Algebra|1965|Seth Warner|prev = Isomorphism between Gaussian Integer Units and Integers Modulo 4 under Addition/Proof 2|next = Group of Gaussian Integer Units/Cayley Table}}: $\S 6$: Example $6.2$ * {{BookReference|Introduction to Abstract Algebra|1969|C.R.J. Clapham|prev = Modulo Multiplication has Identity|next = Modulo Multiplication/Cayley Table/Modulo 4}}: Chapter $1$: Integral Domains: $\S 6$. The Residue Classes * {{BookReference|A Course in Group Theory|1996|John F. Humphreys|prev = Equivalence Relation/Examples/Non-Equivalence/Sum of Integers is Divisible by 3|next = Multiplicative Group of Reduced Residues Modulo 5/Cayley Table}}: Chapter $2$: Maps and relations on sets: Exercise $5$ Category:Examples of Cayley Tables Category:Cyclic Group of Order 4 Category:Examples of Additive Groups of Integers Modulo m lm0id19hwsdk2j663vq0mtie6ol9llg


Symmetric Group on 3 Letters/Cayley Table 0 38831 464009 385973 2020-04-24T13:57:30Z Prime.mover 59 wikitext text/x-wiki == Cayley Table of Symmetric Group on $3$ Letters == The Cayley table of the symmetric group on $3$ letters can be presented in cycle notation as: <onlyinclude> :$\begin{array}{c|cccccc} \circ & e & (123) & (132) & (23) & (13) & (12) \\ \hline e & e & (123) & (132) & (23) & (13) & (12) \\ (123) & (123) & (132) & e & (13) & (12) & (23) \\ (132) & (132) & e & (123) & (12) & (23) & (13) \\ (23) & (23) & (12) & (13) & e & (132) & (123) \\ (13) & (13) & (23) & (12) & (123) & e & (132) \\ (12) & (12) & (13) & (23) & (132) & (123) & e \\ \end{array}$ </onlyinclude> It can also often be seen presented as: :$\begin{array}{c|cccccc}   & e & p & q & r & s & t \\ \hline e & e & p & q & r & s & t \\ p & p & q & e & s & t & r \\ q & q & e & p & t & r & s \\ r & r & t & s & e & q & p \\ s & s & r & t & p & e & q \\ t & t & s & r & q & p & e \\ \end{array}$ == Sources == * {{BookReference|Sets and Groups|1965|J.A. Green|prev = Center of Symmetric Group is Trivial|next = Modulo Addition/Cayley Table/Modulo 6}}: Tables: $1$. Symmetric Group $\map S 3$ * {{BookReference|Modern Algebra|1965|Seth Warner|prev = Left and Right Operation Closed for All Subsets|next = Centralizer of Group Element is Subgroup}}: Exercise $8.6$ * {{BookReference|Topology: An Introduction with Application to Topological Groups|1967|George McCarty|prev = Definition:Cayley Table|next = Cayley Table for Commutative Operation is Symmetrical about Main Diagonal}}: Chapter $\text{II}$: Groups: Exercise $\text{A}$ * {{BookReference|A Course in Group Theory|1996|John F. Humphreys|prev = Order of Symmetric Group/Examples/Degree 3|next = Symmetry Group of Equilateral Triangle is Symmetric Group}}: Chapter $2$: Maps and relations on sets: Example $2.19$ Category:Examples of Cayley Tables Category:Symmetric Group on 3 Letters or9ffoyl7er55nqxdc1xxcfeakzw7k8


Lower Bound is Upper Bound for Inverse Ordering 0 38951 488978 212640 2020-09-18T23:05:43Z Prime.mover 59 wikitext text/x-wiki == Definition == Let $\struct {S, \preceq}$ be an ordered set. Let $T \subseteq S$. Let $m$ be a lower bound for $\struct {T, \preceq}$. Let $\succeq$ be the dual ordering of $\preceq$. Then $m$ is an upper bound for $\struct {T, \succeq}$. == Proof == Let $m$ be a lower bound for $\struct {T, \preceq}$. That is: :$\forall a \in T: m \preceq a$ By definition of dual ordering, it follows that: :$\forall a \in T: a \succeq m$ That is, $M$ is an upper bound for $\struct {T, \succeq}$. {{qed}} == Also see == * Upper Bound is Lower Bound for Inverse Ordering Category:Order Theory 3mg7i2sfm0uwu3tj54x0onxipf4xey5


Real Addition Identity is Zero/Corollary 0 39204 484166 473670 2020-08-30T13:20:59Z Prime.mover 59 wikitext text/x-wiki == Corollary to Real Addition Identity is Zero == <onlyinclude> :$\forall x, y \in \R: x + y = x \implies y = 0$ </onlyinclude> == Proof == {{begin-eqn}} {{eqn | l = x + y       | r = x       | c =  }} {{eqn | ll= \leadsto       | l = \paren {-x} + \paren {x + y}       | r = \paren {-x} + x       | c =  }} {{eqn | ll= \leadsto       | l = \paren {\paren {-x} + x} + y       | r = \paren {-x} + x       | c = Real Number Axioms: $\R \text A 1$ }} {{eqn | ll= \leadsto       | l = 0 + y       | r = 0       | c = Real Number Axioms: $\R \text A 4$ }} {{eqn | ll= \leadsto       | l = y       | r = 0       | c = Real Addition Identity is Zero }} {{end-eqn}} {{qed}} == Sources == * {{BookReference|Calculus|1967|Michael Spivak|prev = Definition:Additive Inverse of Number|next = Definition:Subtraction}}: Part $\text I$: Prologue: Chapter $1$: Basic Properties of Numbers * {{BookReference|Topology|2000|James R. Munkres|ed = 2nd|edpage = Second Edition|prev = Archimedean Principle|next = Real Zero is Zero Element}}: $1$: Set Theory and Logic: $\S 4$: The Integers and the Real Numbers: Exercise $1 \ \text{(a)}$ Category:Real Addition 0adpu22uwwmjuqy2w48prxp6xac52q9


Multiplication by Negative Real Number/Corollary 0 39213 473617 406778 2020-06-12T08:28:10Z Prime.mover 59 wikitext text/x-wiki == Corollary to Multiplication by Negative Real Number == <onlyinclude> : $\forall x \in \R: \paren {-1} \times x = -x$ </onlyinclude> == Proof == {{begin-eqn}} {{eqn | l = \paren {-1} \times x       | r = -\paren {1 \times x}       | c = Multiplication by Negative Real Number }} {{eqn | r = -x       | c = Real Number Axioms: $\R M 3$: Identity for Multiplication }} {{end-eqn}} {{qed}} == Sources == * {{BookReference|Topology|2000|James R. Munkres|ed = 2nd|edpage = Second Edition|prev = Multiplication by Negative Real Number|next = Multiplication of Real Numbers is Left Distributive over Subtraction/Proof 2}}: $1$: Set Theory and Logic: $\S 4$: The Integers and the Real Numbers: Exercise $1 \ \text{(f)}$ Category:Real Multiplication rr96djmwtyqep33grsxamtcvprif4hj


Cartesian Product of Bijections is Bijection/General Result 0 39315 215582 2015-04-30T08:06:02Z Prime.mover 59 Created page with "== Theorem == <onlyinclude> Let $I$ be an indexing set.  Let $\left\langle{S_i}\right\rangle_{i \mathop \in I}$ and $\left\langle{T_i}\right\rangle..." wikitext text/x-wiki == Theorem == <onlyinclude> Let $I$ be an indexing set. Let $\left\langle{S_i}\right\rangle_{i \mathop \in I}$ and $\left\langle{T_i}\right\rangle_{i \mathop \in I}$ be families of sets indexed by $I$. Let $\displaystyle \mathcal S := \prod_{i \mathop \in I} S_i$ and $\displaystyle \mathcal T := \prod_{i \mathop \in I} T_i$ be the Cartesian products of $\left\langle{S_i}\right\rangle_{i \mathop \in I}$ and $\left\langle{T_i}\right\rangle_{i \mathop \in I}$ respectively. For all $i \in I$, let $f_i: S_i \to T_i$ be a bijection. Let $\displaystyle \mathcal F: \mathcal S \to \mathcal T := \prod_{i \mathop \in I} \left({f_i: S_i \to T_i}\right)$ be the Cartesian product of $\left\langle{f_i}\right\rangle_{i \mathop \in I}$ defined as: :$\displaystyle \forall \mathbf s \in \mathcal S: \mathcal F \left({\mathbf s}\right) := \prod_{i \mathop \in I} \left({f_i \left({s_i}\right)}\right)$ where $\mathbf s := \left\langle{s_i}\right\rangle_{i \mathop \in I}$ is an arbitrary element of $\left\langle{S_i}\right\rangle_{i \mathop \in I}$. Then $\mathcal F$ is a bijection. </onlyinclude> == Proof == Because $f_i$ are bijections, it follows by definition that they are surjections. Let $\mathbf t := \left\langle{t_i}\right\rangle_{i \mathop \in I} \in \mathcal T$. Then as $f_i$ is a surjection: :$\forall i \in I: \exists s_i \in S_i: f_i \left({s_i}\right) = t_i$ Thus: :$\exists \mathbf s \in \mathcal S: \mathcal F \left({\mathbf s}\right) = \mathbf t$ So $\mathcal F$ is a surjection. Because $f_i$ are bijections, it follows by definition that they are injections. Let: :$\mathbf t_1 := \left\langle{t_{i 1} }\right\rangle_{i \mathop \in I} \in \mathcal T$ :$\mathbf t_2 := \left\langle{t_{i 2} }\right\rangle_{i \mathop \in I} \in \mathcal T$ Let: : $\mathcal F \left({\mathbf s_1}\right) = \left({\mathbf t_1}\right), \mathcal F \left({\mathbf s_2}\right) = \left({\mathbf t_2}\right)$ for some $\mathbf s_1, \mathbf s_2 \in \mathcal S$ where: :$\mathbf s_1 := \left\langle{s_{i 1} }\right\rangle_{i \mathop \in I}$ :$\mathbf s_2 := \left\langle{s_{i 2} }\right\rangle_{i \mathop \in I}$ Suppose $\mathbf t_1 = \mathbf t_2$. Then: : $\forall i \in I: t_{i 1} = t_{i 2}$ By definition of $\mathcal F$: :$\forall i \in I: f_i \left({s_{i 1} }\right) = t_{i 1}$ :$\forall i \in I: f_i \left({s_{i 2} }\right) = t_{i 2}$ As $f_i$ is an injection for all $i \in I$: :$\forall i \in I: t_{i 1} = t_{i 2 }\implies s_{i 1} = s_{i 2}$ Thus it follows that: :$\mathbf t_1 = \mathbf t_2 \implies \mathbf s_1 = \mathbf s_2$ and so $\mathcal F$ is an injection. So $\mathcal F$ is a surjection and also an injection. Hence by definition, $\mathcal F$ is a bijection. {{qed}} Category:Bijections Category:Cartesian Product Category:Indexed Families 33dptureruoso9fnoj579ovc3zn9efq


Invertible Element of Monoid is Cancellable 0 39369 330008 215967 2017-11-25T22:26:53Z Prime.mover 59 wikitext text/x-wiki == Corollary to Invertible Element of Associative Structure is Cancellable == <onlyinclude> Let $\left({S, \circ}\right)$ be a monoid whose identity is $e_S$. An element of $\left({S, \circ}\right)$ which is invertible is also cancellable. </onlyinclude> == Proof == By definition of monoid, $\left({S, \circ}\right)$ is an algebraic structure in which $\circ$ is associative. The result follows from Invertible Element of Associative Structure is Cancellable. {{Qed}} Category:Monoids Category:Inverse Elements Category:Cancellability 8yx9w4i5jqf0d66ape2klha7p8t23ge


Group/Examples/inv x = 1 - x/Lemma 1

Group/Examples/inv x = 1 - x/Lemma 1 0 40103 374165 374161 2018-10-31T07:46:24Z Prime.mover 59 wikitext text/x-wiki == Lemma for Group Example: $x^{-1} = 1 - x$ == Define $f: \openint 0 1 \to \R$ by: :$\map f x := \map \ln {\dfrac {1 - x} x}$ and $g: \R \to \openint 0 1$: :$\map g z := \dfrac 1 {1 + \exp z}$ Then: <onlyinclude> :$\map {f \circ g} x = x$ </onlyinclude> == Proof == {{begin-eqn}} {{eqn | l = \map {f \circ g} x       | r = \map f {\map g x}       | c = }} {{eqn | r = \map \ln {\frac {1 - \map g x} {\map g x} }       | c =  }} {{eqn | r = \map \ln {\frac {1 - \frac 1 {1 + \exp x} } {\frac 1 {1 + \exp x} } }       | c =  }} {{eqn | r = \map \ln {\frac {1 + \exp x - 1} 1}       | c = multiplying both numerator and denominator by $1 + \exp x$ }} {{eqn | r = \map \ln {\exp x}       | c = Simplifying }} {{eqn | r = x       | c = Exponential of Natural Logarithm }} {{end-eqn}} {{qed|lemma}} Category:Group/Examples/inv x = 1 - x j1s1au4kihahej6uaqr5f92l1g7qboy


Group/Examples/inv x = 1 - x/Lemma 2

Group/Examples/inv x = 1 - x/Lemma 2 0 40104 436941 374163 2019-11-27T13:33:25Z Prime.mover 59 wikitext text/x-wiki == Lemma for Group Example: $x^{-1} = 1 - x$ == Define $f: \openint 0 1 \to \R$ by: :$\map f x := \map \ln {\dfrac {1 - x} x}$ and $g: \R \to \openint 0 1$: :$\map g z := \dfrac 1 {1 + \exp z}$ Then: <onlyinclude> :$\map {g \circ f} x = x$ </onlyinclude> == Proof == {{begin-eqn}} {{eqn | l = \map {g \circ f} x       | r = \map g {\map f x}       | c = }} {{eqn | r = \frac 1 {1 + \exp \map f x}       | c = }} {{eqn | r = \frac 1 {1 + \map \exp {\ln \dfrac {1 - x} x} }       | c = }} {{eqn | r = \frac 1 {1 + \dfrac {1 - x} x}       | c = Exponential of Natural Logarithm }} {{eqn | r = \frac x {x + 1 - x}       | c = multiplying both numerator and denominator by $x$ }} {{eqn | r = x       | c = simplifying }} {{end-eqn}} {{qed|lemma}} Category:Group/Examples/inv x = 1 - x iqjd5dbz1lzzzm513bbk5nvn3zhg34c


