hcju/proofwikiqa huggingface dataset:
=== Positive definiteness === {{begin-eqn}} {{eqn | l = \norm {x}_\infty       | r = \sup_{t \mathop \in I} \size {\map x t}       | c = {{defof|Supremum Norm on Space of Continuous on Closed Interval Real-Valued Functions}} }} {{eqn | r = \max_{t \mathop \in I} \size {\map x t}       | c = Weierstrass Extreme Value Theorem }} {{eqn | o = \ge       | r = \size {\map x t}       | c = {{defof|Max Operation}} }} {{eqn | o = \ge       | r = 0       | c = Complex Modulus is Non-Negative }} {{end-eqn}} Suppose $\norm {x}_\infty = 0$. Then: {{begin-eqn}} {{eqn | l = 0       | r = \norm {x}_\infty }} {{eqn | r = \sup_{t \mathop \in I} \size {\map x t}       | c = {{defof|Supremum Norm on Space of Continuous on Closed Interval Real-Valued Functions}} }} {{eqn | r = \max_{t \mathop \in I} \size {\map x t}       | c = Weierstrass Extreme Value Theorem }} {{eqn | o = \ge       | r = \size {\map x t}       | c = {{defof|Max Operation}} }} {{eqn | o = \ge        | r = 0       | c = Complex Modulus is Non-Negative }} {{eqn | ll = \leadsto       | l = \size {\map x t}       | r = 0 }} {{eqn | ll = \leadsto       | l = \map x t       | r = 0       | c = Complex Modulus equals Zero iff Zero }} {{end-eqn}} Therefore: :$\displaystyle \forall t \in I : \map x t = 0$ === Positive homogeneity ===  Let $\alpha \in \R$. {{begin-eqn}} {{eqn | l = \norm {\alpha \cdot x}_\infty       | r = \sup_{t \mathop \in I} \size {\map {\paren {\alpha \cdot x} } t}       | c = {{defof|Supremum Norm on Space of Continuous on Closed Interval Real-Valued Functions}} }} {{eqn | r = \max_{t \mathop \in I} \size {\map {\paren {\alpha \cdot x} } t}       | c = Weierstrass Extreme Value Theorem }} {{eqn | r = \max_{t \mathop \in I} \size {\alpha {\map x t} }       | c = {{defof|Pointwise Scalar Multiplication of Real-Valued Functions}} }} {{eqn | r = \max_{t \mathop \in I} \size \alpha \size {\map x t}       | c = Absolute Value of Product }} {{eqn | r = \size \alpha \max_{t \mathop \in I} \size {\map x t} }} {{eqn | r = \size \alpha \sup_{t \mathop \in I} \size {\map x t}       | c = Weierstrass Extreme Value Theorem }} {{eqn | r = \size \alpha \norm {x}_\infty       | c = {{defof|Supremum Norm on Space of Continuous on Closed Interval Real-Valued Functions}} }} {{end-eqn}} === Triangle inequality === {{begin-eqn}} {{eqn | l = \size {\map {\paren {x_1 + x_2} } t}       | r = \size {\map {x_1} t + \map {x_2} t}       | c = {{defof|Pointwise Addition of Real-Valued Functions}} }} {{eqn | o = \le       | r = \size {\map {x_1} t} + \size {\map {x_2} t}       | c = Triangle Inequality for Real Numbers }} {{eqn | o = \le       | r = \max_{t \mathop \in I} \size {\map {x_1} t} + \max_{t \mathop \in I} \size {\map {x_2} t}       | c = {{defof|Max Operation}} }} {{eqn | r = \sup_{t \mathop \in I} \size {\map {x_1} t } + \sup_{t \mathop \in I} \size { \map {x_2} t }       | c = Weierstrass Extreme Value Theorem }} {{eqn | r = \norm {x_1}_\infty + \norm {x_2}_\infty       | c = {{defof|Supremum Norm on Space of Continuous on Closed Interval Real-Valued Functions}} }} {{end-eqn}}
