nyuuzyou/wikis huggingface dataset:
Floor of Negative equals Negative of Ceiling

== Theorem ==
Let $x \in \R$ be a real number.
Let $\floor x$ be the floor of $x$, and $\ceiling x$ be the ceiling of $x$.
Then:
:$\floor {-x} = -\ceiling x$
== Proof ==
From Integer equals Ceiling iff between Number and One More we have:
:$x \le \ceiling x < x + 1$
and so, by multiplying by -1:
:$-x \ge -\ceiling x > -x - 1$
From Integer equals Floor iff between Number and One Less we have:
:$\floor x = n \iff x - 1 < n \le x$
Hence:
:$-x - 1 < -\ceiling x \le -x \implies \floor {-x} = -\ceiling x$
== Also see ==
* Ceiling of Negative equals Negative of Floor
== Sources ==
* : $\S 1.2.4$: Integer Functions and Elementary Number Theory
* : $\S 1.2.4$: Integer Functions and Elementary Number Theory: Exercise $4$
Category:Floor Function
Category:Ceiling Function
Floor of Non-Integer

== Theorem ==
Let $x \in \R$ be a real number.
Let $x \notin \Z$.
Then:
:$\left\lfloor{x}\right\rfloor < x$
where $\left\lfloor{x}\right\rfloor$ denotes the floor of $x$.
== Proof ==
From Floor is between Number and One Less:
:$\left\lfloor{x}\right\rfloor \le x$
From Real Number is Integer iff equals Floor:
:$x = \left \lfloor {x} \right \rfloor \iff x \in \Z$
But we have $x \notin \Z$.
So:
:$\left \lfloor {x} \right \rfloor \ne x$
and so:
:$\left \lfloor {x} \right \rfloor < x$
== Also see ==
* Ceiling of Non-Integer
Category:Floor Function
Floor of Root of Floor equals Floor of Root

== Theorem ==
Let $x \in \R_{\ge 0}$ be a positive real number.
Let $\floor x$ denote the floor of $x$.
Then:
:$\ds \floor {\sqrt {\floor x} } = \floor {\sqrt x}$
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Ceiling of Root of Ceiling equals Ceiling of Root
== Sources ==
* : $\S 1.2.4$: Integer Functions and Elementary Number Theory: Exercise $6 \ \text{(a)}$
Category:Floor Function
Category:Floor of Root of Floor equals Floor of Root
Floor of Root of Floor equals Floor of Root/Proof 2

== Theorem ==
== Proof ==
The square root is defined on the interval $\hointr 0 \to$.
We have that Square Root is Strictly Increasing.
From Continuity of Root Function, the square root is continuous.
Hence the conditions are fulfilled for for McEliece's Theorem (Integer Functions) to be applied:
:$\forall x \in A: \paren {\map f x \in \Z \implies x \in \Z} \iff \floor {\map f x} = \floor {\map f {\floor x} }$
It remains to be proved that $\sqrt x \in \Z \implies x \in \Z$.
Let $y = \sqrt x$ such that $y \in \Z$.
By definition of the square root:
:$x = y^2$.
From Integer Multiplication is Closed:
:$x \in \Z$
as required.
Category:Floor of Root of Floor equals Floor of Root
Floor of Simple Finite Continued Fraction

== Theorem ==
Let $\sequence {a_k}_{k \mathop \ge 0}$ be a simple finite continued fraction of length $n \ge 0$.
Let $x = [a_0, \ldots, a_n]$ be its value.
Then the floor of $x$ is the partial denominator $a_0$:
:$\floor x = a_0$
unless $n = 1$ and $a_1 = 1$, in which case $x = \floor x = a_0 + 1$.
== Proof ==
==== Length 0 ====
If $n = 0$, we have $x = \sqbrk {a_0, \ldots, a_n} = a_0$ by definition of value.
By Real Number is Integer iff equals Floor:
:$\floor x = \floor {a_0} = a_0$
==== Length 1 ====
If $n = 1$, then:
:$x = \sqbrk {a_0, a_1} = a_0 + \dfrac 1 {a_1}$
If $a_1 > 1$, then:
:$a_0 < a_0 + \dfrac 1 {a_1} < a_0 + 1$
By definition of floor function:
:$\floor x = a_0$
If $a_1 = 1$, then $x = a_0 + 1$.
By Real Number is Integer iff equals Floor:
:$\floor x = a_0 + 1$
==== Length at least 2 ====
By Value of Finite Continued Fraction of Real Numbers is at Least First Term:
:$\sqbrk {a_0, \ldots, a_n} > a_0$
:$\sqbrk {a_1, \ldots, a_n} > a_1 \ge 1$
Thus:
:$a_0 < \sqbrk {a_0, \ldots, a_n} = a_0 + \dfrac 1 {\sqbrk {a_1, \ldots, a_n} } < a_0 + 1$
By definition of floor function:
:$\floor x = a_0$
== Also see ==
* Properties of Value of Finite Continued Fraction
Category:Simple Continued Fractions
Floor of m+n-1 over n

