avewright/proofwiki-math huggingface dataset:
Theorem: Difference of Two Powers/General Commutative Ring

Let $\struct {R, +, \circ}$ be a commutative ring whose zero is $0_R$.
Let $a, b \in R$.
Let $n \in \N$ such that $n \ge 2$.
Then:
{{begin-eqn}}
{{eqn | l = a^n - b^n
      | r = \paren {a - b} \circ \sum_{j \mathop = 0}^{n - 1} a^{n - j - 1} \circ b^j
      | c = 
}}
{{eqn | r = \paren {a - b} \circ \paren {a^{n - 1} + a^{n - 2} \circ b + a^{n - 3} \circ b^2 + \dotsb + a \circ b^{n - 2} + b^{n - 1} }
      | c = 
}}
{{end-eqn}}

Proof:
Let $\ds S_n = \sum_{j \mathop = 0}^{n - 1} a^{n - j - 1} \circ b^j$.
This can also be written:
:$\ds S_n = \sum_{j \mathop = 0}^{n - 1} b^j \circ a^{n - j - 1}$
Consider:
:$\ds a \circ S_n = \sum_{j \mathop = 0}^{n - 1} a^{n - j} \circ b^j$
Taking the first term (where $j = 0$) out of the summation, we get:
:$\ds a \circ S_n = \sum_{j \mathop = 0}^{n - 1} a^{n - j} \circ b^j = a^n + \sum_{j \mathop = 1}^{n - 1} a^{n - j} \circ b^j$
Similarly, consider:
:$\ds b \circ S_n = \sum_{j \mathop = 0}^{n - 1} a^j \circ b^{n - j}$
Taking the first term (where $j = 0$) out of the summation:
:$\ds b \circ S_n = \sum_{j \mathop = 0}^{n - 1} a^j \circ b^{n - j} = b^n + \sum_{j \mathop = 1}^{n - 1} a^{n - j} \circ b^j$
This is equal to:
:$\ds b^n + \sum_{j \mathop = 1}^{n - 1} a^j \circ b^{n - j}$
by Permutation of Indices of Summation.
So:
{{begin-eqn}}
{{eqn | l = \paren {a - b} \circ S_n
      | r = a \circ \sum_{j \mathop = 0}^{n - 1} a^{n - j - 1} \circ b^j - b \circ \sum_{j \mathop = 0}^{n - 1} a^j \circ b^{n - j - 1}
      | c = 
}}
{{eqn | r = \sum_{j \mathop = 0}^{n - 1} a^{n - j} \circ b^j - \sum_{j \mathop = 0}^{n - 1} a^j \circ b^{n - j}
      | c = 
}}
{{eqn | r = a^n + \sum_{j \mathop = 1}^{n - 1} a^{n - j} \circ b^j - \sum_{j \mathop = 1}^{n - 1} a^{n - j} \circ b^j - b^n
      | c = 
}}
{{eqn | r = a^n - b^n
      | c = 
}}
{{end-eqn}}
{{qed}}
Category:Commutative Rings
Category:Polynomial Theory
Category:Difference of Two Powers
od0fbb4mthzs401ycfl16uptruxj1gf
Theorem: Sequence of Powers of Reciprocals is Null Sequence/Real Index

Let $r \in \R_{>0}$ be a strictly positive real number.
Let $\sequence {x_n}$ be the sequence in $\R$ defined as:
: $x_n = \dfrac 1 {n^r}$
Then $\sequence {x_n}$ is a null sequence.

Proof:
Let $\epsilon > 0$.
We need to show that:
:$\exists N \in \N: n > N \implies \size {\dfrac 1 {n^r} } < \epsilon$
That is, that $n^r > 1 / \epsilon$.
Let us choose $N = \ceiling {\paren {1 / \epsilon}^{1/r} }$.
By Reciprocal of Strictly Positive Real Number is Strictly Positive and power of positive real number is positive, it follows that:
:$\paren {\dfrac 1 \epsilon}^{1/r} \gt 0$
Then by Positive Power Function on Non-negative Reals is Strictly Increasing:
:$\forall n > N: n^r > N^r \ge 1 / \epsilon$
{{qed}}
Theorem: Odd Order Derivative of Even Function Vanishes at Zero

Let $X$ be a symmetric subset of $\R$ containing $0$. 
Let $n$ be a positive integer. 
Let $f:X \to \R$ be an even function. 
Let $f$ be at least $\paren{2 n + 1}$-times differentiable.
Then:
:$\map {f^{\paren {2 n + 1} } } 0 = 0$

Proof:
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
Theorem: Reciprocal of Null Sequence/Corollary

:$x_n \to \infty$ as $n \to \infty$ {{iff}} $\size {\dfrac 1 {x_n} } \to 0$ as $n \to \infty$

Proof:
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}}
Theorem: Unbounded Monotone Sequence Diverges to Infinity/Increasing

Let $\sequence {x_n}$ be increasing and unbounded above.
Then $x_n \to +\infty$ as $n \to \infty$.