Non-Abelian Order 8 Group with One Order 2 Element is Quaternion Group/Lemma 2 0 40263 370920 248745 2018-10-13T20:39:30Z Prime.mover 59 wikitext text/x-wiki == Lemma == <onlyinclude> :$\paren {\pm a}^2 = \paren {\pm b}^2 = \paren {\pm c}^2 = -1$ </onlyinclude> == Proof == {{WLOG}}, $a$ is checked. The proofs for other $5$ elements are similar. {{begin-eqn}} {{eqn | l = \paren {a^2}^2       | r = a^4       | c = Powers of Group Elements: Product of Indices }} {{eqn | r = 1       | c = Definition of Order of Group Element }} {{end-eqn}} So $a^2 = 1$ or $a^2 = -1$. As the order of $a = 4$: :$a^2 \ne 1$ Hence: :$a^2 = -1$ as required. {{qed}} Category:Non-Abelian Order 8 Group with One Order 2 Element is Quaternion Group cf4jnmridoq9xljcikvp5r2l4rtp8qr


Euler Triangle Formula/Lemma

Euler Triangle Formula/Lemma 0 43708 359230 238850 2018-06-18T20:55:31Z Prime.mover 59 wikitext text/x-wiki == Lemma to Euler Triangle Formula == <onlyinclude> :400px 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: :$AP = BP = IP$ </onlyinclude> == Proof == {{WLOG}}, it will be demonstrated that $BP = IP$. Let $CP$ be the bisector of $\angle ACB$. We have therefore that: :$\angle ACP = \angle ICB$ 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: {{begin-eqn}} {{eqn | l = \angle IBP       | r = \angle IBA + \angle ABP       | c =  }} {{eqn | r = \angle IBC + \angle ICB       | c =  }} {{end-eqn}} 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: {{begin-eqn}} {{eqn | l = \angle PIB       | r = \angle IBC + \angle ICB       | c =  }} {{eqn | r = \angle IBP       | c =  }} {{end-eqn}} and from Triangle with Two Equal Angles is Isosceles: :$IP = BP$ The proof that $IP = AP$ follows the same lines. {{qed}} Category:Triangles t682guz08q3t6tq3k9v2gbiq9n6yll4


Cassini's Identity/Lemma

Cassini's Identity/Lemma 0 43712 359832 359831 2018-06-30T09:25:43Z Prime.mover 59 wikitext text/x-wiki == Lemma for Cassini's Identity == <onlyinclude> :$\forall n \in \Z_{>1}: \begin{bmatrix} F_{n + 1} & F_n \\ F_n       & F_{n - 1} \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}^n$ </onlyinclude> == Proof == === Basis for the Induction === :$\begin{bmatrix} F_2 & F_1     \\ F_1 & F_0 \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}^1$ === Induction Hypothesis === For $k \in \Z_{>1}$, it is assumed that: :$\begin{bmatrix} F_{k + 1} & F_k \\ F_k       & F_{k - 1} \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}^k$ It remains to be shown that: :$\begin{bmatrix} F_{k + 2} & F_{k + 1}  \\ F_{k + 1} & F_k \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}^{k + 1}$ === Induction Step === The induction step follows from conventional matrix multiplication: {{begin-eqn}} {{eqn | l = \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}^{k+1}       | r = \begin{bmatrix} F_{k + 1} & F_k \\ F_k       & F_{k - 1} \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}       | c = by the induction hypothesis }} {{eqn | r = \begin{bmatrix} F_{k + 1} + F_k & F_{k + 1} \\ F_k + F_{k - 1} & F_k \end{bmatrix}       | c = Definition of Matrix Product }} {{eqn | r = \begin{bmatrix} F_{k + 2} & F_{k + 1} \\ F_{k + 1} & F_k \end{bmatrix} }} {{end-eqn}} So by induction: :$\begin{bmatrix} F_{n + 1} & F_n \\ F_n       & F_{n - 1} \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}^n$ {{qed}} == Sources == * {{BookReference|The Art of Computer Programming: Volume 1: Fundamental Algorithms|1997|Donald E. Knuth|ed = 3rd|edpage = Third Edition|prev = Fibonacci Numbers which equal the Square of their Index|next = Prime Fibonacci Number has Prime Index except for 3}}: $\S 1.2.8$: Fibonacci Numbers: Exercise $6$ Category:Cassini's Identity 6ycdp5tauruav9p4rowavwxc8r7nfmy


Real Numbers are Uncountable/Proof 2 using Ternary Notation/Lemma 0 43940 301553 301548 2017-06-17T07:00:16Z Prime.mover 59 wikitext text/x-wiki == Lemma to Real Numbers are Uncountable: Proof 2 using Ternary Notation == <onlyinclude> Let $\left\langle{d_n}\right\rangle$ and $\left\langle{e_n}\right\rangle$ be infinite sequences in $\left\{{0, 1}\right\}$ such that: :$\exists m \in \N: d_m \ne e_m$ That is, the sequences $\left\langle{d_n}\right\rangle$ and $\left\langle{e_n}\right\rangle$ are different in at least one term. Then the ternary representations $D = 0.d_1 d_2 \ldots$ and $E = 0.e_1 e_2 \dots$ represent distinct real numbers. </onlyinclude> == Proof == Let $\left\langle{d_n}\right\rangle \ne \left\langle{e_n}\right\rangle$. By the Well-Ordering Principle, there is a smallest $n \in \N_{>0}$ such that $d_n \ne e_n$. {{WLOG}}, suppose that $d_n = 0$ and $e_n = 1$. Let $K = 0.d_1 d_2 \ldots d_{n-1} = \sum_{i \mathop = 1}^{n-1} d_i 3^{-i}$. Let: : $\displaystyle D := K + \sum_{i \mathop = n+1}^\infty d_i 3^{-i}$ : $\displaystyle E := K + 3^{-n} + \sum_{i \mathop = n+1}^\infty e_i 3^{-i} \ge K + 3^{-n}$ But then: {{begin-eqn}} {{eqn | l = D \le K + \sum_{i \mathop = n + 1}^\infty 3^{-i}       | r = K + 3^{-n-1} \sum_{i \mathop = 0}^\infty 3^{-i}       | c =  }} {{eqn | r = K + 3^{-n - 1} \dfrac 3 2       | c =  }} {{eqn | r = K + \frac {3^{-n} } 2       | c =  }} {{end-eqn}} Thus $D < E$, so $D \ne E$. {{qed}} Category:Real Numbers are Uncountable 6hhvrfh9ocg3rm55ze949678cd1ywcl


Complex Sequence is Cauchy iff Convergent/Lemma 1 0 43973 364168 364167 2018-09-03T06:22:47Z Prime.mover 59 wikitext text/x-wiki {{refactor|Extract this into a second definition of Definition:Complex Cauchy Sequence.}} == Lemma for Complex Sequence is Cauchy iff Convergent == <onlyinclude> Let $\sequence {z_n}$ be a complex sequence. Let $\mathcal N$ be the domain of $\sequence {z_n}$. Let $x_n = \Re \paren {z_n}$ for every $n \in \mathcal N$. Let $y_n = \Im \paren {z_n}$ for every $n \in \mathcal N$. Then $\sequence {z_n}$ is a (complex) Cauchy sequence {{iff}} $\sequence {x_n}$ and $\sequence {y_n}$ are (real) Cauchy sequences. </onlyinclude> == Proof == === Necessary Condition === Let $\sequence {z_n}$ be a Cauchy sequence. This means that, for a given $\epsilon > 0$: :$\exists N: \forall m, n \in \mathcal N: m, n \ge N: \cmod {z_n - z_m} < \epsilon$ We have, for every $m, n \ge N$: {{begin-eqn}} {{eqn | l = \cmod {x_n − x_m}       | r = \cmod {\Re \paren {z_n − z_m} }       | c =  }} {{eqn | o = \le       | r = \cmod {z_n − z_m}       | c =  }} {{eqn | o = <       | r = \epsilon       | c =  }} {{end-eqn}} Thus $\sequence {x_n}$ is a Cauchy sequence by definition. A similar argument shows that $\sequence {y_n}$ is a Cauchy sequence. {{qed|lemma}} === Sufficient Condition === Let $\sequence {x_n}$ and $\sequence {y_n}$ be Cauchy sequences. This means for $\sequence {x_n}$ that, for a given $\epsilon > 0$: : $\exists N_1: \forall m, n \in \mathcal N: m, n \ge N_1: \cmod {x_n - x_m} < \dfrac \epsilon 2$ Also, for $\sequence {y_n}$: : $\exists N_2: \forall m, n \in \mathcal N: m, n \ge N_2: \cmod {y_n - y_m} < \dfrac \epsilon 2$ Let $N = \max \paren {N_1, N_2}$. Let $i = \sqrt {-1}$ denote the imaginary unit. We have, for every $m, n \ge N$: {{begin-eqn}} {{eqn | l = \cmod {z_n − z_m}       | r = \cmod {x_n + i y_n − \paren {x_m + i y_m} }       | c = as $z_n = \Re \paren {z_n} + i \Im \paren {z_n}$ and $z_m = \Re \paren {z_m} + i \Im \paren {z_m}$ }} {{eqn | r = \cmod {x_n − x_m + i \paren {y_n − y_m} }       | c =  }} {{eqn | o = \le       | r = \cmod {x_n − x_m} + \cmod {i\paren {y_n − y_m} }       | c = Triangle Inequality for Complex Numbers }} {{eqn | r = \cmod {x_n − x_m} + \cmod {y_n − y_m}       | c = {{Defof|Complex Modulus}} }} {{eqn | o = <       | r = \frac \epsilon 2 + \cmod {y_n − y_m}       | c = since $\cmod {x_n - x_m} < \dfrac \epsilon 2$ }} {{eqn | o = <       | r = \frac \epsilon 2 + \frac \epsilon 2       | c = since $\cmod {y_n - y_m} < \dfrac \epsilon 2$ }} {{eqn | r = \epsilon       | c =  }} {{end-eqn}} Thus $\sequence {z_n}$ is a Cauchy sequence by definition. {{qed}} Category:Complex Sequence is Cauchy iff Convergent d0t0livlp8jmcvldtfykw3pbyrvay3l


Field Adjoined Set/Corollary

Field Adjoined Set/Corollary 0 43999 352118 330667 2018-05-01T06:07:37Z Prime.mover 59 wikitext text/x-wiki == Corollary to Field Adjoined Set == Let $F$ be a field. Let $S \subseteq F$ be a subset of $F$. Ket $K \le F$ be a subfield of $F$. <onlyinclude> Let $A = K \left[{X_1, \ldots, X_n}\right]$ be the ring of polynomial functions in $n$ indeterminates over $K$.  Let $B = K \left({X_1, \ldots, X_n}\right)$ be the field of rational functions in $n$ indeterminates over $K$.  Let $\alpha_1, \ldots, \alpha_n \in F$. Then: :$(1): \quad x \in K \left[{\alpha_1, \ldots, \alpha_n}\right] \iff x = f \left({\alpha_1, \ldots, \alpha_n}\right)$ for some $f \in A$ :$(2): \quad x \in K \left({\alpha_1, \ldots, \alpha_n}\right) \iff x = f \left({\alpha_1, \ldots, \alpha_n}\right)$ for some $f \in B$ :$(3): \quad x \in K \left[{S}\right] \iff x \in K \left[{\alpha_1, \ldots, \alpha_n}\right]$ for some $\alpha_1, \ldots, \alpha_n \in S$ :$(4): \quad x \in K \left({S}\right) \iff x \in K \left({\alpha_1, \ldots, \alpha_n}\right)$ for some $\alpha_1, \ldots, \alpha_n \in S$ </onlyinclude> == Proof == {{proof wanted}} Category:Field Extensions 9btb2at5tqm0twod4kmx52gpfrxehxx


Rule of Conjunction/Proof Rule

Rule of Conjunction/Proof Rule 0 44547 493660 413771 2020-10-10T12:45:45Z Prime.mover 59 wikitext text/x-wiki == Proof Rule == The Rule of Conjunction is a valid deduction sequent in propositional logic. As a proof rule it is expressed in the form: <onlyinclude> :If we can conclude both $\phi$ and $\psi$, we may infer the compound statement $\phi \land \psi$. </onlyinclude> It can be written: :$\ds {\phi \qquad \psi \over \phi \land \psi} \land_i$ === Tableau Form === {{:Rule of Conjunction/Proof Rule/Tableau Form}} == Explanation == {{:Rule of Conjunction/Explanation}} == Also known as == {{:Rule of Conjunction/Also known as}} == Also see == * This is a rule of inference of the following proof system: ** Definition:Hilbert Proof System/Instance 2 * Rule of Simplification == Technical Note == {{:Rule of Conjunction/Proof Rule/Technical Note}} == Sources == * {{BookReference|Introduction to Symbolic Logic|1959|A.H. Basson|author2 = D.J. O'Connor|ed = 3rd|edpage = Third Edition|prev = Modus Ponendo Ponens/Proof Rule|next = Definition:Axiom (Formal Systems)}}: $\S 4.2$: The Construction of an Axiom System: $RST \, 4$ * {{BookReference|Logic: Techniques of Formal Reasoning|1964|Donald Kalish|author2 = Richard Montague|prev = Rule of Simplification/Proof Rule|next = Rule of Addition/Proof Rule}}: $\text{II}$: 'AND', 'OR', 'IF AND ONLY IF': $\S 3$ * {{BookReference|Beginning Logic|1965|E.J. Lemmon|prev = Definition:Disjunct|next = Rule of Conjunction/Sequent Form/Formulation 1/Proof 1}}: $\S 1.3$: Conjunction and Disjunction * {{BookReference|Symbolic Logic|1973|Irving M. Copi|ed = 4th|edpage = Fourth Edition|prev = Rule of Simplification|next = Rule of Addition/Proof Rule}}: $3.1$: Formal Proof of Validity * {{BookReference|Elementary Logic|1980|D.J. O'Connor|author2 = Betty Powell|prev = Rule of Addition/Proof Rule|next = Definition:Logical Equivalence}}: $\S \text{II}$: The Logic of Statements $(2): \ 1$: Decision procedures and proofs: $9$ * {{BookReference|Logic in Computer Science: Modelling and reasoning about systems|2000|Michael R.A. Huth|author2 = Mark D. Ryan|prev = Definition:Valid Argument|next = Rule of Simplification/Proof Rule}}: $\S 1.2.1$: Rules for natural deduction Category:Proof Rules Category:Rule of Conjunction qu0o2hcxjq6fbts7tehnh9grskpfnbl


