hcju/proofwikiqa huggingface dataset:
=== Case 1 === Let $\norm{\mathbf a}_p \le 1$. From Equivalence Class in P-adic Integers Contains Unique P-adic Expansion, $\mathbf a$ has exactly one representative that is a $p$-adic expansion of the form: :$\displaystyle \sum_{n \mathop = 0}^\infty d_n p^n$ From P-adic Expansion Representative of P-adic Number is Unique: :$\displaystyle \sum_{i \mathop = 0}^\infty e_i p^i$ is the only $p$-adic expansion that represents $\mathbf a$ {{qed|lemma}} === Case 2 === Let $\norm{\mathbf a}_p > 1$. Let $\textbf p$ denote the equivalence class that is identified with the prime number $p$.  By definition of the $p$-adic numbers as a quotient of Cauchy sequences:  :the constant sequence $\tuple {p, p, p, \dotsc}$ represents $\mathbf p \in \Q_p$. From P-adic Number times P-adic Norm is P-adic Unit, :$\exists m \in \Z: \mathbf p^m \textbf a \in \Z_p^\times$ where $\Z_p^\times$ denotes the $p$-adic units. By definition a $p$-adic unit is a $p$-adic integer. From Equivalence Class in P-adic Integers Contains Unique P-adic Expansion: :$\mathbf p^m \textbf a$ has exactly one representative that is a $p$-adic expansion of the form: ::$\displaystyle \sum_{n \mathop = 0}^\infty d_n p^n$ From P-adic Unit has Norm Equal to One: :$\norm {\mathbf p^m \textbf a}_p = 1 = p^0$ From P-adic Norm of P-adic Expansion is determined by First Nonzero Coefficient: :$d_0 \neq 0$ Let $\sequence{x_n}$ be a Cauchy sequence of rational numbers that represents $\mathbf a \in \Q_p$. By definition of the $p$-adic numbers as a quotient of Cauchy sequences:  :the constant sequence $\tuple {p^m, p^m, p^m, \dotsc}$ represents $\mathbf p^m \in \Q_p$. By definition of the product in a quotient ring: :the Cauchy sequence $\sequence{p^m x_n}$ is a representative of $\mathbf p^m \mathbf a \in \Q_p$. From Representatives of same P-adic Number iff Difference is Null Sequence: :$\sequence {p^m x_n - \displaystyle \sum_{i \mathop = 0}^n d_i p^i}$ is a null sequence From Multiple Rule for Cauchy Sequences in Normed Division Ring: :$\sequence {x_n - \displaystyle p^{-m} \sum_{i \mathop = 0}^n d_i p^i}$ is a null sequence From Representatives of same P-adic Number iff Difference is Null Sequence: :$\sequence {\displaystyle p^{-m} \sum_{i \mathop = 0}^\infty d_i p^i}$ is a representative of $\mathbf a$. That is: :$\displaystyle \sum_{i \mathop = -m}^\infty e_i p^i$ is a representative of $\mathbf a$ where: :$\forall i \ge -m: e_i = d_{i + m}$ By definition of a $p$-adic expansion; :$\forall i \ge 0: 0 \le d_i < p$ Then: :$\forall i \ge -m: 0 \le e_i = d_{i + m} < p$ Since $d_0 \neq 0$, then $e_{-m} = d_0 \neq 0$. By definition of a $p$-adic expansion; :$\displaystyle \sum_{i \mathop = -m}^\infty e_i p^i$ is a $p$-adic expansion that represents $\mathbf a$ From P-adic Expansion Representative of P-adic Number is Unique: :$\displaystyle \sum_{i \mathop = -m}^\infty e_i p^i$ is the only $p$-adic expansion that represents $\mathbf a$ {{qed}}
