hcju/proofwikips huggingface dataset:
Fibonacci's Greedy Algorithm
== Algorithm == Let $\dfrac p q$ denote a proper fraction expressed in canonical form. '''Fibonacci's greedy algorithm''' is a greedy algorithm which calculates a sequence of distinct unit fractions which together sum to $\dfrac p q$: {{begin-eqn}} {{eqn | l = \dfrac p q       | r = \sum_{\substack {1 \mathop \le k \mathop \le m \\ n_j \mathop \le n_{j + 1} } } \dfrac 1 {n_k}       | c =  }} {{eqn | r = \dfrac 1 {n_1} + \dfrac 1 {n_2} + \dotsb + \dfrac 1 {n_m}       | c =  }} {{end-eqn}}  '''Fibonacci's Greedy Algorithm''' is as follows: :$(1) \quad$ Let $p = x_0$ and $q = y_0$ and set $k = 0$. :$(2) \quad$ Is $x_k = 1$? If so, the algorithm has finished. :$(3) \quad$ Find the largest unit fraction $\dfrac 1 {m_k}$ less than $\dfrac {x_k} {y_k}$.  :$(4) \quad$ Calculate $\dfrac {x_{k + 1} } {y_{k + 1} } = \dfrac {x_k} {y_k} - \dfrac 1 {m_k}$ expressed in canonical form. :$(5) \quad$ Go to step $(2)$.  == Also see == * Proper Fraction can be Expressed as Finite Sum of Unit Fractions/Fibonacci's Greedy Algorithm, which proves that Fibonacci's Greedy Algorithm works as expected {{Namedfor|Leonardo Fibonacci|cat = Fibonacci}} == Sources == * {{BookReference|Liber Abaci|1202|Leonardo Fibonacci}} * {{BookReference|Curious and Interesting Puzzles|1992|David Wells|prev = Proper Fraction can be Expressed as Finite Sum of Unit Fractions|next = Proper Fraction can be Expressed as Finite Sum of Unit Fractions/Fibonacci's Greedy Algorithm}}: Egyptian Fractions
Largest Number not Expressible as Sum of Multiples of Coprime Integers/Generalization
Largest Number not Expressible as Sum of Multiples of Coprime Integers/Generalization 0 89295 479146 2020-07-22T04:02:15Z RandomUndergrad 3904 Created page with "== Theorem ==   Let $a, b$ be integers greater than $1$.  Let $d = \gcd \set {a, b}$.  Then the largest multiple of $d$ not expressible as..." wikitext text/x-wiki == Theorem ==  Let $a, b$ be integers greater than $1$. Let $d = \gcd \set {a, b}$. Then the largest multiple of $d$ not expressible as a sum of multiples of $a$ and $b$ is the number: :$\dfrac {a b} d - a - b$  == Proof == By Integers Divided by GCD are Coprime: :$\dfrac a d \perp \dfrac b d$ By Largest Number not Expressible as Sum of Multiples of Coprime Integers, the largest number not expressible as a sum of multiples of $\dfrac a d$ and $\dfrac b d$ is the number: :$\dfrac {a b} {d^2} - \dfrac a d - \dfrac b d$ Let $k d$ be a multiple of $d$ expressible as a sum of multiples of $a$ and $b$: :$\exists s, t \in \N: s a + t b = k d$ Then: :$s \dfrac a d + t \dfrac b d = k$ showing that $k$ is a sum of multiples of $\dfrac a d$ and $\dfrac b d$. This argument reverses. Hence the largest multiple of $d$ not expressible as a sum of multiples of $a$ and $b$ is the number: :$d \paren {\dfrac {a b} {d^2} - \dfrac a d - \dfrac b d} = \dfrac {a b} d - a - b$
Inscribing Equilateral Triangle inside Square with a Coincident Vertex/Construction 4
== Construction for Inscribing Equilateral Triangle inside Square with a Coincident Vertex == {{:Inscribing Equilateral Triangle inside Square with a Coincident Vertex}} == Construction ==  :300px By Construction of Equilateral Triangle, let an equilateral triangle $\triangle ABN$ be constructed on $AB$ such that $N$ is inside $\Box ABCD$. Let $DN$ be produced to cut $BC$ at $H$. Construct $H$ on $BC$ such that $DH = DG$. Then $DGH$ is the required equilateral triangle.  == Proof == First a lemma: === Lemma === {{:Inscribing Equilateral Triangle inside Square with a Coincident Vertex/Lemma}}{{qed|lemma}} Because $\triangle ABN$ is equilateral: :$AB = AN$ :$\angle BAN = 60 \degrees$ Thus $\triangle ADN$ is isosceles with apex at $A$. Then $\angle DAN = 90 \degrees - 60 \degrees = 30 \degrees$. From Sum of Angles of Triangle equals Two Right Angles: :$\angle ADN + \angle AND = 180 \degrees - 30 \degrees = 150 \degrees$ From Isosceles Triangle has Two Equal Angles: :$\angle ADN = \angle AND = \dfrac {150 \degrees} 2 = 75 \degrees$ Thus: :$\angle CDH = 15 \degrees$ The result follows from the lemma.
