hcju/proofwikiqa huggingface dataset:
We calculate the derivative of $x$ {{WRT|Differentiation}} $t$: {{begin-eqn}} {{eqn | l = \dfrac {\d x} {\d t}       | r = \map {\dfrac \d {\d t} } {\dfrac {3 a t} {1 + t^3} }       | c =  }} {{eqn | r = \dfrac {\paren {1 + t^3} \times 3 a - 3 a t \paren {3 t^2} } {\paren {1 + t^3}^2}       | c = Quotient Rule for Derivatives }} {{eqn | r = \dfrac {3 a - 6 a t^3} {\paren {1 + t^3}^2}       | c = simplifying }} {{end-eqn}} Thus $x$ is stationary when: {{begin-eqn}} {{eqn | l = 3 a - 6 a t^3       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = t       | r = \paren {\dfrac 1 2}^{1/3}       | c =  }} {{end-eqn}} From Behaviour of Parametric Equations for Folium of Descartes according to Parameter, it is clear from the geometry that $x$ is a local maximum for this value of $t$. Then we have: {{begin-eqn}} {{eqn | l = x       | r = \dfrac {3 a \times \paren {1/2}^{1/3} } {1 + \paren {\paren {1/2}^{1/3} }^3}       | c =  }} {{eqn | r = \dfrac {3 a \times \paren {1/2}^{1/3} } {1 + 1/2}       | c =  }} {{eqn | r = 2 a \times \paren {\dfrac 1 2}^{1/3}       | c =  }} {{eqn | r = \paren {\dfrac {2^3} 2}^{1/3} a       | c =  }} {{eqn | r = 2^{2/3} a       | c =  }} {{end-eqn}} and: {{begin-eqn}} {{eqn | l = y       | r = \dfrac {3 a \times \paren {\paren {1/2}^{1/3} }^2} {1 + \paren {\paren {1/2}^{1/3} }^3}       | c =  }} {{eqn | r = \dfrac {3 a \times \paren {1/2}^{2/3} } {1 + 1/2}       | c =  }} {{eqn | r = 2 a \times \paren {\dfrac 1 2}^{2/3}       | c =  }} {{eqn | r = \paren {\dfrac {2^3} {2^2} }^{1/3} a       | c =  }} {{eqn | r = 2^{1/3} a       | c =  }} {{end-eqn}} Hence the result. {{qed}} Category:Folium of Descartes 36vvg81m23h2dtjw42t9p2jjtb49jzn
{{begin-eqn}} {{eqn | l = \map {D^3_x} {\ln x}       | r = \map {D_x} {\map {D^2_x} {\ln x} }       | c =  }} {{eqn | r = \map {D_x} {-\dfrac 1 {x^2} }       | c = Second Derivative of Natural Logarithm Function }} {{eqn | r = -\paren {-2 \paren {\dfrac 1 {x^3} } }       | c = Power Rule for Derivatives: Integer Index }} {{eqn | r = \dfrac 2 {x^3}       | c =  }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \sqrt {a + \sqrt b}       | r = \sqrt {c + \sqrt d} }} {{eqn | ll = \leadsto       | l = a + \sqrt b       | r = c + \sqrt d }} {{eqn | ll = \leadsto       | l = a - c       | r = \sqrt d - \sqrt b }} {{eqn | ll = \leadsto       | l = \sqrt d - \sqrt b       | o = \in       | r = \Q       | c = Rational Subtraction is Closed }} {{end-eqn}} {{AimForCont}} $b \ne d$. Then: {{begin-eqn}} {{eqn | l = \sqrt d - \sqrt b       | o = \ne       | r = 0 }} {{eqn | l = d - b       | r = \paren {\sqrt d - \sqrt b} \paren {\sqrt d + \sqrt b}       | c = Difference of Two Squares }} {{eqn | ll = \leadsto       | l = \sqrt d + \sqrt b       | r = \dfrac {d - b} {\sqrt d - \sqrt b} \in \Q       | c = Rational Division is Closed }} {{eqn | ll = \leadsto       | l = 2 \sqrt b       | r = \paren {\sqrt d + \sqrt b} - \paren {\sqrt d - \sqrt b} \in \Q       | c = Rational Subtraction is Closed }} {{eqn | ll = \leadsto       | l = \sqrt b       | o = \in       | r = \Q       | c = Rational Division is Closed }} {{end-eqn}} But this contradicts our assertion that $\sqrt b \notin \Q$. Hence our supposition that $b \ne d$ must be false. Therefore we must have $b = d$.  