Rule of Simplification/Proof Rule

Rule of Simplification/Proof Rule 0 44549 493659 244443 2020-10-10T12:45:26Z Prime.mover 59 wikitext text/x-wiki == Proof Rule == The Rule of Simplification is a valid deduction sequent in propositional logic. As a proof rule it is expressed in either of the two forms: <onlyinclude> :$(1): \quad$ If we can conclude $\phi \land \psi$, then we may infer $\phi$. :$(2): \quad$ If we can conclude $\phi \land \psi$, then we may infer $\psi$. </onlyinclude> It can be written: :$\ds {\phi \land \psi \over \phi} \land_{e_1} \qquad \qquad {\phi \land \psi \over \psi} \land_{e_2}$ === Tableau Form === {{:Rule of Simplification/Proof Rule/Tableau Form}} == Explanation == {{:Rule of Simplification/Explanation}} == Also known as == {{:Rule of Simplification/Also known as}} == Also see == * Rule of Conjunction == Technical Note == {{:Rule of Simplification/Proof Rule/Technical Note}} == Sources == * {{BookReference|Introduction to Logic and to the Methodology of Deductive Sciences|1946|Alfred Tarski|ed = 2nd|edpage = Second Edition|prev = Definition:Statement Variable|next = Law of Identity}}: $\S \text{II}.12$: Laws of sentential calculus * {{BookReference|Logic: Techniques of Formal Reasoning|1964|Donald Kalish|author2 = Richard Montague|prev = Definition:Language of Propositional Logic|next = Rule of Conjunction/Proof Rule}}: $\text{II}$: 'AND', 'OR', 'IF AND ONLY IF': $\S 3$ * {{BookReference|Beginning Logic|1965|E.J. Lemmon|prev = Rule of Exportation/Forward Implication/Formulation 1/Proof|next = Rule of Simplification/Sequent Form/Formulation 1/Form 1/Proof 1}}: $\S 1.3$: Conjunction and Disjunction * {{BookReference|Symbolic Logic|1973|Irving M. Copi|ed = 4th|edpage = Fourth Edition|prev = Destructive Dilemma/Formulation 2|next = Rule of Conjunction/Proof Rule}}: $3.1$: Formal Proof of Validity * {{BookReference|Elementary Logic|1980|D.J. O'Connor|author2 = Betty Powell|prev = Hypothetical Syllogism/Formulation 1|next = Rule of Addition}}: $\S \text{II}$: The Logic of Statements $(2): \ 1$: Decision procedures and proofs: $7$ * {{BookReference|Logic in Computer Science: Modelling and reasoning about systems|2000|Michael R.A. Huth|author2 = Mark D. Ryan|prev = Rule of Conjunction/Proof Rule|next = Definition:Tableau Proof (Formal Systems)}}: $\S 1.2.1$: Rules for natural deduction Category:Proof Rules Category:Rule of Simplification 62l6x8nvxht42wkim5k8bl4c5aeviun


Quadratic Residue/Examples/11

Quadratic Residue/Examples/11 0 44728 433459 431422 2019-11-01T13:30:08Z Prime.mover 59 wikitext text/x-wiki == Example of Quadratic Residues == <onlyinclude> The set of quadratic residues modulo $11$ is: :$\set {1, 3, 4, 5, 9}$ </onlyinclude> {{OEIS|A010375}} == Proof == To list the quadratic residues of $11$ it is enough to work out the squares $1^2, 2^2, \ldots, 10^2$ modulo $11$. {{begin-eqn}} {{eqn | l = 1^2       | o = \equiv       | r = 1       | rr= \pmod {11} }} {{eqn | l = 2^2       | o = \equiv       | r = 4       | rr= \pmod {11} }} {{eqn | l = 3^2       | o = \equiv       | r = 9       | rr= \pmod {11} }} {{eqn | l = 4^2       | o = \equiv       | r = 5       | rr= \pmod {11} }} {{eqn | l = 5^2       | o = \equiv       | r = 3       | rr= \pmod {11} }} {{eqn | l = 6^2       | o = \equiv       | r = 3       | rr= \pmod {11} }} {{eqn | l = 7^2       | o = \equiv       | r = 5       | rr= \pmod {11} }} {{eqn | l = 8^2       | o = \equiv       | r = 9       | rr= \pmod {11} }} {{eqn | l = 9^2       | o = \equiv       | r = 4       | rr= \pmod {11} }} {{eqn | l = 10^2       | o = \equiv       | r = 1       | rr= \pmod {11} }} {{end-eqn}} So the set of quadratic residues modulo $11$ is: :$\set {1, 3, 4, 5, 9}$ The set of quadratic non-residues of $11$ therefore consists of all the other non-zero least positive residues: :$\set {2, 6, 7, 8, 10}$ {{qed}} == Sources == * {{BookReference|Number Theory|1971|George E. Andrews|prev = Definition:Quadratic Residue|next = Quadratic Residue/Examples/3}}: $\text {3-5}$ The Use of Computers in Number Theory: Exercise $6$ * {{BookReference|Taming the Infinite|2008|Ian Stewart|prev = Law of Quadratic Reciprocity|next = Definition:Quadratic Residue}}: Chapter $7$: Patterns in Numbers: Gauss Category:Examples of Quadratic Residues Category:11 td5wn2d2avl2viu77ndm01a48raj3jj


Kepler's Laws of Planetary Motion/First Law 0 44852 447517 446166 2020-02-08T19:19:48Z Prime.mover 59 wikitext text/x-wiki == Physical Law == '''Kepler's first law of planetary motion''' is one of the three physical laws of celestial mechanics deduced by {{AuthorRef|Johannes Kepler}}: <onlyinclude> : Planets move around the sun in elliptical orbits, with the sun at one focus. </onlyinclude> == Proof == Consider a planet $p$ of mass $m$ moving around the sun in the plane under the influence of the force $\mathbf F$ imparted by the gravitational field which the two bodies give rise to. Let the position of $p$ at time $t$ be given in polar coordinates as $\polar {r, \theta}$. Let $\mathbf F$ be expressed as: :$\mathbf F = F_r \mathbf u_r + F_\theta \mathbf u_\theta$ where: :$\mathbf u_r$ is the unit vector in the direction of the radial coordinate of $p$ :$\mathbf u_\theta$ is the unit vector in the direction of the angular coordinate of $p$ :$F_r$ and $F_\theta$ are the magnitudes of the components of $\mathbf F$ in the directions of $\mathbf u_r$ and $\mathbf u_\theta$ respectively. Let $\mathbf r$ be the radius vector from the origin to $p$. From Motion of Particle in Polar Coordinates, the second order ordinary differential equations governing the motion of $m$ under the force $\mathbf F$ are: {{begin-eqn}} {{eqn | n = 1       | l = F_\theta       | r = m \paren {r \dfrac {\d^2 \theta} {\d t^2} + 2 \dfrac {\d r} {\d t} \dfrac {\d \theta} {\d t} }       | c =  }} {{eqn | n = 2       | l = F_r       | r = m \paren {\dfrac {\d^2 r} {\d t^2} - r \paren {\dfrac {\d \theta} {\d t} }^2}       | c =  }} {{end-eqn}} By Newton's Law of Universal Gravitation, $\mathbf F$ is a central force of value: :$\mathbf F = -G \dfrac {M m} {r^3} \mathbf r$ where: :$G$ is the gravitational constant :$M$ and $m$ are the masses of the bodies :the minus sign indicates that the central force is in the opposite direction to that of the radius vector $\mathbf r$. :600px Hence we have: :$F_\theta = 0$ :$F_r = -G \dfrac {M m} {r^2} \dfrac {\size {\mathbf r} } r = -G \dfrac {M m} {r^2}$ As $G$ and $M$ are both constants, so we can express this as: :$F_r = -\dfrac {k m} {r^2}$ So from $(2)$: {{begin-eqn}} {{eqn | l = -\dfrac {k m} {r^2}       | r = m \paren {\dfrac {\d^2 r} {\d t^2} - r \paren {\dfrac {\d \theta} {\d t} }^2}       | c =  }} {{eqn | n = 3       | ll= \leadsto       | l = \dfrac {\d^2 r} {\d t^2} - r \paren {\dfrac {\d \theta} {\d t} }^2       | r = -\dfrac k {r^2}       | c =  }} {{end-eqn}} Let $z = \dfrac 1 r$. Then: {{begin-eqn}} {{eqn | l = \dfrac {\d r} {\d t}       | r = \dfrac \d {\d t} \dfrac 1 z       | c =  }} {{eqn | r = -\dfrac 1 {z^2} \dfrac {\d z} {\d t}       | c = Power Rule for Derivatives }} {{eqn | r = -\dfrac 1 {z^2} \dfrac {\d z} {\d \theta} \dfrac {\d \theta} {\d t}       | c = Chain Rule for Derivatives }} {{eqn | r = -\dfrac 1 {z^2} \dfrac {\d z} {\d \theta} \dfrac h {r^2}       | c = Derivative of Angular Component under Central Force }} {{eqn | r = -h \dfrac {\d z} {\d \theta}       | c = as $r = \dfrac 1 z$ }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \dfrac {\d^2 r} {\d t^2}       | r = -h \map {\dfrac \d {\d t} } {\dfrac {\d z} {\d \theta} }       | c =  }} {{eqn | r = -h \map {\dfrac \d {\d \theta} } {\dfrac {\d z} {\d \theta} } \dfrac {\d \theta} {\d t}       | c = Chain Rule for Derivatives }} {{eqn | r = -h \dfrac {\d^2 z} {\d \theta^2} \dfrac {\d \theta} {\d t}       | c = Chain Rule for Derivatives }} {{eqn | r = -h \dfrac {\d^2 z} {\d \theta^2} \dfrac h {r^2}       | c = Derivative of Angular Component under Central Force }} {{eqn | n = 4       | r = -h^2 z^2 \dfrac {\d^2 z} {\d \theta^2}       | c = as $r = \dfrac 1 z$ }} {{end-eqn}} Substituting from $(4)$ into $(3)$: {{begin-eqn}} {{eqn | l = -h^2 z^2 \dfrac {\d^2 z} {\d \theta^2} - r \paren {\dfrac {\d \theta} {\d t} }^2       | r = -\dfrac k {r^2}       | c =  }} {{eqn | ll= \leadsto       | l = -h^2 z^2 \dfrac {\d^2 z} {\d \theta^2} - r \paren {\dfrac h {r^2} }^2       | r = -\dfrac k {r^2}       | c = Derivative of Angular Component under Central Force }} {{eqn | ll= \leadsto       | l = -h^2 z^2 \dfrac {\d^2 z} {\d \theta^2} - \frac 1 z \paren {\dfrac h {z^2} }^2       | r = -k z^2       | c = substituting $r = \dfrac 1 z$ }} {{eqn | n = 5       | ll= \leadsto       | l = \dfrac {\d^2 z} {\d \theta^2} + z       | r = \dfrac k {h^2}       | c = simplifying }} {{end-eqn}} From Linear Second Order ODE: $y'' + y = K$, $(5)$ has the general solution: :$(6): \quad z = A \sin \theta + B \cos \theta + \dfrac k {h^2}$ {{WLOG}}, we shift the polar axis so as to make $r$ a minimum when $\theta = 0$. That is, when $\theta = 0$, $p$ is at its closest point to the origin. This means that $z = \dfrac 1 z$ is a  maximum at this point. Thus at $\theta = 0$: From Derivative at Maximum or Minimum: :$\dfrac {\d z} {\d \theta} = 0$ From Second Derivative of Strictly Concave Real Function is Strictly Negative: :$\dfrac {\d^2 z} {\d \theta^2} < 0$ Thus: :$A = 0$ :$B > 0$ Substituting these and $z = \dfrac 1 r$ back into $(6)$: :$r = \dfrac 1 {B \cos \theta + \dfrac k {h^2} }$ {{begin-eqn}} {{eqn | l = r       | r = \dfrac 1 {B \cos \theta + k / h^2}       | c =  }} {{eqn | r = \dfrac {h^2 / k} {1 + \paren {B h^2 / k} \cos \theta}       | c =  }} {{end-eqn}} Setting $B h^2 / k = e$: :$r = \dfrac {h^2 / k} {1 + e \cos \theta}$ where $e \in \R_{>0}$ and is constant. This can be expressed as: :$(7): \quad r = \dfrac {p e} {1 + e \cos \theta}$ by setting $p = \dfrac 1 B$. From Equation of Conic Section in Polar Form, $(7)$ is the equation of a conic section with one focus at the origin. As the planets remain in the solar system it follows that their orbits are stable and elliptical. {{qed}} == Also see == * Kepler's Second Law of Planetary Motion * Kepler's Third Law of Planetary Motion {{Namedfor|Johannes Kepler|cat = Kepler}} == Historical Note == {{:Kepler's Laws of Planetary Motion/Historical Note}} == Sources == * {{BookReference|Men of Mathematics|1937|Eric Temple Bell|prev = Kepler's Laws of Planetary Motion|next = Definition:Equidistance Property of Ellipse}}: Chapter $\text{VI}$: On the Seashore * {{BookReference|Differential Equations|1972|George F. Simmons|prev = Definition:Gravitational Constant|next = Definition:Eccentricity of Conic Section}}: $\S 3.21$: Newton's Law of Gravitation * {{BookReference|Calculus Gems|1992|George F. Simmons|prev = Definition:Focus/Linguistic Note|next = Kepler's Laws of Planetary Motion/Historical Note}}: Chapter $\text {A}.10$: Kepler ({{DateRange|1571|1630}}) * {{BookReference|Calculus Gems|1992|George F. Simmons|prev = Definition:Gravitational Constant|next = Kepler's Third Law of Planetary Motion}}: Chapter $\text {B}.25$: Kepler's Laws and Newton's Law of Gravitation * {{BookReference|The Penguin Dictionary of Mathematics|1998|David Nelson|ed = 2nd|edpage = Second Edition|prev = Kepler's Laws of Planetary Motion/Historical Note|next = Kepler's Second Law of Planetary Motion|entry = Kepler's laws}} * {{BookReference|The Penguin Dictionary of Mathematics|2008|David Nelson|ed = 4th|edpage = Fourth Edition|prev = Kepler's Laws of Planetary Motion/Historical Note|next = Kepler's Second Law of Planetary Motion|entry = Kepler's laws}} * {{BookReference|Taming the Infinite|2008|Ian Stewart|prev = Kepler's Laws of Planetary Motion|next = Kepler's Second Law of Planetary Motion}}: Chapter $8$: The System of the World: Kepler * {{BookReference|The Concise Oxford Dictionary of Mathematics|2014|Christopher Clapham|author2 = James Nicholson|ed = 5th|edpage = Fifth Edition|prev = Kepler's Laws of Planetary Motion|next = Kepler's Second Law of Planetary Motion|entry = Kepler's laws of planetary motion|subentry = i}} Category:Kepler's Laws of Planetary Motion geui0bvlqyjp7nwlo7jqqf5vx3pd8cm