By definition, the plastic constant $P$ is the real root of the cubic: :$x^3 - x - 1 = 0$ Recall Cardano's Formula: {{:Cardano's Formula}} Here we have: {{begin-eqn}} {{eqn | l = a       | r = 1 }} {{eqn | l = b       | r = 0 }} {{eqn | l = c       | r = -1 }} {{eqn | l = d       | r = -1 }} {{end-eqn}} Hence: {{begin-eqn}} {{eqn | l = Q       | r = \dfrac {3 \times 1 \times \paren {-1} - 0^2} {9 \times 1^2}       | c =  }} {{eqn | r = \dfrac {-3} 9       | c =  }} {{eqn | r = -\dfrac 1 3       | c =  }} {{eqn | l = R       | r = \dfrac {9 \times 1 \times 0 \times \paren {-1} - 27 \times 1^2 \times \paren {-1} - 2 \times 0^3} {54 \times 1^3}       | c =  }} {{eqn | r = \dfrac {27} {54}       | c =  }} {{eqn | r = \dfrac 1 2       | c =  }} {{end-eqn}} and so: {{begin-eqn}} {{eqn | l = \sqrt {Q^3 + R^2}       | r = \sqrt {\paren {-\dfrac 1 3}^3 + \paren {\dfrac 1 2}^2}       | c =  }} {{eqn | r = \sqrt {\dfrac 1 4 - \dfrac 1 {27} }       | c =  }} {{eqn | r = \sqrt {\dfrac {27 - 4} {108} }       | c =  }} {{eqn | r = \sqrt {\dfrac {3 \times 23} {3 \times 2^2 \times 3^3} }       | c =  }} {{eqn | r = \sqrt {\dfrac {69} {18^2} }       | c =  }} {{eqn | r = \dfrac {\sqrt {69} } {18}       | c =  }} {{eqn | ll= \leadsto       | l = S = \sqrt [3] {R + \sqrt {Q^3 + R^2} }       | r = \sqrt [3] {\dfrac 1 2 + \dfrac {\sqrt {69} } {18} }       | c =  }} {{eqn | r = \sqrt [3] {\dfrac {9 + \sqrt {69} } {18} }       | c =  }} {{eqn | l = T = \sqrt [3] {R - \sqrt {Q^3 + R^2} }       | r = \sqrt [3] {\dfrac 1 2 - \dfrac {\sqrt {69} } {18} }       | c =  }} {{eqn | r = \sqrt [3] {\dfrac {9 - \sqrt {69} } {18} }       | c =  }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = S + T - \dfrac b {3 a}       | r = \sqrt [3] {\dfrac {9 + \sqrt {69} } {18} } + \sqrt [3] {\dfrac {9 - \sqrt {69} } {18} } - \dfrac 0 {3 \times 1}       | c =  }} {{eqn | r = \sqrt [3] {\dfrac {9 + \sqrt {69} } {18} } + \sqrt [3] {\dfrac {9 - \sqrt {69} } {18} }       | c =  }} {{end-eqn}} The number can then be calculated. Since $S \ne T$, the other two roots $x_2, x_3$ has non-zero imaginary parts $\pm \dfrac {i \sqrt 3} 2 \paren {S - T}$. Hence the root above is the only real root. {{qed}}
Let $i \in \N$. From Integer Arbitrarily Close to Rational in Valuation Ring of P-adic Norm: :$\exists \mathop {\alpha'} \in \Z: \norm{x - \alpha'}_p \le p^{-i}$ By Integer is Congruent to Integer less than Modulus, then there exists $\alpha \in \Z$: :$\alpha \equiv \alpha' \pmod {p^i}$. :$0 \le \alpha \le p^i - 1$ Then $\norm {\alpha' - \alpha}_p \le p^{-i}$ Hence: {{begin-eqn}} {{eqn | l = \norm {x - \alpha}_p       | r = \norm {\paren {x - \alpha'} + \paren {\alpha' - \alpha} }_p }} {{eqn | o = \le       | r = \max \set {\norm {x - \alpha'}_p, \norm {\alpha' - \alpha}_p }       | c = {{NormAxiom|4}} }} {{eqn | o = \le       | r = p^{-i} }} {{end-eqn}} Now suppose $\beta \in \Z$ satisfies: :$(\text a): \quad 0 \le \beta \le p^i - 1$ :$(\text b): \quad \norm {x -\beta}_p \le p^{-i}$ Then: {{begin-eqn}} {{eqn | l = \norm {\alpha - \beta}_p       | r = \norm {\paren{\alpha - x} + \paren {x - \beta} }_p }} {{eqn | o = \le       | r = \max \set {\norm{\alpha - x}_p, \: \norm {x - \beta}_p}       | c = {{NormAxiom|4}} }} {{eqn | o = \le       | r = \max \set {\norm {x - \alpha}_p, \: \norm {x - \beta}_p}       | c = Norm of Negative }} {{eqn | o = \le       | r = p^{-i} }} {{end-eqn}} Hence $p^i \divides \alpha - \beta$, or equivalently, $\alpha \equiv \beta \pmod {p^i}$ By Initial Segment of Natural Numbers forms Complete Residue System then $\alpha = \beta$. The result follows. {{qed}}
:500px Let $\psi_1$ and $\psi_2$ be the angles that $L_1$ and $L_2$ make with the $x$-axis respectively. Then from Slope of Straight Line is Tangent of Angle with Horizontal: {{begin-eqn}} {{eqn | l = \tan \psi_1       | r = m_1 }} {{eqn | l = \tan \psi_2       | r = m_2 }} {{end-eqn}} and so: {{begin-eqn}} {{eqn | l = \tan \psi       | r = \map \tan {\psi_2 - \psi_1}       | c =  }} {{eqn | r = \dfrac {\tan \psi_2 - \tan \psi_1} {1 + \tan \psi_1 \tan \psi_2}       | c = Tangent of Difference }} {{eqn | r = \dfrac {m_2 - m_1} {1 + m_1 m_2}       | c = Definition of $m_1$ and $m_2$ }} {{end-eqn}} {{qed}}
Let $L_1$ and $L_2$ be embedded in a cartesian plane, given by the equations: {{begin-eqn}} {{eqn | ll= L_1:       | l = y       | r = m_1 x + c_1 }} {{eqn | ll= L_2:       | l = y       | r = m_2 x + c_2 }} {{end-eqn}} Let $\phi_1$ and $\phi_2$ be the angles that $L_1$ and $L_2$ make with the $x$-axis respectively. Then from Slope of Straight Line is Tangent of Angle with Horizontal: {{begin-eqn}} {{eqn | l = \tan \psi_1       | r = m_1 }} {{eqn | l = \tan \psi_2       | r = m_2 }} {{end-eqn}} === Necessary Condition === Let $m_1 = m_2$. Then: :$\tan \psi_1 = \tan \psi_2$ and so: :$\psi_1 = \psi_2 + n \pi$ The multiple of $\pi$ makes no difference. Thus from Equal Corresponding Angles implies Parallel Lines, $L_1$ and $L_2$ are parallel. {{qed|lemma}} === Sufficient Condition === Suppose $L_1 \parallel L_2$. Then: {{begin-eqn}} {{eqn | l = \phi_1       | r = \phi_2       | c = Parallelism implies Equal Corresponding Angles }} {{eqn | ll= \leadsto       | l = \tan \phi_1       | r = \tan \phi_2       | c =  }} {{eqn | ll= \leadsto       | l = m_1       | r = m_2       | c =  }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \paren {x^2 + y^2}^2       | r = 2 a^2 \paren {x^2 - y^2}       | c = {{Defof|Lemniscate of Bernoulli|subdef = Cartesian Definition}} }} {{eqn | ll= \leadsto       | l = \map {\dfrac \d {\d x} } {\paren {x^2 + y^2}^2}       | r = \map {\dfrac \d {\d x} } {2 a^2 \paren {x^2 - y^2} }       | c = differentiating {{WRT|Differentiation}} $x$ }} {{eqn | ll= \leadsto       | l = 2 \paren {x^2 + y^2} \paren {2 x + 2 y \dfrac {\d y} {\d x} }       | r = 2 a^2 \paren {2 x - 2 y \dfrac {\d y} {\d x} }       | c = Derivative of Power and Chain Rule for Derivatives }} {{eqn | ll= \leadsto       | l = 4 y \paren {x^2 + y^2 + a^2} \dfrac {\d y} {\d x}       | r = 4 x \paren {a^2 - \paren {x^2 + y^2} }       | c = gathering terms }} {{eqn | ll= \leadsto       | l = \dfrac {\d y} {\d x}       | r = \dfrac {x \paren {a^2 - \paren {x^2 + y^2} } } {y \paren {x^2 + y^2 + a^2} }       | c = gathering terms }} {{end-eqn}} {{finish|Probably requires L'Hopital here, it's a bit tedious to contemplate though}} {{qed}}
