Dihedral Group D4/Center

Dihedral Group D4/Center 0 73863 387887 387886 2019-01-15T21:08:09Z Prime.mover 59 wikitext text/x-wiki == Center of the Dihedral Group $D_4$ == Let $D_4$ denote the dihedral group $D_4$, whose group presentation is given as: {{:Group Presentation of Dihedral Group D4}} <onlyinclude> The center of $D_4$ is given by: :$\map Z {D_4} = \set {e, a^2}$ </onlyinclude> == Proof == From Center of Dihedral Group: :$\map Z {D_n} = \begin{cases} e & : n \text { odd} \\ \set {e, \alpha^{n / 2} } & : n \text { even} \end{cases}$ Hence the result. {{qed}} == Sources == * {{BookReference|A Course in Group Theory|1996|John F. Humphreys|prev = Symmetric Group on 3 Letters/Center|next = Definition:Group Homomorphism}}: Chapter $7$: Normal subgroups and quotient groups: Exercise $5$ Category:Centers of Groups Category:Dihedral Group D4 cswd14eda2ndj39sr0wh4es0d3hluuv


Group Action of Symmetric Group on Complex Vector Space/Orbit/Examples/Example 1 0 73994 417080 388751 2019-08-07T15:42:10Z Prime.mover 59 wikitext text/x-wiki == Example of Orbit of Group Action of Symmetric Group on Complex Vector Space == Let $S_n$ denote the symmetric group on $n$ letters. Let $V$ denote a vector space over the complex numbers $\C$. Let $V$ have a basis: :$\mathcal B := \set {v_1, v_2, \ldots, v_n}$ Let $*: S_n \times V \to V$ be a group action of $S_n$ on $V$ defined as: :$\forall \tuple {\rho, v} \in S_n \times V: \rho * v := \lambda_1 v_{\map \rho 1} + \lambda_2 v_{\map \rho 2} + \dotsb + \lambda_n v_{\map \rho n}$ where: :$v = \lambda_1 v_1 + \lambda_2 v_2 + \dotsb + \lambda_n v_n$ <onlyinclude> Let $n = 4$. Let $v = v_1 + v_2 + v_3 + v_4$. The orbit of $v$ is: :$\Orb v = \set v$ </onlyinclude> == Proof == {{begin-eqn}} {{eqn | l = \Orb v       | r = \set {w \in V: \exists \rho \in S_4: w = \rho * v}       | c = {{Defof|Orbit (Group Theory)|Orbit}} }} {{eqn | r = \set {w \in V: \exists \rho \in S_4: w = \rho * \sum_{k \mathop = 1}^4 v_k}       | c = Definition of $v$ }} {{eqn | r = \set {w \in V: \exists \rho \in S_4: w = \sum_{k \mathop = 1}^4 v_\map \rho k}       | c = Definition of $*$ }} {{eqn | r = \set {w \in V: \exists \rho \in S_4: w = \sum_{k \mathop = 1}^4 v_k}       | c = Permutation of Indices of Summation }} {{eqn | r = \set {w \in V: \exists \rho \in S_4: w = v}       | c = Definition of $v$ }} {{eqn | r = \set v       | c =  }} {{end-eqn}} {{qed}} == Sources == * {{BookReference|A Course in Group Theory|1996|John F. Humphreys|prev = Group Action of Symmetric Group on Complex Vector Space/Stabilizer|next = Group Action of Symmetric Group on Complex Vector Space/Stabilizer/Examples/Example 1}}: Chapter $10$: The Orbit-Stabiliser Theorem: Exercise $1 \ \text {(a)}$ Category:Group Action of Symmetric Group on Complex Vector Space bvxiiee6p1aslbmoipu19e3qdvvc2ia


Sylow Theorems/Examples/Sylow 3-Subgroups in Group of Order 12 0 74043 388972 2019-01-21T21:07:43Z Prime.mover 59 Created page with "== Example of Use of Sylow Theorems == <onlyinclude> In a group of order $12$, there are either $1$ or $4$ Definition:..." wikitext text/x-wiki == Example of Use of Sylow Theorems == <onlyinclude> In a group of order $12$, there are either $1$ or $4$ Sylow $3$-subgroups. </onlyinclude> == Proof == Let $G$ be a group of order $12$. Let $n_3$ be the number of Sylow $3$-subgroups in $G$. From the Fourth Sylow Theorem, $n_3$ is congruent to $1$ modulo $3$, that is, in $\set {1, 4, 7, \ldots}$ Let $H$ be a Sylow $3$-subgroup of $G$. We have that: :$12 = 4 \times 3$ and so the order of $H$ is $3$. Thus: {{begin-eqn}} {{eqn | l = \index G H       | r = \dfrac {12} {3}       | c =  }} {{eqn | r = 4       | c =  }} {{end-eqn}} From the Fifth Sylow Theorem: :$n_3 \divides 4$ where $\divides$ denotes divisibility. Thus there may be $1$ or $4$ Sylow $3$-subgroups of $G$. {{qed}} == Sources == * {{BookReference|A Course in Group Theory|1996|John F. Humphreys|prev = Sylow Theorems/Examples/Sylow 2-Subgroups in Group of Order 12|next = Direct Product of Normal Subgroups is Normal}}: Chapter $11$: The Sylow Theorems: Exercise $1 \ \text{(e)}$ Category:Sylow Theorems Category:Groups of Order 12 i082pikg0cnwnvas2h0g72fnb5xdgqw


Pullback is Subgroup

Pullback is Subgroup 0 74142 389466 389454 2019-01-24T16:18:20Z Prime.mover 59 wikitext text/x-wiki {{disambig}} * Pullback of Quotient Group Isomorphism is Subgroup Category:Group Theory 1bgtgm1hui7kmw21di56qb30x84d37i


Pullback of Quotient Group Isomorphism/Examples/Subgroups of Index 2 0 74155 389510 389500 2019-01-24T21:43:39Z Prime.mover 59 wikitext text/x-wiki == Example of Pullback of Quotient Group Isomorphism == <onlyinclude> Let $G$ and $H$ be groups. Let $N$ and $K$ be normal subgroups of $G$ and $H$ respectively such that: :their quotient groups $G / N$ and $H / K$ are isomorphic :their indices are $2$: ::$\index G N = \index H K = 2$ Let $\theta: G / N \to H / K$ be an isomorphism. The pullback of $G$ and $H$ by $\theta$ is a subset of $G \times H$ of the form: :$G \times^\theta H = \set {\tuple {g, h}: \paren {g \in N, h \in K} \text { or } \paren {g \notin N, h \notin K} }$ </onlyinclude> == Proof == As $\index G N = \index H K = 2$, it follows that: :$\order {G / N} = \order {H / K} = 2$ and they are the cyclic group of order $2$. Let: : $x \in G: x \notin N$ : $y \in H: y \notin K$ Then: :$G / N = \gen {x N}$ :$H / K = \gen {y K}$ and we have: {{begin-eqn}} {{eqn | l = \map \theta N       | r = K }} {{eqn | l = \map \theta {x N}       | r = y K }} {{end-eqn}} Let $\tuple {g, h} \in G \times^\theta H$. By definition: :$G \times^\theta H = \set {\tuple {g, h}: \map \theta {g N} = h K}$ Let $g \in N$. Then $g N = N$ and so: : $\map \theta {g N} = h K = K$ So $g \in N \implies h \in K$. Let $g \notin N$. Then $g N = x N$ and so: : $\map \theta {g N} = h K = x K$ So $g \notin N \implies h \notin K$. Hence the result. {{qed}} == Sources == * {{BookReference|A Course in Group Theory|1996|John F. Humphreys|prev = Pullback of Quotient Group Isomorphism is Subgroup|next = Pullback of Quotient Group Isomorphism/Examples/Alternating Subgroups of Symmetric Groups}}: Chapter $13$: Direct products: Example $13.12$ Category:Examples of Pullbacks of Quotient Group Isomorphisms tgclnb62ygmyfrq9ftv21h2wws2sqvg


Arcsin as an Integral/Lemma 1

Arcsin as an Integral/Lemma 1 0 74302 390719 390718 2019-01-30T14:50:47Z Pelliott 3564 sin_A inverse arcsin_a wikitext text/x-wiki == Lemma == <onlyinclude> Let $sin_A$ be the analytic sine function for real numbers, the one defined by Definition:Sine/Real Numbers. $\arcsin_A$ is the inverse of this function. :$\displaystyle \map {\arcsin_A} x = \int_0^x \frac {\d x} {\sqrt {1 - x^2} }$ </onlyinclude> == Proof == For this proof only, let $\sin_A$ be the analytic sine function from Definition:Sine/Real Numbers. Consider: :$\displaystyle \int_0^x \frac {\d x} {\sqrt {1 - x^2} }$ Let: :$x = \sin_A \theta \iff x = \map {\arcsin_A} \theta$ Then: {{begin-eqn}} {{eqn | l = \d x       | r = \cos_A \theta \rd \theta       | c = Derivative of Sine Function }} {{eqn | l = \int \frac {\d x} {\sqrt {1 - x^2} }       | r = \int \frac {\d x} {\cos_A \theta} \cos_A \theta \rd \theta       | c = Integration by Substitution }} {{eqn | r = \int 1 \rd \theta }} {{eqn | r = \theta + C }} {{eqn | r = \map {\arcsin_A} x + C }} {{eqn | ll= \leadsto       | l = \int_0^x \frac {\d x} {\sqrt {1 - x^2} }       | r = \map {\arcsin_A} x       | c = Fundamental Theorem of Calculus: Second Part }} {{end-eqn}} {{qed|lemma}} chldcq0gp73np0qqonhd2zb3nt5sh13


Arcsin as an Integral/Lemma 2

Arcsin as an Integral/Lemma 2 0 74304 390774 390773 2019-01-31T06:27:22Z Prime.mover 59 Reverted edits by Prime.mover (talk) to last revision by Pelliott wikitext text/x-wiki == Lemma == <onlyinclude> Let, $\sin_G$ be the Geometric Sine from Definition:Sine/Definition from Circle. $\arcsin_G$ is the inverse of this function. :$\displaystyle \map {\arcsin_G} x = \int_0^x \frac {\d x} {\sqrt {1 - x^2} }$ </onlyinclude> == Proof == This result will be used in proving Derivative of Sine Function in the geometric case. So we can not use the same reasoning as Arcsin as an Integral/Lemma 1 because our logic would be circular. :640px {{improve|Might be worth revisiting the diagram to make all the text the same size}} Let $\theta$ be the length of the arc associated with the angle on the circle of radius $1$. By definition of arcsine: :$y = \sin \theta \iff \theta = \arcsin y$ We have that arc length is always positive.  For negative $y$, the $\arcsin$ function is defined as being the negative of the arc length. This makes the $\arcsin$ function and the $\sin$ function odd, and puts us in line with mathematical convention: :Inverse Sine is Odd Function. :Sine Function is Odd Without this convention, the derivative of the $\sin$ function would not be continuous.  Now: {{begin-eqn}} {{eqn | l = x^2 + y^2       | r = 1       | c = Equation of Circle       | n = 1 }} {{eqn | l = \dfrac {\d x} {\d y}       | r = -\dfrac y x       | c = Implicit Differentiation }} {{eqn | r = -\dfrac y {\sqrt {1 - y^2} }       | c = substituting for $x$ }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \arcsin_G y       | r = \int_0^y \sqrt {1 + \paren {\dfrac {\d x} {\d y} }^2} \rd y       | c = {{Defof|Arc Length}} }} {{eqn | r = \int_0^y \sqrt {1 + \paren {-\dfrac y x}^2}       | c = substituting for $\dfrac {\d x} {\d y}$ }} {{eqn | r = \int_0^y \sqrt {1 + \dfrac {y^2} {x^2} } \rd y }} {{eqn | r = \int_0^y \sqrt {\dfrac {x^2} {x^2} + \dfrac {y^2} {x^2} } \rd y       | c = rewriting $1$ to create common denominator }} {{eqn | r = \int_0^y \sqrt {\dfrac {x^2 + y^2} {x^2} } \rd y       | c = combining terms with common denominator }} {{eqn | r = \int_0^y \sqrt {\dfrac 1 {x^2} } \rd y       | c = Equation of Circle $(1)$ }} {{eqn | r = \int_0^y \dfrac 1 x \rd y       | c = in Quadrant $\text I$ and Quadrant $\text {IV}$ }} {{eqn | r = \int_0^y \dfrac 1 {\sqrt {1 - y^2} } \rd y       | c = substituting for $x$ in Quadrant $\text I$ and Quadrant $\text {IV}$ }} {{end-eqn}} {{qed|lemma}} l2gxb81370tgmmd6hllwvmdl8u05l8q


Modulo Arithmetic/Examples/n(n^2-1)(3n-2) Modulo 24 0 74319 390579 2019-01-29T21:13:22Z Prime.mover 59 Created page with "== Example of Modulo Arithmetic == <onlyinclude> :$n \paren {n^2 - 1} \paren {3 n + 2} \equiv 0 \pmod {24}$ </onlyinclude>  == Proof ==  The p..." wikitext text/x-wiki == Example of Modulo Arithmetic == <onlyinclude> :$n \paren {n^2 - 1} \paren {3 n + 2} \equiv 0 \pmod {24}$ </onlyinclude> == Proof == The proof proceeds by induction. For all $n \in \Z_{> 0}$, let $\map P n$ be the proposition: :$n \paren {n^2 - 1} \paren {3 n + 2} \equiv 0 \pmod {24}$ === Basis for the Induction === $\map P 1$ is the case: {{begin-eqn}} {{eqn | l = 1 \paren {1^2 - 1} \paren {3 \times 1 + 2}       | r = 0       | c =  }} {{eqn | o = \equiv       | r = 0       | rr= \pmod {24}       | c =  }} {{end-eqn}} Thus $\map P 1$ is seen to hold. This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true. So this is the induction hypothesis: :$k \paren {k^2 - 1} \paren {3 k + 2} \equiv 0 \pmod {24}$ from which it is to be shown that: :$\paren {k + 1} \paren {\paren {k + 1}^2 - 1} \paren {3 \paren {k + 1} + 2} \equiv 0 \pmod {24}$ === Induction Step === This is the induction step: {{begin-eqn}} {{eqn | l = \paren {k + 1} \paren {\paren {k + 1}^2 - 1} \paren {3 \paren {k + 1} - 2}       | r = k \paren {k + 1} \paren {k + 2} \paren {3 k + 5}       | c =  }} {{eqn | r = k \paren {k + 1} \paren {k - 1 + 3} \paren {3 k + 2 + 3}       | c =  }} {{eqn | r = k \paren {k + 1} \paren {k - 1} \paren {3 k + 2 + 3} + 3 k \paren {k + 1} \paren {3 k + 2 + 3}       | c =  }} {{eqn | r = k \paren {k + 1} \paren {k - 1} \paren {3 k + 2} + 3 k \paren {k + 1} \paren {k - 1} + 3 k \paren {k + 1} \paren {3 k + 5}       | c =  }} {{eqn | r = k \paren {k^2 - 1} \paren {3 k + 2} + 3 k \paren {k + 1} \paren {k - 1 + 3 k + 5}       | c =  }} {{eqn | r = k \paren {k^2 - 1} \paren {3 k + 2} + 3 k \paren {k + 1} \paren {4 k + 4}       | c =  }} {{eqn | r = k \paren {k^2 - 1} \paren {3 k + 2} + 12 k \paren {k + 1}^2       | c =  }} {{end-eqn}} By the induction hypothesis: :$k \paren {k^2 - 1} \paren {3 k + 2} = 24 r$ for some $r \in \Z$. Take $12 k \paren {k + 1}^2$. If $k$ is even, then $12 k$ and so $12 k \paren {k + 1}^2$ is divisible by $24$. If $k$ is odd, then $k + 1$ is even and so $12 k \paren {k + 1}^2$ is again divisible by $24$. Thus: :$12 k \paren {k + 1}^2 = 24 s$ for some $s \in \Z$. Thus: :$k \paren {k^2 - 1} \paren {3 k + 2} + 12 k \paren {k + 1}^2 = 24 \paren {r + s}$ So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\forall n \in \Z_{>0}: n \paren {n^2 - 1} \paren {3 n + 2} \equiv 0 \pmod {24}$ {{qed}} == Sources == * {{BookReference|Number Theory|1971|George E. Andrews|prev = Sum of Sequence of Product of Lucas Numbers with Powers of 2|next = Sum of Odd Positive Powers}}: $\text {1-1}$ Principle of Mathematical Induction: Exercise $18$ Category:Examples of Modulo Arithmetic 040thy9kquhouj6rzuegnhscy352s0e