Inscribing Equilateral Triangle inside Square with a Coincident Vertex/Lemma
== Lemma ==  Let $\Box ABCD$ be a square. Let $\triangle DGH$ be an isosceles triangle inscribed within $\Box ABCD$ such that the apex $D$ of $\triangle DGH$ coincides with vertex $D$ of $\Box ABCD$. :300px Then: :$\triangle DGH$ is equilateral triangle : :$\angle ADG = 15 \degrees \text { or } \angle CDH = 15 \degrees$ (and in fact both are the case).  == Proof == First note that $\triangle DGH$ is isosceles. First we note that: {{begin-eqn}} {{eqn | l = CD       | r = AD       | c = as they are the sides of a square }} {{eqn | l = DH       | r = DG       | c = as $\triangle DGH$ is isosceles }} {{eqn | l = \angle DCH       | r = \angle DAG = 90 \degrees       | c = as $\Box ABCD$ is a square }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = CH^2       | r = DH^2 - CD^2       | c = Pythagoras's Theorem }} {{eqn | r = DG^2 - AD^2       | c =  }} {{eqn | r = AG^2       | c =  }} {{eqn | ll= \leadsto       | l = CH       | r = AG       | c =  }} {{end-eqn}} So by Triangle Side-Side-Side Equality: :$\triangle GAD = \triangle CDH$ and in particular: :$\angle GAD = \angle CDH$ === Necessary Condition === Let $\angle GAD = \angle CDH = 15 \degrees$. Then: :$\angle GDH = 90 \degrees - 2 \times 15 \degrees = 60 \degrees$ We have that $\triangle DGH$ is isosceles. Hence from Isosceles Triangle has Two Equal Angles: :$\angle DGH = \angle DHG$ From Sum of Angles of Triangle equals Two Right Angles it follows that: :$\angle DGH + \angle DHG = 180 \degrees - 60 \degrees = 120 \degrees$ from which: :$\angle DGH = \angle DHG = 60 \degrees$ Hence all the vertices of $\triangle DGH$ equal $60 \degrees$. It follows from Equiangular Triangle is Equilateral that $\triangle DGH$ is equilateral. {{qed|lemma}} === Sufficient Condition === Let $\triangle DGH$ be equilateral. From Internal Angle of Equilateral Triangle: :$\angle GDH = 60 \degrees$ Because $\Box ABCD$ is a square: :$\angle ADC = 90 \degrees$ Thus: {{begin-eqn}} {{eqn | l = \angle GAD + \angle GDH + \angle CDH       | r = 90 \degrees       | c =  }} {{eqn | ll= \leadsto       | l = \angle GAD + \angle CDH       | r = 90 \degrees - \angle GDH       | c =  }} {{eqn | r = 90 \degrees - 60 \degrees       | c =  }} {{eqn | r = 30 \degrees       | c =  }} {{end-eqn}} But we have that: :$\angle GAD = \angle CDH$ and so: :$\angle GAD = \angle CDH = 15 \degrees$
Returning Explorer Puzzle
