nyuuzyou/wikis huggingface dataset:
Bernoulli's Equation/x y^2 y' + y^3 = x cosine x

== Theorem ==
The first order ODE:
:$(1): \quad x y^2 y' + y^3 = x \cos x$
has the general solution:
:$y^3 = 3 \sin x + \dfrac {9 \cos x} x - \dfrac {18 \sin x} {x^2} - \dfrac {18 \cos x} {x^3} + \dfrac C {x^3}$
== Proof ==
Let $(1)$ be rearranged as:
:$(2): \quad \dfrac {\d y} {\d x} + \dfrac 1 x y = \dfrac {\cos x} {y^2}$
It can be seen that $(2)$ is in the form:
:$\dfrac {\d y} {\d x} + \map P x y = \map Q x y^n$
where:
:$\map P x = \dfrac 1 x$
:$\map Q x = \cos x$
:$n = -2$
and so is an example of Bernoulli's equation.
By Solution to Bernoulli's Equation it has the general solution:
:$\ds (3): \quad \frac {\map \mu x} {y^{n - 1} } = \paren {1 - n} \int \map Q x \, \map \mu x \rd x + C$
where:
:$\map \mu x = e^{\paren {1 - n} \int \map P x \rd x}$
Thus $\map \mu x$ is evaluated:
and so substituting into $(3)$:
Hence the general solution to $(1)$ is:
:$y^3 = 3 \sin x + \dfrac {9 \cos x} x - \dfrac {18 \sin x} {x^2} - \dfrac {18 \cos x} {x^3} + \dfrac C {x^3}$
== Sources ==
* : $\S 2.10$: Problem $3 \ \text{(b)}$
Category:Examples of Bernoulli's Equation
Bernoulli's Equation/y' - (1 over x + 2 x^4) y = x^3 y^2

== Theorem ==
The first order ODE:
:$(1): \quad y' - \paren {\dfrac 1 x + 2 x^4} y = x^3 y^2$
has the general solution:
:$y = \dfrac {2 x} {C \map \exp {-\dfrac {2 x^5} 5} - 1}$
== Proof ==
It can be seen that $(1)$ is in the form:
:$\dfrac {\d y} {\d x} + \map P x y = \map Q x y^n$
where:
:$\map P x = -\paren {\dfrac 1 x + 2 x^4}$
:$\map Q x = x^3$
:$n = 2$
and so is an example of Bernoulli's equation.
By Solution to Bernoulli's Equation it has the general solution:
:$(3): \quad \ds \frac {\map \mu x} {y^{n - 1} } = \paren {1 - n} \int \map Q x \, \map \mu x \rd x + C$
where:
:$\map \mu x = e^{\paren {1 - n} \int \map P x \rd x}$
Thus $\map \mu x$ is evaluated:
and so substituting into $(3)$:
Hence the general solution to $(1)$ is:
:$y = \dfrac {2 x} {C \map \exp {-\dfrac {2 x^5} 5} - 1}$
Category:Examples of Bernoulli's Equation
Bernoulli's Equation/y^2 dx = (x^3 - x y) dy

== Theorem ==
The first order ODE:
:$(1): \quad y^2 \rd x = \paren {x^3 - x y} \rd y$
has the general solution:
:$3 y = 2 x^2 + C x^2 y^2$
== Proof ==
Dividing $(1)$ by $y^2$ and rearranging:
:$(2): \quad \dfrac {\d x} {\d y} + \dfrac x y = \dfrac {x^3} {y^2}$
It can be seen that $(2)$ is in the form:
:$\dfrac {\d x} {\d y} + \map P y x = \map Q y x^n$
where:
:$\map P y = \dfrac 1 y$
:$\map Q y = \dfrac 1 {y^2}$
:$n = 3$
and so is an example of Bernoulli's equation.
By Solution to Bernoulli's Equation it has the general solution:
:$(3): \quad \ds \frac {\map \mu y} {x^{n - 1} } = \paren {1 - n} \int \map Q y \, \map \mu y \rd y + C$
where:
:$\map \mu y = e^{\paren {1 - n} \int \map P y \rd y}$
Thus $\map \mu y$ is evaluated:
and so substituting into $(3)$:
== Sources ==
* : Miscellaneous Problems for Chapter $2$: Problem $5$
Category:Examples of Bernoulli's Equation
Bernoulli's Hanging Chain Problem