Euclidean Algorithm/Examples/341 and 527 0 74394 493846 391022 2020-10-11T06:57:10Z Prime.mover 59 wikitext text/x-wiki == Examples of Use of Euclidean Algorithm == <onlyinclude> The GCD of $341$ and $527$ is found to be: :$\gcd \set {341, 527} = 31$ ==== Integer Combination ==== {{:Euclidean Algorithm/Examples/341 and 527/Integer Combination}}</onlyinclude> == Proof == {{begin-eqn}} {{eqn | n = 1       | l = 527       | r = 1 \times 341 + 186 }} {{eqn | n = 2       | l = 341       | r = 1 \times 186 + 155 }} {{eqn | n = 3       | l = 186       | r = 1 \times 155 + 31 }} {{eqn | n =        | l = 155       | r = 5 \times 31 }} {{end-eqn}} Thus: :$\gcd \set {341, 527} = 31$ {{qed}} == Sources == * {{BookReference|Number Theory|1971|George E. Andrews|prev = GCD of Integer and Divisor|next = Euclidean Algorithm}}: $\text {2-2}$ Divisibility: Example $\text {2-7}$ Category:Examples of Euclidean Algorithm 1tp7629hnw1wa04yq9vndlb9a8x1e6g


Lowest Common Multiple/Examples/n and n+1 0 74418 391109 391100 2019-02-02T13:17:31Z Prime.mover 59 wikitext text/x-wiki == Example of Lowest Common Multiple of Integers == Let $n \in \Z_{>0}$ be a strictly positive integer. <onlyinclude> The lowest common multiple of $n$ and $n + 1$ is: :$\lcm \set {n, n + 1} = n \paren {n + 1}$ </onlyinclude> == Proof == We find the greatest common divisor of $n$ and $n + 1$ using the Euclidean Algorithm: {{begin-eqn}} {{eqn | n = 1       | l = n + 1       | r = 1 \times n + 1 }} {{eqn | n = 2       | l = n       | r = n \times 1 }} {{end-eqn}} Thus $\gcd \set {n, n + 1} = 1$. Hence by definition $n$ and $n + 1$ are coprime. The result follows from LCM of Coprime Integers. {{qed}} == Sources == * {{BookReference|Number Theory|1971|George E. Andrews|prev = Lowest Common Multiple/Examples/28 and 29|next = Lowest Common Multiple of Consecutive Odd Integers}}: $\text {2-2}$ Divisibility: Exercise $5 \ \text {(e)}$ Category:Examples of Lowest Common Multiples tenqrtcuqhx6ex6zgf2ko269dorzlak


Area of Triangle in Determinant Form with Vertex at Origin 0 74481 391400 391394 2019-02-04T07:47:41Z Prime.mover 59 wikitext text/x-wiki == Example of Area of Triangle in Determinant Form == <onlyinclude> Let $A = \tuple {0, 0}, B = \tuple {b, a}, C = \tuple {x, y}$ be points in the Cartesian plane. Let $T$ the triangle whose vertices are at $A$, $B$ and $C$. Then the area $\mathcal A$ of $T$ is: :$\map \Area T = \dfrac {\size {b y - a x} } 2$ </onlyinclude> == Proof == {{begin-eqn}} {{eqn | l = \map \Area T       | r = \dfrac 1 2 \size {\paren {\begin{vmatrix} 0 & 0 & 1 \\ b & a & 1 \\ x & y & 1 \end{vmatrix} } }       | c = Area of Triangle in Determinant Form }} {{eqn | r = \dfrac 1 2 \size {b y - a x}       | c = {{Defof|Determinant of Order 3}} }} {{end-eqn}} {{qed}} == Example == {{:Area of Triangle in Determinant Form with Vertex at Origin/Example}} == Sources == * {{BookReference|Number Theory|1971|George E. Andrews|prev = Linear Diophantine Equation/Examples/15x + 51y = 41/Graph|next = Area of Triangle in Determinant Form with Vertex at Origin/Example}}: $\text {2-3}$ The Linear Diophantine Equation: Exercise $4$ Category:Area of Triangle in Determinant Form e4vesh61jg84w2k8j6f36tt46abd0bs


Square Modulo 3/Corollary 3

Square Modulo 3/Corollary 3 0 74617 476278 462461 2020-06-28T15:53:22Z RandomUndergrad 3904 wikitext text/x-wiki == Corollary to Square Modulo 3 == <onlyinclude> Let $n \in \Z$ be an integer such that: :$3 \nmid n$ where $\nmid$ denotes non-divisibility. Then: :$3 \divides n^2 - 1$ where $\divides$ denotes divisibility. </onlyinclude> == Proof == From Square Modulo 3: :$n \equiv 0 \pmod 3 \iff n^2 \equiv 0 \pmod 3$ Hence also from Square Modulo 3: :$n \not \equiv 0 \pmod 3 \iff n^2 \equiv 1 \pmod 3$ That is: $3 \nmid n \iff 3 \divides n^2 - 1$ {{qed}} == Sources == * {{BookReference|Number Theory|1971|George E. Andrews|prev = Integer and Fifth Power have same Last Digit|next = Sufficient Condition for 5 to divide n^2+1}}: $\text {3-2}$ Fermat's Little Theorem: Exercise $5$ Category:Square Modulo 3 n39xgma4sjtzzqmqv79rj9wyejujslo


Quadratic Residue/Examples/3

Quadratic Residue/Examples/3 0 74687 392366 392356 2019-02-14T07:45:47Z Prime.mover 59 wikitext text/x-wiki == Example of Quadratic Residues == <onlyinclude> There exists exactly $1$ quadratic residue modulo $3$, and that is $1$. </onlyinclude> == Proof == To list the quadratic residues of $3$ it is enough to work out the squares $1^2$ and $2^2$ modulo $3$. {{begin-eqn}} {{eqn | l = 1^2       | o = \equiv       | r = 1       | rr= \pmod 3 }} {{eqn | l = 2^2       | o = \equiv       | r = 1       | rr= \pmod 3 }} {{end-eqn}} So the set of quadratic residues modulo $3$ is: :$\set 1$ The set of quadratic non-residues of $3$ therefore consists of all the other non-zero least positive residues: :$\set 2$ {{qed}} == Sources == * {{BookReference|Number Theory|1971|George E. Andrews|prev = Quadratic Residue/Examples/11|next = Quadratic Residue/Examples/5}}: $\text {3-5}$ The Use of Computers in Number Theory: Exercise $6$ Category:Examples of Quadratic Residues Category:3 hychsymwxsahklszp3zsz49zhllthfr


Quadratic Residue/Examples/5

Quadratic Residue/Examples/5 0 74688 392365 392355 2019-02-14T07:45:36Z Prime.mover 59 wikitext text/x-wiki == Example of Quadratic Residues == <onlyinclude> The set of quadratic residues modulo $5$ is: :$\set {1, 4}$ </onlyinclude> == Proof == To list the quadratic residues of $5$ it is enough to work out the squares $1^2, 2^2, 3^2, 4^2$ modulo $5$. {{begin-eqn}} {{eqn | l = 1^2       | o = \equiv       | r = 1       | rr= \pmod 5 }} {{eqn | l = 2^2       | o = \equiv       | r = 4       | rr= \pmod 5 }} {{eqn | l = 3^2       | o = \equiv       | r = 4       | rr= \pmod 5 }} {{eqn | l = 4^2       | o = \equiv       | r = 1       | rr= \pmod 5 }} {{end-eqn}} So the set of quadratic residues modulo $5$ is: :$\set {1, 4}$ The set of quadratic non-residues of $5$ therefore consists of all the other non-zero least positive residues: :$\set {2, 3}$ {{qed}} == Sources == * {{BookReference|Number Theory|1971|George E. Andrews|prev = Quadratic Residue/Examples/3|next = Quadratic Residue/Examples/7}}: $\text {3-5}$ The Use of Computers in Number Theory: Exercise $6$ Category:Examples of Quadratic Residues Category:5 siq4yv3zvjp16uejhkb1053p13l75s9


Quadratic Residue/Examples/7

Quadratic Residue/Examples/7 0 74689 392364 392354 2019-02-14T07:45:22Z Prime.mover 59 wikitext text/x-wiki == Example of Quadratic Residues == <onlyinclude> The set of quadratic residues modulo $7$ is: :$\set {1, 2, 4}$ </onlyinclude> == Proof == To list the quadratic residues of $7$ it is enough to work out the squares $1^2, 2^2, \dotsc, 6^2$ modulo $7$. {{begin-eqn}} {{eqn | l = 1^2       | o = \equiv       | r = 1       | rr= \pmod 7 }} {{eqn | l = 2^2       | o = \equiv       | r = 4       | rr= \pmod 7 }} {{eqn | l = 3^2       | o = \equiv       | r = 2       | rr= \pmod 7 }} {{eqn | l = 4^2       | o = \equiv       | r = 2       | rr= \pmod 7 }} {{eqn | l = 5^2       | o = \equiv       | r = 4       | rr= \pmod 7 }} {{eqn | l = 6^2       | o = \equiv       | r = 1       | rr= \pmod 7 }} {{end-eqn}} So the set of quadratic residues modulo $7$ is: :$\set {1, 2, 4}$ The set of quadratic non-residues of $7$ therefore consists of all the other non-zero least positive residues: :$\set {3, 5, 6}$ {{qed}} == Sources == * {{BookReference|Number Theory|1971|George E. Andrews|prev = Quadratic Residue/Examples/5|next = Quadratic Residue/Examples/17}}: $\text {3-5}$ The Use of Computers in Number Theory: Exercise $6$ Category:Examples of Quadratic Residues Category:7 3igdqrcv9cp30opd1bnv3momeyleese


Quadratic Residue/Examples/17

Quadratic Residue/Examples/17 0 74690 433460 392363 2019-11-01T13:30:35Z Prime.mover 59 wikitext text/x-wiki == Example of Quadratic Residues == <onlyinclude> The set of quadratic residues modulo $17$ is: :$\set {1, 2, 4, 8, 9, 13, 15, 16}$ </onlyinclude> {{OEIS|A010379}} == Proof == To list the quadratic residues of $17$ it is enough to work out the squares $1^2, 2^2, \dotsc, 16^2$ modulo $17$. {{begin-eqn}} {{eqn | l = 1^2       | o = \equiv       | r = 1       | rr= \pmod {17} }} {{eqn | l = 2^2       | o = \equiv       | r = 4       | rr= \pmod {17} }} {{eqn | l = 3^2       | o = \equiv       | r = 9       | rr= \pmod {17} }} {{eqn | l = 4^2       | o = \equiv       | r = 16       | rr= \pmod {17} }} {{eqn | l = 5^2       | o = \equiv       | r = 8       | rr= \pmod {17} }} {{eqn | l = 6^2       | o = \equiv       | r = 2       | rr= \pmod {17} }} {{eqn | l = 7^2       | o = \equiv       | r = 15       | rr= \pmod {17} }} {{eqn | l = 8^2       | o = \equiv       | r = 13       | rr= \pmod {17} }} {{eqn | l = 9^2       | o = \equiv       | r = 13       | rr= \pmod {17} }} {{eqn | l = 10^2       | o = \equiv       | r = 15       | rr= \pmod {17} }} {{eqn | l = 11^2       | o = \equiv       | r = 2       | rr= \pmod {17} }} {{eqn | l = 12^2       | o = \equiv       | r = 8       | rr= \pmod {17} }} {{eqn | l = 13^2       | o = \equiv       | r = 16       | rr= \pmod {17} }} {{eqn | l = 14^2       | o = \equiv       | r = 9       | rr= \pmod {17} }} {{eqn | l = 15^2       | o = \equiv       | r = 4       | rr= \pmod {17} }} {{eqn | l = 16^2       | o = \equiv       | r = 1       | rr= \pmod {17} }} {{end-eqn}} So the set of quadratic residues modulo $17$ is: :$\set {1, 2, 4, 8, 9, 13, 15, 16}$ The set of quadratic non-residues of $17$ therefore consists of all the other non-zero least positive residues: :$\set {3, 5, 6, 7, 10, 11, 12, 14}$ {{OEIS|A028730}}{{qed}} == Sources == * {{BookReference|Number Theory|1971|George E. Andrews|prev = Quadratic Residue/Examples/7|next = Quadratic Residue/Examples/29}}: $\text {3-5}$ The Use of Computers in Number Theory: Exercise $6$ Category:Examples of Quadratic Residues Category:17 5bdaz6nnpoy6ye4vre6u1ec5wxq45oc


Quadratic Residue/Examples/29

Quadratic Residue/Examples/29 0 74691 433461 392370 2019-11-01T13:31:01Z Prime.mover 59 wikitext text/x-wiki == Example of Quadratic Residues == <onlyinclude> The set of quadratic residues modulo $29$ is: :$\set {1, 4, 5, 6, 7, 9, 13, 16, 20, 22, 23, 24, 25, 28}$ </onlyinclude> {{OEIS|A010391}} == Proof == From Square Modulo n Congruent to Square of Inverse Modulo n, to list the quadratic residues of $29$ it is sufficient to work out the squares $1^2, 2^2, \dotsc, \paren {\dfrac {28} 2}^2$ modulo $29$. So: {{begin-eqn}} {{eqn | l = 1^2       | o = \equiv       | r = 1       | rr= \pmod {29} }} {{eqn | l = 2^2       | o = \equiv       | r = 4       | rr= \pmod {29} }} {{eqn | l = 3^2       | o = \equiv       | r = 9       | rr= \pmod {29} }} {{eqn | l = 4^2       | o = \equiv       | r = 16       | rr= \pmod {29} }} {{eqn | l = 5^2       | o = \equiv       | r = 25       | rr= \pmod {29} }} {{eqn | l = 6^2       | o = \equiv       | r = 7       | rr= \pmod {29} }} {{eqn | l = 7^2       | o = \equiv       | r = 20       | rr= \pmod {29} }} {{eqn | l = 8^2       | o = \equiv       | r = 6       | rr= \pmod {29} }} {{eqn | l = 9^2       | o = \equiv       | r = 23       | rr= \pmod {29} }} {{eqn | l = 10^2       | o = \equiv       | r = 13       | rr= \pmod {29} }} {{eqn | l = 11^2       | o = \equiv       | r = 5       | rr= \pmod {29} }} {{eqn | l = 12^2       | o = \equiv       | r = 28       | rr= \pmod {29} }} {{eqn | l = 13^2       | o = \equiv       | r = 24       | rr= \pmod {29} }} {{eqn | l = 14^2       | o = \equiv       | r = 22       | rr= \pmod {29} }} {{end-eqn}} So the set of quadratic residues modulo $29$ is: :$\set {1, 4, 5, 6, 7, 9, 13, 16, 20, 22, 23, 24, 25, 28}$ The set of quadratic non-residues of $29$ therefore consists of all the other non-zero least positive residues: :$\set {2, 3, 8, 10, 11, 12, 14, 15, 17, 18, 19, 21, 26, 27}$ {{OEIS|A028742}}{{qed}} == Sources == * {{BookReference|Number Theory|1971|George E. Andrews|prev = Quadratic Residue/Examples/17|next = Quadratic Residue/Examples/61}}: $\text {3-5}$ The Use of Computers in Number Theory: Exercise $6$ Category:Examples of Quadratic Residues Category:29 fwkl4anq98hwx4k3p3b5ma1cxx5z5p1


