Elementary Symmetric Function/Examples/Recursion 0 82170 493064 493063 2020-10-07T14:06:14Z Prime.mover 59 wikitext text/x-wiki == Example of Elementary Symmetric Function: Recursion == <onlyinclude> Let $\set {z_1, z_2, \ldots, z_{n + 1} }$ be a set of $n + 1$ values, duplicate values permitted. Then for $1 \le m \le n$: :$\map {e_m} {\set {z_1, \ldots, z_n, z_{n + 1} } } = z_{n + 1} \map {e_{m - 1} } {\set {z_1, \ldots, z_n} } + \map {e_m} {\set {z_1, \ldots, z_n} }$ </onlyinclude> == Proof == Case $m = 1$ holds because $e_0$ is $1$ and $e_1$ is the sum of the elements. Assume $2 \le m \le n$. Define four sets: :$A = \set {\set {p_1, \ldots, p_m} : 1 \le p_1 < \cdots < p_m \le n + 1}$ :$B = \set {\set {p_1, \ldots, p_m} : 1 \le p_1 < \cdots < p_{m - 1} \le n, p_m = n + 1}$ :$C = \set {\set {p_1, \ldots, p_m} : 1 \le p_1 < \cdots < p_m \le n}$ :$D = \set {\set {p_1, \ldots, p_{m - 1} } : 1 \le p_1 < \cdots < p_{m - 1} \le n}$ Then $A = B \cup C$ and $B \cap C = \O$ implies: :$\ds \sum_A z_{p_1} \cdots z_{p_m} = \sum_B z_{p_1} \cdots z_{p_m} + \sum_C z_{p_1} \cdots z_{p_m}$ Simplify: :$\ds \sum_B z_{p_1} \cdots z_{p_m} = z_{n + 1} \sum_D z_{p_1} \cdots z_{p_{m - 1} }$ Notation for Elementary Symmetric Functions: :$\ds \map {e_m} {\set {z_1, \ldots, z_n, z_{n + 1} } } = \sum_A z_{p_1} \cdots z_{p_m}$ :$\ds \sum_D z_{p_1} \cdots z_{p_{m - 1} } = \map {e_{m - 1} } {\set {z_1, \ldots, z_n} }$ :$\ds \sum_C z_{p_1} \cdots z_{p_m} = \map {e_m} {\set {z_1, \ldots, z_n} }$ Assemble the preceding equations: {{begin-eqn}} {{eqn | l = \map {e_m} {\set {z_1, \ldots, z_n, z_{n + 1} } }        | r = \sum_A z_{p_1} \cdots z_{p_m} }} {{eqn | r = \sum_B z_{p_1} \cdots z_{p_m} + \sum_C z_{p_1} \cdots z_{p_m} }} {{eqn | r = z_{n + 1} \sum_D z_{p_1} \cdots z_{p_m} + \sum_C z_{p_1} \cdots z_{p_m} }} {{eqn | r = z_{n + 1} \map {e_{m - 1} } {\set {z_1, \ldots, z_n} } + \map {e_m} {\set {z_1, \ldots, z_n} } }} {{end-eqn}} {{qed}} Category:Elementary Symmetric Functions mlp1kflfs6ugu7nw9q8rvovs64gt3mb


Inverse of Vandermonde Matrix/Corollary

Inverse of Vandermonde Matrix/Corollary 0 82308 443085 436852 2020-01-09T20:59:23Z Prime.mover 59 wikitext text/x-wiki {{refactor}} == Corollary to Inverse of Vandermonde Matrix == <onlyinclude> Define for variables $\set {y_1,\ldots, y_k}$ elementary symmetric functions: {{begin-eqn}} {{eqn | l = \map {e_m} {\set {y_1, \ldots, y_k} }       | r = \sum_{1 \mathop \le j_1 \mathop < j_2 \mathop < \mathop \cdots \mathop < j_m \mathop \le k } y_{j_1} y_{j_2} \cdots y_{j_m}       | c =  for $m = 0, 1, \ldots, k$ }} {{end-eqn}} Let $\set {x_1, \ldots, x_n}$ be a set of distinct values. Let $W_n$ and $V_n$ be Vandermonde matrices of order $n$: :$W_n = \begin{bmatrix}   1         &  x_1      & \cdots & x_1^{n-1} \\   1         & x_2       & \cdots & x_2^{n-1} \\ \vdots      & \vdots    & \ddots & \vdots    \\   1         & x_1^{n-1} & \cdots & x_n^{n-1} \\ \end{bmatrix}, \quad V_n = \begin{bmatrix}   x_1   & x_2    & \cdots & x_n    \\   x_1^2 & x_2^2  & \cdots & x_n^2  \\ \vdots  & \vdots & \ddots & \vdots \\   x_1^n & x_2^n  & \cdots & x_n^n  \\ \end{bmatrix}$ Let their matrix inverses be written as $W_n^{-1} = \begin{bmatrix} b_{ij} \end{bmatrix}$ $V_n^{-1} = \begin{bmatrix} c_{ij} \end{bmatrix}$. Then: {{begin-eqn}} {{eqn   | l = b_{ij}   | r = \dfrac {\paren {-1}^{n - i} \map {e_{n - i} } {\set {x_1, \ldots, x_n} \setminus \set {x_j} } } {\prod_{m \mathop = 1, m \mathop \ne j }^n \paren {x_j - x_m} }   | c = for $i, j = 1, \ldots, n$ }} {{eqn   | l = c_{ij}   | r = \dfrac 1 {x_i} \, b_{j i}   | c = for $i, j = 1, \ldots, n$ }} {{end-eqn}} </onlyinclude> == Proof == The details appear in Inverse of Vandermonde Matrix/Proof 1, same notation.{{qed}} Category:Inverse of Vandermonde Matrix 2g40bipql82wprdkeuo5qloqoohp17t


Equation of Cardioid/Parametric

Equation of Cardioid/Parametric 0 82574 435490 2019-11-17T00:32:52Z Prime.mover 59 Created page with "== Theorem == <onlyinclude> Let $C$ be a cardioid embedded in a Cartesian coordinate plane such that: :its Definition:..." wikitext text/x-wiki == Theorem == <onlyinclude> Let $C$ be a cardioid embedded in a Cartesian coordinate plane such that: :its stator of radius $a$ is positioned with its center at $\tuple {a, 0}$ :there is a cusp at the origin. Then $C$ can be expressed by the parametric equation: :$\begin {cases} x = 2 a \cos t \paren {1 + \cos t} \\ y = 2 a \sin t \paren {1 + \cos t} \end {cases}$ </onlyinclude> == Proof == :525px Let $P = \polar {x, y}$ be an arbitrary point on $C$. From Polar Form of Equation of Cardioid, $C$ is expressed as a polar equation as: :$r = 2 a \paren {1 + \cos \theta}$ We have that: :$x = r \cos \theta$ :$y = r \sin \theta$ Replace $\theta$ with $t$ and the required parametric equation is the result. {{qed}} == Sources == * {{MathWorld|Cardioid|Cardioid}} Category:Cardioids 6d8qpudg46rkavewi2qwje22robgjgh


Irrational Number divided by Rational Number is Irrational 0 82682 435876 2019-11-20T16:53:26Z Caliburn 3218 Created page with "== Theorem == <onlyinclude> Let $x$ be a irrational number.   Let $y$ be a non-zero Definition:Rational Number|rational..." wikitext text/x-wiki == Theorem == <onlyinclude> Let $x$ be a irrational number.  Let $y$ be a non-zero rational number. Then:  :$\dfrac x y$ is irrational. </onlyinclude> == Proof == {{AimForCont}} $\dfrac x y$ is rational number. Then there exists an integer $p_1$ and a natural number $q_1$ such that:  :$\dfrac x y = \dfrac {p_1} {q_1}$ That is:  :$x = \dfrac {p_1} {q_1} y$ From the fact that $y$ is rational, we similarly have that there exists an integer $p_2$ and a natural number $q_2$ such that: :$y = \dfrac {p_2} {q_2}$ Then:  :$x = \dfrac {p_1 p_2} {q_1 q_2}$ From Integer Multiplication is Closed, we have that $p_1 p_2$ is an integer.  From Natural Number Multiplication is Closed, we have that $q_1 q_2$ is a natural number. Let $p_3 = p_1 p_2$ and $q_3 = q_1 q_2$. Then $x$ is expressible in the form:  :$x = \dfrac {p_3} {q_3}$ where $p_3$ is an integer and $q_3$ is a natural number. This, however, implies that $x$ is rational, which is a contradiction. By Proof by Contradiction, we conclude that $\dfrac x y$ is irrational. {{qed}} Category:Rational Numbers Category:Irrational Numbers 8lpmk0l0uozjamjcnib2a455dg7oqoi


Equation of Confocal Ellipses/Formulation 1 0 82816 436936 436930 2019-11-27T12:50:20Z Prime.mover 59 wikitext text/x-wiki == Definition == <onlyinclude> The equation: :$(1): \quad \dfrac {x^2} {a^2 + \lambda} + \dfrac {y^2} {b^2 + \lambda} = 1$ where: :$\tuple {x, y}$ denotes an arbitrary point in the cartesian plane :$a$ and $b$ are (strictly) positive constants such that $a^2 > b^2$ :$\lambda$ is a (strictly) positive parameter such that $b^2 > -\lambda$ defines the set of all confocal ellipses whose foci are at $\tuple {\pm \sqrt {a^2 - b^2}, 0}$. </onlyinclude> == Proof == Let $a$ and $b$ be arbitrary (strictly) positive real numbers fulfilling the constraints as defined. Let $E$ be the locus of the equation: :$(1): \quad \dfrac {x^2} {a^2 + \lambda} + \dfrac {y^2} {b^2  + \lambda} = 1$ As $b^2 > -\lambda$ it follows that: :$b^2 + \lambda > 0$ and as $a^2 > b^2$: :$a^2 + \lambda > 0$ Thus $(1)$ is in the form: :$\dfrac {x^2} {r^2} + \dfrac {y^2} {s^2} = 1$ where: :$r^2 = a^2 + \lambda$ :$s^2 = b^2 + \lambda$ From Equation of Ellipse in Reduced Form, this is the equation of an ellipse in reduced form. It follows that: :$\tuple {\pm \sqrt {a^2 + \lambda}, 0}$ are the positions of the vertices of $E$ :$\tuple {0, \pm \sqrt {b^2 + \lambda} }$ are the positions of the covertices of $E$ From Focus of Ellipse from Major and Minor Axis, the positions of the foci of $E$ are given by: :$\paren {a^2 + \lambda} - \paren {b^2 + \lambda} = c^2$ where $\tuple {\pm c, 0}$ are the positions of the foci of $E$. Thus we have: {{begin-eqn}} {{eqn | l = c^2       | r = \paren {a^2 + \lambda} - \paren {b^2 + \lambda}       | c = Focus of Ellipse from Major and Minor Axis }} {{eqn | r = a^2 - b^2       | c =  }} {{end-eqn}} Hence the result. {{qed}} == Also see == * Equation of Confocal Conics * Equation of Confocal Hyperbolas Category:Ellipses Category:Confocal Conics 1sp1wxwjfbglclrywxkh8n4hhl7u2v3


Equation of Confocal Hyperbolas/Formulation 1 0 82817 436938 436937 2019-11-27T13:01:49Z Prime.mover 59 wikitext text/x-wiki == Definition == <onlyinclude> The equation: :$(1): \quad \dfrac {x^2} {a^2 + \lambda} + \dfrac {y^2} {b^2 + \lambda} = 1$ where: :$\tuple {x, y}$ denotes an arbitrary point in the cartesian plane :$a$ and $b$ are (strictly) positive constants such that $a^2 > b^2$ :$\lambda$ is a (strictly) positive parameter such that $b^2 < -\lambda < a^2$ defines the set of all confocal hyperbolas whose foci are at $\tuple {\pm \sqrt {a^2 + b^2}, 0}$. </onlyinclude> == Proof == Let $a$ and $b$ be arbitrary (strictly) positive real numbers fulfilling the constraints as defined. Let $E$ be the locus of the equation: :$(1): \quad \dfrac {x^2} {a^2 + \lambda} + \dfrac {y^2} {b^2 + \lambda} = 1$ As $b^2 < -\lambda$ it follows that: :$b^2 + \lambda < 0$ and as $-\lambda < a^2$: :$a^2 + \lambda > 0$ Thus $(1)$ is in the form: :$\dfrac {x^2} {r^2} - \dfrac {y^2} {s^2} = 1$ where: :$r^2 = a^2 + \lambda$ :$s^2 = -\lambda + b^2$ From Equation of Hyperbola in Reduced Form, this is the equation of an hyperbola in reduced form. It follows that: :$\tuple {\pm \sqrt {a^2 + \lambda}, 0}$ are the positions of the vertices of $E$ :$\tuple {0, \pm \sqrt {b^2 - \lambda} }$ are the positions of the covertices of $E$ From Focus of Hyperbola from Transverse and Conjugate Axis, the positions of the foci of $E$ are given by: :$\paren {a^2 + \lambda} + \paren {b^2 - \lambda} = c^2$ where $\tuple {\pm c, 0}$ are the positions of the foci of $E$. Thus we have: {{begin-eqn}} {{eqn | l = c^2       | r = \paren {a^2 + \lambda} + \paren {b^2 - \lambda}       | c = Focus of Hyperbola from Transverse and Conjugate Axis }} {{eqn | r = a^2 + b^2       | c =  }} {{end-eqn}} Hence the result. {{qed}} == Also see == * Equation of Confocal Conics * Equation of Confocal Ellipses Category:Hyperbolas Category:Confocal Conics 87l6gatjd3w8a9fg9zqn0l1bxm9d1ra


Method of Undetermined Coefficients/Exponential of Sine and Cosine 0 82927 439147 437659 2019-12-10T21:32:12Z Prime.mover 59 wikitext text/x-wiki == Proof Technique == Consider the nonhomogeneous linear second order ODE with constant coefficients: :$(1): \quad y'' + p y' + q y = \map R x$ Let $\map R x$ be of the form: :$\map R x = e^{a x} \paren {\alpha \sin b x + \beta \cos b x}$ The '''Method of Undetermined Coefficients''' can be used to solve $(1)$ in the following manner. == Method and Proof == Let $\map {y_g} x$ be the general solution to: :$(2): \quad y'' + p y' + q y = 0$ From Solution of Constant Coefficient Homogeneous LSOODE, $\map {y_g} x$ can be found systematically. Let $\map {y_p} x$ be a particular solution to $(1)$. Then from General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution: :$\map {y_g} x + \map {y_p} x$ is the general solution to $(1)$. It remains to find $\map {y_p} x$. <onlyinclude> Substitute a trial solution of similar form, either: :$e^{a x} \paren {A \sin b x + B \cos b x}$ or replace the {{RHS}} of $(1)$ by: :$\paren {\alpha - i \beta} e^{i \paren {a + i b} x}$ find a solution, and take the real part. If $e^{a x} \sin b x$ and $e^{a x} \cos b x$ appear in the general solution to $(2)$, then insert a factor of $x$: :$x e^{a x} \paren {A \sin b x + B \cos b x}$ or: :$x \paren {\alpha - i \beta} e^{i \paren {a + i b} x}$ </onlyinclude>{{qed}} == Sources == * {{BookReference|Elementary Differential Equations & Operators|1958|G.E.H. Reuter|prev = Linear Second Order ODE/y'' + 4 y = 3 sin 2 x|next = Method of Undetermined Coefficients/Product of Polynomial and Exponential}}: Chapter $1$: Linear Differential Equations with Constant Coefficients: $\S 2$. The second order equation: $\S 2.6$ Particular solution: some further cases $\text{(i)}$ Category:Method of Undetermined Coefficients jdbghwqd5u95hfnwjk7vhou2369z49v


