hcju/proofwikiqa huggingface dataset:
First we have: {{begin-eqn}} {{eqn | l = \lim_{x \mathop \to 0} \map \ln {\sin x} \sin 2 n x       | r = \lim_{x \mathop \to 0} \frac {\map \ln {\sin x} } {\csc 2 n x}       | c = {{Defof|Cosecant}} }} {{eqn | r = \lim_{x \mathop \to 0} \frac {\cot x} {- 2 n \cot 2 n x \csc 2 n x}       | c = L'Hôpital's Rule/Corollary 2 }} {{eqn | r = \lim_{x \mathop \to 0} \frac {\cos x} {- 2 n \cos 2 n x} \frac {\sin^2 2 n x} {\sin x}       | c = {{Defof|Cosecant}}, {{Defof|Cotangent}} }} {{eqn | l = \lim_{x \mathop \to 0} \frac {\sin^2 2 n x} {\sin x}       | r = \lim_{x \mathop \to 0} \frac {4 n \sin 2 n x \cos 2 n x} {\cos x}       | c = L'Hôpital's Rule }} {{eqn | r = 0 }} {{end-eqn}} hence $\displaystyle \lim_{x \mathop \to 0} \map \ln {\sin x} \sin 2 n x = 0$. Thus: {{begin-eqn}} {{eqn | l = \int_0^{\pi/2} \map \ln {\sin x} \cos 2 n x \ \d x       | r = \frac 1 {2 n} \int_0^{\pi/2} \map \ln {\sin x} \ \map \d {\sin 2 n x}       | c = Primitive of Cosine Function/Corollary }} {{eqn | r = \frac 1 {2 n} \paren {\bigintlimits {\map \ln {\sin x} \sin 2 n x} 0 {\pi/2} - \int_0^{\pi/2} \sin n x \ \map \d {\map \ln {\sin x} } }       | c = Integration by Parts }} {{eqn | r = -\frac 1 {2 n} \int_0^{\pi/2} \sin n x \ \map \d {\map \ln {\sin x} }       | c = From above }} {{eqn | r = -\frac 1 {2 n} \int_0^{\pi/2} \sin 2 n x \frac {\cos x} {\sin x} \ \d x       | c = Primitive of Cotangent Function }} {{eqn | r = -\frac 1 {2 n} \int_0^{\pi/2} \frac {\sin \paren {2 n + 1} x + \sin \paren {2 n - 1} x} {2 \sin x} \ \d x       | c = Simpson's Formulas/Sine by Cosine }} {{eqn | r = -\frac 1 {4 n} \int_0^{\pi} \frac {\sin \paren {\paren {2 n + 1} u/2} + \sin \paren {\paren {2 n - 1} u/2} } {2 \sin \paren {u/2} } \ \d u       | c = Substituting $u = 2 x$ }} {{eqn | r = -\frac 1 {4 n} \int_0^{\pi} \paren {\frac 1 2 + \sum_{k \mathop = 1}^n \map \cos {k u} + \frac 1 2 + \sum_{k \mathop = 1}^{n - 1} \map \cos {k u} } \ \d u       | c = Sum of Cosines of Multiples of Angle }} {{eqn | r = -\frac 1 {4 n} \int_0^{\pi} 1 \ \d u       | c = All integrals involving $\cos k u$ evaluates to $0$ }} {{eqn | r = -\frac \pi {4 n}       | c = Integral of Constant/Definite }} {{end-eqn}} {{qed}} Category:Definite Integrals involving Logarithm Function Category:Definite Integrals involving Sine Function swrdggg8dsb6vrtuf3lswb34vohkf7i
{{begin-eqn}} {{eqn | l = \set {a, b}       | o = \notin       | r = \mathscr I }} {{eqn | ll= \leadstoandfrom       | l = \set a \cup \set b       | o = \notin       | r = \mathscr I       | c = Union of Disjoint Singletons is Doubleton }} {{eqn | ll= \leadstoandfrom       | l = a       | o = \in       | r = \map \sigma {\set b}       | c = Element Depends on Independent Set iff Union with Singleton is Dependent }} {{eqn | lo= \land       | l = b       | o = \in       | r = \map \sigma {\set a}       | c = Element Depends on Independent Set iff Union with Singleton is Dependent }} {{end-eqn}} {{qed}} Category:Distinct Matroid Elements are Parallel iff Each is in Closure of Other a3szsfqgb984wsvqqb2yy8i472rnz2j
