nyuuzyou/wikis huggingface dataset:
Length of Arc of Cycloid/Proof 1

== Theorem ==
== Proof ==
Let $L$ be the length of one arc of the cycloid.
From Arc Length for Parametric Equations:
:$\ds L = \int_0^{2 \pi} \sqrt {\paren {\frac {\d x} {\d \theta} }^2 + \paren {\frac {\d y} {\d \theta} }^2} \rd \theta$
where, from Equation of Cycloid:
:$x = a \paren {\theta - \sin \theta}$
:$y = a \paren {1 - \cos \theta}$
we have:
Thus:
Thus:
So $L = 8 a$ where $a$ is the radius of the generating circle.
== Sources ==
* : Chapter $\text {B}.21$: The Cycloid: Example $2$
Category:Length of Arc of Cycloid
Length of Arc of Cycloid/Proof 2

== Theorem ==
== Proof ==
Consider the tangent $PQ$ to both the generating circle and the cycloid itself.
:
By :
:$PR = 2 PQ$
In the limit, where $P$ is at the cusp, the tangent $PQ$ is perpendicular to the straight line along which the generating circle rolls.
At this point:
:$PQ = 2 a$.
Thus at this point:
:$PR = 4 a$
But $4 a$ is half the length of one arc of $C$.
Hence the result.
== Historical Note ==
== Sources ==
* : Chapter $\text {A}.15$: Torricelli ()
Category:Length of Arc of Cycloid
Length of Arc of Cycloid/Proof 3

== Theorem ==
== Proof ==
Descartes approximated the Cycloid by subtituting a polygon for the generating circle. For example, a hexagon ($n=6$) produces 5 arches or lobes ($n-1$). 
Each arch is generated by turning through the external angle of the polygon. For the hexagon, this is $\dfrac {2 \pi} 6$.
The arches have arms of different lengths. The first is simply the side of the hexagon. The arm of the second arch is the chord which forms a triangle with two adjacent sides of the polygon, etc. 
We can obtain the lengths of those sides by considering the angles that subtend them. The central angle subtending a single side of the hexagon is 
It is convenient to let the sum range from 0 to 6 (the two extra terms don't contribute anything).
In addition, everywhere $6$ appears we can substitute $n$ for the polygon of $n$ sides.
where $\theta = \dfrac \pi n$. We use one of Lagrange's trigonometric identities. This can be derived from the Werner Formula for Sine by Sine
by substituting $A = \sin k \theta$ and $B = \frac 1 2$, and taking the sum of both sides, noting that the infinite sum on the right-hand side is a telescoping series. The identity we need is
Now let $n \rightarrow \infty$. Since $\theta = \dfrac \pi n$, the angle for the first term in the numerator becomes
$\dfrac 1 2 \dfrac \pi n = 0$
so we have for the first term $\cos 0 = 1$.
The second term becomes $\cos 0 + \pi = -1$ so the numerator is just $2$, which cancels the $2$ in the denominator. 
Use the Power Series Expansion for Sine Function (since $\pi/n \rightarrow 0$). The denominator is
$\sin \dfrac 1 2 \theta \approx \dfrac 1 2 \dfrac \pi n$
The expression for the total arc length $L$ becomes
Category:Length of Arc of Cycloid
Length of Arc of Deltoid