Method of Undetermined Coefficients/Product of Polynomial and Exponential 0 82928 437658 437656 2019-12-03T22:24:07Z Prime.mover 59 wikitext text/x-wiki == Proof Technique == Consider the nonhomogeneous linear second order ODE with constant coefficients: :$(1): \quad y'' + p y' + q y = \map R x$ Let $\map R x$ be of the form: :$\map R x = e^{a x} \paren {\map f x}$ where $\map f x$ is a real polynomial function. The '''Method of Undetermined Coefficients''' can be used to solve $(1)$ in the following manner. == Method and Proof == Let $\map {y_g} x$ be the general solution to: :$(2): \quad y'' + p y' + q y = 0$ From Solution of Constant Coefficient Homogeneous LSOODE, $\map {y_g} x$ can be found systematically. Let $\map {y_p} x$ be a particular solution to $(1)$. Then from General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution: :$\map {y_g} x + \map {y_p} x$ is the general solution to $(1)$. It remains to find $\map {y_p} x$. <onlyinclude> Substitute a trial solution of similar form: :$e^{a x} \paren {\map g x}$ where $\map g x$ is a real polynomial function with undetermined coefficients of as high a degree as $f$. Then: :differentiate twice {{WRT|Differentiation}} $x$ :establish a set of simultaneous equations by equating powers :solve these equations for the coefficients. If $e^{a x} \paren {\map g x}$ appears in the general solution to $(2)$, then add a further degree to $g$. The last step may need to be repeated if that last polynomial also appears as a general solution to $(2)$. </onlyinclude>{{qed}} == Sources == * {{BookReference|Elementary Differential Equations & Operators|1958|G.E.H. Reuter|prev = Method of Undetermined Coefficients/Exponential of Sine and Cosine|next = Method of Undetermined Coefficients/Product of Polynomial and Function of Sine and Cosine}}: Chapter $1$: Linear Differential Equations with Constant Coefficients: $\S 2$. The second order equation: $\S 2.6$ Particular solution: some further cases $\text{(ii)}$ Category:Method of Undetermined Coefficients mctg71208riib0jzabmvpl29c71c0n5


Method of Undetermined Coefficients/Product of Polynomial and Function of Sine and Cosine 0 82929 437677 437657 2019-12-03T22:44:13Z Prime.mover 59 wikitext text/x-wiki == Proof Technique == Consider the nonhomogeneous linear second order ODE with constant coefficients: :$(1): \quad y'' + p y' + q y = \map R x$ Let $\map R x$ be of the form: :$\map R x = \paren {\alpha \cos b x + \beta \sin b x} \paren {\map f x}$ where $\map f x$ is a real polynomial function. The '''Method of Undetermined Coefficients''' can be used to solve $(1)$ in the following manner. == Method and Proof == Let $\map {y_g} x$ be the general solution to: :$(2): \quad y'' + p y' + q y = 0$ From Solution of Constant Coefficient Homogeneous LSOODE, $\map {y_g} x$ can be found systematically. Let $\map {y_p} x$ be a particular solution to $(1)$. Then from General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution: :$\map {y_g} x + \map {y_p} x$ is the general solution to $(1)$. It remains to find $\map {y_p} x$. <onlyinclude> Substitute a trial solution of similar form, either: :$\paren {\alpha \cos b x + \beta \sin b x} \paren {\map g x}$ or replace the {{RHS}} of $(1)$ by: :$\paren {\alpha - i \beta} e^{i \paren {a + i b} x} \paren {\map g x}$ find a solution, and take the real part. In the above, $\map g x$ is a real polynomial function with undetermined coefficients of as high a degree as $f$. Then: :differentiate twice {{WRT|Differentiation}} $x$ :establish a set of simultaneous equations by equating powers :solve these equations for the coefficients. If $\paren {\alpha \cos b x + \beta \sin b x} \paren {\map g x}$ appears in the general solution to $(2)$, then add a further degree to $g$. The last step may need to be repeated if that last polynomial also appears as a general solution to $(2)$. </onlyinclude>{{qed}} == Sources == * {{BookReference|Elementary Differential Equations & Operators|1958|G.E.H. Reuter|prev = Method of Undetermined Coefficients/Product of Polynomial and Exponential|next = Method of Undetermined Coefficients/Sum of Several Terms}}: Chapter $1$: Linear Differential Equations with Constant Coefficients: $\S 2$. The second order equation: $\S 2.6$ Particular solution: some further cases $\text{(iii)}$ Category:Method of Undetermined Coefficients b3b0v1ep33jfqbzqz6q6wud2gav5a6p


Solution of Constant Coefficient Linear nth Order ODE 0 82960 437885 437884 2019-12-06T06:33:33Z Prime.mover 59 wikitext text/x-wiki {{refactor|Extract the homogeneous case into its own page and improve its rigour and its structure. Then address the particular case by means of Method of Undetermined Coefficients as per usual.}} == Proof Technique == Consider the linear second order ODE with constant coefficients: :$(1): \quad \displaystyle \sum_{k \mathop = 0}^n a_k \dfrac {\d^k y} {d x^k} = \map R x$ where $a_k$ is a constant for $0 \le k \le n$ and $\map R x$ is a function of $x$. The general solution to $(1)$ can be found as follows. ;Find the roots $m_1, m_2, \ldots, m_n$ of the auxiliary equation: :$(2): \quad \displaystyle \sum_{k \mathop = 0}^n a_k m^k = 0$. If $(2)$ has distinct roots, then the general solution $\map {y_g} x$ is of the form: :$y_g = \sum_{k \mathop = 0}^n A_k e^{m_k x}$ If there are repeated roots of $(2)$, further needs to be done. Let $m_j$ be a repeated root of $(2)$ with multiplicity $r$. Then the {{LHS}} of $(2)$ can be written: :$\map P m \paren {m - m_j}^r$ where $\map P m$ is a polynomial which does not contain the factor $m - m_j$. Then the $r$ instances of $m_j$ give rise to the solutions: :$(4): \quad y = A_{j_0} e^{m_j x} + A_{j_1} x e^{m_j x} + \dotsb + A_{j_{r - 1} } x^{r - 1} e^{m_j x}$ == Proof == Let the reduced equation of $(1)$ be written: :$(3): \quad \paren {a_n D^n + a_{n - 1} D^{n - 1} + \dotsb + a_1 D + a_0} y = 0$ By factoring the {{LHS}} we get: :$\map P D \paren {D - m_j}^r y = 0$ It remains to be shown that $(4)$ is a solution to $(3)$. Each of the terms is of the form $p e^{m_j x} u$, where $u$ is a power of $x$ and $p$ is a constant. So: {{begin-eqn}} {{eqn | l = \map {\paren {D - m_j} } {p e^{m_j x} u}       | r = \map D {p e^{m_j x} u} - m_j p e^{m_j x} u       | c =  }} {{eqn | r = p e^{m_j x} D u       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \map {\paren {D - m_j}^2} {p e^{m_j x} u}       | r = \map {\paren {D - m_j} } {p e^{m_j x} D u}       | c =  }} {{eqn | r = p e^{m_j x} \map D {D u}       | c =  }} {{eqn | r = p e^{m_j x} \map {D^2} u       | c =  }} {{end-eqn}} and so on until: {{begin-eqn}} {{eqn | l = \map {\paren {D - m_j}^r} {p e^{m_j x} u}       | r = p e^{m_j x} D^r u       | c =  }} {{eqn | r = 0       | c =  }} {{end-eqn}} because $D^r u = 0$ when $u = 1, x, x^2, \ldots, x^{r - 1}$. {{finish}} == Sources == * {{BookReference|Elementary Differential Equations & Operators|1958|G.E.H. Reuter|prev = Definition:Reduced Equation of Linear ODE with Constant Coefficients|next = ODE/(D^4 - 1) y = sin x}}: Chapter $1$: Linear Differential Equations with Constant Coefficients: $\S 3$. Equations of higher order and systems of first order equations: $\S 3.1$ The $n$th order equation Category:Linear ODEs f3keoplpt0ksewiw6w9xeg7ybcn2q48


Lindelöf's Lemma/Lemma/Lemma/Lemma

Lindelöf's Lemma/Lemma/Lemma/Lemma 0 83033 443077 438449 2020-01-09T12:14:37Z Prime.mover 59 wikitext text/x-wiki {{rename|This is such a basic result it really should not be hidden as obscurely as a lemma of a lemma of a lemma of a lemma.}} == Lemma == <onlyinclude> Let $S$ be countable set. Let $T$ be a set. Let $T$ be in one-to-one correspondence with $S$. Then $T$ is countable. </onlyinclude> == Proof == $S$ is countable. Therefore, $S$ is in one-to-one correspondence with a subset of the natural numbers by a definition of countable set. $T$ is in one-to-one correspondence with $S$. Therefore, $T$ is in one-to-one correspondence with a subset of the natural numbers by Composite of Bijections is Bijection. Accordingly, $T$ is countable by a definition of countable. {{qed}} Category:Real Analysis sj6s0ge2aqhc7uapyw70f858jj3xbgs


Lindelöf's Lemma/Lemma/Lemma

Lindelöf's Lemma/Lemma/Lemma 0 83034 443076 439243 2020-01-09T12:12:24Z Prime.mover 59 wikitext text/x-wiki == Lemma == <onlyinclude> Let $R$ be a set of real intervals with rational numbers as endpoints. Let every interval in $R$ be of the same type of which there are four: $\openint \ldots \ldots$, $\closedint \ldots  \ldots$, $\hointr \ldots \ldots$, and $\hointl \ldots \ldots$. Then $R$ is countable. </onlyinclude> == Proof == === Lemma 2 === {{:Lindelöf's Lemma/Lemma/Lemma/Lemma|Lemma 2}} {{qed|lemma}} By Rational Numbers are Countably Infinite, the rationals are countable. By Subset of Countably Infinite Set is Countable, a subset of the rationals is countable. The endpoint of an interval in $R$ is characterized by a rational number as every interval in $R$ is of the same type. Therefore, the set consisting of the left hand endpoints of every interval in $R$ is countable. Also, the set consisting of the right hand endpoints of every interval in $R$ is countable. The cartesian product of countable sets is countable. Therefore, the cartesian product of the sets consisting of the respectively left hand and  right hand endpoints of every interval in $R$ is countable. A subset of this cartesian product is in one-to-one correspondence with $R$. This subset is countable by Subset of Countably Infinite Set is Countable. $R$ is countable by Lemma 2 as $R$ is in one-to-one correspondence with a countable set. {{qed}} Category:Real Analysis t5y1vlhdt2cj7xhliqjwgn4fq4etsog


Absolute Value of Cut is Greater Than or Equal To Zero Cut 0 83472 441244 441243 2019-12-28T20:44:19Z Prime.mover 59 wikitext text/x-wiki == Definition == Let $\alpha$ be a cut. Let $\size \alpha$ denote the '''absolute value of $\alpha$'''. Then: :$\size \alpha \ge 0^*$ where: :$0^*$ denotes the rational cut associated with the (rational) number $0$ :$\ge$ denotes the ordering on cuts. == Proof == Let $\alpha \ge 0^*$. Then by definition $\size \alpha = \alpha \ge 0^*$. Let $\alpha < 0^*$. Then: :$\exists \beta: \beta + \alpha = 0^*$ Thus: :$\alpha = -\beta$ and it follows that $\beta > 0^*$. The result follows. {{qed}} == Sources == * {{BookReference|Principles of Mathematical Analysis|1964|ed = 2nd|edpage = Second Edition|Walter Rudin|prev = Definition:Absolute Value of Cut|next = Absolute Value of Cut is Zero iff Cut is Zero}}: Chapter $1$: The Real and Complex Number Systems: Dedekind Cuts: $1.24$. Definition Category:Cuts Category:Absolute Value Function 5zo6m47unf76etmjv8jv6zu8jj659cg


Absolute Value of Cut is Zero iff Cut is Zero 0 83473 441246 441245 2019-12-28T20:47:06Z Prime.mover 59 wikitext text/x-wiki == Definition == Let $\alpha$ be a cut. Let $\size \alpha$ denote the '''absolute value of $\alpha$'''. Then: :$\size \alpha = 0^*$ {{iff}} $\alpha = 0^*$ where $0^*$ denotes the rational cut associated with the (rational) number $0$. == Proof == Let $\alpha 0^*$. Then by definition $\size \alpha = \alpha = 0^*$. Let $\alpha \ne 0^*$. Then either: :$\alpha > 0^*$ in which case $\size \alpha = \alpha > 0^*$ or: :$\alpha < 0^*$ in which case $\size \alpha = -\alpha > 0^*$ In either case $\size \alpha \ne 0^*$. The result follows. {{qed}} == Sources == * {{BookReference|Principles of Mathematical Analysis|1964|ed = 2nd|edpage = Second Edition|Walter Rudin|prev = Absolute Value of Cut is Greater Than or Equal To Zero Cut|next = Definition:Multiplication of Cuts}}: Chapter $1$: The Real and Complex Number Systems: Dedekind Cuts: $1.24$. Definition Category:Cuts Category:Absolute Value Function nlxiqqazy04xx2adz93ayrrs0e4pfko


Dedekind's Theorem/Corollary

Dedekind's Theorem/Corollary 0 83566 441930 2020-01-01T22:30:08Z Prime.mover 59 Created page with "== Theorem == <onlyinclude> Let $\tuple {L, R}$ be a Dedekind cut of the set of real numbers $\R$.   Then either $L$ con..." wikitext text/x-wiki == Theorem == <onlyinclude> Let $\tuple {L, R}$ be a Dedekind cut of the set of real numbers $\R$. Then either $L$ contains a largest number or $R$ contains a smallest number. </onlyinclude> == Proof == From Dedekind's Theorem, there exists a unique real number such that: :$l \le \gamma$ for all $l \in L$ :$\gamma \le r$ for all $r \in R$. Let $\gamma \in L$. Then by definition $\gamma$ is the largest number in $L$. Let $\gamma \in R$. Then by definition $\gamma$ is the smallest number in $R$. By the definition of Dedekind cut, $\tuple {L, R}$ is a partition of $\R$. Hence $\gamma$ is either in $L$ or $R$, but not both. That is, $\gamma$ is either: :the largest number in $L$ or: :the smallest number in $R$. Hence the result. {{Qed}} == Sources == * {{BookReference|Principles of Mathematical Analysis|1964|ed = 2nd|edpage = Second Edition|Walter Rudin|prev = Dedekind's Theorem|next = Dedekind's Theorem/Proof 3}}: Chapter $1$: The Real and Complex Number Systems: Real Numbers: $1.32$. Corollary Category:Real Analysis Category:Dedekind's Theorem i5bjas4p73f66davhsz1m87fy9g50kb


