hcju/proofwikiqa huggingface dataset:
From the definition of distance: :$\forall x, y \in \R: \map d {x, y} = \size {x - y}$ Thus: :$\displaystyle \map d {x, S} = \map {\inf_{y \mathop \in S} } {\size {x - y} }$ As $I \ne \O$ and $I \ne \R$ it follows that one of the following applies: {{begin-eqn}} {{eqn | lo= \exists a, b \in \R:       | l = I       | r = \openint a b }} {{eqn | lo= \exists a \in \R:       | l = I       | r = \openint a \to }} {{eqn | lo= \exists b \in \R:       | l = I       | r = \openint \gets b }} {{end-eqn}} It follows by the definition of open real interval that $I$ has either an infimum $a$, or a supremum $b$, or both.   Thus the required value of $x$ is either $a$ or $b$. {{Qed}}
We have that $\Z_{>0}$ contains only (strictly) positive integers. So it follows from Reciprocal of Strictly Positive Real Number is Strictly Positive that $S$ contains only (strictly) positive real numbers. Hence $0$ is a lower bound for $S$. {{AimForCont}} that $0$ is not the infimum of $S$. Then $\exists h \in \R_{>0}$ such that $h$ is a lower bound for $S$. Then: :$\forall n \in \Z_{>0}: \dfrac 1 n \ge h$ Hence from Reciprocal Function is Strictly Decreasing: :$\forall n \in \Z_{>0}: n \le \dfrac 1 h$ where $\dfrac 1 h \in \R$. But from the Archimedean Principle: :$\exists n \in \Z_{>0}: n > \dfrac 1 h$ This is a contradiction. It follows by Proof by Contradiction that no such $h > 0$ exists. Hence $0$ is the infimum of $S$. {{qed}}
We have that: :$\dfrac {n - 1} n = 1 - \dfrac 1 n$ As $n > 0$ it follows from Reciprocal of Strictly Positive Real Number is Strictly Positive that $\dfrac 1 n > 0$. Thus $1 - \dfrac 1 n < 1$ and so $S$ is bounded above by $1$. Next it is to be shown that $1$ is the supremum of $S$. Suppose $x$ is the supremum of $S$ such that $x < 1$. Then: :$1 - x = \epsilon$ where $\epsilon \in \R_{>0}$. By the Archimedean Principle we have that: :$\exists m \in \Z_{>0}: n > \dfrac 1 \epsilon$ and so from Reciprocal Function is Strictly Decreasing: :$\exists m \in \Z_{>0}: \dfrac 1 n < \epsilon$ Thus: :$1 - \dfrac 1 m > 1 - \epsilon = x$ and so $x$ is not the supremum of $S$ after all. Thus $\sup S$ cannot be less than $1$. It follows that: :$\sup S = 1$ Next it is noted that: :$\forall n \in \Z_{>0}: \dfrac 1 n > 0$ and so: :$\forall x \in S: x < 1$ Thus as $\sup S \notin S$ it follows from Greatest Element is Supremum that $S$ has no greatest element. {{qed}}
We have that: :$\forall r \in S: 0 < r$ and: :$\forall r \in S: r < 1$ Hence $0$ and $1$ are lower and upper bounds of $S$ respectively. Let $s \in S$. Then $s \in \Q: 0 < s < 1$. {{AimForCont}} $s$ is the greatest element of $S$. But then we have: :$0 < s < \dfrac {s + 1} 2 < 1$ and so $\dfrac {s + 1} 2 \in S$ but $s < \dfrac {s + 1} 2$. This contradicts our assertion that $s$ is the greatest element of $S$. {{AimForCont}} $s$ is the smallest element of $S$. But then we have: :$0 < \dfrac s 2 < s < 1$ and so $\dfrac s 2 \in S$ but $\dfrac s 2 < s$. This contradicts our assertion that $s$ is the smallest element of $S$. So $S$ cannot have either a greatest or smallest element. Let $H$ be such that $0 < H < 