We proceed by Proof by Contraposition. Let $A \cup \set x$ be independent. By matroid axiom $( \text I 2)$: :$A$ is independent We have: {{begin-eqn}} {{eqn | l = \map \rho {A \cup \set x}        | r = \size {A \cup \set x}       | c = Rank of Independent Subset Equals Cardinality }} {{eqn | r = \size A + \size{\set x}       | c = Corollary to Cardinality of Set Union }} {{eqn | r = \size A + 1       | c = Cardinality of Singleton }} {{eqn | o = >       | r = \size A }} {{eqn | r = \map \rho A       | c = Rank of Independent Subset Equals Cardinality }} {{end-eqn}} Then $x$ does not depend on $A$ by definition. The theorem holds by the Rule of Transposition. {{qed}} Category:Matroid Theory 1imatcs3l1aayw7qezhwfd1eluhtu15
=== Necessary Condition === Let $x \in \map \sigma X$. By definition of the closure: :$x$ depends on $X$. From Union with Disjoint Singleton is Dependent if Element Depends on Subset: :$X \cup \set x$ is dependent. {{qed|lemma}} === Sufficient Condition === Let $X \cup \set x$ be dependent. ==== Lemma ==== {{:Element Depends on Independent Set iff Union with Singleton is Dependent/Lemma}}{{qed|lemma}} From Max yields Supremum of Operands: :$\map \rho {X \cup \set x} = \max \set {\size A : A \subseteq X \cup \set x \land A \in \mathscr I} \le \size X$ By assumption: :$X \in \mathscr I$ From Max yields Supremum of Operands: :$\size X \le \map \rho {X \cup \set x}$ Thus: :$\size X = \map \rho {X \cup \set x}$ From Rank of Independent Subset Equals Cardinality: :$\map \rho X = \size X$ Thus: :$\map \rho X = \map \rho {X \cup \set x}$ It follows that $x$ depends on $X$ by definition. So: :$x \in \map \sigma X$ {{qed}} Category:Matroid Theory Category:Element Depends on Independent Set iff Union with Singleton is Dependent jzb8a7ulf7qe7hmju1wbvtzkkjojfjb
==== Case 1: $x \in A$ ==== Let $x \in A$. We have: {{begin-eqn}} {{eqn | l = A \setminus \set x       | o = \subseteq       | r = \paren {X \cup \set x} \setminus \set x       | c = Set Difference over Subset }} {{eqn | r = \paren {X \setminus \set x} \cup \paren {\set x \setminus \set x}       | c = Set Difference is Right Distributive over Union }} {{eqn | r = X \cup \paren {\set x \setminus \set x}       | c = Set Difference with Disjoint Set }} {{eqn | r = X \cup \O       | c = Set Difference with Superset is Empty Set }} {{eqn | r = X       | c = Union with Empty Set }} {{end-eqn}} {{AimForCont}}: :$A \setminus \set x = X$ Then: {{begin-eqn}} {{eqn | l = X \cup \set x       | r = \paren {A \setminus \set x} \cup \set x       | c =  }} {{eqn | r = A       | c = Set Difference Union Second Set is Union }} {{end-eqn}} So: :$X \cup \set x$ is independent. This contradicts: :$X \cup \set x$ is dependent. So: :$A \setminus \set x \subsetneq X$ Then: {{begin-eqn}} {{eqn | l = \size X       | o = >       | r = \size {A \setminus \set x}       | c = Cardinality of