Proof:
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}}
Theorem: Index of Subsequence not Less than its Index

Let $\sequence {x_n}_{n \mathop \ge 1}$ be a sequence in a set $S$.
Let $\sequence {x_{n_r} }$ be a subsequence of $\sequence {x_n}$.
Then:
:$\forall n \in \N_{>0}: n_r \ge r$

Proof:
The proof proceeds by induction.
For all $r \in \Z_{\ge 1}$, let $\map P r$ be the proposition:
:$n_r \ge r$
Theorem: Propagation of Light in Inhomogeneous Medium

Let $v: \R^3 \to \R$ be a real function.
Let $M$ be a 3-dimensional Euclidean space.
Let $\gamma:t \in \R \to M$ be a smooth curve embedded in $M$, where $t$ is time.
Denote its derivative {{WRT}} time by $v$.
Suppose $M$ is filled with an optically inhomogeneous medium such that at each point speed of light is $v = \map v {x, y, z}$
Suppose $\map y x$ and $\map z x$ are real functions.
Let the light move according to Fermat's principle.
Then equations of motion have the following form:
{{begin-eqn}}
{{eqn | l = \dfrac {\partial v} {\partial y} \dfrac {\sqrt {1 + y'^2 + z'^2} } {v^2} + \dfrac \d {\d x} \dfrac {y'} {v \sqrt {1 + y'^2 + z'^2} }
      | r = 0
}}
{{eqn | l = \dfrac {\partial v} {\partial z} \dfrac {\sqrt {1 + y'^2 + z'^2} } {v^2} + \dfrac \d {\d x} \dfrac {z'} {v \sqrt {1 + y'^2 + z'^2} }
      | r = 0
}}
{{end-eqn}}

Proof:
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}}
Theorem: Even Order Derivative of Odd Function Vanishes at Zero

Let $X$ be a symmetric subset of $\R$ containing $0$. 
Let $n$ be a positive integer. 
Let $f:X \to \R$ be an odd function. 
Let $f$ be at least $\paren {2 n}$-times differentiable.
Then: 
:$\map {f^{\paren {2 n} } } 0 = 0$

Proof:
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
Theorem: Geometric Mean of two Positive Real Numbers is Between them

Let $a, b \in \R$ be real numbers such that $0 < a < b$.
Let $\map G {a, b}$ denote the geometric mean of $a$ and $b$.
Then:
:$a < \map G {a, b} < b$

Proof:
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
Theorem: Geometric Mean of Reciprocals is Reciprocal of Geometric Mean

Let $x_1, x_2, \ldots, x_n \in \R_{> 0}$ be strictly positive real numbers.
Let $G_n$ denote the geometric mean of $x_1, x_2, \ldots, x_n$.
Let ${G_n}'$ denote the geometric mean of their reciprocals $\dfrac 1 {x_1}, \dfrac 1 {x_2}, \ldots, \dfrac 1 {x_n}$.
Then:
:${G_n}' = \dfrac 1 {G_n}$

Proof:
{{begin-eqn}}
{{eqn | l = {G_n}'
      | r = \paren {\prod_{1 \mathop = k}^n \dfrac 1 {x_k} }^{1/n}
      | c = {{Defof|Geometric Mean}}
}}
{{eqn | r = \paren {\dfrac 1 {\prod_{1 \mathop = k}^n x_k} }^{1/n}
      | c = 
}}
{{eqn | r = \dfrac 1 {\paren {\prod_{1 \mathop = k}^n x_k}^{1/n} }
      | c = 
}}
{{eqn | r = \dfrac 1 {G_n}
      | c = {{Defof|Geometric Mean}}
}}
{{end-eqn}}
{{qed}}
Category:Geometric Mean
Category:Reciprocals
0yyp90q5hxvtmo032pg6qbo1to3ds61
Theorem: GM-HM Inequality

Let $x_1, x_2, \ldots, x_n \in \R_{> 0}$ be strictly positive real numbers.
Let $G_n$ be the geometric mean of $x_1, x_2, \ldots, x_n$.
Let $H_n$ be the harmonic mean of $x_1, x_2, \ldots, x_n$.
Then:
:$G_n \ge H_n$

Proof:
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:
:$\ds \dfrac 1 {H_n} := \frac 1 n \paren {\sum_{k \mathop = 1}^n \frac 1 {x_k} }$
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}}
Theorem: GM-HM Inequality

Let $x_1, x_2, \ldots, x_n \in \R_{> 0}$ be strictly positive real numbers.
Let $G_n$ be the geometric mean of $x_1, x_2, \ldots, x_n$.
Let $H_n$ be the harmonic mean of $x_1, x_2, \ldots, x_n$.
Then:
:$G_n \ge H_n$