== Theorem ==
Let $m, n \in \Z$ such that $n > 0$.
Then:
:$\floor {\dfrac {m + n - 1} n} = \ceiling {\dfrac m n}$
The identity does not necessarily apply for $n < 0$.
=== Example ===
== Proof ==
First let $n > 0$ as stated.
Suppose $n \divides m$. 
Then $m = k n$ for some $k \in \Z$.
It follows that:
:$\floor {\dfrac {m + n - 1} n} = \floor {k + 1 - \dfrac 1 n} = k$
and:
:$\ceiling {\dfrac m n} = k$
Now suppose $n \nmid m$.
Since $n > 0$, we have $m = k n + r$ for some $k \in\Z$ and $r \in \N$, $0 < r < n$.
Therefore:
:$\floor {\dfrac {m + n - 1} n} = \floor {k + 1 + \dfrac {r - 1} n} = k + 1$
and:
:$\ceiling {\dfrac m n} = k + 1$
Setting $m = 1, n = -2$ we have:
Thus, as stated, it is confirmed that the identity does not hold for $n < 0$.
It is noted that when $n = 0$ the expressions on either side are not defined.
== Sources ==
* : $\S 1.2.4$: Integer Functions and Elementary Number Theory: Exercise $48 \ \text{(a)}$
Category:Floor Function
Floor of m+n-1 over n/Example 1

== Example of use of Floor of $\frac {m + n - 1} n$ ==
Let $n \in \Z$.
Then:
:$\floor {\dfrac {n + 2 - \floor {n / 25} } 3} = \floor {\dfrac {8 n + 24} {25} }$
== Proof ==
== Sources ==
* : $\S 1.2.4$: Integer Functions and Elementary Number Theory: Exercise $48 \ \text{(b)}$
Category:Floor Function
Floor of x+m over n

== Theorem ==
Let $m, n \in \Z$ such that $n > 0$.
Let $x \in \R$.
Then:
:$\floor {\dfrac {x + m} n} = \floor {\dfrac {\floor x + m} n}$
where $\floor x$ denotes the floor of $x$.
=== Corollary ===
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $\S 1.2.4$: Integer Functions and Elementary Number Theory: Exercise $35$
Category:Floor Function
Category:Floor of x+m over n
Floor of x+m over n/Corollary

== Corollary to Floor of $\frac {x + m} n$ ==
Let $n \in \Z$ such that $n > 0$.
Let $x \in \R$.
Then:
:$\floor {\dfrac x n} = \floor {\dfrac {\floor x} n}$
where $\floor x$ denotes the floor of $x$.
== Proof ==
This is a special case of Floor of $\dfrac {x + m} n$:
:$\floor {\dfrac {x + m} n} = \floor {\dfrac {\floor x + m} n}$
where $m = 0$.
== Sources ==
* : $\S 1.2.4$: Integer Functions and Elementary Number Theory: Exercise $35$
Category:Floor Function
Category:Floor of x+m over n
Floor of x+m over n/Proof 1

== Theorem ==
== Proof ==
Let:
:$y = x - \floor x$
:$M = \floor x + m$
We now have:
:$(1): \quad 0 \le y < 1$
and thus:
:$\floor y = 0$
By Division Theorem, we can write:
:$(2): \quad M = k n + r$
with $k \in \Z$ and $0 \le r \le n - 1$.
By $(1)$ and $(2)$:
:$(3): \quad 0 \le y + r < 1 + n - 1 = n$
We have:
Substituting $y$ and $M$, we obtain:
and the result follows.
== Sources ==
* : $\S 1.2.4$: Integer Functions and Elementary Number Theory: Exercise $35$ (Answers to Exercises)
Category:Floor of x+m over n
Floor of x+m over n/Proof 2