Uniqueness of Positive Root of Positive Real Number/Negative Exponent 0 83583 442142 2020-01-03T21:18:38Z Prime.mover 59 Created page with "== Theorem == <onlyinclude> Let $x \in \R$ be a real number such that $x > 0$.  Let $n \in \Z$ be an integer such that $n < 0..." wikitext text/x-wiki == Theorem == <onlyinclude> Let $x \in \R$ be a real number such that $x > 0$. Let $n \in \Z$ be an integer such that $n < 0$. Then there is at most one $y \in \R: y \ge 0$ such that $y^n = x$. </onlyinclude> == Proof == Let $m = -n$. Let $g$ be the real function defined on $\hointr 0 \to$ defined by: :$\map g y = y^m$ From the definition of power: :$\map g y = \dfrac 1 {\map f y}$ Hence $\map g y$ is strictly decreasing. {{explain|Needs to invoke a result about reciprocals inverting the sense.}} From Strictly Monotone Mapping with Totally Ordered Domain is Injective: :there is at most one $y \in \R: y \ge 0$ such that $y^n = x$. {{qed}} Category:Uniqueness of Positive Root of Positive Real Number fgjw4321lcbb33b8bbthpra853zansn


Russell's Paradox/Corollary

Russell's Paradox/Corollary 0 83748 444615 444300 2020-01-23T20:40:50Z Prime.mover 59 wikitext text/x-wiki == Corollary to Russell's Paradox == <onlyinclude> :$\not \exists x: \forall y: \paren {\map \RR {x, y} \iff \neg \map \RR {y, y} }$ </onlyinclude> Given a relation $\RR$, there cannot exist an element $x$ that bears $\RR$ to all $y$ that do not bear $\RR$ to $y$. == Proof == {{AimForCont}} there does exist such an $x$. Let $\RR$ be such that $\map \RR {x, x}$. Then $\neg \map \RR {x, x}$. Hence it cannot be the case that $\map \RR {x, x}$. Now suppose that $\neg \map \RR {x, x}$. Then by definition of $x$ it follows that $\map \RR {x, x}$. In both cases a contradiction results. Hence there can be no such $x$. {{qed}} == Sources == * {{BookReference|Set Theory and the Continuum Problem|2010|Raymond M. Smullyan|author2 = Melvin Fitting|ed = revised|edpage = Revised Edition|prev = Barber Paradox|next = Axiom:Axiom of Specification/Set Theory}}: Chapter $1$: General Background: $\S 8$ Russell's paradox Category:Russell's Paradox 4m9sguygtvjcki18ilwru953leixhdu


Axiom:Axiom of Extension/Class Theory

Axiom:Axiom of Extension/Class Theory 100 83823 444630 444266 2020-01-23T20:48:15Z Prime.mover 59 Prime.mover moved page Axiom:Axiom of Extension/Classes to Axiom:Axiom of Extension/Class Theory wikitext text/x-wiki == Axiom == <onlyinclude> Let $A$ and $B$ be classes. Then: :$\forall x: \paren {x \in A \iff x \in B} \iff A = B$ </onlyinclude> Hence the order in which the elements are listed in the sets is immaterial. == Also known as == {{:Axiom:Axiom of Extension/Also known as}} == Also see == * Definition:Class Equality * Definition:Equals == Linguistic Note == {{:Axiom:Axiom of Extension/Linguistic Note}} == Sources == * {{BookReference|Set Theory and the Continuum Problem|2010|Raymond M. Smullyan|author2 = Melvin Fitting|ed = revised|edpage = Revised Edition|prev = Definition:Subclass|next = Definition:Class Equality/Definition 2}}: Chapter $2$: Some Basics of Class-Set Theory: $\S 1$ Extensionality and separation Extension Extension Extension d49c7lcjvs32dhy04hu4mqht7odovxl


Axiom:Axiom of Extension/Set Theory

Axiom:Axiom of Extension/Set Theory 100 83836 497668 491976 2020-11-03T06:13:01Z Prime.mover 59 wikitext text/x-wiki == Axiom == <onlyinclude> Let $A$ and $B$ be sets. The '''axiom of extension''' states that $A$ and $B$ are equal {{iff}} they contain the same elements. That is, {{iff}}: :every element of $A$ is also an element of $B$ and: :every element of $B$ is also an element of $A$. This can be formulated as follows: </onlyinclude> === Formulation 1 === <onlyinclude>{{:Axiom:Axiom of Extension/Set Theory/Formulation 1}}</onlyinclude> === Formulation 2 === In set theories that define $=$ instead of admitting it as a primitive, the '''axiom of extension''' can be formulated as: {{:Axiom:Axiom of Extension/Set Theory/Formulation 2}} The order of the elements in the sets is immaterial. Hence a set is completely and uniquely determined by its elements. == Also known as == {{:Axiom:Axiom of Extension/Also known as}} == Also see == * Definition:Set Equality * Definition:Equals * Axiom:Axiom of Extension (Classes) == Linguistic Note == {{:Axiom:Axiom of Extension/Linguistic Note}} == Sources == * {{BookReference|Naive Set Theory|1960|Paul R. Halmos|prev = Definition:Set Equality/Definition 1|next = Definition:Subset}}: $\S 1$: The Axiom of Extension * {{BookReference|Abstract Algebra|1964|W.E. Deskins|prev = Definition:Uniqueness of Set Elements|next = Axiom:Axiom of Extension/Also known as}}: Chapter $1$: A Common Language: $\S 1.1$ Sets * {{BookReference|Elements of Abstract Algebra|1966|Richard A. Dean|prev = Definition:Set Equality/Definition 1|next = Definition:Explicit Set Definition}}: $\S 0.2$. Sets * {{BookReference|The Joy of Sets: Fundamentals of Contemporary Set Theory|1993|Keith Devlin|ed = 2nd|edpage = Second Edition|prev = Definition:Set Equality/Definition 1|next = Axiom:Axiom of Extension/Set Theory/Formulation 1}}: $\S 1$: Naive Set Theory: $\S 1.1$: What is a Set? Extension Extension Extension Extension ezzwxmd0yb8s2uo5azlvo4angx795ly


Axiom:Axiom of Specification/Class Theory 100 83846 444613 444609 2020-01-23T20:39:57Z Prime.mover 59 wikitext text/x-wiki == Axiom == The '''axiom of specification''' is an axiom schema which can be formally stated as follows: <onlyinclude> Let $\map \phi {A_1, A_2, \ldots, A_n, x}$ be a function of propositional logic such that: :$A_1, A_2, \ldots, A_n$ are a finite number of free variables whose domain ranges over all classes :$x$ is a free variable whose domain ranges over all sets. Then the '''axiom of specification''' gives that: :$\forall A_1, A_2, \ldots, A_n: \exists B: \forall x: \paren {x \in B \iff \paren {x \in B \land \phi {A_1, A_2, \ldots, A_n, x} } }$ where each of $B$ ranges over arbitrary classes. </onlyinclude> This means that for any finite number $A_1, A_2, \ldots, A_n$ of subclasses of the universal class $V$, the class  $B$ exists (or can be formed) of all sets $x \in V$ that satisfy the function $\map \phi {A_1, A_2, \ldots, A_n, x}$. == Also known as == {{:Axiom:Axiom of Specification/Also known as}} == Also see == * Axiom:Axiom of Specification/Set Theory == Historical Note == {{:Axiom:Axiom of Specification/Historical Note}} {{Languages|Axiom of specification}} {{language|German|Aussonderungsaxiom|lit = axiom of segregation}} {{end-languages}} == Sources == * {{BookReference|Set Theory and the Continuum Problem|2010|Raymond M. Smullyan|author2 = Melvin Fitting|ed = revised|edpage = Revised Edition|prev = Axiom:Axiom of Specification/Also known as|next = Not Every Class is a Set/Proof 1}}: Chapter $2$: Some Basics of Class-Set Theory: $\S 1$ Extensionality and separation Specification Specification Specification 4ucgl8faz73dvwupsgvw7quikszyn4c


Axiom:Axiom of Transitivity

Axiom:Axiom of Transitivity 100 83863 449994 449988 2020-02-19T10:27:16Z Prime.mover 59 wikitext text/x-wiki == Axiom == Let $V$ be a basic universe. <onlyinclude> :$V$ is a transitive class. </onlyinclude> That is, every set $S$ which is an element of $V$ is a subclass of $V$. Briefly: :Every set is a class. == Sources == * {{BookReference|Set Theory and the Continuum Problem|2010|Raymond M. Smullyan|author2 = Melvin Fitting|ed = revised|edpage = Revised Edition|prev = Definition:Basic Universe Axioms|next = Axiom:Axiom of Swelledness}}: Chapter $2$: Some Basics of Class-Set Theory: $\S 2$ Transitivity and supercompleteness Transitivity scdhzlpwz76qf6x9m0n8b032aanphlz


Axiom:Axiom of Swelledness

Axiom:Axiom of Swelledness 100 83864 449996 449987 2020-02-19T10:27:52Z Prime.mover 59 wikitext text/x-wiki == Axiom == Let $V$ be a basic universe. <onlyinclude> :$V$ is a swelled class. </onlyinclude> That is, every subclass of a set which is an element of $V$ is a set in $V$. Briefly: :Every subclass of a set is a set. == Sources == * {{BookReference|Set Theory and the Continuum Problem|2010|Raymond M. Smullyan|author2 = Melvin Fitting|ed = revised|edpage = Revised Edition|prev = Axiom:Axiom of Transitivity|next = Basic Universe is Supercomplete}}: Chapter $2$: Some Basics of Class-Set Theory: $\S 2$ Transitivity and supercompleteness Swelledness qts7d0jpsbn35fi3ipb1mrhwouvkz41


Axiom:Axiom of Empty Set/Class Theory

Axiom:Axiom of Empty Set/Class Theory 100 83881 449995 449989 2020-02-19T10:27:32Z Prime.mover 59 wikitext text/x-wiki == Axiom == Let $V$ be a basic universe. <onlyinclude> The empty class $\O$ is a set, that is: :$\O \in V$ </onlyinclude> == Sources == * {{BookReference|Set Theory and the Continuum Problem|2010|Raymond M. Smullyan|author2 = Melvin Fitting|ed = revised|edpage = Revised Edition|prev = Empty Class is Supercomplete|next = Basic Universe is not Empty}}: Chapter $2$: Some Basics of Class-Set Theory: $\S 3$ Axiom of the empty set Empty Set Empty Set Empty Set 5q3ryuct4vv49oes7o7j8g6p3i4z23x


Axiom:Axiom of Pairing/Set Theory/Formulation 1 100 83902 491372 444586 2020-09-28T07:53:10Z Prime.mover 59 wikitext text/x-wiki == Axiom == <onlyinclude> For any two sets, there exists a set to which only those two sets are elements: :$\forall A: \forall B: \exists x: \forall y: \paren {y \in x \iff y = A \lor y = B}$ Thus it is possible to create a set that contains as elements any two sets that have already been created. </onlyinclude> == Also known as == {{:Axiom:Axiom of Pairing/Also known as}} == Also see == * Equivalence of Definitions of Axiom of Pairing == Sources == * {{BookReference|Point Set Topology|1964|Steven A. Gaal|prev = Set Definition by Predicate/Examples/Sums of Two Squares|next = Definition:Doubleton}}: Introduction to Set Theory: $1$. Elementary Operations on Sets * {{MathWorld|Zermelo-Fraenkel Axioms|Zermelo-FraenkelAxioms}} * {{MathWorld|Axiom of the Unordered Pair|AxiomoftheUnorderedPair}} Pairing 0vps8rk1gp0i13mvkkqa7sam5szgj7u


Axiom:Axiom of Pairing/Set Theory/Formulation 2 100 83903 444588 444570 2020-01-23T20:23:15Z Prime.mover 59 Prime.mover moved page Axiom:Axiom of Pairing/Sets/Formulation 2 to Axiom:Axiom of Pairing/Set Theory/Formulation 2 wikitext text/x-wiki == Axiom == <onlyinclude> For any two sets, there exists a set containing those two sets as elements: :$\forall A: \forall B: \exists x: \forall y: \paren {y \in x \implies y = A \lor y = B}$ Thus it is possible to create a set that contains as elements '''at least''' two sets that have already been created. </onlyinclude> == Also known as == {{:Axiom:Axiom of Pairing/Also known as}} == Also see == * Equivalence of Definitions of Axiom of Pairing == Sources == * {{BookReference|Naive Set Theory|1960|Paul R. Halmos|prev = Empty Set is Subset of All Sets/Proof 2|next = Definition:Doubleton}}: $\S 3$: Unordered Pairs Pairing 4kdr8tfqdpzc7yqjhyixitdy2i6nkbp


Axiom:Axiom of Pairing/Class Theory

Axiom:Axiom of Pairing/Class Theory 100 83907 449997 449990 2020-02-19T10:28:05Z Prime.mover 59 wikitext text/x-wiki == Axiom == === Formulation 1 === <onlyinclude>{{:Axiom:Axiom of Pairing/Class Theory/Formulation 1}}</onlyinclude> === Formulation 2 === {{:Axiom:Axiom of Pairing/Class Theory/Formulation 2}} == Also known as == {{:Axiom:Axiom of Pairing/Also known as}} == Also see == * Equivalence of Definitions of Axiom of Pairing for Classes * Definition:Doubleton Class * Doubleton Class can be Formed from Two Sets, which demonstrates the fact that the class $\set {a, b}$ can be created in the first place Pairing Pairing Pairing Pairing 0dvx4rpabm6uigw4fmmjdmfferroicn


Axiom:Axiom of Pairing/Class Theory/Formulation 1 100 83923 444674 444667 2020-01-23T23:19:24Z Prime.mover 59 wikitext text/x-wiki == Axiom == <onlyinclude> Let $a$ and $b$ be sets. Then the class $\set {a, b}$ is likewise a set. </onlyinclude> == Also known as == {{:Axiom:Axiom of Pairing/Also known as}} == Also see == * Equivalence of Definitions of Axiom of Pairing for Classes == Sources == * {{BookReference|Set Theory and the Continuum Problem|2010|Raymond M. Smullyan|author2 = Melvin Fitting|ed = revised|edpage = Revised Edition|prev = Doubleton Class of Equal Sets is Singleton Class|next = Axiom:Axiom of Pairing/Class Theory/Formulation 2}}: Chapter $2$: Some Basics of Class-Set Theory: $\S 4$ The pairing axiom Pairing b0jm1h7muszglmtaayg9uzvuujle2at


Axiom:Axiom of Pairing/Class Theory/Formulation 2 100 83924 444676 444675 2020-01-23T23:20:12Z Prime.mover 59 wikitext text/x-wiki == Axiom == <onlyinclude> Let $a$ and $b$ be sets. Then there exists a set $c$ such that $a \in c$ and $b \in c$. </onlyinclude> == Also known as == {{:Axiom:Axiom of Pairing/Also known as}} == Also see == * Equivalence of Definitions of Axiom of Pairing for Classes == Sources == * {{BookReference|Set Theory and the Continuum Problem|2010|Raymond M. Smullyan|author2 = Melvin Fitting|ed = revised|edpage = Revised Edition|prev = Axiom:Axiom of Pairing/Class Theory/Formulation 1|next = Equivalence of Definitions of Axiom of Pairing for Classes}}: Chapter $2$: Some Basics of Class-Set Theory: $\S 4$ The pairing axiom Pairing b0z2difchl5tiyxr25667hu45jlh9bi