Consequently: :$a - c = \sqrt d - \sqrt b = 0$ This implies $a = c$. {{qed}} Category:Condition for Denesting of Square Root iaj113b7dbp6i6rtut377iudo3xs66h
This first order ODE is in the form: :$\dfrac {\d y} {\d x} + k y = 0$ where $k = 1$. From First Order ODE: $\d y = k y \rd x$, this has the solution: :$y = C e^{-x}$
{{begin-eqn}} {{eqn | l = y''       | r = \dfrac 1 {1 - x^2}       | c =  }} {{eqn | ll= \leadsto       | l = \int \dfrac {\d^2 y} {\d x^2} \rd x       | r = \int \frac 1 {1 - x^2} \rd x       | c =  }} {{eqn | ll= \leadsto       | l = \dfrac {\d y} {\d x}       | r = \tanh^{-1} x + C       | c = Primitive of $\dfrac 1 {a^2 - x^2}$: Inverse Hyperbolic Tangent Form }} {{eqn | ll= \leadsto       | l = \int \dfrac {\d y} {\d x} \rd x       | r = \int \paren {\tanh^{-1} x + C} \rd x       | c =  }} {{eqn | ll= \leadsto       | l = y       | r = x \tanh^{-1} x + \map \ln {1 - x^2} + C x + D       | c = Primitive of $\tanh^{-1} x$ }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = a \preceq_1 b       | o = \leadstoandfrom       | r = a \circ b = b       | c = Definition of the ordering $\,\preceq_1$ }} {{eqn | o = \leadsto       | r = \map \phi { a \circ b} = \map \phi b }} {{eqn | o = \leadstoandfrom       | r = \map \phi a * \map \phi b = \map \phi b       | c = {{Defof|Semilattice Homomorphism}} }} {{eqn | o = \leadstoandfrom       | r = \map \phi a \preceq_2 \map \phi b       | c = Definition of the ordering $\,\preceq_2$ }} {{end-eqn}} {{qed}}
Let $S = \set{a, b, c}$. Let $\circ : S \times S \to S$ be defined by: :$a \circ b = b \circ a = c$ :$a \circ c = c \circ a = c$ :$b \circ c = c \circ b = c$ :$a \circ a = a$ :$b \circ b = b$ :$c \circ c = c$ Then a manual check shows that $\circ$ is: :*closed :*associative :*commutative :*idempotent So $\struct{S, \circ}$ is a semilattice with ordering :$\preceq_1 = \set{\tuple{a, a}, \tuple{b, b}, \tuple{c, c}, \tuple{a, c}, \tuple{b,c}}$ Let $T = \set{p, q, r, s}$. Let $* : S \times S \to S$ be defined by: :$p * q = q * p = r$ :$p * r = r * p = r$ :$q * r = r * q = r$ :$p * s = s * p = s$ :$q * s = s * q = s$ :$r * s = s * r = s$ :$p * p = p$ :$q * q = q$ :$r * r = r$ :$s * s = s$ Then a manual check shows that $*$ is: :*closed :*associative :*commutative :*idempotent So $\struct{T, *}$ is a semilattice with ordering :$\preceq_2 = \set{\tuple{p, p}, \tuple{q, q}, \tuple{r, r}, \tuple{s, s}, \tuple{p, r}, \tuple{q,r}, \tuple{p, s}, \tuple{q,s}, \tuple{r, s}}$ Let $\phi : S \to T$ be defined by: :$\map \phi a = p$ :$\map \phi b = q$ :$\map \phi c = s$ Then a manual check shows that $\phi$ is order-preserving. Now: {{begin-eqn}} {{eqn | l = \map \phi a * \map \phi b       | r = p * q }} {{eqn | r = r }} {{eqn | o = \ne       | r = s }} {{eqn | r = \map \phi c }} {{eqn | r = \map \phi {a \circ b} }} {{end-eqn}} Hence $\phi$ is not a semilattice homomorphism by definition. {{qed}}