== Theorem ==
== Proof ==
Let $f: \R \to \R$ be the real function defined as:
:$\forall x \in \R: \map f x = \dfrac {x + m} n$
It is clear that $f$ is both strictly increasing and continuous on the whole of $\R$.
Let $\dfrac {x + m} n \in \Z$.
Then:
Thus:
:$\forall x \in \R: \map f x \in \Z \implies x \in \Z$
So the conditions are fulfilled for McEliece's Theorem (Integer Functions) to be applied:
:$\floor {\map f x} = \floor {\map f {\floor x} } \iff \paren {\map f x \in \Z \implies x \in \Z}$
Hence the result.
== Sources ==
* : $\S 1.2.4$: Integer Functions and Elementary Number Theory: Exercise $35$ (Answers to Exercises)
Category:Floor of x+m over n
Floquet's Theorem

== Theorem ==
Let $\mathbf A \left({t}\right)$ be a continuous matrix function with period $T$.
Let $\Phi \left({t}\right)$ be a fundamental matrix of the Floquet system $\mathbf x' = \mathbf A \left({t}\right) \mathbf x$.
Then $\Phi \left({t + T}\right)$ is also a fundamental matrix.
Moreover, there exists:
: A nonsingular, continuously differentiable matrix function $\mathbf P \left({t}\right)$ with period $T$
: A constant (possibly complex) matrix $\mathbf B$ such that:
::$\Phi \left({t}\right) = \mathbf P \left({t}\right) e^{\mathbf Bt}$
== Proof 1 ==
== Proof 2 ==
== Also see ==
	
* General Fundamental Matrix
Category:Linear Algebra
Category:Differential Equations
Category:Floquet's Theorem
Floquet's Theorem/Proof 1

== Theorem ==
== Proof ==
We assume the two hypotheses of the theorem.
We have that:
So the first implication of the theorem holds, that is: that $\map \Phi {t + T}$ is a fundamental matrix.
Because $\map \Phi t$ and $\map \Phi {t + T}$ are both fundamental matrices, there must exist some matrix $\mathbf C$ such that:
:$\map \Phi {t + T} = \map \Phi t \mathbf C$
Hence by the existence of the matrix logarithm, there exists a matrix $\mathbf B$ such that:
:$\mathbf C = e^{\mathbf BT}$
Defining $\map {\mathbf P} t = \map \Phi t e^{-\mathbf B t}$, it follows that:
and hence $\map {\mathbf P} t$ is periodic with period $T$.
As $\map \Phi t = \map {\mathbf P} t e^{\mathbf B t}$, the second implication also holds.
Category:Floquet's Theorem
Floquet's Theorem/Proof 2

== Theorem ==
== Proof ==
Let $\map S t = \map \Phi {t + T} {\map \Phi T}^{-1}$.
Then:
So $\map S t$ is a fundamental matrix and:
:$\map S 0 = Id$
Then:
:$\map S t = \map \Phi t$
which means that:
:$\map \Phi {t + T} = \map \Phi t \map \Phi T$
Hence by the existence of the matrix logarithm, there exists a matrix $\mathbf B$ such that:
:$\map \Phi T = e^{\mathbf B T}$
Defining $\map {\mathbf P} t = \map \Phi t e^{-\mathbf B t}$, it follows that:
and hence $\map {\mathbf P} t$ is a periodic real function with period $T$.
As $\map \Phi t = \map {\mathbf P} t e^{\mathbf B t}$, the second implication also holds.
Category:Floquet's Theorem
Flos/Problems/Problem 1

== Problem from by ==
:Find a rational number such that $5$ added to, or subtracted from, its square, is also a square.
== Solution ==
:$\dfrac {41} {12}$
== Proof ==
We need to find $3$ squares in arithmetic sequence whose common difference is of the form $5 p^2$.
That is:
:$\tuple {x^2 - 5 p^2, x^2, x^2 + 5 p^2}$
We can find:
and:
Hence:
== Sources ==
* 
* : Liber Abaci: $86$
Category:Flos
Flos/Problems/Problem 2

