nyuuzyou/wikis huggingface dataset:
Bernstein's Theorem on Unique Global Solution to y''=F(x,y,y')/Lemma 1

== Lemma for Bernstein's Theorem on Unique Global Solution to $y = \map F {x, y, y'}$==
Let $F$ and its partial derivatives $F_y, F_{y'}$ be real functions, defined on the closed interval $I = \closedint a b$.
Let $F, F_y, F_{y'} $ be continuous at every point $\tuple {x, y}$ for all finite $y'$.
Suppose there exists a constant $k > 0$ such that:
:$\map {F_y} {x, y, y'} > k$
Suppose there exist real functions $\alpha = \map \alpha {x, y} \ge 0$, $\beta = \map \beta {x, y}\ge 0$ bounded in every bounded region of the plane such that:
:$\size {\map F {x, y, y'} } \le \alpha y'^2 + \beta$
Let $L$ be an integral curve of equation $y = \map F {x, y, y'}$ which passes through the two points $\tuple {a, A}$ and $\tuple {b, B}$ such that $a \ne b$.
Then $L$ is unique.
== Proof ==
 there are two integral curves $y = \map {\phi_1} x$ and $y = \map {\phi_2} x$ such that:
:$\map {y} x = \map F {x, y, y'}$
Define a mapping $\delta: I \to S \subset \R$:
:$\map \delta x = \map {\phi_2} x - \map {\phi_1} x$
From definition it follows that:
:$\map \delta a = \map \delta b = 0$
Then the second derivative of $\delta$ yields:
where:
:$F_y^* = \map {F_y} {x, \phi_1 + \theta \delta, \phi_1' + \theta \delta'}$
:$F_{y'}^* = \map {F_{y'} } {x, \phi_1 + \theta \delta, \phi_1' + \theta \delta'}$
and $0 < \theta < 1$.
Suppose:
:$\forall x \in I: \map {\phi_2} x \ne \map {\phi_1} x$
Then there are two possibilities for $\delta$:
:$\map \delta x$ attains a positive maximum within $\tuple {a, b}$
:$\map \delta x$ attains a negative minimum within $\tuple {a, b}$.
Denote this point by $\xi$.
Suppose $\xi$ denotes a maximum.
Then:
:$\map {\delta} \xi \le 0$
:$\map \delta \xi > 0$
:$\map {\delta'} \xi = 0$
These, together with $(1)$, imply that $F_y^* \le 0$. 
This is in contradiction with assumption.
For the minimum the inequalities are reversed, but the last equality is the same.
Therefore, it must be the case that:
:$\map {\phi_1} x = \map {\phi_2} x$
== Sources ==
* 
* : $\S 1.9$: A Theorem of Bernstein
* : $\S 1.4$: The Simplest Variational Problem. Euler's Equation
* 
Category:Bernstein's Theorem on Unique Global Solution to y=F(x,y,y')
Bernstein's Theorem on Unique Global Solution to y''=F(x,y,y')/Lemma 2