Reduced Residue System/Examples/Modulo 18/Least Positive Residues 0 74833 392896 392895 2019-02-20T07:49:41Z Prime.mover 59 wikitext text/x-wiki == Examples of Reduced Residue Systems == <onlyinclude> The least positive reduced residue system of $18$ is the set of positive integers: :$\set {1, 5, 7, 11, 13, 17}$ </onlyinclude> == Proof == The least positive residues of $18$ are: :$S := \set {0, 1, 2, \dotsc, 17}$ We have that: :$18 = 2 \times 3^2$ so the least positive reduced residue system of $18$ is the set of elements of $S$ which have neither $2$ or $3$ as a prime factor. That is: :$S \setminus \set {0, 2, 3, 4, 6, 8, 9, 10, 12, 14, 15, 16}$ Hence the result. {{Qed}} == Sources == * {{BookReference|Number Theory|1971|George E. Andrews|prev = Reduced Residue System/Examples/Modulo 18/Square Numbers|next = Congruence Modulo Power of p as Linear Combination of Congruences Modulo p}}: $\text {4-2}$ Residue Systems: Exercise $2 \ \text {(d)}$ Category:Examples of Residue Systems 48gahudxbhrdrfntwdnfmgl02jw3rqk


Binomial Theorem/Examples/Cube of Difference 0 74934 393424 393370 2019-02-23T11:47:30Z Prime.mover 59 wikitext text/x-wiki == Example of Use of Binomial Theorem == <onlyinclude> :$\paren {x - y}^3 = x^3 - 3 x^2 y + 3 x y^2 - y^3$ </onlyinclude> == Proof == Follows directly from the Binomial Theorem: :$\displaystyle \forall n \in \Z_{\ge 0}: \paren {x + \paren {-y} }^n = \sum_{k \mathop = 0}^n \binom n k x^{n - k} \paren {-y}^k$ putting $n = 3$. {{qed}} == Sources == * {{BookReference|Mathematical Handbook of Formulas and Tables|1968|Murray R. Spiegel|prev = Cube of Sum|next = Fourth Power of Sum}}: $\S 2$: Special Products and Factors: $2.4$ * {{BookReference|Mathematical Handbook of Formulas and Tables|1968|Murray R. Spiegel|prev = Square of Sum/Algebraic Proof 2|next = Fourth Power of Sum}}: $\S 20$: Binomial Series: $20.6$ Category:Examples of Use of Binomial Theorem Category:Third Powers slkcwun4hpi3tv9f8xfnyqmchcwkcuv


Binomial Theorem/Examples/4th Power of Difference 0 74937 393380 393379 2019-02-23T11:05:01Z Prime.mover 59 Prime.mover moved page Fourth Power of Difference to Binomial Theorem/Examples/4th Power of Difference wikitext text/x-wiki == Example of Use of Binomial Theorem == <onlyinclude> :$\paren {x - y}^4 = x^4 - 4 x^3 y + 6 x^2 y^2 - 4 x y^3 + y^4$ </onlyinclude> == Proof == Follows directly from the Binomial Theorem: :$\displaystyle \forall n \in \Z_{\ge 0}: \paren {x + \paren {-y} }^n = \sum_{k \mathop = 0}^n \binom n k x^{n - k} \paren {-y}^k$ putting $n = 4$. {{qed}} == Sources == * {{BookReference|Mathematical Handbook of Formulas and Tables|1968|Murray R. Spiegel|prev = Fourth Power of Sum|next = Fifth Power of Sum}}: $\S 2$: Special Products and Factors: $2.6$ Category:Examples of Use of Binomial Theorem Category:Fourth Powers mfi4lv1t7avb4bdvyt7g8fh69w8a8ku


Square Root of Complex Number in Cartesian Form/Examples/-15-8i 0 75200 394768 2019-03-05T20:37:56Z Prime.mover 59 Created page with "== Example of Square Root of Complex Number in Cartesian Form == <onlyinclude> :$\sqrt {-15 - 8 i} = \pm \paren {1 - 4 i}$ </onlyinclude>  == Square Root of Complex Numb..." wikitext text/x-wiki == Example of Square Root of Complex Number in Cartesian Form == <onlyinclude> :$\sqrt {-15 - 8 i} = \pm \paren {1 - 4 i}$ </onlyinclude> == Proof 1 == {{:Square Root of Complex Number in Cartesian Form/Examples/-15-8i/Proof 1}} == Proof 2 == {{:Square Root of Complex Number in Cartesian Form/Examples/-15-8i/Proof 2}} Category:Examples of Square Roots 98oypb55b6zaacw4n4kdz9vczrzctm0


Complex Dot Product/Examples/3-4i dot -4+3i 0 75209 394832 394815 2019-03-05T23:07:31Z Prime.mover 59 wikitext text/x-wiki == Examples of Complex Dot Product == <onlyinclude> Let: :$z_1 = 3 - 4 i$ :$z_2 = -4 + 3 i$ Then: :$z_1 \circ z_2 = -24$ where $\circ$ denotes (complex) dot product. </onlyinclude> == Proof 1 == {{:Complex Dot Product/Examples/3-4i dot -4+3i/Proof 1}} == Proof 2 == {{:Complex Dot Product/Examples/3-4i dot -4+3i/Proof 2}} Category:Examples of Complex Dot Product eg5l3le7w8vqt9naxty6h6qiylu358n


Condition for Points in Complex Plane to form Isosceles Triangle 0 75372 395335 2019-03-11T07:44:51Z Prime.mover 59 Created page with "== Theorem == <onlyinclude> Let $A = z_1 = x_1 + i y_1$, $B = z_2 = x_2 + i y_2$ and $C = z_3 = x_3 + i y_3$ represent on the complex plane the ..." wikitext text/x-wiki == Theorem == <onlyinclude> Let $A = z_1 = x_1 + i y_1$, $B = z_2 = x_2 + i y_2$ and $C = z_3 = x_3 + i y_3$ represent on the complex plane the vertices of a triangle. Then $\triangle ABC$ is isosceles, where $A$ is the apex, {{iff}}: :${x_2}^2 + {y_2}^2 - 2 \paren {x_1 x_2 + y_1 y_2} = {x_3}^2 + {y_3}^2 - 2 \paren {x_1 x_3 + y_1 y_3}$ </onlyinclude> == Proof == By definition of isosceles triangle: :$\triangle ABC$ is isosceles, where $A$ is the apex, {{iff}} $AB = AC$. Hence: {{begin-eqn}} {{eqn | l = \cmod {z_1 - z_2}       | r = \cmod {z_1 - z_3}       | c =  }} {{eqn | ll= \leadstoandfrom       | l = \cmod {x_1 + i y_1 - x_2 + i y_2}^2       | r = \cmod {x_1 + i y_1 - x_3 + i y_3}^2       | c =  }} {{eqn | ll= \leadstoandfrom       | l = \paren {x_1 - x_2}^2 + \paren {y_1 - y_2}^2       | r = \paren {x_1 - x_3}^2 + \paren {y_1 - y_3}^2       | c = {{Defof|Complex Modulus}} }} {{eqn | ll= \leadstoandfrom       | l = {x_1}^2 - 2 x_1 x_2 + {x_2}^2 + {y_1}^2 - 2 y_1 y_2 + {y_2}^2       | r = {x_1}^2 - 2 x_1 x_3 + {x_3}^2 + {y_1}^2 - 2 y_1 y_3 + {y_3}^2       | c =  }} {{eqn | ll= \leadstoandfrom       | l = {x_2}^2 + {y_2}^2 - 2 \paren {x_1 x_2 + y_1 y_2}       | r = {x_3}^2 + {y_3}^2 - 2 \paren {x_1 x_3 + y_1 y_3}       | c =  }} {{end-eqn}} {{qed}} == Examples == {{:Condition for Points in Complex Plane to form Isosceles Triangle/Examples}} Category:Isosceles Triangles Category:Geometry of Complex Plane 0lnphteume90914gxuq2b9xzokin3qx


Equation of Hyperbola in Complex Plane

Equation of Hyperbola in Complex Plane 0 75404 395426 2019-03-12T06:57:38Z Prime.mover 59 Created page with "== Theorem == <onlyinclude> Let $\C$ be the complex plane.  Let $H$ be a hyperbola in $\C$ whose Definition:Major Axis..." wikitext text/x-wiki == Theorem == <onlyinclude> Let $\C$ be the complex plane. Let $H$ be a hyperbola in $\C$ whose major axis is $d \in \R_{>0}$ and whose foci are at $\alpha, \beta \in \C$. Then $C$ may be written as: :$\cmod {z - \alpha} - \cmod {z - \beta} = d$ where $\cmod {\, \cdot \,}$ denotes complex modulus. </onlyinclude> == Proof == By definition of complex modulus: :$\cmod {z - \alpha}$ is the distance from $z$ to $\alpha$ :$\cmod {z - \beta}$ is the distance from $z$ to $\beta$. Thus $\cmod {z - \alpha} - \cmod {z - \beta}$ is the difference of the distance from $z$ to $\alpha$ and from $z$ to $\beta$. This is precisely the equidistance property of the hyperbola. From Equidistance of Hyperbola equals Transverse Axis, the constant distance $d$ is equal to the transverse axis of $H$. {{qed}} == Examples == {{:Equation of Hyperbola in Complex Plane/Examples}} Category:Hyperbolas Category:Geometry of Complex Plane Category:Equation of Hyperbola in Complex Plane d6w7jvqz2etr8cp7z27p3nl5n2t38kf


Complex Algebra/Examples/z^4 - 3z^2 + 1 = 0 0 75476 441412 441410 2019-12-29T23:47:35Z Prime.mover 59 wikitext text/x-wiki == Example of Complex Algebra == <onlyinclude> The roots of the equation: :$z^4 - 3z^2 + 1 = 0$ are: :$2 \cos 36 \degrees, 2 \cos 72 \degrees, 2 \cos 216 \degrees, 2 \cos 252 \degrees$ </onlyinclude> == Proof == {{begin-eqn}} {{eqn | l = z^4 - 3z^2 + 1       | r = z^4 - 2z^2 + 1 - z^2       | c = Separating the $z^2$ term }} {{eqn | r = (z^2 - 1)^2 - z^2       | c = Completing the Square }} {{eqn | r = (z^2 - 1 - z)(z^2 - 1 + z)       | c = Difference of Two Squares }} {{end-eqn}} From the Quadratic Formula applied to each of the above quadratic factors we can easily see that the four roots are: :$\dfrac {\pm 1 \pm \sqrt 5} 2$ $360 \degrees = 2 \pi \radians$, so $72 \degrees = 2 \pi / 5 \radians$ From De Moivre's Formula, the roots of $x^5 - 1 = 0$ are: :$(\cos(2 n \pi) + i \sin(2 n \pi))^{1/5}= \cos \dfrac {2 n \pi} 5 + i \sin \dfrac {2 n \pi} 5$ However, the coefficient of $x^4$ is $0$ and therefore, by Viète's Formulas, the sum of the roots of $x^5 - 1 = 0$ are also $0$ which means that the sum of the real parts of the roots are also $0$: {{begin-eqn}} {{eqn | l = 0       | r = \cos 0 + \cos \dfrac {2 \pi} 5 + \cos \dfrac {4 \pi} 5 + \cos \dfrac {6 \pi} 5 + \cos \dfrac {8 \pi} 5       | c }} {{eqn | r = 1 + \cos \dfrac {2 \pi} 5 + \cos \dfrac {4 \pi} 5 + \cos \dfrac {6 \pi} 5 + \cos \dfrac {8 \pi} 5       | c }} {{end-eqn}} Now: {{begin-eqn}} {{eqn | l = \cos(\pi + x)       | r = -\cos(x)       | c = Cosine of Angle plus Straight Angle }} {{eqn | r = \cos(-x)       | c = Cosine Function is Even }} {{eqn | r = -\cos(\pi - x)       | c = Cosine of Angle plus Straight Angle }} {{eqn | r = \cos(\pi - x)       | c = Cosine Function is Even }} {{end-eqn}} Hence: {{begin-eqn}} {{eqn | l = \cos \dfrac {6 \pi} 5       | r = \map {\cos} {\pi + \dfrac \pi 5}       | c = }} {{eqn | r = \map {\cos} {\pi - \dfrac \pi 5}       | c = }} {{eqn | r = \cos \dfrac {4 \pi} 5       | c = }} {{end-eqn}} and {{begin-eqn}} {{eqn | l = \cos \dfrac {8 \pi} 5       | r = \map {\cos} {\pi + \dfrac {3 \pi} 5}       | c = }} {{eqn | r = \map {\cos} {\pi - \dfrac {3 \pi} 5}       | c = }} {{eqn | r = \cos \dfrac {2 \pi} 5       | c = }} {{end-eqn}} We can now simplify the sum of the real parts of the roots of $x^5 - 1 = 0$: {{begin-eqn}} {{eqn | l = 0       | r = 1 + \cos \dfrac {2 \pi} 5 + \cos \dfrac {4 \pi} 5 + \cos \dfrac {6 \pi} 5 + \cos \dfrac {8 \pi} 5       | c = }} {{eqn | r = 1 + 2 \cos \dfrac {2 \pi} 5 + 2 \cos \dfrac {4 \pi} 5       | c = $\cos(\pi + x) = \cos(\pi - x)$ }} {{eqn | r = 1 + 2 \cos \dfrac {2 \pi} 5 + 4 \cos^2 \dfrac {2 \pi} 5 - 2       | c = Double Angle Formula for Cosine, $\cos 2 \theta = 2 \cos^2 \theta - 1$ }} {{eqn | r = 4 \cos^2 \dfrac {2 \pi} 5 + 2 \cos \dfrac {2 \pi} 5 - 1       | c = simplifying }} {{end-eqn}} From the Quadratic Formula we then have two '''potential''' values: :$\cos \dfrac {2 \pi} 5 = \dfrac {-1 \pm \sqrt 5} 4$ $0 < 2 \pi / 5 < \pi / 2$, so we know that $\cos \dfrac {2 \pi} 5 > 0$, hence: :$2 \cos 72 \degrees = 2 \cos \dfrac {2 \pi} 5 = \dfrac {-1 + \sqrt 5} 2$ From Cosine of Angle plus Straight Angle, $\map \cos {x + 180 \degrees} = -\cos x$, hence: :$2 \cos 252 \degrees = \dfrac {1 - \sqrt 5} 2$ Now: {{begin-eqn}} {{eqn | l = 2 \cos 36 \degrees       | r = 2 \cos \dfrac \pi 5       | c = }} {{eqn | r = 2 \map {\cos} {\dfrac {- \pi} 5}       | c = Cosine Function is Even }} {{eqn | r = - 2 \map {\cos} {\pi - \dfrac \pi 5}       | c = Cosine of Angle plus Straight Angle }} {{eqn | r = - 2 \map {\cos} {\dfrac {4 \pi} 5}       | c = simplifying }} {{eqn | r = 1 + 2 \cos \dfrac {2 \pi} 5       | c = Sum of the real parts, $1 + 2 \cos \dfrac {2 \pi} 5 + 2 \cos \dfrac {4 \pi} 5 = 0$ }} {{eqn | r = 1 + \dfrac {-1 + \sqrt 5} 2       | c =  }} {{eqn | r = \dfrac {1 + \sqrt 5} 2       | c =  }} {{end-eqn}} From Cosine of Angle plus Straight Angle, $\map \cos {x + 180 \degrees} = -\cos x$, hence: :$2 \cos 216 \degrees = \dfrac {- 1 - \sqrt 5} 2$ We have therefore shown that the four roots of $z^4 - 3z^2 + 1 = 0$ are $\dfrac {\pm 1 \pm \sqrt 5} 2$ and that these four values are also equal to $2 \cos 36 \degrees, 2 \cos 72 \degrees, 2 \cos 216 \degrees, 2 \cos 252 \degrees$ {{qed}} == Sources == * {{BookReference|Theory and Problems of Complex Variables|1981|Murray R. Spiegel|ed = SI|edpage = SI (Metric) Edition|prev = Triple Angle Formula for Cosine|next = Cosine of 36 Degrees/Proof 2}}: $1$: Complex Numbers: Supplementary Problems: De Moivre's Theorem: $91$ Category:Examples of Complex Algebra khrbdqo2y3dxrz51j3mg5370ewlxrmp