We prove the result by induction on $n$. === Basis for the Induction === The case $n = 7$ is verified by direct calculation: :$12! \times 13! > 20!$ This is the basis for the induction. === Induction Hypothesis === We suppose for some $k \ge 7$, we have: :$\paren {2 k - 2}! \, \paren {2 k - 1}! > \paren {3 k - 1}!$ This is our induction hypothesis. === Induction Step === This is our induction step: {{begin-eqn}} {{eqn | l = \paren {2 k}! \, \paren {2 k + 1}!       | o = >       | r = \paren {2 k} \paren {2 k - 1} \paren {2 k + 1} \paren {2 k} \paren {3 k - 1}!       | c = Induction Hypothesis }} {{eqn | o = >       | r = k \paren {2 k - 1} \paren {2 k + 1} \paren {3 k} \paren {3 k - 1}!       | c = because $4 k > 3 k$ }} {{eqn | o = >       | r = 4 \paren {2 k - 1} \paren {2 k + 1} \paren {3 k} \paren {3 k - 1}!       | c = because $k > 4$ }} {{eqn | r = \paren {4 k - 2} \paren {4 k + 2} \paren {3 k} \paren {3 k - 1}! }} {{eqn | o = \ge       | r = \paren {3 k + 5} \paren {3 k + 9} \paren {3 k} \paren {3 k - 1}!       | c = because $k \ge 7$ }} {{eqn | o = >       | r = \paren {3 k + 1} \paren {3 k + 2} \paren {3 k} \paren {3 k - 1}! }} {{eqn | r = \paren {3 k + 2}!       | c = {{Defof|Factorial}} }} {{end-eqn}} The result follows by the Principle of Mathematical Induction. {{qed}} Category:Factorials Category:Factorial as Product of Consecutive Factorials 7puvni7rhyv8vm98h7ysu1va71syvvp
Let $g : \C \to \C$ be a complex function with: :$\displaystyle \map g z = e^{\map f z}$ By Derivative of Complex Composite Function, $g$ is entire with derivative: :$\displaystyle \map {g'} z = \map {f'} z e^{\map f z}$ We have: {{begin-eqn}} {{eqn	| l = \cmod {\map g z} 	| r = e^{\map \Re {\map f z} } 	| c = Modulus of Positive Real Number to Complex Power is Positive Real Number to Power of Real Part }} {{eqn	| o = \le 	| r = e^{\cmod {\map \Re {\map f z} } } 	| c = Exponential is Strictly Increasing }} {{eqn	| o = < 	| r = e^M 	| c = Exponential is Strictly Increasing }} {{end-eqn}} So $g$ is a bounded entire function, and is therefore constant by Liouville's Theorem. We therefore have, by Derivative of Constant: Complex: :$\map {g'} z = 0$ for all $z \in \C$. That is: :$\map {f'} z e^{\map f z} = 0$ Since the exponential function is non-zero, we must have:  :$\map {f'} z = 0$ for all $z \in \C$. From Zero Derivative implies Constant Complex Function, we then have that $f$ is constant on $\C$.  {{qed}} Category:Complex Analysis f7l5zu0b42iwt5f29gltw9r91dmlohc