Fix $a$ and set:  :$\displaystyle \map I m = \int_0^\infty \frac {\sin m x} {x \paren {x^2 + a^2} } \rd x$ for $m \ge 0$.  We have: {{begin-eqn}} {{eqn	| l = \map {I'} m 	| r = \frac \d {\d m} \int_0^\infty \frac {\sin m x} {x \paren {x^2 + a^2} } \rd x }} {{eqn	| r = \int_0^\infty \frac \partial {\partial m} \paren {\frac {\sin m x} {x \paren {x^2 + a^2} } } \rd x 	| c = Definite Integral of Partial Derivative }} {{eqn	| r = \int_0^\infty \frac {\cos m x} {x^2 + a^2} \rd x 	| c = Derivative of $\sin a x$ }} {{eqn	| r = \frac \pi {2 a} e^{-m a} 	| c = Definite Integral to Infinity of $\dfrac {\cos m x} {x^2 + a^2}$ }} {{end-eqn}} So by Primitive of $e^{a x}$: :$\displaystyle \map I m = -\frac \pi {2 a^2} e^{-m a} + C$ for some constant $C \in \R$. We have:  {{begin-eqn}} {{eqn	| l = \map I 0 	| r = \int_0^\infty \frac {\sin 0 x} {x^2 + a^2} \rd x }} {{eqn	| r = \int_0^\infty 0 \rd x 	| c = Sine of Zero is Zero }} {{eqn	| r = 0 }} {{end-eqn}} On the other hand:  {{begin-eqn}} {{eqn	| l = \map I 0 	| r = -\frac \pi {2 a^2} e^0 + C }} {{eqn	| r = -\frac \pi {2 a^2} + C 	| c = Exponential of Zero }} {{end-eqn}} So: :$C = \dfrac \pi {2 a^2}$ giving:  :$\displaystyle \map I m = \int_0^\infty \frac {\sin m x} {x \paren {x^2 + a^2} } \rd x = \frac \pi {2 a^2} \paren {1 - e^{-m a} }$ {{qed}}
From Fourier Series of $\map \ln {\sin x}$ from $0$ to $\pi$: :$\displaystyle \map \ln {\sin x} = -\ln 2 - \sum_{n \mathop = 1}^\infty \frac {\cos 2 n x} n$ Then, by Parseval's Theorem: {{begin-eqn}} {{eqn	| l = \frac 2 \pi \int_0^\pi \paren {\map \ln {\sin x} }^2 \rd x 	| r = 2 \paren {\ln 2}^2 + \sum_{n = 1}^\infty \frac 1 {n^2} }} {{eqn	| r = 2 \paren {\ln 2}^2 + \frac {\pi^2} 6 	| c = Basel Problem }} {{end-eqn}} We then have:  {{begin-eqn}} {{eqn	| l = \int_0^\pi \paren {\map \ln {\sin x} }^2 \rd x 	| r = \int_0^{\pi/2} \paren {\map \ln {\sin x} }^2 \rd x + \int_{\pi/2}^\pi \paren {\map \ln {\sin x} }^2 \rd x 	| c = Sum of Integrals on Adjacent Intervals for Integrable Functions }} {{eqn	| r = \int_0^{\pi/2} \paren {\map \ln {\sin x} }^2 \rd x + \int_0^{\pi/2} \paren {\map \ln {\map \sin {\pi - x} } }^2 \rd x }} {{eqn	| r = 2 \int_0^{\pi/2} \paren {\map \ln {\sin x} }^2 \rd x 	| c = Sine of Supplementary Angle }} {{end-eqn}} So we have:  :$\displaystyle \frac 4 \pi \int_0^{\pi/2} \paren {\map \ln {\sin x} }^2 \rd x = 2 \paren {\ln 2}^2 + \frac {\pi^2} 6$ multiplying by $\dfrac \pi 4$ we have:  :$\displaystyle \int_0^{\pi/2} \paren {\map \ln {\sin x} }^2 \rd x = \frac \pi 2 \paren {\ln 2}^2 + \frac {\pi^3} {24}$ {{qed}}
{{begin-eqn}} {{eqn	| l = \int_0^{\pi/2} \paren {\map \ln {\cos x} }^2 \rd x 	| r = \int_0^{\pi/2} \paren {\map \ln {\map \cos {\frac \pi 2 - x} } }^2 \rd x }} {{eqn	| r = \int_0^{\pi/2} \paren {\map \ln {\sin x} }^2 \rd x 	| c = Cosine of Complement equals Sine }} {{eqn	| r = \frac \pi 2 \paren {\ln 2}^2 + \frac {\pi^3} {24} 	| c = Definite Integral from $0$ to $\dfrac \pi 2$ of $\paren {\map \ln {\sin x} }^2$ }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \int_0^{2 \pi} \map \ln {a + b \cos x} \rd x       | r = \int_0^\pi \map \ln {a + b \cos x} \rd x + \int_\pi^{2 \pi} \map \ln {a + b \cos x} \rd x       | c = Sum of Integrals on Adjacent Intervals for Integrable Functions }} {{eqn | r = \int_0^\pi \map \ln {a + b \cos x} \rd x - \int_\pi^0 \map \ln {a + b \map \cos {2 \pi - x} } \rd x       | c = substituting $x \mapsto 2 \pi - x$ }} {{eqn | r = \int_0^\pi \map \ln {a + b \cos x} \rd x + \int_0^\pi \map \ln {a + b \map \cos {2 \pi - x} } \rd x       | c = Reversal of Limits of Definite Integral }} {{eqn | r = \int_0^\pi \map \ln {a + b \cos x} \rd x + \int_0^\pi \map \ln {a + b \cos x} \rd x       | c = Cosine of Conjugate Angle }} {{eqn | r = 2 \int_0^\pi \map \ln {a + b \cos x} \rd x }} {{eqn | r = 2 \pi \map \ln {\frac {a + \sqrt {a^2 - b^2} } 2}       | c = Definite Integral from 0 to Pi of Logarithm of a plus b Cosine x }} {{end-eqn}} {{qed}}
Note that by Difference of Logarithms:  :$\displaystyle \int_0^{\pi/2} \sec x \map \ln {\frac {1 + b \cos x} {1 + a \cos x} } \rd x = \int_0^{\pi/2} \sec x \map \ln {1 + b \cos x} \rd x - \int_0^{\pi/2} \sec \map \ln {1 + a \cos x} \rd x$ For each $\alpha \in \openint {-1} 1$, set:  :$\displaystyle \map I \alpha = \int_0^{\pi/2} \sec x \map \ln {1 + \alpha \cos x} \rd x$ Then:  :$\displaystyle \int_0^{\pi/2} \sec x \map \ln {\frac {1 + b \cos x} {1 + a \cos x} } \rd x = \map I b - \map I a$ We have:  {{begin-eqn}} {{eqn	| l = \map {I'} \alpha 	| r = \frac \d {\d \alpha} \int_0^{\pi/2} \sec x \map \ln {1 + \alpha \cos x} \rd x }} {{eqn	| r = \int_0^{\pi/2} \frac \partial {\partial \alpha} \paren {\sec x \map \ln {1 + \alpha \cos x} } \rd x 	| c = Definite Integral of Partial Derivative }} {{eqn	| r = \int_0^{\pi/2} \frac {\cos x \sec x} {1 + \alpha \cos x} \rd x 	| c = Chain Rule, Derivative of Natural Logarithm }} {{eqn	| r = \int_0^{\pi/2} \frac 1 {1 + \alpha \cos x} \rd x }} {{eqn	| r = \frac {\arccos \alpha} {\sqrt {1 - \alpha^2} } 	| c = Definite Integral from $0$ to $\dfrac \pi 2$ of $\dfrac 1 {a + b \cos x}$ }} {{end-eqn}} Then: {{begin-eqn}} {{eqn	| l = \map I b - \map I a 	| r = \int_a^b \map {I'} \alpha \rd \alpha 	| c = Fundamental Theorem of Calculus: Second Part }} {{eqn	| r = \int_a^b \frac {\arccos \alpha} {\sqrt {1 - \alpha^2} } \rd \alpha }} {{end-eqn}} From Derivative of Arccosine Function, we have:  :$\displaystyle \frac \d {\d \alpha} \arccos \alpha = -\frac 1 {\sqrt {1 - \alpha^2} }$ so: {{begin-eqn}} {{eqn	| l = \int_a^b \frac {\arccos \alpha} {\sqrt {1 - \alpha^2} } \rd \alpha 	| r = \int_{\arccos a}^{\arccos b} x \frac {-\sqrt {1 - \alpha^2} } {\sqrt {1 - \alpha^2} } \rd x 	| c = substituting $x = \arccos \alpha$ }} {{eqn	| r = -\int_{\arccos a}^{\arccos b} x \rd x }} {{eqn	| r = \intlimits {-\frac {x^2} 2} {\arccos a} {\arccos b} 	| c = Primitive of Power }} {{eqn	| r = \frac 1 2 \paren {\paren {\arccos a}^2 - \paren {\arccos b}^2} }} {{end-eqn}} so: :$\displaystyle \int_0^{\pi/2} \sec x \map \ln {\frac {1 + b \cos x} {1 + a \cos x} } \rd x = \frac 1 2 \paren {\paren {\arccos a}^2 - \paren {\arccos b}^2}$ as required. {{qed}}