== Theorem ==
The total length of the arcs of a deltoid constructed within a deferent of radius $a$ is given by:
:$\LL = \dfrac {16 a} 3$
== Proof ==
Let $H$ be embedded in a cartesian plane with its center at the origin and one of its cusps positioned at $\tuple {a, 0}$.
:
We have that $\LL$ is $3$ times the length of one arc of the deltoid.
From Arc Length for Parametric Equations:
:$\ds \LL = 3 \int_{\theta \mathop = 0}^{\theta \mathop = 2 \pi/3} \sqrt {\paren {\frac {\d x} {\d \theta} }^2 + \paren {\frac {\d y} {\d \theta} }^2} \rd \theta$
where, from Equation of Deltoid:
:$\begin{cases}
x & = 2 b \cos \theta + b \cos 2 \theta \\
y & = 2 b \sin \theta - b \sin 2 \theta
\end{cases}$
We have:
Thus:
Thus:
:$\sqrt {\paren {\dfrac {\d x} {\d \theta} }^2 + \paren {\dfrac {\d y} {\d \theta} }^2} = 4 b \sin \dfrac \theta 2 \size {1 + 2 \cos \theta}$
In the range $0$ to $2 \pi / 3$, $1 + 2 \cos \theta$ is not less than $0$, and so:
:$\ds \LL = 3 \int_0^{2 \pi / 3} 4 b \sin \dfrac \theta 2 \paren {1 + 2 \cos \theta} \rd \theta$
Put:
:$u = \cos \dfrac \theta 2$
so:
:$2 \dfrac {\d u} {\d \theta} = -\sin \dfrac \theta 2$
As $\theta$ increases from $0$ to $\dfrac {2 \pi} 3$, $u$ decreases from $1$ to $\dfrac 1 2$.
Then:
Substituting:
:$2 \dfrac {\d u} {\d \theta} = -\sin \dfrac \theta 2$
and the limits of integration:
:$u = 1$ for $\theta = 0$
:$u = \dfrac 1 2$ for $\theta = \dfrac {2 \pi} 3$
we obtain, after simplifying the sign:
== Sources ==
* : Chapter $\text {B}.21$: The Cycloid: Problem $11$
Category:Deltoids (Hypocycloids)
Length of Arc of Evolute equals Difference in Radii of Curvature

== Theorem ==
Let $C$ be a curve defined by a real function which is twice differentiable.
Let the curvature of $C$ be non-constant.
The length of arc of the evolute $E$ of $C$ between any two points $Q_1$ and $Q_2$ of $C$ is equal to the difference between the radii of curvature at the corresponding points $P_1$ and $P_2$ of $C$.
== Proof ==
:
Let $P = \tuple {x, y}$ be a general point on $C$.
Let $Q = \tuple {X, Y}$ be the center of curvature of $C$ at $P$.
From the above diagram:
:$(1): \quad \begin {cases} x - X = \pm \rho \sin \psi \\ Y - y = \pm \rho \cos \psi \end {cases}$
where:
:$\rho$ is the radius of curvature of $C$ at $P$
:$\psi$ is the angle between the tangent to $C$ at $P$ and the $x$-axis.
Whether the sign is plus or minus depends on whether the curve is convex or concave.
For simplicity, let it be assumed that the curvature $k$ at each point under consideration on $C$ is positive.
The case for $k < 0$ can then be treated similarly.
Thus we have $k > 0$ and so $(1)$ can be written:
:$(2): \quad \begin {cases} X = x - \rho \sin \psi \\ Y = y +\rho \cos \psi \end {cases}$
By definition of curvature:
:$k = \dfrac {\d \psi} {\d s}$
and:
:$\rho = \dfrac 1 k = \dfrac {\d s} {\d \psi}$
Hence:
and:
Differentiating $(2)$ $\psi$:
and:
Let $S$ be the length of arc of the evolute $E$ from a fixed point $Q_0$ on $E$ to the variable point $Q$ corresponding to $P$.
We have that:
:$\d S^2 = \d X^2 + \d Y^2$
and so:
Choosing the direction of $S$ so that $S$ and $\rho$ increase together, this tells us:
:$\dfrac {\d S} {\d \psi} = \dfrac {\d \rho} {\d \psi}$
Integrating $\psi$ gives:
:$S = \rho + c$
where $c$ is a constant.
== Sources ==
* : Chapter $\text {B}.23$: Evolutes and Involutes. The Evolute of a Cycloid
Category:Evolutes
Length of Arc of Nephroid

