hcju/proofwikiqa huggingface dataset:
Suppose $\sqrt 2$ could be expressed as the root of the linear polynomial: :$a_1 x + a_0 = 0$ for some $a_0, a_1 \in \Q$. Then: :$\sqrt 2 = -\dfrac {a_0} {a_1}$ and would be rational. But as Square Root of 2 is Irrational, this is not the case. However, $\sqrt 2$ is a root of the polynomial of degree $2$: :$x^2 - 2 = 0$ Hence the result by definition of degree of algebraic number. {{qed}}
From the Gelfond-Schneider Theorem: If: :$\alpha$ and $\beta$ are algebraic numbers such that $\alpha \notin \set {0, 1}$ :$\beta$ is either irrational or not wholly real then $\alpha^\beta$ is transcendental. From Rational Number is Algebraic: :$2$ is algebraic. From Square Root of 2 is Algebraic of Degree 2: :$\sqrt 2$ is algebraic. From Square Root of 2 is Irrational: :$\sqrt 2$ is irrational. Hence the result. {{qed}}
By the recurrence relation for the Bernoulli numbers: :$B_n = \begin{cases} 1 & : n = 0 \\ \displaystyle - \sum_{k \mathop = 0}^{n-1} \binom n k \frac {B_k} {n - k + 1} & : n > 0 \end{cases}$ :$B_0$ is rational. :$B_n$ is a finite sum of the products of a binomial coefficient with a Bernoulli number earlier in the sequence divided by an integer. The proof proceeds by induction. For all $n \in \N$, let $P \left({n}\right)$ be the proposition: :$B_n$ is rational. $P \left({0}\right)$ is true, as this just says: :$B_0$ is rational. === Basis for the Induction === $P \left({0}\right)$ is the case: :$B_0$ is rational. This is true, as $B_0 = 1$ which is rational. This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that, if $P \left({j}\right)$ is true, where $0 \le j \le k$, then it logically follows that $P \left({k + 1}\right)$ is true. So this is the induction hypothesis: :$\forall j \in \N: 0 \le j \le k: B_j$ is rational from which it is to be shown that: :$B_{k + 1}$ is rational. === Induction Step === This is the induction step: By the recurrence relation for the Bernoulli numbers: {{begin-eqn}} :$\displaystyle B_{k + 1} = - \sum_{j \mathop = 0}^k \binom {k + 1} j \frac {B_j} {k - j + 2}$ We have that: :From Binomial Coefficient is Integer, $\dbinom {k + 1} j$ is an integer throughout. :From the Induction Hypothesis, $\forall j \in \N: 0 \le j \le k: B_j$ is rational. :$k - j + 2$ is an integer. Thus as $B_{k + 1}$ is the sum of a finite number of rational numbers, it follows that $B_{k + 1}$ it itself rational. The result follows by the Second Principle of Mathematical Induction. Therefore: :$\forall n \in \N: B_n$ is rational. {{qed}}
From Power Series Expansion for Tangent Function: {{begin-eqn}} {{eqn | l = \tan x       | r = \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^{n - 1} 2^{2 n} \paren {2^{2 n} - 1} B_{2 n} } {\paren {2 n}!} x^{2 n - 1}       | c =  }} {{eqn | r = \frac {2^2 \paren {2^2 - 1} B_2} {2!} x - \frac {2^4 \paren {2^4 - 1} B_4} {4!} x^3 + \frac {2^6 \paren {2^6 - 1} B_6} {6!} x^5 - \frac {2^8 \paren {2^8 - 1} B_8} {8!} x^7 + \frac {2^{10} \paren {2^{10} - 1} B_{10} } {10!} x^9 - \cdots       | c =  }} {{end-eqn}} Enumerating the Bernoulli numbers: {{begin-eqn}} {{eqn | l = B_2       | r = \dfrac 1 6 }} {{eqn | l = B_4       | r = -\dfrac 1 {30} }} {{eqn | l = B_6       | r = \dfrac 1 {42} }} {{eqn | l = B_8       | r = -\dfrac 1 {30} }} {{eqn | l = B_{10}       | r = \dfrac 5 {66} }} {{end-eqn}} Thus the appropriate arithmetic is performed on each coefficient: {{begin-eqn}} {{eqn | l = \frac {2^2 \paren {2^2 - 1} B_2} {2!