Let $m_Q$ grams be the mass of one atom of $Q$. Let $m_C$ grams be the mass of one atom of carbon-12. Let $M_Q$ grams be the mass of one mole of $Q$. Let $M_C$ grams be the mass of one mole of carbon-12. The mass of a proton and the mass of a neutron are approximately equal. Together, the protons and neutrons make up the vast majority of the mass of the atom they compose. Hence the mass of one atom is proportional to the total number of its protons and neutrons. There are $6$ protons and $6$ neutrons in one atom of carbon-12. Thus we have: {{begin-eqn}} {{eqn | l = m_Q : m_C       | r = \paren {n + p} : {6 + 6}       | c =  }} {{eqn | ll= \leadsto       | l = \dfrac {m_Q} {m_C}       | r = \dfrac {n + p} {12}       | c =  }} {{eqn | ll= \leadsto       | l = \dfrac {M_Q} {M_C}       | r = \dfrac {n + p} {12}       | c =  }} {{eqn | n = 1       | ll= \leadsto       | l = M_Q       | r = \paren {n + p} \dfrac {M_C} {12}       | c =  }} {{end-eqn}} Before $2018$, the '''mole''' was defined as the number of atoms in $12$ grams of carbon-12. While the '''mole''' is no longer so defined, the mass of $1$ mole of carbon-12 is still $12$ grams, for all practical purposes. Thus we have: {{begin-eqn}} {{eqn | l = M_C       | r = 12 \ \mathrm g       | c =  }} {{eqn | ll= \leadsto       | l = M_Q       | r = \paren {n + p} \dfrac {12 \ \mathrm g} {12}       | c = substituting for $M_C$ in $(1)$ }} {{eqn | ll= \leadsto       | l = M_Q       | r = \paren {n + p} \ \mathrm g       | c =  }} {{end-eqn}} {{Qed}} Category:Substance jkn10vsar0mwnsposvr5qu8eqcvqmd3
Let $m_S$ grams be the mean mass of one molecule of $S$. Let $m_C$ grams be the mass of one atom of carbon-12. Let $M_S$ grams be the mass of one mole of $S$. Let $M_C$ grams be the mass of one mole of carbon-12. Then: {{begin-eqn}} {{eqn | l = W_S       | r = m_S \times \dfrac {12} {m_C}       | c = {{Defof|Molecular Weight}} }} {{eqn | r = M_S \times \dfrac {12} {M_C}       | c =  }} {{eqn | n = 1       | ll= \leadsto       | l = M_S       | r = W_S \times \dfrac {M_C} {12}       | c =  }} {{end-eqn}} Before $2018$, the '''mole''' was defined as the number of atoms in $12$ grams of carbon-12. While the '''mole''' is no longer so defined, the mass of $1$ mole of carbon-12 is still $12$ grams, for all practical purposes. Thus we have: {{begin-eqn}} {{eqn | l = M_C       | r = 12 \ \mathrm g       | c =  }} {{eqn | ll= \leadsto       | l = M_S       | r = W_S \dfrac {12 \ \mathrm g} {12}       | c = substituting for $M_C$ in $(1)$ }} {{eqn | ll= \leadsto       | l = M_S       | r = W_S \ \mathrm g       | c =  }} {{end-eqn}} {{Qed}} Category:Substance nla3didvr4mnna01sqhxbidf7w4wg85
{{begin-eqn}} {{eqn | l = x y'       | r = 2 y       | c =  }} {{eqn | ll= \leadsto       | l = \int \dfrac {\d y} y       | r = 2 \int \dfrac {\d x} x       | c = Separation of Variables }} {{eqn | ll= \leadsto       | l = \ln y       | r = 2 \ln x + \ln C       | c = Primitive of Reciprocal }} {{eqn | ll= \leadsto       | l = \ln y       | r = \map \ln C x^2       | c =  }} {{eqn | ll= \leadsto       | l = y       | r = C x^2       | c =  }} {{end-eqn}} {{qed}}