== Theorem ==
The total length of the arcs of a nephroid constructed around a deferent of radius $a$ is given by:
:$\LL = 12 a$
== Proof ==
Let a nephroid $H$ be embedded in a cartesian plane with its center at the origin and its cusps positioned at $\tuple {\pm a, 0}$.
:
We have that $\LL$ is $2$ times the length of one arc of the nephroid.
From Arc Length for Parametric Equations:
:$\ds \LL = 2 \int_{\theta \mathop = 0}^{\theta \mathop = \pi} \sqrt {\paren {\frac {\d x} {\d \theta} }^2 + \paren {\frac {\d y} {\d \theta} }^2} \rd \theta$
where, from Equation of Nephroid:
:$\begin{cases}
x & = 3 b \cos \theta - b \cos 3 \theta \\
y & = 3 b \sin \theta - b \sin 3 \theta
\end{cases}$
We have:
Thus:
Thus:
:$\sqrt {\paren {\dfrac {\d x} {\d \theta} }^2 + \paren {\dfrac {\d y} {\d \theta} }^2} = 6 b \sin \theta$
So:
== Sources ==
* : Chapter $\text {B}.21$: The Cycloid: Problem $14$
Category:Nephroids
Length of Arc of Small Circle

== Theorem ==
Let $S$ be a sphere.
Let $\bigcirc FCD$ be a small circle on $S$.
Let $C$ and $D$ be the points on $\bigcirc FCD$ such that $CD$ is the arc of $\bigcirc FCD$ whose length is to be determined.
=== Construction ===
Let $P$ and $Q$ be the poles of $\bigcirc FCD$.
Let $\bigcirc PCQ$ and $\bigcirc PDQ$ be great circles on $S$.
Let $\bigcirc EAB$ be the great circle whose poles are $P$ and $Q$.
Let $A$ and $B$ be the points on $\bigcirc EAB$ which intersect $\bigcirc PCQ$ and $\bigcirc PDQ$.
The length of the arc $CD$ of $\bigcirc FCD$ is given by:
:$CD = AB \cos AC$
or:
:$CD = AB \sin PC$
== Proof ==
:
Let $R$ denote the center of $\bigcirc FCD$.
Let $O$ denote the center of $S$, which is also the center of $\bigcirc EAB$.
We have:
:$CD = RC \times \angle CRD$
Similarly:
:$AB = OA \times \angle AOB$
By Circles with Same Poles are Parallel:
:$\bigcirc FCD \parallel \bigcirc EAB$
Hence $RC$ and $RD$ are parallel to $OA$ and $OB$ respectively.
Thus:
We also have that:
and that:
We have that $\angle AOC$ is the (plane) angle subtended at $O$ by the arc $AC$ of $\bigcirc EAB$.
Thus:
Hence the result.
== Sources ==
* : Chapter $\text I$. Spherical Trigonometry: $3$. Length of a small circle arc.
Category:Small Circles
Length of Arch of Sine Function

== Theorem ==
The length of one arch of the sine function:
:$y = \sin x$
is given by:
:$L = 2 \sqrt 2 \map E {\dfrac {\sqrt 2} 2}$
where $E$ denotes the incomplete elliptic integral of the second kind.
== Proof ==
Let $L$ be the length of one arch of $y = \sin x$.
Then:
== Sources ==
* : $1$: The Nature of Differential Equations: $\S 5$: Falling Bodies and Other Rate Problems: Problem $6$
Category:Sine Function
Category:Complete Elliptic Integral of the Second Kind
Length of Basis Representation is Primitive Recursive