}       | r = \frac {4 \times 3} {2} \times \dfrac 1 6 }} {{eqn | r = 1 }} {{end-eqn}} {{begin-eqn}} {{eqn | l = -\frac {2^4 \paren {2^4 - 1} B_4} {4!}       | r = -\dfrac {16 \times 15} {24} \times \dfrac {-1} {30} }} {{eqn | r = \dfrac {\paren {2^4} \times \paren {3 \times 5} } {\paren {2^3 \times 3} \times \paren {2 \times 3 \times 5} } }} {{eqn | r = \dfrac 1 3 }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \frac {2^6 \paren {2^6 - 1} B_6} {6!}       | r = \dfrac {64 \times 63} {720} \times \dfrac 1 {42} }} {{eqn | r = \dfrac {\paren {2^6} \times \paren {3^2 \times 7} } {\paren {2^4 \times 3^2 \times 5} \times \paren {2 \times 3 \times 7} } }} {{eqn | r = \dfrac 2 {15} }} {{end-eqn}} {{begin-eqn}} {{eqn | l = -\frac {2^8 \paren {2^8 - 1} B_8} {8!}       | r = -\dfrac {256 \times 255} {40 \, 320} \times \dfrac {-1} {30} }} {{eqn | r = \dfrac {\paren {2^8} \times \paren {3 \times 5 \times 17} } {\paren {2^7 \times 3^2 \times 5 \times 7} \times \paren {2 \times 3 \times 5} } }} {{eqn | r = \dfrac {17} {3^2 \times 5 \times 7} }} {{eqn | r = \dfrac {17} {315} }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \frac {2^{10} \paren {2^{10} - 1} B_{10} } {10!}       | r = \dfrac {1024 \times 1023} {3 \, 628 \, 800} \times \dfrac 5 {66} }} {{eqn | r = \dfrac {\paren {2^{10} } \times \paren {3 \times 11 \times 31} \times 5} {\paren {2^8 \times 3^4 \times 5^2 \times 7} \times \paren {2 \times 3 \times 11} } }} {{eqn | r = \dfrac {2 \times 31} {\paren {3^4 \times 5 \times 7} } }} {{eqn | r = \dfrac {62} {2835} }} {{end-eqn}} Hence the result. {{qed}}
We have that: :$\cot \pi x = \dfrac {\cos \pi x} {\sin \pi x}$ has a denominator which is $0$ at $x = 0, \pm 1, \pm 2, \ldots$. Hence the limitation on the domain of $x \cot \pi x$ to exclude integer $x$. Having established that, we should be able to express $\cot \pi x$ in the form: :$\cot \pi x = \dfrac a x + \displaystyle \sum_{n \mathop = 1}^\infty \left({\frac {b_n} {x - n} + \frac {c_n} {x + n} }\right)$ using a partial fractions expansion. By evaluating the coefficients $b_n$ and $c_n$ in the usual manner, they are found to be: {{explain|Link to what this usual manner is}} {{finish|Do the work}} :$\forall n \in \N: b_n = c_n = \dfrac 1 \pi$ The result follows. {{qed}}
From the Euler Formula for Sine Function: :$\displaystyle \frac {\sin x} x = \prod_{n \mathop = 1}^\infty \left({1 - \frac {x^2} {n^2 \pi^2}}\right)$ Taking the logarithm of both sides: :$\displaystyle \ln \frac {\sin x} x = \sum_{n \mathop = 1}^\infty \ln \left({1 - \frac {x^2} {n^2 \pi^2}}\right)$ and differentiating {{WRT|Differentiation}} $x$: {{begin-eqn}} {{eqn | l = \cot x - \frac 1 x       | r = \sum_{n \mathop = 1}^\infty \frac {- 2 x} {n^2 \pi^2 - x^2}       | c =  }} {{eqn | ll= \leadsto       | l = \cot x       | r = \frac 1 x + 2 x \sum_{n \mathop = 1}^\infty \frac 1 {x^2 - n^2 \pi^2}       | c =  }} {{eqn | ll= \leadsto       | l = \pi x \cot \pi x       | r = 1 + 2 x^2 \sum_{n \mathop = 1}^\infty \frac 1 {x^2 - n^2 \pi^2}       | c = putting $\pi x$ for $x$ and multiplying by $\pi x$ }} {{end-eqn}} Hence the result. {{qed}}