Axiom:Axiom of Unions/Class Theory

Axiom:Axiom of Unions/Class Theory 100 83953 449993 449991 2020-02-19T10:26:56Z Prime.mover 59 wikitext text/x-wiki == Axiom == <onlyinclude> Let $x$ be a set (of sets). Then its union $\displaystyle \bigcup x$ is also a set. </onlyinclude> == Also known as == {{:Axiom:Axiom of Unions/Also known as}} == Sources == * {{BookReference|Set Theory and the Continuum Problem|2010|Raymond M. Smullyan|author2 = Melvin Fitting|ed = revised|edpage = Revised Edition|prev = Definition:Set Union/General Definition|next = Definition:Intersection of Class}}: Chapter $2$: Some Basics of Class-Set Theory: $\S 5$ The union axiom Unions Unions Unions Unions 1r5hxyqtplv3nii7ppvijkwqih0qqga


Axiom:Axiom of Powers/Class Theory

Axiom:Axiom of Powers/Class Theory 100 83988 473014 449992 2020-06-10T02:05:34Z RandomUndergrad 3904 wikitext text/x-wiki == Axiom == <onlyinclude> Let $x$ be a set. Then its power set $\powerset x$ is also a set. </onlyinclude> == Also known as == {{:Axiom:Axiom of Powers/Also known as}} == Sources == * {{BookReference|Set Theory and the Continuum Problem|2010|Raymond M. Smullyan|author2 = Melvin Fitting|ed = revised|edpage = Revised Edition|prev = Power Set Exists and is Unique|next = Axiom:Axiom of Powers/Set Theory}}: Chapter $2$: Some Basics of Class-Set Theory: $\S 6$ The power axiom :::''although there exists a misprint: it is referred to as the '''Power aet axiom'''.'' Powers Powers Powers Powers lbb7ba64npqzd7u1kflb2nwjvfmvn82


Chain Rule

Chain Rule 0 84211 447653 447652 2020-02-09T01:09:31Z Prime.mover 59 wikitext text/x-wiki {{Disambiguation}} * Chain Rule for Derivatives * Chain Rule for Probability ropx1clmxaww5s5fmzg03to31b4a8kc


Projection from Product Topology is Continuous/General Result/Proof 0 84475 449480 2020-02-16T13:02:03Z Prime.mover 59 Created page with "== Theorem == {{:Projection from Product Topology is Continuous/General Result}}  == Proof == <onlyinclude> By definition of the Definition:Tychonoff Topology|Tychonoff topo..." wikitext text/x-wiki == Theorem == {{:Projection from Product Topology is Continuous/General Result}} == Proof == <onlyinclude> By definition of the Tychonoff topology on $S$: :$\tau$ is the initial topology on $S$ with respect to $\family {\pr_i}_{i \mathop \in I}$ By definition of the Initial Topoplogy:Definition 2: :$\tau$ is the coarsest topology on $S$ such that each $\pr_i: S \to S_i$ is a $\struct{\tau, \tau_i}$-continuous. </onlyinclude>{{qed}} Category:Projection from Product Topology is Open and Continuous t3mxbrerdivq2amckxjvljr3exiz6zm


Projection from Product Topology is Open/General Result/Proof 0 84476 449481 2020-02-16T13:08:46Z Prime.mover 59 Created page with "== Theorem == {{:Projection from Product Topology is Open/General Result}}  == Proof == <onlyinclude> Let $U \in \tau$.  It follows from the definition of Definition:Tychono..." wikitext text/x-wiki == Theorem == {{:Projection from Product Topology is Open/General Result}} == Proof == <onlyinclude> Let $U \in \tau$. It follows from the definition of Tychonoff topology that $U$ can be expressed as: :$\displaystyle U = \bigcup_{j \mathop \in J} \bigcap_{k \mathop = 1}^{n_j} \map {\pr_{i_{k, j} }^{-1} } {U_{k, j} }$ where: :$J$ is an arbitrary index set :$n_j \in \N$ :$i_{k, j} \in I$ :$U_{k, j} \in \tau_{i_{k, j} }$. For all $i' \in I$, define $V_{i', k, j} \in \tau_{i'}$ by: :$V_{i', k, j} = \begin {cases} U_{k, j} & : i' = i_{k, j} \\ S_{i'} & : i' \ne i_{k, j} \end {cases}$ For all $i \in I$, we have: {{begin-eqn}} {{eqn | l = \map {\pr_i} U       | r = \bigcup_{j \mathop \in J} \map {\pr_i} {\bigcap_{k \mathop = 1}^{n_j} \map {\pr_{i_{k,j} }^{-1} } { U_{k,j} } }       | c = Image of Union under Relation: Family of Sets }} {{eqn | r = \bigcup_{j \mathop \in J} \map {\pr_i} {\bigcap_{k \mathop = 1}^{n_j} \prod_{i' \mathop \in I} V_{i', k, j} }       | c = {{Defof|Projection (Mapping Theory)|Projection}} }} {{eqn | r = \bigcup_{j \mathop \in J} \map {\pr_i} {\prod_{i' \mathop \in I} \bigcap_{k \mathop = 1}^{n_j} V_{i', k, j} }       | c = Cartesian Product of Intersections: General Case }} {{eqn | r = \bigcup_{j \mathop \in J} \bigcap_{k \mathop = 1}^{n_j} V_{i,k,j}       | c = {{Defof|Projection (Mapping Theory)|Projection}} }} {{end-eqn}} As: :$\displaystyle \bigcup_{j \mathop \in J} \bigcap_{k \mathop = 1}^{n_j} V_{i, k, j} \in \tau_i$ it follows that $\pr_i$ is open. </onlyinclude>{{qed}} Category:Projection from Product Topology is Open and Continuous id4pkftu0oqbob7rnlrhlpcasyrdq3a


Axiom:Axiom of Infinity/Class Theory

Axiom:Axiom of Infinity/Class Theory 100 84547 449998 449984 2020-02-19T10:28:41Z Prime.mover 59 wikitext text/x-wiki == Axiom == === Formulation 1 === <onlyinclude>{{:Axiom:Axiom of Infinity/Class Theory/Formulation 1}}</onlyinclude> === Formulation 2 === {{:Axiom:Axiom of Infinity/Class Theory/Formulation 2}} === Formulation 3 === {{:Axiom:Axiom of Infinity/Class Theory/Formulation 3}} == Also see == * Equivalence of Formulations of Axiom of Infinity for Zermelo Universe Infinity Infinity Infinity klx7xc3gld1khpe2k745q44yanxdsls


Axiom:Axiom of Infinity/Class Theory/Formulation 1 100 84548 450002 449985 2020-02-19T10:36:44Z Prime.mover 59 wikitext text/x-wiki == Axiom == <onlyinclude> Let $\omega$ be the class of natural numbers as constructed by the Von Neumann construction: {{begin-eqn}} {{eqn | l = 0       | o = :=       | r = \O       | c =  }} {{eqn | l = 1       | o = :=       | r = 0 \cup \set 0       | c =  }} {{eqn | l = 2       | o = :=       | r = 1 \cup \set 1       | c =  }} {{eqn | l = 3       | o = :=       | r = 2 \cup \set 2       | c =  }} {{eqn | o = \vdots       | c =  }} {{eqn | l = n + 1       | o = :=       | r = n \cup \set n       | c =  }} {{eqn | o = \vdots       | c =  }} {{end-eqn}} Then $\omega$ is a set. </onlyinclude> == Also see == * Equivalence of Formulations of Axiom of Infinity for Zermelo Universe == Sources == * {{BookReference|Set Theory and the Continuum Problem|2010|Raymond M. Smullyan|author2 = Melvin Fitting|ed = revised|edpage = Revised Edition|prev = Definition:Inductive Set Definition for Natural Numbers|next = Definition:Zermelo Universe}}: Chapter $3$: The Natural Numbers: $\S 2$ Definition of the Natural Numbers: Axiom $A_7$ Infinity Infinity mhsvoa9flwdf1bidqd73qx0ix0q0dsl


Axiom:Axiom of Infinity/Class Theory/Formulation 2 100 84551 450010 450004 2020-02-19T11:04:10Z Prime.mover 59 wikitext text/x-wiki == Axiom == <onlyinclude> There exists an inductive set. </onlyinclude> == Also see == * Equivalence of Formulations of Axiom of Infinity for Zermelo Universe == Sources == * {{BookReference|Set Theory and the Continuum Problem|2010|Raymond M. Smullyan|author2 = Melvin Fitting|ed = revised|edpage = Revised Edition|prev = Definition:Zermelo Universe|next = Axiom:Axiom of Infinity/Class Theory/Formulation 3/Historical Note}}: Chapter $3$: The Natural Numbers: $\S 2$ Definition of the Natural Numbers Infinity Infinity lu67jeskhv591puclw4kt2ka86u3h0s


Axiom:Axiom of Infinity/Class Theory/Formulation 3 100 84554 450009 450007 2020-02-19T11:03:55Z Prime.mover 59 wikitext text/x-wiki == Axiom == <onlyinclude> Not every set is a natural number. </onlyinclude> == Also see == * Equivalence of Formulations of Axiom of Infinity for Zermelo Universe == Historical Note == {{:Axiom:Axiom of Infinity/Class Theory/Formulation 3/Historical Note}} == Sources == * {{BookReference|Set Theory and the Continuum Problem|2010|Raymond M. Smullyan|author2 = Melvin Fitting|ed = revised|edpage = Revised Edition|prev = Axiom:Axiom of Infinity/Class Theory/Formulation 3/Historical Note|next = Equivalence of Formulations of Axiom of Infinity for Zermelo Universe}}: Chapter $3$: The Natural Numbers: $\S 2$ Definition of the Natural Numbers Infinity Infinity 5rohtd602vj2snp31dz1wzeorlmvcax


Condition for Expectation of Non-Negative Random Variable to be Zero 0 85335 455455 2020-03-18T11:29:40Z Caliburn 3218 Created page with "== Theorem == <onlyinclude> Let $X$ be a random variable.   Let:   :$\map \Pr {X \ge 0} = 1$   Then $\expect X = 0$ {{iff}} $\map \Pr {X = 0} =..." wikitext text/x-wiki == Theorem == <onlyinclude> Let $X$ be a random variable.  Let:  :$\map \Pr {X \ge 0} = 1$ Then $\expect X = 0$ {{iff}} $\map \Pr {X = 0} = 1$. </onlyinclude> == Proof == {{ProofWanted}} Category:Expectation fnf3epu60to075znwj0nr2hoeytp6ot


Covariance of Linear Combination of Random Variables with Another 0 85343 455483 2020-03-18T12:17:31Z Caliburn 3218 Created page with "== Theorem == <onlyinclude> Let $X, Y, Z$ be random variables.   Let $a, b$ be real numbers.   Then:   :$\cov {a X +..." wikitext text/x-wiki == Theorem == <onlyinclude> Let $X, Y, Z$ be random variables.  Let $a, b$ be real numbers. Then:  :$\cov {a X + b Y, Z} = a \cov {X, Z} + b \cov {Y, Z}$ </onlyinclude> == Proof == {{begin-eqn}} {{eqn	| l = \cov {a X + b Y, Z} 	| r = \expect {\paren {a X + b Y} Z} - \expect {a X + b Y} \expect Z 	| c = Covariance as Expectation of Product minus Product of Expectations }} {{eqn	| r = a \expect {X Z} + b \expect {Y Z} - \paren {a \expect X + b \expect Y} \expect Z 	| c = Linearity of Expectation Function }} {{eqn	| r = a \paren {\expect {X Z} - \expect X \expect Z} + b \paren {\expect {Y Z} - \expect Y \expect Z} }} {{eqn	| r = a \cov {X, Z} + b \cov {Y, Z} 	| c = Covariance as Expectation of Product minus Product of Expectations }} {{end-eqn}} {{qed}} Category:Covariance clomckc88pxcsw54tysvlx7zl4j66xw


Covariance of Sums of Random Variables/Lemma 0 85345 455505 2020-03-18T14:14:08Z Caliburn 3218 Created page with "== Theorem == <onlyinclude> Let $n$ be a strictly positive integer.  Let $\sequence {X_i}_{1 \le i \le n}$ be a Definition:Sequence|..." wikitext text/x-wiki == Theorem == <onlyinclude> Let $n$ be a strictly positive integer. Let $\sequence {X_i}_{1 \le i \le n}$ be a sequence of random variables. Let $Y$ be a random variable. Then:  :$\displaystyle \cov {\sum_{i \mathop = 1}^n X_i, Y} = \sum_{i \mathop = 1}^n \cov {X_i, Y}$ </onlyinclude>  == Proof == Proof by induction: For all $n \in \N$, let $\map P n$ be the proposition:  :$\displaystyle \cov {\sum_{i \mathop = 1}^n X_i, Y} = \sum_{i \mathop = 1}^n \cov {X_i, Y}$ === Basis for the Induction === We have that:  :$\displaystyle \cov {\sum_{i \mathop = 1}^n X_i, Y} = \cov {X_1, Y} = \sum_{i \mathop = 1}^1 \cov {X_i, Y}$ We therefore have that $\map P 1$ is true.  This is our base case. === Induction Hypothesis === Suppose that $\map P n$ is true for some fixed $n \in \N$.  That is:  :$\displaystyle \cov {\sum_{i \mathop = 1}^n X_i, Y} = \sum_{i \mathop = 1}^n \cov {X_i, Y}$ We aim to show that it logically follows that $\map P {n + 1}$ is true.  That is:  :$\displaystyle \cov {\sum_{i \mathop = 1}^{n + 1} X_i, Y} = \sum_{i \mathop = 1}^{n + 1} \cov {X_i, Y}$ === Induction Step === This is our induction step: We have:  {{begin-eqn}} {{eqn	| l = \cov {\sum_{i \mathop = 1}^{n + 1} X_i, Y} 	| r = \cov {\sum_{i \mathop = 1}^n X_i + X_{n + 1}, Y} 	| c = splitting up the sum }} {{eqn	| r = \cov {\sum_{i \mathop = 1}^n X_i, Y} + \cov {X_{n + 1}, Y} 	| c = Covariance of Linear Combination of Random Variables with Another }} {{eqn	| r = \sum_{i \mathop = 1}^n \cov {X_i, Y} + \cov {X_{n + 1}, Y} 	| c = induction hypothesis }} {{eqn	| r = \sum_{i \mathop = 1}^{n + 1} \cov {X_i, Y} }} {{end-eqn}} Hence the result by induction. {{qed}} Category:Covariance Category:Covariance of Sums of Random Variables di2bhly3lry6nhiskt1rq3obzfzf9i3


Expectation of Linear Transformation of Random Variable 0 85348 455529 2020-03-18T16:10:35Z Caliburn 3218 Created page with "== Theorem == <onlyinclude> Let $X$ be a random variable.   Let $a, b$ be real numbers.    Then we have:   :$\expect..." wikitext text/x-wiki == Theorem == <onlyinclude> Let $X$ be a random variable.  Let $a, b$ be real numbers.  Then we have:  :$\expect {a X + b} = a \expect X + b$ where $\expect X$ denotes the expectation of $X$.  </onlyinclude> == Proof == === Discrete Random Variable === {{:Expectation of Linear Transformation of Random Variable/Discrete}} === Continuous Random Variable === {{:Expectation of Linear Transformation of Random Variable/Continuous}} Category:Expectation Category:Expectation of Linear Transformation of Random Variable s4z0ah7c7vdt0d5628c128pjokk1yvv