== Puzzle == An explorer walks: :one mile due south,  :one mile due east, :and one mile due north. He finds himself back where he started. He shoots a {{WP|Bear|bear}}. What colour is that {{WP|Bear|bear}}? === Variant === {{:Returning Explorer Puzzle/Variant}} == Solution == White. == Proof == The explorer starts from the North Pole. Every direction from the North Pole is due south. Walking due east keeps him the same distance (one mile) from the North Pole. Walking due north takes him back to the North Pole again. The bear, as a consequence, must be a {{WP|Polar_bear|polar bear}}.  == Sources == * {{BookReference|Mathematical Puzzles and Diversions|1965|Martin Gardner|prev = |next = Returning Explorer Puzzle/Variant}}: Nine Problems: $1$ * {{BookReference|Hexaflexagons and Other Mathematical Diversions|1988|Martin Gardner|prev = |next = Returning Explorer Puzzle/Variant}}: Nine Problems: $1$
Rationals are Everywhere Dense in Reals/Topology
Rationals are Everywhere Dense in Reals/Topology 0 90798 486361 2020-09-08T09:33:47Z Julius 3095 Created page with "== Theorem ==  Let $\struct {\R, \tau_d}$ denote the real number line with the usual (Euclidean) topology...." wikitext text/x-wiki == Theorem ==  Let $\struct {\R, \tau_d}$ denote the real number line with the usual (Euclidean) topology. Let $\Q$ be the set of rational numbers. Then $\Q$ is everywhere dense in $\struct {\R, \tau_d}$.  == Proof == Let $x \in \R$. Let $U \subseteq \R$ be an open set of $\struct {\R, \tau_d}$ such that $x \in U$. From Basis for Euclidean Topology on Real Number Line, there exists an open interval $V = \openint {x - \epsilon} {x + \epsilon} \subseteq U$ for some $\epsilon > 0$ such that $x \in V$. Now consider the open interval $\openint x {x + \epsilon} \subseteq V$. By Subset Relation is Transitive it follows that $\openint x {x + \epsilon} \subseteq U$. Note that $x \notin \openint x {x + \epsilon}$. From Between two Real Numbers exists Rational Number, there exists $y \in \Q: y \in \openint x {x + \epsilon}$. As $x \notin \openint x {x + \epsilon}$, it must be the case that $x \ne y$. That is, $V$ is an open set of $\struct {\R, \tau_d}$ containing $x$ which also contains an element of $\Q$ other than $x$. As $V$ is arbitrary, it follows that every open set of $\struct {\R, \tau_d}$ containing $x$ also contains an element of $\Q$ other than $x$. That is, $x$ is by definition a limit point of $\Q$. As $x$ is arbitrary, it follows that all elements of $\R$ are limit points of $\Q$. The result follows from the definition of everywhere dense.