== Theorem ==
Consider a uniform chain $C$ whose physical properties are as follows:
:$C$ is of length $l$
:The mass per unit length of $C$ is $m$
:$C$ is of zero stiffness.
Let $C$ be suspended in a vertical line from a fixed point and otherwise free to move.
Let $C$ be slightly disturbed in a vertical plane from its position of stable equilibrium.
Let $\map y t$ be the horizontal displacement at time $t$ from its position of stable equilibrium of a particle of $C$ which is a vertical distance $x$ from its point of attachment.
The $2$nd order ODE describing the motion of $y$ is:
:$\dfrac {\d^2 y} {\d t^2} = g \paren {l - x} \dfrac {\d^2 y} {\d x^2} - g \dfrac {\d y} {\d x}$
== Proof ==
== Historical Note ==
== Sources ==
* : Chapter $\text{I}$: Introductory: $\S 1$. Bernoulli's Problem
Category:Second Order ODEs
Bernoulli's Hanging Chain Problem/Historical Note

== Historical Note on Bernoulli's Hanging Chain Problem ==
This problem was first discussed by around the year $1732$.
In $1781$ took up the problem, and found a solution for the special case: $u \dfrac {\d^2 y} {\d u^2} + \dfrac {\d y} {\d u} + y = 0$
== Sources ==
* 
* 
* : Chapter $\text{I}$: Introductory: $\S 1$. Bernoulli's Problem
Category:Historical Notes
Bernoulli's Inequality

== Theorem ==
Let $x \in \R$ be a real number such that $x > -1$.
Let $n \in \Z_{\ge 0}$ be a positive integer.
Then:
:$\paren {1 + x}^n \ge 1 + n x$
=== Corollary ===
== Proof 1 ==
== Proof 2 ==
Category:Real Analysis
Category:Inequalities
Category:Bernoulli's Inequality
Bernoulli's Inequality/Corollary

== Theorem ==
Let $x \in \R$ be a real number such that $0 < x < 1$.
Let $n \in \Z_{\ge 0}$ be a positive integer.
Then:
:$\paren {1 - x}^n \ge 1 - n x$
=== General Result ===
== Proof 1 ==
== Proof 2 ==
Category:Real Analysis
Category:Inequalities
Category:Bernoulli's Inequality
Bernoulli's Inequality/Corollary/General Result

== Theorem ==
For all $n \in \Z_{\ge 0}$:
:$\ds \prod_{j \mathop = 1}^n \paren {1 - a_j} \ge 1 - \sum_{j \mathop = 1}^n a^j$
where $0 < a_j < 1$ for all $j$.
== Proof ==
The proof proceeds by induction.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$\ds \prod_{j \mathop = 1}^n \paren {1 - a_j} \ge 1 - \sum_{j \mathop = 1}^n a^j$
$\map P 0$ is the case:
Thus $\map P 0$ is seen to hold.
=== Basis for the Induction ===
$\map P 1$ is the case:
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:
:$\ds \prod_{j \mathop = 1}^k \paren {1 - a_j} \ge 1 - \sum_{j \mathop = 1}^k a^j$
from which it is to be shown that:
:$\ds \prod_{j \mathop = 1}^{k + 1} \paren {1 - a_j} \ge 1 - \sum_{j \mathop = 1}^{k + 1} a^j$
=== Induction Step ===
This is the induction step:
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\ds \forall n \in \Z_{\ge 0}: \prod_{j \mathop = 1}^n \paren {1 - a_j} \ge 1 - \sum_{j \mathop = 1}^n a^j$
== Sources ==
* : $\S 1.2.3$: Sums and Products: Exercise $27$
Category:Bernoulli's Inequality
Category:Summations
Category:Continued Products
Bernoulli's Inequality/Corollary/Proof 1

== Theorem ==
== Proof ==
Let $0 < x < 1$.
Let $y = -x$.
Then $y > -1$ and by Bernoulli's Inequality:
:$\paren {1 + y}^n \ge 1 + n y$
Thus:
:$\paren {1 + \paren {-x} }^n \ge 1 + n \paren {-x}$
Hence the result.
Category:Bernoulli's Inequality
Bernoulli's Inequality/Corollary/Proof 2