By the definition of the lemniscate of Bernoulli, we have that the polar equation of $M$ is: :$r^2 = 2 a^2 \cos 2 \theta$ Let $\AA$ denote the area of one lobe of $M$. The boundary of the right hand lobe of $M$ is traced out where $-\dfrac \pi 2 \le 2 \theta \le \dfrac \pi 2$. Thus: {{begin-eqn}} {{eqn | l = \AA       | r = \int_{-\pi / 4}^{\pi / 4} \dfrac {\map {r^2} \theta} 2 \rd \theta       | c = Area between Radii and Curve in Polar Coordinates }} {{eqn | r = \int_{-\pi / 4}^{\pi / 4} \dfrac {2 a^2 \cos 2 \theta} 2 \rd \theta       | c = {{Defof|Lemniscate of Bernoulli|subdef = Polar Definition}} }} {{eqn | r = a^2 \int_{-\pi / 4}^{\pi / 4} \cos 2 \theta \rd \theta       | c = simplifying }} {{eqn | r = a^2 \intlimits {\dfrac {\sin 2 \theta} 2} {-\pi / 4} {\pi / 4}       | c = Primitive of $\cos a x$ }} {{eqn | r = \dfrac {a^2} 2 \paren {\sin \dfrac \pi 2 - \map \sin {-\dfrac \pi 2} }       | c =  }} {{eqn | r = \dfrac {a^2} 2 \paren {1 - \paren {-1} }       | c = Sine of Right Angle etc. }} {{eqn | r = a^2       | c =  }} {{end-eqn}} {{qed}}
:525px Let $P = \polar {r, \theta}$ be an arbitrary point on $C$. Let $A$ and $B$ be the centers of the stator and rotor respectively. Let $Q$ be the point where the stator and rotor touch. By definition of the method of construction of $C$, we have that the arc $OQ$ of the stator equals the arc $PQ$ of the rotor. Thus: :$\angle OAQ = \angle PBQ$ and it follows that $AB$ is parallel to $OP$. With reference to the diagram above, we have: {{begin-eqn}} {{eqn | l = r       | r = OR + RS + SP       | c =  }} {{eqn | r = a \cos \theta + 2 a + a \cos \theta       | c =  }} {{eqn | r = 2 a \paren {1 + \cos \theta}       | c =  }} {{end-eqn}} and the result follows. {{qed}}
Let $\AA$ denote the area inside $C$. The boundary of $C$ is traced out where $-\pi \le \theta \le \pi$. Thus: {{begin-eqn}} {{eqn | l = \AA       | r = \int_{-\pi}^\pi \dfrac {\map {r^2} \theta} 2 \rd \theta       | c = Area between Radii and Curve in Polar Coordinates }} {{eqn | r = \int_{-\pi}^\pi \dfrac {\paren {2 a \paren {1 + \cos \theta} }^2} 2 \rd \theta       | c = Equation of Cardioid }} {{eqn | r = 2 a^2 \int_{-\pi}^\pi \paren {1 + 2 \cos \theta + \cos^2 \theta} \rd \theta       | c = simplifying }} {{eqn | r = 2 a^2 \intlimits {\theta + 2 \sin \theta + \frac \theta 2 + \frac {\sin 2 \theta} 4} {-\pi} \pi       | c = Primitive of $\cos a x$, Primitive of Square of Cosine Function }} {{eqn | r = 2 a^2 \paren {\paren {\pi + 2 \sin \pi + \dfrac \pi 2 + \frac {\sin 2 \pi} 4} - \paren {-\pi + 2 \, \map \sin {-\pi} + \dfrac {-\pi} 2 + \frac {\map \sin {-2 \pi} } 4} }       | c = evaluation between $-\pi$ and $\pi$ }} {{eqn | r = 2 a^2 \paren {\pi + \dfrac \pi 2 - \paren {-\pi} - \paren {\dfrac {-\pi} 2} }       | c = Sine of Multiple of Pi }} {{eqn | r = 6 \pi a^2       | c =  }} {{end-eqn}} {{qed}}