Kepler's Laws of Planetary Motion/Second Law 0 44856 446167 431497 2020-02-04T09:51:30Z Prime.mover 59 wikitext text/x-wiki == Physical Law == '''Kepler's second law of planetary motion''' is one of the three physical laws of celestial mechanics deduced by {{AuthorRef|Johannes Kepler}}: <onlyinclude> : The line joining the sun and a planet sweeps out equal areas in equal times. :500px </onlyinclude> == Proof == Consider a planet $p$ of mass $m$ moving around the sun in the plane under the influence of the force $\mathbf F$ imparted by the gravitational field which the two bodies give rise to. Let the position of $p$ at time $t$ be given in polar coordinates as $\polar {r, \theta}$. By definition of the gravitational field, $\mathbf F$ is a central force. From Derivative of Angular Component under Central Force: :$r^2 \dfrac {\d \theta} {\d t} = h$ for some constant $h$. {{WLOG}}, assume that $h > 0$, which means that $p$ is travelling in the direction of positive $\theta$. Let $\map A t$ be the area swept out by $\mathbf r$ in time $t$ relative to some fixed point of reference. For a small angle $\delta \theta$, the area $\delta A$ can be approximated to the area of a sector of a circle. Thus: :$\delta A = \dfrac {r^2 \delta \theta} 2$ and so in the limit: {{begin-eqn}} {{eqn | l = \dfrac {\d A} {\d \theta}       | r = \dfrac {r^2} 2       | c =  }} {{eqn | ll= \leadsto       | l = \int \rd A       | r = \dfrac 1 2 \int r^2 \rd \theta       | c =  }} {{eqn | r = \dfrac 1 2 \int r^2 \dfrac {\d \theta} {\d t} \rd t       | c =  }} {{eqn | r = \dfrac 1 2 \int h \rd t       | c =  }} {{eqn | ll= \leadsto       | l = \int_{t_1}^{t_2} \rd A       | r = \dfrac 1 2 \int_{t_1}^{t_2} h \rd t       | c = integrating between times $t_1$ and $t_2$ }} {{eqn | ll= \leadsto       | l = \map A {t_2} - \map A {t_1}       | r = \dfrac h 2 \paren {t_2 - t_1}       | c =  }} {{end-eqn}} That is, given a time interval $t_2 - t_1$, the area $\map A {t_2} - \map A {t_1}$ is the same, whatever the physical position of $p$. {{qed}} == Also see == * Kepler's First Law of Planetary Motion * Kepler's Third Law of Planetary Motion {{Namedfor|Johannes Kepler|cat = Kepler}} == Historical Note == {{:Kepler's Laws of Planetary Motion/Historical Note}} == Sources == * {{BookReference|Men of Mathematics|1937|Eric Temple Bell|prev = Definition:Equidistance Property of Ellipse|next = Kepler's Third Law of Planetary Motion}}: Chapter $\text{VI}$: On the Seashore * {{BookReference|Differential Equations|1972|George F. Simmons|prev = Derivative of Angular Component under Central Force|next = Kepler's Laws of Planetary Motion/Historical Note}}: $\S 3.21$: Newton's Law of Gravitation: $(10)$ * {{BookReference|Calculus Gems|1992|George F. Simmons|prev = Book:Johannes Kepler/Astronomia Nova|next = Book:Johannes Kepler/Epitome Astronomiae Copernicanae}}: Chapter $\text {A}.10$: Kepler ({{DateRange|1571|1630}}) * {{BookReference|Calculus Gems|1992|George F. Simmons|prev = Derivative of Angular Component under Central Force|next = Definition:Gravitational Constant}}: Chapter $\text {B}.25$: Kepler's Laws and Newton's Law of Gravitation * {{BookReference|The Penguin Dictionary of Mathematics|1998|David Nelson|ed = 2nd|edpage = Second Edition|prev = Kepler's First Law of Planetary Motion|next = Kepler's Third Law of Planetary Motion|entry = Kepler's laws}} * {{BookReference|The Penguin Dictionary of Mathematics|2008|David Nelson|ed = 4th|edpage = Fourth Edition|prev = Kepler's First Law of Planetary Motion|next = Kepler's Third Law of Planetary Motion|entry = Kepler's laws}} * {{BookReference|Taming the Infinite|2008|Ian Stewart|prev = Kepler's First Law of Planetary Motion|next = Kepler's Third Law of Planetary Motion}}: Chapter $8$: The System of the World: Kepler * {{BookReference|The Concise Oxford Dictionary of Mathematics|2014|Christopher Clapham|author2 = James Nicholson|ed = 5th|edpage = Fifth Edition|prev = Kepler's First Law of Planetary Motion|next = Kepler's Third Law of Planetary Motion|entry = Kepler's laws of planetary motion|subentry = ii}} Category:Kepler's Laws of Planetary Motion e5e7gaohjfdqoc3w21h4niy5irqebyk


Kepler's Laws of Planetary Motion/Third Law 0 44857 446168 431498 2020-02-04T09:52:56Z Prime.mover 59 wikitext text/x-wiki == Physical Law == '''Kepler's third law of planetary motion''' is one of the three physical laws of celestial mechanics deduced by {{AuthorRef|Johannes Kepler}}: <onlyinclude> :The square of the period of the orbit of a planet around the sun is proportional to the cube of its average distance from the sun. </onlyinclude> == Proof == Consider a planet $p$ of mass $m$ orbiting a star $S$ of mass $M$ under the influence of the gravitational field which the two bodies give rise to. From Kepler's First Law of Planetary Motion, $p$ travels in an elliptical orbit around $S$: :$(1): \quad r = \dfrac {h^2 / k} {1 + e \cos \theta}$ where $k = G M$. From Equation of Ellipse in Reduced Form: Cartesian Frame, the equation of the orbit can also be given as: :$(2): \quad \dfrac {x^2} {a^2} + \dfrac {y^2} {b^2} = 1$ where the foci are placed at $\tuple {\pm c, 0}$. From Focus of Ellipse from Major and Minor Axis: :$a^2 - b^2 = c^2$ and also: :$e = \dfrac c a$ {{explain|The above needs to be established -- I'm not sure if this has been done}} Thus: {{begin-eqn}} {{eqn | l = e^2       | r = \dfrac {a^2 - b^2} {a^2}       | c =  }} {{eqn | n = 3       | ll= \leadsto       | l = b^2       | r = a^2 \paren {1 - e^2}       | c =  }} {{end-eqn}} :400px Then mean distance $a$ of $p$ from the focus $F$ is half the sum of the least and greatest values of $r$. So $(1)$ and $(3)$ give: {{begin-eqn}} {{eqn | l = a       | r = \dfrac 1 2 \paren {\dfrac {h^2 / k} {1 + e} + \dfrac {h^2 / k} {1 - e} }       | c =  }} {{eqn | r = \dfrac {h^2} {k \paren {1 - e^2} }       | c =  }} {{eqn | r = \dfrac {h^2 a^2} {k b^2}       | c =  }} {{eqn | n = 4       | ll= \leadsto       | l = b^2       | r = \dfrac {h^2 a} k       | c =  }} {{end-eqn}} Let $T$ be the orbital period of $p$. From Area of Ellipse, the area $\mathcal A$ of the orbit is given by: :$\mathcal A = \pi a b$ From Kepler's Second Law of Planetary Motion it follows that: :$\dfrac {h T} 2 = \pi a b$ and so from $(4)$: {{begin-eqn}} {{eqn | l = T^2       | r = \dfrac {4 \pi^2 a^2 b^2} {h^2}       | c =  }} {{eqn | r = \paren {\dfrac {4 \pi^2} k} a^3       | c =  }} {{end-eqn}} We have that: :$k = G M$ where $G$ is the gravitational constant and $M$ is the mass of $S$. Hence the result. {{qed}} == Astronomical Units == {{:Kepler's Laws of Planetary Motion/Third Law/Astronomical Units}} == Examples == {{:Kepler's Laws of Planetary Motion/Third Law/Examples}} == Also see == * Kepler's First Law of Planetary Motion * Kepler's Second Law of Planetary Motion {{Namedfor|Johannes Kepler|cat = Kepler}} == Historical Note == {{:Kepler's Laws of Planetary Motion/Historical Note}} == Sources == * {{BookReference|Men of Mathematics|1937|Eric Temple Bell|prev = Kepler's Second Law of Planetary Motion|next = Newton's Law of Universal Gravitation}}: Chapter $\text{VI}$: On the Seashore * {{BookReference|Differential Equations|1972|George F. Simmons|prev = Eccentricity of Orbit indicates its Total Energy|next = Mathematician:Leonhard Paul Euler}}: $\S 3.21$: Newton's Law of Gravitation: $(22)$ * {{BookReference|Calculus Gems|1992|George F. Simmons|prev = Book:Johannes Kepler/Epitome Astronomiae Copernicanae|next = Book:Johannes Kepler/Harmonices Mundi}}: Chapter $\text {A}.10$: Kepler ({{DateRange|1571|1630}}) * {{BookReference|Calculus Gems|1992|George F. Simmons|prev = Kepler's First Law of Planetary Motion|next = Definition:Astronomical Unit}}: Chapter $\text {B}.25$: Kepler's Laws and Newton's Law of Gravitation * {{BookReference|The Penguin Dictionary of Mathematics|1998|David Nelson|ed = 2nd|edpage = Second Edition|prev = Kepler's Second Law of Planetary Motion|next = Definition:Kernel of Group Homomorphism|entry = Kepler's laws}} * {{BookReference|The Penguin Dictionary of Mathematics|2008|David Nelson|ed = 4th|edpage = Fourth Edition|prev = Kepler's Second Law of Planetary Motion|next = Definition:Kernel of Group Homomorphism|entry = Kepler's laws}} * {{BookReference|Taming the Infinite|2008|Ian Stewart|prev = Kepler's Second Law of Planetary Motion|next = Book:Johannes Kepler/Mysterium Cosmographicum}}: Chapter $8$: The System of the World: Kepler * {{BookReference|The Concise Oxford Dictionary of Mathematics|2014|Christopher Clapham|author2 = James Nicholson|ed = 5th|edpage = Fifth Edition|prev = Kepler's Second Law of Planetary Motion|next = Symbols:K/kg|entry = Kepler's laws of planetary motion|subentry = iii}} Category:Kepler's Laws of Planetary Motion jnhutz4crv3ed9uuctowyv7hxt26i2t


Equation of Catenary/Formulation 1

Equation of Catenary/Formulation 1 0 44943 494367 248173 2020-10-13T06:10:39Z Prime.mover 59 wikitext text/x-wiki == Curve == Consider a '''catenary'''. Let a cartesian plane be arranged so that the y-axis passes through the lowest point of the catenary. <onlyinclude> The '''catenary''' is described by the equation: :$y = \dfrac {e^{ax} + e^{-ax}} {2 a} = \dfrac {\cosh a x} a$ where $a$ is a constant. The lowest point of the catenary is at $\left({0, \dfrac 1 a}\right)$. </onlyinclude> == Proof == {{:Equation of Catenary/Formulation 1/Proof}} == Historical Note == {{:Definition:Catenary/Historical Note}} == Lingustic Note == {{:Definition:Catenary/Linguistic Note}} Category:Catenary f4loomt5a2ca6k6pdwyboub26r0czn9


Principle of Conservation of Energy

Principle of Conservation of Energy 0 45101 486181 254735 2020-09-07T21:34:55Z Prime.mover 59 wikitext text/x-wiki == Physical Law == Let $B$ be a body in a force field. Let $B$ be subject only to forces within that force field. Let $P$ be the potential energy of $B$. Let $K$ be the kinetic energy of $B$. Then: :$P + K = \text{constant}$ == Historical Note == {{:Principle of Conservation of Energy/Historical Note}} == Sources == * {{BookReference|Differential Equations|1972|George F. Simmons|prev = Speed of Body under Free Fall from Height/Proof 2|next = Kinetic Energy of Motion}}: $1$: The Nature of Differential Equations: $\S 5$: Falling Bodies and Other Rate Problems Category:Physics Category:Mechanics cw4ntztqn5pw7dyz6wefsk4lseopbpj


Kinetic Energy of Motion

Kinetic Energy of Motion 0 45105 486182 330581 2020-09-07T21:35:10Z Prime.mover 59 wikitext text/x-wiki == Physical Law == Let $B$ be a body of mass $m$ moving with speed $v$ such that $v$ is very much less than the speed of light. Then its kinetic energy $K$ is given by: :$K = \dfrac {m v^2} 2$ == Proof == {{ProofWanted}} == Sources == * {{BookReference|Differential Equations|1972|George F. Simmons|prev = Principle of Conservation of Energy|next = Potential Energy of Position}}: $1$: The Nature of Differential Equations: $\S 5$: Falling Bodies and Other Rate Problems Category:Physics Category:Mechanics 8yqgf2igf92s7zdgpimbft443q0b75q