From Slope of Tangent to Cycloid, the slope of the tangent to $C$ at the point $\tuple {x, y}$ is given by: :$\dfrac {\d y} {\d x} = \cot \dfrac \theta 2$ At the cusps, $\theta = 2 n \pi$ for $n \in \Z$. Thus at the cusps, the slope of the tangent to $C$ is $\cot n \pi$. From Shape of Cotangent Function: :$\displaystyle \lim_{\theta \mathop \to n \pi} \cot \theta \to \infty$ Hence the result by definition of vertical tangent line. {{qed}}
From Tangent to Cycloid, the equation for the tangent to $C$ at a point $P = \tuple {x, y}$ is given by: :$(1): \quad y - a \paren {1 - \cos \theta} = \dfrac {\sin \theta} {1 - \cos \theta} \paren {x - a \theta + a \sin \theta}$ From Equation of Cycloid, the point at the top of the generating circle of $C$ has coordinates $\tuple {2 a, a \theta}$. Substituting $x = 2 a$ in $(1)$: {{begin-eqn}} {{eqn | l = y - a \paren {1 - \cos \theta}       | r = \dfrac {\sin \theta} {1 - \cos \theta} \paren {a \theta - a \theta + a \sin \theta}       | c =  }} {{eqn | ll= \leadsto       | l = y       | r = a \paren {1 - \cos \theta} + \dfrac {\sin \theta} {1 - \cos \theta} a \sin \theta       | c =  }} {{eqn | r = \frac {a \paren {1 - \cos \theta}^2 + a \sin^2 \theta} {1 - \cos \theta}       | c =  }} {{eqn | r = \frac {a \paren {1 - 2 \cos \theta + \cos^2 \theta} + a \sin^2 \theta} {1 - \cos \theta}       | c =  }} {{eqn | r = \frac {a \paren {1 - 2 \cos \theta + 1} } {1 - \cos \theta}       | c =  }} {{eqn | r = \frac {a \paren {2 - 2 \cos \theta} } {1 - \cos \theta}       | c =  }} {{eqn | r = 2 a       | c =  }} {{end-eqn}} That is, the tangent to $C$ passes through $\tuple {a \theta, 2 a}$ as was required. {{qed}}
From Slope of Tangent to Cycloid, the slope of the tangent to $C$ at the point $\left({x, y}\right)$ is given by: :$\dfrac {\mathrm d y} {\mathrm d x} = \cot \dfrac \theta 2$ This tangent to $C$ also passes through the point $\left({a \left({\theta - \sin \theta}\right),  a \left({1 - \cos \theta}\right)}\right)$. {{Finish|Find (or write) the result which gives the equation of a line from its slope and a point through which it passes.}} Hence the result. {{qed}}
:500px Let $C_1$ have rolled so that the line $OB$ through the radii of $C_1$ and $C_2$ is at angle $\theta$ to the $x$-axis. Let $C_1$ have turned through an angle $\phi$ to reach that point. By definition of sine and cosine, $P = \tuple {x, y}$ is defined by: :$x = \paren {a - b} \cos \theta + b \map \cos {\phi - \theta}$ :$y = \paren {a - b} \sin \theta - b \map \sin {\phi - \theta}$ The arc of $C_1$ between $P$ and $B$ is the same as the arc of $C_2$ between $A$ and $B$. Thus by Arc Length of Sector: :$ a \theta = b \phi$ Thus: :$\phi - \theta = \paren {\dfrac {a - b} b} \theta$ whence the result. {{qed}}
By Meet is Intersection in Set of Ideals: :$I_1 \wedge I_2 = I_1 \cap I_2$ === ($\subseteq$) === Let $x \in I_1 \cap I_2$. By definition of intersection: :$x \in I_1 \land x \in I_2$ By Meet is Idempotent: :$x = x \wedge x$ Thus :$x \in \left\{ {i_1 \wedge i_2: i_1 \in I_1, i_2 \in I_2}\right\}$ {{qed|lemma}} === ($\supseteq$) === Let $x \in \left\{ {i_1 \wedge i_2: i_1 \in I_1, i_2 \in I_2}\right\}$. Then :$\exists i_1 \in I_1, i_2 \in I_2: x = i_1 \wedge i_2$ By Meet Precedes Operands: :$x \preceq i_1 \land x \preceq i_2$ By definition of lower set: :$x \in I_1 \land x \in I_2$ Thus by definition of intersection: :$x \in I_1 \cap I_2$ {{qed}}