Proof:
For $p \in \R$, let $\map {M_p} {x_1, x_2, \ldots, x_n}$ denote the '''Hölder mean with exponent $p$ of $x_1, x_2, \ldots, x_n$'''.
By definition of Hölder Mean with $p = 0$:
:$\map {M_0} {x_1, x_2, \ldots, x_n} = \map G {x_1, x_2, \ldots, x_n}$
From Hölder Mean for Exponent -1 is Harmonic Mean:
:$\map {M_{-1} } {x_1, x_2, \ldots, x_n} = \map G {x_1, x_2, \ldots, x_n}$
The result follows from Inequality of Hölder Means:
:$-1 < 0 \implies \map {M_{-1} } {x_1, x_2, \ldots, x_n} < \map {M_0} {x_1, x_2, \ldots, x_n}$
unless $x_k = 0$ for some $k \in \set {1, 2, \ldots, n}$ or $x_1 = x_2 = \cdots = x_n$, in which case:
:$\map {M_{-1} } {x_1, x_2, \ldots, x_n} = \map {M_0} {x_1, x_2, \ldots, x_n}$
Hence the result.
{{qed}}
Theorem: Harmonic Mean of two Positive Real Numbers is Between them

Let $a, b \in \R_{\gt 0}$ be (strictly) positive real numbers such that $a < b$.
Let $\map H {a, b}$ denote the harmonic mean of $a$ and $b$.
Then:
:$a < \map H {a, b} < b$

Proof:
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 = {{hypothesis}}
}}
{{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
gnoj6c78ptarrum9af7e8m6x0zkp9ed
Theorem: Arithmetic Mean of two Real Numbers is Between them

Let $a, b \in \R_{\ne 0}$ be real numbers such that $a < b$.
Let $\map A {a, b}$ denote the arithmetic mean of $a$ and $b$.
Then:
:$a < \map A {a, b} < b$

Proof:
By definition of arithmetic mean:
:$\map A {a, b} := \dfrac {a + b} 2$
Thus:
{{begin-eqn}}
{{eqn | l = a
      | o = <
      | r = b
      | c = {{hypothesis}}
}}
{{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 = {{hypothesis}}
}}
{{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
hm8jub088vj6kj3a1tdk2brw2vhu91i
Theorem: Kurtosis of Normal Distribution

Let $X$ be a continuous random variable with a normal distribution with parameters $\mu$ and $\sigma^2$ for some $\mu \in \R$ and $\sigma \in \R_{> 0}$:
:$X \sim \Gaussian \mu {\sigma^2}$
Then the kurtosis $\alpha_4$ of $X$ is equal to $3$.
That is, $\Gaussian \mu {\sigma^2}$ is mesokurtic.

Proof:
From the definition of kurtosis, we have: 
:$\alpha_4 = \expect {\paren {\dfrac {X - \mu} \sigma}^4}$
where: 
:$\mu$ is the expectation of $X$. 
:$\sigma$ is the standard deviation of $X$.
By Expectation of Normal Distribution, we have: 
:$\mu = \mu$
By Variance of Normal Distribution, we have: 
:$\sigma = \sigma$ 
So: 
{{begin-eqn}}
{{eqn | l = \alpha_4
      | r = \dfrac {\expect {X^4} - 4 \mu \expect {X^3} + 6 \mu^2 \expect {X^2} - 3 \mu^4} {\sigma^4}
      | c = Kurtosis in terms of Non-Central Moments
}}
{{eqn | r = \frac {\expect {X^4} - 4 \mu \paren {3 \mu \sigma^2 + \mu^3} + 6 \mu^2 \paren {\sigma^2 + \mu^2} - 3 \mu^4} {\sigma^4}
      | c = Skewness of Normal Distribution: Proof 2, Variance of Normal Distribution: Proof 2
}}
{{end-eqn}}
To calculate $\alpha_4$, we must calculate $\expect {X^4}$.
We find this using the moment generating function of $X$, $M_X$. 
From Moment in terms of Moment Generating Function:
:$\expect {X^4} = \map { {M_X}^{\paren 4} } 0$
From Moment Generating Function of Normal Distribution: Fourth Moment: 
:$\map { {M_X}^{\paren 4} } t = \paren {3 \sigma^4 + 6 \sigma^2 \paren {\mu + \sigma^2 t}^2 + \paren {\mu + \sigma^2 t}^4} \map \exp {\mu t + \dfrac 1 2 \sigma^2 t^2}$
Setting $t = 0$ and from Exponential of Zero, we have:
{{begin-eqn}}
{{eqn | l = \expect {X^4}
      | r = \paren {3 \sigma^4 + 6 \sigma^2 \paren {\mu + \sigma^2 0}^2 + \paren {\mu + \sigma^2 0}^4} \map \exp {\mu 0 + \dfrac 1 2 \sigma^2 0^2}
}}
{{eqn | r = \mu^4 + 6 \mu^2 \sigma^2 + 3 \sigma^4
      | c = Exponential of Zero
}}
{{end-eqn}}
Plugging this result back into our equation above:
{{begin-eqn}}
{{eqn | l = \alpha_4
      | r = \frac {\expect {X^4} - 4 \mu \paren {3 \mu \sigma^2 + \mu^3} + 6 \mu^2 \paren {\sigma^2 + \mu^2} - 3 \mu^4} {\sigma^4}
      | c = 
}}
{{eqn | r = \frac {\paren {\mu^4 + 6 \mu^2 \sigma^2 + 3 \sigma^4} - 4 \mu \paren {3 \mu \sigma^2 + \mu^3} + 6 \mu^2 \paren {\sigma^2 + \mu^2} - 3 \mu^4} {\sigma^4}
      | c = 
}}
{{eqn | r = \frac {\paren {1 - 4 + 6 - 3 } \mu^4 + \paren {6 - 12 + 6 } \mu^2 \sigma^2 + 3 \sigma^4 } {\sigma^4}
}}
{{eqn | r = \frac {3 \sigma^4 } {\sigma^4}
}}
{{eqn | r = 3
}}
{{end-eqn}}
{{qed}}
Theorem: Raw Moment of Bernoulli Distribution