Proper Subset of Finite Set }} {{eqn | r = \size A - \size {\set x}       | c = Cardinality of Set Difference with Subset }} {{eqn | r = \size A - 1       | c = Cardinality of Singleton }} {{end-eqn}} So: :$\size A \le \size X$ {{qed|lemma}} ==== Case 2: $x \notin A$ ==== Let $x \notin A$. Then: {{begin-eqn}} {{eqn | l = A       | r = \paren {X \cup \set x} \cap A       | c = Intersection with Subset is Subset }} {{eqn | r = \paren {X \cap A} \cup \paren {\set x \cap A}       | c = Intersection Distributes over Union }} {{eqn | r = \paren {X \cap A}  \cup \O       | c = Intersection With Singleton is Disjoint if Not Element }} {{eqn | r = X \cap A       | c = Union with Empty Set }} {{end-eqn}} From Intersection with Subset is Subset: :$A \subseteq X$ From Cardinality of Subset of Finite Set: :$\size A \le \size X$ {{qed|lemma}} In either case: :$\size A \le \size X$ {{qed}} Category:Element Depends on Independent Set iff Union with Singleton is Dependent 10jyue4m0c7k6n9v64pm847l618e8c9
{{begin-eqn}} {{eqn | l = \map \rho X       | r = \max \set {\size Y : Y \subseteq X \land X \in \mathscr I}       | c = {{Defof|Rank Function (Matroid)}} }} {{eqn | r = \max \set {\size Y : Y \in \powerset X \land X \in \mathscr I}       | c = {{Defof|Power Set}} of $X$ }} {{eqn | r = \max \set {\size Y : Y \in \powerset X \cap \mathscr I}       | c = {{Defof|Intersection}} }} {{eqn | r = \max \set {\size Y : Y \in \powerset X}       | c = Matroid axiom $(\text I 2)$ }} {{eqn | r = \size X       | c = Cardinality of Proper Subset of Finite Set }} {{end-eqn}} {{qed}} Category:Matroid Theory k9pv9s0dp4n4bglwvc5wk3460w4qt6l
Let the form of $\map G z$ be assumed as: {{begin-eqn}} {{eqn | l = \map G z       | r = \sum_{k \mathop \ge 0} L_k z^k       | c =  }} {{eqn | r = L_0 + L_1 z + L_2 z^2 + L_3 z^3 + L_4 z^4 + \cdots       | c =  }} {{eqn | r = 2 + z + 3 z^2 + 4 z^3 + 7 z^4 + \cdots       | c =  }} {{end-eqn}} where $L_n$ denotes the $n$th Lucas number. Then: {{begin-eqn}} {{eqn | l = z \map G z       | r = L_0 z + L_1 z^2 + L_2 z^3 + L_3 z^4 + L_4 z^5 + \cdots       | c =  }} {{eqn | l = z^2 \map G z       | r = L_0 z^2 + L_1 z^3 + L_2 z^4 + L_3 z^5 + L_4 z^6 + \cdots       | c =  }} {{end-eqn}} and so: {{begin-eqn}} {{eqn | l = \paren {1 - z - z^2} \map G z       | r = L_0 + \paren {L_1 - L_0} z + \paren {L_2 - L_1 - L_0} z^2 + \paren {L_3 - L_2 - L_1} z^3 + \cdots       | c =  }} {{eqn | r = L_0 + \paren {L_1 - L_0} z       | c = {{Defof|Lucas Number}}: $L_n = L_{n - 1} + L_{n - 2}$ }} {{eqn | r = 2 - z       | c = {{Defof|Lucas Number}}: $L_0 = 2, L_1 = 1$ }} {{end-eqn}} Hence the result: :$\map G z = \dfrac {2 - z} {1 - z - z^2}$ {{qed}} Category:Generating Functions Category:Lucas Numbers 1w3pj53p6rjf7zepkedhxtc72mut7w4