Fix $p$ and set:  :$\displaystyle \map I \alpha = \int_0^\infty \frac {e^{-\alpha x} } {x \sec p x} \rd x$ for all $\alpha \ge 0$. Then:  :$\displaystyle \int_0^\infty \frac {e^{-a x} - e^{-b x} } {x \sec p x} \rd x = \map I a - \map I b$ We have:  {{begin-eqn}} {{eqn	| l = \map {I'} \alpha 	| r = \frac \d {\d \alpha} \int_0^\infty \frac {e^{-\alpha x} } {x \sec p x} \rd x }} {{eqn	| r = \int_0^\infty \frac \partial {\partial \alpha} \paren {\frac {e^{-\alpha x} } {x \sec p x} } \rd x 	| c = Definite Integral of Partial Derivative }} {{eqn	| r = -\int_0^\infty e^{-\alpha x} \cos p x \rd x 	| c = Derivative of $e^{a x}$, {{Defof|Secant Function}} }} {{eqn	| r = -\frac \alpha {\alpha^2 + p^2} 	| c = Definite Integral to Infinity of $e^{-a x} \cos b x$ }} {{end-eqn}} so: {{begin-eqn}} {{eqn	| l = \map I \alpha 	| r = -\int \frac \alpha {\alpha^2 + p^2} \rd \alpha }} {{eqn	| r = -\frac 1 2 \map \ln {\alpha^2 + p^2} + C 	| c = Primitive of $\dfrac x {x^2 + a^2}$ }} {{end-eqn}} for all $\alpha \ge 0$, for some constant $C \in \R$. We then have:  {{begin-eqn}} {{eqn	| l = \int_0^\infty \frac {e^{-a x} - e^{-b x} } {x \sec p x} \rd x 	| r = \map I a - \map I b }} {{eqn	| r = \paren {-\frac 1 2 \map \ln {a^2 + p^2} + C} - \paren {-\frac 1 2 \map \ln {b^2 + p^2} + C} }} {{eqn	| r = \frac 1 2 \paren {\map \ln {b^2 + p^2} - \map \ln {a^2 + p^2} } }} {{eqn	| r = \frac 1 2 \map \ln {\frac {b^2 + p^2} {a^2 + p^2} } 	| c = Difference of Logarithms }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn	| l = \int_0^\infty \frac {\cos p x - \cos q x} {x^2} \rd x 	| r = \int_0^\infty \frac {1 - \cos q x - \paren {1 - \cos p x} } {x^2} \rd x }} {{eqn	| r = \int_0^\infty \frac {1 - \cos q x} {x^2} \rd x - \int_0^\infty \frac {1 - \cos p x} {x^2} \rd x }} {{eqn	| r = \frac \pi 2 \size q - \frac \pi 2 \size p 	| c = Integral to Infinity of $\dfrac {1 - \cos p x} {x^2}$ }} {{eqn	| r = \frac {\pi \paren {\size q - \size p} } 2 }} {{end-eqn}} {{qed}}
Fix $p$ and set:  :$\displaystyle \map I \alpha = \int_0^\infty \frac {e^{-\alpha x} } {x \csc p x} \rd x$ for all $\alpha \ge 0$. Then:  :$\displaystyle \int_0^\infty \frac {e^{-a x} - e^{-b x} } {x \csc p x} \rd x = \map I a - \map I b$ We have:  {{begin-eqn}} {{eqn	| l = \map {I'} \alpha 	| r = \frac \d {\d \alpha} \int_0^\infty \frac {e^{-\alpha x} } {x \csc p x} \rd x }} {{eqn	| r = \int_0^\infty \frac \partial {\partial \alpha} \paren {\frac {e^{-\alpha x} } {x \csc p x} } \rd x 	| c = Definite Integral of Partial Derivative }} {{eqn	| r = -\int_0^\infty e^{-\alpha x} \sin p x \rd x 	| c = Derivative of $e^{a x}$, {{Defof|Cosecant/Real Function|Cosecant}} }} {{eqn	| r = -\frac p {\alpha^2 + p^2} 	| c = Definite Integral to Infinity of $e^{-a x} \sin b x$ }} {{end-eqn}} so: {{begin-eqn}} {{eqn	| l = \map I \alpha 	| r = -p \int \frac 1 {\alpha^2 + p^2} \rd \alpha }} {{eqn	| r = -\arctan \frac \alpha p + C 	| c = Primitive of $\dfrac 1 {x^2 + a^2}$ }} {{end-eqn}} for all $\alpha \ge 0$ and constant $C \in \R$. We therefore have:  {{begin-eqn}} {{eqn	| l = \int_0^\infty \frac {e^{-a x} - e^{-b x} } {x \csc p x} \rd x 	| r = \map I a - \map I b }} {{eqn	| r = \paren {-\arctan \frac a p + C} - \paren {-\arctan \frac b p + C} }} {{eqn	| r = \arctan \frac b p - \arctan \frac a p }} {{end-eqn}} {{qed}}