Definite Integral of Odd Function/Corollary 0 85650 457201 2020-03-23T11:39:22Z Prime.mover 59 Created page with "== Corollary to Definite Integral of Odd Function == <onlyinclude> Let $f$ be an odd function with a Definition:Primitive (Calculus)|primitiv..." wikitext text/x-wiki == Corollary to Definite Integral of Odd Function == <onlyinclude> Let $f$ be an odd function with a primitive on the open interval $\openint {-a} a$, where $a > 0$. Then the improper integral of $f$ on $\openint {-a} a$ is: :$\displaystyle \int_{\mathop \to -a}^{\mathop \to a} \map f x \rd x = 0$ </onlyinclude> == Proof == {{begin-eqn}} {{eqn | l = \int_{\mathop \to -a}^{\mathop \to a} \map f x \rd x       | r = \lim_{y \mathop \to a} \int_{-y}^y \map f x \rd x       | c = {{Defof|Improper Integral over Open Interval}} }} {{eqn | r = \lim_{y \mathop \to a} 0       | c = Definite Integral of Odd Function }} {{eqn | r = 0       | c =  }} {{end-eqn}} {{qed}} Category:Integral Calculus Category:Odd Functions Category:Improper Integrals ds0u8xk48wh9kabypnu744covo100ww


Coulomb's Law of Electrostatics

Coulomb's Law of Electrostatics 0 86443 462481 462451 2020-04-17T07:38:30Z Prime.mover 59 wikitext text/x-wiki == Physical Law == :400pxrightthumbForce Between two Like Charges Let $a$ and $b$ be stationary particles in a vacuum, each carrying an electric charge of $q_a$ and $q_b$ respectively. Then $a$ and $b$ exert a force upon each other whose magnitude and direction are given by '''Coulomb's law (of electrostatics)''': :$\mathbf F_{a b} \propto \dfrac {q_a q_b {\mathbf r_{a b} } } {r^3}$ where: :$\mathbf F_{a b}$ is the force exerted on $b$ by the electric charge on $a$ :$\mathbf r_{a b}$ is the displacement vector from $a$ to $b$ :$r$ is the distance between $a$ and $b$. :the constant of proportion is defined as being positive. By exchanging $a$ and $b$ in the above, it is seen that $b$ exerts the same force on $a$ as $a$ does on $b$, but in the opposite direction. == SI Units == {{:Coulomb's Law of Electrostatics/SI Units}} Thus the equation becomes: :$\mathbf F_{a b} = \dfrac 1 {4 \pi \varepsilon_0} \dfrac {q_a q_b {\mathbf r_{a b} } } {r^3}$ == Also presented as == :$\mathbf F_{a b} \propto \dfrac {q_a q_b \hat {\mathbf r}_{a b} } {r^2}$ where $\hat {\mathbf r}_{a b}$ is the unit vector in the direction from $a$ to $b$. == Also known as == '''Coulomb's Law of Electrostatics''' is also known as just '''Coulomb's Law'''. {{Namedfor|Charles-Augustin de Coulomb|cat = Coulomb}} == Historical Note == {{:Coulomb's Law of Electrostatics/Historical Note}} == Sources == * {{BookReference|Electromagnetism|1990|I.S. Grant|author2 = W.R. Phillips|ed = 2nd|edpage = Second Edition|prev = Newton's Law of Universal Gravitation|next = Newton's Law of Universal Gravitation/Historical Note}}: Chapter $1$: Force and energy in electrostatics * {{BookReference|Electromagnetism|1990|I.S. Grant|author2 = W.R. Phillips|ed = 2nd|edpage = Second Edition|prev = Definition:Negative Electric Charge|next = Definition:Unit Vector}}: Chapter $1$: Force and energy in electrostatics: $1.1$ Electric Charge Category:Electrostatics Category:Coulomb's Law of Electrostatics 4l3qloy2d1vr2or0tyutli4spkychjd


Bijection/Examples/Real Cube Function

Bijection/Examples/Real Cube Function 0 86552 463120 2020-04-20T09:42:42Z Prime.mover 59 Created page with "== Example of Bijection == <onlyinclude> Let $f: \R \to \R$ be the mapping defined on the Definition:Real Numbers|set of real..." wikitext text/x-wiki == Example of Bijection == <onlyinclude> Let $f: \R \to \R$ be the mapping defined on the set of real numbers as: :$\forall x \in \R: \map f x = x^3$ Then $f$ is a bijection. </onlyinclude> == Proof == A direct application of Integer Power Function is Bijective iff Index is Odd. {{qed}} == Sources == * {{BookReference|Topology: An Introduction with Application to Topological Groups|1967|George McCarty|prev = Inverse of Identity Mapping|next = Inverse Mapping/Examples/Real Cube Function}}: Chapter $\text{I}$: Sets and Functions: Composition of Functions Category:Examples of Bijections r50xzx9y89yeysojmrvmxud5nr5raov


Limit of Subsequence equals Limit of Sequence/Normed Vector Space 0 86622 463604 2020-04-22T17:22:12Z Julius 3095 Created page with "== Theorem == <onlyinclude> Let $\struct {X, \norm {\, \cdot \,} }$ be a normed vector space.  Let $\sequence {x_n}$ be a Definition:Seque..." wikitext text/x-wiki == Theorem == <onlyinclude> Let $\struct {X, \norm {\, \cdot \,} }$ be a normed vector space. Let $\sequence {x_n}$ be a sequence in $X$. Let $\sequence {x_n}$ be convergent in the norm $\norm {\, \cdot \,}$ to the following limit: :$\displaystyle \lim_{n \mathop \to \infty} x_n = l$ Let $\sequence {x_{n_r} }$ be a subsequence of $\sequence {x_n}$. Then: :$\sequence {x_{n_r} }$ is convergent and $\displaystyle \lim_{r \mathop \to \infty} x_{n_r} = l$ </onlyinclude> That is, the limit of a convergent sequence equals the limit of a subsequence of it. == Proof == Let $\epsilon > 0$. Since $\displaystyle \lim_{n \mathop \to \infty} x_n = l$, it follows from the definition of limit that: :$\exists N \in \N : \forall n \in \N : n > N \implies \norm{x_n - l} < \epsilon$ Now let $R = N$. Then from Strictly Increasing Sequence of Natural Numbers: : $\forall r > R: n_r \ge r$ Thus $n_r > N$ and so: :$\norm {x_n - l} < \epsilon$ The result follows. {{qed}} Category:Normed Vector Spaces Category:Convergence Category:Limits of Sequences 8oblsbs01cotfach8fs1lr7skim9xsd


Natural Numbers under Addition do not form Group/Corollary 0 86692 492381 464126 2020-10-03T15:43:14Z Prime.mover 59 wikitext text/x-wiki == Corollary to Natural Numbers under Addition do not form Group == <onlyinclude> The algebraic structure $\struct {\Z_{\ge 0}, +}$ consisting of the set of non-negative integers $\Z_{\ge 0}$ under addition $+$ does not form a subgroup of the additive group of integers. </onlyinclude> == Proof == By Natural Numbers are Non-Negative Integers, $\struct {\Z_{\ge 0}, +}$ and $\struct {\N, +}$ are the same (or if not exactly the same, at least isomorphic). The result follows from Natural Numbers under Addition do not form Group. {{qed}} == Sources == * {{BookReference|Topology: An Introduction with Application to Topological Groups|1967|George McCarty|prev = Group Product Identity therefore Inverses/Part 2/Proof 2|next = Subgroups of Additive Group of Integers/Examples/Even Integers}}: Chapter $\text{II}$: Groups: Exercise $\text{E i}$ Category:Natural Numbers 8drkvaruqw9i95ilcv9v2r7eibuglwc


Existence and Uniqueness of Direct Limit of Sequence of Groups/Lemma 1 0 87099 466013 465985 2020-05-05T09:53:58Z Prime.mover 59 wikitext text/x-wiki {{MissingLinks}} == Lemma == On $\widehat G_\infty := \displaystyle \coprod_{n \mathop \in \N} G_n$ the relation: :$\tuple {x_n, n} \sim \tuple {y_m, m} \iff \exists k \ge n, m: \map {g_{n, k} } {x_n} = \map {g_{m, k} } {y_m}$ is an equivalence relation. == Proof == === Reflexivity === Since $g_{n,n} = \mathop {Id}_{G_n}$ we have: :$\forall \tuple {x_n, n} \in \widehat G_\infty: \map {g_{n, n} } {x_n} = \map {g_{n, n} } {x_n}$ Hence: :$\tuple {x_n, n} \sim \tuple {x_n, n}$ {{qed|lemma}} === Symmetry === Let $\tuple {x_n, n} \sim \tuple {y_m, m}$. Then there exists a $k \ge n, m$ such that: :$\map {g_{n, k} } {x_n} = \map {g_{m, k} } {x_m}$ Hence also: :$\map {g_{m, k} } {x_m} = \map {g_{n, k} } {x_n}$ That is: :$\tuple {y_m, m} \sim \tuple {x_n, n}$ {{qed|lemma}} === Transitivity === Let $\tuple {x_n, n} \sim \tuple {y_m, m}$ and  $\tuple {y_m, m} \sim \tuple {z_r, r}$. Then there exist $k \ge m, n$ and $l \ge n, r$ such that: :$\map {g_{n, k} } {x_n} = \map {g_{m, k} } {y_m}$ :$\map {g_{m, l} } {y_m} = \map {g_{r, l} } {z_r}$ Let $q:= \max \set {k, l}$. Then we have: {{begin-eqn}} {{eqn | l = \map {g_{n, q} } {x_n}       | r = \map {g_{k, q} } {\map {g_{m, k} } {y_m} }       | c =  }} {{eqn | r = \map {g_{m, q} } {y_m}       | c =  }} {{eqn | r = \map {g_{l,q} } {\map {g_{m, l} } {y_m} }       | c =  }} {{eqn | r = \map {g_{l, q} } {\map {g_{r, l} } {z_r} }       | c =  }} {{eqn | r = \map {g_{r,q} } {z_r}       | c =  }} {{end-eqn}} that is: :$\tuple {x_n, n} \sim \tuple {z_r, r}$ {{qed}} Category:Existence and Uniqueness of Direct Limit of Sequence of Groups fir5rrdz7eqpwgerchypsxfad5j9us8


Existence and Uniqueness of Direct Limit of Sequence of Groups/Lemma 2 0 87100 466052 466043 2020-05-05T12:42:39Z Prime.mover 59 wikitext text/x-wiki {{tidy}} {{MissingLinks}} == Lemma == The following defines a group structure on $G_\infty$:  Let $\struct {G_\infty, \cdot}$ be the algebraic structure defined as follows. Let $\eqclass {\tuple {x_n, n} } {}, \eqclass {\tuple {y_m, m} } {} \in G_\infty$ be arbitrary elements of $G_\infty$. Let $l := \max \set {m, n}$. Let the operation $\cdot$ on $G_\infty$ be defined as: :$\tuple {\eqclass {\tuple {x_n, n} } {} \cdot \eqclass {\tuple {y_m, m} } {} } := \eqclass {\tuple {\map {g_{n l} } {x_n} \map {g_{m l} } {y_m}, l} } {}$ Then $\struct {G_\infty, \cdot}$ is a group. == Proof == === Well-Definedness === The definition depends on the choice $\tuple {x_n, n}$ and $\tuple {y_m, m}$ of representatives of $\eqclass {\tuple {x_n, n} } {}$ and $\eqclass {\tuple {y_m, m} } {}$.  We have to show that the product element is independent of this choice. Let $\tuple {x_{n'}, n'}$ and $\tuple {y_{m'}, m'}$ be different representatives of the chosen equivalence classes.  Let $l' := \max \set {n', m'}$. {{WLOG}}, suppose that $l' \ge l$. We have that: :$\tuple {x_n, n} \sim \tuple {x_{n'}, n'}$ and: :$\tuple {y_m, m} \sim \tuple {y_{m'}, m'}$ 		 and so: :$\map {g_{n, l'} } {x_n} = \map {g_{n', l'} } {x_{n'} }$ and: :$\map {g_{m, l'} } {y_m} = \map {g_{m', l'} } {y_{m'} }$ 		 Then we have, since all our maps are group homomorphisms: {{begin-eqn}} {{eqn | l = \map {g_{ l, l'} } {\map {g_{n, l} } {x_n} \map {g_{m, l} } {y_m} }       | r = \map {g_{ l, l'} } {\map {g_{n, l} } {x_n} } \map{ g_{ l,l' } } { \map{ g_{ m,l } } {y_m} }       | c = }} {{eqn | r = \map {g_{n, l'} } {x_n} \map {g_{m, l'} } {y_m}        | c =  }} {{eqn | r = \map{ g_{ n',l' } } { x_{ n' } } \map{ g_{m',l'} } { y_{ m' } }       | c = }} 		 {{end-eqn}}	 that is: :$\map {g_{n, l} } {x_n} \map {g_{m,l} } {y_m} \sim \map {g_{ n', l'} } {x_{n'} } \map {g_{ m',l' } } {y_{m'} }$ This proves that our definition is independent of the choice of representative. {{qed|lemma}} === Group Axioms === By the definition of the group operation, we may assume, without loss of generality, that the representatives are always in the same group $G_l \in \sequence {G_n}_{n \mathop \in \N}$. To see this we note that we always consider a finite collection of group elements  :$\{ \eqclass{ \tuple{ x_{n_1}, {n_1} } }{}, \dots, \eqclass{ \tuple{ x_{n_k}, {n_k} } }{} \} \subset G_\infty$. Define $l:= \max\{n_1,\dots, n_k\}$. Then  :$\forall i \in \{1,\dots, k\} :  \map{ g_{n_i,l} } { x_{n_i} } \in G_n \land \tuple{ x_{n_1}, {n_1} } \sim \tuple{ \map{ g_{n_i,l}  } { x_{n_i} }, l}$ ==== $\text G 1$: Associativity ==== Let $\eqclass{\tuple{x_n, n}}{},\eqclass{\tuple{y_m, m}}{},\eqclass{\tuple{y_n, n}}{},\eqclass{\tuple{z_n, n}}{} \in G_\infty$.  Then:  {{begin-eqn}} {{eqn | l = \paren{\eqclass{\tuple{x_n, n} }{} \cdot \eqclass{ \tuple{y_n, n} }{} } \cdot \eqclass{ \tuple{z_n, n} }{}       | r = \eqclass{ \tuple{ x_n y_n, n} }{} \cdot \eqclass{ \tuple{z_n, n} }{}          | c =  }} {{eqn | r = \eqclass{ \tuple{ \paren{x_n y_n}z_n, n} }{}       | c =  }} {{eqn | r = \eqclass{ \tuple{ x_n \paren{y_n z_n}, n} }{}       | c = $G_n$ is a group }} {{eqn | r =\eqclass{ \tuple{ x_n, n} }{} \cdot \eqclass{ \tuple{ y_n z_n, n} }{}        | c =  }} {{eqn | r = \eqclass{ \tuple{ x_n, n} }{} \cdot \paren{\eqclass{ \tuple{ y_n, n} }{} \cdot \eqclass{ \tuple{ z_n, n} }{} }       | c =  }} {{end-eqn}} ==== $\text G 2$: Identity ==== Let $\eqclass{\tuple{x_n, n}}{} \in G_\infty$ and let $1_n$ be the identity of $G_n$.  Note that :$\forall k, n \in \N : \paren{ k \ge n \implies  \map{ g_{nk} }{1_n} = 1_k}$ because the maps $g_{nk}$ are group homomorphisms. Then:  {{begin-eqn}} {{eqn | l = \eqclass{\tuple{x_n, n} }{} \cdot \eqclass{ \tuple{1_n, n} }{}        | r = \eqclass{\tuple{x_n 1_n, n} }{}        | c =  }} {{eqn | r = \eqclass{\tuple{x_n, n} }{}       | c =  }} {{end-eqn}} Similarly we also find that $\eqclass{\tuple{1_n, n} }{} \cdot \eqclass{\tuple{x_n, n} }{} = \eqclass{\tuple{x_n, n} }{}$. Thus $\eqclass{\tuple{1_n, n} }{}$ is the identity of $G_\infty$. ==== $\text G 3$: Inverses ==== Let $\eqclass{\tuple{x_n, n}}{} \in G_\infty$. Then : {{begin-eqn}} {{eqn | l = \eqclass{\tuple{x_n, n} }{} \cdot \eqclass{ \tuple{x_n^{-1}, n} }{}        | r = \eqclass{\tuple{x_n x^{-1}_n, n} }{}        | c =  }} {{eqn | r = \eqclass{\tuple{1_n, n} }{}       | c =  }} {{end-eqn}} Similarly we also find that $\eqclass{\tuple{x^{-1}_n, n} }{} \cdot \eqclass{\tuple{x_n, n} }{} = \eqclass{\tuple{1_n, n} }{}$. Thus $\eqclass{\tuple{x_n, n} }{}$ has an inverse, that is: :$\eqclass{\tuple{x_n^{-1}, n} }{}$ in $G_\infty$. {{qed}} Category:Existence and Uniqueness of Direct Limit of Sequence of Groups 656h564i43fjh6iurxncjg9e44p4vb4