== Theorem ==
== Proof ==
Proof by induction:
Let $0 < x < 1$.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$\paren {1 - x}^n \ge 1 - n x$
=== Basis for the Induction ===
$\map P 0$ is the case:
:$\paren {1 - x}^0 = 1 \ge 1 - 0 x = 1$
so $\map P 0$ holds.
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $\map P k$ is true, where $k \ge 0$, then it logically follows that $\map P {k + 1}$ is true.
So this is our induction hypothesis:
:$\paren {1 - x}^k \ge 1 - k x$
We need to show that:
:$\paren {1 - x}^{k + 1} \ge 1 - \paren {k + 1} x$
=== Induction Step ===
This is our induction step:
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
== Sources ==
* : $\S 1.2.1$: Mathematical Induction: Exercise $9$
Category:Bernoulli's Inequality
Bernoulli's Inequality/Proof 1

== Theorem ==
== Proof ==
Proof by induction:
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$\paren {1 + x}^n \ge 1 + nx$
=== Basis for the Induction ===
$\map P 0$ is the case:
:$\paren {1 + x}^0 \ge 1$
so $\map P 0$ holds.
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $\map P k$ is true, where $k \ge 0$, then it logically follows that $\map P {k + 1}$ is true.
So this is our induction hypothesis:
:$\left({1 + x}\right)^k \ge 1 + kx$
We need to show that:
:$\paren {1 + x}^{k + 1} \ge 1 + \paren {k + 1} x$
=== Induction Step ===
This is our induction step:
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Category:Bernoulli's Inequality
Bernoulli's Inequality/Proof 2

== Theorem ==
== Proof ==
Let $y = 1 + x$.
Then $y \ge 0$, and:
:$\paren {1 + x}^n = 1 + \paren {y^n - 1}$
If $y \ge 1$, then by Sum of Geometric Sequence:
:$\ds y^n - 1 = \paren {y - 1} \sum_{k \mathop = 0}^{n - 1} y^k \ge n \paren {y - 1} = n x$
If $y < 1$, then by Sum of Geometric Sequence:
:$\ds y^n - 1 = -\paren {1 - y} \sum_{k \mathop = 0}^{n - 1} y^k \ge -n \paren {1 - y} = n x$
Hence the result.
Category:Bernoulli's Inequality
Bernoulli's Theorem

== Theorem ==
Let the probability of the occurrence of an event be $p$.
Let $n$ independent trials be made, with $k$ successes.
Then:
:$\ds \lim_{n \mathop \to \infty} \frac k n = p$
== Proof ==
Let the random variable $k$ have the binomial distribution with parameters $n$ and $p$, that is:
:$k \sim \Binomial n p$
where $k$ denotes the number of successes of the $n$ independent trials of the event with probability $p$.
From Expectation of Binomial Distribution:
:$\expect k = n p \leadsto \dfrac 1 n \expect k = p$
Expectation is Linear gives:
:$ \expect {\dfrac k n} = p =: \mu$
Similarly, from Variance of Binomial Distribution:
:$\var k = n p \paren {1 - p} \leadsto \dfrac 1 {n^2} \var k = \dfrac {p \paren {1 - p} } n$
From Variance of Linear Combination of Random Variables:
:$\var {\dfrac k n} = \dfrac {p \paren {1 - p} } n =: \sigma^2$
By applying the Bienaymé-Chebyshev Inequality to $\dfrac k n$, we have for any $l > 0$:
:$\map \Pr {\size {\dfrac k m - \mu} \ge l \sigma} \le \dfrac 1 {l^2}$
Now, let $\epsilon > 0$ and choose $l = \dfrac \epsilon \sigma$, to get:
:$\map \Pr {\size {\dfrac k m - \mu} \ge \dfrac \epsilon \sigma \cdot \sigma} \le \dfrac {\sigma^2} {\epsilon^2}$
Simplifying and plugging in the values of $\mu$ and $\sigma^2$ defined above yields:
:$\map \Pr {\size {\dfrac k n - p} \ge \epsilon} \le \dfrac {p \paren {1 - p} } {n \epsilon^2}$
Scaling both sides by $-1$ and adding $1$ to both sides yields:
:$1 - \map \Pr {\size {\dfrac k n - p} \ge \epsilon} \ge 1 - \dfrac {p \paren {1 - p} } {n \epsilon^2}$
Applying Union of Event with Complement is Certainty to the :
:$\map \Pr {\size {\dfrac k n - p} \le \epsilon} \ge 1 - \dfrac {p \paren {1 - p} } {n\epsilon^2}$
Taking the limit as $n$ approaches infinity on both sides, we have:
:$\ds \lim_{n \mathop \to \infty} \map \Pr {\size {\frac k n - p} < \epsilon} = 1$
== Also presented as ==
== Also known as ==
== Also see ==
* Khinchin's Law, otherwise known as the Weak Law of Large Numbers
* Kolmogorov's Law, otherwise known as the Strong Law of Large Numbers
== Historical Note ==
== Sources ==
* : $1$: The Nature of Differential Equations: $\S 6$: The Brachistochrone. Fermat and the Bernoullis
* : Chapter $\text {A}.20$: The Bernoulli Brothers
* 
* 
* 
Category:Bernoulli's Theorem
Category:Laws of Large Numbers
Category:Probability Theory
Bernoulli's Theorem/Also known as