== Lemma for Bernstein's Theorem on Unique Global Solution to $y = \map F {x, y, y'}$==
Let $F$ and its partial derivatives $F_y, F_{y'}$ be real functions, defined on the closed interval $I = \closedint a b$.
Let $F, F_y, F_{y'} $ be continuous at every point $\tuple {x, y}$ for all finite $y'$.
Suppose there exists a constant $k > 0$ such that:
:$\map {F_y} {x, y, y'} > k$
Suppose there exist real functions $\alpha = \map \alpha {x, y} \ge 0$, $\beta = \map \beta {x, y}\ge 0$ bounded in every bounded region of the plane such that:
:$\size {\map F {x, y, y'} } \le \alpha y'^2 + \beta$
Suppose that:
:$\map {y} x = \map F {x, y, y'}$
for all $x \in \closedint a c$, where:
:$\map y a = a_1$
:$\map y c = c_1$
Then the following bound holds:
:$\size {\map y x - \dfrac {a_1 \paren {c - x} + c_1 \paren {x - a} } {c - a} } \le \dfrac 1 k \max \limits_{a \mathop \le x \mathop \le b} \size {\map F {x, \dfrac {a_1 \paren {c - x} + c_1 \paren {x - a} } {c - a}, \dfrac {c_1 - a_1} {c - a} } }$
== Proof ==
As a consequence of $y = \map F {x, y, y'}$ we have:
where:
:$\psi = \dfrac {a_1 \paren {c - x} + c_1 \paren {x - a} } {c - a} + \theta \paren {\map y x - \dfrac {a_1 \paren {c - x} + c_1 \paren {x - a} } {c - a} }$
and:
:$0 < \theta < 1$
Note that the term $\chi$, defined as:
:$\chi = \map y x - \dfrac {a_1 \paren {c - a} + c_1 \paren {x - a} } {c - a}$ 
vanishes at $x = a$ and $x = c$.
Unless $\chi$ is identically zero, there exists a point $\xi \in \openint a b$ such one of the following holds:
:$\chi$ attains a positive maximum at $\xi$
:$\chi$ attains a negative minimum at $\xi$. 
In the first case:
:$\map {y} \xi \le 0$,
:$\map {y'} \xi = \dfrac {c_1 - a} {c - a}$ 
which implies:
Hence:
:$\map y \xi - \dfrac {a_1 \paren {c - \xi} + c_1 \paren {\xi - a} } {c - a} \le -\dfrac 1 k \map F {\xi, \dfrac {a_1 \paren {c - \xi} + c_1 \paren {\xi - a} } {c - a}, \dfrac {c_1 - a_1} {c - a} }$
The negative minimum part is proven analogously.
Hence, the following holds: 
:$\size {\map y x - \dfrac {a_1 \paren {c - x} + c_1 \paren {x - a} } {c - a} } \le \dfrac 1 k \max \limits_{a \mathop \le x \mathop \le b} \size {\map F {x, \dfrac {a_1 \paren {c - x} + c_1 \paren {x - a} } {c - a}, \dfrac {c_1 - a_1} {c - a} } }$
== Sources ==
* 
* : $\S 1.9$: A Theorem of Bernstein
* : $\S 1.4$: The Simplest Variational Problem. Euler's Equation
* 
Category:Bernstein's Theorem on Unique Global Solution to y=F(x,y,y')
Bernstein's Theorem on Unique Global Solution to y''=F(x,y,y')/Lemma 3

== Lemma for Bernstein's Theorem on Unique Global Solution to $y = \map F {x, y, y'}$==
Let $F$ and its partial derivatives $F_y, F_{y'}$ be real functions, defined on the closed interval $I = \closedint a b$.
Let $F, F_y, F_{y'} $ be continuous at every point $\tuple {x, y}$ for all finite $y'$.
Suppose there exists a constant $k > 0$ such that:
:$\map {F_y} {x, y, y'} > k$
Suppose there exist real functions $\alpha = \map \alpha {x, y} \ge 0$, $\beta = \map \beta {x, y}\ge 0$ bounded in every bounded region of the plane such that:
:$\size {\map F {x, y, y'} } \le \alpha y'^2 + \beta$
Suppose that for $x \in I$:
:$\map {y} x = \map F {x, y, y'}$
where:
:$\map y a = a_1$
:$\map y c = c_1$
Then:
:$\forall x \in I: \exists M \in \R: \size {\map {y'} x - \dfrac {c_1 - a_1} {c - a} } \le M$
== Proof ==
Let $\AA$ and $\BB$ be the least upper bounds of $\map \alpha {x, y}$ and $\map \beta {x, y}$ respectively in the rectangle $a \le x \le c$, $\size y \le m + max \set {\size a_1, \size c_1}$
where:
:$m = \dfrac 1 k \max \limits_{a \mathop \le x \mathop \le b} \size {\map F {x, \dfrac {a_1 \paren {c - x} + c_1 \paren {x - a} } {c - a}, \dfrac {c_1 - a_1} {c - a} } }$
Suppose that $\AA \ge 1$.
Let $u, v$ be real functions such that:
From Lemma 2, for $x \in I$ the s of $(3)$ and $(4)$ are not negative. 
Thus:
:$\forall x \in I: u, v \ge 1$
Differentiate equations $(3)$ and $(4)$ $x$:
Differentiate again: 
 