Convergent Sequence is Cauchy Sequence/Metric Space
Convergent Sequence is Cauchy Sequence/Metric Space 0 91017 487795 2020-09-14T11:31:55Z Julius 3095 Created page with "== Theorem ==  Let $M = \struct {A, d}$ be a metric space.   Every Definition:Convergent Sequence in Metric Space|convergent sequenc..." wikitext text/x-wiki == Theorem ==  Let $M = \struct {A, d}$ be a metric space. Every convergent sequence in $M$ is a Cauchy sequence.  == Proof == Let $\sequence {x_n}$ be a sequence in $A$ that converges to the limit $l \in A$. Let $\epsilon > 0$.  Then also $\dfrac \epsilon 2 > 0$. Because $\sequence {x_n}$ converges to $l$, we have: :$\exists N: \forall n > N: \map d {x_n, l} < \dfrac \epsilon 2$ So if $m > N$ and $n > N$, then: {{begin-eqn}} {{eqn | l = \map d {x_n, x_m}       | o = \le       | r = \map d {x_n, l} + \map d {l, x_m}       | c = Triangle Inequality }} {{eqn | o = <       | r = \frac \epsilon 2 + \frac \epsilon 2       | c = (by choice of $N$) }} {{eqn | r = \epsilon       | c =  }} {{end-eqn}} Thus $\sequence {x_n}$ is a Cauchy sequence.  == Also see == * Definition:Complete Metric Space, where the converse is true. * Real Convergent Sequence is Cauchy Sequence * Complex Sequence is Cauchy iff Convergent == Sources == * {{BookReference|Principles of Mathematical Analysis|1953|Walter Rudin|next = Compact Metric Space is Complete}}: $3.11a$ * {{BookReference|Introduction to Metric and Topological Spaces|1975|W.A. Sutherland|prev = Definition:Cauchy Sequence (Metric Space)|next = Subsequence of Sequence in Metric Space with Limit}}: $7.2$: Sequential compactness: Proposition $7.2.4$ * {{BookReference|Counterexamples in Topology|1978|Lynn Arthur Steen|author2 = J. Arthur Seebach, Jr.|ed = 2nd|edpage = Second Edition|prev = Definition:Cauchy Sequence (Metric Space)|next = Cauchy Sequence in Metric Space is not necessarily Convergent}}: Part $\text I$: Basic Definitions: Section $5$: Metric Spaces: Complete Metric Spaces
Cauchy Sequence is Bounded/Metric Space
Cauchy Sequence is Bounded/Metric Space 0 91058 488083 2020-09-15T11:26:19Z Julius 3095 Created page with "== Theorem ==  Let $M = \struct {A, d}$ be a metric space.   Then every Cauchy sequence..." wikitext text/x-wiki == Theorem ==  Let $M = \struct {A, d}$ be a metric space. Then every Cauchy sequence in $M$ is bounded.  == Proof == Let $\sequence {x_n}$ be a Cauchy sequence in $M$. By definition: :$\forall \epsilon > 0: \exists N \in \N: \forall m, n > N: \map d {x_n, x_m} < \epsilon$ Particularly, setting $\epsilon = 1$: :$\exists N_1: \forall m, n > N_1: \map d {x_n, x_m} < 1$ Note that since $N_1 \ge N_1$, this means that: :$\forall n \ge N_1: \map d {x_n, x_{N_1} } < 1$ To show $\sequence {x_n}$ is bounded, we need to show that there exists $a \in A$ and $K \in \R$ such that $\map d {x_n, a} \le K$ for all $x_n \in \sequence {x_n}$. Let $K'$ be the maximum distance from $x_{N_1}$ to any of the earlier terms in the sequence. That is, $K' = \max \set {\map d {x_{N_1}, x_1}, \map d {x_{N_1},  x_2}, \ldots, \map d {x_{N_1}, x_{N_1 - 1} } }$ Then: :Each $x_n$ for $n \ge N_1$ satisfies $\map d {x_{N_1}, x_n} \le 1$ by choice of $N_1$ as mentioned above :Each $x_n$ for $n < N_1$ satisfies $\map d {x_{N_1}, x_n} \le K'$ by choice of $K'$. Thus, taking $a = x_{N_1}$ and $K = \max \set {K', 1}$, we have shown that each $x_n$ satisfies $\map d {a, x_n} \le K$. So, $\sequence {x_n}$ is bounded.  == Sources == * {{BookReference|Mathematical Analysis|1957|Tom M. Apostol|prev = Real Convergent Sequence is Cauchy Sequence|next = Unbounded Sequence is Divergent}}: $\S 12$-$2$: Convergent and divergent sequences