== Bernoulli's Theorem: Also known as ==
Bernoulli's Theorem is also popularly known as the Law of Large Numbers.
It also needs to be noted that it is a special case of the Weak Law of Large Numbers, also known as Khinchin's Law, and so also goes under that name.
== Sources ==
* 
* : $1$: The Nature of Differential Equations: $\S 6$: The Brachistochrone. Fermat and the Bernoullis
* : Chapter $\text {A}.20$: The Bernoulli Brothers
Category:Bernoulli's Theorem
Bernoulli's Theorem/Also presented as

== Bernoulli's Theorem: Also presented as ==
Bernoulli's Theorem can also be presented in the form:
:$\forall \epsilon \in \R_{>0}: \ds \lim_{n \mathop \to \infty} \map \Pr {\size {\frac k n - p} < \epsilon} = 1$
== Sources ==
* 
* 
* 
Category:Bernoulli's Theorem
Bernoulli Number in terms of Euler Numbers

== Theorem ==
Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Then:
where:
:$B_n$ denotes the $n$th Bernoulli number
:$E_n$ denotes the $n$th Euler number.
== Proof ==
The coefficient $a_n$ of $x^{2 n - 2}$ is:
:$\ds a_n = \frac {\paren {-1}^{n - 1} 2^{2 n} \paren {2^{2 n} - 1} B_{2 n} } {2 n \paren {2 n - 2}!} $
From Derivative of Tangent Function, we have:
:$\map {\dfrac \d {\d x} } {\tan x} = \sec^2 x$
The coefficient $a_n$ of $x^{2 n - 2}$ above is:
:$\ds a_n = \sum_{k \mathop = 0}^{n - 1} \frac 1 {\paren {2 k}! \paren {2 n - 2 k - 2}!} E_{2 k} E_{2 n - 2 k - 2}$
Equating coefficients gives us:
== Also presented as ==
== Sources ==
* : $\S 21$: Relationships of Bernoulli and Euler Numbers: $21.7$
* : Chapter $5$. Bernoulli Numbers and the Riemann Zeta-Function
* : $\S 23$: Bernoulli and Euler Numbers: Relationships of Bernoulli and Euler Numbers: $23.7.$
Category:Bernoulli Number in terms of Euler Numbers
Category:Euler Numbers
Category:Bernoulli Numbers
Category:Examples of Equating Coefficients
Bernoulli Number in terms of Euler Numbers/Also presented as

== Bernoulli Number in terms of Euler Numbers: Also presented as == 
Bernoulli Number in terms of Euler Numbers can also be presented in the form:
where:
:${B_n}^*$ denotes the archaic form of the Bernoulli numbers
:${E_n}^*$ denotes the alternative form of the Euler numbers.
== Sources ==
* : $\S 21$: Relationships of Bernoulli and Euler Numbers: $21.7$
* : $\S 23$: Bernoulli and Euler Numbers: Relationships of Bernoulli and Euler Numbers: $23.7.$
Category:Bernoulli Number in terms of Euler Numbers
Bernoulli Numbers are Rational