For $n \ge 0$: {{begin-eqn}} {{eqn | l = \int_\alpha^{\alpha + n L} \map f x \d x       | r = \int_\alpha^0 \map f x \d x + \sum_{k \mathop = 0}^{n - 1} \int_{k L}^{\paren {k + 1} L} \map f x \d x + \int_{n L}^{\alpha + n L} \map f x \d x       | c = Sum of Integrals on Adjacent Intervals for Integrable Functions/Corollary }} {{eqn | r = \int_\alpha^0 \map f x \d x + \sum_{k \mathop = 0}^{n - 1} \int_{k L}^{\paren {k + 1} L} \map f {x - k L} \d x + \int_{n L}^{\alpha + n L} \map f {x - n L} \d x       | c = General Periodicity Property }} {{eqn | r = \int_\alpha^0 \map f x \d x + \sum_{k \mathop = 0}^{n - 1} \int_0^L \map f x \d x + \int_0^\alpha \map f x \d x       | c = Integration by Substitution }} {{eqn | r = n \int_0^L \map f x \d x       | c = Reversal of Limits of Definite Integral }} {{end-eqn}} For $n < 0$: {{begin-eqn}} {{eqn | l = \int_\alpha^{\alpha + n L} \map f x \d x       | r = -\int_{\alpha + n L}^\alpha \map f x \d x       | c = Reversal of Limits of Definite Integral }} {{eqn | r = -\int_{\alpha + n L}^{\alpha + n L + \paren {-n L} } \map f x \d x }} {{eqn | r = -\paren {-n \int_0^L \map f x \d x}       | c = by the above; $-n > 0$ }} {{eqn | r = n \int_0^L \map f x \d x }} {{end-eqn}} Hence the result. {{qed}} Category:Definite Integrals Category:Periodic Functions bewj8ikxf8sp0alog7mr5srk702adv3
By definition of the rank function on $M$: :$\size X \le \map \rho A$ === Case 1 : $\size X = \map \rho A$ === Let $\size X = \map \rho A$. Let $Y = X$ and the result follows. {{qed|lemma}}   === Case 2 : $\size X < \map \rho A$ ===   Let $\size X < \map \rho A$. By definition of the rank function on $M$: :$\map \rho A = \max \set{\size I : I \subseteq A \land I \in \mathscr I}$ From Max Equals an Operand: :$\exists Z \in \mathscr I : Z \subseteq A$ and $\size Z = \map \rho A$ From Independent Set can be Augmented by Larger Independent Set: :$\exists Y' \subseteq Z \setminus X: \size {X \cup Y'} = \size Z$ and $X \cup Y' \in \mathscr I$ We have: {{begin-eqn}} {{eqn | l = Y'       | o = \subseteq       | r = Z \setminus X }} {{eqn | o = \subseteq       | r = Z       | c = Set Difference is Subset }} {{eqn | o = \subseteq       | r = A       | c = by choice of $Z$ }} {{end-eqn}} From Union of Subsets is Subset: :$X \cup Y' \subseteq A$ Let $Y = X \cup Y'$ and the result follows. {{qed|lemma}} In either case, the result follows. {{qed}} Category:Matroid Theory 6ujcur7ahw1f3k9r25liuy0a1cduvu2