:500px Let $\triangle OAB$ be the isosceles triangle $T$ formed by the legs $OA$ and $OB$. Thus the apex of $T$ is at $O$. Let $\theta$ be the angle $\angle AOB$. We see that by keeping $OA$ fixed, $B$ can range over the semicircle $AOB$. Thus $\theta$ can range from $0$ to $180 \degrees$, that is, $2$ right angles. From Area of Triangle in Terms of Two Sides and Angle, the area $\AA$ of $T$ is: :$\AA = \dfrac 1 2 a^2 \sin \theta$ This is a maximum when $\sin \theta = 1$, that is, when $\theta$ is a right angle. The result follows. {{qed}}
As $DE = CD = DA$, the points $A$, $C$ and $E$ all lie on a circle of radius $AC$. $CE$ is a straight line through the centers of circle $ACE$ and so is a diameter of circle $ACE$. Hence by Thales' Theorem, $\angle CAE$ is a right angle {{qed}}
By Gregory Series: :$\displaystyle \arctan t = \sum_{m \mathop = 0}^\infty \frac {\paren {-1}^m t^{2 m + 1} } {2 m + 1}$ Let $t = \dfrac x {\sqrt {1 - x^2} }$. Let $y = \arcsin x$. Then: {{begin-eqn}} {{eqn | l = t       | r = \frac {\sin y} {\sqrt {1 - \sin^2 y} } }} {{eqn | r = \frac {\sin y} {\cos y}       | c = Sum of Squares of Sine and Cosine }} {{eqn | r = \tan y }} {{end-eqn}} Hence $\arctan t = \arcsin x$. We have: {{begin-eqn}} {{eqn | l = \frac {2 x \arcsin x} {\sqrt {1 - x^2} }       | r = 2 t \arctan t }} {{eqn | r = 2 t \sum_{m \mathop = 0}^\infty \frac {\paren {-1}^m t^{2 m + 1} } {2 m + 1}       | c = Gregory Series }} {{eqn | r = 2 t \sum_{m \mathop = 1}^\infty \frac {\paren {-1}^{m - 1} t^{2 m - 1} } {2 m - 1}       | c = Translation of Index Variable of Summation }} {{eqn | r = 2 \sum_{m \mathop = 1}^\infty \frac {\paren {-1}^{m - 1} t^{2 m} } {2 m - 1} }} {{eqn | r = 2 \sum_{m \mathop = 1}^\infty \frac {\paren {-1}^{m - 1} x^{2 m} } {\paren {2 m - 1} \paren {1 - x^2}^m} }} {{eqn | r = 2 \sum_{m \mathop = 1}^\infty \frac {\paren {-1}^{m - 1} x^{2 m} } {2 m - 1} \sum_{k \mathop = 0}^\infty \dbinom {m + k - 1} {m - 1} x^{2 k}       | c = Binomial Theorem for Negative Index and Negative Parameter }} {{end-eqn}} It remains to show the the coefficient of $x^{2 n}$ on the {{RHS}} is equal to $\dfrac {2^{2 n} } {n \dbinom {2 n} n}$, that is: :$\displaystyle 2 \sum_{r \mathop = 1}^n \frac {\paren {-1}^{r - 1} } {2 r - 1} \dbinom {r + n - r - 1} {r - 1} = \frac {2^{2 n} } {n \dbinom {2 n} n}$ The {{LHS}} above is generated by picking, for each $m > 0$, the corresponding $k = n - m$ from the right sum $\displaystyle \sum_{k \mathop = 0}^\infty \dbinom {m + k - 1} {m - 1} x^{2 k}$. We have: {{begin-eqn}} {{eqn | r = 2 n \dbinom {2 n} n \sum_{r \mathop = 1}^n \frac {\paren {-1}^{r - 1} } {2 r - 1} \dbinom {n - 1} {r - 1}       | o = }} {{eqn | r = 2 n \dbinom {2 n} n \sum_{r \mathop = 0}^{n - 1} \frac {\paren {-1}^r} {2 r + 1} \dbinom {n - 1} r       | c = Translation of Index Variable of Summation }} {{eqn | r = 2 n \dbinom {2 n} n \int_0^1 \sum_{r \mathop = 0}^{n - 1} \paren {-1}^r \dbinom {n - 1} r y^{2 r} \d y }} {{eqn | r = 2 n \dbinom {2 n} n \int_0^1 \paren {1 - y^2}^{n - 1} \d y       | c = Binomial Theorem }} {{eqn | r = 2 n \dbinom {2 n} n \int_{\frac \pi 2}^0 \sin^{2 n - 2} \theta \, \frac {\d y} {\d \theta} \d \theta       | c = by substitution of $y = \cos \theta$ }} {{eqn | r = 2 n \dbinom {2 n} n \int_0^{\frac \pi 2} \sin^{2 n - 1} \theta \, \d \theta }} {{eqn | r = 2 n \dbinom {2 n} n \frac {\paren {2^{n - 1} \paren {n - 1}!