Let $X$ be a discrete random variable with a Bernoulli distribution with parameter $p$.
Let $n$ be a strictly positive integer. 
Then the $n$th raw moment $\expect {X^n}$ of $X$ is given by: 
:$\expect {X^n} = p$

Proof:
From the definition of expectation:
:$\ds \expect {X^n} = \sum_{x \mathop \in \Img X} x^n \map \Pr {X = x}$
From the definition of the Bernoulli distribution:
:$\ds \expect {X^n} = 1^n \times p + 0^n \times \paren {1 - p} = p$
{{qed}}
Theorem: Equivalence of Definitions of Local Basis

Let $T = \struct {S, \tau}$ be a topological space.
Let $x$ be an element of $S$.
{{TFAE|def = Local Basis}}

Proof:
=== 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}}
Theorem: Equivalence of Definitions of Local Basis/Local Basis for Open Sets Implies Neighborhood Basis of Open Sets

Let $\BB$ be a set of open neighborhoods of $x$ such that:
:$\forall U \in \tau: x \in U \implies \exists H \in \BB: H \subseteq U$

Proof:
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.
Theorem: Equivalence of Definitions of Local Basis/Neighborhood Basis of Open Sets Implies Local Basis for Open Sets

Let $\BB$ be a set of open neighborhoods of $x$ such that:
:every neighborhood of $x$ contains a set in $\BB$.

Proof:
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.
Theorem: Local Basis Generated from Neighborhood Basis

Let $T = \struct {S, \tau}$ be a topological space.
Let $x$ be an element of $S$.
Let $\BB$ be a neighborhood basis of $x$.
For any subset $A \subseteq S$, let $A^\circ$ denote the interior of $A$.
Then the set:
:$\BB' = \set {H^\circ: H \in \BB}$
is a local basis of $x$.

Proof:
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
0sr2qlpuqsczcs7cv7ydwddjta4w824
Theorem: Arcsine of Zero is Zero

:$\arcsin 0 = 0$
where $\arcsin$ is the arcsine function.

Proof:
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
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Theorem: Arcsine of One is Half Pi

:$\arcsin 1 = \dfrac \pi 2$
where $\arcsin$ is the arcsine function.

Proof:
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
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Theorem: Real Sequence (1 + x over n)^n is Convergent

The sequence $\sequence {s_n}$ defined as:
:$s_n = \paren {1 + \dfrac x n}^n$
is convergent.

Proof:
From Cauchy's Mean Theorem:
:$(1): \quad \ds \paren {\prod_{k \mathop = 1}^n a_k}^{1/n} \le \frac 1 n \paren {\sum_{k \mathop = 1}^n a_k}$
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 Power Series Expansion, 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}}
Theorem: Raw Moment of Poisson Distribution

Let $X$ be a discrete random variable with the Poisson distribution with parameter $\lambda$.
Let $n$ be a strictly positive integer. 
Then the $n$th raw moment $\expect {X^n}$ of $X$ is given by: 
:$\ds \expect {X^n} = \sum_{k \mathop = 0}^n \lambda^k {n \brace k}$
where $\ds {n \brace k}$ is a Stirling number of the second kind.

Proof:
{{ProofWanted}}
Category:Poisson Distribution
Category:Raw Moments
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