By assumption:
 
where:
:$\BB_1 = \BB + 2 \AA \paren {\dfrac {c_1 - a_1} {c - a} }^2$
Then:
Multiply the inequality by $2 \AA u$ and simplify:
 
Similarly: 
Multiply the inequality by $-2 \AA v$ and simplify:
Note that $\map u a = \map u c$ and $\map v a = \map v c$.
From Intermediate Value Theorem it follows that
:$\exists K \subset I: \forall x_0 \in K: \map {y'} {x_0} - \dfrac {c_1 - a_1} {c - a} = 0$
Points $x_0$ divide $I$ into subintervals.
Due to $(5)$ and $(6)$ both $\map {u'} x$ and $\map {v'} x$ maintain sign in the subintervals and vanish at one or both endpoints of each subinterval.
Let $J$ be one of the subintervals.
Let functions $\map {u'} x$, $\map {v'} x$ be zero at $\xi$, the right endpoint. 
The quantity:
:$\map {y'} x - \dfrac {c_1 - a_1} {c - a}$
has to be either positive or negative.
Suppose it is positive in $J$. 
From $(5)$, $u'$ is not negative.
Multiply both sides of $(10)$ by $u'$: 
:$u u' \ge - 2 \AA \BB_1 u u'$
Integrating this from $x \in J$ to $\xi$, together with $\map {u'} \xi = 0$, yields:
:$-\map {u'^2} x \ge - 2 \AA \BB_1 \paren {\map {u^2} \xi - \map {u^2} x}$
Then:
Hence:
:$\forall x \in J: \size {\map {y'} x - \dfrac {c_1 - a_1} {c - a} } \le \sqrt {\dfrac {\BB_1} {2 \AA} } e^{4 \AA m}$
Similar arguments for aforementioned quantity being negative.
== Sources ==
* 
* : $\S 1.9$: A Theorem of Bernstein
* : $\S 1.4$: The Simplest Variational Problem. Euler's Equation
* 
Category:Bernstein's Theorem on Unique Global Solution to y=F(x,y,y')
Berry Paradox

== Paradox ==
Every number can be defined by a sentence in natural language.
For the purpose of this argument, let that natural language be .
It is assumed without proof that has a finite number of words.
Let $n$ be an integer such that $n \ge 15$.
Then the cardinality of the set of integers that can be defined in no more than $n$ words is finite.
Consider the integer that is defined as:
:the smallest integer which cannot be defined by a sentence of at most fifteen words.
Let this number be $N$.
That is $N$ cannot be defined by a sentence of at most fifteen words.
But that very sentence itself has fifteen words.
So $N$ has been demonstrated to be definable in a fifteen-word sentence.
So: can it or can't it?
== Resolution ==
The expression:
:the smallest integer which cannot be defined by a sentence of at most fifteen words
is self-contradictory, as any integer it defines can be defined in at most fifteen words.
The problem arises because the definition itself contains the word "defined". 
To formalize a statement like this, it would be necessary to first define the word "defined". 
== Also known as ==
== Also see ==
This paradox is related to Gödel's Incompleteness Theorems, specifically Gödel numbers.
== Historical Note ==
== Sources ==
* 
* : Chapter $1$: Sets and mappings: $\S 1.2$: Sets: Exercise $7$
* : $111,777$
* 
* 
* 
Category:Berry Paradox
Category:Antinomies
Category:Semantic Paradoxes
Bertrand's Theorem