}^2} {\paren {2 n - 1}!}       | c = Definite Integral from 0 to Half Pi of Odd Power of Sine x }} {{eqn | r = 2 n \paren {\frac {\paren {2 n}!} {n! \, n!} } \paren {\frac {\paren {2^{n - 1} \paren {n - 1}!}^2} {\paren {2 n - 1}!} }       | c = {{Defof|Binomial Coefficient}} }} {{eqn | r = 2 n \paren {\frac {2 n} {n^2} } \paren {2^{2 n - 2} } }} {{eqn | r = 2^{2 n} }} {{end-eqn}} Hence the result. {{qed}}
Fix $a \in \R$ and define the function $f_a : \R \to \R$ by:  :$\map {f_a} x = \map \exp {a - x} \exp x$ for all $x \in \R$. We aim to establish that:  :$\map {f_a} x = \map \exp {a - x} \exp x = \exp a$ for all $a, x \in \R$.  Then, we can fix $x, y \in \R$ and set $a = x + y$ to obtain:  :$\map {f_a} x = \exp y \exp x = \map \exp {x + y}$ which is the claim. Note that $f_a$ is differentiable and we have:  {{begin-eqn}} {{eqn	| l = \map {f_a'} x 	| r = \map {\frac \d {\d x} } {\map \exp {a - x} } \exp x + \map {\frac \d {\d x} } {\exp x} \map \exp {a - x} 	| c = Product Rule }} {{eqn	| r = -\map \exp {a - x} \exp x + \exp x \map \exp {a - x} 	| c = Chain Rule, Derivative of Exponential Function }} {{eqn	| r = 0 }} {{end-eqn}} From Zero Derivative implies Constant Function, $f_a$ is constant. That is, there exists $C \in \R$ such that:  :$\map {f_a} x = C$ for all $x \in \R$. We have:  {{begin-eqn}} {{eqn	| l = \map {f_a} a 	| r = \map \exp {a - a} \exp a }} {{eqn	| r = \exp 0 \exp a }} {{eqn	| r = \exp a 	| c = Exponential of Zero }} {{end-eqn}} So:  :$\map {f_a} x = \exp a$ for all $a, x \in \R$.  That is:  :$\map \exp {a - x} \map \exp x = \exp a$ for all $a, x \in \R$.  Hence the claim. {{qed}}
Let:  :$\theta = \arccos x$ Then: :$x = \cos \theta$ and:  :$0 \le \theta < \pi$ by the definition of arccosine. Then:  {{begin-eqn}} {{eqn	| l = 2 \map \arctan {\sqrt {\frac {1 - x} {1 + x} } } 	| r = 2 \map \arctan {\sqrt {\frac {1 - \cos \theta} {1 + \cos \theta} } } }} {{eqn	| r = 2 \map \arctan {\sqrt {\frac {2 \sin^2 \frac \theta 2} {2 \cos^2 \frac \theta 2} } } 	| c = Double Angle Formulas: Cosine: Corollary 1, Double Angle Formulas: Cosine: Corollary 2 }} {{eqn	| r = 2 \map \arctan {\tan \frac \theta 2} 	| c = for $0 \le \theta < \dfrac \pi 2$ we have $\sin \theta \ge 0$ and $\cos \theta > 0$ }} {{eqn	| r = \theta 	| c = {{Defof|Arctangent}} }} {{eqn	| r = \arccos x }} {{end-eqn}} {{qed}} Category:Arccosine Function Category:Arctangent Function 9w15qwluzce1ugsmhchgoisx49c76w6
Since $a > b > 0$, we have $a^2 > b^2$. So: {{begin-eqn}} {{eqn	| l = \int_0^{\pi/2} \frac 1 {a + b \cos x} \rd x 	| r = \intlimits {\frac 2 {\sqrt {a^2 - b^2} } \map \arctan {\sqrt {\frac {a - b} {a + b} } \tan \frac x 2} } 0 1 	| c = Primitive of $\dfrac 1 {p + q \cos x}$ }} {{eqn	| r = \frac 1 {\sqrt {a^2 - b^2} } \paren {2 \map \arctan {\sqrt {\frac {a - b} {a + b} } } } }} {{eqn	| r = \frac 1 {\sqrt {a^2 - b^2} } \paren {2 \map \arctan {\sqrt {\frac {1 - \frac b a} {1 + \frac b a} } } } }} {{eqn	| r = \frac 1 {\sqrt {a^2 - b^2} } \map \arccos {\frac b a} 	| c = Arccosine in terms of Arctangent }} {{end-eqn}} {{qed}}