== Theorem ==
Define $\operatorname{basislen} : \N^2 \to \N$ as:
:$\map {\operatorname{basislen}} {b, n} = \begin{cases}m + 1 & : b > 1 \land n > 0 \\n & : b = 1 \\0 & : b = 0 \lor n = 0\end{cases}$
where $\sqbrk {r_m r_{m - 1} \dotsm r_1 r_0}_b$ is the base-$b$ representation of $n$.
Then $\operatorname{basislen}$ is primitive recursive.
== Proof ==
By:
* Definition by Cases is Primitive Recursive
* Ordering Relations are Primitive Recursive
* Equality Relation is Primitive Recursive
* Constant Function is Primitive Recursive
* Set Operations on Primitive Recursive Relations
all that needs to be shown is that $m + 1$ can be computed from $n$ in the case that $b > 1$ and $n > 0$.
By the definition of base-$b$ representation:
:$\ds n = \sum_{i \mathop = 0}^m r_i b^i$
Let $m' \le m$ be arbitrary.
Since $r_m > 0$, it follows that:
:$b^{m'} \le r_m b^m \le n$
Therefore:
:$m' \le m \implies b^{m'} \le n$
Now, let $m' > m$ be arbitrary.
Then:
Therefore:
:$m' > m \implies b^{m'} > n$
Thus, $m + 1$ is the smallest $k \in \N$ such that:
:$b^k > n$
The result follows from:
* Exponentiation is Primitive Recursive
* Ordering Relations are Primitive Recursive
* Bounded Minimization is Primitive Recursive
assuming we can find some function $\map g n$ such that:
:$\map g n \ge m + 1$
for every $n$.
By Basis Representation is No Longer than Number:
:$\map g n = n$
suffices.
Category:Number Bases
Category:Primitive Recursive Functions
Length of Binary Sequence Code is Primitive Recursive

== Theorem ==
Let the function $\map {\operatorname{bincodelen} } n$ be defined as the least $m$ such that:
:For every $i \ge m$, $\map {\operatorname{bincode} } {n, i} = 0$
where $\operatorname{bincode}$ is defined as in Binary Sequence Codes are Primitive Recursive.
Then, $\operatorname{bincodelen}$ is primitive recursive.
== Proof ==
The base-$2$ representation of $n$ is never longer than $n$ digits.
Therefore, there are at most $n$ digits that are $0$.
By definition of $\operatorname{bincode}$, there are thus at most $n$ nonzero values.
Thus, the following definition is correct:
:$\map {\operatorname{bincodelen} } n = \map {\mu m \le n} {\map {\mu i \le n} {i \ge m \land \map {\operatorname{bincode} } {n, i} \ne 0} > n}$
For, the inner minimization attempts to find some counterexample $i$ to the proposition that $m$ is the value of the function, as defined.
If no such counterexample is found, by the definition of bounded minimization, the result is $n + 1 > n$.
It follows from:
* Binary Sequence Codes are Primitive Recursive
* Set Operations on Primitive Recursive Relations
* Ordering Relations are Primitive Recursive
* Bounded Minimization is Primitive Recursive
* Equality Relation is Primitive Recursive
that $\operatorname{bincodelen}$ is primitive recursive.
Category:Primitive Recursive Functions
Length of Chord Projected from Point on Intersecting Circle

== Theorem ==
Let $C_1$ and $C_2$ be two circles which intersect at $A$ and $B$.
Let $T$ be a point on $C_1$.
Let $P$ and $Q$ be the points $TA$ and $TB$ intersect $C_2$.
:
Then $PQ$ is constant, wherever $T$ is positioned on $C_1$.
== Proof ==
Outline of proof:
By angle in the same segment, $\angle APB$ and $\angle ATB$ are constant.
Hence $\angle PBQ$ is constant.
This forces $PQ$ to be constant.
However this must be divided into multiple cases as sometimes $P, Q, T$ are contained in $C_1$.
The limit cases $T = A$ and $T = B$ can also be proven by interpreting $AT, BT$ as tangents at $T$. 
== Sources ==
* 
Category:Circles
Length of Chord of Circle