Potential Energy of Position

Potential Energy of Position 0 45106 486183 329808 2020-09-07T21:35:22Z Prime.mover 59 wikitext text/x-wiki == Physical Law == Let $B$ be a body of mass $m$ in a gravitational field. Let the Acceleration Due to Gravity be of constant magnitude $g$. Let $B$ be at a height $h$ above a reference level $h_0$. Then the potential energy $P$ of $B$ due to the position of $B$ relative to $h_0$ is given by: :$P = m g h$ == Proof == {{proof wanted}} == Sources == * {{BookReference|Differential Equations|1972|George F. Simmons|prev = Kinetic Energy of Motion|next = Speed of Body under Free Fall from Height/Proof 3}}: $1$: The Nature of Differential Equations: $\S 5$: Falling Bodies and Other Rate Problems Category:Physics Category:Mechanics r2pi2yt5umi2omx3f08map4eewwzjb1


Equation of Confocal Ellipses/Formulation 2 0 45235 486234 486230 2020-09-07T22:14:04Z Prime.mover 59 wikitext text/x-wiki == Definition == <onlyinclude> The equation: :$(1): \quad \dfrac {x^2} {a^2} + \dfrac {y^2} {a^2 - c^2} = 1$ where: :$\tuple {x, y}$ denotes an arbitrary point in the cartesian plane :$c$ is a (strictly) positive constant :$a$ is a (strictly) positive parameter such that $a > c$ defines the set of all confocal ellipses whose foci are at $\tuple {\pm c, 0}$. </onlyinclude> == Proof == Let $a$ and $c$ be arbitrary (strictly) positive real numbers fulfilling the constraints as defined. Let $E$ be the locus of the equation: :$(1): \quad \dfrac {x^2} {a^2} + \dfrac {y^2} {a^2 - c^2} = 1$ As $a > c$ it follows that: :$a^2 > c^2$ and so: :$a^2 - c^2 > 0$ Thus $(1)$ is in the form: :$\dfrac {x^2} {a^2} + \dfrac {y^2} {b^2} = 1$ From Equation of Ellipse in Reduced Form, this is the equation of an ellipse in reduced form. Thus: :$\tuple {\pm a, 0}$ are the positions of the vertices of $E$ :$\tuple {0, \pm b}$ are the positions of the covertices of $E$ From Focus of Ellipse from Major and Minor Axis: :$\tuple {\pm c, 0}$ are the positions of the foci of $E$. Hence the result. {{qed}} == Also see == * Equation of Confocal Conics * Equation of Confocal Hyperbolas == Sources == * {{BookReference|Differential Equations|1972|George F. Simmons|prev = Equation of Confocal Conics/Formulation 2|next = Equation of Confocal Hyperbolas/Formulation 2}}: $1$: The Nature of Differential Equations: Miscellaneous Problems for Chapter $1$: $6$ Category:Ellipses Category:Confocal Conics jkeyk4cycjfs0pi68zeyfamhyjypwpb


Equation of Confocal Hyperbolas/Formulation 2 0 45236 486231 436926 2020-09-07T22:13:17Z Prime.mover 59 wikitext text/x-wiki == Definition == <onlyinclude> The equation: :$\dfrac {x^2} {a^2} + \dfrac {y^2} {a^2 - c^2} = 1$ where: :$\tuple {x, y}$ denotes an arbitrary point in the cartesian plane :$c$ is a (strictly) positive constant :$a$ is a (strictly) positive parameter such that $a < c$ defines the set of all confocal hyperbolas whose foci are at $\tuple {\pm c, 0}$. </onlyinclude> == Proof == Let $a$ and $c$ be arbitrary (strictly) positive real numbers fulfilling the constraints as defined. Let $H$ be the locus of the equation: :$(1): \quad \dfrac {x^2} {a^2} + \dfrac {y^2} {a^2 - c^2} = 1$ As $a < c$ it follows that: :$a^2 < c^2$ and so: :$a^2 - c^2 < 0$ Thus $(1)$ is in the form: :$\dfrac {x^2} {a^2} - \dfrac {y^2} {b^2} = 1$ Thus from Equation of Hyperbola in Reduced Form, $H$ defines an hyperbola where: :$\tuple {\pm a, 0}$ are the positions of the vertices of $H$ :the transverse axis of $H$ has length $2 a$ :the conjugate axis of $H$ has length $2 b$ From Focus of Hyperbola from Transverse and Conjugate Axis :$\tuple {\pm c, 0}$ are the positions of the foci of $H$. Hence the result. {{qed}} == Also see == * Equation of Confocal Conics * Equation of Confocal Ellipses == Sources == * {{BookReference|Differential Equations|1972|George F. Simmons|prev = Equation of Confocal Ellipses/Formulation 2|next = Confocal Conics are Self-Orthogonal}}: $1$: The Nature of Differential Equations: : Miscellaneous Problems for Chapter $1$: $6$ Category:Hyperbolas Category:Confocal Conics dx6hs47nfc8tr32sy0wo19x7sehutn3


Solution to Linear First Order Ordinary Differential Equation/Solution by Integrating Factor 0 45466 437267 437259 2019-11-30T20:50:23Z Prime.mover 59 wikitext text/x-wiki == Proof Technique == <onlyinclude> The technique to solve a linear first order ordinary differential equation in the form: :$\dfrac {\d y} {\d x} + \map P x y = \map Q x$ It immediately follows from Integrating Factor for First Order ODE that: :$e^{\int \map P x \rd x}$ is an integrating factor for $(1)$. Multiplying it by: :$e^{\int \map P x \rd x}$ to reduce it to a form: :$\dfrac {\d y} {\d x} e^{\int \map P x \rd x} y = e^{\int \map P x \rd x} \map Q x$ is known as '''Solution by Integrating Factor'''. It is remembered by the procedure: :''Multiply by $e^{\int \map P x \rd x}$ and integrate.'' </onlyinclude> == Examples == {{:Solution by Integrating Factor/Examples}} == Sources == * {{BookReference|Differential Equations|1972|George F. Simmons|prev = Linear First Order ODE/y' + (y over x) = 3 x|next = Solution to Linear First Order Ordinary Differential Equation/Proof 1}}: $\S 2.10$ Category:Linear First Order ODEs Category:Proof Techniques 3y48xcqvjezllsl3jz1rl5t4z8aeo2a


Ohm's Law

Ohm's Law 0 45555 462359 462358 2020-04-16T13:42:51Z Prime.mover 59 wikitext text/x-wiki == Physical Law == Let a resistor $r$ oppose a current $I$ with resistance $R$. Then the drop in electromotive force produced by $r$ is given by: :$E_R = R I$ {{Namedfor|Georg Simon Ohm|cat = Ohm, Georg}} == Sources == * {{BookReference|Differential Equations|1972|George F. Simmons|prev = Pursuit Curve of Boat in River|next = Mathematician:Georg Simon Ohm}}: $\S 2.13$: Simple Electric Circuits Category:Electromagnetism Category:Resistance 3wcowasbdslr0qjjpyb19ewzpck91me


Drop in EMF caused by Inductance is proportional to Rate of Change of Current 0 45562 462380 462360 2020-04-16T16:19:13Z Prime.mover 59 wikitext text/x-wiki == Physical Law == The drop in EMF $E_L$ caused by an inductor is proportional to the rate of change of current $I$ through the inductor: :$E_L = L \dfrac {\d I} {\d t}$ The constant of proportion $L$ is known as the inductance of the inductor. == Sources == * {{BookReference|Differential Equations|1972|George F. Simmons|prev = Mathematician:Georg Simon Ohm|next = Drop in EMF caused by Capacitance is proportional to Accumulated Charge}}: $\S 2.13$: Simple Electric Circuits Category:Inductance 7wueewrc5dmtep0aoebrrfbcrres0s5


Drop in EMF caused by Capacitance is proportional to Accumulated Charge 0 45564 462382 462381 2020-04-16T16:19:51Z Prime.mover 59 wikitext text/x-wiki == Physical Law == The drop in EMF $E_C$ caused by a capacitor is proportional to the electric charge $Q$ that has accumulated on the capacitor: :$E_C = \dfrac 1 C Q$ The reciprocal of the constant of proportion $\dfrac 1 C$ is known as the capacitance of the capacitor. == Sources == * {{BookReference|Differential Equations|1972|George F. Simmons|prev = Drop in EMF caused by Inductance is proportional to Rate of Change of Current|next = Definition:Electric Current}}: $\S 2.13$: Simple Electric Circuits Category:Capacitance qufcm0jhh59pteff0or2dg18loelq44


Kirchhoff's Voltage Law

Kirchhoff's Voltage Law 0 45566 495622 495621 2020-10-20T22:07:45Z Prime.mover 59 wikitext text/x-wiki == Physical Law == The algebraic sum of the electromotive forces around a closed electrical circuit is equal to zero. {{Namedfor|Gustav Robert Kirchhoff|cat = Kirchhoff}} == Sources == * {{BookReference|Differential Equations|1972|George F. Simmons|prev = Definition:Electric Current|next = Mathematician:Gustav Robert Kirchhoff}}: $\S 2.13$: Simple Electric Circuits Category:Electromagnetism q40s0ki9ihl3qyiorjjrbr749hfnyu6


Non-Abelian Order 8 Group with One Order 2 Element is Quaternion Group/Lemma 1 0 45715 370919 321896 2018-10-13T20:38:35Z Prime.mover 59 wikitext text/x-wiki == Lemma == <onlyinclude> :$\set {\pm 1}$ is a normal subgroup of $G$. </onlyinclude> == Proof == {{begin-eqn}} {{eqn| ll= \forall g \in G:      | l = g \circ 1 \circ g^{-1}      | r = g \circ g^{-1}      | c = {{Defof|Identity Element}} }} {{eqn| r = 1      | c = {{Defof|Inverse Element}} }} {{end-eqn}} {{begin-eqn}} {{eqn| ll= \forall g \in G:      | l = \paren {g \circ \paren {-1} \circ g^{-1} }^2      | r = g \circ \paren {-1} \circ g^{-1} \circ g \circ \paren {-1} \circ g^{-1}      | c =  }} {{eqn| r = g \circ \paren {-1} \circ \paren {-1} \circ g^{-1}      | c = {{Defof|Inverse Element}} }} {{eqn| r = g \circ g^{-1}      | c = Order of $-1$ is $2$ }} {{eqn| r = 1      | c = {{Defof|Inverse Element}} }} {{end-eqn}} So: : $g \circ \paren {-1} \circ g^{-1} = 1$ or: : $g \circ \paren {-1} \circ g^{-1} = -1$ Suppose $g \circ \paren {-1} \circ g^{-1} = 1$. Then: {{begin-eqn}} {{eqn| l = g \circ \paren {-1} \circ g^{-1}      | r = 1      | c =  }} {{eqn| l = g^{-1} \circ g \circ \paren {-1} \circ g^{-1} \circ g      | r = g^{-1} \circ 1 \circ g      | c = applying same action on both sides }} {{eqn| l = -1      | r = 1      | c = {{Defof|Inverse Element}}, {{Defof|Identity Element}} }} {{end-eqn}} which is a contradiction. So it is the other case: : $g \circ \paren {-1} \circ g^{-1} = -1$ Therefore: : $\forall g \in G: g \circ \set {\pm 1} \circ g^{-1} = \set {\pm 1}$ By definition, $\set {\pm 1}$ is a normal subgroup. {{qed}} Category:Non-Abelian Order 8 Group with One Order 2 Element is Quaternion Group qmac4lua36vm0o22dwomik7p1fcutrh