Existence and Uniqueness of Direct Limit of Sequence of Groups/Lemma 3 0 87101 466053 466031 2020-05-05T12:44:48Z Prime.mover 59 wikitext text/x-wiki {{MissingLinks|It is important for all pages to be self-contained, so we need to include on this page the required definitions from its parent page.}} == Lemma == Let $h_\infty: G_\infty \to H$ be the mapping defined as: :$\eqclass {\tuple {x_n, n} } {} \mapsto \map {h_n} {x_n}$ Then $h_\infty$ is a well-defined group homomorphism. == Proof == === Well-Definedness of $h_\infty$ === Let $\tuple {x_n, n}, \tuple{x_{n'}, n'} \in \eqclass {\tuple {x_n, n} } {}$. {{WLOG}}, let $n' \ge n$.  Then we have: :$\map {g_{n, n'} } {x_n} = x_{n'}$ and: {{begin-eqn}} {{eqn | l = \map {h_{n'} } {x_{n'} }        | r = \map {h_{n'} } {\map {g_{n,n'} } {x_n} } }} {{eqn | r = \map {\paren {h_{n'} \circ g_{n, n'} } } {x_n}        | c = because $h_{n'} \circ g_{n, n'} = h_n$ }} {{eqn | r = \map {h_n} {x_n} }} {{end-eqn}} This proves that $h_\infty$ is independent of the representative chosen. That is, $h_\infty$ is well-defined. {{qed|lemma}} === Homomorphism Property === Let $\eqclass {\tuple{x_n, n} } {}, \eqclass {\tuple {y_m, m} } {} \in G_\infty$. By the definition of the group operation, we may assume, without loss of generality, that $n = m$. See Lemma 2 for details. {{handwaving|Can that be demonstrated?}} It follows that: {{begin-eqn}} {{eqn | l = \map {h_{\infty} } {\eqclass {\tuple {x_n, n} } {} \circ \eqclass {\tuple {y_n, n} } {} }       | r = \map {h_n} {x_n y_n}       | c = Definition of $h_\infty$ }} {{eqn | r = \map {h_n} {x_n} \map {h_n} {y_n}       | c = $h_n$ is a homomorphism }} {{eqn | r = \map {h_\infty} {\eqclass {\tuple {x_n, n} } {} } \circ \map {h_\infty} {\eqclass {\tuple {y_n, n} } {} }       | c = Definition of $h_\infty$ }} {{end-eqn}} Thus $h_\infty$ is a homomorphism. {{qed}} Category:Existence and Uniqueness of Direct Limit of Sequence of Groups nswa1slfide05n79yjxj9m0v5r6vdcj


Schwarz's Lemma/Lemma

Schwarz's Lemma/Lemma 0 87118 466065 2020-05-05T14:18:32Z Caliburn 3218 Created page with "== Theorem == <onlyinclude> Let $D$ be the unit disk centred at $0$.  Let $g : D \to \C$ be a complex function with:..." wikitext text/x-wiki == Theorem == <onlyinclude> Let $D$ be the unit disk centred at $0$. Let $g : D \to \C$ be a complex function with:  :$\map g z = \begin{cases}\frac {\map f z} z & z \ne 0 \\ \map {f'} 0 & z = 0\end{cases}$ Then $g$ is holomorphic on $D$. </onlyinclude> == Proof == By Differentiable Function is Continuous, $f$ is continuous, so by Combination Theorem for Continuous Functions: Quotient Rule: :$g$ is continuous on $D \setminus \set 0$. We aim to show that $f$ is continuous on $D$.  Note that since $f$ is holomorphic on $D$ and $0 \in D$ we have, by the definition of the complex derivative:  :$\displaystyle \lim_{z \mathop \to 0} \frac {\map f z - \map f 0} z = \map {f'} 0 \in \C$ Since $\map f 0 = 0$, we furthermore have:  :$\displaystyle \map {f'} 0 = \lim_{z \mathop \to 0} \frac {\map f z} z$ That is:  :$\displaystyle \map g 0 = \lim_{z \mathop \to 0} \map g z$ so $g$ is continuous at $0$. Since $f$ is holomorphic on $D$, by the Combination Theorem for Complex Derivatives: Quotient Rule: :$g$ is differentiable on $D \setminus \set 0$. It remains to show that $g$ is differentiable at $0$.  Take $z \ne 0$ and consider:  :$\dfrac {\map g z - \map g 0} z$ We have:  {{begin-eqn}} {{eqn	| l = \frac {\map g z - \map g 0} z 	| r = \frac {\frac {\map f z} z - \map {f'} 0} z 	| c = as $\map g z = \dfrac {\map f z} z$ for $z \ne 0$ and $\map g 0 = \map {f'} 0$ }} {{eqn	| r = \frac {\map f z - z \map {f'} 0} {z^2} }} {{end-eqn}} Since $f$ is holomorphic on $D$, by Holomorphic Function is Analytic, there exists a positive real number $R$ such that the series:  :$\displaystyle \sum_{n \mathop = 0}^\infty \frac {\map {f^{\paren n} } 0} {n!} z^n$ converges to $\map f z$ on $\cmod z < R$. Note that since $\map f 0 = 0$, the first term of this series is zero. With that, we have:  {{begin-eqn}} {{eqn	| l = \frac {\map f z - z \map {f'} 0} {z^2} 	| r = \frac {\sum_{n \mathop = 1}^\infty \frac {\map {f^{\paren n} } 0} {n!} z^n - z \map {f'} 0} {z^2} }} {{eqn	| r = \frac {z \map {f'} 0 + \frac {z^2} 2 \map {f''} 0 + \sum_{n \mathop = 3}^\infty \frac {\map {f^{\paren n} } 0} {n!} z^n - z \map {f'} 0} {z^2} }} {{eqn	| r = \frac 1 2 \map {f''} 0 + \sum_{n \mathop = 3}^\infty \frac {\map {f^{\paren n} } 0} {n!} z^{n - 2} }} {{end-eqn}} Taking $z \to 0$ we have:  :$\displaystyle \lim_{z \mathop \to 0} \frac {\map g z - \map g 0} z = \frac 1 2 \map {f''} 0$ so $g$ is indeed differentiable at $0$ and hence holomorphic on $D$. {{qed}} Category:Complex Analysis Category:Schwarz's Lemma 1o8sfwvysfvchvd5xuw4lup45d07t6p


Partial Derivative/Examples/x^(x y)/wrt x 0 87256 466578 2020-05-08T07:34:37Z Prime.mover 59 Created page with "== Example of Partial Derivative ==  Let $\map f {x, y} = x^{x y}$ be a Definition:Real Function of Two Variables|real function of $2$ vari..." wikitext text/x-wiki == Example of Partial Derivative == Let $\map f {x, y} = x^{x y}$ be a real function of $2$ variables such that $x, y \in \R_{>0}$. Then: :$\dfrac {\partial f} {\partial x} = x^{x y} \paren {y \ln x + y}$ == Proof == <onlyinclude> By definition, the partial derivative {{WRT|Differentiation}} $x$ is obtained by holding $y$ constant. Hence Derivative of $x^{a x}$ can be directly used: :$\dfrac \d {\d x} x^{y x} = y x^{y x} \paren {\ln x + 1}$ The result can then be rearranged to match the form given. {{qed}} </onlyinclude>  == Sources == * {{BookReference|Advanced Calculus|1961|David V. Widder|ed = 2nd|edpage = Second Edition|prev = Partial Derivative/Examples/x^(x y)|next = Partial Derivative/Examples/x^(x y)/wrt y}}: $1$ Partial Differentiation: $\S 1$. Introduction: $1.1$ Partial Derivatives: Example $\text A$ Category:Examples of Partial Derivatives okref889tzskah8eegtcifw5thw7tbe


Partial Derivative/Examples/x^(x y)/wrt y 0 87257 466680 466583 2020-05-08T14:23:04Z Prime.mover 59 wikitext text/x-wiki == Example of Partial Derivative == Let $\map f {x, y} = x^{x y}$ be a real function of $2$ variables such that $x, y \in \R_{>0}$. Then: :$\dfrac {\partial f} {\partial y} = x^{x y + 1} \ln x$ == Proof == <onlyinclude> By definition, the partial derivative {{WRT|Differentiation}} $y$ is obtained by holding $x$ constant. From Derivative of Power of Constant: :$\map {D_y} {x^y} = x^y \ln x$ for constant $a$. Then: {{begin-eqn}} {{eqn | l = \map {D_y} {x^{x y} }       | r = x \map {D_{x y} } {x^{x y} }       | c = Derivative of Function of Constant Multiple }} {{eqn | r = x \paren {x^{x y} } \ln x       | c =  }} {{eqn | r = x^{x y + 1} \ln x       | c =  }} {{end-eqn}} {{qed}} </onlyinclude>  == Sources == * {{BookReference|Advanced Calculus|1961|David V. Widder|ed = 2nd|edpage = Second Edition|prev = Partial Derivative/Examples/x^(x y)/wrt x|next = Partial Derivative/Examples/x sine y z}}: $1$ Partial Differentiation: $\S 1$. Introduction: $1.1$ Partial Derivatives: Example $\text A$ Category:Examples of Partial Derivatives lleozosukca2zilh8np4wdog51r2uuu


Partial Derivative/Examples/u + ln u = x y 0 87272 466715 2020-05-08T22:40:08Z Prime.mover 59 Created page with "== Example of Partial Derivative == <onlyinclude> Let $u + \ln u = x y$ be an implicit function.  Th..." wikitext text/x-wiki == Example of Partial Derivative == <onlyinclude> Let $u + \ln u = x y$ be an implicit function. Then: {{begin-eqn}} {{eqn | l = \dfrac {\partial u} {\partial x}       | r = \dfrac {u y} {u + 1}       | c =  }} {{eqn | l = \dfrac {\partial u} {\partial y}       | r = \dfrac {u x} {u + 1}       | c =  }} {{end-eqn}} </onlyinclude> == Proof == {{begin-eqn}} {{eqn | l = u + \ln u       | r = x y       | c =  }} {{eqn | ll= \leadsto       | l = \dfrac \partial {\partial x} u + \dfrac \partial {\partial x} \ln u       | r = \dfrac \partial {\partial x} x y       | c =  }} {{eqn | ll= \leadsto       | l = \dfrac {\partial u} {\partial x} + \dfrac 1 u \dfrac {\partial u} {\partial x}       | r = y       | c =  }} {{eqn | ll= \leadsto       | l = \dfrac {\partial u} {\partial x} \paren {1 + \dfrac 1 u}       | r = y       | c =  }} {{eqn | ll= \leadsto       | l = \dfrac {\partial u} {\partial x}       | r = \dfrac {u y} {u + 1}       | c =  }} {{end-eqn}} and: {{begin-eqn}} {{eqn | l = u + \ln u       | r = x y       | c =  }} {{eqn | ll= \leadsto       | l = \dfrac \partial {\partial y} u + \dfrac \partial {\partial y} \ln u       | r = \dfrac \partial {\partial y} x y       | c =  }} {{eqn | ll= \leadsto       | l = \dfrac {\partial u} {\partial y} + \dfrac 1 u \dfrac {\partial u} {\partial y}       | r = x       | c =  }} {{eqn | ll= \leadsto       | l = \dfrac {\partial u} {\partial y} \paren {1 + \dfrac 1 u}       | r = x       | c =  }} {{eqn | ll= \leadsto       | l = \dfrac {\partial u} {\partial y}       | r = \dfrac {u x} {u + 1}       | c =  }} {{end-eqn}} {{qed}} == Sources == * {{BookReference|Advanced Calculus|1961|David V. Widder|ed = 2nd|edpage = Second Edition|prev = Partial Derivative/Examples/u^2 + x^2 + y^2 = a^2|next = Partial Derivative/Examples/v + ln u = x y, u + ln v = x - y}}: $1$ Partial Differentiation: $\S 1$. Introduction: $1.2$ Implicit Functions Category:Examples of Partial Derivatives fvlsrhhtykoswl8uglatw8x5ai8pi4y


Function of Bounded Variation is Bounded 0 87489 467886 2020-05-14T12:35:35Z Caliburn 3218 Created page with "== Theorem == <onlyinclude> Let $a, b$ be real numbers with $a < b$.  Let $f : \closedint a b \to \R$ be a function of..." wikitext text/x-wiki == Theorem == <onlyinclude> Let $a, b$ be real numbers with $a < b$. Let $f : \closedint a b \to \R$ be a function of bounded variation. Then $f$ is bounded. </onlyinclude> == Proof == We use the notation from the definition of bounded variation. Since $f$ is of bounded variation, there exists $M \in \R$ such that: :$\map {V_f} P \le M$ for all finite subdivisions $P$ of $\closedint a b$. Let $x$ be a real number with: :$a < x < b$ Then $\set {a, x, b}$ is a finite subdivision of $\closedint a b$. We have:  :$\map {V_f} {\set {a, x, b} } = \size {\map f x - \map f a} + \size {\map f b - \map f x}$ Since $x \in \openint a b$ was arbitrary, we therefore have:  :$\size {\map f x - \map f a} + \size {\map f b - \map f x} \le M$ for all $x \in \openint a b$. We have: {{begin-eqn}} {{eqn	| l = \size {\map f x - \map f a} + \size {\map f b - \map f x} 	| o = \ge 	| r = \size {\map f x - \map f a} }} {{eqn	| o = \ge 	| r = \size {\size {\map f x} - \size {\map f a} } 	| c = Reverse Triangle Inequality: Real and Complex Fields }} {{eqn	| o = \ge 	| r = \size {\map f x} - \size {\map f a} }} {{end-eqn}} So for all $x \in \openint a b$, we have:  :$\size {\map f x} \le \size {\map f a} + M$ Since $M \ge 0$, this inequality is also satisfied for $x = a$.  We therefore have:  :$\size {\map f x} \le \map \max {\size {\map f a} + M, \size {\map f b} }$ for all $x \in \closedint a b$. So $f$ is is bounded. {{qed}} == Sources == * {{BookReference|Mathematical Analysis|1973|Tom M. Apostol|prev = Differentiable Function with Bounded Derivative is of Bounded Variation|next = Definition:Total Variation|ed = 2nd|edpage = Second Edition}}: $\S 6.3$: Functions of Bounded Variation: Theorem $6.7$ Category:Bounded Variation atydoyyyc1561qo943yi4zocd8iv4p0