== Problem from by ==
:Solve the cubic:
::$x^3 + 2 x^2 + 10 x = 20$
== Solution ==
:$1 \cdotp 3688081075 \dots$
== Proof ==
Express the equation as:
:$x^3 + 2 x^2 + 10 x - 20 = 0$
We apply Cardano's Formula, with $a = 1$, $b = 2$, $c = 10$ and $d = -20$.
Thus:
Then:
The other two roots are complex.
== Sources ==
* 
* : Liber Abaci
Category:Flos
Flos/Problems/Problem 3

== Problem from by ==
:$3$ men possess a pile of money, their shares being $\dfrac 1 2$, $\dfrac 1 3$ and $\dfrac 1 6$.
:Each man takes some money from the pile until nothing is left.
:The first man returns $\dfrac 1 2$ of what he took,
:the second $\dfrac 1 3$
:and the third $\dfrac 1 6$.
:When the total so returned is divided equally among the men it is found that each then possess what he is entitled to.
:How much money was in the original pile, and how much did each man take from the pile?
== Solution ==
The total amount $T$ cannot be determined by the information given.
All that can be done is determine the proportions of $T$ that each man took.
Hence we can offer an arbitrary number for $T$.
In order for all numbers in question to be integers, we select this number to be $282$.
Then the first man took $198$, the second man took $78$ and the third man took $6$.
== Proof ==
Let $A$, $B$ and $C$ denote the man whose share is $\dfrac 1 2$, $\dfrac 1 3$, and $\dfrac 1 6$ respectively.
Let $a$, $b$ and $c$ be the amounts taken by $A$, $B$ and $C$ respectively.
The total in the original pile is then $T = a + b + c$.
Let $3 x$ be the sum returned equally to $A$, $B$ and $C$.
Before they received a third of this, they had $\dfrac T 2 - x$, $\dfrac T 3 - x$ and $\dfrac T 6 - x$ respectively.
Because these were what they had after they had put back $\dfrac 1 2$, $\dfrac 1 3$ and $\dfrac 1 6$ respectively, we have::
Thus we have:
This is indeterminate.
 picked the simplest integer solution: $T = 47$ and $x = 7$.
This leads to:
It is neater to multiply everything by $6$, so as to make it so that the amount given and taken at each stage is an integer.
Checking that the numbers are correct and consistent, we check them against the conditions of the question.
We have that:
So there was $99 + 26 + 1 = 126$ to be shared out.
Thus each got $\dfrac {126} 3 = 42$ back, giving them:
== Sources ==
* 
* : Lecture $15$: The Blockhead
* : Liber Abaci: $87$
Category:Flos
Flow/Examples/Arbitrary Example 1

== Example of Flow ==
Let the position of a point $x$ at time $t$ be denoted $\map x t$.
The equation $\map x t = t^2 + 1$ defines a flow on the real number line which is a solution to the differential equation $\map {x'} t = 2 t$.
== Sources ==
* 
* 
Category:Examples of Dynamical Systems
Flow/Examples/Arbitrary Example 1/Mistake

== Source Work ==
:
:dynamical system
== Mistake ==
:For example, if the position of a point $x$ at time $t$ is denoted by $\map x t$, the equation $\map x t = t^2 + 1$ defines a flow on the real line which is a solution to the equation $\map x t = 2 t$.
== Correction ==
A glitch has crept in since the second edition, which shows the correct sentence:
:For example, if the position of a point $x$ at time $t$ is denoted by $\map x t$, the equation $\map x t = t^2 + 1$ defines a flow on the real line which is a solution to the equation $\map {\dot x} t = 2 t$.
As can be seen, there is a dot missing off the final $x$ of the sentence.
== Sources ==
* 
Category:Mistakes/The Penguin Dictionary of Mathematics
Focal Property of Conic Section

== Theorem ==
Let $\KK$ be a conic section.
Let $F_1$ and $F_2$ be the foci of $\KK$.
Let $P$ be a point on $\KK$.
Let $\TT$ be the tangent to $\KK$ at $P$.
Let $\LL_1$ be the straight line through $F_1$ to $P$.
Let $\LL_2$ be the straight line through $P$ which makes the same angle with $\TT$ as does $\LL_1$.
Then $\LL_2$ passes through $F_2$.
=== Focal Property of Parabola ===
In the case of the parabola, the focus $F_2$ is the point at infinity:
== Also known as ==
== Sources ==
* 
* 
Category:Focal Property of Conic Section
Category:Conic Sections
Focal Property of Conic Section/Also known as