Take $a$ constant and define:  :$\displaystyle \map I b = \int_0^\infty \frac {e^{-a x} \sin b x} x \rd x$ We have:  {{begin-eqn}} {{eqn	| l = \map {I'} b 	| r = \frac \d {\d b} \int_0^\infty \frac {e^{-a x} \sin b x} x \rd x }} {{eqn	| r = \int_0^\infty \frac \partial {\partial b} \paren {\frac {e^{-a x} \sin b x} x} \rd x 	| c = Definite Integral of Partial Derivative }} {{eqn	| r = \int_0^\infty e^{-a x} \cos b x \rd x 	| c = Derivative of $\cos a x$ }} {{eqn	| r = \frac a {a^2 + b^2} 	| c = Definite Integral to Infinity of $e^{-a x} \cos b x$ }} {{end-eqn}} so: {{begin-eqn}} {{eqn	| l = \map I b 	| r = a \int \frac 1 {b^2 + a^2} \rd b }} {{eqn	| r = \frac a a \arctan \frac b a + C 	| c = Primitive of $\dfrac 1 {x^2 + a^2}$ }} {{eqn	| r = \arctan \frac b a + C }} {{end-eqn}} for some constant $C \in \R$. We have:  {{begin-eqn}} {{eqn	| l = \map I 0 	| r = \int_0^\infty \frac {e^{-a x} \sin 0} x \rd x }} {{eqn	| r = \int_0^\infty 0 \rd x }} {{eqn	| r = 0 }} {{end-eqn}} on the other hand: {{begin-eqn}} {{eqn	| l = \map I b 	| r = \arctan 0 + C }} {{eqn	| r = C }} {{end-eqn}} so: :$C = 0$ So we have:  :$\displaystyle \int_0^\infty \frac {e^{-a x} \sin b x} x \rd x = \map \arctan {\frac b a}$ as required. {{qed}}
We have: {{begin-eqn}} {{begin-eqn}} {{eqn	| l = \int_0^\infty \frac {\sin m x} {e^{2 \pi x} - 1} \rd x 	| r = \int_0^\infty \frac {e^{-2 \pi x} \sin m x} {1 - e^{-2 \pi x} } \rd x }} {{eqn	| r = \int_0^\infty e^{-2 \pi x} \sin m x \paren {\sum_{k = 0}^\infty e^{-2 \pi k x} } \rd x 	| c = Sum of Infinite Geometric Sequence }} {{eqn	| r = \sum_{k = 0}^\infty \paren {\int_0^\infty e^{-x \paren {2 \pi \paren {1 + k} } } \sin m x \rd x} }} {{eqn	| r = \sum_{k = 0}^\infty \frac m {\paren {2 \pi \paren {1 + k} }^2 + m^2} 	| c = Definite Integral to Infinity of $e^{-a x} \sin b x$ }} {{eqn	| r = \frac 1 {2 \pi} \sum_{k = 0}^\infty \frac {\frac m {2 \pi} } {\paren {k + 1}^2 + \paren {\frac m {2 \pi} }^2} 	| c = dividing by $\paren {2 \pi}^2$ }} {{eqn	| r = \frac 1 {2 \pi} \sum_{k = 1}^\infty \frac {\frac m {2 \pi} } {k^2 + \paren {\frac m {2 \pi} }^2} 	| c = shifting the index }} {{end-eqn}} By Mittag-Leffler Expansion for Hyperbolic Cotangent Function we have:  :$\displaystyle \sum_{k = 1}^\infty \frac z {k^2 + z^2} = \frac \pi 2 \map \coth {\pi z} - \frac 1 {2 z}$ for all $z \in \C$ where $z$ is not an integer multiple of $i$.  Since all variables concerned in this instance all real-valued, we can apply this identity.  Setting $z = \dfrac m {2 \pi}$ in the above we obtain: {{begin-eqn}} {{eqn	| l = \sum_{k = 1}^\infty \frac {\frac m {2 \pi} } {k^2 + \paren {\frac m {2 \pi} }^2} 	| r = \frac \pi 2 \map \coth {\pi \times \frac m {2 \pi} } - \frac 1 {2 \times \frac m {2 \pi} } }} {{eqn	| r = \frac \pi 2 \coth \frac m 2 - \frac \pi m }} {{end-eqn}} So: :$\displaystyle \int_0^\infty \frac {\sin m x} {e^{2 \pi x} - 1} \rd x = \frac 1 {2 \pi} \paren {\frac \pi 2 \coth \frac m 2 - \frac \pi m} = \frac 1 4 \coth \frac m 2 - \frac 1 {2 m}$ as required. {{qed}}