Let $\mathcal L$ denote the length of the perimeter of $C$. The boundary of the $C$ is traced out where $-\pi \le \theta \le \pi$. From Arc Length for Polar Curve: :$\displaystyle \mathcal L = \int_{\theta \mathop = -\pi}^{\theta \mathop = \pi} \sqrt {r^2 + \paren {\frac {\d r} {\d \theta} }^2} \rd \theta$ where: :$r = 2a \paren {1 + \cos \theta}$ Note that we have:  {{begin-eqn}} {{eqn	| l = \frac {\d r} {\d \theta} 	| r = 2 a \frac \d {\d \theta} \paren {1 + \cos \theta} }} {{eqn	| r = -2 a \sin \theta 	| c = Sum Rule for Derivatives, Derivative of Cosine Function }} {{end-eqn}} We therefore have:  {{begin-eqn}} {{eqn	| l = \mathcal L 	| r = \int_{-\pi}^\pi \sqrt {4 a^2 \paren {1 + \cos \theta}^2 + 4 a^2 \sin^2 \theta} \rd \theta }} {{eqn	| r = 2 a \int_{-\pi}^\pi \sqrt {1 + 2 \cos \theta + \cos^2 \theta + \sin^2 \theta} \rd \theta 	| c = extracting a factor of $\sqrt {4 a^2} = 2 a$, noting that $a \ge 0$ }} {{eqn	| r = 2 a \int_{-\pi}^\pi \sqrt {2 + 2 \cos \theta} \rd \theta 	| c = Sum of Squares of Sine and Cosine }} {{eqn	| r = 4 a \int_0^\pi \sqrt {4 \cos^2 \frac \theta 2} \rd \theta 	| c = Definite Integral of Even Function, Double Angle Formula for Cosine: Corollary 1 }} {{eqn	| r = 8 a \int_0^\pi \size {\cos \frac \theta 2} \rd \theta }} {{eqn	| r = 8 a \int_0^\pi \cos \frac \theta 2 \rd \theta 	| c = {{Defof|Absolute Value}}, noting that $\cos \theta \ge 0$ for $0 \le \theta \le \dfrac \pi 2$ }} {{eqn	| r = 16 a \intlimits {\sin \frac \theta 2} 0 \pi 	| c = Primitive of Cosine Function: Corollary }} {{eqn	| r = 16 a \paren {\sin \frac \pi 2 - \sin 0} }} {{eqn	| r = 16 a 	| c = Sine of Right Angle, Sine of Zero is Zero }} {{end-eqn}} {{qed}}
Let $R$ be defined by $r = a \cos n \theta$. The tips of each of the petals of $R$ occur when $\cos n \theta = \pm 1$. This happens whenever $n \theta \in \set {0, \pi}$. Thus for $0 \le \theta < 2 \pi$, the tips occur at: :$\theta \in \set {\dfrac {2 k \pi} {2 n}: k \in \set {0, 1, \ldots, 2 n - 1} }$ Let $k < n$. Then each of $\dfrac {2 k \pi} {2 n}$ is a distinct value between $0$ and $\pi$. Hence there are at least $n$ petals, one for each $k \in \set {0, 1, \ldots, n - 1}$. Now consider $\theta \in \hointr \pi {2 \pi}$. From Cosine of Angle plus Straight Angle, we have that: :$\cos \theta = -\map \cos {\theta - \pi}$ But as $\theta$ is on the opposite side of the circle to $\theta - \pi$, the points they describe are exactly the same. Hence the petals for $\theta \ge \pi$ are in the same place as that for $\theta < \pi$. So for $\pi \le \theta < 2 \pi$ the pattern will be traced out again. Hence the result. {{qed}} {{handwaving|Anyone care to make this a bit more rigorous?}}