== Theorem ==
The Bernoulli numbers are rational.
== Proof ==
By the recurrence relation for the Bernoulli numbers:
:$B_n = \begin{cases} 1 & : n = 0 \\
\ds - \sum_{k \mathop = 0}^{n-1} \binom n k \frac {B_k} {n - k + 1} & : n > 0
\end{cases}$
:$B_0$ is rational.
:$B_n$ is a finite sum of the products of a binomial coefficient with a Bernoulli number earlier in the sequence divided by an integer.
The proof proceeds by induction.
For all $n \in \N$, let $\map P n$ be the proposition:
:$B_n$ is rational.
=== Basis for the Induction ===
$\map P 0$ is the case:
:$B_0$ is rational.
This is true, as $B_0 = 1$ which is rational.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that, if $\map P j$ is true, where $0 \le j \le k$, then it logically follows that $\map P {k + 1}$ is true.
So this is the induction hypothesis:
:$\forall j \in \N: 0 \le j \le k: B_j$ is rational
from which it is to be shown that:
:$B_{k + 1}$ is rational.
=== Induction Step ===
This is the induction step:
By the recurrence relation for the Bernoulli numbers:
:$\ds B_{k + 1} = - \sum_{j \mathop = 0}^k \binom {k + 1} j \frac {B_j} {k - j + 2}$
We have that:
:From Binomial Coefficient is Integer, $\dbinom {k + 1} j$ is an integer throughout.
:From the Induction Hypothesis, $\forall j \in \N: 0 \le j \le k: B_j$ is rational.
:$k - j + 2$ is an integer.
Thus as $B_{k + 1}$ is the sum of a finite number of rational numbers, it follows that $B_{k + 1}$ it itself rational.
The result follows by the Second Principle of Mathematical Induction.
Therefore:
:$\forall n \in \N: B_n$ is rational.
== Sources ==
* : Chapter $\text {B}.20$: The Bernoulli Numbers and some Wonderful Discoveries of Euler
Category:Bernoulli Numbers
Bernoulli Process as Binomial Distribution

== Theorem ==
Let $\sequence {X_i}$ be a finite Bernoulli process of length $n$ such that each of the $X_i$ in the sequence is a Bernoulli trial with parameter $p$.
Then the number of successes in $\sequence {X_i}$ is modelled by a binomial distribution with parameters $n$ and $p$.
Hence it can be seen that:
:$X \sim \Binomial 1 p$ is the same thing as $X \sim \Bernoulli p$
== Proof ==
Consider the sample space $\Omega$ of all sequences $\sequence {X_i}$ of length $n$.
The $i$th entry of any such sequence is the result of the $i$th trial.
We have that $\Omega$ is finite.
Let us take the event space $\Sigma$ to be the power set of $\Omega$.
As the elements of $\Omega$ are independent, by definition of the Bernoulli process, we have that:
:$\forall \omega \in \Omega: \map \Pr \omega = p^{\map s \omega} \paren {1 - p}^{n - \map s \omega}$
where $\map s \omega$ is the number of successes in $\omega$.
In the same way:
:$\ds \forall A \in \Sigma: \map \Pr A = \sum_{\omega \mathop \in A} \map \Pr \omega$
Now, let us define the discrete random variable $Y_i$ as follows:
:$\map {Y_i} \omega = \begin{cases}
1 & : \text {$\omega_i$ is a success} \\
0 & : \text {$\omega_i$ is a failure}
\end{cases}$
where $\omega_i$ is the $i$th element of $\omega$.
Thus, each $Y_i$ has image $\set {0, 1}$ and a probability mass function:
:$\map \Pr {Y_i = 0} = \map \Pr {\set {\omega \in \Omega: \text {$\omega_i$ is a success} } }$
Thus we have:
Then:
:$\map \Pr {Y_i = 0} = 1 - \map \Pr {Y_i = 1} = 1 - p$
So (by a roundabout route) we have confirmed that $Y_i$ has the Bernoulli distribution with parameter $p$.
Now, let us define the random variable:
:$\ds \map {S_n} \omega = \sum_{i \mathop = 1}^n \map {Y_i} \omega$
By definition:
:$\map {S_n} \omega$ is the number of successes in $\omega$
:$S_n$ takes values in $\set {0, 1, 2, \ldots, n}$ (as each $Y_i$ can be $0$ or $1$).
Also, we have that:
Hence the result.
== Sources ==
* : $\S 2.2$: Examples: Example $11$
Category:Bernoulli Distribution
Category:Binomial Distribution
Bernoulli Process as Geometric Distribution