== Theorem ==
Let $C$ be a circle of radius $r$.
Let $AB$ be a chord which joins the endpoints of the arc $ADB$.
Then:
:$AB = 2 r \sin \dfrac \theta 2$
where $\theta$ is the angle subtended by $AB$ at the center of $C$.
== Proof 1 ==
== Proof 2 ==
== Historical Note ==
== Sources ==
* : Chapter $5$: Eternal Triangles: The origins of trigonometry
Category:Circles
Category:Length of Chord of Circle
Length of Chord of Circle/Proof 1

== Theorem ==
== Proof ==
:
Let $O$ be the center of $C$.
Let $AB$ be bisected by $OD$.
Consider the pair of triangles $\triangle AOE$ and $\triangle BOE$.
We see that:
:$AE = ED$ since $AB$ is bisected by $OD$
:$AO = BO$ since they are radii
:$OE = OE$ since they are common sides.
By Triangle Side-Side-Side Congruence:
:$\triangle AOE = \triangle BOE$
Then we have:
:$\angle AOE = \angle BOE = \dfrac \theta 2$
:$\angle OEA = \angle OEB = \dfrac {180 \degrees} 2 = 90 \degrees$
By :
:$\sin \dfrac \theta 2 = \dfrac {AE} {AO} = \dfrac {\frac 1 2 AB} r$
Rearranging, we get:
:$AB = 2 r \sin \dfrac \theta 2$
as desired.
Category:Length of Chord of Circle
Length of Chord of Circle/Proof 2

== Theorem ==
== Proof ==
We have $AO = BO$ since they are radii.
Therefore $\triangle AOB$ is isosceles.
By Isosceles Triangle has Two Equal Angles:
:$\angle OAB = \angle OBA$
By Sum of Angles of Triangle equals Two Right Angles:
:$\angle OAB + \angle OBA + \theta = 180 \degrees$
Therefore $\angle OAB = \dfrac {180 \degrees - \theta} 2 = 90 \degrees - \dfrac \theta 2$.
Thus:
Category:Length of Chord of Circle
Length of Circumradius of Triangle

== Theorem ==
Let $\triangle ABC$ be a triangle whose sides are of lengths $a, b, c$.
Then the length of the circumradius $R$ of $\triangle ABC$ is given by:
:$R = \dfrac {abc} {4 \sqrt {s \paren {s - a} \paren {s - b} \paren {s - c} } }$
where $s = \dfrac {a + b + c} 2$ is the semiperimeter of $\triangle ABC$.
== Proof ==
:
Let $\AA$ be the area of $\triangle ABC$.
From Area of Triangle in Terms of Circumradius:
:$\AA = \dfrac {a b c} {4 R}$
From Heron's Formula:
:$\AA = \sqrt {s \paren {s - a} \paren {s - b} \paren {s - c} }$
where $s = \dfrac {a + b + c} 2$ is the semiperimeter of $\triangle ABC$.
Hence the result:
:$R = \dfrac {abc} {4 \sqrt {s \paren {s - a} \paren {s - b} \paren {s - c} } }$
== Sources ==
* : $\S 4$: Geometric Formulas: $4.16$
* : $\S 7$: Geometric Formulas: Radius of Circle Circumscribing a Triangle of Sides $a, b, c$: $7.16.$
Category:Circumradii
Category:Circumcircles of Triangles
Category:Triangles
Length of Concatenation

== Theorem ==
Let $S$ and $T$ be words, and let $ST$ be their concatenation.
Then:
:$\size {S T}\ = \size S + \size T$
where $\size S$ denotes the length of $S$.
== Proof ==
Because of the unique readability of $ST$, we can determine for each symbol $s$ that is part of $S T$, whether:
:$s$ is part of $S$
:$s$ is part of $T$
and furthermore, precisely one of these options occurs.
There are $\size S$ symbols in $S$, and $\size T$ symbols in $T$.
In total, then, $S T$ is seen to consist of $\size S + \size T$ symbols.
Category:Collations
Length of Contour is Well-Defined