Complex Dot Product/Examples/2+5i dot 3-i 0 75561 395995 395983 2019-03-17T12:19:53Z Prime.mover 59 wikitext text/x-wiki == Example of Complex Dot Product == <onlyinclude> Let: :$z_1 = 2 + 5 i$ :$z_2 = 3 - i$ Then: :$z_1 \circ z_2 = 1$ where $\circ$ denotes (complex) dot product. </onlyinclude> == Proof == {{begin-eqn}} {{eqn | l = z_1 \circ z_2       | r = \paren {2 + 5 i} \circ \paren {3 - i}       | c =  }} {{eqn | r = 2 \times 3 + 5 \times \paren {-1}       | c = {{Defof|Dot Product|subdef = Complex|index = 1}} }} {{eqn | r = 6 - 5       | c =  }} {{eqn | r = 1       | c =  }} {{end-eqn}} {{qed}} == Sources == * {{BookReference|Theory and Problems of Complex Variables|1981|Murray R. Spiegel|ed = SI|edpage = SI (Metric) Edition|prev = Complex Algebra/Examples/(1+z)^5 = (1-z)^5|next = Complex Cross Product/Examples/2+5i cross 3-i}}: $1$: Complex Numbers: Supplementary Problems: The Dot and Cross Product: $110 \ \text {(a)}$ Category:Examples of Complex Dot Product 0r746nc4d54hjyo2e5id9idadr1g0o9


Complex Cross Product/Examples/2+5i cross 3-i 0 75562 395984 2019-03-17T11:32:37Z Prime.mover 59 Created page with "== Example of Complex Cross Product == <onlyinclude> Let: :$z_1 = 2 + 5 i$ :$z_2 = 3 - i$ Then: :$z_1 \times z_2 = -17$ where $\times$ den..." wikitext text/x-wiki == Example of Complex Cross Product == <onlyinclude> Let: :$z_1 = 2 + 5 i$ :$z_2 = 3 - i$ Then: :$z_1 \times z_2 = -17$ where $\times$ denotes (complex) cross product. </onlyinclude> == Proof == {{begin-eqn}} {{eqn | l = z_1 \circ z_2       | r = \paren {2 + 5 i} \times \paren {3 - i}       | c =  }} {{eqn | r = 2 \times \paren {-1} - 3 \times 5       | c = {{Defof|Vector Cross Product|subdef = Complex|index = 1|Complex Cross Product}} }} {{eqn | r = -2 - 15       | c =  }} {{eqn | r = -17       | c =  }} {{end-eqn}} {{qed}} == Sources == * {{BookReference|Theory and Problems of Complex Variables|1981|Murray R. Spiegel|ed = SI|edpage = SI (Metric) Edition|prev = Complex Dot Product/Examples/2+5i dot 3-i|next = Complex Dot Product/Examples/3-i dot 2+5i}}: $1$: Complex Numbers: Supplementary Problems: The Dot and Cross Product: $110 \ \text {(b)}$ Category:Examples of Complex Cross Product 1so6w2l42oop38p1lok9kwclfkrmzy5


Complex Dot Product/Examples/3-i dot 2+5i 0 75563 395985 2019-03-17T11:33:50Z Prime.mover 59 Created page with "== Examples of Complex Dot Product == <onlyinclude> Let: :$z_1 = 3 - i$ :$z_1 = 2 + 5 i$ Then: :$z_1 \circ z_2 = 1$ where $\circ$ denotes ..." wikitext text/x-wiki == Examples of Complex Dot Product == <onlyinclude> Let: :$z_1 = 3 - i$ :$z_1 = 2 + 5 i$ Then: :$z_1 \circ z_2 = 1$ where $\circ$ denotes (complex) dot product. </onlyinclude> == Proof == {{begin-eqn}} {{eqn | l = z_1 \circ z_2       | r = \paren {3 - i} \circ \paren {2 + 5 i}       | c =  }} {{eqn | r = 3 \times 2 + \paren {-1} \times 5       | c = {{Defof|Dot Product|subdef = Complex|index = 1}} }} {{eqn | r = 6 - 5       | c =  }} {{eqn | r = 1       | c =  }} {{end-eqn}} {{qed}} == Sources == * {{BookReference|Theory and Problems of Complex Variables|1981|Murray R. Spiegel|ed = SI|edpage = SI (Metric) Edition|prev = Complex Cross Product/Examples/2+5i cross 3-i|next = Complex Cross Product/Examples/3-i cross 2+5i}}: $1$: Complex Numbers: Supplementary Problems: The Dot and Cross Product: $110 \ \text {(c)}$ Category:Examples of Complex Dot Product jnyt7mr50lf1umwf9f3uzoywtl1p3jo


Complex Cross Product/Examples/3-i cross 2+5i 0 75568 395997 395996 2019-03-17T12:24:37Z Prime.mover 59 wikitext text/x-wiki == Example of Complex Cross Product == <onlyinclude> Let: :$z_1 = 3 - i$ :$z_2 = 2 + 5 i$ Then: :$z_1 \times z_2 = 17$ where $\times$ denotes (complex) cross product. </onlyinclude> == Proof == {{begin-eqn}} {{eqn | l = z_1 \circ z_2       | r = \paren {3 - i} \times \paren {2 + 5 i}       | c =  }} {{eqn | r = 3 \times \paren 5 - \paren {-1} \times 2       | c = {{Defof|Vector Cross Product|subdef = Complex|index = 1|Complex Cross Product}} }} {{eqn | r = 15 - \paren {-2}       | c =  }} {{eqn | r = 17       | c =  }} {{end-eqn}} {{qed}} == Sources == * {{BookReference|Theory and Problems of Complex Variables|1981|Murray R. Spiegel|ed = SI|edpage = SI (Metric) Edition|prev = Complex Dot Product/Examples/3-i dot 2+5i|next = Complex Dot Product/Examples/Size of 2+5i dot 3-i}}: $1$: Complex Numbers: Supplementary Problems: The Dot and Cross Product: $110 \ \text {(d)}$ Category:Examples of Complex Cross Product 3diaq2k5gsoeu2jimbxz9j8ddhrl80x


Property of Group Automorphism which Fixes Identity Only/Corollary 2 0 75766 397082 397081 2019-03-24T13:41:56Z Prime.mover 59 wikitext text/x-wiki == Corollary to Property of Group Automorphism which Fixes Identity Only == Let $G$ be a finite group whose identity is $e$. Let $\phi: G \to G$ be a group automorphism. Let $\phi$ have the property that: :$\forall g \in G \setminus \set e: \map \phi t \ne t$ That is, the only fixed element of $\phi$ is $e$. <onlyinclude> Let: :$\phi^2 = I_G$ where $I_G$ denotes the identity mapping on $G$. Then: :$\forall g \in G: \map \phi g = g^{-1}$ </onlyinclude> == Proof == Let $g \in G$. Then: {{begin-eqn}} {{eqn | lo= \exists x \in G:       | l = \map \phi g       | r = \map \phi {x^{-1} \, \map \phi x}       | c = Corollary 1 }} {{eqn | r = \paren {\map \phi x}^{-1} x       | c =  }} {{eqn | r = g^{-1}       | c =  }} {{end-eqn}} {{qed}} == Sources == * {{BookReference|An Introduction to Abstract Algebra|1978|Thomas A. Whitelaw|prev = Property of Group Automorphism which Fixes Identity Only/Corollary 1|next = Property of Group Automorphism which Fixes Identity Only/Corollary 3}}: Chapter $8$: Homomorphisms, Normal Subgroups and Quotient Groups: Exercise $26$ Category:Property of Group Automorphism which Fixes Identity Only 5gql12bgeelztra6b58ljh7glyqgfui


Ideal of Ring/Examples/Order 2 Matrices with some Zero Entries 0 75836 398716 398710 2019-04-01T06:57:59Z Prime.mover 59 wikitext text/x-wiki == Example of Ideal of Ring == <onlyinclude> Let $R$ be the set of all order $2$ square matrices of the form $\begin{pmatrix} x & y \\ 0 & z \end{pmatrix}$ with $x, y, z \in \R$. Let $S$ be the set of all order $2$ square matrices of the form $\begin{pmatrix} x & y \\ 0 & 0 \end{pmatrix}$ with $x, y \in \R$. Then $R$ is a ring and $S$ is an ideal of $R$. </onlyinclude> === Corollary === {{:Ideal of Ring/Examples/Order 2 Matrices with some Zero Entries/Corollary}} == Proof 1 == {{:Ideal of Ring/Examples/Order 2 Matrices with some Zero Entries/Proof 1}} == Proof 2 == {{:Ideal of Ring/Examples/Order 2 Matrices with some Zero Entries/Proof 2}} Category:Examples of Ideals of Rings 2k524j7dyy588n7efhx04zowmioa2vr


Cyclotomic Ring/Examples/5th

Cyclotomic Ring/Examples/5th 0 76116 399083 399078 2019-04-05T05:01:24Z Prime.mover 59 wikitext text/x-wiki == Examples of Cyclotomic Rings == <onlyinclude> The '''$5$th cyclotomic ring''' is the algebraic structure: :$\struct {\Z \sqbrk {i \sqrt 5}, +, \times}$ where $\Z \sqbrk {i \sqrt 5}$ is the set $\set {a + i b \sqrt 5: a, b \in \Z}$. $\struct {\Z \sqbrk {i \sqrt 5}, +, \times}$ is a ring. </onlyinclude> == Proof == We have that $\Z \sqbrk {i \sqrt 5}$ is a subset of the Field of Complex Numbers $\struct {\C, +, \times}$. So to prove that $\struct {\Z \sqbrk {i \sqrt 5}, +, \times}$ is a ring it is sufficient to demonstrate that $\struct {\Z \sqbrk {i \sqrt 5}, +, \times}$ fulfils the conditions of the Subring Test. First we note that setting $a = 1, b = 0$ we have that $1 + 0 i \in \Z \sqbrk {i \sqrt 5}$ and so $\Z \sqbrk {i \sqrt 5} \ne \O$. Let $z_1 = a_1 + i b_1 \sqrt 5$ and $z_2 = a_2 + i b_2 \sqrt 5$ be arbitrary elements of $\Z \sqbrk {i \sqrt 5}$ Then: {{begin-eqn}} {{eqn | l = z_1 - z_2       | r = \paren {a_1 + i b_1 \sqrt 5} - \paren {a_2 + i b_2 \sqrt 5}       | c =  }} {{eqn | r = \paren {a_1 - a_2} + i \paren {b_1 - b_2} \sqrt 5       | c = {{Defof|Complex Addition}} }} {{eqn | o = \in       | r = \Z \sqbrk {i \sqrt 5}       | c = as $a_1 - a_2$ and $b_1 - b_2$ are both integers }} {{end-eqn}} and: {{begin-eqn}} {{eqn | l = z_1 z_2       | r = \paren {a_1 + i b_1 \sqrt 5} \paren {a_2 + i b_2 \sqrt 5}       | c =  }} {{eqn | r = \paren {a_1 a_2 - 5 b_1 b_2} + i \sqrt 5 \paren {a_1 b_2 + a_2 b_1}       | c = {{Defof|Complex Multiplication}} }} {{eqn | o = \in       | r = \Z \sqbrk {i \sqrt 5}       | c = as $a_1 a_2 - 5 b_1 b_2$ and $a_1 b_2 + a_2 b_1$ are both integers }} {{end-eqn}} The Subring Test is satisfied, and so $\struct {\Z \sqbrk {i \sqrt 5}, +, \times}$ is a ring. {{qed}} == Sources == * {{BookReference|An Introduction to Abstract Algebra|1978|Thomas A. Whitelaw|prev = Field Norm of Complex Number is Multiplicative Function|next = Field Norm on 5th Cyclotomic Ring}}: Chapter $9$: Rings: Exercise $19$ Category:Cyclotomic Rings png51kxp7hzzwt12dzlxh9d5ya3oj6z


Matrix Entrywise Addition/Examples/Real 2 x 2 0 76177 399417 2019-04-07T13:43:52Z Prime.mover 59 Created page with "== Example of Matrix Entrywise Addition == <onlyinclude> Let $\mathbf A = \begin {pmatrix} p & q \\ r & s \end {pmatrix}$ and $\mathbf..." wikitext text/x-wiki == Example of Matrix Entrywise Addition == <onlyinclude> Let $\mathbf A = \begin {pmatrix} p & q \\ r & s \end {pmatrix}$ and $\mathbf B = \begin {pmatrix} w & x \\ y & z \end {pmatrix}$ be order $2$ square matrices over the real numbers. Then the matrix sum of $\mathbf A$ and $\mathbf B$ is given by: :$\mathbf A + \mathbf B = \begin {pmatrix} p + w & q + x \\ r + y & s + z \end {pmatrix}$ </onlyinclude> == Sources == * {{BookReference|Introduction to Abstract Algebra|1969|C.R.J. Clapham|prev = Square Matrix/Examples/Real 2 x 2|next = Matrix Product (Conventional)/Examples/2 x 2 Real Matrices}}: Chapter $1$: Integral Domains: $\S 3$. Definition of an Integral Domain: Example $3$ Category:Examples of Matrix Entrywise Addition 4chl61eqf35mue4u9kw7nimhso3gh3z


Matrix Product (Conventional)/Examples/2 x 2 Real Matrices 0 76178 399456 399418 2019-04-07T22:37:11Z Prime.mover 59 wikitext text/x-wiki == Example of (Conventional) Matrix Product == <onlyinclude> Let $\mathbf A = \begin {pmatrix} p & q \\ r & s \end {pmatrix}$ and $\mathbf B = \begin {pmatrix} w & x \\ y & z \end {pmatrix}$ be order $2$ square matrices over the real numbers. Then the matrix product of $\mathbf A$ with $\mathbf B$ is given by: :$\mathbf A \mathbf B = \begin {pmatrix} p w + q y & p x + q z \\ r w + s y & r x + s z \end {pmatrix}$ </onlyinclude> == Sources == * {{BookReference|Introduction to Abstract Algebra|1969|C.R.J. Clapham|prev = Matrix Entrywise Addition/Examples/Real 2 x 2|next = Ring of Square Matrices over Real Numbers/Examples/2 x 2}}: Chapter $1$: Integral Domains: $\S 3$. Definition of an Integral Domain: Example $3$ Category:Examples of Matrix Product 8z73wwbi4jtocrp4s7psqg6r3kraskx


Irreducible Polynomial/Examples/8 x^3 - 6 x - 1 in Rationals 0 76645 402320 402308 2019-04-23T15:08:49Z Prime.mover 59 wikitext text/x-wiki == Examples of Irreducible Polynomials == <onlyinclude> Consider the polynomial: :$\map P x = 8 x^3 - 6 x - 1$ over the ring of polynomials $\Q \sqbrk X$ over the rational numbers. Then $\map P x$ is irreducible. </onlyinclude> == Proof == {{AimForCont}} $\map P x$ has proper factors. Then one of these has to be of degree $1$. Thus from Factors of Polynomial with Integer Coefficients have Integer Coefficients we have: :$8 x^3 - 6 x - 1 = \paren {a x + b} \paren {c^2 + d x + e}$ for some $a, b, c, d, e \in \Z$. Hence: {{begin-eqn}} {{eqn | l = a c       | r = 8       | c =  }} {{eqn | ll= \leadsto       | l = a       | o = \divides       | r = 8       | c =  }} {{eqn | l = b e       | r = -1       | c =  }} {{eqn | ll= \leadsto       | l = b       | o = \divides       | r = 1       | c =  }} {{end-eqn}} The only possible degree $1$ factors with integer coefficients are: :$x \pm 1, 2 x \pm 1, 4 x \pm 1, 8 x \pm 1$ By trying each of these possibilities, it is determined that no integer value of $d$ gives the correct values. Hence the result. {{qed}} == Sources == * {{BookReference|Introduction to Abstract Algebra|1969|C.R.J. Clapham|prev = Polynomial which is Irreducible over Integers is Irreducible over Rationals|next = Square Root of Prime is Irrational/Proof 2}}: Chapter $6$: Polynomials and Euclidean Rings: $\S 31$. Polynomials with Integer Coefficients: Example $59$ Category:Examples of Irreducible Polynomials iid2qj8cu0183a2m7m4cpq3hogfbkko