== Focal Property of Conic Section: Also known as ==
The focal property of conic section is also known as:
:the reflection property of conic section, as it explains how light and sound are reflected by a reflector the shape of a conic section.
Hence also:
:the optical property of conic section, from the way light behaves
:the acoustical property of conic section, from the exploitation of the phenomenon in the science of acoustics.
== Sources ==
* 
* 
* 
* 
* 
* 
* 
* 
Category:Focal Property of Conic Section
Focal Property of Ellipse

== Theorem ==
Let $\EE$ be an ellipse.
Let $P$ be an arbitrary point on $\EE$.
Let $APB$ be a tangent to $\EE$ at $P$.
Then:
:$\angle APF_1 = \angle BPF_2$
where $F_1$ and $F_2$ are the foci of $\EE$.
:
== Proof ==
== Also known as ==
== Sources ==
* 
* 
Category:Focal Property of Ellipse
Category:Foci of Ellipses
Category:Ellipses
Focal Property of Hyperbola

== Theorem ==
Let $\KK$ be a hyperbola.
Let $APB$ be a tangent to $\KK$ at $P$.
Then:
:$\angle APF_1 = \angle BPF_2$
where $F_1$ and $F_2$ are the foci of $\KK$.
== Proof ==
== Also known as ==
== Sources ==
* 
* 
Category:Focal Property of Hyperbola
Category:Foci of Hyperbolas
Category:Hyperbolas
Focal Property of Parabola

== Theorem ==
Let $\KK$ be a parabola whose focus is at $F$ and whose directrix is $D$.
Let $P$ be an arbitrary point on $\KK$.
The tangent to $\KK$ at $P$ bisects the angle between the straight line $PF$ and the straight line through $P$ perpendicular to $D$. 
:
== Proof ==
Let $BQ$ be the straight line constructed perpendicular to $D$ passing through $P$.
Let $AT$ be the tangent to $\KK$ at $P$.
Let $\angle APF$ be constructed equal to $\angle BPT$ where $F$ is on the axis of $\KK$.
It is to be shown that $F$ is the focus of $\KK$.
From Two Straight Lines make Equal Opposite Angles:
:$\angle BPT = \angle APQ$
and so as $\angle APF = \angle BPT$ by construction:
:$\angle APQ = \angle APF$
Consider the two triangles $\triangle APQ$ and $\triangle APF$.
We have that:
:$AP$ is common
:$\angle APQ = \angle APF$
From Triangle Side-Angle-Side Congruence $\triangle APQ$ and $\triangle APF$ will be congruent $PF = PQ$.
From the focus-directrix property of parabola $P$ is on the locus of the points that are equidistant from $F$ and $D$.
Hence the point $F$ is the focus.
== Also known as ==
== Sources ==
* : $\S 2.9$: Integrating Factors: Example $2$
* 
* 
Category:Focal Property of Parabola
Category:Foci of Parabolas
Category:Parabolas
Focus of Ellipse from Major and Minor Axis

== Theorem ==
Let $K$ be an ellipse whose major axis is $2 a$ and whose minor axis is $2 b$.
Let $c$ be the distance of the foci of $K$ from the center.
Then:
:$a^2 = b^2 + c^2$
== Proof ==
:
Let the foci of $K$ be $F_1$ and $F_2$.
Let the vertices of $K$ be $V_1$ and $V_2$.
Let the covertices of $K$ be $C_1$ and $C_2$.
Let $P = \tuple {x, y}$ be an arbitrary point on the locus of $K$.
From the equidistance property of $K$ we have that:
:$F_1 P + F_2 P = d$
where $d$ is a constant for this particular ellipse.
This is true for all points on $K$.
In particular, it holds true for $V_2$, for example.
Thus:
It also holds true for $C_2$:
:$F_1 C_2 + F_2 C_2 = d$
Then:
and:
Thus:
== Also presented as ==
This result is also seen presented as:
:$c^2 = a^2 - b^2$
Category:Ellipses
Focus of Hyperbola from Transverse and Conjugate Axis

== Theorem ==
Let $K$ be a hyperbola whose transverse axis is $2 a$ and whose conjugate axis is $2 b$.
Let $c$ be the distance of the foci of $K$ from the center.
Then:
:$c^2 = a^2 + b^2$
== Proof ==
 