== Theorem ==
Let $U: \R_{>0} \to \R$ be analytic for $r > 0$.
Let $M > 0$ be a nonvanishing angular momentum such that a stable circular orbit exists.
Suppose that every orbit sufficiently close to the circular orbit is closed.
Then $U$ is either $k r^2$ or $-\dfrac k r$ (for $k > 0$) up to an additive constant.
=== Preliminary Lemma ===
== Non-Perturbative Proof ==
== Proof based on Asymptotic Behaviour ==
== Sources ==
* : $\S 2.\text D$, problems $1$. to $6$.
* 
* A non-perturbative proof of Bertrand's theorem - Santos, F C et al - arXiv:0704.0575
Category:Classical Mechanics
Category:Bertrand's Theorem
Bertrand's Theorem/Asymptotic Proof

== Theorem ==
== Proof ==
=== Preliminary Lemma ===
The substitution $x = M / r$ gives:
:$\ds \Phi = \sqrt 2 \int_{x_\min}^{x_\max} \frac {\d x} {\sqrt {E - \map W x}}$
where $\map W x \equiv \map U {\dfrac M x} + \dfrac 1 2 x^2$.
In general, the frequency of oscillations around a stable equilibrium at $x = x_0$ for a particle of mass $m$ in a potential $V$ is given by:
:$\omega = \sqrt {\dfrac {\map {V} {x_0} } m}$
which means that:
:$\Phi \approx 2 \pi \dfrac M {x_0^2} \sqrt {\map {W} {x_0} } = 2 \pi \sqrt {\dfrac {U'} {3 U' + x_0 U} }$
Suppose $\Phi$ were constant.
Then the solutions are:
:$\map U r = k r^\alpha, \alpha \ge - 2, \alpha \ne 0$
and:
:$\map U r = b \log r$
Therefore:
: $\Phi = \dfrac {2 \pi} {\sqrt{\alpha + 2}}$
with $\alpha = 0$ corresponding to the logarithmic solution.
This is discarded, since in that case $\Phi$ is not commensurable with $\pi$.
Now consider $\alpha > 0$, so that $\map U r \to \infty$ as $r \to \infty$.
The substitution $x = y x_\max$ reduces $\Phi$ to:
:$\ds \Phi = \sqrt 2 \int_{y_\min}^1 \frac {\d y} {\sqrt {\map W 1 - \map W y} }$
where:
:$\map W y \equiv \dfrac {y^2} 2 + \dfrac 1 { {x_\max}^2} \map U {\dfrac M {y x_\max} }$
Therefore, when $E \to \infty$:
:$x_\max \to \infty$ and $y_\min \to 0$
This means that:
:$\ds \lim_{E \mathop \to \infty} \Phi = \pi$
Equating this with the previous result gives:
:$\dfrac {2 \pi} {\sqrt {\alpha + 2}} = \pi$
Therefore:
:$\alpha=2$
Now let $\map U r = -k r^{-\beta}, 0 < \beta < 2$.
A similar approach gives:
:$\ds \lim_{E \mathop \to -0} \Phi = \frac {2 \pi} {2 - \beta}$
so that:
:$\dfrac {2 \pi} {2 + \alpha} = \dfrac {2 \pi} {\sqrt {2 + \alpha} }$
It follows that:
:$\alpha = -1$
and the result follows.
== Sources ==
* : $\S 2.\text D$, problems $1$. to $6$.
Category:Bertrand's Theorem
Bertrand's Theorem/Lemma

== Theorem ==
== Preliminary Lemma ==
For simplicity we set $m = 1$, so that the effective potential becomes:
:$U_M = U + \dfrac {M^2} {2 r^2}$
Consider the apsidial angle:
:$\ds \Phi = \sqrt 2 \int_{r_\min}^{r_\max} \frac {M \rd r} {r^2 \sqrt {E - U_M} }$
where:
:$E$ is the energy
:$r_\min, r_\max$ are solutions to $\map {U_M} r = E$. 
By definition, this is the angle between adjacent apocenters (pericenters).
Recall that if $\Phi$ is commensurable with $\pi$, then an orbit is closed.
== Sources ==
* : $\S 2.\text D$, problems $1$. to $6$.
Category:Bertrand's Theorem
Bertrand's Theorem/Non-Perturbative Proof