Vector Space over Division Subring/Examples/Real Numbers in Complex Numbers 0 76773 403683 403680 2019-05-01T11:44:59Z Prime.mover 59 wikitext text/x-wiki == Example of Vector Space over Division Subring == <onlyinclude> Consider the field of complex numbers $\struct {\C, +, \times}$, which is a ring with unity whose unity is $1$. Consider the field of real numbers $\struct {\R, +, \times}$, which is a division subring of $\struct {\C, +, \circ}$ such that $1 \in \R$. Then $\struct {\C, +, \times_\R}_\R$ is an $\R$-vector space, where $\times_\R$ is the restriction of $\times$ to $\R \times \C$. $\struct {\C, +, \times_\R}_\R$ is of dimension $2$. The set $\paren {1 + 0 i, 0 + i}$ forms a basis of $\struct {\C, +, \times_\R}_\R$, as do any two complex numbers which are not real multiples of each other. </onlyinclude> == Sources == * {{BookReference|Introduction to Abstract Algebra|1969|C.R.J. Clapham|prev = Vector Space over Division Subring is Vector Space|next = Definition:Generator of Vector Space}}: Chapter $7$: Vector Spaces: $\S 32$. Definition of a Vector Space: Example $64$ * {{BookReference|Introduction to Abstract Algebra|1969|C.R.J. Clapham|prev = Dimension of Vector Space of Polynomial Functions|next = Linearly Independent Set is Contained in some Basis/Proof 2}}: Chapter $7$: Vector Spaces: $\S 34$. Dimension: Example $68$ Category:Examples of Vector Spaces hhi02na2gim992gybj174abzqorm2nq


Smallest Field containing Subfield and Complex Number/Examples/Numbers of Type Rational a plus b root 2 0 76845 458060 458058 2020-03-28T20:26:37Z Prime.mover 59 wikitext text/x-wiki == Example of Smallest Field containing Subfield and Complex Number == <onlyinclude> Let $\Q \sqbrk {\sqrt 2}$ denote the set: :$\Q \sqbrk {\sqrt 2} := \set {a + b \sqrt 2: a, b \in \Q}$ that is, all numbers of the form $a + b \sqrt 2$ where $a$ and $b$ are rational numbers. Then $\Q \sqbrk {\sqrt 2}$ is the smallest field containing $\Q$ and $\sqrt 2$. </onlyinclude> {{refactor|The following material needs to go on a page explaining splitting fields.}} Formally, $\Q \sqbrk {\sqrt 2}$ is the field extension of $\Q$ for the minimal polynomial of $\sqrt 2$, the second-degree polynomial $x^2 - 2$. Therefore, $\Q \sqbrk {\sqrt 2}$ is the vector space of dimension $2$ isomorphic to $\Q \sqbrk x / \left\langle x^2 - 2 \right\rangle$. == Sources == * {{BookReference|Introduction to Abstract Algebra|1969|C.R.J. Clapham|prev = Definition:Smallest Field containing Subfield and Complex Number/General Definition|next = Smallest Field containing Subfield and Complex Number/Examples/Complex Numbers}}: Chapter $8$: Field Extensions: $\S 36$. The Degree of a Field Extension: Example $72$ * {{BookReference|Contemporary Abstract Algebra|2017|Joseph A. Gallian|ed = 9th|edpage = Ninth Edition|prev = Definition:Vector Space|next = Definition:Algebraic Field Extension}}: Chapter $20$: Extension Fields: $\S 1$. Splitting Fields Category:Examples of Field Extensions ftu2qn8d0qoqq44yx3cl24kh3zvomzr


Algebraic Number/Examples/Root of (2 plus Root 3) 0 76862 403850 2019-05-02T17:55:58Z Prime.mover 59 Created page with "== Example of Algebraic Number == <onlyinclude> :$\sqrt {2 + \sqrt 3}$ is an algebraic number. </onlyinclude>..." wikitext text/x-wiki == Example of Algebraic Number == <onlyinclude> :$\sqrt {2 + \sqrt 3}$ is an algebraic number. </onlyinclude> == Proof == Let $x = \sqrt {2 + \sqrt 3}$. We have that: {{begin-eqn}} {{eqn | l = x^2       | r = 2 + \sqrt 3       | c =  }} {{eqn | ll= \leadsto       | l = \paren {x^2 - 2}^2       | r = 3       | c =  }} {{eqn | ll= \leadsto       | l = x^4 - 4 x^2 + 4       | r = 3       | c =  }} {{eqn | ll= \leadsto       | l = x^4 - 4 x^2 + 1       | r = 0       | c =  }} {{end-eqn}} Thus $\sqrt {2 + \sqrt 3}$ is a root of $x^4 - 4 x^2 + 1 = 0$. Hence the result by definition of algebraic number. {{qed}} </onlyinclude> == Sources == * {{BookReference|Introduction to Abstract Algebra|1969|C.R.J. Clapham|prev = Algebraic Number/Examples/Imaginary Unit|next = Pi is Transcendental}}: Chapter $8$: Field Extensions: $\S 38$. Simple Algebraic Extensions: Example $76$ Category:Examples of Algebraic Numbers qzt6n7kxbx8nk1c3ui3cbzc2ln82ngy


Group of Order 30 is not Simple

Group of Order 30 is not Simple 0 76938 404179 2019-05-04T14:22:49Z Prime.mover 59 Created page with "== Theorem == <onlyinclude> Let $G$ be a group of order $30$.  Then $G$ is not simple. </onlyinc..." wikitext text/x-wiki == Theorem == <onlyinclude> Let $G$ be a group of order $30$. Then $G$ is not simple. </onlyinclude> == Proof == From Group of Order 30 has Normal Cyclic Subgroup of Order 15, $G$ has a normal subgroup of order $15$. Hence the result, by definition of simple group. {{qed}} == Sources == * {{BookReference|Elements of Abstract Algebra|1971|Allan Clark|prev = Prime Group is Simple|next = Group of Order p^2 q is not Simple}}: Chapter $2$: The Sylow Theorems: $\S 59 \epsilon$ Category:Groups of Order 30 Category:Simple Groups qegxmxy6exuousudjz43olsur3diedn


Bessel Function of the First Kind/Instances/Order 0 0 77298 405970 405969 2019-05-30T08:42:23Z Prime.mover 59 wikitext text/x-wiki == Specific Instance of Bessel Functions of the First Kind == Let $\map {J_n} x$ denote the Bessel function of the first kind of order $n$. <onlyinclude> {{begin-eqn}} {{eqn | l = \map {J_0} x       | r = \sum_{k \mathop = 0}^\infty \dfrac {\paren {-1}^k} {\paren {k!}^2} \paren {\dfrac x 2}^{2 k}       | c =  }} {{eqn | r = 1 - \dfrac {x^2} {2^2} + \dfrac {x^4} {2^2 \times 4^2} - \dfrac {x^6} {2^2 \times 4^2 \times 6^2} + \dotsb       | c =  }} {{end-eqn}} </onlyinclude> == Proof == From Series Expansion of Bessel Function of the First Kind: {{begin-eqn}} {{eqn | l = \map {J_n} x       | r = \dfrac {x^n} {2^n \, \map \Gamma {n + 1} } \paren {1 - \dfrac {x^2} {2 \paren {2 n + 2} } + \dfrac {x^4} {2 \times 4 \paren {2 n + 2} \paren {2 n + 4} } - \cdots}       | c =  }} {{eqn | r = \sum_{k \mathop = 0}^\infty \dfrac {\paren {-1}^k} {k! \, \map \Gamma {n + k + 1} } \paren {\dfrac x 2}^{n + 2 k}       | c =  }} {{end-eqn}} where $n$ is not a (strictly) negative integer. $0$ fits that category, and so: {{begin-eqn}} {{eqn | l = \map {J_0} x       | r = \sum_{k \mathop = 0}^\infty \dfrac {\paren {-1}^k} {k! \, \map \Gamma {0 + k + 1} } \paren {\dfrac x 2}^{0 + 2 k}       | c =  }} {{eqn | r = \sum_{k \mathop = 0}^\infty \dfrac {\paren {-1}^k} {k! \, \map \Gamma {k + 1} } \paren {\dfrac x 2}^{2 k}       | c =  }} {{eqn | r = \sum_{k \mathop = 0}^\infty \dfrac {\paren {-1}^k} {\paren {k!}^2} \paren {\dfrac x 2}^{2 k}       | c = Gamma Function Extends Factorial }} {{end-eqn}} Or working directly upon the terms themselves: {{begin-eqn}} {{eqn | l = \map {J_0} x       | r = \dfrac {x^0} {2^0 \, \map \Gamma {0 + 1} } \paren {1 - \dfrac {x^2} {2 \paren {2 \times 0 + 2} } + \dfrac {x^4} {2 \times 4 \paren {2 \times 0 + 2} \paren {2 \times 0 + 4} } - \cdots}       | c =  }} {{eqn | r = \dfrac 1 {1 \times \map \Gamma 1} \paren {1 - \dfrac {x^2} {2 \times 2} + \dfrac {x^4} {2 \times 4 \times 2 \times 4} - \cdots}       | c =  }} {{eqn | r = 1 - \dfrac {x^2} {2^2} + \dfrac {x^4} {2^2 \times 4^2} - \cdots       | c =  }} {{end-eqn}} {{qed}} == Sources == * {{BookReference|Mathematical Handbook of Formulas and Tables|1968|Murray R. Spiegel|prev = Bessel Function of the First Kind for Negative Non-Integer Order is Unbounded|next = Bessel Function of the First Kind/Instances/Order 1}}: $\S 24$: Bessel Functions: Bessel Function of the First Kind of Order $n$: $24.5$ Category:Examples of Bessel Functions 5wp5vajmwmg214g25lzd0sog675f43v


Laplace Transform of Dirac Delta Function/Lemma 0 77407 406282 2019-06-03T06:43:29Z Prime.mover 59 Created page with "== Lemma for Laplace Transform of Dirac Delta Function == <onlyinclude> Let $F_\epsilon: \R \to \R$ be the real function defined as:  :$\map {..." wikitext text/x-wiki == Lemma for Laplace Transform of Dirac Delta Function == <onlyinclude> Let $F_\epsilon: \R \to \R$ be the real function defined as: :$\map {F_\epsilon} t = \begin{cases} 0 & : x < 0 \\ \dfrac 1 \epsilon & : 0 \le t \le \epsilon \\ 0 & : t > \epsilon \end{cases}$ Then: :$\laptrans {\map {F_\epsilon} t} = \dfrac {1 - e^{-s \epsilon} } {\epsilon s}$ </onlyinclude> == Proof == {{begin-eqn}} {{eqn | l = \laptrans {\map {F_\epsilon} t}       | r = \int_0^\infty e^{-s t} \map {F_\epsilon} t \rd t       | c =  }} {{eqn | r = \int_0^\epsilon e^{-s t} \map {F_\epsilon} t \rd t + \int_\epsilon^\infty e^{-s t} \map {F_\epsilon} t \rd t       | c = Sum of Integrals on Adjacent Intervals for Integrable Functions }} {{eqn | r = \int_0^\epsilon e^{-s t} \dfrac 1 \epsilon \rd t + \int_\epsilon^\infty  e^{-s t} \times 0 \rd t       | c = Definition of $F_\epsilon$ }} {{eqn | r = \dfrac 1 \epsilon \int_0^\epsilon e^{-s t} \rd t       | c = simplification }} {{eqn | r = \dfrac 1 \epsilon \intlimits {\dfrac {e^{-s t} } {-s} } 0 \epsilon       | c = Primitive of $e^{a x}$ }} {{eqn | r = \dfrac 1 \epsilon \paren {\dfrac {e^{-s \epsilon} - e^{-s \times 0} } {-s} }       | c =  }} {{eqn | r = \dfrac {1 - e^{-s \epsilon} } {\epsilon s}       | c = simplification }} {{end-eqn}} {{qed}} == Sources == * {{BookReference|Theory and Problems of Laplace Transforms|1965|Murray R. Spiegel|prev = Laplace Transform of Heaviside Step Function/Proof 2|next = Laplace Transform of Dirac Delta Function/Proof 2}}: Chapter $1$: The Laplace Transform: Solved Problems: Impulse Functions. The Dirac Delta Function: $41$ Category:Laplace Transform of Dirac Delta Function dec9evt703r3ni8ww9fpol3yp13c8wl


Infimum of Subset of Real Numbers/Examples/Example 3 0 77566 407912 407276 2019-06-14T14:53:18Z Prime.mover 59 wikitext text/x-wiki == Example of Infimum of Subset of Real Numbers == <onlyinclude> The subset $V$ of the real numbers $\R$ defined as: :$V := \set {x \in \R: x > 0}$ admits an infimum: :$\inf V = 0$ </onlyinclude> == Sources == * {{BookReference|Mathematical Analysis: A Straightforward Approach|1977|K.G. Binmore|prev = Supremum of Subset of Real Numbers/Examples/Example 3|next = Definition:Supremum of Subset of Real Numbers}}: $\S 2$: Continuum Property: $\S 2.5$: Examples: $\text{(iii)}$ Category:Examples of Infima erold9knxsoe1w6d6sh0ju16cv8vx7x


Real Null Sequence/Examples/n^alpha x^n/Lemma 0 77733 408456 408444 2019-06-16T09:09:59Z Prime.mover 59 wikitext text/x-wiki == Lemma for Real Null Sequence: $n^\alpha x^n$ == Let $\alpha \in \Q$ be a (strictly) positive rational number. Let $x \in \R$ be a real number such that $\size x < 1$. <onlyinclude> There exists $N \in \N$ such that: :$\paren {1 + \dfrac 1 N}^{\alpha + 1} \, \size x \le 1$ </onlyinclude> == Proof == {{AimForCont}}: :$\forall n \in \N: \paren {1 + \dfrac 1 n}^{\alpha + 1} \, \size x > 1$ Then: {{begin-eqn}} {{eqn | lo= \forall n \in \N:       | l = \dfrac 1 n       | o = >       | r = \paren {\dfrac 1 {\size x} }^{1 / \paren {\alpha + 1} } - 1       | c =  }} {{eqn | o = >       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = \paren {\dfrac 1 {\size x} }^{1 / \paren {\alpha + 1} }       | o = >       | r = 1       | c = as $\dfrac 1 n > 0$ for all $n \in \N$ }} {{end-eqn}} But this contradicts Sequence of Powers of Reciprocals is Null Sequence. Hence by Proof by Contradiction: :$\exists N \in \N: \paren {1 + \dfrac 1 N}^{\alpha + 1} \, \size x \le 1$ {{qed}} == Sources == * {{BookReference|Mathematical Analysis: A Straightforward Approach|1977|K.G. Binmore|prev = Power over Factorial|next = Real Null Sequence/Examples/n^alpha x^n}}: $\S 4$: Convergent Sequences: Exercise $\S 4.20 \ (5)$ Category:Real Null Sequence/Examples/n^alpha x^n 9krbowsncd6pnacbratgp3py4jh6cx1