1$. Suppose $H$ is a smaller upper bound of $S$ than $1$. Then from Between two Real Numbers exists Rational Number there exists a rational number $r$ such that $H < r < 1$. But then $r$ is a rational number between $0$ and $1$ which is greater than $H$. Hence $H$ cannot be an upper bound of $S$. Thus, by definition, $1$ is the supremum of $S$ Suppose $H$ is a greater lower bound of $S$ than $0$. Then from Between two Real Numbers exists Rational Number there exists a rational number $r$ such that $0 < r < H$. But then $r$ is a rational number between $0$ and $1$ which is smaller than $H$. Hence $H$ cannot be a lower bound of $S$. Thus, by definition, $0$ is the infimum of $S$. {{Qed}}
From Number of Type Rational r plus s Root 2 is Irrational we have that real numbers of the form $r \sqrt 2$, where $r \ne 0$ is rational, is irrational. From Between two Real Numbers exists Rational Number there exists a rational number $r$ such that: :$\dfrac a {\sqrt 2} < r < \dfrac b {\sqrt 2}$ and so: :$a < r \sqrt 2 < b$ giving us $\xi = r \sqrt 2$ as the irrational number we seek. If $a < 0$ and $b > 0$, it may be that the $r$ that we find happens to be zero Then we locate a second rational number $r$ such that: :$0 < s < \dfrac b {\sqrt 2}$ which leads us to: :$a < 0 < s \sqrt 2 < b$ and so in this case $\xi = s \sqrt 2$ is the irrational number we seek. {{qed}}
From the definition of an even function, for all $x \in X$ we have:  :$\map f x = \map f {-x}$  Differentiating $2 n + 1$ times, we have, by the Chain Rule for Derivatives:  :$\map {f^{\paren {2 n + 1} } } x = \paren {-1}^{2 n + 1} \map {f^{\paren {2 n + 1} } } {-x} = -\map {f^{\paren {2 n + 1} } } {-x}$ Setting $x = 0$ gives:  :$\map {f^{\paren {2 n + 1} } } 0 = -\map {f^{\paren {2 n + 1} } } {-0}$ That is:  :$2 \map {f^{\paren {2 n + 1 } } } 0 = 0$ The result follows on division by $2$.  {{qed}} Category:Differential Calculus Category:Even Functions 37zgmcyk8vtm61938r3vxa7x3erswr8
Let $\sequence {y_n}$ be the sequence in $\R$ defined as: :$y_n = \size {\dfrac 1 {x_n} }$ From Reciprocal of Null Sequence: :$y_n \to 0$ as $n \to \infty$ {{iff}} $\size {\dfrac 1 {y_n} } \to \infty$ as $n \to \infty$ That is: :$\size {\dfrac 1 {x_n} } \to 0$ as $n \to \infty$ {{iff}} $x_n \to \infty$ as $n \to \infty$ {{qed}}
Let $H > 0$. As $\sequence {x_n}$ is unbounded above: :$\exists N: x_N > H$ As $\sequence {x_n}$ is increasing: :$\forall n \ge N: x_n \ge x_N > H$ It follows from the definition of divergence to $+\infty$ that $x_n \to +\infty$ as $n \to \infty$. {{qed}}
The proof proceeds by induction. For all $r \in \Z_{\ge 1}$, let $\map P r$ be the proposition: :$n_r \ge r$ === Basis for the Induction === The first term of $\sequence {x_{n_r} }$ by definition cannot be less than $1$. That is: :$n_1 \ge 1$ Thus $\map P 1$ is seen to hold. This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true. So this is the induction hypothesis: :$n_k \ge k$ from which it is to be shown that: :$n_{k + 1} \ge {k + 1}$ === Induction Step === This is the induction step: {{begin-eqn}} {{eqn | l = n_{k + 1}       | o = >       | r = n_k       | c = {{Defof|Strictly Increasing Sequence}} }} {{eqn | o = >       | r = k       | c = Induction Hypothesis }} {{eqn | o = \ge       | r = k + 1       | c = as $k$ is an integer }} {{end-eqn}} So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\forall r \in \Z_{\ge 1}: n_r \ge r$ {{qed}}