Method of Undetermined Coefficients/Exponential 0 45825 439206 437485 2019-12-10T23:01:16Z Prime.mover 59 wikitext text/x-wiki == Proof Technique == Consider the nonhomogeneous linear second order ODE with constant coefficients: :$(1): \quad y'' + p y' + q y = \map R x$ Let $\map R x$ be of the form of an exponential function: :$\map R x = K e^{a x}$ The '''Method of Undetermined Coefficients''' can be used to solve $(1)$ in the following manner. == Method and Proof == Let $\map {y_g} x$ be the general solution to: :$y'' + p y' + q y = 0$ From Solution of Constant Coefficient Homogeneous LSOODE, $\map {y_g} x$ can be found systematically. Let $\map {y_p} x$ be a particular solution to $(1)$. Then from General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution: :$\map {y_g} x + \map {y_p} x$ is the general solution to $(1)$. It remains to find $\map {y_p} x$. <onlyinclude> Let $\map R x = K e^{a x}$. Consider the auxiliary equation to $(1)$: :$(2): \quad m^2 + p m + q = 0$ There are three cases to consider. ;$a$ is not a root of $(2)$ Assume that there is a particular solution to $(1)$ of the form: :$y_p = A e^{a x}$ We have: {{begin-eqn}} {{eqn | l = \frac {\d} {\d x} y_p       | r = a A e^{a x}       | c = Derivative of Exponential Function }} {{eqn | l = \frac {\d^2} {\d x^2} y_p       | r = a^2 A e^{a x}       | c = Derivative of Exponential Function }} {{end-eqn}} Inserting into $(1)$: {{begin-eqn}} {{eqn | l = a^2 A e^{a x} + a p A e^{a x} + q A e^{a x}       | r = K e^{a x}       | c =  }} {{eqn | ll= \leadsto       | l = A \left({a^2 + p a + q}\right) e^{a x}       | r = K e^{a x}       | c =  }} {{eqn | ll= \leadsto       | l = A       | r = \frac K {a^2 + p a + q}       | c =  }} {{end-eqn}} Hence: :$y_p = \dfrac {K e^{a x} } {a^2 + p a + q}$ From Solution of Constant Coefficient Homogeneous LSOODE, $y_g$ depends on whether $(2)$ has equal or unequal roots. Let $m_1$ and $m_2$ be the roots of $(2)$. Then: :$y = \begin{cases} C_1 e^{m_1 x} + C_2 e^{m_2 x} + \dfrac {K e^{a x} } {a^2 + p a + q} & : m_1 \ne m_2: m_1, m_2 \in \R \\ C_1 e^{m_1 x} + C_2 x e^{m_1 x} + \dfrac {K e^{a x} } {a^2 + p a + q} & : m_1 = m_2 \\ e^{r x} \paren {C_1 \cos s x + C_2 \sin s x} + \dfrac {K e^{a x} } {a^2 + p a + q} & : m_1 = r + i s, m_2 = r - i s \end{cases}$ is the general solution to $(1)$. {{qed|lemma}} ;$a$ is a root of $(2)$ If $a$ is a root of $(2)$, then $a^2 + p a + q = 0$ and so $\dfrac {K e^{a x} } {a^2 + p a + q}$ is undefined. Let the auxiliary equation to $(2)$ have two unequal real roots $a$ and $b$. Assume that there is a particular solution to $(1)$ of the form: :$y_p = A x e^{a x}$ We have: {{begin-eqn}} {{eqn | l = \frac {\d} {\d x} y_p       | r = a A x e^{a x} + A e^{a x}       | c = Product Rule and Derivative of Exponential Function }} {{eqn | l = \frac {\d^2} {\d x^2} y_p       | r = a^2 A x e^{a x} + a A e^{a x} + a A e^{a x}       | c = Product Rule and Derivative of Exponential Function }} {{eqn | r = a^2 A x e^{a x} + 2 a A e^{a x}       | c =  }} {{end-eqn}} Inserting into $(1)$: {{begin-eqn}} {{eqn | l = a^2 A x e^{a x} + 2 a A e^{a x} + p \paren {a A x e^{a x} + A e^{a x} } + q A x e^{a x}       | r = K e^{a x}       | c =  }} {{eqn | ll= \leadsto       | l = A \paren {a^2 + p a + q} x e^{a x} + A \paren {2 a + p} e^{a x}       | r = K e^{a x}       | c =  }} {{eqn | ll= \leadsto       | l = A \paren {2 a + p} e^{a x}       | r = K e^{a x}       | c = as $a^2 + p a + q = 0$ }} {{eqn | ll= \leadsto       | l = A       | r = \frac K {2 a + p}       | c =  }} {{end-eqn}} Hence: :$y_p = \dfrac {K x e^{a x} } {2 a + p}$ and so from Solution of Constant Coefficient Homogeneous LSOODE: :$y = C_1 e^{a x} + C_2 e^{b x} + \dfrac {K x e^{a x} } {2 a + p}$ is the general solution to $(1)$. {{qed|lemma}} ;$a$ is a repeated root of $(2)$ If the auxiliary equation to $(2)$ has two equal real roots $a$, then: :$a = - \dfrac p 2$ and so not only: :$a^2 + p a + q = 0$ but also: :$2 a + p = 0$ and so neither of the above expressions involving $e^{a x}$ and $x e^{a x}$ will work as particular solution to $(1)$. So, assume that there is a particular solution to $(1)$ of the form: :$y_p = A x^2 e^{a x}$ We have: {{begin-eqn}} {{eqn | l = \frac {\d} {\d x} y_p       | r = a A x^2 e^{a x} + 2 A x e^{a x}       | c = Product Rule and Derivative of Exponential Function }} {{eqn | l = \frac {\d^2} {\d x^2} y_p       | r = a^2 A x^2 e^{a x} + 2 a x A e^{a x} + 2 a x A e^{a x} + 2 A e^{a x}       | c = Product Rule and Derivative of Exponential Function }} {{eqn | r = a^2 A x e^{a x} + 4 a x A e^{a x} + 2 A e^{a x}       | c = Derivative of Exponential Function }} {{end-eqn}} Inserting into $(1)$: {{begin-eqn}} {{eqn | l = a^2 A x e^{a x} + 4 a x A e^{a x} + 2 A e^{a x} + p \paren {a A x^2 e^{a x} + 2 A x e^{a x} } + q A x^2 e^{a x}       | r = K e^{a x}       | c =  }} {{eqn | ll= \leadsto       | l = A \paren {a^2 + p a + q} x e^{a x} + 2 A \paren {2 a + p} x e^{a x} + 2 A e^{a x}       | r = K e^{a x}       | c =  }} {{eqn | ll= \leadsto       | l = 2 A e^{a x}       | r = K e^{a x}       | c = as $a^2 + p a + q = 0$ and $2 a + p = 0$ }} {{eqn | ll= \leadsto       | l = A       | r = \frac K 2       | c =  }} {{end-eqn}} Hence: :$y_p = \dfrac {K x^2 e^{a x} } 2$ and so: :$y = y_g + \dfrac {K x^2 e^{a x} } 2$ and so from Solution of Constant Coefficient Homogeneous LSOODE: :$y = C_1 e^{a x} + C_2 x e^{a x} + \dfrac {K x^2 e^{a x} } 2$ is the general solution to $(1)$. </onlyinclude>{{qed}} == Sources == * {{BookReference|Elementary Differential Equations & Operators|1958|G.E.H. Reuter|prev = Method of Undetermined Coefficients/Polynomial|next = Linear Second Order ODE/y'' - 3 y' + 2 y = 5 exp 3 x}}: Chapter $1$: Linear Differential Equations with Constant Coefficients: $\S 2$. The second order equation: $\S 2.4$ Particular solution: exponential $\map f x$ * {{BookReference|Differential Equations|1972|George F. Simmons|prev = Method of Undetermined Coefficients|next = Method of Undetermined Coefficients/Sine and Cosine}}: $\S 3.18$: The Method of Undetermined Coefficients Category:Method of Undetermined Coefficients tp4r9q6kqfv0pe8z3y74t7jauq0mdu1


Method of Undetermined Coefficients/Sine and Cosine 0 45827 439029 437648 2019-12-10T20:58:06Z Prime.mover 59 wikitext text/x-wiki == Proof Technique == Consider the nonhomogeneous linear second order ODE with constant coefficients: :$(1): \quad y'' + p y' + q y = \map R x$ Let $\map R x$ be a linear combination of sine and cosine: :$\map R x = \alpha \sin b x + \beta \cos b x$ The '''Method of Undetermined Coefficients''' can be used to solve $(1)$ in the following manner. == Method and Proof == Let $\map {y_g} x$ be the general solution to: :$y'' + p y' + q y = 0$ From Solution of Constant Coefficient Homogeneous LSOODE, $\map {y_g} x$ can be found systematically. Let $\map {y_p} x$ be a particular solution to $(1)$. Then from General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution: :$\map {y_g} x + \map {y_p} x$ is the general solution to $(1)$. It remains to find $\map {y_p} x$. <onlyinclude> Let $\map R x = \alpha \sin b x + \beta \cos b x$. Consider the auxiliary equation to $(1)$: :$(2): \quad m^2 + p m + q = 0$ There are two cases which may apply. === $i b$ is not Root of Auxiliary Equation === First we investigate the case where $i b$ is not a root of the auxiliary equation to $(1)$. {{:Method of Undetermined Coefficients/Sine and Cosine/Particular Solution/i b is not Root of Auxiliary Equation}} === $i b$ is Root of Auxiliary Equation === Now suppose that $(1)$ is of the form $y'' + b^2 y = A \sin b x + B \cos b x$. Thus one of the $i b$ is one of the roots of the auxiliary equation to $(1)$. From Linear Second Order ODE: $y'' + k^2 y = 0$ the general solution to $(2)$ is: :$y = C_1 \sin b x + C_2 \cos b x$ and it can be seen that an expression of the form $A \sin b x + B \cos b x$ is already a particular solution of $(2)$. Thus we have: :$\paren {q - b^2}^2 + b^2 p^2 = 0$. But using the Method of Undetermined Coefficients in the above manner, this would result in an attempt to calculate: :$\dfrac {\alpha \paren {q - b^2} + \beta b p} {\paren {q - b^2}^2 + b^2 p^2}$ and: :$\dfrac {\beta \paren {q - b^2} - \alpha b p} {\paren {q - b^2}^2 + b^2 p^2}$ both of which are are undefined. {{:Method of Undetermined Coefficients/Sine and Cosine/Particular Solution/i b is Root of Auxiliary Equation}}</onlyinclude>{{qed}} == Sources == * {{BookReference|Differential Equations|1972|George F. Simmons|prev = Method of Undetermined Coefficients/Exponential|next = Method of Undetermined Coefficients/Sine and Cosine/Particular Solution/i b is not Root of Auxiliary Equation/Trigonometric Form}}: $\S 3.18$: The Method of Undetermined Coefficients Category:Method of Undetermined Coefficients g6izvwvjf8ft04tg89p1m64x74am18j


Method of Undetermined Coefficients/Polynomial 0 45828 439212 438979 2019-12-10T23:02:46Z Prime.mover 59 wikitext text/x-wiki == Proof Technique == Consider the nonhomogeneous linear second order ODE with constant coefficients: :$(1): \quad y'' + p y' + q y = \map R x$ Let $\map R x$ be a polynomial in $x$: :$\displaystyle \map R x = \sum_{j \mathop = 0}^n a_j x^j$ The '''Method of Undetermined Coefficients''' can be used to solve $(1)$ in the following manner. == Method and Proof == Let $\map {y_g} x$ be the general solution to: :$y'' + p y' + q y = 0$ From Solution of Constant Coefficient Homogeneous LSOODE, $\map {y_g} x$ can be found systematically. Let $\map {y_p} x$ be a particular solution to $(1)$. Then from General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution: :$\map {y_g} x + \map {y_p} x$ is the general solution to $(1)$. It remains to find $\map {y_p} x$. <onlyinclude> Let $\displaystyle \map R x = \sum_{j \mathop = 0}^n a_j x^j$. Assume that there is a particular solution to $(1)$ of the form: :$\displaystyle y_p = \sum_{j \mathop = 0}^n A_j x^j$ We have: {{begin-eqn}} {{eqn | l = \frac {\d} {\d x} y_p       | r = \sum_{j \mathop = 1}^n j A_j x^{j - 1}       | c = Power Rule for Derivatives }} {{eqn | l = \frac {\d^2} {\d x^2} y_p       | r = \sum_{j \mathop = 2}^n j \paren {j - 1} A_j x^{j - 2}       | c = Power Rule for Derivatives }} {{end-eqn}} Inserting into $(1)$: {{begin-eqn}} {{eqn | l = \sum_{j \mathop = 2}^n j \paren {j - 1} A_j x^{j - 2} + p \sum_{j \mathop = 1}^n j A_j x^{j - 1} + q \sum_{j \mathop = 0}^n A_j x^j       | r = \sum_{j \mathop = 0}^n a_j x^j       | c =  }} {{end-eqn}} The coefficients $A_0$ to $A_n$ can hence be solved in terms of $a_0$ to $a_n$ using the techniques of simultaneous equations. The general form is tedious and unenlightening to evaluate. </onlyinclude>{{qed}} == Sources == * {{BookReference|Elementary Differential Equations & Operators|1958|G.E.H. Reuter|prev = Linear Second Order ODE/y'' - 7 y' - 5 y = x^3 - 1|next = Method of Undetermined Coefficients/Exponential}}: Chapter $1$: Linear Differential Equations with Constant Coefficients: $\S 2$. The second order equation: $\S 2.3$ Particular solution: polynomial $\map f x$ * {{BookReference|Differential Equations|1972|George F. Simmons|prev = Method of Undetermined Coefficients/Sine and Cosine/Particular Solution/i b is Root of Auxiliary Equation/Trigonometric Form|next = Linear Second Order ODE/y'' + 3 y' - 10 y = 6 exp 4 x}}: $\S 3.18$: The Method of Undetermined Coefficients Category:Method of Undetermined Coefficients bshbbr9yjc5lemdexfw8a5git12pmve