Autocorrelation of Strictly Stationary Stochastic Process
== Example of Strictly Stationary Stochastic Process == Let $S$ be a strictly stationary stochastic process giving rise to a time series $T$.  It is necessary that: :The autocorrelation between every two observations $z_t, z_{t + k}$ separated by a given lag $k$  is the same as: :the autocorrelation between every other two observations $z_{t + m}, z_{t + m + k}$ separated by a given lag $k$  For such a strictly stationary stochastic process: :$\rho_k = \dfrac {\gamma_k} {\gamma_0}$ where $\gamma_k$ denotes the autocovariance of $S$. == Proof == The autocorrelation is defined as: :$\rho_k := \dfrac {\expect {\paren {z_t - \mu} \paren {z_{t + k} - \mu} } } {\sqrt {\expect {\paren {z_t - \mu}^2} \expect {\paren {z_{t + k} - \mu}^2} } }$ The autocovariance is defined as: :$\gamma_k := \expect {\paren {z_t - \mu} \paren {z_{t - k} - \mu} }$ Hence: :$\rho_k := \dfrac {\gamma_k} {\sqrt {\expect {\paren {z_t - \mu}^2} \expect {\paren {z_{t + k} - \mu}^2} } }$ Then we have that for a strictly stationary stochastic process: :$\expect {\paren {z_t - \mu}^2} = \sigma_z^2$ where $\sigma_z^2$ is the variance of $S$ and, for a strictly stationary stochastic process, is constant. Thus: :$\rho_k := \dfrac {\gamma_k} {\sigma_t^2}$ Then from Autocovariance at Zero Lag for Strictly Stationary Stochastic Process is Variance: :$\sigma_z^2 = \gamma_0$ Hence: :$\rho_k = \dfrac {\gamma_k} {\gamma_0}$  == Sources == * {{BookReference|Time Series Analysis: Forecasting and Control|1994|George E.P. Box|author2 = Gwilym M. Jenkins|author3 = Gregory C. Reinsel|ed = 3rd|edpage = Third Edition|prev = Definition:Autocorrelation|next = Autocovariance at Zero Lag for Strictly Stationary Stochastic Process is Variance}}: ::Part $\text {I}$: Stochastic Models and their Forecasting: :::$2$: Autocorrelation Function and Spectrum of Stationary Processes: ::::$2.1$ Autocorrelation Properties of Stationary Models: :::::$2.1.2$ Stationary Stochastic Processes: Autocovariance and autocorrelation coefficients: $(2.1.6)$
Autocovariance at Zero Lag for Strictly Stationary Stochastic Process is Variance
== Example of Strictly Stationary Stochastic Process == Let $S$ be a strictly stationary stochastic process giving rise to a time series $T$. Then the autocovariance at zero lag is given by: :$\gamma_0 = \sigma_z^2$ where $\sigma_z^2$ is the variance of $S$. == Proof == By definition, the '''autocovariance''' of $S$ at lag $k$ is defined as: :$\gamma_k := \cov {z_t, z_{t + k} } = \expect {\paren {z_t - \mu} \paren {z_{t - k} - \mu} }$ where: :$z_t$ is the observation at time $t$ :$\mu$ is the mean of $S$ :$\expect \cdot$ is the expectation. For a strictly stationary stochastic process: :$\expect {\paren {z_t - \mu}^2} = \sigma_z^2$ where: :$\mu$ is the constant mean level of $S$ :$\expect {\paren {z_t - \mu}^2}$ is the expectation of $\paren {z_t - \mu}^2$ :$\sigma_z^2$ is the variance of $S$ and, for a strictly stationary stochastic process, is constant. Hence we have that: {{begin-eqn}} {{eqn | l = \gamma_0       | r = \expect {\paren {z_t - \mu} \paren {z_{t + 0} - \mu} }       | c =  }} {{eqn | r = \expect {\paren {z_t - \mu}^2}       | c =  }} {{eqn | r = \sigma_z^2       | c =  }} {{end-eqn}}  == Sources == * {{BookReference|Time