Note that either $\set x = X$ or $\set x \subset X$. Suppose, $X = \set x$. Then: :$X \setminus \set x = \empty$ We have that Empty Set is Open in Normed Vector Space. $\set x$ is a complement of $\empty$ in $X$. By definition, $x$ is closed. Suppose, $X \ne \set x$, i.e. $\set x \subset X$. Define $U := X \setminus \set x$. Let $y \in U$. Let $\epsilon := \norm {x - y}$. Suppose, $\epsilon > 0$. By Normed Vector Space is Open in Itself: :$\forall x \in X : \exists \epsilon \in \R_{>0} : \map {B_\epsilon} x \subseteq X$ Suppose, $z \in \map {B_\epsilon} y$. Then: :$\norm {y - z} < \epsilon$. We have that: {{begin-eqn}} {{eqn | l = \norm {z - x}       | o = \ge       | r = \norm {x - y} - \norm {y - z}       | c = Triangle inequality }} {{eqn | o = \ge       | r = \epsilon - \norm {y - z} }} {{eqn | o = >       | r = \epsilon - \epsilon }} {{eqn | r = 0}}  {{end-eqn}} Hence, $z \ne x$. Then $z$ is in the complement of $\set x$ in $X$, i.e. $z \in U$. In other words: :$\forall y \in U : \exists \epsilon = \norm {x - y} > 0 : \map {B_\epsilon} x \subseteq U$ So $U$ is open, and $\set x = X \setminus U$ is closed.  {{qed}}
Let $x \in \Q$. By definition of nowhere dense, we need to show that: :$\paren {\set x^-}^\circ = \O$ where $S^-$ denotes the closure of a set $S$ and $S^\circ$ denotes its interior. By Real Number is Closed in Real Number Line, $\set x$ is closed in $\struct {\R, \tau_d}$. $\struct {\Q, \tau_d}$ is a subspace of $\struct {\R, \tau_d}$. Thus by Closed Set in Topological Subspace, $\set x$ is closed in $\struct {\Q, \tau_d}$. From Closed Set Equals its Closure, it follows that: :$\set x^- = \set x$ From Interior of Singleton in Real Number Line is Empty: :$\set x^\circ = \O$ Hence: :$\paren {\set x^-}^\circ = \set x^\circ = \O$ {{qed}} Category:Rational Number Space Category:Singletons 3uqnuszvgv6ohcaybfa1xu4xee4jd3l
Suppose, $F$ is empty. By Empty Set is Closed in Normed Vector Space, $F$ is closed. Suppose, for some $n \in \N$: :$\displaystyle F = \bigcup_{i \mathop = 1}^n \set {x_i}$ We have that Singleton in Normed Vector Space is Closed. Hence, $F$ is a finite union of closed sets. By Finite Union of Closed Sets is Closed in Normed Vector Space , $F$ is closed. {{qed}}
Let $\displaystyle \bigcup_{i \mathop = 1}^n F_i$ be the union of a finite number of closed sets of $M$. By definition of closed set, each of the $X \setminus F_i$ is by definition open in $M$. Then from De Morgan's laws: :$\displaystyle X \setminus \bigcup_{i \mathop = 1}^n F_i = \bigcap_{i \mathop = 1}^n \paren {X \setminus F_i}$ We have that $\displaystyle \bigcap_{i \mathop = 1}^n \paren {X \setminus F_i}$ is the intersection of a finite number of open sets of $M$. By Finite Intersection of Open Sets of Normed Vector Space is Open, $\displaystyle \bigcap_{i \mathop = 1}^n \paren {X \setminus F_i} = X \setminus \bigcup_{i \mathop = 1}^n F_i$ is likewise open in $M$. Then by definition of closed set, $\displaystyle \bigcup_{i \mathop = 1}^n F_i$ is closed in $M$. {{qed}}