By assumption, $\map y x$ and $\map z x$ are real functions. This allows us to $x$ instead of $t$ to parameterize the curve. This reduces the number of equations of motion to $2$, that is: $\map y x$ and $\map z x$. The time it takes to traverse the curve $\gamma$ equals: {{begin-eqn}} {{eqn | l = T       | r = \int_{t_a}^{t_b} \rd t }} {{eqn | r = \int_{s_a}^{s_b} \dfrac {\d t} {\d s} \rd s       | c = Chain Rule for Derivatives, $\d s$ - arc length element }} {{eqn | r = \int_a^b \dfrac 1 v \frac {\d s} {\d x} \rd x       | c = Chain Rule for Derivatives, $v = \dfrac {\d s} {\d t}$ }} {{eqn | r = \int_a^b \frac {\sqrt{1 + y'^2 + z'^2} } {\map v {x, y, z} } \rd x       | c = {{Defof|Arc Length}} in 3-dimensional Euclidean space }} {{end-eqn}} According to Fermat's principle, light travels along the trajectory of least time. Therefore, this integral has to be minimized {{WRT}} $\map y x$ and $\map z x$. It holds that: :$\dfrac \partial {\partial y} \dfrac {\sqrt {1 + x'^2 + y'^2} } {\map v {x, y, z} } = -\dfrac {\sqrt {1 + x'^2 + y'^2} } {\map {v^2} {x, y, z} } \dfrac {\partial v} {\partial y}$ Also: :$\dfrac \d {\d x} \dfrac \partial {\partial y'} \dfrac {\sqrt{1 + x'^2 + y'^2} } {\map v {x, y, z} } = \dfrac \d {\d x} \dfrac {y'} {v \sqrt {1 + x'^2 + y'^2} }$ Analogous relations hold for $\map z x$. Then, by Euler's Equations and multiplication by $-1$, the desired result follows. {{qed}}
From the definition of an odd function, for all $x \in X$ we have: :$\map f x = -\map f {-x}$ Differentiating $2 n$ times, we have, by the Chain Rule for Derivatives: :$\map {f^{\paren {2 n} } } x = -\paren {-1}^{2 n} \map {f^{\paren {2 n} } } {-x} = -\map {f^{\paren {2 n} } } {-x}$ Setting $x = 0$ gives:  :$\map {f^{\paren {2 n} } } 0 = -\map {f^{\paren {2 n} } } {- 0}$  That is: :$2 \map {f^{\paren {2 n} } } 0 = 0$ The result follows on division by $2$.  {{qed}} Category:Differential Calculus Category:Odd Functions qbsi7rhhu6s36h28r9qshapx3vh2uxj
By definition of geometric mean: :$\map G {a, b} := \sqrt {a b}$ where $\sqrt {a b}$ specifically denotes the positive square root of $a$ and $b$. Thus: {{begin-eqn}} {{eqn | l = a       | o = <       | r = b       | c =  }} {{eqn | ll= \leadsto       | l = a^2       | o = <       | r = a b       | c =  }} {{eqn | ll= \leadsto       | l = a       | o = <       | r = \sqrt {a b}       | c =  }} {{end-eqn}} and: {{begin-eqn}} {{eqn | l = a       | o = <       | r = b       | c =  }} {{eqn | ll= \leadsto       | l = a b       | o = <       | r = b^2       | c =  }} {{eqn | ll= \leadsto       | l = \sqrt {a b}       | o = <       | r = b       | c =  }} {{end-eqn}} {{qed}} Category:Geometric Mean 77u50syj2nokq1ehyncqw0we4y7k9mq