Fix $a$ and define:  :$\displaystyle \map I b = \int_0^\infty e^{-a x^2} \cos b x \rd x$ for all $b \in \R$.  Then, we have:  {{begin-eqn}} {{eqn	| l = \map {I'} b 	| r = \frac \d {\d b} \paren {\int_0^\infty e^{-a x^2} \cos b x \rd x} }} {{eqn	| r = \int_0^\infty \frac \partial {\partial b} \paren {e^{-a x^2} \cos b x} \rd x 	| c = Definite Integral of Partial Derivative }} {{eqn	| r = -\int_0^\infty \paren {x e^{-a x^2} } \sin b x \rd x 	| c = Derivative of $\cos a x$ }} {{eqn	| r = -\paren {\intlimits {-\frac 1 {2 a} e^{-a x^2} \sin b x} 0 \infty - b \int_0^\infty \paren {-\frac 1 {2 a} e^{-a x^2} } \cos b x \rd x} 	| c = Integration by Parts }} {{end-eqn}} Note that:  {{begin-eqn}} {{eqn	| l = \size {\frac 1 {2 a} e^{-a x^2} \sin b x} 	| o = \le 	| r = \frac 1 {2 a} e^{-a x^2} 	| c = noting that $\size {\sin x} \le 1$ }} {{eqn	| o = \to 	| r = 0 	| c = Exponential Tends to Zero and Infinity }} {{end-eqn}} So: {{begin-eqn}} {{eqn	| l = -\paren {\intlimits {-\frac 1 {2 a} e^{-a x^2} \sin b x} 0 \infty - b \int_0^\infty \paren {-\frac 1 {2 a} e^{-a x^2} } \cos b x \rd x} 	| r = -\frac b {2 a} \int_0^\infty e^{-a x^2} \cos b x \rd x }} {{eqn	| r = -\frac b {2 a} \map I b }} {{end-eqn}} We then have:  :$\displaystyle \frac {\map {I'} b} {\map I b} = -\frac b {2 a}$ Integrating, by Primitive of Function under its Derivative and Primitive of Constant: :$\displaystyle \ln \size {\map I b} = -\frac {b^2} {4 a} + C$ for some $C \in \R$. {{finish|This obviously only gives us an expression for $\ln \size {\map I b}$, we then need to determine that $\map I b > 0$ which seems nontrivial}} So:  :$\displaystyle \map I b = A \map \exp {-\frac {b^2} {4 a} }$ for some $A \in \R$. We have:  {{begin-eqn}} {{eqn	| l = \map I 0 	| r = \int_0^\infty e^{-a x^2} \rd x }} {{eqn	| r = \frac 1 2 \sqrt {\frac \pi a} 	| c = Definite Integral to Infinity of $e^{-a x^2}$ }} {{end-eqn}} on the other hand we have:  {{begin-eqn}} {{eqn	| l = \map I 0 	| r = A \map \exp 0 }} {{eqn	| r = A 	| c = Exponential of Zero }} {{end-eqn}} So we have:  $\displaystyle \map I b = \int_0^\infty e^{-a x^2} \cos b x \rd x = \frac 1 2 \sqrt {\frac \pi a} \map \exp {-\frac {b^2} {4 a} }$ for all $b \in \R$ as required. {{qed}}
We find the Half-Range Fourier Cosine Series over $\openint 0 {\dfrac \pi 2}$ for $\map \ln {\sin x}$. By definition: :$\displaystyle \map \ln {\sin x} \sim \frac {a_0} 2 + \sum_{n \mathop = 1}^\infty a_n \cos 2 n x$ where for all $n \in \Z_{\ge 0}$: :$\displaystyle a_n = \frac 4 \pi \int_0^{\pi/2} \map \ln {\sin x} \cos 2 n x \ \d x$ By Definite Integral from 0 to Half Pi of Logarithm of Sine x: :$a_0 = \dfrac 4 \pi \paren {- \dfrac \pi 2 \ln 2} = -2 \ln 2$ By Definite Integral from 0 to Half Pi of Logarithm of Sine x by Cosine of 2nx: :$a_n = \dfrac 4 \pi \paren {- \dfrac \pi {4 n}} = -\dfrac 1 n$ Therefore: {{begin-eqn}} {{eqn | l = \map \ln {\sin x}       | o = \sim       | r = \frac {a_0} 2 + \sum_{n \mathop = 1}^\infty a_n \cos 2 n x }} {{eqn | r = - \ln 2 - \sum_{n \mathop = 1}^\infty \frac {\cos 2 n x} n }} {{end-eqn}} {{qed}} Category:Examples of Fourier Series 90o3kbsvaey3is713tdddlykcy0litm