== Theorem ==
Let $C_1, \ldots, C_n$ be directed smooth curves.
Let $C_k$ be parameterized by the smooth path $\gamma_k: \closedint {a_k} {b_k} \to \C$ for all $k \in \set {1, \ldots, n}$.
Let $C$ be the contour defined by the finite sequence $C_1, \ldots, C_n$.
Suppose that $\sigma_k: \closedint {c_k} {d_k} \to \C$ is a reparameterization of $C_k$ for all $k \in \set {1, \ldots, n}$
Then:
:$\ds \sum_{k \mathop = 1}^n \int_{a_k }^{b_k} \size {\map {\gamma_k'} t} \rd t = \sum_{k \mathop = 1}^n \int_{c_k}^{d_k} \size {\map {\sigma_k'} t} \rd t$
and all real integrals in the equation are defined.
== Proof ==
From the definition of directed smooth curve, it follows that $\sigma_k = \gamma_k \circ \phi_k$ for all $k \in \set {1, \ldots, n}$.
Here, $\phi_k: \closedint {c_k} {d_k} \to \closedint {a_k} {b_k}$ is a bijective differentiable strictly increasing function. 
For all $k \in \set {1, \ldots, n}$, $\gamma_k$ and $\sigma_k$ are continuous.
From Complex Modulus Function is Continuous and Continuity of Composite Mapping: Corollary, it follows that $\cmod {\gamma_k'}$ and $\cmod {\sigma_k'}$ are continuous.
From Continuous Real Function is Darboux Integrable, we find that $\ds \sum_{k \mathop = 1}^n \int_{a_k}^{b_k} \cmod {\map {\gamma_k'} t} \rd t$ and $\ds \sum_{k \mathop = 1}^n \int_{c_k}^{d_k} \cmod {\map {\sigma_k'} t} \rd t$ are defined.
Hence, all real integrals in the theorem are defined.
Then:
Category:Complex Contour Integrals
Category:Contour Integrals
Length of Diagonal of Unit Square

== Theorem ==
The length of a diagonal of a square of side length $1$ is $\sqrt 2$) (the square root of $2$).
== Proof ==
:
Two adjacent sides $AB$, $BC$ and the diagonal $AC$ of square $ABCD$ form a right triangle.
The hypotenuse of triangle $\triangle ABC$ can be found by using Pythagoras's Theorem:
:$AC^2 = AB^2 + BC^2$
from which:
:$AC^2 = 2$
and so:
:$AC = \sqrt 2$
== Sources ==
* : $1 \cdotp 41421 \, 35623 \, 73095 \, 04880 \, 16887 \, 24209 \, 69807 \, 85697 \ldots$
* : $2$. Graphs: Paradox
* : $1 \cdotp 41421 \, 35623 \, 73095 \, 04880 \, 16887 \, 24209 \, 69807 \, 85697 \ldots$
Category:Squares
Length of Element of Arc in Orthogonal Curvilinear Coordinates