=== Directed Subset === Let $x, y \in I_1 \cap I_2$. By definition of intersection: :$x, y \in I_1$ and $x, y \in I_2$ By definition of directed subset: :$\exists z_1 \in I_1: x \preceq z_1 \land y \preceq z_1$ and :$\exists z_2 \in I_2: x \preceq z_2 \land y \preceq z_2$ By Meet Precedes Operands: :$z_1 \wedge z_2 \preceq z_1$ and $z_1 \wedge z_2 \preceq z_2$ By definition of lower set: :$z_1 \wedge z_2 \in I_1$ and $z_1 \wedge z_2 \in I_2$ By definition of intersection: :$z_1 \wedge z_2 \in I_1 \cap I_2$ By definition of meet: :$z_1 \wedge z_2 = \inf \set {z_1, z_2}$ By definition of infimum: :$x \preceq z_1 \wedge z_2$ and $y \preceq z_1 \wedge z_2$ Thus by definition: :$I_1 \cap I_2$ is directed. {{qed|lemma}} === Lower Set === Let $x, y \in S$ such that: :$x \in I_1 \cap I_2$ and $y \preceq x$ By definition of intersection: :$x \in I_1$ and $x \in I_2$ By definition of lower set: :$y \in I_1$ and $y \in I_2$ By definition of intersection: :$y \in I_1 \cap I_2$ Thus by definition: :$I_1 \cap I_2$ is a lower set. {{qed|lemma}} === Non-Empty Set === By assumption: :$I_1 \ne \O$ and $I_2 \ne \O$ By definition of non-empty: :$\exists x: x \in I_1$ and $\exists y: y \in I_2$ By Meet Precedes Operands: :$x \wedge y \preceq x$ and $x \wedge y \preceq y$ By definition of lower set: :$x \wedge y \in I_1$ and $x \wedge y \in I_2$ By definition of intersection: :$x \wedge y \in I_1 \cap I_2$ Thus by definition: :$I_1 \cap I_2$ is non-empty. {{qed|lemma}} Thus by definition: :$I_1 \cap I_2$ is ideal in $\struct {S, \preceq}$. {{qed}}
The length of the arc of $C_2$ between two adjacent cusps of $H$ is $2 \pi b$. The total length of the circumference of $C_1$ is $2 \pi a$. Thus the total number of cusps of $H$ is: :$\dfrac {2 \pi a} {2 \pi b} = \dfrac {2 \pi n b} {2 \pi b} = n$ {{qed}}
{{begin-eqn}} {{eqn | l = \cos 3 \theta       | r = \cos \paren {2 \theta + \theta} }} {{eqn | r = \cos 2 \theta \cos \theta - \sin 2 \theta \sin \theta       | c = Cosine of Sum }} {{eqn | r = \paren {2 \cos^2 \theta - 1} \cos \theta - 2 \sin^2 \theta \cos \theta       | c = Double Angle Formula for Cosine and Double Angle Formula for Sine }} {{eqn | r = \paren {2 \cos^2 \theta - 1 - 2 \paren {1 - 2 \cos^2 \theta} } \cos \theta       | c = Sum of Squares of Sine and Cosine }} {{eqn | r = 4 \cos^3 \theta - 3 \cos \theta       | c = gathering terms }} {{end-eqn}} {{qed}}
:400px From Equation of Astroid, $H$ can be expressed as: :$\begin{cases} x & = a \cos^3 \theta \\ y & = a \sin^3 \theta \end{cases}$ Thus the slope of the tangent to $H$ at $\tuple {x, y}$ is: {{begin-eqn}} {{eqn | l = \frac {\d y} {\d x}       | r = \frac {3 a \sin^2 \theta \cos \theta \rd \theta} {-3 a \cos^2 \theta \sin \theta \rd \theta}       | c =  }} {{eqn | r = -\tan \theta       | c =  }} {{end-eqn}} Thus the equation of the tangent to $H$ is given by: :$y - a \sin^3 \theta = -\tan \theta \paren {x - a \cos^3 \theta}$ {{explain|Find, or post up, the equation of a line of given tangent passing through point $\tuple {x, y}$ as this is what is needed here}} The $x$-intercept is found by setting $y = 0$ and solving for $x$: {{begin-eqn}} {{eqn | l = x       | r = a \cos^3 \theta + a \sin^2 \theta \cos \theta       | c =  }} {{eqn | r = a \cos \theta \paren {\cos^2 \theta + \sin^2 \theta}       | c =  }} {{eqn | r = a \cos \theta       | c = Sum of Squares of Sine and Cosine }} {{end-eqn}} Similarly, the $y$-intercept is found by setting $x = 0$ and solving for $y$, which gives: :$y = a \sin \theta$ The length of the part of the tangent to $H$ between the $x$-axis and the $y$-axis is given by: {{begin-eqn}} {{eqn | l = \sqrt {a^2 \cos^2 \theta + a^2 \sin^2 \theta}       | r = a \sqrt {\cos^2 \theta + \sin^2 \theta}       | c =  }} {{eqn | r = a       | c = Sum of Squares of Sine and Cosine }} {{end-eqn}} which is constant. {{qed}}
From Evolute of Cycloid is Cycloid, the evolute of a cycloid is another cycloid. From Curve is Involute of Evolute, the involute of a cycloid is another cycloid as well. But by the definition of involute, the path defined by the pendulum as described is the involute of the cycloid. Hence the result. {{qed}}
By Lower Closure is Closure Operator: :$D_1 \subseteq D_1^\preceq$ and $D_2 \subseteq D_2^\preceq$ Then :$\left\{ {i_1 \wedge i_2: i_1 \in D_1, i_2 \in D_2}\right\} \subseteq \left\{ {i_1 \wedge i_2: i_1 \in D_1^\preceq, i_2 \in D_2^\preceq}\right\}$ By Lower Closure is Closure Operator: :$\left\{ {i_1 \wedge i_2: i_1 \in D_1, i_2 \in D_2}\right\}^\preceq \subseteq \left\{ {i_1 \wedge i_2: i_1 \in D_1^\preceq, i_2 \in D_2^\preceq}\right\}^\preceq$ By definition of set equality it remains to prove that :$\left\{ {i_1 \wedge i_2: i_1 \in D_1^\preceq, i_2 \in D_2^\preceq}\right\}^\preceq \subseteq \left\{ {i_1 \wedge i_2: i_1 \in D_1, i_2 \in D_2}\right\}^\preceq$ Let $x \in \left\{ {i_1 \wedge i_2: i_1 \in D_1^\preceq, i_2 \in D_2^\preceq}\right\}^\preceq$. By definition of lower closure of set: :$\exists y \in \left\{ {i_1 \wedge i_2: i_1 \in D_1^\preceq, i_2 \in D_2^\preceq}\right\} \land x \preceq y$ Then: :$\exists i_1 \in D_1^\preceq, i_2 \in D_2^\preceq: y = i_1 \wedge i_2$ By definition of lower closure of set: :$\exists d_1 \in D_1: i_1 \preceq d_1$ and :$\exists d_2 \in D_2: i_2 \preceq d_2$ By Meet Semilattice is Ordered Structure: :$y \preceq d_1 \wedge d_2$ By definition of transitivity: :$x \preceq d_1 \wedge d_2$ :$d_1 \wedge d_2 \in \left\{ {i_1 \wedge i_2: i_1 \in D_1, i_2 \in D_2}\right\}$ Thus by definition of lower closure of set: :$x \in \left\{ {i_1 \wedge i_2: i_1 \in D_1, i_2 \in D_2}\right\}^\preceq$ Thus the result. {{qed}}
From Equation of Cycloid, the point $P = \tuple {x, y}$ is described by the equations: :$x = a \paren {\theta - \sin \theta}$ :$y = a \paren {1 - \cos \theta}$ Expressing $\theta$ and $\sin \theta$ in terms of $y$: {{begin-eqn}} {{eqn | l = \cos \theta       | r = 1 - \frac y a       | c =  }} {{eqn | ll= \leadsto       | l = \sin \theta       | r = \sqrt {1 - \paren {1 - \frac y a}^2}       | c =  }} {{eqn | r = \sqrt {\frac {2 y} a - \frac {y^2} {a^2} }       | c =  }} {{eqn | r = \frac {\sqrt {2 a y - y^2} } a        | c =  }} {{eqn | ll= \leadsto       | l = \theta       | r = \sin^{-1} \paren {\frac {\sqrt {2 a y - y^2} } a}       | c =  }} {{end-eqn}} Substituting for $\theta$ and $\sin \theta$ in the expression for $x$: {{begin-eqn}} {{eqn | l = x       | r = a \paren {\sin^{-1} \paren {\frac {\sqrt {2 a y - y^2} } a} - \frac 1 a \sqrt {2 a y - y^2} }       | c =  }} {{eqn | ll= \leadsto       | l = a \sin^{-1} \paren {\frac {\sqrt {2 a y - y^2} } a}       | r = \sqrt {2 a y - y^2} + x       | c =  }} {{end-eqn}} {{qed}}