We have: {{begin-eqn}} {{eqn | l = A \cap \paren {S \setminus T}        | r = \paren{A \cap S} \setminus T       | c = Intersection with Set Difference is Set Difference with Intersection }} {{eqn | r = A \setminus T       | c = Intersection with Subset is Subset }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | ll =        | o =       | r = A \subseteq \paren{ S \setminus T}       | c = }} {{eqn | ll = \leadstoandfrom       | o =       | r = A = A \cap \paren{ S \setminus T}       | c = Intersection with Subset is Subset }} {{eqn | ll = \leadstoandfrom       | o =       | r = A = A \setminus T       | c = As $A \cap \paren {S \setminus T} = A \setminus T$ }} {{eqn | ll = \leadstoandfrom       | o =       | r = A \cap T = \O       | c = Set Difference with Disjoint Set }} {{end-eqn}} {{qed}} Category:Set Difference Category:Disjoint Sets 20ifr7cnqwnyfplovl2gvqlufalz85n
{{begin-eqn}} {{eqn | l = \set{x, y} \setminus \set x       | r = \big \{ z: z \in \set{x, y} \land z \notin \set x \big \}       | c = {{Defof|Set Difference}} }} {{eqn | r = \big \{ z: \paren{ z = x \lor z = y} \land z \notin \set x \big \}       | c = {{Defof|Doubleton}} }} {{eqn | r = \big \{ z: \paren{ z = x \lor z = y} \land z \ne x \big \}       | c = {{Defof|Singleton}} }} {{eqn | r = \big \{ z: \paren{z = x \land z \ne x} \lor \paren {z = y \land z \ne x}  \big \}       | c = Conjunction Distributes over Disjunction }} {{eqn | r = \big \{ z: \bot \lor \paren {z = y \land z \ne x} \big \}       | c = {{Defof|Contradiction}} }} {{eqn | r = \big \{ z: \paren {z = y \land z \ne x} \big \}       | c = Disjunction with Contradiction }} {{eqn | r = \big \{ z : z = y \big \}       | c = Rule of Simplification }} {{eqn | r = \set y       | c = {{Defof|Singleton}} }} {{end-eqn}} {{qed}} Category:Singletons Category:Doubletons 4jshkpfs1zzd6deak4twch4o3ox2k2a
{{AimForCont}} there is a solution. Then: {{begin-eqn}} {{eqn | l = a^n       | r = 2^m - 1 }} {{eqn | o = \equiv       | r = -1       | rr = \pmod 4       | c = as $m > 1$ }} {{end-eqn}} $a$ is immediately seen to be odd. By Square Modulo 4, $n$ must also be odd. Now: {{begin-eqn}} {{eqn | l = 2^m       | r = a^n + 1 }} {{eqn | r = \paren {a + 1} \sum_{k \mathop = 0}^{n - 1} \paren {-1}^k a^{n - k - 1}       | c = Sum of Two Odd Powers }} {{end-eqn}} The latter sum has $n$ powers of $a$, which sums to an odd number. The only odd divisor of $2^m$ is $1$. However, if the sum is $1$, we have: :$a^n + 1 = a + 1$ giving $n = 1$, contradicting our constraint $n > 1$. Hence the result by Proof by Contradiction. {{qed}} Category:Number Theory ex9a8rtweepg9xdh7umpt71f6nn8kuk