== Theorem ==
Let $\tuple {q_1, q_2, q_3}$ denote a set of orthogonal curvilinear coordinates.
Let the relation between those orthogonal curvilinear coordinates and Cartesian coordinates be expressed as:
where $\tuple {x, y, z}$ denotes the Cartesian coordinates.
Let $S$ be an infinitesimal arc.
Let $\d s$ be the length of $S$
Then:
where:
:$\d q_i$ is the projection of $S$ onto the unit normal to the curvilinear coordinate surface determined by $q_i$, for $i \in \set {1, 3}$
:${h_i}^2 = \paren {\dfrac {\partial x} {\partial q_i} }^2 + \paren {\dfrac {\partial y} {\partial q_i} }^2 + \paren {\dfrac {\partial z} {\partial q_i} }^2$
== Proof ==
By definition of the metric of $\tuple {q_1, q_2, q_3}$:
From Value of Curvilinear Coordinate Metric:
:$\forall i, j \in \set {1, 2, 3}: {h_{i j} }^2 = \dfrac {\partial x} {\partial q_i} \dfrac {\partial x} {\partial q_j} + \dfrac {\partial y} {\partial q_i} \dfrac {\partial y} {\partial q_j} + \dfrac {\partial z} {\partial q_i} \dfrac {\partial z} {\partial q_j}$
But we have that $\tuple {q_1, q_2, q_3}$ is orthogonal.
From Definition 1 of orthogonal curvilinear coordinates:
:$\dfrac {\partial x} {\partial q_i} \dfrac {\partial x} {\partial q_j} + \dfrac {\partial y} {\partial q_i} \dfrac {\partial y} {\partial q_j} + \dfrac {\partial z} {\partial q_i} \dfrac {\partial z} {\partial q_j} = 0$
wherever $i \ne j$.
Hence we have:
Elements $h_{i i}$ are those elements of the metric which do not vanish when $\tuple {q_1, q_2, q_3}$ is orthogonal.
To streamline notation, we rename them $h_1$, $h_2$ and $h_3$.
Hence:
where:
:${h_i}^2 = \paren {\dfrac {\partial x} {\partial q_i} }^2 + \paren {\dfrac {\partial y} {\partial q_i} }^2 + \paren {\dfrac {\partial z} {\partial q_i} }^2$
== Also see ==
* Definition:Scale Factor for Orthogonal Curvilinear Coordinates
== Sources ==
* : Chapter $\text I$: Introduction: $\S 1$. The origin of special functions: $(1.3)$
* : Chapter $2$ Coordinate Systems $2.1$ Curvilinear Coordinates: $(2.8)$
Category:Orthogonal Curvilinear Coordinates
Length of Fibonacci String is Fibonacci Number

== Theorem ==
Let $S_n$ denote the $n$th Fibonacci string.
Let $\map \len {S_n}$ denote the length of $S_n$.
Then:
:$\map \len {S_n} = F_n$
where $F_n$ denotes the $n$th Fibonacci number.
== Proof ==
The proof proceeds by strong induction.
For all $n \in \Z_{>0}$, let $\map P n$ be the proposition:
:$\map \len {S_n} = F_n$
=== Basis for the Induction ===
$\map P 1$ is the case:
Thus $\map P 1$ is seen to hold.
$\map P 2$ 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 $1 \le j \le k$, then it logically follows that $\map P {k + 1}$ is true.
This is the induction hypothesis:
:$\map \len {S_k} = F_k$
and:
:$\map \len {S_{k - 1} } = F_{k - 1}$
from which it is to be shown that:
:$\map \len {S_{k + 1} } = F_{k + 1}$
=== Induction Step ===
This is the induction step:
So $\map P k \implies \map P {k + 1}$ and the result follows by the Second Principle of Mathematical Induction.
Therefore:
:$\forall n \in \Z_{>0}: \map \len {S_n} = F_n$
== Sources ==
* : $\S 1.2.8$: Fibonacci Numbers: Exercise $36$
Category:Fibonacci Strings
Length of Inradius of Triangle

== Theorem ==
Let $\triangle ABC$ be a triangle whose sides are of lengths $a, b, c$.
Then the length of the inradius $r$ of $\triangle ABC$ is given by:
:$r = \dfrac {\sqrt {s \paren {s - a} \paren {s - b} \paren {s - c} } } s$
where $s = \dfrac {a + b + c} 2$ is the semiperimeter of $\triangle ABC$.
== Proof ==
:
Let $\AA$ be the area of $\triangle ABC$.
From Area of Triangle in Terms of Inradius:
:$\AA = r s$
where $s = \dfrac {a + b + c} 2$ is the semiperimeter of $\triangle ABC$.
From Heron's Formula:
:$\AA = \sqrt {s \paren {s - a} \paren {s - b} \paren {s - c} }$
where $s = \dfrac {a + b + c} 2$ is the semiperimeter of $\triangle ABC$.
Hence the result:
:$r = \dfrac {\sqrt {s \paren {s - a} \paren {s - b} \paren {s - c} } } s$
== Sources ==
* : $\S 4$: Geometric Formulas: Radius of Circle Inscribed in a Triangle of Sides $a, b, c$: $4.15$
* : $\S 7$: Geometric Formulas: Radius of Circle Inscribed in a Triangle of Sides $a, b, c$: $7.15.$
Category:Incircles of Triangles
Category:Triangles
Length of Latus Rectum of Ellipse