Let ${G_n}'$ denotes the geometric mean of the reciprocals of $x_1, x_2, \ldots, x_n$. By definition of harmonic mean, we have that: :$\dfrac 1 {H_n} = \displaystyle \sum_{k \mathop = 0}^n \dfrac 1 {x_n}$ That is, $\dfrac 1 {H_n}$ is the arithmetic mean of the reciprocals of $x_1, x_2, \ldots, x_n$. Then: {{begin-eqn}} {{eqn | l = \dfrac 1 {H_n}       | o = \ge       | r = {G_n}'       | c = Cauchy's Mean Theorem }} {{eqn | r = \dfrac 1 {G_n}       | c = Geometric Mean of Reciprocals is Reciprocal of Geometric Mean }} {{eqn | ll= \leadsto       | l = H_n       | o = \le       | r = G_n       | c = Reciprocal Function is Strictly Decreasing }} {{end-eqn}} {{qed}}
By definition of harmonic mean: :$\dfrac 1 {\map H {a, b} } := \dfrac 1 2 \paren {\dfrac 1 a + \dfrac 1 b}$ Thus: {{begin-eqn}} {{eqn | l = a       | o = <       | r = b       | c = by assumption }} {{eqn | ll= \leadsto       | l = \dfrac 1 b       | o = <       | r = \dfrac 1 a       | c = Reciprocal Function is Strictly Decreasing }} {{end-eqn}} But $\dfrac 1 {\map H {a, b} }$ is the arithmetic mean of $\dfrac 1 b$ and $\dfrac 1 a$. Hence from Arithmetic Mean of two Real Numbers is Between them: :$\dfrac 1 b < \dfrac 1 {\map H {a, b} } < \dfrac 1 a$ So by Reciprocal Function is Strictly Decreasing: :$b > \map H {a, b} > a$ Hence the result. {{qed}} Category:Harmonic Mean olcbzb7rwg72t0jt7eetaj0tn443n6x
By definition of arithmetic mean: :$\map A {a, b} := \dfrac {a + b} 2$ Thus: {{begin-eqn}} {{eqn | l = a       | o = <       | r = b       | c = by assumption }} {{eqn | ll= \leadsto       | l = 2 a       | o = <       | r = a + b       | c = adding $a$ to both sides }} {{eqn | ll= \leadsto       | l = a       | o = <       | r = \dfrac {a + b} 2       | c = dividing both sides by $2$ }} {{eqn | r = \map A {a, b}       | c =  }} {{end-eqn}} and: {{begin-eqn}} {{eqn | l = b       | o = >       | r = a       | c = by assumption }} {{eqn | ll= \leadsto       | l = 2 b       | o = >       | r = a + b       | c = adding $b$ to both sides }} {{eqn | ll= \leadsto       | l = b       | o = >       | r = \dfrac {a + b} 2       | c = dividing both sides by $2$ }} {{eqn | r = \map A {a, b}       | c =  }} {{end-eqn}} Hence the result. {{qed}} Category:Arithmetic Mean 4ajd99dupmufyzeyczpdcz9yohll759
From the definition of expectation: :$\displaystyle \expect {X^n} = \sum_{x \mathop \in \Img X} x^n \, \map \Pr {X = x}$ From the definition of the Bernoulli distribution: :$\displaystyle \expect {X^n} = 1^n \times p + 0^n \times \paren {1 - p} = p$ {{qed}}
=== Local Basis for Open Sets Implies Neighborhood Basis of Open Sets === {{:Equivalence of Definitions of Local Basis/Local Basis for Open Sets Implies Neighborhood Basis of Open Sets}}{{qed|lemma}} === Neighborhood Basis of Open Sets Implies Local Basis for Open Sets === {{:Equivalence of Definitions of Local Basis/Neighborhood Basis of Open Sets Implies Local Basis for Open Sets}}{{qed}} Category:Local Bases Category:Equivalence of Definitions of Local Basis sfbgo6j2xd7qirp9qifnx0htsqrs3l0
Let $N$ be a neighborhood of $x$. Then there exists $U \in \tau$ such that $x \in U$ and $U \subseteq N$ by definition. By assumption, there exists $H \in \BB$ such that $H \subseteq U$. From Subset Relation is Transitive, $H \subseteq N$. The result follows.