Method of Variation of Parameters

Method of Variation of Parameters 0 45895 439155 438423 2019-12-10T21:43:59Z Prime.mover 59 wikitext text/x-wiki == Proof Technique == The '''method of variation of parameters''' is a technique for finding a particular solution to a nonhomogeneous linear second order ODE: :$(1): \quad y'' + \map P x y' + \map Q x y = \map R x$ provided that the general solution of the corresponding homogeneous linear second order ODE: :$(2): \quad y'' + \map P x y' + \map Q x y = 0$ is already known. == Method == Let the general solution of $(2)$ be: :$y = C_1 \, \map {y_1} x + C_2 \, \map {y_2} x$ Then a particular solution of $(1)$ is: :$\displaystyle y = y_1 \int -\frac {\map {y_2} x \, \map R x} {\map W {y_1, y_2} } \rd x + y_2 \int \frac {\map {y_1} x \, \map R x} {\map W {y_1, y_2} } \rd x$ where $\map W {y_1, y_2}$ denotes the Wronskian of $\map {y_1} x$ and $\map {y_2} x$. == Proof == Let the general solution of $(2)$ be: :$(3): \quad y = C_1 \, \map {y_1} x + C_2 \, \map {y_2} x$ Let the arbitrary constants $C_1$ and $C_2$ be replaced by functions $\map {v_1} x$ and $\map {v_2} x$. It is required that $v_1$ and $v_2$ be determined so as to make: :$(4): \quad y = \map {v_1} x \, \map {y_1} x + \map {v_2} x \, \map {y_2} x$ a particular solution of $(1)$. Then: {{begin-eqn}} {{eqn | l = y'       | r = \paren {v_1 {y_1}' + {v_1}' y_1} + \paren {v_2 {y_2}' + {v_2}' y_2}       | c = Product Rule for Derivatives }} {{eqn | n = 5       | r = \paren {v_1 {y_1}' + v_2 {y_2}'} + \paren { {v_1}' y_1 + {v_2}' y_2}       | c =  }} {{end-eqn}} Suppose ${v_1}' y_1 + {v_2}' y_2$ were made to vanish: :$(6): \quad {v_1}' y_1 + {v_2}' y_2 = 0$ Then: {{begin-eqn}} {{eqn | n = 7       | l = y'       | r = v_1 {y_1}' + v_2 {y_2}'       | c =  }} {{eqn | n = 8       | ll= \leadsto       | l = y''       | r = \paren {v_1 {y_1}'' + {v_1}' {y_1}'} + \paren {v_2 {y_2}'' + {v_2}' {y_2}'}       | c = Product Rule for Derivatives }} {{end-eqn}} Hence: {{begin-eqn}} {{eqn | l = y'' + \map P x y' + \map Q x y       | r = \map R x       | c = $(1):$ given }} {{eqn | ll= \leadsto       | l = \paren {v_1 {y_1}'' + {v_1}' {y_1}'} + \paren {v_2 {y_2}'' + {v_2}' {y_2}'} + \map P x y' + \map Q x y       | r = \map R x       | c = substituting from $(8)$ }} {{eqn | ll= \leadsto       | l = \paren {v_1 {y_1}'' + {v_1}' {y_1}'} + \paren {v_2 {y_2}'' + {v_2}' {y_2}'} + \map P x \paren {v_1 {y_1}' + v_2 {y_2}'} + \map Q x y       | r = \map R x       | c = substituting from $(7)$ }} {{eqn | ll= \leadsto       | l = \paren {v_1 {y_1}'' + {v_1}' {y_1}'} + \paren {v_2 {y_2}'' + {v_2}' {y_2}'} + \map P x \paren {v_1 {y_1}' + v_2 {y_2}'} + \map Q x \paren {v_1 y_1 + v_2 y_2}       | r = \map R x       | c = substituting from $(4)$ }} {{eqn | n = 9       | ll= \leadsto       | l = v_1 \paren { {y_1}'' + \map P x {y_1}' + \map Q x y_1} + v_2 \paren { {y_2}'' + \map P x {y_2}' + \map Q x y_2} + {v_1}' {y_1}' + {v_2}' {y_2}'       | r = \map R x       | c = rearranging }} {{end-eqn}} Because $y_1$ and $y_2$ are both particular solutions of $(2)$: :${y_1}'' + \map P x {y_1}' + \map Q x y_1 = {y_2}'' + \map P x {y_2}' + \map Q x y_2 = 0$ and so from $(9)$: :$(10): \quad {v_1}' {y_1}' + {v_2}' {y_2}' = \map R x$ In summary: {{begin-eqn}} {{eqn | n = 6       | l = {v_1}' y_1 + {v_2}' y_2       | r = 0 }} {{eqn | n = 10       | l = {v_1}' {y_1}' + {v_2}' {y_2}'       | r = \map R x }} {{eqn | ll= \leadsto       | l = {v_1}'       | r = \frac {y_2 \map R x} {y_2 {y_1}' - y_1 {y_2}'}       | c =  }} {{eqn | l = {v_2}'       | r = \frac {y_1 \map R x} {y_1 {y_2}' - y_2 {y_1}'}       | c =  }} {{eqn | n = 11       | ll= \leadsto       | l = {v_1}'       | r = -\frac {y_2 \map R x} {\map W {y_1, y_2} }       | c =  }} {{eqn | l = {v_2}'       | r = \frac {y_1 \map R x} {\map W {y_1, y_2} }       | c =  }} {{end-eqn}} We started with the assumption that: :$(3): \quad y = C_1 \, \map {y_1} x + C_2 \, \map {y_2} x$ and so $y_1$ and $y_2$ are linearly independent. Thus by Zero Wronskian of Solutions of Homogeneous Linear Second Order ODE iff Linearly Dependent: :$\map W {y_1, y_2} \ne 0$ and so $(11)$ is defined. Thus: {{begin-eqn}} {{eqn | l = v_1       | r = \int -\frac {y_2 \map R x} {\map W {y_1, y_2} } \rd x       | c =  }} {{eqn | l = v_2       | r = \int \frac {y_1 \map R x} {\map W {y_1, y_2} } \rd x       | c =  }} {{end-eqn}} and so as required: :$\displaystyle y = y_1 \int -\frac {\map {y_2} x \map R x} {\map W {y_1, y_2} } \rd x + y_2 \int \frac {\map {y_1} x \map R x} {\map W {y_1, y_2} } \rd x$ {{qed}} == Source of Name == The name '''method of variation of parameters''' derives from the method of operation: the parameters $C_1$ and $C_2$ are made to '''vary''' by replacing them with the functions $\map {v_1} x$ and $\map {v_2} x$. == Sources == * {{BookReference|Differential Equations|1972|George F. Simmons|prev = Linear Second Order ODE/y'' + 4 y = 4 cosine 2 x + 6 cosine x + 8 x^2 - 4 x|next = Linear Second Order ODE/y'' + y = cosecant x}}: $\S 3.19$: The Method of Variation of Parameters Category:Linear Second Order ODEs Category:Proof Techniques 049iojp3clzezm3eijf0cdbu0mri6xq


6

6 0 46741 481579 481569 2020-08-13T21:08:55Z Prime.mover 59 wikitext text/x-wiki {{NumberPageLink|prev = 5|next = 7}} == Number == $6$ ('''six''') is: :$2 \times 3$ :The only triangular number with less than $660$ digits whose square is also triangular: ::$6^2 = 36 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = \dfrac {8 \paren {8 + 1} } 2$ :The only positive integer which is the sum of exactly $3$ of its distinct coprime divisors === $1$st Term === :The smallest positive integer which can be expressed as the sum of $2$ odd primes in $1$ way: ::$6 = 3 + 3$ :The $1$st triangular number which can be expressed as the product of $3$ consecutive integers: ::$6 = T_3 = 1 \times 2 \times 3$ :The $1$st power of $6$ after the zeroth $1$: ::$6 = 6^1$ :The $1$st: :: perfect number :: semiperfect number :: primitive semiperfect number :and the only number which is the sum and product of the same $3$ distinct positive integers: ::$6 = 1 + 2 + 3 = 1 \times 2 \times 3$ :The $1$st unitary perfect number: ::$6 = 1 + 2 + 3$ === $2$nd Term === :The $2$nd Ore number after $1$: ::$\dfrac {6 \times \map \tau 6} {\map \sigma 6} = 2$ :and the $2$nd after $1$ whose divisors also have an arithmetic mean which is an integer: ::$\dfrac {\map \sigma 6} {\map \tau 6} = 3$ :The $2$nd semiprime after $4$: ::$6 = 2 \times 3$ :The $2$nd primorial after $1$, $2$ (counting $1$ as the zeroth): ::$6 = p_2 \# = 3 \# = 2 \times 3$ :The $2$nd hexagonal number after $1$: ::$6 = 1 + 5 = 2 \paren {2 \times 2 - 1}$ :The $2$nd pentagonal pyramidal number after $1$: ::$6 = 1 + 5 = \dfrac {2^2 \paren {2 + 1} } 2$ :The $2$nd composite number, and the first with distinct prime factors: ::$6 = 2 \times 3$ :Hence the $1$st positive integer after $1$ which is not the power of a prime number. {{WIP|Set up the above as a sequence in its own right}} :The $2$nd primorial which can be expressed as the product of consecutive integers: ::$3 \# = 6 = 2 \times 3$ :The $2$nd central binomial coefficient after $2$: ::$6 = \dbinom {2 \times 2} 2 := \dfrac {4!} {\paren {2!}^2}$ :The $2$nd of the $3$rd pair of consecutive integers whose product is a primorial: ::$5 \times 6 = 30 = 5 \#$ === $3$rd Term === :The $3$rd special highly composite number after $1$, $2$ :The $3$rd factorial after $1$, $2$: ::$6 = 3! = 3 \times 2 \times 1$ :The $3$rd triangular number after $1$, $3$: ::$6 = 1 + 2 + 3 = \dfrac {3 \paren {3 + 1} } 2$ :The $3$rd automorphic number after $1$, $5$: ::$6^2 = 3 \mathbf 6$ :The $3$rd even number after $2$, $4$ which cannot be expressed as the sum of $2$ composite odd numbers :The index (after $2, 3$) of the $3$rd Woodall prime: ::$6 \times 2^6 - 1 = 383$ :The $3$rd positive integer which is not the sum of $1$ or more distinct squares: ::$2$, $3$, $6$, $\ldots$ :The $3$rd integer $m$ after $3$, $4$ such that $m! - 1$ (its factorial minus $1$) is prime: ::$6! - 1 = 720 - 1 = 719$ === $4$th Term === :The $4$th highly composite number after $1$, $2$, $4$: ::$\map \tau 6 = 4$ :The $4$th superabundant number after $1$, $2$, $4$: ::$\dfrac {\map \sigma 6} 6 = \dfrac {12} 6 = 2$ :The $4$th trimorphic number after $1$, $4$, $5$: ::$6^3 = 21 \mathbf 6$ :The $4$th palindromic triangular number after $0$, $1$, $3$ :The $4$th palindromic triangular number after $0$, $1$, $3$ whose index is itself palindromic: ::$6 = T_3$ :The $4$th integer $n$ after $3$, $4$, $5$ such that $m = \displaystyle \sum_{k \mathop = 0}^{n - 1} \paren {-1}^k \paren {n - k}! = n! - \paren {n - 1}! + \paren {n - 2}! - \paren {n - 3}! + \cdots \pm 1$ is prime: ::$6! - 5! + 4! - 3! + 2! - 1! = 619$ === $5$th Term === :The $5$th highly abundant number after $1$, $2$, $3$, $4$: ::$\map \sigma 6 = 12$ :The $5$th Ulam number after $1$, $2$, $3$, $4$: ::$6 = 2 + 4$ :The $5$th (strictly) positive integer after $1$, $2$, $3$, $4$ which cannot be expressed as the sum of exactly $5$ non-zero squares. :The $5$th positive integer after $1$, $2$, $3$, $4$ such that all smaller positive integers coprime to it are prime :The $5$th after $1$, $2$, $4$, $5$ of the $24$ positive integers which cannot be expressed as the sum of distinct non-pythagorean primes === $6$th Term === :The $6$th of the (trivial $1$-digit) pluperfect digital invariants after $1$, $2$, $3$, $4$, $5$: ::$6^1 = 6$ :The $6$th of the (trivial $1$-digit) Zuckerman numbers after $1$, $2$, $3$, $4$, $5$: ::$6 = 1 \times 6$ :The $6$th of the (trivial $1$-digit) harshad numbers after $1$, $2$, $3$, $4$, $5$: ::$6 = 1 \times 6$ === $7$th Term === :The $7$th integer $n$ after $0$, $1$, $2$, $3$, $4$, $5$ such that $2^n$ contains no zero in its decimal representation: ::$2^6 = 64$ :The $7$th integer $n$ after $0$, $1$, $2$, $3$, $4$, $5$ such that $5^n$ contains no zero in its decimal representation: ::$5^6 = 15 \, 625$ :The $7$th integer $n$ after $0$, $1$, $2$, $3$, $4$, $5$ such that both $2^n$ and $5^n$ have no zeroes: ::$2^6 = 64$, $5^6 = 15 \, 625$ :The $7$th integer after $0$, $1$, $2$, $3$, $4$, $5$ which is (trivially) the sum of the increasing powers of its digits taken in order: ::$6^1 = 6$ === Miscellaneous === :$6 = \sqrt {1^3 + 2^3 + 3^3}$ :The number of faces of a cube :The number of vertices of its dual, the regular octahedron :The number of edges of a tetrahedron :The area and semiperimeter of the $3-4-5$ triangle: ::$6 = \dfrac {3 \times 4} 2 = \dfrac {3 + 4 + 5} 2$ {{ArithmeticFunctionTable|n = 6|sigma = 12|tau = 4|phi = 2}} == Also see == * Perfect Number is Sum of Successive Odd Cubes except 6 * Prime equals Plus or Minus One modulo 6 * Only Number which is Sum of 3 Factors is 6 * Triangular Number whose Square is Triangular === Previous in Sequence: $1$ === * {{NumberPageLink|prev = 1|next = 15|type = Hexagonal Number|cat = Hexagonal Numbers}} * {{NumberPageLink|prev = 1|next = 18|type = Pentagonal Pyramidal Number|cat = Pyramidal Numbers}} * {{NumberPageLink|prev = 1|next = 28|type = Ore Number|cat = Ore Numbers}} * {{NumberPageLink|prev = 1|next = 36|result = Sequence of Powers of 6|cat = Powers of 6}} * {{NumberPageLink|prev = 1|next = 140|result = Sequence of Numbers with Integer Arithmetic and Harmonic Means of Divisors}} === Previous in Sequence: $2$ === * {{NumberPageLink|prev = 2|next = 12|type = Special Highly Composite Number|cat = Special Highly Composite Numbers}} * {{NumberPageLink|prev = 2|next = 20|type = Central Binomial Coefficient|cat = Central Binomial Coefficients}} * {{NumberPageLink|prev = 2|next = 24|type = Factorial|cat = Factorials}} * {{NumberPageLink|prev = 2|next = 30|type = Primorial|cat = Primorials}} * {{NumberPageLink|prev = 2|next = 30|result = Primorials which are Product of Consecutive Integers}} === Previous in Sequence: $3$ === * {{NumberPageLink|prev = 3|next = 7|result = Numbers not Sum of Distinct Squares}} * {{NumberPageLink|prev = 3|next = 10|type = Triangular Number|cat = Triangular Numbers}} * {{NumberPageLink|prev = 3|next = 30|type = Woodall Prime|cat = Woodall Primes}} * {{NumberPageLink|prev = 3|next = 55|result = Palindromic Triangular Numbers}} * {{NumberPageLink|prev = 3|next = 66|result = Palindromic Triangular Numbers with Palindromic Index}} === Previous in Sequence: $4$ === * {{NumberPageLink|prev = 4|next = 7|result = Sequence of Integers whose Factorial minus 1 is Prime}} * {{NumberPageLink|prev = 4|next = 7|result = Integer not Expressible as Sum of 5 Non-Zero Squares}} * {{NumberPageLink|prev = 4|next = 8|type = Ulam Number|cat = Ulam Numbers}} * {{NumberPageLink|prev = 4|next = 8|type = Highly Abundant Number|cat = Highly Abundant Numbers}} * {{NumberPageLink|prev = 4|next = 8|result = Positive Even Integers not Expressible as Sum of 2 Composite Odd Numbers}} * {{NumberPageLink|prev = 4|next = 8|result = Integers such that all Coprime and Less are Prime}} * {{NumberPageLink|prev = 4|next = 9|type = Semiprime Number|cat = Semiprimes}} * {{NumberPageLink|prev = 4|next = 12|type = Highly Composite Number|cat = Highly Composite Numbers}} * {{NumberPageLink|prev = 4|next = 12|type = Superabundant Number|cat = Superabundant Numbers}} === Previous in Sequence: $5$ === * {{NumberPageLink|prev = 5|next = 7|type = Pluperfect Digital Invariant|cat = Pluperfect Digital Invariants}} * {{NumberPageLink|prev = 5|next = 7|type = Zuckerman Number|cat = Zuckerman Numbers}} * {{NumberPageLink|prev = 5|next = 7|type = Harshad Number|cat = Harshad Numbers}} * {{NumberPageLink|prev = 5|next = 7|result = Powers of 2 with no Zero in Decimal Representation}} * {{NumberPageLink|prev = 5|next = 7|result = Powers of 5 with no Zero in Decimal Representation}} * {{NumberPageLink|prev = 5|next = 7|result = Powers of 2 and 5 without Zeroes}} * {{NumberPageLink|prev = 5|next = 7|result = Numbers which are Sum of Increasing Powers of Digits}} * {{NumberPageLink|prev = 5|next = 7|result = Sum of Sequence of Alternating Positive and Negative Factorials being Prime}} * {{NumberPageLink|prev = 5|next = 8|result = Positive Integers Not Expressible as Sum of Distinct Non-Pythagorean Primes}} * {{NumberPageLink|prev = 5|next = 9|type = Trimorphic Number|cat = Trimorphic Numbers}} * {{NumberPageLink|prev = 5|next = 14|result = Consecutive Integers whose Product is Primorial}} * {{NumberPageLink|prev = 5|next = 25|type = Automorphic Number|cat = Automorphic Numbers}} === Next in Sequence: $10$ and above === * {{NumberPageLink|next = 10|result = Smallest Positive Integer which is Sum of 2 Odd Primes in n Ways}} * {{NumberPageLink|next = 12|type = Semiperfect Number|cat = Primitive Semiperfect Numbers}} * {{NumberPageLink|next = 20|type = Primitive Semiperfect Number|cat = Semiperfect Numbers}} * {{NumberPageLink|next = 28|type = Perfect Number|cat = Perfect Numbers}} * {{NumberPageLink|next = 60|type = Unitary Perfect Number|cat = Unitary Perfect Numbers}} * {{NumberPageLink|next = 120|result = Triangular Numbers which are Product of 3 Consecutive Integers}} * {{NumberPageLink|next = 140|result = Sequence of Numbers with Integer Arithmetic and Harmonic Means of Divisors}} == Historical Note == {{:6/Historical Note}} == Linguistic Note == {{:6/Linguistic Note}} == Sources == * {{BookReference|Curious and Interesting Numbers|1986|David Wells|prev = Definition:Automorphic Number/Sequence|next = Prime equals Plus or Minus One modulo 6}}: $5$ * {{BookReference|Curious and Interesting Numbers|1986|David Wells|prev = Volume of Unit Hypersphere/Sequence|next = 6/Historical Note}}: $6$ * {{BookReference|Curious and Interesting Numbers|1997|David Wells|ed = 2nd|edpage = Second Edition|prev = Definition:Automorphic Number/Sequence|next = Prime equals Plus or Minus One modulo 6}}: $5$ * {{BookReference|Curious and Interesting Numbers|1997|David Wells|ed = 2nd|edpage = Second Edition|prev = Volume of Unit Hypersphere/Sequence|next = 6/Historical Note}}: $6$ Category:Specific Numbers Category:6 ejp9al0qz8hkby09qu128kg0lw3tj1v