:
Let the foci of $K$ be $F_1$ and $F_2$.
Let the vertices of $K$ be $V_1$ and $V_2$.
Let the covertices of $K$ be $C_1$ and $C_2$.
Let $P = \tuple {x, y}$ be an arbitrary point on the locus of $K$.
From the equidistance property of $K$ we have that:
:$\size {F_1 P - F_2 P} = d$
where $d$ is a constant for this particular hyperbola.
This is true for all points on $K$.
In particular, it holds true for $V_2$, for example.
Thus:
Category:Hyperbolas
Fontené Theorems/First

== Theorem ==
Let $\triangle ABC$ be a triangle.
Let $P$ be an arbitrary point in the same plane as $\triangle ABC$.
Let $A_1$, $B_1$ and $C_1$ be the midpoints of $BC$, $CA$ and $AB$ respectively.
Let $A_2 B_2 C_2$ be the pedal triangle of $P$ $\triangle A B C$.
Let $X, Y, Z$ be the intersections of $B_1 C_1$ and $B_2 C_2$, $A_1 C_1$ and $A_2 C_2$, and $A_1 B_1$ and $A_2 B_2$ respectively.
Then $A_2 X$, $B_2 Y$ and $C_2 Z$ concur at the intersection of the circle through $A_1, B_1, C_1$ and the circle through $A_2, B_2, C_2$.
== Proof ==
== Also see ==
* Second Fontené Theorem
* Third Fontené Theorem
== Sources ==
* 
Category:Fontené Theorems
Fontené Theorems/Second

== Theorem ==
Let $\triangle ABC$ be a triangle.
:
Let $P$ be a point moving on a fixed straight line through the circumcenter $O$ of $\triangle ABC$.
Then the pedal circle of $P$ with respect to passes through a fixed point $F$ on the Feuerbach circle of $\triangle ABC$.
== Proof ==
By the First Fontené Theorem, the point of contact $Q$ of the circle $E$ and the circle $O'$ is the reflection of a point $F$ which lies on $O P$ with respect to the line $B_1 C_1$.
It is easy to show that $O$ is the orthocenter of triangle $A_1 B_1 C_1$
thus $Q$ is the anti-Steiner point of $d$.
Therefore $Q$ is fixed.
== Also known as ==
== Also see ==
* First Fontené Theorem
* Third Fontené Theorem
== Sources ==
* 
Category:Fontené Theorems
Category:Pedal Circles
Category:Feuerbach Circles
Fontené Theorems/Third

== Theorem ==
Let $\triangle ABC$ be a triangle.
Let $P$ be an arbitrary point in the plane of $\triangle ABC$.
Let the isogonal conjugate of $P$ to $\triangle ABC$ be denoted $P^{-1}$.
Let $O$ be the circumcenter of $\triangle ABC$.
Then the pedal circle of $P$ is tangent to the Feuerbach circle of $\triangle ABC$ $O$, $P$, $P^{-1}$ are collinear.
== Proof ==
By the Second Fontené Theorem we can prove that the second intersection $Q'$ of the circle $O'$ and the circle $E$ is the anti-Steiner point of $OP^{-1}$.
This means $Q' = Q$ $O P = O P^{-1}$
That is: 
:$O$, $P$ and $P^{-1}$ are collinear.
== Also see ==
* First Fontené Theorem
* Second Fontené Theorem
== Sources ==
* 
Category:Fontené Theorems
Category:Feuerbach Circles
Category:Pedal Circles
For Complete Ritz Sequence Continuous Functional approaches its Minimal Value