== Theorem ==
== Proof ==
=== Preliminary Lemma ===
In general $U_M$ is not monotonic on $\openint {r_\min} {r_\max}$.
Therefore a unique inverse $\map r {U_M}$ does not exist.
However, suppose it is possible to construct separate inverse functions $r_{1, 2}$ for the intervals $\openint {r_0} {r_\min}$ and $\openint {r_0} {r_\max}$, where $r_0$ is the minimum of $U_M$ on $\openint {r_\min} {r_\max}$.
Note that the orbit corresponding to $r_0$ is the stable circular orbit.
Therefore this will be possible for orbits sufficiently close to it. 
Now write:
:$\ds \Phi = \int_{U_0}^E \map F {U_M} \frac {\d U_{eff} } {\sqrt {E - U_M} }$
where:
:$F = \sqrt 2 M \, \map {\dfrac {\d} {\d U_M} } {\dfrac 1 {r_1} - \dfrac 1 {r_2} }$
:$U_0 \equiv \map {U_M} {r_0}$
This is Abel's Integral Equation which can be solved for $\map F {U_M}$, giving:
:$\ds \map F U = \dfrac 1 \pi \int_{U_0}^{U_M} \frac {\Phi \rd E} {\sqrt {U_M - E} }$
$\Phi$ may have energy dependence.
However, we require:
:$\map \Phi E = 2 \pi \map q E$
such that $q: \R \to \Q$ is continuous, and a rational continuous function must be constant.
Finally, integrating gives:
:$\dfrac 1 {r_1} - \dfrac 1 {r_2} = \gamma \sqrt{U_M - U_0}, \gamma = \dfrac {\sqrt 2 \Phi} {\pi M}$
Now consider defining the to be a single function $\dfrac 1 {\map r U}$, meromorphic with a simple pole at $r = 0$.
Write:
:$\dfrac 1 r = \gamma \sqrt {U_M - U_0} + \map \Omega {U_0, U}$
where $\Omega$ is an analytic function in $U$ in an open neighborhood of $U_0$ such that:
:$\map \Omega {U_0, U_0} = \dfrac 1 {r_0}$
Explicitly:
:$\sqrt {U_M - U_0} = \sqrt {U + \dfrac {M^2} {2 r^2} - U_0} = \dfrac 1 r \sqrt {\dfrac {M^2} 2 + \paren ({U - U_0} r^2}$
It can be seen that the also has a simple pole at $r=0$.
However, it also has a branching point at $r = r_0$.
To avoid this, the radicand must be the square of an analytic function with a zero at $r_0$.
The only possibilities are:
:$U \propto r^2, -\dfrac 1 {r^2}$
== Sources ==
* : $\S 2.\text D$, problems $1$. to $6$.
Category:Bertrand's Theorem
Bertrand-Chebyshev Theorem

== Theorem == 
For all $n \in \N_{>0}$, there exists a prime number $p$ with $n < p \le 2 n$.
== Proof 1 ==
== Proof 2 ==
== Also known as ==
== Historical Note ==
== Sources ==
* 
* 
* 
* 
* 
Category:Bertrand-Chebyshev Theorem
Category:Number Theory
Bertrand-Chebyshev Theorem/Also known as

== Bertrand-Chebyshev Theorem: Also known as == 
The Bertrand-Chebyshev Theorem is also known as Bertrand's Postulate or Bertrand's Conjecture.
Some sources give this as Chebyshev's theorem (in number theory) to distinguish it from a theorem in statistics.
== Sources ==
* 
* 
* 
* 
Category:Number Theory
Category:Bertrand-Chebyshev Theorem
Bertrand-Chebyshev Theorem/Historical Note