Hero's Method/Lemma 1

Hero's Method/Lemma 1 0 77808 419809 419795 2019-08-23T07:17:28Z Prime.mover 59 wikitext text/x-wiki == Lemma for Hero's Method == Let $a \in \R$ be a real number such that $a > 0$. Let $x_1 \in \R$ be a real number such that $x_1 > 0$. Let $\sequence {x_n}$ be the sequence in $\R$ defined recursively by: :$\forall n \in \N_{>0}: x_{n + 1} = \dfrac {x_n + \dfrac a {x_n} } 2$ Then: <onlyinclude> :$\forall n \in \N_{>0}: x_n > 0$ </onlyinclude> == Proof == The proof proceeds by induction. For all $n \in \Z_{>0}$, let $\map P n$ be the proposition: :$x_n > 0$ === Basis for the Induction === $\map P 1$ is the case: :$x_1 > 0$ which is assumed. Thus $\map P 1$ is seen to hold. This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true. So this is the induction hypothesis: :$x_k > 0$ from which it is to be shown that: :$x_{k + 1} > 0$ === Induction Step === This is the induction step: We have that: :$x_{k + 1} = \dfrac {x_k + \dfrac a {x_k} } 2$ But as $x_k > 0$ and $a > 0$, it follows that: :$\dfrac a {x_k} > 0$ Then as $x_k > 0$ and $\dfrac a {x_k} > 0$, it follows that: :$\dfrac 1 2 \paren {x_k + \dfrac a {x_k} } > 0$ So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\forall n \in \N_{>0}: x_n > 0$ {{qed}} == Sources == * {{BookReference|Mathematical Analysis: A Straightforward Approach|1977|K.G. Binmore|prev = Hero's Method|next = Hero's Method/Lemma 2}}: $\S 5$: Subsequences: $\S 5.5$: Example Category:Hero's Method 6oibp2a6jgh5au897um27rh5vca257g


Hero's Method/Lemma 2

Hero's Method/Lemma 2 0 77814 419810 419797 2019-08-23T07:18:13Z Prime.mover 59 wikitext text/x-wiki == Lemma for Hero's Method == Let $a \in \R$ be a real number such that $a > 0$. Let $x_1 \in \R$ be a real number such that $x_1 > 0$. Let $\sequence {x_n}$ be the sequence in $\R$ defined recursively by: :$\forall n \in \N_{>0}: x_{n + 1} = \dfrac {x_n + \dfrac a {x_n} } 2$ Then: <onlyinclude> :$\forall n \ge 2: x_n \ge \sqrt a$ </onlyinclude> == Proof == === Lemma 1 === {{:Hero's Method/Lemma 1}}{{qed|lemma}} We have: {{begin-eqn}} {{eqn | l = x_{n + 1}       | r = \frac {x_n + \dfrac a {x_n} } 2       | c =  }} {{eqn | ll= \leadstoandfrom       | l = 2 x_n x_{n + 1}       | r = x_n^2 + a       | c =  }} {{eqn | ll= \leadstoandfrom       | l = x_n^2 - 2 x_n x_{n + 1} + a       | r = 0       | c =  }} {{end-eqn}} This is a quadratic equation in $x_n$. We know that this equation must have a real solution with respect to $x_n$, because $x_n$ has been explicitly constructed by the iterative process. Thus its discriminant is $b^2 - 4 a c \ge 0$, where: :$a = 1$ :$b = -2 x_{n + 1}$ :$c = a$ Thus $x_{n + 1}^2 \ge a$. From Lemma 1: :$x_{n + 1} > 0$  It follows that: :$\forall n \ge 1: x_{n + 1} \ge \sqrt a$ for $n \ge 1$ Thus: :$\forall n \ge 2: x_n \ge \sqrt a$ for $n \ge 2$ {{qed}} == Sources == {{SourceReview|username = Prime.mover}} * {{BookReference|Mathematical Analysis: A Straightforward Approach|1977|K.G. Binmore|prev = Hero's Method/Lemma 1|next = Hero's Method/Proof 1}}: $\S 5$: Subsequences: $\S 5.5$: Example Category:Hero's Method fnm4qakm9gr0gsrk7pwre9y0zlkta6c


Convergent Real Sequence/Examples/x (n+1) = x n^2 + k/Lemma 1 0 77817 408882 408871 2019-06-18T14:21:24Z Prime.mover 59 wikitext text/x-wiki == Example of Convergent Real Sequence == Let $\sequence {x_n}$ be the real sequence defined as: :$x_n = \begin {cases} h & : n = 1 \\ {x_{n - 1} }^2 + k & : n > 1 \end {cases}$ where: :$0 < k < \dfrac 1 4$ :$a < h < b$, where $a$ and $b$ are the roots of the quadratic equation $x^2 - x + k = 0$. Then: <onlyinclude> :$\forall n \in \N_{>0}: a < x_n < b$ </onlyinclude> == Proof == The proof proceeds by induction. For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition: :$a < x_n < b$ === Basis for the Induction === $\map P 1$ is the case: :$a < x_1 < b$ By assertion: :$a < h < b$ and: :$x_1 = h$ Thus $\map P 1$ is seen to hold. This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true. So this is the induction hypothesis: :$a < x_k < b$ from which it is to be shown that: :$a < x_{k + 1} < b$ === Induction Step === This is the induction step: {{begin-eqn}} {{eqn | l = x_{k + 1} - a       | r = {x_k}^2 + k - a       | c = Definition of $x_{k + 1}$ }} {{eqn | o = >       | r = a^2 - a + k       | c = as $x_k > a$ }} {{eqn | r = 0       | c = Definition of $a$ }} {{eqn | ll= \leadsto       | l = x_{k + 1}       | o = >       | r = a       | c =  }} {{end-eqn}} Similarly: {{begin-eqn}} {{eqn | l = x_{k + 1} - b       | r = {x_k}^2 + k - b       | c = Definition of $x_{k + 1}$ }} {{eqn | o = <       | r = b^2 - b + k       | c = as $x_k < b$ }} {{eqn | r = 0       | c = Definition of $b$ }} {{eqn | ll= \leadsto       | l = x_{k + 1}       | o = <       | r = b       | c =  }} {{end-eqn}} So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\forall n \in \N_{>0}: a < x_n < b$ {{qed}} == Sources == * {{BookReference|Mathematical Analysis: A Straightforward Approach|1977|K.G. Binmore|prev = Convergent Real Sequence/Examples/x (n+1) = x n^2 + k|next = Convergent Real Sequence/Examples/x (n+1) = x n^2 + k/Lemma 2}}: $\S 5$: Subsequences: Exercise $\S 5.7 \ (2)$ Category:Convergent Real Sequence/Examples/x (n+1) = x n^2 + k a8stp465ywvsogvr8u3ug937ibusiqx


Convergent Real Sequence/Examples/x (n+1) = x n^2 + k/Lemma 2 0 77819 408885 408864 2019-06-18T14:29:03Z Prime.mover 59 wikitext text/x-wiki == Example of Convergent Real Sequence == <onlyinclude> Let $a$ and $b$ be the roots of the quadratic equation: :$(1): \quad x^2 - x + k = 0$ Let: :$0 < k < \dfrac 1 4$ Then $a$ and $b$ are both strictly positive real numbers. </onlyinclude> == Proof == First we investigate the consequences of the condition $k < \dfrac 1 4$. By Solution to Quadratic Equation with Real Coefficients: :In order for the quadratic equation $a x^2 + b x + c$ to have real roots, its discriminant $b^2 - 4 a c$ needs to be strictly positive. The discriminant $D$ of $(1)$ is: {{begin-eqn}} {{eqn | l = D       | r = \paren {-1}^2 - 4 \times 1 \times k       | c =  }} {{eqn | r = 1 - 4 k       | c =  }} {{end-eqn}} Thus: {{begin-eqn}} {{eqn | l = k       | o = <       | r = \dfrac 1 4       | c =  }} {{eqn | ll= \leadsto       | l = 1 - 4 k       | o = >       | r = 0       | c =  }} {{end-eqn}} and so when $k < \dfrac 1 4$, $(1)$ has real roots. {{qed|lemma}} Next we investigate the consequences of the condition $0 < k$. By Solution to Quadratic Equation: {{begin-eqn}} {{eqn | l = x       | r = \dfrac {-b \pm \sqrt {b^2 - 4 a c} } {2 a}       | c = where $b = -1$, $a = 1$, $c = k$ in $(1)$ }} {{eqn | r = \dfrac {-\paren {-1} \pm \sqrt {\paren {-1}^2 - 4 \times 1 \times k} } {2 \times 1}       | c =  }} {{eqn | r = \dfrac {1 \pm \sqrt {1 - 4 k} } 2       | c =  }} {{end-eqn}} We have that: {{begin-eqn}} {{eqn | l = 0       | o = <       | r = k       | c =  }} {{eqn | ll= \leadsto       | l = 1       | o = >       | r = 1 - 4 k       | c =  }} {{eqn | ll= \leadsto       | l = 1       | o = >       | r = +\sqrt {1 - 4 k}       | c =  }} {{eqn | l = -1       | o = <       | r = -\sqrt {1 - 4 k}       | c =  }} {{eqn | ll= \leadsto       | l = \dfrac {1 \pm \sqrt {1 - 4 k} } 2       | o = >       | r = 0       | c =  }} {{end-eqn}} That is, when $0 < k$ both roots of $(1)$ are strictly positive. {{qed|lemma}} Hence when $0 < k < \dfrac 1 4$, both roots of $(1)$ are strictly positive real numbers. {{qed}} == Sources == * {{BookReference|Mathematical Analysis: A Straightforward Approach|1977|K.G. Binmore|prev = Convergent Real Sequence/Examples/x (n+1) = x n^2 + k/Lemma 1|next = Convergent Real Sequence/Examples/x (n+1) = k over 1 + x n}}: $\S 5$: Subsequences: Exercise $\S 5.7 \ (2)$ Category:Convergent Real Sequence/Examples/x (n+1) = x n^2 + k 47tbne0b2az5ehu854zf2el1ldm973l


Convergent Real Sequence/Examples/x (n+1) = k over 1 + x n/Lemma 1 0 77823 408895 2019-06-18T15:50:09Z Prime.mover 59 Created page with "== Example of Convergent Real Sequence ==  Let $h, k \in \R_{>0}$.  Let $\sequence {x_n}$ be the Definition:Real Sequence|real sequen..." wikitext text/x-wiki == Example of Convergent Real Sequence == Let $h, k \in \R_{>0}$. Let $\sequence {x_n}$ be the real sequence defined as: :$x_n = \begin {cases} h & : n = 1 \\ \dfrac k {1 + x_{n - 1} } & : n > 1 \end {cases}$ Then: <onlyinclude> :$\forall n \in \N_{>1}: k > x_n > 0$ </onlyinclude> == Proof == The proof proceeds by induction. For all $n \in \Z_{>1}$, let $\map P n$ be the proposition: :$k > x_n > 0$ === Basis for the Induction === $\map P 2$ is the case: :$k > x_2 > 0$ We have: {{begin-eqn}} {{eqn | l = x_2       | r = \dfrac k {1 + x_1}       | c =  }} {{eqn | o = <       | r = \dfrac k 1       | c = as $x_1 > 0$ }} {{eqn | r = k       | c =  }} {{end-eqn}} Also, as $k > 0$ and $x_1 > 0$ we have that: :$\dfrac k {1 + x_1} > 0$ Thus $\map P 2$ is seen to hold. This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that if $\map P r$ is true, where $r \ge 2$, then it logically follows that $\map P {r + 1}$ is true. So this is the induction hypothesis: :$k > x_r > 0$ from which it is to be shown that: :$k > x_{r + 1} > 0$ === Induction Step === This is the induction step: {{begin-eqn}} {{eqn | l = x_{r + 1}       | r = \dfrac k {1 + x_r}       | c =  }} {{eqn | o = <       | r = \dfrac k 1       | c = Induction Hypothesis: $x_r > 0$ }} {{eqn | r = k       | c =  }} {{end-eqn}} Also, as $k > 0$ and $x_r > 0$ we have that: :$\dfrac k {1 + x_r} > 0$ So $\map P r \implies \map P {r + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\forall n \in \N_{>1}: k > x_n > 0$ {{qed}} == Sources == * {{BookReference|Mathematical Analysis: A Straightforward Approach|1977|K.G. Binmore|prev = Convergent Real Sequence/Examples/x (n+1) = k over 1 + x n|next = Convergent Real Sequence/Examples/x (n+1) = k over 1 + x n/Mistake}}: $\S 5$: Subsequences: Exercise $\S 5.7 \ (3)$ Category:Convergent Real Sequence/Examples/x (n+1) = k over 1 + x n bb2tocxnxw2uu7d3q7cfbgoncgs8myz


Convergent Real Sequence/Examples/x (n+1) = k over 1 + x n/Lemma 2 0 77831 408946 408930 2019-06-18T22:26:35Z Prime.mover 59 wikitext text/x-wiki == Example of Convergent Real Sequence == Let $h, k \in \R_{>0}$. Let $\sequence {x_n}$ be the real sequence defined as: :$x_n = \begin {cases} h & : n = 1 \\ \dfrac k {1 + x_{n - 1} } & : n > 1 \end {cases}$ <onlyinclude> Consider the subsequences $\sequence {x_{2 n} }$ and $\sequence {x_{2 n - 1} }$. One of them is strictly increasing and the other is strictly decreasing. </onlyinclude> == Proof == We have that: {{begin-eqn}} {{eqn | l = x_{n + 1} - x_{n - 1}       | r = \dfrac k {1 + x_n} - \dfrac k {1 + x_{n - 2} }       | c =  }} {{eqn | r = \dfrac {k \paren {x_{n - 2} - x_n} } {\paren {1 + x_n} \paren {1 + x_{n - 2} } }       | c =  }} {{end-eqn}} and so $x_{n + 1} - x_{n - 1}$ has the opposite sign to $x_{n - 2} - x_n$. It can be proved by induction that one of the sequences $\sequence {x_{2 n} }$ and $\sequence {x_{2 n - 1} }$ increases and one decreases. {{finish|Provide the workings for the above}} In fact: :$\sequence {x_{2 n - 1} }$ is strictly increasing {{iff}} $x_3 > x_1$ and is strictly decreasing {{iff}} $x_3 < x_1$. {{finish|Prove the above as well.}} == Sources == * {{BookReference|Mathematical Analysis: A Straightforward Approach|1977|K.G. Binmore|prev = Convergent Real Sequence/Examples/x (n+1) = k over 1 + x n/Mistake|next = Convergent Real Sequence/Examples/x n = root x n-1 y n-1, 1 over y n = half (1 over x n + 1 over y n-1)}}: $\S 5$: Subsequences: Exercise $\S 5.7 \ (3)$ Category:Convergent Real Sequence/Examples/x (n+1) = k over 1 + x n 50e8ta82z9l1obkx7f6th0hiv29yzvx


Square of Chi Random Variable has Chi-Squared Distribution 0 77952 409556 2019-06-22T10:16:19Z Caliburn 3218 Created page with "== Theorem ==  <onlyinclude> Let $n$ be a strictly positive integer.  Let $X \sim \chi_n$ where $\chi_n$ is the Definition:Chi Distr..." wikitext text/x-wiki == Theorem ==  <onlyinclude> Let $n$ be a strictly positive integer. Let $X \sim \chi_n$ where $\chi_n$ is the chi distribution with $n$ degrees of freedom.  Then $X^2 \sim \chi^2_n$ where $\chi^2_n$ is the chi-squared distribution with $n$ degrees of freedom. </onlyinclude> == Proof ==  Let $Y \sim \chi^2_n$.  We aim to show that:  :$\map \Pr {Y < x^2} = \map \Pr {X < x}$ for all $x \in \hointr 0 \infty$. We have: {{begin-eqn}} {{eqn	| l = \map \Pr {Y < x^2} 	| r = \int_0^{x^2} \frac 1 {2^{n / 2} \map \Gamma {n / 2} } t^{\paren {n / 2} - 1} e^{- t / 2} \rd t 	| c = {{Defof|Chi-Squared Distribution}} }} {{eqn	| r = \frac 2 {2^{n / 2} \map \Gamma {n / 2} } \int_0^x u \paren {u^2}^{\paren {n / 2} - 1} e^{- u^2 / 2} \rd u 	| c = substituting $t = u^2$ }} {{eqn	| r = \frac 1 {2^{\paren {n / 2} - 1} \map \Gamma {n / 2} } \int_0^x u u^{2 \paren {\paren {n / 2} - 1} } e^{- u^2 / 2} \rd u }} {{eqn	| r = \frac 1 {2^{\paren {n / 2} - 1} \map \Gamma {n / 2} } \int_0^x u^{n - 1} e^{- u^2 / 2} \rd u }} {{eqn	| r = \map \Pr {X < x} 	| c = {{Defof|Chi Distribution}} }} {{end-eqn}} {{qed}} Category:Chi Distribution Category:Chi-Squared Distribution 4plq49xte1ivxn4xpdvm9dx8g2ibaq9