== Theorem ==
Let $J$ be a continuous functional.
Let $\sequence {\phi_n}$ be a complete Ritz sequence.
Then:
:$\ds \lim_{n \mathop \to \infty} \mu_n = \mu$
where $\ds \mu = \inf_y J \sqbrk y$.
== Proof ==
Let $y^*: \R \to \R$ be such that:
:$\forall \epsilon > 0: J \sqbrk {y^*} < \mu + \epsilon$
By assumption of continuity of $J$:
:$\forall \epsilon > 0: \exists \map \delta \epsilon > 0: \paren {\size {y - y^*} < \delta} \implies \paren {\size {J \sqbrk y - J \sqbrk {y^*} } < \epsilon}$
Let $\eta_n = \bsalpha \bsphi$, such that:
:$\exists n \in\ N: \exists N \in \N: \paren {n > N} \implies \paren {\size {\eta_n - y^*} < \epsilon}$
where $\bsalpha$ is an $n$-dimensional real vector.
Let $y_n = \bsalpha \bsphi$, where $\bsalpha$ is such that it minimises $J \sqbrk {y_n}$.
Hence:
:$J \sqbrk {y_n} < J \sqbrk {\eta_n} + \epsilon < \mu + 2 \epsilon$
On the other hand, by the definition of $y_n$:
:$\mu \le J \sqbrk {y_n} \le J \sqbrk {\eta_n}$
Thus:
:$\mu \le J \sqbrk {y_n} < \mu + 2 \epsilon$
If $J \sqbrk {y_n} = \mu$, then the inequality is satisfied, and $n$ is finite.
If $ \mu < J \sqbrk {y_n}$, then subtract $\mu$ from all the terms:
:$0 < J \sqbrk {y_n} - \mu < 2 \epsilon$
We have that:
:$\paren {\epsilon > 0} \implies \paren {-2 \epsilon < 0}$
Hence:
:$-2 \epsilon < J \sqbrk {y_n} - \mu < 2 \epsilon$ 
or:
:$\size {J \sqbrk {y_n} - \mu} < 2 \epsilon$
Since $\epsilon$ and $n$ inherits their constraints from the definitions of $\eta_n$ and the continuity of $J$, these together with the last inequality imply:
:$\ds \lim_{n \mathop \to \infty} J \sqbrk {y_n} = \mu$
where $J \sqbrk {y_n} = \mu_n$.
== Sources ==
* : $\S 8.40 $: The Ritz Method and the Method of Finite Differences
Category:Calculus of Variations
Force Polygon/Examples/Arbitrary Example 1

== Example of Force Polygon ==
The following is an example of a force polygon of a system of forces consisting of the forces $\mathbf F_1, \mathbf F_2, \mathbf F_3, \mathbf F_4$:
:
The resultant can be seen to be represented by the directed line segment $A_0 A_4$.
== Sources ==
* 
Category:Examples of Force Polygons
Force between Infinite Parallel Straight Conductors carrying Steady Current

== Theorem ==
Let $s_1$ and $s_2$ be wires in a vacuum carrying steady currents $I_1$ and $I_2$.
Let $s_1$ and $s_2$ be parallel and (effectively) infinitely long.
Then the force between $s_1$ and $s_2$ is given by:
:$\mathbf F = \dfrac {\mu_0 I_1 I_2} {2 \pi r}$
where:
:$r$ denotes the distance between $s_1$ and $s_2$
:$\mu_0$ denotes the vacuum permeability.
== Proof ==
Category:Electromagnetism
Category:Electric Current
Forced Vibration of Cart attached to Wall by Spring

== Theorem ==
=== Problem Definition ===
Then the motion of $C$ is described by the second order ODE:
:$m \dfrac {\d^2 \mathbf x} {\ d t^2} + c \dfrac {\d \mathbf x} {\d t} + k \mathbf x = \mathbf F_0 \cos \omega t$
== Proof ==
Let $\mathbf F_s$ be the force on $C$ exerted by the spring $S$.
Let $\mathbf F_d$ be the damping force.
Let $\mathbf F_e$ be the external force on $C$.
By Newton's Second Law of Motion, the total force on $C$ equals its mass times its acceleration:
:$\mathbf F_s + \mathbf F_d + \mathbf F_e = m \mathbf a$
By Acceleration is Second Derivative of Displacement with respect to Time:
:$\mathbf a = \dfrac {\d^2 \mathbf x} {\d t^2}$
By definition, the velocity $\mathbf v$ is defined as:
:$\mathbf v = \dfrac {\d \mathbf x} {\d t}$
By Hooke's Law:
:$\mathbf F = -k \mathbf x$
By definition of the damping force:
:$\mathbf F_d = - c \mathbf v$
By definition of the external force:
:$\mathbf F_e = \mathbf F_0 \cos \omega t$
So:
== Sources ==
* : $\S 3.20$: Vibrations in Mechanical Systems: $(25)$
Category:Mechanics