Let $U \in \tau$ such that $x \in U$. From Set is Open iff Neighborhood of all its Points then $U$ is a neighborhood of $x$. By assumption, there exists $H \in \BB$ such that $H \subseteq U$. The result follows.
First it must be shown that $\BB'$ is a set of open neighborhoods of $x$. From the definition of the interior of a subset, $\BB'$ is a set of open sets. Let $H \in \BB$.  By assumption, $H$ is a neighborhood of $x$. From the definition of a neighborhood: :$\exists U \in \tau : x \in U \subseteq H$ From the definition of the interior of a subset: :$U \subseteq H^\circ$ Hence $x \in H^\circ$. It follows that $\BB'$ is a set of open neighborhoods of $x$. It remains to show that $\BB'$ is a neighborhood basis. Let $N$ be a neighborhood of $x$. Since $\BB$ is a neighborhood basis of $x$ then: :$\exists H \in \BB : H \subseteq N$ By definition of the interior, $H^\circ \subseteq H$. From Subset Relation is Transitive $H^\circ \subseteq N$. Hence $\BB'$ is a neighborhood basis. The result follows. {{qed}} Category:Local Bases Category:Neighborhood Bases 300tidwlpc2w2obngrytl9f1utl4dha
By definition, $\arcsin$ is the inverse of the restriction of the sine function to $\closedint {-\dfrac \pi 2} {\dfrac \pi 2}$. Therefore, if:  :$\sin x = 0$ and $-\dfrac \pi 2 \le x \le \dfrac \pi 2$, then $\arcsin 0 = x$.  From Sine of Zero is Zero, we have that:  :$\sin 0 = 0$ We have $-\dfrac \pi 2 < 0 < \dfrac \pi 2$, so:  :$\arcsin 0 = 0$ {{qed}} Category:Arcsine Function 11wtbduj29rils37p0iwqxcsqlyu7xl
By definition, $\arcsin$ is the inverse of the restriction of the sine function to $\closedint {-\dfrac \pi 2} {\dfrac \pi 2}$. Therefore, if:  :$\sin x = 1$ and $-\dfrac \pi 2 \le x \le \dfrac \pi 2$, then $\arcsin 1 = x$. From Sine of Right Angle, we have that:  :$\sin \dfrac \pi 2 = 1$ We therefore have:  :$\arcsin 1 = \dfrac \pi 2$  {{qed}} Category:Arcsine Function i39n3ea00d11w1xi5kjdkcu1pix48ix
From Cauchy's Mean Theorem: :$(1): \quad \displaystyle \paren {\prod_{k \mathop = 1}^n a_k}^{1/n} \le \frac 1 n \paren {\sum_{k \mathop = 1}^n a_k}$ for $r_1, r_2, \ldots, r_n$. Setting: :$a_1 = a_2 = \ldots = a_{n - 1} := 1 + \dfrac x {n - 1}$ and: :$a_n = 1$ Substituting for $a_1, a_2, \ldots, a_n$ into $(1)$ gives: {{begin-eqn}} {{eqn | l = \paren {1 + \dfrac x {n - 1} }^{\frac {n - 1} n}       | o = \le       | r = \dfrac {\paren {n - 1} \paren {1 + \frac x {n - 1} } + 1} n       | c =  }} {{eqn | r = 1 + \dfrac x n       | c =  }} {{eqn | ll= \leadsto       | l = \paren {1 + \dfrac x {n - 1} }^{n - 1}       | o = \le       | r = \paren {1 + \dfrac x n}^n       | c =  }} {{end-eqn}} The above is valid only if $a_1, a_2, \ldots, a_n$ are positive. Hence we have shown that $\sequence {s_n}$ is increasing when: :$1 + \dfrac x {n - 1} \ge 0$ that is, when: :$n \ge 1 - x$ From Equivalence of Definitions of Real Exponential Function: Limit of Sequence implies Sum of Series, we have: :$(2): \quad \paren {1 + \dfrac x n}^n \le 1 + \size x + \dfrac {\size x^2} {2!