== Theorem ==
Let $K$ be a ellipse such that:
:$a$ denotes the length of semi-major axis of $K$
:$b$ denotes the length of semi-minor axis of $K$.
Let $L$ denote the length of the latus rectum of $K$.
Then:
:$L = \dfrac {2 b^2} a$
== Proof ==
== Sources ==
* 
* 
Category:Latus Rectum of Ellipse
Category:Ellipses
Length of Latus Rectum of Hyperbola

== Theorem ==
Let $\KK$ be a hyperbola such that:
:$a$ denotes the length of the semitransverse axis of $\KK$
:$b$ denotes the length of the semiconjugate axis of $\KK$
Let $L$ denote the length of the latus rectum of $\KK$.
Then:
:$L = \dfrac {2 b^2} a$
== Proof ==
== Sources ==
* 
* 
Category:Latus Rectum of Hyperbola
Category:Hyperbolas
Length of Latus Rectum of Parabola

== Theorem ==
Let $\KK$ be a parabola such that the distance between the focus and directrix of $\KK$ is $2 a$.
Let $L$ denote the length of the latus rectum of $\KK$.
Then:
:$L = 4 a$
== Proof ==
== Sources ==
* 
* 
Category:Latus Rectum of Parabola
Category:Parabolas
Length of Legendre Polynomial

== Theorem ==
Let $\map {P_n} x$ denote the Legendre polynomial of order $n$.
Let $\norm {\map {P_n} x}$ denote the length of $\map {P_n} x$.
Then:
:$\norm {\map {P_n} x} := \sqrt {\dfrac 2 {2 n + 1} }$
== Proof ==
Applying Bonnet's Recursion Formula for $n - 1$:
:$n \map {P_n} x = \paren {2 n - 1} x \map {P_{n - 1} } x - \paren {n - 1} \map {P_{n - 2} } x$
so:
:$\map {P_n} x = \dfrac {2 n - 1} n x \map {P_{n - 1} } x - \dfrac {n - 1} n \map {P_{n - 2} } x$
Substituting for $\map {P_n} x$:
From Orthogonality of Legendre Polynomials:
:$\ds \int_{-1}^1 \map {P_n} x \map {P_m} x \rd x = 0 \iff n \ne m$
so:
:$\ds (1): \quad \norm {\map {P_n} x}^2 = \frac {2 n - 1} n \int_{-1}^1 x \map {P_n} x \map {P_{n - 1} } x \rd x$
From Bonnet's Recursion Formula:
:$x \map {P_n} x = \dfrac {n + 1} {2 n + 1} \map {P_{n + 1} } x + \dfrac n {2 n + 1} \map {P_{n - 1} } x$
Substituting for $x \map {P_n} x$ in $(1)$:
Thus:
Most of this cancels out, leaving:
:$\norm {\map {P_n} x}^2 = \dfrac {\norm {\map {P_0} x}^2} {2 n + 1}$
It remains to compute the length of the first Legendre polynomial:
Thus:
:$\norm {\map {P_n} x}^2 = \dfrac 2 {2 n + 1}$
and so taking the square root:
:$\norm {\map {P_n} x} = \sqrt {\dfrac 2 {2 n + 1} }$
Category:Legendre Polynomials