== Historical Note on Bertrand-Chebyshev Theorem == 
The Bertrand-Chebyshev Theorem was first postulated by in $1845$. He verified it for $n < 3 \, 000 \, 000$.
It became known as Bertrand's Postulate.
The first proof was given by in $1850$ as a by-product of his work attempting to prove the Prime Number Theorem.
After this point, it no longer being a postulate, Bertrand's Postulate was referred to as the Bertrand-Chebyshev Theorem.
In $1919$, gave a simpler proof based on the Gamma function.
In $1932$, gave an even simpler proof based on basic properties of binomial coefficients. That proof is the one which is presented here.
== Sources ==
* : Chapter $\text {A}.31$: Chebyshev ()
* 
* 
Category:Historical Notes
Bertrand-Chebyshev Theorem/Lemma 1

== Lemma ==
For all $n \in \N$:
:$\dbinom {2 n} n \ge \dfrac {2^{2 n}} {2 n + 1}$
where $\dbinom {2 n} n$ denotes a binomial coefficient.
== Proof ==
From Sequence of Binomial Coefficients is Strictly Increasing to Half Upper Index, $\dbinom n k$ increases for $k < \dfrac n 2$.
From Sequence of Binomial Coefficients is Strictly Decreasing from Half Upper Index, $\dbinom n k$ decreases for $k > \dfrac n 2$.
Therefore, $\dbinom {2 n} n$ is the largest term in the sequence $\dbinom {2 n} 0, \dbinom {2 n} 1, \ldots, \dbinom {2 n} {2 n}$.
Finally, observe that the mean of the terms in the sequence is $\dfrac {2^{2 n}} {2 n + 1}$, by Sum of Binomial Coefficients over Lower Index.
Category:Bertrand-Chebyshev Theorem
Bertrand-Chebyshev Theorem/Lemma 2

== Lemma for Bertrand-Chebyshev Theorem ==
For all $m \in \N$:
:$\ds \prod_{p \mathop \le m} p \le 2^{2 m}$
where the continued product is taken over all prime numbers $p \le m$.
== Proof ==
The proof proceeds by strong induction.
For all $m \in \Z_{\ge 0}$, let $\map P m$ be the proposition:
:$\ds \prod_{p \mathop \le m} p \le 2^{2 m}$
$\map P 0$ and $\map P 1$ are the vacuous cases:
=== Basis for the Induction ===
$\map P 2$ is the case:
Thus $\map P 2$ is seen to hold.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that if $\map P j$ is true, for all $j$ such that $0 \le j \le k$, then it logically follows that $\map P {k + 1}$ is true.
This is the induction hypothesis:
:$\ds \prod_{p \mathop \le k} p \le 2^{2 k}$
from which it is to be shown that:
:$\ds \prod_{p \mathop \le k + 1} p \le 2^{2 \paren {k + 1} }$
=== Induction Step ===
This is the induction step:
Let $m > 2$ be even.
Then:
and $\map P k$ seen to hold.
Let $m = 2 k + 1$ be odd.
By Sum of Binomial Coefficients over Lower Index:
:$\ds \sum_{r \mathop = 0}^{2 k + 1} \binom {2 k + 1} r = 2^{2 k + 1}$
Therefore:
Let $p$ be a prime number such that:
:$k + 2 \le p \le 2 k + 1$
Consider:
:$q = \dbinom {2 k + 1} k = \dfrac {\paren {2 k + 1}!} {k! \paren {k + 1}!}$
by definition of binomial coefficient.
We note that $p$ is greater than all of the divisors of the denominator of $q$.
Hence $p$ is a divisor of the numerator.
Hence:
:$p \divides q$
Thus:
So $\map P k \implies \map P {k + 1}$ and the result follows by the Second Principle of Mathematical Induction.
Therefore:
:$\forall m \in \N: \ds \prod_{p \mathop \le m} p \le 2^{2 m}$
Category:Bertrand-Chebyshev Theorem
Bertrand-Chebyshev Theorem/Lemma 3

== Lemma ==
Let $p$ be a prime number.
Let $p^k \divides \dbinom {2 n} n$.
Then $p^k \le 2 n$.
== Proof ==
Let $l$ be the largest power of $p$ with $p^l \le 2 n$.
By De Polignac's Formula, the largest power of $p$ dividing $n!$ is $\ds \sum_{i \mathop \ge 1} \floor {\frac n {p^i} }$.
So:
Category:Bertrand-Chebyshev Theorem
Bertrand-Chebyshev Theorem/Lemma 4