} + \dotsb + \dfrac {\size x^n} {n!}$ Since there exists $N$ such that: :$\forall n > N: \dfrac {\size x^n} {n!} \le \paren {\dfrac 1 2}^n$ it follows from $(2)$ that: {{begin-eqn}} {{eqn | l = \paren {1 + \dfrac x n}^n       | o = \le       | r = 1 + \size x + \dfrac {\size x^2} {2!} + \dotsb + \dfrac {\size x^N} {N!} + \paren {\dfrac 1 2}^{N + 1} + \dotsb + \paren {\dfrac 1 2}^n       | c =  }} {{eqn | o = \le       | r = C + \dfrac {1 + \paren {\frac 1 2}^{n + 1} } {1 + \frac 1 2}       | c = where $C$ is some constant }} {{eqn | o = <       | r = C + 2       | c =  }} {{end-eqn}} Hence we have that $\sequence {s_n}$ is strictly increasing and bounded above. So by the Monotone Convergence Theorem (Real Analysis), $\sequence {s_n}$ is convergent. As $1 + \dfrac x n$ is positive when $n$ is large enough, it follows that the limit of $\sequence {s_n}$ is positive. {{qed}}
From Moment Generating Function of Exponential Distribution, the moment generating function of $X$, $M_X$, is given by: :$\displaystyle \map {M_X} t = \frac 1 {1 - \beta t}$ By Moment in terms of Moment Generating Function: :$\displaystyle \expect {X^n} = \map {M^{\paren n}_X} 0$ We have:  {{begin-eqn}} {{eqn	| l = \map {M^{\paren n}_X} t 	| r = \frac {\d^n} {\d t^n} \paren {\frac 1 {1 - \beta t} } }} {{eqn	| r = -\frac 1 \beta \frac {\d^n} {\d t^n} \paren {\frac 1 {t - \frac 1 \beta} } }} {{eqn	| r = -\frac {\paren {-1}^n n!} {\beta \paren {t - \frac 1 \beta}^{n + 1} } 	| c = Nth Derivative of Reciprocal of Mth Power: Corollary, Chain Rule for Derivatives }} {{eqn	| r = \frac {\paren {-1}^{n + 1} n!} {\frac \beta {\beta^{n + 1} } \paren {\beta t - 1}^{n + 1} } }} {{eqn	| r = \frac {\paren {-1}^{n + 1} n! \beta^n} {\paren {\beta t - 1}^{n + 1} }  }} {{end-eqn}} Setting $t = 0$ gives: {{begin-eqn}} {{eqn	| l = \expect {X^n} 	| r = \frac {\paren {-1}^{n + 1} n! \beta^n} {\paren {-1}^{n + 1} } }} {{eqn	| r = n! \beta^n }} {{end-eqn}} {{qed}} Category:Exponential Distribution Category:Raw Moments q3xldfva9k84pjlpe8sn6sl6jh0a0j4
From definition of central moment we have: :$\mu_n = \expect {\paren {x - \mu}^n}$ By Expectation of Exponential Distribution we have: :$\mu = \beta$ So:  {{begin-eqn}} {{eqn	| l = \mu_2 	| r = \expect {\sum_{k \mathop = 0}^n \binom n k X^{n - k} \paren {-\beta}^k} 	| c = Binomial Theorem }} {{eqn	| r = \sum_{k \mathop = 0}^n \binom n k \paren {-\beta}^k \expect {X^{n - k} } 	| c = Linearity of Expectation Function }} {{eqn	| r = \sum_{k \mathop = 0}^n \beta^k \binom n k \paren {-1}^k \paren {\paren {n - k}! \beta^{n - k} } 	| c = Raw Moment of Exponential Distribution }} {{eqn	| r = \sum_{k \mathop = 0}^n \beta^n \paren {\frac {n!} {k! \paren {n - k}!} } \paren {-1}^k \paren {n - k}!         | c = {{Defof|Binomial Coefficient}} }} {{eqn	| r = n! \beta^n \sum_{k \mathop = 0}^n \frac {\paren {-1}^k} {k!} }} {{end-eqn}} {{qed}} Category:Exponential Distribution Category:Central Moments kkahcwjz56a43ojpmxejzm73ynfk1m4