Multiple of Function of Bounded Variation is of Bounded Variation 0 87633 469076 2020-05-19T14:59:04Z Caliburn 3218 Created page with "== Theorem == <onlyinclude> Let $a, b, k$ be real numbers with $a < b$.  Let $f : \closedint a b \to \R$ be a functions..." wikitext text/x-wiki == Theorem == <onlyinclude> Let $a, b, k$ be real numbers with $a < b$. Let $f : \closedint a b \to \R$ be a functions of bounded variation. Let the total variations of $f$ be $V_f$. Then $k f$ is of bounded variation with:  :$V_{k f} = \size k V_f$ where $V_{k f}$ is the total variation of $k f$. </onlyinclude> == Proof == For each finite subdivision $P$ of $\closedint a b$, write:  :$P = \set {x_0, x_1, \ldots, x_n }$ with: :$a = x_0 < x_1 < x_2 < \cdots < x_{n - 1} < x_n = b$ Then: {{begin-eqn}} {{eqn	| l = \map {V_{k f} } P 	| r = \sum_{i \mathop = 1}^n \size {k \map f {x_i} - k \map f {x_{i - 1} } } 	| c = using the notation from the definition of bounded variation }} {{eqn	| r = \size k \sum_{i \mathop = 1}^n \size {\map f {x_i} - \map f {x_{i - 1} } } }} {{eqn	| r = \size k \map {V_f} P }} {{end-eqn}} Since $f$ is of bounded variation, there exists $M \in \R$ such that:  :$\map {V_f} P \le M$ for all finite subdivisions $P$.  So: :$\map {V_{k f} } P \le \size k M$ So $k f$ is of bounded variation. We then have:  {{begin-eqn}} {{eqn	| l = V_{k f} 	| r = \sup_P \paren {\map {V_{k f} } P} 	| c = {{Defof|Total Variation}} }} {{eqn	| r = \sup_P \paren {\size k \map {V_f} P} }} {{eqn	| r = \size k \sup_P \paren {\map {V_f} P} 	| c = Multiple of Supremum }} {{eqn	| r = \size k V_f }} {{end-eqn}} {{qed}} Category:Bounded Variation Category:Total Variation tm2b1jjxz079gs06iegfka06ihtvebc


Constant Real Function is Absolutely Continuous 0 87716 470257 2020-05-24T14:39:47Z Caliburn 3218 Created page with "== Theorem == <onlyinclude> Let $I \subseteq \R$ be a real interval.  Let $f : I \to \R$ be an Definition:Constant Mapping|constant real functio..." wikitext text/x-wiki == Theorem == <onlyinclude> Let $I \subseteq \R$ be a real interval. Let $f : I \to \R$ be an constant real function. Then $f$ is absolutely continuous. </onlyinclude> == Proof == Let $\delta, \varepsilon$ be positive real numbers. Let $\closedint {a_1} {b_1}, \dotsc, \closedint {a_n} {b_n} \subseteq I$ be a collection of disjoint closed real intervals with:  :$\displaystyle \sum_{i \mathop = 1}^n \paren {b_i - a_i} < \delta$ Since $f$ is constant, for all $i \in \set {1, 2, \ldots, n}$ we have: :$\size {\map f {b_i} - \map f {a_i} } = 0$ so: :$\displaystyle \sum_{i = 1}^n \size {\map f {b_i} - \map f {a_i} } = 0 < \varepsilon$ Since $\varepsilon$ was arbitrary:  :$f$ is absolutely continuous. {{qed}} Category:Absolutely Continuous Functions gxiz5ue31jl2hsjw5qjzh2h1iwsifgs


Multiple of Absolutely Continuous Function is Absolutely Continuous 0 87718 470270 2020-05-24T15:56:02Z Caliburn 3218 Created page with "== Theorem == <onlyinclude> Let $k$ be a real number.  Let $I \subseteq \R$ be a real interval.  Let $f : I \to \R$ be..." wikitext text/x-wiki == Theorem == <onlyinclude> Let $k$ be a real number. Let $I \subseteq \R$ be a real interval. Let $f : I \to \R$ be absolutely continuous function. Then $k f$ is absolutely continuous. </onlyinclude> == Proof == Note that if $k = 0$, then $k f$ is constant.  Hence, by Constant Real Function is Absolutely Continuous:  :$k f$ is absolutely continuous if $k = 0$. Take now $k \ne 0$. Let $\varepsilon$ be a positive real number. Since $f$ is absolutely continuous, there exists real $\delta > 0$ such that for all collections of disjoint closed real intervals $\closedint {a_1} {b_1}, \dotsc, \closedint {a_n} {b_n} \subseteq I$ with:  :$\displaystyle \sum_{i \mathop = 1}^n \paren {b_i - a_i} < \delta$ we have:  :$\displaystyle \sum_{i \mathop = 1}^n \size {\map f {b_i} - \map f {a_i} } < \frac {\varepsilon} {\size k}$ Then: {{begin-eqn}} {{eqn	| l = \sum_{i \mathop = 1}^n \size {\map {\paren {k f} } {b_i} - \map {\paren {k f} } {a_i} } 	| r = \size k \sum_{i \mathop = 1}^n \size {\map f {b_i} - \map f {a_i} } }} {{eqn	| o = < 	| r = \size k \times \frac {\varepsilon} {\size k} }} {{eqn	| r = \varepsilon }} {{end-eqn}} whenever: :$\displaystyle \sum_{i \mathop = 1}^n \paren {b_i - a_i} < \delta$ Since $\varepsilon$ was arbitrary:  :$k f$ is absolutely continuous if $k \ne 0$. Therefore:  :$k f$ is absolutely continuous for all $k \in \R$. {{qed}} Category:Absolutely Continuous Functions 0fnxe36o4cbzvbtdwfv9uhnt7ari2tz


Elementary Row Operation/Examples/r3 + 2r2 0 88375 476229 476182 2020-06-26T21:46:13Z Prime.mover 59 wikitext text/x-wiki == Examples of Elementary Row Operations == <onlyinclude> Consider the elementary row operation $e$ defined as: :$e := r_3 \to r_3 + 2 r_2$ acting on a matrix space $\map \MM {3, n}$ for some $n \in \Z_{>0}$. The elementary row matrix corresponding to $e$ is: :$\begin {pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 2 & 1 \end {pmatrix}$ </onlyinclude> == Proof == Let $\mathbf E$ denote the elementary row matrix corresponding to $e$. From Elementary Matrix corresponding to Elementary Row Operation: Scale Row and Add: :$E_{a b} = \delta_{a b} + 2 \delta_{a 3} \cdot \delta_{2 b}$ where: :$E_{a b}$ denotes the element of $\mathbf E$ whose indices are $\tuple {a, b}$ :$\delta_{a b}$ is the Kronecker delta: ::$\delta_{a b} = \begin {cases} 1 & : \text {if $a = b$} \\ 0 & : \text {if $a \ne b$} \end {cases}$ That is: :When $a \ne 3$ and $b \ne 2$, elements of $\mathbf E$ are $0$ except for those on the main diagonal :When $a = 3$ and $b = 2$, $E_{a b}$ equals $2 + \delta_{a b}$. But as $a \ne b$ it follows that $\delta_{a b} = 0$. Hence $\mathbf E$ can be constructed as described. {{qed}} == Sources == * {{BookReference|Linear Algebra|1998|Richard Kaye|author2 = Robert Wilson|prev = Elementary Row Operations as Matrix Multiplications|next = Elementary Row Operation/Examples/lambda r2}}: Part $\text I$: Matrices and vector spaces: $1$ Matrices: $1.5$ Row and column operations: $1$ Category:Examples of Elementary Row Operations pze5znfymx9cgsoxx9krw861l622ojf


Elementary Row Operation/Examples/lambda r2 0 88376 476177 474167 2020-06-26T19:39:05Z Prime.mover 59 wikitext text/x-wiki == Examples of Elementary Row Operations == <onlyinclude> Consider the elementary row operation $e$ defined as: :$e := r_2 \to \lambda r_2$ acting on a matrix space $\map \MM {3, n}$ for some $n \in \Z_{>0}$. The elementary row matrix corresponding to $e$ is: :$\begin {pmatrix} 1 & 0 & 0 \\ 0 & \lambda & 0 \\ 0 & 0 & 1 \end {pmatrix}$ </onlyinclude> == Proof == Let $\mathbf E$ denote the elementary row matrix corresponding to $e$. From Elementary Matrix corresponding to Elementary Row Operation: Scale Row and Add: :$E_{a b} = \begin {cases} \delta_{a b} & : a \ne 2 \\ \lambda \cdot \delta_{a b} & : a = 2 \end{cases}$ where: :$E_{a b}$ denotes the element of $\mathbf E$ whose indices are $\tuple {a, b}$ :$\delta_{a b}$ is the Kronecker delta: ::$\delta_{a b} = \begin {cases} 1 & : \text {if $a = b$} \\ 0 & : \text {if $a \ne b$} \end {cases}$ That is: :When $a \ne 2$, the elements of $\mathbf E$ are $0$ except for those on the main diagonal, when they are $1$ :When $a = 2$, the elements of $\mathbf E$ are $0$ except for those on the main diagonal, when they are $\lambda$. Hence $\mathbf E$ can be constructed as described. {{qed}} == Sources == * {{BookReference|Linear Algebra|1998|Richard Kaye|author2 = Robert Wilson|prev = Elementary Row Operation/Examples/r3 + 2r2|next = Elementary Row Operation/Examples/Swap r1 and r2}}: Part $\text I$: Matrices and vector spaces: $1$ Matrices: $1.5$ Row and column operations: $2$ Category:Examples of Elementary Row Operations aqxgu67irgyx1epnqekbhtdfw41zxwb


Elementary Row Operation/Examples/Swap r1 and r2 0 88377 476183 474168 2020-06-26T19:42:13Z Prime.mover 59 wikitext text/x-wiki == Examples of Elementary Row Operations == <onlyinclude> Consider the elementary row operation $e$ defined as: :$e := r_1 \leftrightarrow r_2$ acting on a matrix space $\map \MM {3, n}$ for some $n \in \Z_{>0}$. The elementary row matrix corresponding to $e$ is: :$\begin {pmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end {pmatrix}$ </onlyinclude> == Proof == By definition, the elementary row matrix corresponding to $e$ is found by applying $e$ to the unit matrix. By definition of unit matrix: :$\mathbf I = \begin {pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end {pmatrix}$ Let $\mathbf E$ denote the elementary row matrix corresponding to $e$. $\mathbf E$ is constructed by exchanging row $1$ with row $2$. {{qed}} == Sources == * {{BookReference|Linear Algebra|1998|Richard Kaye|author2 = Robert Wilson|prev = Elementary Row Operation/Examples/lambda r2|next = Definition:Row Operation}}: Part $\text I$: Matrices and vector spaces: $1$ Matrices: $1.5$ Row and column operations: $3$ Category:Examples of Elementary Row Operations ge9rfuj2jkwfytemgp4pnri8bevnl66


Matrix is Row Equivalent to Echelon Matrix/Examples/Arbitrary Matrix 1 0 88443 474449 2020-06-17T06:35:30Z Prime.mover 59 Created page with "== Examples of Use of Matrix is Row Equivalent to Echelon Matrix == <onlyinclude> Let $\mathbf A = \begin {bmatrix} 0 & 1 & 1 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \\ \end {bmatrix}$..." wikitext text/x-wiki == Examples of Use of Matrix is Row Equivalent to Echelon Matrix == <onlyinclude> Let $\mathbf A = \begin {bmatrix} 0 & 1 & 1 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \\ \end {bmatrix}$ This can be converted into the echelon form: :$\mathbf E = \begin {bmatrix} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end {bmatrix}$ </onlyinclude> == Proof == Using Row Operation to Clear First Column of Matrix we obtain: :$\mathbf B = \begin {bmatrix} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 1 & 1 \\ \end {bmatrix}$ which happens on the first step, exchanging row $1$ with row $3$. Then we investigate the submatrix: :$\mathbf B' = \begin {bmatrix} 1 & 0 \\ 1 & 1 \\ \end {bmatrix}$ Using Row Operation to Clear First Column of Matrix we obtain: :$\mathbf C' = \begin {bmatrix} 1 & 0 \\ 0 & 1 \\ \end {bmatrix}$ which is obtained by adding $-1$ of row $1$ of $\mathbf B'$ to row $2$ of $\mathbf B'$. This is the same as adding $-1$ of row $2$ of $\mathbf B$ to row $3$ of $\mathbf B$. Thus we are left with: :$\mathbf E = \begin {bmatrix} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end {bmatrix}$ {{qed}} == Sources == * {{BookReference|Linear Algebra|1998|Richard Kaye|author2 = Robert Wilson|prev = Matrix is Row Equivalent to Echelon Matrix/Examples|next = Matrix is Row Equivalent to Echelon Matrix/Examples/Arbitrary Matrix 2}}: Part $\text I$: Matrices and vector spaces: $1$ Matrices: $1.5$ Row and column operations: Exercise $1.4 \ \text {(a)}$ Category:Matrix is Row Equivalent to Echelon Matrix q1s1xe6u108ddrmiocao6mc40fo9ieu


Matrix is Row Equivalent to Echelon Matrix/Examples/Arbitrary Matrix 2 0 88444 474450 2020-06-17T06:39:44Z Prime.mover 59 Created page with "== Examples of Use of Matrix is Row Equivalent to Echelon Matrix == <onlyinclude> Let $\mathbf A = \begin {bmatrix} 1 & 1 & 1 & 1 \\ 2 & 3 & 4 & 5 \\ 3 & 4 & 5 & 6 \\ \end..." wikitext text/x-wiki == Examples of Use of Matrix is Row Equivalent to Echelon Matrix == <onlyinclude> Let $\mathbf A = \begin {bmatrix} 1 & 1 & 1 & 1 \\ 2 & 3 & 4 & 5 \\ 3 & 4 & 5 & 6 \\ \end {bmatrix}$ This can be converted into the echelon form: :$\mathbf E = \begin {bmatrix} 1 & 1 & 1 & 1 \\ 0 & 1 & 2 & 3 \\ 0 & 0 & 0 & 0 \\ \end {bmatrix}$ </onlyinclude> == Proof == Using Row Operation to Clear First Column of Matrix we obtain: :$\mathbf B = \begin {bmatrix} 1 & 1 & 1 & 1 \\ 0 & 1 & 2 & 3 \\ 0 & 1 & 2 & 3 \\ \end {bmatrix}$ which is obtained by: :adding $-2$ of row $1$ to row $2$ :adding $-3$ of row $1$ to row $3$. Then we investigate the submatrix: :$\mathbf B' = \begin {bmatrix} 1 & 2 & 3 \\ 1 & 2 & 3 \\ \end {bmatrix}$ Using Row Operation to Clear First Column of Matrix we obtain: :$\mathbf C' = \begin {bmatrix} 1 & 2 & 3 \\ 0 & 0 & 0 \\ \end {bmatrix}$ which is obtained by adding $-1$ of row $1$ of $\mathbf B'$ to row $2$ of $\mathbf B'$. Thus we are left with: :$\mathbf E = \begin {bmatrix} 1 & 1 & 1 & 1 \\ 0 & 1 & 2 & 3 \\ 0 & 0 & 0 & 0 \\ \end {bmatrix}$ {{qed}} == Sources == * {{BookReference|Linear Algebra|1998|Richard Kaye|author2 = Robert Wilson|prev = Matrix is Row Equivalent to Echelon Matrix/Examples/Arbitrary Matrix 1|next = Matrix is Row Equivalent to Echelon Matrix/Examples/Arbitrary Matrix 3}}: Part $\text I$: Matrices and vector spaces: $1$ Matrices: $1.5$ Row and column operations: Exercise $1.4 \ \text {(b)}$ Category:Matrix is Row Equivalent to Echelon Matrix 4umk6y8jvs2i4g6ijtc9oo6yuahan7o