Up-Complete Product/Lemma 1

Up-Complete Product/Lemma 1 0 47356 256329 2016-05-19T16:27:54Z GrzegorzBancerek 2860 Created page with "== Theorem == {{:Up-Complete Product}} <onlyinclude> Let $X$ be a directed subset of $S$.  Let $Y$ be a Definition:Directed Subset|directed su..." wikitext text/x-wiki == Theorem == {{:Up-Complete Product}} <onlyinclude> Let $X$ be a directed subset of $S$. Let $Y$ be a directed subset of $T$. Then $X \times Y$ is also a directed subset of $S \times T$. </onlyinclude> == Proof == Let $\left({s_1, t_1}\right)$, $\left({s_2, t_2}\right) \in X \times Y$. By definition of Cartesian product: :$s_1, s_2 \in X$ and $t_1, t_2 \in Y$ By definition of directed subset: :$\exists h_1 \in X: s_1 \preceq_1 h_1 \land s_2 \preceq_1 h_1$ and :$\exists h_2 \in X: t_1 \preceq_2 h_2 \land t_2 \preceq_2 h_2$ By definition of Cartesian product of ordered sets: :$\exists \left({h_1, h_2}\right) \in X \times Y: \left({s_1, t_1}\right) \preceq \left({h_1, h_2}\right) \land \left({s_2, t_2}\right) \preceq \left({h_1, h_2}\right)$ Thus by definition: :$X \times Y$ is a directed subset of $S \times T$. {{qed}} == Sources == * {{BookReference|A Compendium of Continuous Lattices|1980|G. Gierz|author2 = K.H. Hofmann|author3 = K. Keimel|author4 = J.D. Lawson|author5 = M.W. Mislove|author6 = D.S. Scott}} * {{Mizar|link = yellow_3|sublink = FC9|display = YELLOW_3:funcreg 9}} Category:Order Theory jul4rr7gh926ejar25yrngrnpac5iyr


Dirichlet's Theorem on Arithmetic Sequences/Lemma 1 0 47367 456516 455650 2020-03-19T22:26:35Z Prime.mover 59 wikitext text/x-wiki == Lemma for Dirichlet's Theorem on Arithmetic Sequences == Let $a, q$ be coprime integers. Let $\PP_{a, q}$ be the set of primes $p$ such that $p \equiv a \pmod q$. {{MissingLinks|trivial character}} <onlyinclude> Let $\chi$ be a Dirichlet character modulo $q$. Let: :$\displaystyle \map f s = \sum_p \map \chi p p^{-s}$ If $\chi$ is non-trivial then $\map f s$ is bounded as $s \to 1$. If $\chi$ is the trivial character then: :$\map f s \sim \map \ln {\dfrac 1 {s - 1} }$ as $s \to 1$. </onlyinclude> == Proof == By Logarithm of Dirichlet L-Functions: :$(1):\quad \displaystyle \sum_p \map \chi p p^{-s} = \ln \map L {s, \chi} - \sum_p \sum_{n \mathop \ge 2} \frac {\map \chi p^n} {n p^{n s} }$ If $\chi$ is non-trivial, then by L-Function does not Vanish at One, $\ln \map L {s, \chi}$ is bounded as $s \to 1$. If $\chi$ is trivial, then by Analytic Continuation of Dirichlet L-Function, $\map L {s, \chi}$ has a simple pole at $s = 1$. Therefore, in this case: :$\map L {s, \chi} \sim \dfrac \lambda {s - 1}$ where $\lambda$ is the residue of $\map L {s, \chi}$ at $1$, and: :$\ln \map L {s, \chi} \sim \map \ln {\dfrac \lambda {s - 1} } \sim \map \ln {\dfrac 1 {s - 1} }$ Thus if we can show that the second term of $(1)$ is bounded, the result holds. On $\map \Re s > 1$: {{begin-eqn}} {{eqn | l = \size {\sum_p \sum_{n \mathop \ge 2} \frac {\map \chi p^n} {n p^{n s} } }       | o = \le       | r = \sum_p \sum_{n \mathop \ge 2} \frac 1 {\size {p^s}^n} }} {{eqn | r = \sum_p \frac 1 {\size {p^{2 s} } \, \size {p^s - 1}^n}       | c = Sum of Infinite Geometric Sequence }} {{eqn | o = \le       | r = \sum_p \frac 1 {p^2}       | c = because $\map \Re s > 1$ }} {{eqn | o = \le       | r = \sum_n \frac 1 {n^2} }} {{end-eqn}} This last is $\map \zeta 2$ where $\zeta$ is the Riemann zeta function, so is finite. {{qed}} Category:Dirichlet's Theorem on Arithmetic Sequences ax5uhccn3jldi6qc30a9840yrqwyec9


Dirichlet's Theorem on Arithmetic Sequences/Lemma 2 0 47370 455651 455645 2020-03-18T17:55:54Z Prime.mover 59 wikitext text/x-wiki == Lemma for Dirichlet's Theorem on Arithmetic Sequences == Let $a, q$ be coprime integers. Let $\PP_{a, q}$ be the set of primes $p$ such that $p \equiv a \pmod q$. Define: :$\eta_{a, q} : n \mapsto \begin{cases} 1 & : n \equiv a \pmod q\\ 0 & : \text{otherwise} \end{cases}$ <onlyinclude> Let $G = \paren {\Z / q \Z}^\times$. Let $G^*$ be the dual group of characters on $G$. {{explain|Explain the notation $\paren {\Z / q \Z}^\times$}} Then for all $n \in \N$: :$\displaystyle \map {\eta_{a, q} } n = \sum_{\chi \mathop \in G^*} \frac {\map {\overline \chi} a} {\map \phi q} \, \map \chi n$ </onlyinclude> == Proof == There is only one $x \in G$ such that $\map \eta x \ne 0$, and this equals $\map \eta a = 1$. {{explain|Back up the above statement by a link or an explanation. Are we sure that $\map \eta x$ means $\map {\eta_{a, q} } x$ here?}} So: :$\displaystyle \sum_{x \mathop \in G} \map {\eta_{a, q} } x \, \map {\overline \chi} x = \map {\overline \chi} a$ Therefore, by Discrete Fourier Transform on Abelian Group we have for all $x \in G$: :$\displaystyle \map \eta x = \frac 1 {\map \phi q} \sum_{\chi \mathop \in G^*} \map {\overline \chi} a \, \map \chi x$ as required. {{qed}} Category:Dirichlet's Theorem on Arithmetic Sequences djw33ie4ddkrld71wuqv6baweck1a68


Squaring the Circle

Squaring the Circle 0 47413 452698 275657 2020-03-05T16:15:35Z Prime.mover 59 wikitext text/x-wiki == Classic Problem == '''Squaring the Circle''' is creating a square with the same area as that of a given circle. == Solution == The problem of '''squaring the circle''' reduces to: :starting with a line segment of given length ($1$, say) :constructing a line segment of length $\sqrt \pi$ of the first. == Fallacious Proofs == {{:Squaring the Circle/Fallacious Proofs}} == Also see == * Squaring the Circle by Archimedean Spiral * Squaring the Circle by Compass and Straightedge Construction is Impossible * Trisecting the Angle * Doubling the Cube == Historical Note == {{:Squaring the Circle/Historical Note}} == Sources == * {{BookReference|Men of Mathematics|1937|Eric Temple Bell|prev = Doubling the Cube|next = Trisecting the Angle/Historical Note}}: Chapter $\text{II}$: Modern Minds in Ancient Bodies * {{BookReference|Calculus Gems|1992|George F. Simmons|prev = Definition:Compass and Straightedge Construction|next = Squaring the Circle by Compass and Straightedge Construction is Impossible/Historical Note}}: Chapter $\text {B}.18$: Algebraic and Transcendental Numbers. $e$ is Transcendental * {{BookReference|The Concise Oxford Dictionary of Mathematics|2014|Christopher Clapham|author2 = James Nicholson|ed = 5th|edpage = Fifth Edition|prev = Definition:Positive Square Root|next = Squaring the Circle/Historical Note|entry = squaring the circle}} Category:Classic Problems Category:Euclidean Geometry Category:Circles Category:Squaring the Circle eflxogcb0bd0glkrrarijc17w1kk9xd


Liouville's Constant is Transcendental/Corollary 0 47418 468551 441910 2020-05-17T13:07:44Z RandomUndergrad 3904 wikitext text/x-wiki == Corollary to Liouville's Constant is Transcendental == <onlyinclude> All real numbers of the form: {{begin-eqn}} {{eqn | l = \sum_{n \mathop \ge 1} \frac {a_n} {10^{n!} }       | r = \frac {a_1} {10^1} + \frac {a_2} {10^2} + \frac {a_3} {10^6} + \frac {a_4} {10^{24} } + \cdots       | c =  }} {{end-eqn}} where :$a_1, a_2, a_3, \ldots \in \set {1, 2, \ldots, 9}$ are transcendental. </onlyinclude> == Proof == Let $n \in \N$. For $n = 1$, let $p = a_1$ and $q = 10$. Then: :$\size {L - \dfrac p q} = \displaystyle \sum_{k \mathop \ge 2} \dfrac {a_k} {10^{k!} } < \dfrac 1 {10} = \dfrac 1 q$ For $n > 1$, let $q = 10^{n!}$ and write: :$\displaystyle L = \frac p q + \sum_{k \mathop = n + 1}^\infty \frac {a_k} {10^{k!} }$ for some suitable $p \in \Z$. Then: {{begin-eqn}} {{eqn | l = \size {L - \frac p q}       | r = \sum_{k \mathop = n + 1}^\infty \frac {a_k} {10^{k!} }       | c =   }} {{eqn | r = \sum_{k \mathop = n + 1}^\infty \frac 9 {10^{k!} }       | o = \le       | c =  }} {{eqn | r = \frac {18} {10^{\paren {n + 1}!} }       | o = \le       | c =  }} {{eqn | r = \frac {18} {q^{n + 1} }       | c =  }} {{eqn | r = \frac 1 {q^n}       | o = <       | c = as $q \ge 100$ for all $n > 1$ }} {{end-eqn}} Thus, by definition, $L$ is a Liouville number. Therefore, by Liouville's Theorem, $L$ is transcendental. {{qed}} == Sources == * {{BookReference|Calculus Gems|1992|George F. Simmons|prev = Liouville's Constant is Transcendental|next = Euler's Number is Transcendental/Historical Note}}: Chapter $\text {B}.18$: Algebraic and Transcendental Numbers. $e$ is Transcendental Category:Transcendental Number Theory Category:Transcendental Numbers 0alyacskwwju8zw0k0p1qhj80ivn46c