== Lemma ==
For all $n \ge 1$:
:$\dbinom {2 n} n \ge \dfrac {2^{2 n}} {2 n}$
where $\dbinom {2 n} n$ denotes a binomial coefficient.
== Proof ==
From Sequence of Binomial Coefficients is Strictly Increasing to Half Upper Index, $\dbinom n k$ increases for $k < \dfrac n 2$.
From Sequence of Binomial Coefficients is Strictly Decreasing from Half Upper Index, $\dbinom n k$ decreases for $k > \dfrac n 2$.
Hence for all $0 \le k \le n$:
:$\dbinom n k \le \dbinom n {\floor {n / 2} }$
where $\floor {n / 2}$ denotes the floor function.
So, for all $0 \le k \le 2 n$:
:$(1): \quad \dbinom {2 n} k \le \dbinom {2 n} n$
Hence:
Category:Bertrand-Chebyshev Theorem
Bertrand-Chebyshev Theorem/Proof 1

== Theorem == 
== Proof ==
We will first prove the theorem for the case $n \le 2047$. 
Consider the following sequence of prime numbers:
:$2, 3, 5, 7, 13, 23, 43, 83, 163, 317, 631, 1259, 2503$
Each of these prime number is smaller than twice the previous one.
Hence every interval $\set {x: n < x \le 2 n}$, with $n \le 2047$, contains one of these prime numbers.
=== ===
=== ===
=== ===
From :
:if $p > \sqrt {2 n}$, then $p$ appears at most once in $\dbinom {2 n} n$.
For $n \ge 3$, there is no prime factor $p$ with $\dfrac {2 n} 3 < p \le n$, because such a prime number divides $n!$ exactly once and $\paren {2 n}!$ exactly twice.
Therefore, by :
:$\ds \frac {2^{2 n} } {2 n + 1} \le \dbinom {2 n} n \le \prod_{p \mathop \le \sqrt {2 n} } 2 n \prod_{\sqrt {2 n} \mathop < p \mathop \le \frac {2 n} 3} p \prod_{n \mathop < p \mathop \le 2 n} p$
for $n \ge 3$.
 there is no prime number $p$ with $n < p \le 2 n$. 
Then we have:
This is a contradiction for sufficiently large $n$.
Indeed, we have:
:$2^{2 n / 3} \le \paren {2 n + 1} \paren {2 n}^{\sqrt {2 n} }$
Now:
:$2 n + 1 \le \paren {2 n}^2 \le \paren {2 n}^{\sqrt {2 n} / 3}$
for $n \ge 18$.
So:
:$2^{2 n} \le \paren {2 n}^{4 \sqrt {2 n} }$
Put $r = \sqrt {2 n}$.
Then:
:$2^{r^2} \le r^{8 r}$
or equivalently:
:$2^r \le r^8$
This fails when $r = 2^6 = 64$.
It fails thereafter, since $2^r$ increases faster than $r^8$.
So our proof works if:
:$n \ge 2^{11} = 2048$
and the examples show it is true for smaller $n$.
== Also known as ==
== Historical Note ==
Category:Bertrand-Chebyshev Theorem
Bertrand-Chebyshev Theorem/Proof 2