From the definition of excess kurtosis, we have:  :$\gamma_2 = \expect {\paren {\dfrac {X - \mu} \sigma}^4} - 3$ where:  :$\mu$ is the expectation of $X$.  :$\sigma$ is the standard deviation of $X$. By Expectation of Bernoulli Distribution, we have:  :$\mu = p$ By Variance of Bernoulli Distribution, we have:  :$\sigma = \sqrt {p q}$  So:  {{begin-eqn}} {{eqn	| l = \gamma_2 	| r = \frac {\expect {X^4 - 4 X^3 p + 6 X^2 p^2 - 4 X p^3 + p^4} } {\paren {p q}^2} - 3 	| c = Fourth Power of Difference, Linearity of Expectation Function }} {{eqn	| r = \frac {\expect {X^4} - 4 p \expect {X^3} + 6 p^2 \expect {X^2} - 4 p^3 \expect X + p^4} {\paren {p q}^2} - 3 	| c = Linearity of Expectation Function }} {{eqn	| r = \frac {p - 4p^2 + 6p^3 - 4p^4 + p^4} {\paren {p q}^2} - 3 	| c = Raw Moment of Bernoulli Distribution }} {{eqn	| r = \frac {p \paren {1 - 4p + 6p^2 - 3p^3} } {\paren {p q}^2} - 3 }} {{eqn	| r = \frac {p \paren {1 - p} \paren {3p^2 - 3p + 1} } {\paren {p q}^2} - 3 }} {{eqn	| r = \frac {p q \paren {1 - 3 p \paren {1 - p} } - 3 \paren {p q}^2} {\paren {p q}^2} }} {{eqn	| r = \frac {1 - 3 p q - 3 p q} {p q} }} {{eqn	| r = \frac {1 - 6 p q} {p q} }} {{end-eqn}} {{qed}} Category:Kurtosis Category:Bernoulli Distribution dqk58n4kbdras3rag8q487nw9a55z32
From Real Sequence $\paren {1 + \dfrac x n}^n$ is Convergent, $\sequence {a_n}$ is convergent. Setting $x \to -x$, it follows that $\sequence {\paren {1 + \dfrac {\paren {-x} } n}^n} = \sequence {b_n}$ is also convergent. Then: {{begin-eqn}} {{eqn | l = \paren {1 - \dfrac x n}^n \paren {1 - \dfrac x n}^n       | r = \paren {\paren {1 - \dfrac x n} \paren {1 - \dfrac x n} }^n       | c =  }} {{eqn | r = \paren {1 - \dfrac {x^2} {n^2} }^n       | c = Difference of Two Squares }} {{eqn | r = \paren {\paren {1 - \dfrac {x^2} {n^2} }^{n^2} }^{1/n}       | c =  }} {{end-eqn}} From Real Sequence $\paren {1 + \dfrac x n}^n$ is Convergent, $\sequence {\paren {1 - \dfrac {x^2} {n^2} }^{n^2} }$ is convergent to a positive limit. It follows that: :$\exists a, b \in \R_{>0}: a < \paren {1 - \dfrac {x^2} {n^2} }^{n^2} < b$ for sufficiently large $n$. Hence: :$a^{1/n} < \paren {1 - \dfrac x n}^n \paren {1 - \dfrac x n}^n < b^{1/n}$ The result follows from Limit of Root of Positive Real Number and the Squeeze Theorem for Real Sequences. {{qed}}