Series Analysis: Forecasting and Control|1994|George E.P. Box|author2 = Gwilym M. Jenkins|author3 = Gregory C. Reinsel|ed = 3rd|edpage = Third Edition|prev = Autocorrelation of Strictly Stationary Stochastic Process|next = Autocorrelation at Zero Lag for Strictly Stationary Stochastic Process is 1}}: ::Part $\text {I}$: Stochastic Models and their Forecasting: :::$2$: Autocorrelation Function and Spectrum of Stationary Processes: ::::$2.1$ Autocorrelation Properties of Stationary Models: :::::$2.1.2$ Stationary Stochastic Processes: Autocovariance and autocorrelation coefficients
Determinant of Autocorrelation Matrix is Strictly Positive/Examples/Order 2
Determinant of Autocorrelation Matrix is Strictly Positive/Examples/Order 2 0 91368 490102 2020-09-22T10:34:28Z Prime.mover 59 Created page with "== Example of Use of Determinant of Autocorrelation Matrix is Strictly Positive ==  Let $\rho_1$ be the autocorrelation of a ..." wikitext text/x-wiki == Example of Use of Determinant of Autocorrelation Matrix is Strictly Positive ==  Let $\rho_1$ be the autocorrelation of a strictly stationary stochastic process $S$ at lag $1$. Then: :$-1 < \rho_1 < 1$  == Proof == Consider the autocorrelation matrix of order $2$: {{begin-eqn}} {{eqn | l = \map \det {\mathbf P_2}       | o = >       | r = 0       | c = Determinant of Autocorrelation Matrix is Strictly Positive }} {{eqn | l = \begin {vmatrix} 1 & \rho_1 \\ \rho_1 & 1 \end {vmatrix}       | o = >       | r = 0       | c = {{Defof|Autocorrelation Matrix}} }} {{eqn | l = 1 - \rho_1^2       | o = >       | r = 0       | c = {{Defof|Determinant}} }} {{eqn | ll= \leadsto       | l = \rho_1^2       | o = <       | r = 1       | c =  }} {{eqn | ll= \leadsto       | l = \size {\rho_1}       | o = <       | r = 1       | c =  }} {{end-eqn}}  == Sources == * {{BookReference|Time Series Analysis: Forecasting and Control|1994|George E.P. Box|author2 = Gwilym M. Jenkins|author3 = Gregory C. Reinsel|ed = 3rd|edpage = Third Edition|prev = Determinant of Autocorrelation Matrix is Strictly Positive|next = Determinant of Autocorrelation Matrix is Strictly Positive/Examples/Order 3}}: ::Part $\text {I}$: Stochastic Models and their Forecasting: :::$2$: Autocorrelation Function and Spectrum of Stationary Processes: ::::$2.1$ Autocorrelation Properties of Stationary Models: :::::$2.1.3$ Positive Definiteness and the Autocovariance Matrix: Conditions satisfied by the autocorrelations of a stationary process
Negative of Logarithm of x plus Root x squared plus a squared/Corollary
Negative of Logarithm of x plus Root x squared plus a squared/Corollary 0 91523 490958 2020-09-26T13:22:45Z Prime.mover 59 Created page with "== Theorem ==  Let $x \in \R: \size x > 1$.  Let $x > 1$.  Then:  :$-\map \ln {x + \sqrt {x^2 + a^2} } = \ln \size {x - \sqrt {x^2 + a^2} } - \map \ln {a^2}$ </on..." wikitext text/x-wiki == Theorem == Let $x \in \R: \size x > 1$. Let $x > 1$. Then:  :$-\map \ln {x + \sqrt {x^2 + a^2} } = \ln \size {x - \sqrt {x^2 + a^2} } - \map \ln {a^2}$  == Proof == We have that $\sqrt {x^2 + a^2} > x$ for all $x$. Hence for all $x$: :$-x + \sqrt {x^2 + a^2} > 0$ and so: :$x - \sqrt {x^2 + a^2} < 0$ Hence: :$-x + \sqrt {x^2 + a^2} = \size {x - \sqrt {x^2 + a^2} }$ Then we have: {{begin-eqn}} {{eqn | l = -\map \ln {x + \sqrt {x^2 - a^2} }       | r = \map \ln {-x + \sqrt {x^2 + a^2} } - \map \ln {a^2}       | c = Negative of Logarithm of x plus Root x squared plus a squared }} {{eqn | ll= \leadsto       | l = -\map \ln {x + \sqrt {x^2 - a^2} }       | r = \ln \size {x - \sqrt {x^2 + a^2} } - \map \ln {a^2}       | c =  }} {{end-eqn}}  == Also see == * Negative of Logarithm of x plus Root x squared plus a squared