Let $C$ have rolled so that the radius to the point $P = \tuple {x, y}$ is at angle $\theta$ to the vertical. The center of $C$ is at $\tuple {a \theta, a}$. Then it follows from the definition of sine and cosine that: :$x = a \theta - b \sin \theta$ :$y = a - b \cos \theta$ whence the result. {{qed}}
Let $P = \tuple {x, y}$ and $A = \tuple {d, y}$. Let $\theta$ be the angle that $ON$ makes with the horizontal. We have by definition of cotangent: :$\dfrac {OM} {MN} = \dfrac {2 a} x = \cot \theta$ By Thales' Theorem $\angle OAM$ is a right angle. Hence $\angle OMA = \theta$ and so: :$OA = 2 a \cos \theta$ Thus: :$2 a - y = 2 a \cos^2 \theta$ {{begin-eqn}} {{eqn | l = 2 a - y       | r = 2 a \cos^2 \theta       | c =  }} {{eqn | ll= \leadsto       | l = y       | r = 2 a \paren {1 - \dfrac {\cos 2 \theta + 1} 2}       | c =  }} {{eqn | r = a \paren {2 - \paren {\cos 2 \theta + 1} }       | c =  }} {{eqn | r = a \paren {1 - \cos 2 \theta}       | c =  }} {{end-eqn}} {{qed}}
Let $F$ be a folium of Descartes expressed in parametric form. Then by definition: {{:Definition:Folium of Descartes/Parametric Form}} Hence: {{begin-eqn}} {{eqn | l = x^3 + y^3       | r = \dfrac {\paren {3 a t}^3} {\paren {1 + t^3}^3} + \dfrac {\paren {3 a t^2}^3} {\paren {1 + t^3}^3}       | c =  }} {{eqn | r = \dfrac {\paren {3 a t}^3 \paren {1 + t^3} } {\paren {1 + t^3}^3}       | c =  }} {{eqn | r = \dfrac {3 a \paren {3 a t} \paren {3 a t^2} } {\paren {1 + t^3}^2}       | c =  }} {{eqn | r = 3 a \dfrac {3 a t} {1 + t^3} \dfrac {3 a t^2} {1 + t^3}       | c =  }} {{eqn | r = 3 a x y       | c =  }} {{end-eqn}} Thus $F$ is a folium of Descartes expressed in Cartesian form. {{qed}} Category:Folium of Descartes fp0vatx66okswol1xfxj3auxzftkdgf
From Behaviour of Parametric Equations for Folium of Descartes according to Parameter we have that the loop is traversed for $0 \le t < +\infty$. We convert the parametric equation to polar form: {{begin-eqn}} {{eqn | l = r^2       | r = x^2 + y^2       | c =  }} {{eqn | r = \dfrac {\paren {3 a t}^2} {\paren {1 + t^3}^2} + \dfrac {\paren {3 a t^2}^2} {\paren {1 + t^3}^2}       | c =  }} {{eqn | r = \dfrac {\paren {3 a t}^2 \paren {1 + t^2} } {\paren {1 + t^3}^2}       | c =  }} {{eqn | l = \tan \theta       | r = \dfrac y x       | c =  }} {{eqn | r = \dfrac {3 a t^2} {1 + t^3} \dfrac {1 + t^3} {3 a t}       | c =  }} {{eqn | r = t       | c =  }} {{eqn | ll= \leadsto       | l = \theta       | r = \arctan t       | c =  }} {{eqn | ll= \leadsto       | l = \dfrac {\d \theta} {\d t}       | r = \dfrac 1 {1 + t^2}       | c = Derivative of Arctangent Function }} {{end-eqn}} Then we have: {{begin-eqn}} {{eqn | l = \AA       | r = \dfrac 1 2 \int_{t \mathop = 0}^{t \mathop \to \infty} r^2 \rd \theta       | c = Area between Radii and Curve in Polar Coordinates }} {{eqn | r = \dfrac 1 2 \int_{t \mathop = 0}^{t \mathop \to \infty} \dfrac {\paren {3 a t}^2 \paren {1 + t^2} } {\paren {1 + t^3}^2} \dfrac 1 {1 + t^2} \rd t       | c = Integration by Substitution }} {{eqn | r = \dfrac {3 a^2} 2 \int_{t \mathop = 0}^{t \mathop \to \infty} \dfrac {3 t^2 \rd t} {\paren {1 + t^3}^2}       | c = Integration by Substitution }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = u       | r = 1 + t^3       | c =  }} {{eqn | ll= \leadsto       | l = \dfrac {\d u} {\d t}       | r = 3 t^2       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = t       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = u       | r = 1       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = t       | o = \to       | r = +\infty       | c =  }} {{eqn | ll= \leadsto       | l = u       | o = \to       | r = +\infty       | c =  }} {{end-eqn}} which leads us to: {{begin-eqn}} {{eqn | l = \AA       | r = \dfrac {3 a^2} 2 \int_{t \mathop = 0}^{t \mathop \to \infty} \dfrac {3 t^2 \rd t} {\paren {1 + t^3}^2}       | c =  }} {{eqn | r = \dfrac {3 a^2} 2 \int_{u \mathop = 1}^{u \mathop \to \infty} \dfrac {\d u} {u^2}        | c = Integration by Substitution }} {{eqn | r = \dfrac {3 a^2} 2 \intlimits {-\dfrac 1 u} 1 {+\infty}       | c = Primitive of Power }} {{eqn | r = \dfrac {3 a^2} 2 \paren {-0 - \paren {-1} }       | c = evaluating limits: $u \to +\infty \implies \dfrac 1 u \to 0$ }} {{eqn | r = \dfrac {3 a^2} 2       | c =  }} {{end-eqn}} Hence the result. {{qed}}
First we note that from Behaviour of Parametric Equations for Folium of Descartes according to Parameter: :when $t = 0$ we have that $x = y = 0$ :when $t \to \pm \infty$ we have that $x \to 0$ and $y \to 0$ :when $t \to -1^+$ we have that $1 + t^3 \to 0+$, and so: ::$x \to -\infty$ ::$y \to +\infty$ :when $t \to -1^-$ we have that $1 + t^3 \to 0-$, and so: ::$x \to +\infty$ ::$y \to -\infty$ We have that: {{begin-eqn}} {{eqn | l = x + y       | r = \dfrac {3 a t} {1 + t^3} + \dfrac {3 a t^2} {1 + t^3}       | c =  }} {{eqn | r = \dfrac {3 a t \paren {1 + t} } {\paren {1 + t} \paren {1 - t + t^2} }       | c = Sum of Two Cubes }} {{eqn | r = \dfrac {3 a t} {1 - t + t^2}       | c = simplifying }} {{end-eqn}} So setting $t = -1$: {{begin-eqn}} {{eqn | l = x + y       | r = \dfrac {3 a \times \paren {-1} } {1 - \paren {-1} + \paren {-1}^2}       | c =  }} {{eqn | r = \dfrac {-3 a} {1 + 1 + 1}       | c =  }} {{eqn | r = -a       | c =  }} {{eqn | ll= \leadsto       | l = x + y + a       | r = 0       | c =  }} {{end-eqn}} The result follows. {{qed}}
By definition the involute of $C$ is described by the endpoint of a string unwinding from $C$. Let that endpoint start at $\tuple {a, 0}$ on the circumference of $C$. :500px Let $P = \tuple {x, y}$ be an arbitrary point on $V$. Let $Q$ be the point at which the cord is tangent to $C$. Then $PQ$ equals the arc of $C$ from which the cord has unwound. Thus: :$PQ = a \theta$ where $\theta$ is the angle of $OQ$ to the $x$-axis. Thus: {{begin-eqn}} {{eqn | l = x       | r = OQ \cos \theta + PQ \, \map \cos {\theta - \dfrac \pi 2}       | c =  }} {{eqn | r = a \cos \theta + a \theta \paren {\cos \theta \cos \dfrac \pi 2 + \sin \theta \sin \dfrac \pi 2}       | c = Cosine of Difference }} {{eqn | r = a \paren {\cos \theta + \theta \sin \theta}       | c = various trigonometrical identities }} {{end-eqn}} {{begin-eqn}} {{eqn | l = y       | r = OQ \sin \theta + PQ \, \map \sin {\theta - \dfrac \pi 2}       | c =  }} {{eqn | r = a \sin \theta + a \theta \paren {\sin \theta \cos \dfrac \pi 2 - \cos \theta \sin \dfrac \pi 2}       | c = Sine of Difference }} {{eqn | r = a \paren {\sin \theta - \theta \cos \theta}       | c = various trigonometrical identities }} {{end-eqn}} Hence the result. {{qed}}
The '''ovals of Cassini''' are the loci of points $M$ in the plane such that: :$P_1 M \times P_2 M = b^2$ Let $b$ be chosen. Let $P = \tuple {x, y}$ be an arbitrary point of $M$. We have: {{begin-eqn}} {{eqn | l = P_1 P       | r = \sqrt {\paren {x - a}^2 + y^2}       | c = Distance Formula }} {{eqn | l = P_2 P       | r = \sqrt {\paren {x + a}^2 + y^2}       | c = Distance Formula }} {{eqn | ll= \leadsto       | l = \paren {P_1 P} \paren {P_2 P}       | r = \sqrt {\paren {x + a}^2 + y^2} \sqrt {\paren {x - a}^2 + y^2}       | c = {{Defof|Ovals of Cassini}} }} {{eqn | r = b^2       | c =  }} {{eqn | ll= \leadsto       | l = b^4       | r = \paren {\paren {x + a}^2 + y^2} \paren {\paren {x - a}^2 + y^2}       | c = squaring both sides }} {{eqn | r = \paren {x^2 + y^2 + a^2 + 2 a x} \paren {x^2 + y^2 + a^2 - 2 a x}       | c = multiplying out and rearranging }} {{eqn | r = \paren {x^2 + y^2 + a^2}^2 - 4 a^2 x^2       | c = Difference of Two Squares }} {{end-eqn}} Hence the result. {{qed}}
The '''ovals of Cassini''' are the loci of points $M$ in the plane such that: :$P_1 M \times P_2 M = b^2$ Let $b$ be chosen. Let $P = \tuple {x, y}$ be an arbitrary point of $M$. Let $d_1 = \size {P_1 P}$ and $d_2 = \size {P_2 P}$. We have: {{begin-eqn}} {{eqn | l = b^2       | r = d_1 d_2       | c = {{Defof|Ovals of Cassini}} }} {{eqn | r = \sqrt {r^2 + a^2 - 2 a r \cos \theta} \times \sqrt {r^2 + a^2 - 2 a r \, \map \cos {\pi - \theta} }       | c = Cosine Rule }} {{eqn | ll= \leadsto       | l = \paren {b^2}^2       | r = \paren {r^2 + a^2 - 2 a r \cos \theta} \paren {r^2 + a^2 + 2 a r \cos \theta}       | c = Cosine of Supplementary Angle, and squaring throughout }} {{eqn | r = \paren {r^2 + a^2}^2 - \paren {2 a r \cos \theta}^2       | c = Difference of Two Squares }} {{eqn | r = \paren {r^2}^2 + 2 a^2 r^2 + \paren {a^2}^2 - 4 a^2 r^2 \cos^2 \theta       | c = Square of Sum }} {{eqn | r = r^4 + a^4 - 2 a^2 r^2 \paren {2 \cos^2 \theta - 1}       | c = simplifying }} {{eqn | r = r^4 + a^4 - 2 a^2 r^2 \cos 2 \theta       | c = Double Angle Formula for Cosine }} {{end-eqn}} Hence the result. {{qed}}
The ovals of Cassini are defined as follows: {{:Definition:Ovals of Cassini}} The lemniscate of Bernoulli is defined geometrically as: {{:Definition:Lemniscate of Bernoulli/Geometric Definition}} It follows that the lemniscate of Bernoulli is an oval of Cassini where $b = a$. {{qed}}