== Theorem == 
== Proof ==
We will first prove the theorem for the case $n \le 426$. 
Consider the following sequence of prime numbers:
:$2, 3, 5, 7, 13, 23, 43, 83, 163, 317, 631$
Each of these prime number is smaller than twice the previous one.
Hence every interval $\set {x: n < x \le 2 n}$, with $n \le 426$, contains one of these prime numbers.
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=== ===
=== ===
From :
:if $p > \sqrt {2 n}$, then $p$ appears at most once in $\dbinom {2 n} n$.
For $n \ge 3$, there is no prime factor $p$ with $\dfrac 2 3 n < p \le n$, because such a prime number divides $n!$ exactly once and $\paren {2 n}!$ exactly twice.
For $n \ge 5$:
:$ \sqrt {2 n} < \dfrac {2 n} 3$
Therefore, by :
:$\ds \frac {2^{2 n} } {2 n} \le \dbinom {2 n} n \le \paren{\prod_{p \mathop \le \sqrt {2 n} } 2 n} \paren{\prod_{\sqrt {2 n} \mathop < p \mathop \le \frac 2 3 n} p} \paren{\prod_{n \mathop < p \mathop \le 2 n} p}$
for $n \ge 3$.
Let $\map \pi n$ denote the prime-counting function.
Then:
:$\forall n \ge 1: \map \pi n \le n - 1$
as $1$ is neither prime nor composite.
 there is no prime number $p$ such that $n < p \le 2 n$.
Then we have:
This is a contradiction for sufficiently large $n$.
Indeed, we have:
:$2^{2 n / 3} \le \paren {2 n}^{\sqrt {2 n} }$
So:
:$2^{2 n} \le \paren {2 n}^{3 \sqrt {2 n} }$
Let $r = \sqrt {2 n}$.
Then:
:$2^{r^2} \le r^{6 r}$
or equivalently:
:$2^r \le r^6$
This fails when $r = 30$.
It fails thereafter, since $2^r$ increases faster than $r^6$.
So our proof works if:
:$n \ge 427$
and the examples show it is true for smaller $n$.
== Also known as ==
== Historical Note ==
Category:Bertrand-Chebyshev Theorem
Bessel's Correction

== Theorem ==
Let $X_1, X_2, \ldots, X_n$ form a random sample from a population with mean $\mu$ and variance $\sigma^2$.
Let: 
:$\ds \bar X = \frac 1 n \sum_{i \mathop = 1}^n X_i$
Then:
:$\ds \hat {\sigma^2} = \frac 1 {n - 1} \sum_{i \mathop = 1}^n \paren {X_i - \bar X}^2$
is an unbiased estimator of $\sigma^2$. 
== Proof ==
We know that $\hat {\sigma^2}$ is an unbiased estimator of $\sigma^2$ : 
:$\expect {\hat {\sigma^2} } = \sigma^2$
In Bias of Sample Variance, it is shown that: 
:$\ds \expect {\frac 1 n \sum_{i \mathop = 1}^n \paren {X_i - \bar X}^2} = \paren {1 - \frac 1 n} \sigma^2$
By Expectation is Linear: 
So: 
So $\hat {\sigma^2}$ is an unbiased estimator of $\sigma^2$. 
Category:Inductive Statistics
Bessel's Equation/x^2 y'' + x y' + (x^2 - (1 over 4)) y = 0

== Theorem ==
The special case of Bessel's equation:
:$(1): \quad x^2 y + x y' + \paren {x^2 - \dfrac 1 4} y = 0$
has the general solution:
:$y = C_1 \dfrac {\sin x} {\sqrt x} + C_2 \dfrac {\cos x} {\sqrt x}$
== Proof ==
=== Particular Solution ===
$(1)$ can be expressed as:
:$(2): \quad y + \dfrac 1 x y' + \paren {1 - \dfrac 1 {4 x^2} } y = 0$
which is in the form:
:$y + \map P x y' + \map Q x y = 0$
where:
:$\map P x = \dfrac 1 x$
:$\map Q x = 1 - \dfrac 1 {4 x^2}$
From Particular Solution to Homogeneous Linear Second Order ODE gives rise to Another:
:$\map {y_2} x = \map v x \, \map {y_1} x$
where:
:$\ds v = \int \dfrac 1 { {y_1}^2} e^{-\int P \rd x} \rd x$
is also a particular solution of $(1)$.
We have that:
Hence:
and so:
From Two Linearly Independent Solutions of Homogeneous Linear Second Order ODE generate General Solution:
:$y = C_1 \dfrac {\sin x} {\sqrt x} + C_2 \dfrac {\cos x} {\sqrt x}$
== Sources ==
* : $\S 3.16$: Problem $6$
Category:Examples of Bessel's Equation