It is taken as a condition that $a \ne b \ne c \ne a$. We have that: :$\dfrac 1 a + \dfrac 1 b + \dfrac 1 c + \dfrac 1 n = 1$ and so we need to investigate the solutions to the above equations. From Sum of 4 Unit Fractions that equals 1, we have that the only possible solutions are: {{begin-eqn}} {{eqn | l = \dfrac 1 2 + \dfrac 1 3 + \dfrac 1 7 + \dfrac 1 {42}       | r = 1 }} {{eqn | l = \dfrac 1 2 + \dfrac 1 3 + \dfrac 1 8 + \dfrac 1 {24}       | r = 1 }} {{eqn | l = \dfrac 1 2 + \dfrac 1 3 + \dfrac 1 9 + \dfrac 1 {18}       | r = 1 }} {{eqn | l = \dfrac 1 2 + \dfrac 1 3 + \dfrac 1 {10} + \dfrac 1 {15}       | r = 1 }} {{eqn | l = \dfrac 1 2 + \dfrac 1 3 + \dfrac 1 {12} + \dfrac 1 {12}       | r = 1 }} {{eqn | l = \dfrac 1 2 + \dfrac 1 4 + \dfrac 1 5 + \dfrac 1 {20}       | r = 1 }} {{eqn | l = \dfrac 1 2 + \dfrac 1 4 + \dfrac 1 6 + \dfrac 1 {12}       | r = 1 }} {{eqn | l = \dfrac 1 2 + \dfrac 1 4 + \dfrac 1 8 + \dfrac 1 8       | r = 1 }} {{eqn | l = \dfrac 1 2 + \dfrac 1 5 + \dfrac 1 5 + \dfrac 1 {10}       | r = 1 }} {{eqn | l = \dfrac 1 2 + \dfrac 1 6 + \dfrac 1 6 + \dfrac 1 6       | r = 1 }} {{eqn | l = \dfrac 1 3 + \dfrac 1 3 + \dfrac 1 4 + \dfrac 1 {12}       | r = 1 }} {{eqn | l = \dfrac 1 3 + \dfrac 1 3 + \dfrac 1 6 + \dfrac 1 {6}       | r = 1 }} {{eqn | l = \dfrac 1 3 + \dfrac 1 4 + \dfrac 1 4 + \dfrac 1 6       | r = 1 }} {{eqn | l = \dfrac 1 4 + \dfrac 1 4 + \dfrac 1 4 + \dfrac 1 4       | r = 1 }} {{end-eqn}} From these, we can eliminate the following, because it is not the case that $a \ne b \ne c \ne a$: {{begin-eqn}} {{eqn | l = \dfrac 1 2 + \dfrac 1 5 + \dfrac 1 5 + \dfrac 1 {10}       | r = 1 }} {{eqn | l = \dfrac 1 2 + \dfrac 1 6 + \dfrac 1 6 + \dfrac 1 6       | r = 1 }} {{eqn | l = \dfrac 1 3 + \dfrac 1 3 + \dfrac 1 4 + \dfrac 1 {12}       | r = 1 }} {{eqn | l = \dfrac 1 3 + \dfrac 1 3 + \dfrac 1 6 + \dfrac 1 {6}       | r = 1 }} {{eqn | l = \dfrac 1 3 + \dfrac 1 4 + \dfrac 1 4 + \dfrac 1 6       | r = 1 }} {{eqn | l = \dfrac 1 4 + \dfrac 1 4 + \dfrac 1 4 + \dfrac 1 4       | r = 1 }} {{end-eqn}} Then we can see by inspection that the following are indeed solutions to the problem: {{begin-eqn}} {{eqn | l = \dfrac 1 2 + \dfrac 1 3 + \dfrac 1 7 + \dfrac 1 {42}       | r = 1 }} {{eqn | l = \dfrac 1 2 + \dfrac 1 3 + \dfrac 1 8 + \dfrac 1 {24}       | r = 1 }} {{eqn | l = \dfrac 1 2 + \dfrac 1 3 + \dfrac 1 9 + \dfrac 1 {18}       | r = 1 }} {{eqn | l = \dfrac 1 2 + \dfrac 1 3 + \dfrac 1 {12} + \dfrac 1 {12}       | r = 1 }} {{eqn | l = \dfrac 1 2 + \dfrac 1 4 + \dfrac 1 5 + \dfrac 1 {20}       | r = 1 }} {{eqn | l = \dfrac 1 2 + \dfrac 1 4 + \dfrac 1 6 + \dfrac 1 {12}       | r = 1 }} {{eqn | l = \dfrac 1 2 + \dfrac 1 4 + \dfrac 1 8 + \dfrac 1 8       | r = 1 }} {{end-eqn}} The remaining tuple we have is: :$\dfrac 1 2 + \dfrac 1 3 + \dfrac 1 {10} + \dfrac 1 {15} = 1$ But we note that: :$\dfrac 1 2 + \dfrac 1 3 + \dfrac 1 {10} = \dfrac {28} {30}$ which is not in the correct form. Hence the $7$ possible solutions given. {{qed}}