{{begin-eqn}} {{eqn | l = \int x^4 \cos a x \rd x       | r = \frac {\sin a x} a x^4 + \frac {4 \cos a x} {a^2} x^{4 - 1} - \frac {4 \paren {4 - 1} } {a^2} \int x^{4 - 2} \cos a x \rd x + C       | c = Primitive of $x^n \cos a x$ }} {{eqn | r = \frac {\sin a x} a x^4 + \frac {4 \cos a x} {a^2} x^3 - \frac {12} {a^2} \int x^2 \cos a x \rd x + C       | c = Simplifying }} {{eqn | r = \frac {\sin a x} a x^4 + \frac {4 \cos a x} {a^2} x^3 - \frac {12} {a^2} \paren {\frac {2 x \cos a x} {a^2} + \paren {\frac {x^2} a - \frac 2 {a^3} } \sin a x} + C       | c = Primitive of $x^2 \cos a x$ }} {{eqn | r = \frac {\sin a x} a x^4 + \frac {4 \cos a x} {a^2} x^3 - \frac {12 \sin a x} {a^3} x^2 - \frac {24 \cos a x} {a^4} x + \frac {24 \sin a x} {a^5} + C       | c = simplifying }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \int x^6 \cos a x \rd x       | r = \frac {\sin a x} a x^6 + \frac {6 \cos a x} {a^2} x^{6 - 1} - \frac {6 \paren {6 - 1} } {a^2} \int x^{6 - 2} \cos a x \rd x + C       | c = Primitive of $x^n \cos a x$ }} {{eqn | r = \frac {\sin a x} a x^6 + \frac {6 \cos a x} {a^2} x^5 - \frac {30} {a^2} \int x^4 \cos a x \rd x + C       | c = Simplifying }} {{eqn | r = \frac {\sin a x} a x^6 + \frac {6 \cos a x} {a^2} x^5 - \frac {30} {a^2} \paren { \frac {\sin a x} a x^4 + \frac {4 \cos a x} {a^2} x^3 - \frac {12 \sin a x} {a^3} x^2 - \frac {24 \cos a x} {a^4} x + \frac {24 \sin a x} {a^5} } + C       | c = Primitive of $x^4 \cos a x$ }} {{eqn | r = \frac {\sin a x} a x^6 + \frac {6 \cos a x} {a^2} x^5 - \frac {30 \sin a x} {a^3} x^4 - \frac {120 \cos a x} {a^4} x^3 + \frac {360 \sin a x} {a^5} x^2 + \frac {720 \cos a x} {a^6} x - \frac {720 \sin a x} {a^7} + C       | c = simplifying }} {{end-eqn}} {{qed}}
Since $x^6 = \paren {-x}^6$, $x^6$ is an even function. By Fourier Series for Even Function over Symmetric Range, the Fourier series of $\map f x$ can be expressed as: :$x^6 \sim \dfrac {a_0} 2 + \displaystyle \sum_{n \mathop = 1}^\infty a_n \cos n x$ where for all $n \in \Z_{> 0}$: {{begin-eqn}} {{eqn | l = a_n        | r = \dfrac 2 \pi \displaystyle \int_0^\pi x^6 \cos n x \  \d x }} {{eqn | r = \dfrac 2 \pi \paren {\intlimits {\frac {\sin n x} n x^6 + \frac {6 \cos n x} {n^2} x^5 - \frac {30 \sin n x} {n^3} x^4 - \frac {120 \cos n x} {n^4} x^3 + \frac {360 \sin n x} {n^5} x^2 + \frac {720 \cos n x} {n^6} x - \frac {720 \sin n x} {n^7} } {x \mathop = 0} {x \mathop = \pi} }       | c = Primitive of $x^6 \cos a x$ }} {{eqn | r = \dfrac 2 \pi \paren {\frac {6 \pi^5 \cos n \pi} {n^2} - \frac {120 \pi^3 \cos n \pi} {n^4} + \frac {720 \pi \cos n \pi} {n^6}  }       | c =  }} {{eqn | r = \frac {12 n^4 \pi^4 - 240 n^2 \pi^2 + 1440} {n^6} \cos n \pi       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = a_0       | r = \dfrac 2 \pi \int_0^\pi x^6 \  \d x }} {{eqn | r = \dfrac 2 \pi \intlimits {\frac {x^7} 7} {x \mathop = 0} {x \mathop = \pi}       | c = Primitive of Power }} {{eqn | r = \frac {2 \pi^6} 7 }} {{end-eqn}} This gives: :$x^6 \sim \dfrac {\pi^6} 7 + \displaystyle \sum_{n \mathop = 1}^\infty \frac {12 n^4 \pi^4 - 240 n^2 \pi^2 + 1440} {n^6} \cos n \pi \cos n x$ {{qed}}