Matrix is Row Equivalent to Echelon Matrix/Examples/Arbitrary Matrix 3 0 88445 474451 2020-06-17T06:52:35Z Prime.mover 59 Created page with "== Examples of Use of Matrix is Row Equivalent to Echelon Matrix == <onlyinclude> Let $\mathbf A = \begin {bmatrix} 1 & 2 & 3 & 5 \\ 1 & 2 & 3 & 4 \\ 0 & 0 & 1 & 1 \\ \end..." wikitext text/x-wiki == Examples of Use of Matrix is Row Equivalent to Echelon Matrix == <onlyinclude> Let $\mathbf A = \begin {bmatrix} 1 & 2 & 3 & 5 \\ 1 & 2 & 3 & 4 \\ 0 & 0 & 1 & 1 \\ \end {bmatrix}$ This can be converted into the echelon form: :$\mathbf E = \begin {bmatrix} 1 & 2 & 3 & 5 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 \\ \end {bmatrix}$ </onlyinclude> == Proof == Using Row Operation to Clear First Column of Matrix we obtain: :$\mathbf B = \begin {bmatrix} 1 & 2 & 3 & 5 \\ 0 & 0 & 0 & -1 \\ 0 & 0 & 1 & 1 \\ \end {bmatrix}$ which is obtained by adding $-2$ of row $1$ to row $2$. Then we investigate the submatrix: :$\mathbf B' = \begin {bmatrix} 0 & 0 & -1 \\ 0 & 1 & 1 \\ \end {bmatrix}$ Using Row Operation to Clear First Column of Matrix we obtain: :$\mathbf C' = \begin {bmatrix} 0 & 1 & 1 \\ 0 & 0 & 1 \\ \end {bmatrix}$ which is obtained by: :$(1): \quad$ exchanging row $1$ of $\mathbf B'$ with row $2$ of $\mathbf B'$. :$(2): \quad$ multiplying row $2$ of $\mathbf B'$ by $-1$. Thus we are left with: :$\mathbf E = \begin {bmatrix} 1 & 2 & 3 & 5 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 \\ \end {bmatrix}$ {{qed}} == Also presented as == Some sources use the non-unity variant of the echelon matrix. Such sources do not require that the leading coefficients necessarily have to equal to $1$. Hence they consider the final step to convert row $3$ of $\mathbf E$ from $\begin {bmatrix} 0 & 0 & 0 & -1 \end {bmatrix}$ to $\begin {bmatrix} 0 & 0 & 0 & 1 \end {bmatrix}$ as optional. == Sources == * {{BookReference|Linear Algebra|1998|Richard Kaye|author2 = Robert Wilson|prev = Matrix is Row Equivalent to Echelon Matrix/Examples/Arbitrary Matrix 2|next = Matrix is Row Equivalent to Echelon Matrix/Examples/Arbitrary Matrix 4}}: Part $\text I$: Matrices and vector spaces: $1$ Matrices: $1.5$ Row and column operations: Exercise $1.4 \ \text {(c)}$ Category:Matrix is Row Equivalent to Echelon Matrix j5ktt0est6bisyso96lj4ls8hqanuxe


Matrix is Row Equivalent to Echelon Matrix/Examples/Arbitrary Matrix 4 0 88446 474452 2020-06-17T06:57:05Z Prime.mover 59 Created page with "== Examples of Use of Matrix is Row Equivalent to Echelon Matrix == <onlyinclude> Let $\mathbf A = \begin {bmatrix} 1 & 1 & 2 & 3 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 3 & 3 \\ \end..." wikitext text/x-wiki == Examples of Use of Matrix is Row Equivalent to Echelon Matrix == <onlyinclude> Let $\mathbf A = \begin {bmatrix} 1 & 1 & 2 & 3 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 3 & 3 \\ \end {bmatrix}$ This can be converted into the echelon form: :$\mathbf E = \begin {bmatrix} 1 & 1 & 2 & 3 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 \\ \end {bmatrix}$ </onlyinclude> == Proof == It is noted that $\mathbf A$ is already most of the way there. It remains to use the elementary row operation: :$e := r_3 \to r_3 - 3 r_2$ to convert $\mathbf A$ to the form: :$\mathbf E = \begin {bmatrix} 1 & 1 & 2 & 3 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 \\ \end {bmatrix}$ {{qed}} == Sources == * {{BookReference|Linear Algebra|1998|Richard Kaye|author2 = Robert Wilson|prev = Matrix is Row Equivalent to Echelon Matrix/Examples/Arbitrary Matrix 3|next = Matrix is Row Equivalent to Echelon Matrix/Examples/Arbitrary Matrix 5}}: Part $\text I$: Matrices and vector spaces: $1$ Matrices: $1.5$ Row and column operations: Exercise $1.4 \ \text {(d)}$ Category:Matrix is Row Equivalent to Echelon Matrix gjjde47xvgm9q7oghx3f8qxbkms6pr4


Matrix is Row Equivalent to Echelon Matrix/Examples/Arbitrary Matrix 5 0 88447 475261 474454 2020-06-21T21:00:54Z Prime.mover 59 wikitext text/x-wiki == Examples of Use of Matrix is Row Equivalent to Echelon Matrix == <onlyinclude> Let $\mathbf A = \begin {bmatrix} -1 & 0 & 1 & 2 & 3 \\ 0 & 1 & 2 & 3 & 4 \\ -1 & -2 & -3 & -4 & -5 \\ \end {bmatrix}$ This can be converted into the echelon form: :$\mathbf E = \begin {bmatrix} 1 & 0 & -1 & -2 & -3 \\ 0 & 1 & 2 & 3 & 4 \\ 0 & 0 & 0 & 0 & 0 \\ \end {bmatrix}$ </onlyinclude> == Proof == This matrix can more easily be handled by direct application of elementary row operations, as follows. Let $e_1$ be the elementary row operation: :$e_1 := r_3 \to r_3 - r_1$ which leaves: $\mathbf A_1 = \begin {bmatrix} -1 & 0 & 1 & 2 & 3 \\ 0 & 1 & 2 & 3 & 4 \\ 0 & -2 & -4 & -6 & -8 \\ \end {bmatrix}$ Let $e_2$ be the elementary row operation: :$e_2 := r_3 \to r_3 + 2 r_2$ which leaves: :$\mathbf A_2 = \begin {bmatrix} -1 & 0 & 1 & 2 & 3 \\ 0 & 1 & 2 & 3 & 4 \\ 0 & 0 & 0 & 0 & 0 \\ \end {bmatrix}$ $\mathbf A_2$ is in non-unity echelon form. It remains to perform the elementary row operation $e_3$ to convert it into echelon form: :$e_3 := r_1 \to -r_1$ which leaves: :$\mathbf E = \begin {bmatrix} 1 & 0 & -1 & -2 & -3 \\ 0 & 1 & 2 & 3 & 4 \\ 0 & 0 & 0 & 0 & 0 \\ \end {bmatrix}$ {{qed}} == Also presented as == Some sources use the non-unity variant of the echelon matrix. Such sources do not require that the leading coefficients necessarily have to equal to $1$. Hence they consider the final step to convert row $1$ of $\mathbf E$ from $\begin {bmatrix} -1 & 0 & 1 & 2 & 3 \end {bmatrix}$ to $\begin {bmatrix} 1 & 0 & -1 & -2 & -3 \end {bmatrix}$ as optional. == Sources == * {{BookReference|Linear Algebra|1998|Richard Kaye|author2 = Robert Wilson|prev = Matrix is Row Equivalent to Echelon Matrix/Examples/Arbitrary Matrix 4|next = Definition:Rank/Matrix/Definition 2}}: Part $\text I$: Matrices and vector spaces: $1$ Matrices: $1.5$ Row and column operations: Exercise $1.4 \ \text {(e)}$ Category:Matrix is Row Equivalent to Echelon Matrix 5x7co12j9kvmvmkvqmyiia2t825x5kb


Simultaneous Linear Equations/Examples/Arbitrary System 1 0 88460 475448 474529 2020-06-23T06:30:57Z Prime.mover 59 wikitext text/x-wiki == Example of Simultaneous Linear Equations == <onlyinclude> The system of simultaneous linear equations: {{begin-eqn}} {{eqn | n = 1       | l = x_1 - 2 x_2 + x_3       | r = 1 }} {{eqn | n = 2       | l = 2 x_1 - x_2 + x_3       | r = 2 }} {{eqn | n = 3       | l = 4 x_1 + x_2 - x_3       | r = 1 }} {{end-eqn}} has as its solution set: {{begin-eqn}} {{eqn | l = x_1       | r = -\dfrac 1 2 }} {{eqn | l = x_2       | r = \dfrac 1 2 }} {{eqn | l = x_3       | r = \dfrac 3 2 }} {{end-eqn}} </onlyinclude> == Proof == Subtract $2 \times$ equation $(1)$ from equation $(2)$. Subtract $4 \times$ equation $(1)$ from equation $(3)$. This gives us: {{begin-eqn}} {{eqn | n = 1       | l = x_1 - 2 x_2 + x_3       | r = 1 }} {{eqn | n = 2'       | l = 3 x_2 - x_3       | r = 0 }} {{eqn | n = 3'       | l = 9 x_2 - 5 x_3       | r = -3 }} {{end-eqn}} Divide equation $(2')$ by $3$ to get $(2'')$. Add $2 \times$ equation $(2'')$ to equation $(1)$. Subtract $9 \times$  equation $(2'')$ from equation $(3')$. This gives us: {{begin-eqn}} {{eqn | n = 1'       | l = x_1 + \dfrac {x_3} 3       | r = 1 }} {{eqn | n = 2''       | l = x_2 - \dfrac {x_3} 3       | r = 0 }} {{eqn | n = 3''       | l = -2 x_3       | r = -3 }} {{end-eqn}} From equation $(3'')$ we have directly that $x_3 = \dfrac 3 2$. Substituting for $x_3$ in equation $(1')$ and equation $(2'')$ gives the single solution: {{begin-eqn}} {{eqn | l = x_1       | r = -\dfrac 1 2 }} {{eqn | l = x_2       | r = \dfrac 1 2 }} {{eqn | l = x_3       | r = \dfrac 3 2 }} {{end-eqn}} {{qed}} == Sources == * {{BookReference|Linear Algebra: An Introduction|1982|A.O. Morris|ed = 2nd|edpage = Second Edition|next = Simultaneous Linear Equations/Examples/Arbitrary System 2}}: Chapter $1$: Linear Equations and Matrices: $1.1$ Introduction: Example $\text {(i)}$ Category:Examples of Simultaneous Linear Equations 13dg15ivlcdbesrvf4ppbezrsjuf304


Simultaneous Linear Equations/Examples/Arbitrary System 2 0 88461 475449 474528 2020-06-23T06:31:10Z Prime.mover 59 wikitext text/x-wiki == Example of Simultaneous Linear Equations == <onlyinclude> The system of simultaneous linear equations: {{begin-eqn}} {{eqn | n = 1       | l = x_1 + x_2       | r = 2 }} {{eqn | n = 2       | l = 2 x_1 + 2 x_2       | r = 3 }} {{end-eqn}} has no solutions. </onlyinclude> == Proof == {{AimForCont}} $(1)$ and $(2)$ together have a solution. Subtract $2 \times$ equation $(1)$ from equation $(2)$. {{begin-eqn}} {{eqn | n = 1       | l = x_1 - 2 x_2 + x_3       | r = 1 }} {{eqn | n = 2'       | l = 0       | r = -1 }} {{end-eqn}} which is an inconsistency. Hence there is no such solution. {{qed}} == Sources == * {{BookReference|Linear Algebra: An Introduction|1982|A.O. Morris|ed = 2nd|edpage = Second Edition|prev = Simultaneous Linear Equations/Examples/Arbitrary System 1|next = Simultaneous Linear Equations/Examples/Arbitrary System 3}}: Chapter $1$: Linear Equations and Matrices: $1.1$ Introduction: Example $\text {(ii)}$ Category:Examples of Simultaneous Linear Equations 93lvo8ybsxkb39742mc7kad62bzvkfx


Elementary Row Operation on Augmented Matrix leads to Equivalent System of Simultaneous Linear Equations/Corollary 1 0 88536 476404 476375 2020-06-30T06:03:14Z Prime.mover 59 wikitext text/x-wiki == Corollary to Elementary Row Operation on Augmented Matrix leads to Equivalent System of Simultaneous Linear Equations == <onlyinclude> Let $S$ be a system of simultaneous linear equations: :$\displaystyle \forall i \in \set {1, 2, \ldots, m}: \sum_{j \mathop = 1}^n \alpha_{i j} x_j = \beta_i$ Let $\begin {pmatrix} \mathbf A & \mathbf b \end {pmatrix}$ denote the augmented matrix of $S$. Let $\begin {pmatrix} \mathbf R & \mathbf s \end {pmatrix}$ be a reduced echelon matrix derived from $\begin {pmatrix} \mathbf A & \mathbf b \end {pmatrix}$. Let $S'$ be the system of simultaneous linear equations: :$\displaystyle \forall i \in \set {1, 2, \ldots, m}: \sum_{j \mathop = 1}^n \rho_{i j} x_j = \sigma_i$ whose augmented matrix is $\begin {pmatrix} \mathbf R & \mathbf s \end {pmatrix}$. Then $S$ and $S'$ are equivalent. </onlyinclude> == Proof == By Matrix is Row Equivalent to Reduced Echelon Matrix, $\begin {pmatrix} \mathbf R & \mathbf s \end {pmatrix}$ can be obtained from $\begin {pmatrix} \mathbf A & \mathbf b \end {pmatrix}$ by means of a finite sequence of elementary row operations. By Elementary Row Operation on Augmented Matrix leads to Equivalent System of Simultaneous Linear Equations, the system of simultaneous linear equations whose augmented matrix is obtained from $\begin {pmatrix} \mathbf A & \mathbf b \end {pmatrix}$ in this way is equivalent to $S$ after each elementary row operation. Hence the entire sequence of such systems of simultaneous linear equations are equivalent to $S$. In particular, this applies to $S'$. {{qed}} == Sources == * {{BookReference|Linear Algebra: An Introduction|1982|A.O. Morris|ed = 2nd|edpage = Second Edition|prev = Definition:Inverse of Elementary Row Operation|next = Elementary Row Operation on Augmented Matrix leads to Equivalent System of Simultaneous Linear Equations/Corollary 2}}: Chapter $1$: Linear Equations and Matrices: $1.3$ Applications to Linear Equations: Corollary $1$ Category:Elementary Row Operation on Augmented Matrix leads to Equivalent System of Simultaneous Linear Equations oam0mx3c7tnxpnhhn57ngl7tev13wwu