hcju/proofwikips huggingface dataset:
Definite Integral of Odd Function/Corollary
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 ==  Let $f$ be an odd function with a Definition:Primitive (Calculus)|primitiv..." wikitext text/x-wiki == Corollary to Definite Integral of Odd Function ==  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: :$\displaystyle \int_{\mathop \to -a}^{\mathop \to a} \map f x \rd x = 0$  == 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}}
Coulomb's Law of Electrostatics
== 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
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 ==  Let $f: \R \to \R$ be the mapping defined on the Definition:Real Numbers|set of real..." wikitext text/x-wiki == Example of Bijection ==  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.  == Proof == A direct application of Integer Power Function is Bijective iff Index is Odd.  == 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
Limit of Subsequence equals Limit of Sequence/Normed Vector Space
Limit of Subsequence equals Limit of Sequence/Normed Vector Space 0 86622 463604 2020-04-22T17:22:12Z Julius 3095 Created page with "== Theorem ==  Let $\struct {X, \norm {\, \cdot \,} }$ be a normed vector space.  Let $\sequence {x_n}$ be a Definition:Seque..." wikitext text/x-wiki == Theorem ==  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 $\displaystyle \lim_{r \mathop \to \infty} x_{n_r} = l$  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.
Natural Numbers under Addition do not form Group/Corollary
== Corollary to Natural Numbers under Addition do not form Group ==  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.  == 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.  == 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 $\text{E i}$
Existence and Uniqueness of Direct Limit of Sequence of Groups/Lemma 1
 == 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: :$\tuple {x_n, n} \sim \tuple {z_r, r}$
Existence and Uniqueness of Direct Limit of Sequence of Groups/Lemma 3
{{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 $\tuple {x_n, n}, \tuple{x_{n'}, n'} \in \eqclass {\tuple {x_n, n} } {}$. , 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.
Schwarz's Lemma/Lemma
Schwarz's Lemma/Lemma 0 87118 466065 2020-05-05T14:18:32Z Caliburn 3218 Created page with "== Theorem ==  Let $D$ be the unit disk centred at $0$.  Let $g : D \to \C$ be a complex function with:..." wikitext text/x-wiki == Theorem ==  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 $D$.  == 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 $D$.
Partial Derivative/Examples/x^(x y)/wrt x
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 ==  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.    == 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 $\text A$
Partial Derivative/Examples/x^(x y)/wrt y
== 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 ==  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}}    == 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 $\text A$
Partial Derivative/Examples/u + ln u = x y
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 ==  Let $u + \ln u = x y$ be an implicit function.  Th..." wikitext text/x-wiki == Example of Partial Derivative ==  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}}  == 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}}  == 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
Function of Bounded Variation is Bounded
Function of Bounded Variation is Bounded 0 87489 467886 2020-05-14T12:35:35Z Caliburn 3218 Created page with "== Theorem ==  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 ==  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.  == 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.  == 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 $6.7$
Multiple of Function of Bounded Variation is of Bounded Variation
Multiple of Function of Bounded Variation is of Bounded Variation 0 87633 469076 2020-05-19T14:59:04Z Caliburn 3218 Created page with "== Theorem ==  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 ==  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 $k f$.  == 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}}
Constant Real Function is Absolutely Continuous
Constant Real Function is Absolutely Continuous 0 87716 470257 2020-05-24T14:39:47Z Caliburn 3218 Created page with "== Theorem ==  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 ==  Let $I \subseteq \R$ be a real interval. Let $f : I \to \R$ be an constant real function. Then $f$ is absolutely continuous.  == 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.
Multiple of Absolutely Continuous Function is Absolutely Continuous
Multiple of Absolutely Continuous Function is Absolutely Continuous 0 87718 470270 2020-05-24T15:56:02Z Caliburn 3218 Created page with "== Theorem ==  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 ==  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.  == 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 $k \in \R$.
Elementary Row Operation/Examples/r3 + 2r2
== Examples of Elementary Row Operations ==  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: :$\begin {pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 2 & 1 \end {pmatrix}$  == 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.  == 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: $1$