We observe that: {{begin-eqn}} {{eqn | l = \frac 1 j - \frac 2 {j + 1} + \frac 1 {j + 2}       | r = \frac {\paren {j + 1} \paren {j + 2} - 2 j \paren {j + 2} + j \paren {j + 1} } {j \paren {j + 1} \paren {j + 2} } }} {{eqn | r = \frac {j^2 + 3 j + 2 - 2 j^2 - 4 j + j^2 + j} {j \paren {j + 1} \paren {j + 2} } }} {{eqn | r = \frac 2 {j \paren {j + 1} \paren {j + 2} } }} {{end-eqn}} Hence: {{begin-eqn}} {{eqn | l = \sum_{j \mathop = 1}^n \frac 1 {j \paren {j + 1} \paren {j + 2} }       | r = \frac 1 2 \sum_{j \mathop = 1}^n \paren {\frac 1 j - \frac 2 {j + 1} + \frac 1 {j + 2} } }} {{eqn | r = \frac 1 2 \paren {\sum_{j \mathop = 1}^n \frac 1 j - \sum_{j \mathop = 1}^n \frac 2 {j + 1} + \sum_{j \mathop = 1}^n \frac 1 {j + 2} } }} {{eqn | r = \frac 1 2 \paren {\sum_{j \mathop = 1}^n \frac 1 j - \sum_{j \mathop = 2}^{n + 1} \frac 2 j + \sum_{j \mathop = 3}^{n + 2} \frac 1 j}       | c = Translation of Index Variable of Summation }} {{eqn | r = \frac 1 2 \paren {\frac 1 1 + \frac 1 2 + \sum_{j \mathop = 3}^n \frac 1 j - \frac 2 2 - \frac 2 {n + 1} - \sum_{j \mathop = 3}^n \frac 2 j + \frac 1 {n + 1} + \frac 1 {n + 2} + \sum_{j \mathop = 3}^n \frac 1 j} }} {{eqn | r = \frac 1 2 \paren {\frac 1 1 + \frac 1 2 - \frac 2 2 - \frac 2 {n + 1} + \frac 1 {n + 1} + \frac 1 {n + 2} } }} {{eqn | r = \frac 1 2 \paren {\frac 1 2 - \frac 1 {n + 1} + \frac 1 {n + 2} } }} {{eqn | r = \frac {\paren {n + 1} \paren {n + 2} - 2 \paren {n + 2} + 2 \paren {n + 1} } {4 \paren {n + 1} \paren {n + 2} } }} {{eqn | r = \frac {n^2 + 3 n + 2 - 2 n - 4 + 2 n + 2} {4 \paren {n + 1} \paren {n + 2} } }} {{eqn | r = \frac {n^2 + 3 n} {4 \paren {n + 1} \paren {n + 2} } }} {{eqn | r = \frac {n \paren {n + 3} } {4 \paren {n + 1} \paren {n + 2} } }} {{end-eqn}} {{qed}} Category:Sums of Sequences Category:Reciprocals Category:Sum of Sequence of Products of 3 Consecutive Reciprocals qkvjhla4ruv1a7xa74143x5lni4gbfy
{{begin-eqn}} {{eqn | l = \sum_{j \mathop = 1}^\infty \frac 1 {j \paren {j + 1} \paren {j + 2} }       | r = \lim_{n \mathop \to \infty} \sum_{j \mathop = 1}^n \frac 1 {j \paren {j + 1} \paren {j + 2} } }} {{eqn | r = \lim_{n \mathop \to \infty} \frac {n \paren {n + 3} } {4 \paren {n + 1} \paren {n + 2} }       | c = Sum of Sequence of Products of 3 Consecutive Reciprocals }} {{eqn | r = \lim_{n \mathop \to \infty} \frac {1 + \frac 3 n} {4 \paren {1 + \frac 1 n} \paren {1 + \frac 2 n} }       | c = dividing top and bottom by $n^2$ }} {{eqn | r = \frac 1 4       | c = Basic Null Sequences }} {{end-eqn}} {{qed}} Category:Limits of Series Category:Reciprocals Category:Sum of Sequence of Products of 3 Consecutive Reciprocals 5rbergex7jbt15a4pgmcgfn3s1ft25m
Let $\triangle ABC$ be a triangle whose sides $a$, $b$ and $c$ are rational in length. Let $\AA$ be the area of $\triangle ABC$. Let $h_a$, $h_b$ and $h_c$ be the altitudes from $A$, $B$ and $C$ respectively. From Area of Triangle in Terms of Side and Altitude, we have: :$\AA = \dfrac {c \cdot h_c} 2 = \dfrac {b \cdot h_b} 2 = \dfrac {a \cdot h_a} 2$ We are given that $a$, $b$ and $c$ are rational. It follows from Rational Multiplication is Closed and Rational Division is Closed that $\AA$ is rational {{iff}} $h_a$, $h_b$ and $h_c$ are rational. Hence the result. {{qed}}