We have the following subsequences of $F$ which are all convergent to a different limit: :$\dfrac 1 2, \dfrac 2 4, \dfrac 3 6, \dfrac 4 8 \to \dfrac 1 2$ as $n \to \infty$ :$\dfrac 1 2, \dfrac 1 3, \dfrac 1 4, \dfrac 1 5 \to 0$ as $n \to \infty$ :$\dfrac 1 2, \dfrac 2 3, \dfrac 3 4, \dfrac 4 5 \to 1$ as $n \to \infty$ It follows from Limit of Subsequence equals Limit of Real Sequence that $F$ cannot be convergent. {{qed}}
{{begin-eqn}} {{eqn | l = \gamma_1       | r = \frac {\expect {X^3} - 3 \mu \sigma^2 - \mu^3} {\sigma^3}       | c = Skewness in terms of Non-Central Moments }} {{eqn | r = \paren {\paren {\prod_{r \mathop = 0}^2 \frac {\alpha + r} {\alpha + \beta + r} } - \frac {3 \alpha} {\alpha + \beta} \cdot \frac {\alpha \beta} {\paren {\alpha + \beta}^2 \paren {\alpha + \beta + 1} } - \paren {\frac \alpha {\alpha + \beta} }^3} \paren {\frac {\alpha \beta} {\paren {\alpha + \beta}^2 \paren {\alpha + \beta + 1} } }^{-\frac 3 2}       | c = Raw Moment of Beta Distribution, Expectation of Beta Distribution, Variance of Beta Distribution }} {{eqn | r = \paren {\frac {\alpha \paren {\alpha + 1} \paren {\alpha + 2} \paren {\alpha + \beta}^2 - 3 \alpha^2 \beta \paren {\alpha + \beta + 2} - \alpha^3 \paren {\alpha + \beta + 1} \paren {\alpha + \beta + 2} } {\paren {\alpha + \beta}^3 \paren {\alpha + \beta + 1} \paren {\alpha + \beta + 2} } } \paren {\frac {\alpha \beta} {\paren {\alpha + \beta}^2 \paren {\alpha + \beta + 1} } }^{-\frac 3 2} }} {{eqn | r = \paren {\frac {\alpha \paren {\alpha + 1} \paren {\alpha + 2} \paren {\alpha + \beta}^2 - 3 \alpha^2 \beta \paren {\alpha + \beta + 2} - \alpha^3 \paren {\alpha + \beta + 1} \paren {\alpha + \beta + 2} } {\alpha + \beta + 2} } \sqrt {\frac {\alpha + \beta + 1} {\alpha^3 \beta^3} } }} {{eqn | r = \paren {\frac {2 \alpha \beta \paren {\beta - \alpha} } {\alpha + \beta + 2} } \sqrt {\frac {\alpha + \beta + 1} {\alpha^3 \beta^3} } }} {{eqn | r = \frac {2 \paren {\beta - \alpha} \sqrt {\alpha + \beta + 1} } {\paren {\alpha + \beta + 2} \sqrt {\alpha \beta} } }} {{end-eqn}} The fifth step is justified by: {{begin-eqn}} {{eqn | o =        | r = \alpha \paren {\alpha + 1} \paren {\alpha + 2} \paren {\alpha + \beta}^2 - 3 \alpha^2 \beta \paren {\alpha + \beta + 2} - \alpha^3 \paren {\alpha + \beta + 1} \paren {\alpha + \beta + 2} }} {{eqn | r = \alpha^5 + 3 \alpha^4 + 2 \alpha^3 + 2 \alpha^4 \beta + 6 \alpha^3 \beta + 4 \alpha^2 \beta + \alpha^3 \beta^2 + 3 \alpha^2 \beta^2 + 2 \alpha \beta^2 - 3 \alpha^3 \beta - 3 \alpha^2 \beta^2 - 6 \alpha^2 \beta - \alpha^5 - 2 \alpha^4 \beta - \alpha^3 \beta^2 - 3 \alpha^4 - 3 \alpha^3 \beta - 2 \alpha^3 }} {{eqn | r = - 2 \alpha^2 \beta + 2 \alpha \beta^2 }} {{eqn | r = 2 \alpha \beta \paren {\beta - \alpha} }} {{end-eqn}} {{qed}} Category:Skewness Category:Beta Distribution bfiv2jnhw00t0p2dxdf5osxfcgl4a4n