Multiple of Chi-Squared Random Variable has Gamma Distribution 0 77989 409843 2019-06-23T13:22:53Z Caliburn 3218 Created page with "== Theorem ==  <onlyinclude> Let $n$ be a strictly positive integer.   Let $k > 0$ be a real number.   Let..." wikitext text/x-wiki == Theorem ==  <onlyinclude> Let $n$ be a strictly positive integer.  Let $k > 0$ be a real number.  Let $X \sim \chi^2_n$ where $\chi^2_n$ is the chi-squared distribution with $n$ degrees of freedom.  Then:  :$k X \sim \map \Gamma {\dfrac n 2, \dfrac 1 {2 k}}$ where $\map \Gamma {\dfrac n 2, \dfrac 1 {2 k}}$ is the gamma distribution with parameters $\dfrac n 2$ and $\dfrac 1 {2 k}$. </onlyinclude> == Proof == Let: :$Y \sim \map \Gamma {\dfrac n 2, \dfrac 1 {2 k}}$ We aim to show that:  :$\map \Pr {Y < k x} = \map \Pr {X < x}$ for all real $x \ge 0$. We have:  {{begin-eqn}} {{eqn	| l = \map \Pr {Y < k x} 	| r = \frac 1 {\map \Gamma {n / 2} } \paren {\frac 1 {2 k} }^{n / 2} \int_0^{k x} t^{\paren {n / 2} - 1} e^{-\paren {1 / k} t / 2} \rd t 	| c = {{Defof|Gamma Distribution}} }} {{eqn	| r = \frac 1 {\map \Gamma {n / 2} } \paren {\frac 1 {2 k} }^{n / 2} \int_0^x k \paren {k u}^{\paren {n / 2} - 1} e^{-u / 2} \rd u 	| c = substituting $t = k u$ }} {{eqn	| r = \frac {k^{n / 2} } {\map \Gamma {n / 2} \paren {2 k}^{n / 2} } \int_0^x u^{\paren {n / 2} - 1} e^{-u / 2} \rd u }} {{eqn	| r = \frac 1 {2^{n / 2} \map \Gamma {n / 2} } \int_0^x u^{\paren {n / 2} - 1} e^{-u / 2} \rd u  }} {{eqn	| r = \map \Pr {X < x} 	| c = {{Defof|Chi-Squared Distribution}} }} {{end-eqn}} {{qed}} Category:Chi-Squared Distribution Category:Gamma Distribution iemi4sla1j0pqk0tb8sxnp2gkq4hd72


Expectation of Student's t-Distribution

Expectation of Student's t-Distribution 0 78121 410154 2019-06-25T13:45:04Z Caliburn 3218 Created page with "== Theorem == <onlyinclude> Let $k$ be a strictly positive integer.   Let $X \sim t_k$ where $t_k$ is the Definition:Student's t-Dis..." wikitext text/x-wiki == Theorem == <onlyinclude> Let $k$ be a strictly positive integer.  Let $X \sim t_k$ where $t_k$ is the $t$-distribution with $k$ degrees of freedom. Then the expectation of $X$ is equal to $0$ for $k > 1$, and does not exist otherwise. </onlyinclude> == Proof == {{ProofWanted}} Category:Expectation Category:Student's t-Distribution n1kupkw20s6aloggikkrhn7j4h9r356


Raw Moment of Erlang Distribution

Raw Moment of Erlang Distribution 0 78142 410309 2019-06-26T11:05:45Z Caliburn 3218 Created page with "== Theorem == <onlyinclude> Let $n, k$ be strictly positive integers.   Let $\lambda$ be a Definition:Strictly Positive Real Number|..." wikitext text/x-wiki == Theorem == <onlyinclude> Let $n, k$ be strictly positive integers.  Let $\lambda$ be a strictly positive real number.  Let $X$ have a continuous random variable with an Erlang distribution with parameters $k$ and $\lambda$. Then the $n$th raw moment of $X$ is given by: :$\displaystyle \expect {X^n} = \frac 1 {\lambda^n} \prod_{m \mathop = 0}^{n - 1} \paren {k + m}$ </onlyinclude>  == Proof == From the definition of the Erlang distribution, $X$ has probability density function: :$\map {f_X} x = \dfrac {\lambda^k x^{k - 1} e^{- \lambda x} } {\map \Gamma k}$ From the definition of the expected value of a continuous random variable:  :$\displaystyle \expect {X^n} = \int_0^\infty x^n \map {f_X} x \rd x$ So: {{begin-eqn}} {{eqn	| l = \expect {X^n} 	| r = \frac {\lambda^k} {\map \Gamma k} \int_0^\infty x^{n + k - 1} e^{- \lambda x} \rd x }} {{eqn	| r = \frac {\lambda^k} {\lambda \map \Gamma k} \int_0^\infty \paren {\frac u \lambda}^{n + k - 1} e^{- u} \rd u 	| c = substituting $u = \lambda x$ }} {{eqn	| r = \frac {\lambda^k} {\lambda^{n + k} \map \Gamma k} \int_0^\infty u^{n + k - 1} e^{- u} \rd u }} {{eqn	| r = \frac 1 {\lambda^n \map \Gamma k} \map \Gamma {n + k} 	| c = {{Defof|Gamma Function}} }} {{eqn	| r = \frac 1 {\lambda^n} \frac {\map \Gamma k} {\map \Gamma k} \prod_{m \mathop = 0}^{n - 1} \paren {k + m} 	| c = Gamma Difference Equation }} {{eqn	| r = \frac 1 {\lambda^n} \prod_{m \mathop = 0}^{n - 1} \paren {k + m} }} {{end-eqn}} {{qed}} Category:Raw Moments Category:Erlang Distribution jou9tst4u65j3aafpjsdh7jycni2zu7


Linear Transformation of Gaussian Random Variable 0 78298 411601 2019-07-04T13:50:30Z Caliburn 3218 Created page with "== Theorem ==  <onlyinclude> Let $\mu$, $\alpha$ and $\beta$ be real numbers.   Let $\sigma$ be a Definition:Positive Real Number|positive real nu..." wikitext text/x-wiki == Theorem ==  <onlyinclude> Let $\mu$, $\alpha$ and $\beta$ be real numbers.  Let $\sigma$ be a positive real number.  Let $X \sim \Gaussian \mu {\sigma^2}$ where $\Gaussian \mu {\sigma^2}$ is the Gaussian distribution with parameters $\mu$ and $\sigma^2$.  Then:  :$\alpha X + \beta \sim \Gaussian {\alpha \mu + \beta} {\alpha^2 \sigma^2}$ </onlyinclude>  == Proof == Let $Z = \alpha X + \beta$.  Let $M_Z$ be the moment generating function of $Z$.  We aim to show that:  :$Z \sim \Gaussian {\alpha \mu + \beta} {\alpha^2 \sigma^2}$ By Moment Generating Function of Gaussian Distribution and Moment Generating Function is Unique, it is sufficient to show that: :$\displaystyle \map {M_Z} t = \map \exp {\paren {\alpha \mu + \beta} t + \frac 1 2 \alpha^2 \sigma^2 t^2}$ We also have, by Moment Generating Function of Gaussian Distribution, that the moment generating function of $X$, $M_X$, is given by:  :$\displaystyle \map {M_X} t = \map \exp {\mu t + \frac 1 2 \sigma^2 t^2}$ We have: {{begin-eqn}} {{eqn	| l = \map {M_Z} t 	| r = e^{\beta t} \map {M_X} {\alpha t} 	| c = Moment Generating Function of Linear Combination of Independent Random Variables }} {{eqn	| r = e^{\beta t} \map \exp {\alpha \mu t + \frac 1 2 \sigma^2 \paren {\alpha t}^2} }} {{eqn	| r = \map \exp {\paren {\alpha \mu + \beta} t + \frac 1 2 \sigma^2 \alpha^2 t^2} 	| c = Exponential of Sum }} {{end-eqn}} {{qed}} Category:Gaussian Distribution nc5ethpa6uic5hj0lh5och2t610pcii


Quotient of Independent Random Variables with Chi-Squared Distribution Divided by Degrees of Freedom has F-Distribution 0 78327 411751 2019-07-05T11:40:51Z Caliburn 3218 Created page with "== Theorem == <onlyinclude> Let $n$ and $m$ be strictly positive integers.  Let $X$ and $Y$ be Definition:Independent Random Variabl..." wikitext text/x-wiki == Theorem == <onlyinclude> Let $n$ and $m$ be strictly positive integers. Let $X$ and $Y$ be independent random variables.  Let $X \sim \chi^2_n$ where $\chi^2_n$ is the chi-squared distribution with $n$ degrees of freedom.  Let $Y \sim \chi^2_m$ where $\chi^2_m$ is the chi-squared distribution with $m$ degrees of freedom. Then:  :$\dfrac {X / n} {Y / m} \sim F_{n, m}$ where $F_{n, m}$ is the F-distribution with $\tuple {n, m}$ degrees of freedom. </onlyinclude>  == Proof == {{ProofWanted}} Category:Chi-Squared Distribution Category:F-Distribution cta2r07liq22w6ls6deby0820pfi5tc


Derivative of Function plus Constant

Derivative of Function plus Constant 0 78816 414752 2019-07-27T07:52:21Z Prime.mover 59 Created page with "== Theorem == <onlyinclude> Let $f$ be a real function which is differentiable on $\R$.  Let $c \in \R$..." wikitext text/x-wiki == Theorem == <onlyinclude> Let $f$ be a real function which is differentiable on $\R$. Let $c \in \R$ be a constant. Then: :$\map {D_x} {\map f x + c} = \map {D_x} {\map f x}$ </onlyinclude> == Proof == {{begin-eqn}} {{eqn | l = \map {D_x} {\map f x + c}       | r = \map {D_x} {\map f x} + \map f x \, c       | c = Sum Rule for Derivatives }} {{eqn | r = \map {D_x} {\map f x} + 0       | c = Derivative of Constant }} {{eqn | r = \map {D_x} {\map f x}       | c = }} {{end-eqn}} {{qed}} Category:Differential Calculus stxghwu9hz138be4dsaxr1oyx8br1bi


Definite Integral from 0 to Half Pi of Logarithm of Sine x/Lemma 0 78911 415395 2019-07-30T15:31:34Z Caliburn 3218 need this for another page or two wikitext text/x-wiki == Lemma for Definite Integral from 0 to Half Pi of Logarithm of Sine x == <onlyinclude> :$\displaystyle \int_0^\pi \map \ln {\sin x} \rd x = 2 \int_0^{\pi/2} \map \ln {\sin x} \rd x$ </onlyinclude> == Proof == We have:  {{begin-eqn}} {{eqn	| l = \int_{\pi/2}^\pi \map \ln {\sin x} \rd x 	| r = -\int_{\pi/2}^0 \map \ln {\map \sin {\pi - x} } \rd x 	| c = substituting $x \mapsto \pi - x$ }} {{eqn	| r = \int_0^{\pi/2} \map \ln {\sin x} \rd x 	| c = Reversal of Limits of Definite Integral, Sine of Supplementary Angle }} {{end-eqn}} We can therefore write:  {{begin-eqn}} {{eqn	| l = \int_0^\pi \map \ln {\sin x} \rd x 	| r = \int_0^{\pi/2} \map \ln {\sin x} \rd x + \int_{\pi/2}^\pi \map \ln {\sin x} \rd x 	| c = Sum of Integrals on Adjacent Intervals for Integrable Functions }} {{eqn	| r = 2 \int_0^{\pi/2} \map \ln {\sin x} \rd x }} {{end-eqn}} {{qed|lemma}} Category:Definite Integral from 0 to Half Pi of Logarithm of Sine x 5q2g8m27o01zps1mk2jo1crqxcaa6a5


Lowest Common Multiple/Examples/6 and 15 0 79059 416693 2019-08-05T07:01:34Z Prime.mover 59 Created page with "== Example of Lowest Common Multiple of Integers == <onlyinclude> The Definition:Lowest Common Multiple|lowest common multi..." wikitext text/x-wiki == Example of Lowest Common Multiple of Integers == <onlyinclude> The lowest common multiple of $6$ and $15$ is: :$\lcm \set {6, 15} = 30$ </onlyinclude> == Proof == We find the greatest common divisor of $6$ and $15$ using the Euclidean Algorithm: {{begin-eqn}} {{eqn | n = 1       | l = 15       | r = 2 \times 6 + 3 }} {{eqn | n = 2       | l = 6       | r = 2 \times 3 }} {{end-eqn}} Thus $\gcd \set {6, 15} = 3$. Then: {{begin-eqn}} {{eqn | l = \lcm \set {6, 15}       | r = \dfrac {6 \times 15} {\gcd \set {6, 15} }       | c = Product of GCD and LCM }} {{eqn | r = \dfrac {\paren {2 \times 3} \times \paren {3 \times 5} } 3       | c =  }} {{eqn | r = 2 \times 3 \times 5       | c =  }} {{eqn | r = 30       | c =  }} {{end-eqn}} {{qed}} Category:Examples of Lowest Common Multiples Category:30 k9gdjvcsr46fx5lmvholwp81g8j1236


Definite Integral from 0 to Half Pi of Logarithm of Sine x/Proof 1 0 79137 417073 2019-08-07T15:17:48Z Caliburn 3218 Created page with "== Theorem == {{:Definite Integral from 0 to Half Pi of Logarithm of Sine x}}  == Proof == <onlyinclude> By Definite Integral from 0 to Half Pi of Logarithm of Sine x/Lemma|..." wikitext text/x-wiki == Theorem == {{:Definite Integral from 0 to Half Pi of Logarithm of Sine x}} == Proof == <onlyinclude> By Definite Integral from $0$ to $\dfrac \pi 2$ of $\map \ln {\sin x}$: Lemma, we have:  :$\displaystyle \int_0^\pi \map \ln {\sin x} \rd x = 2 \int_0^{\pi/2} \map \ln {\sin x} \rd x$ We also have:  {{begin-eqn}} {{eqn	| l = \int_0^{\pi/2} \map \ln {\sin x} \rd x 	| r = \int_0^{\pi/2} \map \ln {\map \sin {\frac \pi 2 - x} } \rd x 	| c = Integral between Limits is Independent of Direction }} {{eqn	| r = \int_0^{\pi/2} \map \ln {\cos x} \rd x 	| c = Sine of Complement equals Cosine }} {{end-eqn}} giving:  {{begin-eqn}} {{eqn	| l = 2 \int_0^{\pi/2} \map \ln {\sin x} \rd x 	| r = \int_0^{\pi/2} \map \ln {\sin x \cos x} \rd x 	| c = Sum of Logarithms }} {{eqn	| r = \int_0^{\pi/2} \map \ln {\frac 1 2 \sin 2 x} \rd x 	| c = Double Angle Formula for Sine }} {{eqn	| r = \frac \pi 2 \map \ln {\frac 1 2} + \int_0^{\pi/2} \map \ln {\sin 2 x} \rd x 	| c = Primitive of Constant, Sum of Logarithms }} {{eqn	| r = \frac \pi 2 \map \ln {\frac 1 2} + \frac 1 2 \int_0^\pi \map \ln {\sin u} \rd u 	| c = substituting $u = 2 x$ }} {{eqn	| r = -\frac \pi 2 \ln 2 + \int_0^{\pi/2} \map \ln {\sin u} \rd u 	| c = Logarithm of Reciprocal }} {{end-eqn}} Therefore:  :$\displaystyle \int_0^{\pi/2} \map \ln {\sin x} \rd x = -\frac \pi 2 \ln 2$ {{qed}} </onlyinclude> Category:Definite Integral from 0 to Half Pi of Logarithm of Sine x mybzaitrvxdq39bjnc0ju2ey2qdui8b