We note that a Heronian triangle whose sides are all consecutive integers is also known as a Fleenor-Heronian triangle. From Sequence of Fleenor-Heronian Triangles, we have that the smallest such triangles are as follows: :$\tuple {1, 2, 3}$, which has an altitude of $0$ This is the degenerate case where the Heronian triangle is a straight line. While $0, 1, 2, 3$ is a sequence of $4$ consecutive integers, this is not technically a triangle. :$\tuple {3, 4, 5}$ with area $6$. It has altitudes $3$, $4$ and $\dfrac {12} 5$. :$\tuple {13, 14, 15}$ This can be constructed by placing the $2$ Pythagorean triangles $\tuple {5, 12, 13}$ and $\tuple {9, 12, 15}$ together along their common side $12$: :500px Thus the altitude and sides are: :$\tuple {12, 13, 14, 15}$ and this is the Heronian triangle we seek. It has area $84$. The next largest Fleenor-Heronian triangle has sides $\tuple {51, 52, 53}$. Using Heron's Formula, its area is given by: :$\AA = \sqrt {78 \times 25 \times 26 \times 27} = 1170$ Hence its altitudes are: :$45 \frac {45} {51}$, $45$, $44 \frac 8 {53}$ For still larger triangles, the altitudes are never within $1$ unit of the sides: Consider the triangle with sides $\tuple {a - 1, a, a + 1}$. Using Heron's Formula, its area is given by: {{begin-eqn}} {{eqn | l = \AA       | r = \sqrt {s \paren {s - a + 1} \paren {s - a} \paren {s - a - 1} } }} {{eqn | r = \sqrt {\frac 3 2 a \paren {\frac 1 2 a + 1} \paren {\frac 1 2 a} \paren {\frac 1 2 a - 1} } }} {{eqn | r = \frac a 4 \sqrt {3 \paren {a + 2} \paren {a - 2} } }} {{eqn | r = \frac a 4 \sqrt {3 a^2 - 12} }} {{end-eqn}} Its longest altitude is therefore: {{begin-eqn}} {{eqn | l = \frac {2 a} {4 \paren {a - 1} } \sqrt {3 a^2 - 12}       | o = <       | r = \frac {a^2 \sqrt 3} {2 \paren {a - 1} } }} {{end-eqn}} and we have: {{begin-eqn}} {{eqn | l = \frac {a^2 \sqrt 3} {2 \paren {a - 1} }       | o = <       | r = \paren {a - 1} - 1 }} {{eqn | ll= \leadstoandfrom       | l = a^2 \sqrt 3       | o = <       | r = 2 \paren {a - 1}^2 - 2 \paren {a - 1} }} {{eqn | ll= \leadstoandfrom       | l = 2 a^2 - 4 a + 2 - 2 a + 2 - \sqrt 3 a^2       | o = >       | r = 0 }} {{eqn | ll= \leadstoandfrom       | l = \paren {2 - \sqrt 3} a^2 - 6 a + 4       | o = >       | r = 0 }} {{eqn | ll= \leadsto       | l = a       | o = >       | r = \frac {6 + \sqrt {6^2 - 4 \times 4 \paren {2 - \sqrt 3} } } {2 \paren {2 - \sqrt 3} }       | c = Quadratic Formula }} {{eqn | o = \approx       | r = 21.7 }} {{end-eqn}} This shows that for $a \ge 22$, all altitudes of the triangle is less than $a - 2$. Hence there are no more examples. {{qed}}
For a given triangle $T$: :let $\map \AA T$ denote the area of $T$ :let $\map P T$ denote the perimeter of $T$. We are given that $T_1$ is an integer Heronian triangle whose sides are $a$, $b$ and $c$. Let $\map P {T_1} = k \map \AA {T_1}$. Let $T_2$ have sides $k a$, $k b$ and $k c$. Then we have that: {{begin-eqn}} {{eqn | l = \map P {T_2}       | r = k \map P {T_1}       | c =  }} {{eqn | l = \map A {T_2}       | r = k^2 \map A {T_1}       | c =  }} {{eqn | r = k \map P {T_1}       | c =  }} {{eqn | r = \map P {T_2}       | c =  }} {{end-eqn}} {{qed}} Category:Heronian Triangles hfgq1h2x4tu21n5ln7trqjwr8qpy72w