From Equation of Cycloid: :$x = a \paren {\theta - \sin \theta}$ :$y = a \paren {1 - \cos \theta}$ From Slope of Tangent to Cycloid: {{begin-eqn}} {{eqn | l = y'       | r = \cot \dfrac \theta 2       | c = Slope of Tangent to Cycloid }} {{eqn | ll= \leadsto       | l = \dfrac {\d y'} {\d x}       | r = \dfrac {\d} {\d \theta} \cot \dfrac \theta 2 \frac {\d \theta} {\d x}       | c = Chain Rule for Derivatives }} {{eqn | r = -\dfrac 1 2 \csc^2 \dfrac \theta 2 / \dfrac {\d x} {\d \theta}       | c = Derivative of Cotangent Function }} {{eqn | r = -\dfrac 1 2 \csc^2 \dfrac \theta 2 \paren {\dfrac 1 {a \paren {1 - \cos \theta} } }       | c = Derivative of Sine Function }} {{eqn | r = -\dfrac 1 {2 \sin^2 \dfrac \theta 2} \paren {\dfrac 1 {a \paren {1 - \cos \theta} } }       | c = {{Defof|Cosecant}} }} {{eqn | r = -\dfrac 1 {1 - \cos \theta} \paren {\dfrac 1 {a \paren {1 - \cos \theta} } }       | c = Double Angle Formulas for Cosine }} {{eqn | r = -\dfrac a {y^2}       | c = from $y = a \paren {1 - \cos \theta}$ }} {{end-eqn}} {{qed}}
From Second Derivative of Locus of Cycloid: :$y'' = -\dfrac a {y^2}$ As $y \ge 0$ throughout, then $y'' < 0$ wherever $y \ne 0$, which is at the cusps. The result follows from Second Derivative of Concave Real Function is Non-Positive. {{qed}}
From Normal to Cycloid, the equation for the normal to $C$ at a point $P = \left({x, y}\right)$ is given by: {{ProofWanted|We don't even have the definition of a normal to a curve posted up yet, so there's a long way to go here.}}
From Tangent to Cycloid passes through Top of Generating Circle, the tangent to $C$ at a point $P = \left({x, y}\right)$ passes through the top of the generating circle. By definition, the normal to $C$ at $P$ is perpendicular to the tangent to $C$ at $P$. From Thales' Theorem, the normal, the tangent and the diameter of the generating circle form a right triangle. Thus the normal to $C$ at $P$ meets the generating circle at the opposite end of the diameter to the tangent. Hence the result. {{qed}}
Let the center of $C$ be $O$. {{WLOG}}, let $P$ be at the origin at time $t = t_0$. By definition, $P$ traces out a cycloid. From Equation of Cycloid, $P = \tuple {x, y}$ is described by: :$(1): \quad \begin{cases} x & = a \paren {\theta - \sin \theta} \\ y & = a \paren {1 - \cos \theta} \end{cases}$ Let $\tuple {x_c, y_c}$ be the coordinates of $O$ at time $t$. We have that: :$y_c$ is constant: $y_c = a$ From Body under Constant Acceleration: Velocity after Time :$x_c = \mathbf v_0 t$ as the acceleration of $O$ is zero. But $x_c$ is equal to the length of the arc of $C$ that has rolled along the $x$-axis. By Arc Length of Sector: :$x_c = a \theta$ and so: {{begin-eqn}} {{eqn | n = 2       | l = \theta       | r = \frac {\mathbf v_0 t} a       | c =  }} {{eqn | n = 3       | ll= \leadsto       | l = \frac {\d \theta} {\d t}       | r = \frac {\mathbf v_0} a       | c =  }} {{end-eqn}} Thus: {{begin-eqn}} {{eqn | l = x       | r = a \paren {\frac {\mathbf v_0 t} a - \sin \theta}       | c = substituting for $\theta$ from $(2)$ }} {{eqn | l = x       | r = \mathbf v_0 t - a \sin \theta       | c =  }} {{eqn | ll= \leadsto       | l = \frac {\d x} {\d t}       | r = \mathbf v_0 - a \cos \theta \frac {\d \theta} {\d t}       | c = Chain Rule for Derivatives }} {{eqn | r = \mathbf v_0 - a \cos \theta \frac {\mathbf v_0} a       | c = from $(3)$ }} {{eqn | r = \mathbf v_0 \paren {1 - \cos \theta}       | c =  }} {{end-eqn}} and: {{begin-eqn}} {{eqn | l = y       | r = a \paren {1 - \cos \theta}       | c =  }} {{eqn | ll= \leadsto       | l = \frac {\d y} {\d t}       | r = a \sin \theta \frac {\d \theta} {\d t}       | c = Chain Rule for Derivatives }} {{eqn | r = a \sin \theta \frac {\mathbf v_0} a       | c = from $(3)$ }} {{eqn | r = \mathbf v_0 \sin \theta       | c =  }} {{end-eqn}} {{qed}}
From Rate of Change of Cartesian Coordinates of Cycloid, the rate of change of $x$ is given by: :$\dfrac {\d x} {\d t} = \mathbf v_0 \paren {1 - \cos \theta}$ This is a maximum when $1 - \cos \theta$ is a maximum. That happens when $\cos \theta$ is at a minimum. That happens when $\cos \theta = -1$. That happens when $\theta = \pi, 3 \pi, \ldots$ That is, when $\theta = \paren {2 n + 1} \pi$ where $n \in \Z$. That is, when $P$ is at the top of the circle $C$. When $\cos \theta = -1$ we have: {{begin-eqn}} {{eqn | l = \frac {\d x} {\d t}       | r = \mathbf v_0 \paren {1 - \paren {-1} }       | c =  }} {{eqn | r = 2 \mathbf v_0       | c =  }} {{end-eqn}} Hence the result. {{qed}}
From Rate of Change of Cartesian Coordinates of Cycloid, the rate of change of $y$ is given by: :$\dfrac {\d y} {\d t} = \mathbf v_0 \sin \theta$. This is a maximum when $\sin \theta$ is a maximum. That happens when $\sin \theta = 1$. That happens when $\theta = \dfrac \pi 2 + 2 n \pi$ where $n \in \Z$. When $\sin \theta = 1$ we have: :$\dfrac {\d y} {\d t} = \mathbf v_0$ Hence the result. {{qed}}
=== Inflationary === Let $X$ be a subset of $S$. Let $x \in X$. By definition of reflexivity: :$x \preceq x$ Thus by definition of lower closure of set: :$x \in X^\preceq$ Thus by definition of subset: :$X \subseteq X^\preceq$ {{qed|lemma}} === Increasing === Let $X, Y$ be subsets of $S$ such that :$X \subseteq Y$ Let $x \in X^\preceq$. By definition of lower closure of set: :$\exists y \in X: x \preceq y$ By definition of subset: :$y \in Y$ Thus by definition of lower closure of set: :$x \in Y^\preceq$ Thus by definition of subset: :$X^\preceq \subseteq Y^\preceq$ {{qed|lemma}} === Idempotent === Let $X$ be a subset of $S$. Let $x \in \left({X^\preceq}\right)^\preceq$. By definition of lower closure of set: :$\exists y \in X^\preceq: x \preceq y$ By definition of lower closure of set: :$\exists z \in X: y \preceq z$ By definition of transitivity: :$x \preceq z$ Thus by definition of lower closure of set: :$x \in X^\preceq$ Thus by definition of subset: :$\left({X^\preceq}\right)^\preceq \subseteq X^\preceq$ By definition of inflationary: :$X \subseteq X^\preceq$ By definition of increasing: :$X^\preceq \subseteq \left({X^\preceq}\right)^\preceq$ Thus definition of set equality; :$\left({X^\preceq}\right)^\preceq = X^\preceq$ {{qed|lemma}} Thus by definition: :lower closure of set is a closure operator. {{qed}} Category:Lower Closures Category:Closure Operators qovjrkspfmlam6i4h6nxxkjwqgeavjr