Equivalence Class/Examples/Months that Start on the Same Day of the Week 0 79374 418399 418395 2019-08-16T05:31:13Z Prime.mover 59 wikitext text/x-wiki == Examples of Equivalence Class == <onlyinclude> Let $M$ be the set of months of the (calendar) year according to the (usual) Gregorian calendar. Let $\sim$ be the relation on $M$ defined as: :$\forall x, y \in M: x \sim y \iff \text {$x$ and $y$ both start on the same day of the week}$ The set of equivalence classes under $\sim$ depends on whether the year is a leap year. For a non-leap year, the set of equivalence classes is: :$\set {\set {\text {January}, \text {October} }, \set {\text {February}, \text {March}, \text {November} }, \set {\text {April}, \text {July} }, \set {\text {May} }, \set {\text {June} }, \set {\text {August} }, \set {\text {September}, \text {December} } }$ For a leap year, the set of equivalence classes is: :$\set {\set {\text {January}, \text {April}, \text {July} }, \set {\text {February}, \text {August} }, \set {\text {March}, \text {November} }, \set {\text {May} }, \set {\text {June} }, \set {\text {September}, \text {December} }, \set {\text {October} } }$ </onlyinclude> == Proof == We have that: :The months with $30$ days are: ::$\text {April}, \text {June}, \text {September}, \text {November}$ :The months with $31$ days are: ::$\text {January}, \text {March}, \text {May}, \text {July}, \text {August}, \text {October}, \text {December}$ :In a non-leap year, $\text {February}$ has $28$ days :In a leap year, $\text {February}$ has $29$ days. Let month $m$ have $m_d$ days in it. Let month $m$ start on day $d$, where $d$ is in the range $0$ to $6$ (which day of the week corresponds to which number is irrelevant at this stage). Then month $m + 1$ starts on day $\paren {d + m_d} \pmod 7$. For reference: {{begin-eqn}} {{eqn | l = 28       | o = \equiv       | r = 0 \pmod 7       | c =  }} {{eqn | l = 29       | o = \equiv       | r = 0 \pmod 7       | c =  }} {{eqn | l = 30       | o = \equiv       | r = 0 \pmod 7       | c =  }} {{eqn | l = 31       | o = \equiv       | r = 0 \pmod 7       | c =  }} {{end-eqn}} {{WLOG}}, let $\text {January}$ start on day $0$. Then the sequence of the days which are the $1$st of the month are as follows: For a non-leap year: :$\tuple {0, 3, 3, 6, 1, 4, 6, 2, 5, 0, 3, 5}$ {{OEIS|A189915}} For a leap year: :$\tuple {0, 3, 4, 0, 2, 5, 0, 3, 6, 1, 4, 6}$ {{OEIS|A189916}} The result follows. {{qed}} == Sources == * {{BookReference|Set Theory and Abstract Algebra|1975|T.S. Blyth|prev = Equivalence Relation/Examples/Months that Start on the Same Day of the Week|next = Number of Friday 13ths in a Year}}: $\S 6$. Indexed families; partitions; equivalence relations: Exercise $7$ Category:Examples of Equivalence Classes Category:Calendars i37jokwxlf2u7mc5q9xhufclfbsaq5v


Symbols:Abbreviations/W/WFF

Symbols:Abbreviations/W/WFF 104 79554 446989 424930 2020-02-06T23:42:49Z Prime.mover 59 wikitext text/x-wiki == Abbreviation: WFF == <onlyinclude> :Well-formed formula. </onlyinclude> == Sources == * {{BookReference|The Penguin Dictionary of Mathematics|1998|David Nelson|ed = 2nd|edpage = Second Edition|prev = Mathematician:Hermann Klaus Hugo Weyl|next = Mathematician:Alfred North Whitehead|entry = wff}} * {{BookReference|The Penguin Dictionary of Mathematics|2008|David Nelson|ed = 4th|edpage = Fourth Edition|prev = Mathematician:Hermann Klaus Hugo Weyl|next = Mathematician:Alfred North Whitehead|entry = wff}} Category:Symbols/Abbreviations/W oip6gzhhqge2br7jpl3lq08vts6f7ml


Symbols:Number Theory/Does Not Divide

Symbols:Number Theory/Does Not Divide 104 80006 421036 421033 2019-08-26T16:02:36Z Prime.mover 59 wikitext text/x-wiki == Is Not a Divisor == <onlyinclude> :$x \nmid y$ This means '''$x$ is not a divisor of $y$'''. {{LatexFor|for = x \nmid y}} </onlyinclude> == Also denoted as == This symbol is preferable to $x \mathrel {\not \backslash} y$ due to the somewhat confusing appearance of this symbol: {{LatexFor|for = x \mathrel {\not \backslash} y}} Category:Symbols/Number Theory f7u01z8p54adxwuyz5rv8ksyyd8bar0


Principle of Stationary Action

Principle of Stationary Action 0 80401 479412 479410 2020-07-22T16:15:04Z Prime.mover 59 wikitext text/x-wiki == Physical Law == The '''principle of stationary actions''' states that the equations of motion of a physical system can be acquired by finding a stationary point of the action. In other words, the first variation of the action has to vanish. == Also known as == The '''principle of stationary action''' is also known as the '''principle of least action'''. == Notes == {{refactor|Extract the appropriate bit of this into a historical note, and the other bits into separate pages of their own as statement and proof. Hence lose the "notes" section.}} The '''principle of least action''' attained its name due to classical problems of minimization. However, if broken trajectories are allowed, the action can sometimes acquire lower values than for any allowed smooth trajectory. Since smooth trajectories are more realistic, leastness has been weakened to stationarity. Category:Mechanics Category:Physics Category:Applied Mathematics Category:Lagrangian Mechanics 0d1283hbybpfcwj6pcfhzchm2qcxr43


Powers of 3 Modulo 8

Powers of 3 Modulo 8 0 80449 424107 2019-09-08T12:06:35Z Prime.mover 59 Created page with "== Theorem == <onlyinclude> Let $n \in \Z_{\ge 0}$ be a strictly positive integer.  Then: :$3^n \equiv \begin {cases} 1 \pmod 8 & : \t..." wikitext text/x-wiki == Theorem == <onlyinclude> Let $n \in \Z_{\ge 0}$ be a strictly positive integer. Then: :$3^n \equiv \begin {cases} 1 \pmod 8 & : \text {$n$ even} \\ 3 \pmod 8 & : \text {$n$ odd} \end {cases}$ </onlyinclude> == Proof 1 == {{:Powers of 3 Modulo 8/Proof 1}} == Proof 2 == {{:Powers of 3 Modulo 8/Proof 2}} Category:Powers of 3 Category:Modulo Arithmetic Category:Powers of 3 Modulo 8 lxmoxenu4w6om1ahif62euwc57p46is


Common Logarithm/Examples/2.36

Common Logarithm/Examples/2.36 0 80916 432976 426975 2019-10-30T18:28:15Z Prime.mover 59 wikitext text/x-wiki == Example of Common Logarithm == <onlyinclude> The common logarithm of $2 \cdotp 36$ is: :$\log_{10} 2 \cdotp 36 = 0 \cdotp 3729$ </onlyinclude> == Sources == * {{BookReference|Theory and Problems of Statistics|1972|Murray R. Spiegel|author2 = R.W. Boxer|ed = SI|edpage = SI Edition|prev = Range of Common Logarithm of Number between 1 and 10|next = Common Logarithm/Examples/23.6}}: Chapter $1$: Logarithms: '''Example 1.''' Category:Examples of Common Logarithms k8050bu2cizg984eixbrl73qdt96g8s


Jensen's Inequality (Real Analysis)/Corollary 0 81312 429062 429061 2019-10-01T19:51:22Z Caliburn 3218 wikitext text/x-wiki == Corollary to Jensen's Inequality: Real Analysis == <onlyinclude> Let $I$ be a real interval. Let $\phi: I \to \R$ be a concave function. Let $x_1, x_2, \ldots, x_n \in I$. Let $\lambda_1, \lambda_2, \ldots, \lambda_n \ge 0$ be real numbers, at least one of which is non-zero. Then: :$\displaystyle \map \phi {\frac {\sum_{k \mathop = 1}^n \lambda_k x_k} {\sum_{k \mathop = 1}^n \lambda_k} } \ge \frac {\sum_{k \mathop = 1}^n \lambda_k \map \phi {x_k} } {\sum_{k \mathop = 1}^n \lambda_k}$ For $\phi$ strictly concave, equality holds {{iff}} $x_1 = x_2 = \cdots = x_n$. </onlyinclude> == Proof == By Real Function is Concave iff its Negative is Convex, $-\phi: I \to \R$ is a convex function. Therefore, we can apply Jensen's Inequality: Real Analysis with $-\phi$ to obtain: :$\displaystyle -\map \phi {\frac {\sum_{k \mathop = 1}^n \lambda_k x_k} {\sum_{k \mathop = 1}^n \lambda_k} } \le -\frac {\sum_{k \mathop = 1}^n \lambda_k \map \phi {x_k} } {\sum_{k \mathop = 1}^n \lambda_k}$ with equality for $-\phi$ strictly convex {{iff}} $x_1 = x_2 = \cdots = x_n$. That is, for $\phi$ strictly concave, equality holds {{iff}} $x_1 = x_2 = \cdots = x_n$. With that, we have established the equality case. Multiplying through $-1$ in our inequality gives:  :$\displaystyle \map \phi {\frac {\sum_{k \mathop = 1}^n \lambda_k x_k} {\sum_{k \mathop = 1}^n \lambda_k} } \ge \frac {\sum_{k \mathop = 1}^n \lambda_k \map \phi {x_k} } {\sum_{k \mathop = 1}^n \lambda_k}$ as required. {{qed}} Category:Jensen's Inequality (Real Analysis) 8y2kpnwksu76razdzqnd66qcoxwmfd2


Axiom:Axioms of Uncertainty/Axiom 5

Axiom:Axioms of Uncertainty/Axiom 5 100 81786 431708 2019-10-18T00:07:10Z Prime.mover 59 Created page with "== Axiom == {{:Axiom:Axioms of Uncertainty}}  ${H_n}$ fulfils the following axiom: <onlyinclude> :$\map {H_n} {\dfrac 1 n, \dfrac 1 n, \dotsc, \dfrac 1 n}..." wikitext text/x-wiki == Axiom == {{:Axiom:Axioms of Uncertainty}} ${H_n}$ fulfils the following axiom: <onlyinclude> :$\map {H_n} {\dfrac 1 n, \dfrac 1 n, \dotsc, \dfrac 1 n} \le \map {H_{n + 1} } {\dfrac 1 {n + 1}, \dfrac 1 {n + 1}, \dotsc, \dfrac 1 {n + 1} }$ </onlyinclude> Thus, for example, a $2$-horse race is less uncertain than a $3$-horse race. == Sources == * {{BookReference|Codes and Cryptography|1988|Dominic Welsh|prev = Axiom:Axioms of Uncertainty/Axiom 4|next = Axiom:Axioms of Uncertainty/Axiom 6}}: $\S 1$: Entropy = uncertainty = information: $1.1$ Uncertainty Category:Axioms/Axioms of Uncertainty d9zz9602wmis3uhox6g9pqlboesvvc1


Axiom:Axioms of Uncertainty/Axiom 6

Axiom:Axioms of Uncertainty/Axiom 6 100 81787 431709 2019-10-18T00:08:51Z Prime.mover 59 Created page with "== Axiom == {{:Axiom:Axioms of Uncertainty}}  ${H_n}$ fulfils the following axiom: <onlyinclude> :$H_n$ is a Definition:Continuous Real-Valued Function|..." wikitext text/x-wiki == Axiom == {{:Axiom:Axioms of Uncertainty}} ${H_n}$ fulfils the following axiom: <onlyinclude> :$H_n$ is a continuous function of its arguments. </onlyinclude> == Sources == * {{BookReference|Codes and Cryptography|1988|Dominic Welsh|prev = Axiom:Axioms of Uncertainty/Axiom 5|next = Axiom:Axioms of Uncertainty/Axiom 7}}: $\S 1$: Entropy = uncertainty = information: $1.1$ Uncertainty Category:Axioms/Axioms of Uncertainty 93a1kzzyzfb8d1h9vfh8lseoqttcern


Axiom:Axioms of Uncertainty/Axiom 7

Axiom:Axioms of Uncertainty/Axiom 7 100 81788 431710 2019-10-18T00:15:50Z Prime.mover 59 Created page with "== Axiom == {{:Axiom:Axioms of Uncertainty}}  ${H_n}$ fulfils the following axiom: <onlyinclude> :$\map {H_{m n} } {\dfrac 1 {m n}, \dfrac 1 {m n}, \dotsc..." wikitext text/x-wiki == Axiom == {{:Axiom:Axioms of Uncertainty}} ${H_n}$ fulfils the following axiom: <onlyinclude> :$\map {H_{m n} } {\dfrac 1 {m n}, \dfrac 1 {m n}, \dotsc, \dfrac 1 {m n} } = \map {H_m} {\dfrac 1 m, \dfrac 1 m, \dotsc, \dfrac 1 m} + \map {H_n} {\dfrac 1 n, \dfrac 1 n, \dotsc, \dfrac 1 n}$ </onlyinclude> == Sources == * {{BookReference|Codes and Cryptography|1988|Dominic Welsh|prev = Axiom:Axioms of Uncertainty/Axiom 6|next = Axiom:Axioms of Uncertainty/Axiom 8}}: $\S 1$: Entropy = uncertainty = information: $1.1$ Uncertainty Category:Axioms/Axioms of Uncertainty ehf4tp5ajzwggteo2hcaxjxu9x0wfrm


Axiom:Axioms of Uncertainty/Axiom 8

Axiom:Axioms of Uncertainty/Axiom 8 100 81789 431757 431711 2019-10-18T23:15:59Z Prime.mover 59 wikitext text/x-wiki == Axiom == {{:Axiom:Axioms of Uncertainty}} ${H_n}$ fulfils the following axiom: <onlyinclude> Let: :$p = p_1 + p_2 + \dotsb + p_m$ :$q = q_1 + q_2 + \dotsb + q_n$ such that: :each of $p_i$ and $q_j$ are non-negative :$p + q = 1$ Then: :$\map {H_{m + n} } {p_1, p_2, \dotsc, p_m, q_1, q_2, \dotsc q_n} = \map {H_2} {p, q} + p \map {H_m} {\dfrac {p_1} p, \dfrac {p_2} p, \dotsc, \dfrac {p_m} p} + q \map {H_n} {\dfrac {q_1} q, \dfrac {q_2} q, \dotsc, \dfrac {q_n} q}$ </onlyinclude> == Sources == * {{BookReference|Codes and Cryptography|1988|Dominic Welsh|prev = Axiom:Axioms of Uncertainty/Axiom 7|next = Function that Satisfies Axioms of Uncertainty}}: $\S 1$: Entropy = uncertainty = information: $1.1$ Uncertainty Category:Axioms/Axioms of Uncertainty drq6oflopludhg4ygoda2ost1fg9qs1


Complex Exponential Function is Entire

Complex Exponential Function is Entire 0 81953 432464 2019-10-27T15:09:41Z Caliburn 3218 Created page with "== Theorem == <onlyinclude> Let $\exp: \C \to \C$ be the complex exponential function.   Then $\exp$ is Definition:Entire Functio..." wikitext text/x-wiki == Theorem == <onlyinclude> Let $\exp: \C \to \C$ be the complex exponential function.  Then $\exp$ is entire.  </onlyinclude> == Proof == By the definition of the complex exponential function, $\exp$ admits a power series expansion about $0$:  :$\displaystyle \exp z = \sum_{n \mathop = 0}^\infty \frac {z^n} {n!}$ By Complex Function is Entire iff it has Everywhere Convergent Power Series, to show that $\exp$ is entire it suffices to show that this series is everywhere convergent. Note that this power series is of the form:  :$\displaystyle \sum_{n \mathop = 0}^\infty \frac {\paren {z - \xi}^n} {n!}$ with $\xi = 0$.  Therefore, by Radius of Convergence of Power Series over Factorial: Complex Case, we have that the former power series is everywhere convergent.  Hence the result. {{qed}} Category:Exponential Function bs3ajz8lpqth8r3s5pmgrrjvyc98dq8