Primitive of Reciprocal of a squared minus x squared/Logarithm Form/Lemma
Primitive of Reciprocal of a squared minus x squared/Logarithm Form/Lemma 0 91571 491217 2020-09-27T16:21:54Z Prime.mover 59 Created page with "== Lemma ==  Let $a \in \R_{>0}$ be a strictly positive real constant.  Then: :$\map \ln {\df..." wikitext text/x-wiki == Lemma ==  Let $a \in \R_{>0}$ be a strictly positive real constant. Then: :$\map \ln {\dfrac {a + x} {a - x} }$ is defined  $\size x < a$ :$\map \ln {\dfrac {x + a} {x - a} }$ is defined  $\size x > a$  == Proof == We have that the real natural logarithm is defined only on the strictly positive real numbers. Hence: :$\map \ln {\dfrac {a + x} {a - x} }$ is defined  $\dfrac {a + x} {a - x} > 0$ :$\map \ln {\dfrac {x + a} {x - a} }$ is defined  $\dfrac {x + a} {x - a} > 0$ First we note that if $\size x = a$, then either the numerator or denominator of the arguments of the logarithm functions in question are either $0$ or undefined. Hence the expressions have no meaning unless $\size x \ne a$. The following table indicates whether each of $x + a$, $a - x$ and $x - a$ are positive $(+)$ or negative $(-)$ on the domains in question. $\begin {array} {c|ccc|cc} & x + a & a - x & x - a & \dfrac {a + x} {a - x} & \dfrac {x + a} {x - a} \\ \hline  a < x      & + & - & + & + & - \\ 0 < x < a  & + & + & - & - & + \\ -a < x < 0 & + & + & - & - & + \\ x < -a     & - & + & - & + & - \\ \end {array}$ Hence: :$\map \ln {\dfrac {a + x} {a - x} }$ is defined  $-a < x < a$ :$\map \ln {\dfrac {x + a} {x - a} }$ is defined  $x > a$ or $x < -a$ as we were required to show.
Primitive of Reciprocal of x squared minus a squared/Logarithm Form/Lemma
Primitive of Reciprocal of x squared minus a squared/Logarithm Form/Lemma 0 91592 491313 2020-09-27T22:16:21Z Prime.mover 59 Created page with "== Lemma ==  Let $a \in \R_{>0}$ be a strictly positive real constant.  Then: :$\map \ln {\df..." wikitext text/x-wiki == Lemma ==  Let $a \in \R_{>0}$ be a strictly positive real constant. Then: :$\map \ln {\dfrac {x - a} {x + a} }$ is defined  $\size x > a$ :$\map \ln {\dfrac {a - x} {a + x} }$ is defined  $\size x < a$  == Proof == We have that the real natural logarithm is defined only on the strictly positive real numbers. Hence: :$\map \ln {\dfrac {x - a} {x + a} }$ is defined  $\dfrac {x - a} {x + a} > 0$ :$\map \ln {\dfrac {a - x} {a + x} }$ is defined  $\dfrac {a - x} {a + x} > 0$ First we note that if $\size x = a$, then either the numerator or denominator of the arguments of the logarithm functions in question are either $0$ or undefined. Hence the expressions have no meaning unless $\size x \ne a$. The following table indicates whether each of $x + a$, $a - x$ and $x - a$ are positive $(+)$ or negative $(-)$ on the domains in question. $\begin {array} {c|ccc|cc} & x + a & a - x & x - a & \dfrac {x - a} {x + a} & \dfrac {a - x} {a + x} \\ \hline  a < x      & + & - & + & + & - \\ 0 < x < a  & + & + & - & - & + \\ -a < x < 0 & + & + & - & - & + \\ x < -a     & - & + & - & + & - \\ \end {array}$ Hence: :$\map \ln {\dfrac {x - a} {x + a} }$ is defined  $x > a$ or $x < -a$ :$\map \ln {\dfrac {a - x} {a + x} }$ is defined  $-a < x < a$ as we were required to show.