== Theorem ==
Let $\sequence {X_i}$ be a Bernoulli process with parameter $p$.
Let $\EE$ be the experiment which consists of performing the Bernoulli trial $X_i$ until a failure occurs, and then stop.
Let $k$ be the number of successes before a failure is encountered.
Then $k$ is modelled by a geometric distribution with parameter $p$.
=== Shifted Geometric Distribution ===
== Proof ==
Follows directly from the definition of geometric distribution.
Let $X$ be the discrete random variable defined as the number of successes before a failure is encountered.
Thus the last trial (and the last trial only) will be a failure, and the others will be successes.
The probability that $k$ successes are followed by a failure is:
:$\map \Pr {X = k} = p^k \paren {1 - p}$
Hence the result.
== Also presented as ==
This proof uses as its model the first formulation of the geometric distribution.
However, if the second formulation is used, $k$ is then the number of failures before a success is encountered.
Hence this bears similarity to the shifted geometric distribution.
== Sources ==
* 
* 
Category:Bernoulli Distribution
Category:Geometric Distribution
Bernoulli Process as Geometric Distribution/Shifted

== Theorem ==
Let $\sequence {Y_i}$ be a Bernoulli process with parameter $p$.
Let $\EE$ be the experiment which consists of performing the Bernoulli trial $Y_i$ as many times as it takes to achieve a success, and then stop.
Let $k$ be the number of Bernoulli trials to achieve a success.
Then $k$ is modelled by a shifted geometric distribution with parameter $p$.
== Proof ==
Follows directly from the definition of shifted geometric distribution.
Let $Y$ be the discrete random variable defined as the number of trials for the first success to be achieved.
Thus the last trial (and the last trial only) will be a success, and the others will be failures.
The probability that $k-1$ failures are followed by a success is:
:$\map \Pr {Y = k} = \paren {1 - p}^{k - 1} p$
Hence the result.
== Sources ==
* : $\S 2.2$: Examples: Example $14$
* 
* 
Category:Bernoulli Distribution
Category:Geometric Distribution
Bernstein's Theorem on Unique Global Solution to y''=F(x,y,y')

== Theorem ==
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:
:$\size {\map F {x, y, y'} } \le \alpha y'^2 + \beta$
Then one and only one integral curve of the equation $y = \map F {x, y, y'}$ passes through any two points $\tuple {a, A}$ and $\tuple {b, B}$ such that $a \ne b$.
== Proof ==
=== Lemma 1 (Uniqueness) ===
=== Lemma 2 ===
=== Lemma 3 ===
Consider a plane with axes denoted by $x$ and $y$:
:
Put the point $A \tuple {a, a_1}$.
Through this point draw an arc of the integral curve such that $\map {y'} a = 0$.
On this arc put another point $D \tuple {d, d_1}$.
For $x \ge d$ draw the straight line $y = d_1$.
Put the point $B \tuple {b, b_1}$.
For $y \ge d_1$ draw the straight line $x = b_1$.
Denote the intersection of these two straight lines by $Q$.
Then the broken curve $DQB$ connects points $D$ and $B$.
Choose any point of $DQB$ and denote it by $P \tuple {\xi, \xi_1}$.
Consider a family of integral curves $y = \map \phi {x, \alpha}$, passing through the point $A$, where $\alpha = \map {y'} a$.
For $\alpha = 0$ the integral curve concides with $AD$.
Suppose point $P$ is sufficiently close to the point $D$.
By , there exists a unique curve $AP$.
Then, $\alpha$ can be found uniquely from:
:$d_1 = \map \phi {\xi, \alpha}$.
Due to uniqueness and continuity, it follows that $\xi$ is a monotonic function of $\alpha$.
Hence, $\alpha$ is a monotonic function of $\xi$.
Put the point $R$ in between of $D$ and $Q$.
Suppose, that, except for $R$, any point of $DR$ can be reached by the aforementioned procedure.
When $\xi$ approaches the abscissa $r$ of $R$, $\alpha$ monotonically approaches a limit.
If it is different from $\pm \dfrac \pi 2$, point $R$ is attained.
By assumption, $R$ is not attained.
Thus: 
:$\ds \lim_{\xi \mathop \to r} \alpha = \pm \dfrac \pi 2$
In other words, as $P$ approaches $R$, the derivative of $\map y x$ joining $A$ to $P$ will not be bounded at $x = a$.
This contradicts the bounds from and , and the fact that the difference of abscissas of $A$ and $P$ does not approach $0$.
Therefore, $R$ can be reached.
Similar argument can be repeated for the line segment $QB$.
== 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')
Category:Calculus of Variations