Let $y := x^{a x}$. As $x > 0$, we can take the natural logarithm of both sides: {{begin-eqn}} {{eqn | l = \ln y       | r = \ln x^{a x} }} {{eqn | r = a x \ln x       | c = Laws of Logarithms }} {{eqn | l = \map {\frac \d {\d x} } {\ln y}       | r = \map {\frac \d {\d x} } {a x \ln x}       | c = differentiating both sides {{WRT|Differentiation}} $x$ }} {{eqn | l = \frac 1 y \frac {\d y} {\d x}       | r = a \map {\frac \d {\d x} } {x \ln x}       | c = Chain Rule for Derivatives and Derivative of Natural Logarithm Function }} {{eqn | r = a \paren {\map {\frac \d {\d x} } x \cdot \ln x + x \frac {\d} {\d x} \ln x}       | c = Product Rule for Derivatives }} {{eqn | r = a \paren {1 \cdot \ln x + x \cdot \frac 1 x}       | c = Derivative of Identity Function and Derivative of Natural Logarithm Function }} {{eqn | r = a \paren {\ln x + 1}       | c = simplification }} {{eqn | l = \frac {\d y} {\d x}       | r = a x^{a x} \paren {\ln x + 1}       | c = multiplying both sides by $y = x^{a x}$ }} {{end-eqn}} {{qed}} Category:Derivative of x to the x k7bzemqa34muxcyuwp4uuwkipcjckaq
From Closed Form for Polygonal Numbers: :$\map P {m + 2, k} = \dfrac m 2 \paren {k^2 - k} + k = m T_{k - 1} + k$ Where $T_{k - 1}$ are triangular numbers. The first few $\paren {m + 2}$-gonal numbers less than $116 m$ are: :$0, 1, m + 2, 3 m + 3, 6 m + 4, 10 m + 5, 15 m + 6, 21 m + 7, 28 m + 8, 36 m + 9, 45 m + 10, 55 m + 11, 66 m + 12, 78 m + 13, 91 m + 14, 105 m + 15$ We show the expression of the first few numbers explicitly. {{begin-eqn}} {{eqn | l = 1       | r = 1 }} {{eqn | l = 2       | r = 1 + 1 }} {{eqn | o = : }} {{eqn | l = m + 1       | r = 1 + 1 + \dots + 1       | c = There are $m + 1$ $1$'s }} {{eqn | l = m + 2       | r = \paren {m + 2} }} {{eqn | o = : }} {{eqn | l = 2 m + 3       | r = \paren {m + 2} + 1 + 1 + \dots + 1       | c = There are $m + 1$ $1$'s }} {{eqn | l = 2 m + 4       | r = \paren {m + 2} + \paren {m + 2} }} {{eqn | o = : }} {{eqn | l = 3 m + 2       | r = \paren {m + 2} + \paren {m + 2} + 1 + 1 + \dots + 1       | c = There are $m - 2$ $1$'s }} {{eqn | l = 3 m + 3       | r = \paren {3 m + 3} }} {{eqn | o = : }} {{eqn | l = 4 m + 4       | r = \paren {3 m + 3} + 1 + 1 + \dots + 1       | c = There are $m + 1$ $1$'s }} {{eqn | l = 4 m + 5       | r = \paren {3 m + 3} + \paren {m + 2} }} {{eqn | o = : }} {{eqn | l = 5 m + 5       | r = \paren {3 m + 3} + \paren {m + 2} + 1 + 1 + \dots + 1       | c = There are $m$ $1$'s }} {{eqn | l = 5 m + 6       | r = \paren {m + 2} + \paren {m + 2} + \paren {m + 2} + \paren {m + 2} + 1 + 1 + \dots + 1       | c = There are $m - 2$ $1$'s }} {{eqn | l = 5 m + 7       | r = \paren {3 m + 3} + \paren {m + 2} + \paren {m + 2} }} {{eqn | o = : }} {{eqn | l = 6 m + 6       | r = \paren {3 m + 3} + \paren {m + 2} + \paren {m + 2} + 1 + 1 + \dots + 1       | c = There are $m - 1$ $1$'s }} {{eqn | l = 6 m + 7       | r = \paren {6 m + 4} + 1 + 1 + 1 }} {{eqn | o = : }} {{eqn | l = 7 m + 5       | r = \paren {6 m + 4} + 1 + 1 + \dots + 1       | c = There are $m + 1$ $1$'s }} {{eqn | l = 7 m + 6       | r = \paren {6 m + 4} + \paren {m + 2} }} {{eqn | o = : }} {{eqn | l = 8 m + 6       | r = \paren {6 m + 4} + \paren {m + 2} + 1 + 1 + \dots + 1       | c = There are $m$ $1$'s }} {{eqn | l = 8 m + 7       | r = \paren {3 m + 3} + \paren {3 m + 3} + \paren {m + 2} + 1 + 1 + \dots + 1       | c = There are $m - 1$ $1$'s }} {{eqn | l = 8 m + 8       | r = \paren {6 m + 4} + \paren {m + 2} + \paren {m + 2} }} {{eqn | o = : }} {{eqn | l = 9 m + 7       | r = \paren {6 m + 4} + \paren {m + 2} + \paren {m + 2} + 1 + 1 + \dots + 1       | c = There are $m - 1$ $1$'s }} {{eqn | l = 9 m + 8       | r = \paren {6 m + 4} + \paren {3 m + 3} + 1 }} {{eqn | o = : }} {{eqn | l = 10 m + 8       | r = \paren {6 m + 4} + \paren {3 m + 3} + 1 + 1 + \dots + 1       | c = There are $m$ $1$'s }} {{eqn | l = 10 m + 9       | r = \paren {10 m + 5} + 1 + 1 + 1 + 1 }} {{eqn | o = : }} {{eqn | l = 11 m + 10       | r = \paren {10 m + 5} + 1 + 1 + \dots + 1       | c = There are $m + 1$ $1$'s }} {{end-eqn}} and so on. {{qed}} Category:Integer as Sum of Polygonal Numbers augqs1sbxvr8d530qfuide9xk8fwp88
=== $b^2 < 4 a$ === :$b^2 - 4 a = b^2 - 4 \paren {1 - \dfrac 2 m} b - 8 \paren {\dfrac {n - r} m}$ By the Quadratic Formula, $b^2 - 4 a < 0$ when $b$ is between: {{begin-eqn}} {{eqn | o =       | r = \frac 1 2 \paren {4 \paren {1 - \frac 2 m} \pm \sqrt {16 \paren {1 - \frac 2 m}^2 + 32 \paren {\frac {n - r} m} } } }} {{eqn | r = 2 - \frac 4 m \pm \sqrt {\paren {2 - \frac 4 m}^2 + 8 \paren {\frac n m} - 8 \paren {\frac r m} } }} {{end-eqn}} Observing the term in the square root, we have:  :$2 - \dfrac 4 m - \sqrt {\paren {2 - \dfrac 4 m}^2 + 8 \paren {\dfrac n m} - 8 \paren {\dfrac r m} } < 0$ Since $b > 0$ this is satisfied. Also we have: {{begin-eqn}} {{eqn | l = b       | o = <       | r = \frac 2 3 + \sqrt {8 \paren {\frac n m} - 8} }} {{eqn | o = <       | r = 2 - \frac 4 m + \sqrt {8 \paren {\frac n m} - 8 \paren {\dfrac r m} }       | c = $m \ge 3$, $r < m$ }} {{eqn | o = \le       | r = 2 - \frac 4 m + \sqrt {\paren {2 - \frac 4 m}^2 + 8 \paren {\frac n m} - 8 \paren {\dfrac r m} }       | c = Square is nonnegative }} {{end-eqn}} showing that first inequality is satisfied. {{qed|lemma}} === $3 a < b^2 + 2 b + 4$ === :$b^2 + 2 b + 4 - 3 a = b^2 - \paren {1 - \dfrac 6 m} b - 6 \paren {\dfrac {n - r} m} + 4$ By the Quadratic Formula, $b^2 + 2 b + 4 - 3 a > 0$ when $b$ is greater than: {{begin-eqn}} {{eqn | o =       | r = \frac 1 2 \paren {1 - \frac 6 m + \sqrt {\paren {1 - \frac 6 m}^2 + 4 \paren {6 \paren {\frac {n - r} m} - 4} } } }} {{eqn | o = <       | r = \frac 1 2 \paren {1 + \sqrt {24 \paren {\frac n m} - 16} }       | c = $m > 0$, $r \ge 0$, $-1 \le 1 - \dfrac 6 m < 1$ }} {{eqn | r = \frac 1 2 + \sqrt {6 \paren {\frac n m} - \frac {15} 4} }} {{end-eqn}} showing that second inequality is satisfied by any $b > \dfrac 1 2 + \sqrt {6 \paren {\dfrac n m} - 3}$. {{qed}}